Is Distance/space Lorentz contracted?

[PeterDonis]

Austin0: I'm going to respond to parts of both your posts here.

As far as I can see they are somewhat mutually exclusive. Certainly the expected end results appear to be.

The expected end results should be the same for both versions of conservation of momentum. If you get different results by applying the two different versions then there's a mistake somewhere.

Given: A ship with thrust applied to the front and back, with greater thrust at one end.

Which end? It makes a difference. Also, does "thrust" mean the actual force applied (presumably by a rocket engine), or the resultant acceleration? They're not necessarily the same, as I've noted before. See next comment.

#1 The momentum from the front propagates through the ship to the back, while the momentum from the back propagates through to the front and then results in system motion . So the differential is equalized and the net acceleration is also equal at the front and the back.

#2 The momentum propagates locally from the source as motion and results in greater velocity at the end with greater thrust.

I'm not exactly sure what you mean by the above. Rather than try to parse it, let me instead describe how a situation like this one would be analyzed with the tools I'm used to using.

First of all, as I noted above, it makes a difference whether "thrust" means the applied force (which might be combined with other forces, internal to the ship, to obtain the resultant net force--and acceleration--of a given small piece of the ship) or the resultant acceleration (which is a result of the net force, not the rocket thrust alone). I'm going to assume below that "thrust" means the applied force, since that's the usual meaning of that term.

Let's first write down in general terms what both versions of conservation of momentum will look like (applied to the ship as a whole, and applied to each piece):

(A) Treat the ship as a single object. The total momentum transferred to the ship per unit time will be equal to the sum of the thrusts on the two ends. Since we're treating the ship as a single object, we don't care how that force is propagated through the ship, or what it does to the internal stresses or distances within the ship. All we care about is that the momentum transferred to the ship per unit time must be equal to the total applied thrust (sum of both ends). In other words, we have

$$F = T_R + T_F = \frac{d P}{d \tau}$$

for the ship as a whole, where $T_R$ and $T_F$ are the externally applied thrusts at the front and rear ends. That's all there is to it.

(B) Treat the ship as an extended object. For simplicity, we'll treat it here as composed of three segments: R, the rear segment, where the rear thrust is applied; M, the middle segment, where no thrust is applied (but which can exchange internal forces with the other segments), and F, the front segment, where the front thrust is applied.

Then conservation of momentum for the segments individually looks like this:

$$F_R = T_R + S_{MR} = \frac{d P_R}{d \tau_R}$$

$$F_M = S_{RM} + S_{FM} = \frac{d P_M}{d \tau_M}$$

$$F_F = T_F + S_{MF} = \frac{d P_F}{d \tau_F}$$

where we've used T for the thrusts (rear and front) and S for the internal forces between each segment, with the order of subscripts indicating the direction of the internal force (for example, $S_{MR}$ is the internal force exerted by the middle segment on the rear segment). In order to check for consistency with (A) above, we calculate the total momentum gained by the ship per unit time by adding the three forces above:

$$F = F_R + F_M + F_F$$

and check to see that the sum equals the sum given in (A) above, i.e., the sum of the externally applied thrusts.

Let's now suppose that the ship starts out at rest, with all forces zero. Then, at time t = 0 in the initial rest frame, the front and rear thrusts are applied. The internal forces at that moment will be zero, because there has been no time for the ship's internal structure to respond to the applied forces. So we will have

$$F_R = T_R$$

$$F_M = 0$$

$$F_F = T_F$$

when the thrusts are initially applied. This means that, initially, the rear and front segments will gain momentum, but the middle one will not. And you can see from the above that, intially, the sum of the momentum gains by the rear and front segments will be equal to the total momentum gained by the ship, which is equal to the sum of the applied thrusts at front and rear. So momentum is conserved for each individual ship segment, and it's also conserved for the ship as a whole.

In order to predict in detail how the internal forces will develop, we would need a detailed model of the material properties of the ship's structure, which I don't want to discuss here. But in general terms, you can see from the above that, assuming the front and rear thrusts are in the same direction, the initial effect will be that the rear segment will move closer to the middle segment, while the front segment will move further from it. So, after some small time has elapsed, we will have something like this:

$$F_R = T_R - C$$

$$F_M = C + S$$

$$F_F = T_F - S$$

where C is an internal pressure between the rear and middle segments, which arises because the segments are moving closer together and the material is assumed to resist compression (note the signs of the C terms in each force), and S is an internal tension between the middle and front segments, which arises because the segments are moving further apart, and the material is assumed to resist stretching (again, note the signs of the S terms in each force).

Once again, the above equations automatically conserve momentum for each individual segment. But if we add up the total momentum gained by all the segments, we find that, once again, it's just the sum of the external thrusts applied to front and rear; all the internal forces cancel out (because momentum gained by one segment due to each internal force is exactly balanced by momentum lost by the other segment due to the same force). So again, momentum is conserved for each individual ship segment, and it's also conserved for the ship as a whole.

All this will continue to be true as the internal forces vary with time; for example, the above suggests that after some more time has elapsed, the middle segment will move away from the rear segment and towards the front segment, as part of the oscillation we were discussing in earlier posts. Then we will have:

$$F_R = T_R + S$$

$$F_M = - S - C$$

$$F_F = T_F + C$$

where now S is a tension between the rear and middle segments and C is a pressure between the middle and front segments. The same analysis as above still applies: the internal forces are different than before, but they still cancel out when the total momentum is added up. So at each individual instant of time, momentum is conserved individually for each ship segment, and total momentum is conserved for the ship as a whole. Everything is consistent.

I can discuss more specific cases if you want (meaning specific relative values of the front and rear thrusts), but you can already see from the above that, regardless of whether the front or the rear thrust is larger, or whether they're both the same, conservation of momentum will hold as I've stated it. The only difference the specific relative values of the thrusts makes is to determine what kind of final state will be reached--i.e., will the ship end up stretched or compressed, and so forth--but that question is independent of the question of conservation of momentum.

I should note, since I'm going to address your questions about damping of the oscillations below, that the above model doesn't really include damping, except implicitly in whatever model you come up with for determining the internal forces between the segments. However, damping does not affect the correctness of what I've said above; all it does is add extra equations (dealing with the internal energies of each segment) that still will be consistent with the equations I've written above. If you want me to go into more detail about how that could work, let me know and I'll do so in a future post.

Where did this energy come from?

Would it completely stop after initial adjustment to acceleration??

Would the system be getting continually hotter??

The energy that heats the ship's structure as it oscillates in response to an externally applied thrust comes from the energy that fuels the source of the thrust--the rocket engine or whatever. The specific mechanism of how that would occur depends on how you model the internal forces between segments of the ship, as I noted above; but in general, the energy is converted into heat as a result of the relative motion between the internal parts of the ship.

If we assume that the ship reaches some steady-state equilibrium while under acceleration (i.e., a state where the internal parts of the ship are no longer moving relative to one another), then the heating of the ship's structure should stop once that steady-state equilibrium is reached, since at that point there is no longer any relative motion between the ship's parts.

AM I incorrect in my understanding, that kinetic enrgy in its many forms including heat,
results in an increase in inertial mass??
If this is the case, how does this fit into the momentum conservation equation?
It would appear that any part of the directed applied momentum that was internally transformed into kinetic energy, would then contribute to the overall system conservation of momentum as an increase in inertial mass rather than, neccessarily, an increase in system velocity.
Is there some principle I am unaware of that would negate this concept??

It seems like you're trying to separate out momentum from energy, and kinetic energy from other kinds of energy. That's probably not a good way to think of momentum and energy. Relativistically, the important object is the 4-momentum, which includes both momentum and energy and does not distinguish between types of energy--all of them appear in the 4-momentum and they aren't separated out. Trying to separate things out usually causes more problems than it solves.

That said, the quick answers to your questions are:

* Yes, kinetic energy, heat, etc. can all affect inertial mass. See my discussion starting in the paragraph after next.

* The momentum conservation equation deals with 4-momentum, which includes everything, as I noted above. Nothing is left out.

* As you apply constant force to an object, you change its 4-momentum. That's the relativistically invariant way to describe what happens.

In the simplest case, the object's rest mass is constant, and that means that (since 4-momentum is rest mass times 4-velocity) all of the change in the object's 4-momentum, in response to applied force, will be manifested in the object's 4-velocity. Viewed from a specific frame of reference, part of the change in 4-velocity will appear as a change in the object's inertial mass, and part will appear as a change in the object's velocity. The split between the two depends on your frame of reference. If you keep viewing things from the same frame of reference (for example, the initial rest frame of the object to which force is being applied), then as a constant force (thrust) is applied to an object, over time, more and more of the change in 4-velocity will appear as a change in inertial mass, and less and less as a change in velocity (as the object's observed velocity gets closer and closer to the speed of light).

However, an object may have internal degrees of freedom (for example, its individual molecules may vibrate). Energy that is in these internal degrees of freedom--for example, heat--is included, relativistically, in the rest mass of the object. In the example above of the ship oscillating in response to applied thrust, and part of the energy of the oscillation being converted into heat, the effect would be an increase in the ship's rest mass. This is still an increase in 4-momentum, but it would not appear as an increase in velocity; it would appear as an increase in inertial mass, but due to the object's rest mass increasing, not due to its 4-velocity increasing.

Are we going to discuss any of the other examples in my post??

I'd like your feedback on the above before addressing the spheres scenario or responding to your other questions. But I haven't forgotten the other questions.

 

[Austin0]

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=PeterDonis;2384418]

The expected end results should be the same for both versions of conservation of momentum. If you get different results by applying the two different versions then there's a mistake somewhere.

Does this mean that both versions result in equal acceleration throughout the system??

Which end? It makes a difference. Also, does "thrust" mean the actual force applied (presumably by a rocket engine), or the resultant acceleration? They're not necessarily the same, as I've noted before. See next comment.

AS you point out later in this post it doesn't really make any difference which end or degrees of magnitude as far as the principles involved.
Yes I am using thrust as meaning force applied, not resultant acceleration.


I hope you will bear with me if I clarify an earlier concept I presented . For my own education and to check my understanding against your and other's knowledge.

Original AUstin0
Given that the magnitude of force is within the materials ability to transmit it fast enough, applied energy, momentum, propagates through the system, not as motion, but as a reciprocal oscillation.
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As a rough analogy;
assume a hydro-dynamic system within which the liquid velocity is magically limited to a range that is dependant simply on pressure- density.
Given: a pipe of a certain diameter and a variable input (volume/unit time) that starts at zero and incrementally increases.
Would it not be reasonable to suppose that the liquid would flow without increase in pressure up to some threshold of input??
That exceeding that threshold , since the velocity cannot increase , the pressure would .
This increased pressure would then increase the possible velocity, resulting in this high pressure/velocity itself progressing throughout the pipe until equilibrium was re-established throughout the entire system?
That increasing the diameter of the pipe would inevitably raise the threshold of stress/pressureless flow??
Taken into the context of propagation of momentum:
The pipe diameter would be mass distribution.
Given two rods of equal mass but different shape: Two sylinders , one short with a large cross section and one long and thin. Disregarding the system length and propagation time.
With the same force applied to their ends as an incrementally increasing magnitude starting at zero;
wouldn't it be reasonable to expect that the short rod transmitting this force ,distributed through a larger number of individual particles, would have a higher threshold of stressfree propagation than the rod with a smaller cross section ?
Is there some principle that would make this view untenable?

_____________________________________________________________

(A) Treat the ship as a single object. The total momentum transferred to the ship per unit time will be equal to the sum of the thrusts on the two ends. Since we're treating the ship as a single object, we don't care how that force is propagated through the ship, or what it does to the internal stresses or distances within the ship. All we care about is that the momentum transferred to the ship per unit time must be equal to the total applied thrust (sum of both ends). In other words, we have

$$F = T_R + T_F = \frac{d P}{d \tau}$$

for the ship as a whole, where $T_R$ and $T_F$ are the externally applied thrusts at the front and rear ends. That's all there is to it.

Complete agreement here as a generalization.

Original post austin0 #61
I would assume that the overall acceleration of the system would be the sum of force applied at both ends.

(B) Treat the ship as an extended object. For simplicity, we'll treat it here as composed of three segments: R, the rear segment, where the rear thrust is applied; M, the middle segment, where no thrust is applied (but which can exchange internal forces with the other segments), and F, the front segment, where the front thrust is applied.

Then conservation of momentum for the segments individually looks like this:

$$F_R = T_R + S_{MR} = \frac{d P_R}{d \tau_R}$$

$$F_M = S_{RM} + S_{FM} = \frac{d P_M}{d \tau_M}$$

$$F_F = T_F + S_{MF} = \frac{d P_F}{d \tau_F}$$

where we've used T for the thrusts (rear and front) and S for the internal forces between each segment, with the order of subscripts indicating the direction of the internal force (for example, $S_{MR}$ is the internal force exerted by the middle segment on the rear segment). In order to check for consistency with (A) above, we calculate the total momentum gained by the ship per unit time by adding the three forces above:

$$F = F_R + F_M + F_F$$

and check to see that the sum equals the sum given in (A) above, i.e., the sum of the externally applied thrusts.

Let's now suppose that the ship starts out at rest, with all forces zero. Then, at time t = 0 in the initial rest frame, the front and rear thrusts are applied. The internal forces at that moment will be zero, because there has been no time for the ship's internal structure to respond to the applied forces. So we will have

$$F_R = T_R$$

$$F_M = 0$$

$$F_F = T_F$$

when the thrusts are initially applied. This means that, initially, the rear and front segments will gain momentum, but the middle one will not. And you can see from the above that, intially, the sum of the momentum gains by the rear and front segments will be equal to the total momentum gained by the ship, which is equal to the sum of the applied thrusts at front and rear. So momentum is conserved for each individual ship segment, and it's also conserved for the ship as a whole.

I am in complete agreement with this . With one small caveat.
It is agreed that initially the front and rear will unequivocally gain momentum before the middle within the context of momentum as energy. IMHO --Whether that momentum results in actual coordinate translation or not depends on magnitude.
Would you agree that compression is dependant on constraint?
That force applied to one end of a spring requires the other end to be fixed in order for compression to take place? That in this context the restraint is simply system inertia.
Isn't it possible that if the force is transmitted as fast as it is applied it then has no constraint ? It becomes a reciprocal motion with no net motion or compression?

Suppose we have a cyclindrical rod with the force applied to both ends but in opposite directions. What would happen?

A) [Local] The rod is stretched locally from the ends. The cross section decreasing and the length increasing at both ends with the middle retaining its original cross section or at least greater than the ends. Eventually breaking at one end or the other.

B) [Global] The rod stretches initially equally throughout its length, but because the force is applied to the ends they would tend to retain the cross section, so the middle of the rod would be the area of maximal stretching and eventual breakdown.

A or B ?

In order to predict in detail how the internal forces will develop, we would need a detailed model of the material properties of the ship's structure, which I don't want to discuss here.

I completely agree. There is nothing to be gained by trying to create a detailed model.
If we can't come to some conclusions based on basic principles we probably won't at all.

But in general terms, you can see from the above that, assuming the front and rear thrusts are in the same direction, the initial effect will be that the rear segment will move closer to the middle segment, while the front segment will move further from it. So, after some small time has elapsed, we will have something like this:

$$F_R = T_R - C$$

$$F_M = C + S$$

$$F_F = T_F - S$$

where C is an internal pressure between the rear and middle segments, which arises because the segments are moving closer together and the material is assumed to resist compression (note the signs of the C terms in each force), and S is an internal tension between the middle and front segments, which arises because the segments are moving further apart, and the material is assumed to resist stretching (again, note the signs of the S terms in each force).

Once again, the above equations automatically conserve momentum for each individual segment. But if we add up the total momentum gained by all the segments, we find that, once again, it's just the sum of the external thrusts applied to front and rear; all the internal forces cancel out (because momentum gained by one segment due to each internal force is exactly balanced by momentum lost by the other segment due to the same force). So again, momentum is conserved for each individual ship segment, and it's also conserved for the ship as a whole.

I agree with all of this. That overall momentum would be equivelant and that the internal stresses would cancel out.
I think there might be other ways they would cancel out though.

If we assume a large central mass with the propulsion units connected by springs at the front and back then I think your description would be accurate. Compression at the back and expansion at the front.
But given a uniform system isn't it possible that they both progress throughout the system
as waves and the established equilibrium is also essentially uniform throughout?

That given a greater force at one end, the total system acceleration would be the sum, but the internal stress would depend on the difference between the two forces, given that they were in the same direction??
That it would be equivalent to that difference being the only force , applied from one end??

All this will continue to be true as the internal forces vary with time; for example, the above suggests that after some more time has elapsed, the middle segment will move away from the rear segment and towards the front segment, as part of the oscillation we were discussing in earlier posts. Then we will have:

$$F_R = T_R + S$$

$$F_M = - S - C$$

$$F_F = T_F + C$$

where now S is a tension between the rear and middle segments and C is a pressure between the middle and front segments. The same analysis as above still applies: the internal forces are different than before, but they still cancel out when the total momentum is added up. So at each individual instant of time, momentum is conserved individually for each ship segment, and total momentum is conserved for the ship as a whole. Everything is consistent.

I can discuss more specific cases if you want (meaning specific relative values of the front and rear thrusts), but you can already see from the above that, regardless of whether the front or the rear thrust is larger, or whether they're both the same, conservation of momentum will hold as I've stated it. The only difference the specific relative values of the thrusts makes is to determine what kind of final state will be reached--i.e., will the ship end up stretched or compressed, and so forth--but that question is independent of the question of conservation of momentum.

Agreed, there is no gain from going into more specifics as far as the principles involved.


One point I am unclear on : given this equivalence of methods and balance of forces does this mean that the resultant velocity would then be the same at both ends??

If we assume that the applied force at one end of the object continues to be applied, then once the initial wave has traveled the length of the object, it will start to be damped out. This is because as each layer pushes against the next, not all of the energy in the push gets converted into motion of the next layer; some of the energy goes into the internal degrees of freedom in the atoms/molecules--i.e., it gets converted into heat. . While the wave is damping, the object will be oscillating, expanding and compressing with gradually decreasing amplitude. Once the wave is damped out, the object will be in a state (assuming the force continues to be applied at one end) in which there is a compressive stress all along its length, and in which its material is somewhat hotter than it was before the force was applied (because some of the applied energy got converted into heat during the damping

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The energy that heats the ship's structure as it oscillates in response to an externally applied thrust comes from the energy that fuels the source of the thrust--the rocket engine or whatever. The specific mechanism of how that would occur depends on how you model the internal forces between segments of the ship, as I noted above; but in general, the energy is converted into heat as a result of the relative motion between the internal parts of the ship.

Agreed some of the heat will result from the propulsion mechanism itself and the impossibility of 100% efficiency. I assume we agree this is not relevant and simply assumes some infrared radiative dissapation or the ship will eventually cook.
Also I assume we would agree that even if we posit an external force, an ion beam from an inertial frame or a light sail that the principles would still apply but aren't important for our enquiry?

If we assume that the ship reaches some steady-state equilibrium while under acceleration (i.e., a state where the internal parts of the ship are no longer moving relative to one another), then the heating of the ship's structure should stop once that steady-state equilibrium is reached, since at that point there is no longer any relative motion between the ship's parts.

Can we agree that the 2nd law of TD will still operate but that the results would be negligable and irrelevant to the real questions we are addressing??

It seems like you're trying to separate out momentum from energy, and kinetic energy from other kinds of energy. That's probably not a good way to think of momentum and energy. Relativistically, the important object is the 4-momentum, which includes both momentum and energy and does not distinguish between types of energy--all of them appear in the 4-momentum and they aren't separated out. Trying to separate things out usually causes more problems than it solves.

I think in actuality I am trying to integrate momentum and kinetic energy .
That there are three fundamental principles involved here.
Conservation of mass and energy.
Conservation of momentum.
2nd law of thermodynamics.
Correct me if I am wrong :
Newtonian conservation of momentum was based on a conception of instantaneous force and completely invariant mass.
SR evolved this concept in two fundamental ways.
1) Finite propagation speed of momentum [energy]
2) A variable mass in two distinct senses.
A) The relativistic inertial mass increase with velocity.
B) The equivalence and interchangability of energy and mass. This means that not only kinetic energy but the energy of internal stress etc results in increased inertial mass.
I trust we agree that the Lorentz mass increase is not relevant to our discussion.
As I understand the 4 momentum and the equations for conservation of momentum you have presented so far they are basically unchanged from the Newtonian form with regard to internal propagation of momentum and the equivalence of mass and energy.
The 4 momentum differs in its coordinate time component but still defines instantaneous change. Mass/energy is included in the concept of rest mass, invariant mass etc. but a possible change in inertial mass through applied force is not included in the basic formulations. It is still based on a concept of ideal elastic interactions. An assumption that both momentum and kinetic energy will be conserved. I think we would agree that in the real world there are no ideally rigid or elastic bodies.
So how does this get handled with complex systems?
I hope that we agree that the magnitude of effects involved are slight and we can proceed without regarding them further??

That said, the quick answers to your questions are:

* Yes, kinetic energy, heat, etc. can all affect inertial mass. See my discussion starting in the paragraph after next.

* The momentum conservation equation deals with 4-momentum, which includes everything, as I noted above. Nothing is left out.

* As you apply constant force to an object, you change its 4-momentum. That's the relativistically invariant way to describe what happens.

In the simplest case, the object's rest mass is constant, and that means that (since 4-momentum is rest mass times 4-velocity) all of the change in the object's 4-momentum, in response to applied force, will be manifested in the object's 4-velocity. Viewed from a specific frame of reference, part of the change in 4-velocity will appear as a change in the object's inertial mass, and part will appear as a change in the object's velocity. The split between the two depends on your frame of reference. If you keep viewing things from the same frame of reference (for example, the initial rest frame of the object to which force is being applied), then as a constant force (thrust) is applied to an object, over time, more and more of the change in 4-velocity will appear as a change in inertial mass, and less and less as a change in velocity (as the object's observed velocity gets closer and closer to the speed of light).

However, an object may have internal degrees of freedom (for example, its individual molecules may vibrate). Energy that is in these internal degrees of freedom--for example, heat--is included, relativistically, in the rest mass of the object. In the example above of the ship oscillating in response to applied thrust, and part of the energy of the oscillation being converted into heat, the effect would be an increase in the ship's rest mass. This is still an increase in 4-momentum, but it would not appear as an increase in velocity; it would appear as an increase in inertial mass, but due to the object's rest mass increasing, not due to its 4-velocity increasing.



I'd like your feedback on the above before addressing the spheres scenario or responding to your other questions. But I haven't forgotten the other questions

I hope not ;-) it would be good to get on to the really interesting stuff.
Thanks

 

[PeterDonis]

Austin0: Let me start with this from your post because I think it's the most basic point that needs to be clarified:

I am in complete agreement with this . With one small caveat.
It is agreed that initially the front and rear will unequivocally gain momentum before the middle within the context of momentum as energy. IMHO --Whether that momentum results in actual coordinate translation or not depends on magnitude.

This is false. *Any* gain in momentum must result in *some* coordinate translation. More precisely: if, viewed in a given frame, an object gains some 4-momentum, and the 4-momentum gained has a spatial component (i.e., momentum as well as energy) in that frame, then the object must gain *some* coordinate velocity in that frame.

For example, consider a simpler version of the example in my last post. Let's consider a ship with a rocket engine only at the rear, and model it as having just two segments, the rear segment R with the rocket engine, and the middle segment M with no engine. Suppose we turn on the rocket engine at time t = 0 in the ship's initial rest frame. Then, at that instant, we have (same notation as my last post):

$$F_R = T_R$$

$$F_M = 0$$

So after a very small time dt, the rear segment will have acquired 4-momentum $T_R dt$. Since the force is in a particular direction (the "forward" direction of the ship), the 4-force $T_R$ will have a spatial component, so the rear segment will have *some* coordinate velocity in the original rest frame. If we assume that the time dt is short enough that the 4-momentum acquired is small compared to the rear segment's rest mass $M_R$, then the 4-force will have *only* a spatial component, and the coordinate velocity will be approximately the non-relativistic value

$$v = \frac{T_R dt}{M_R}$$

This observation affects a number of the scenarios you've suggested: basically, *any* scenario where an application of a force in a particular direction does not immediately result in the object acquiring a momentum in that direction *cannot* be correct. The momentum may reside in only one part of the object (as above, only the rear segment has a momentum on the initial application of the force), but *some* part of the object must move immediately on application of a force. Given that observation, I'm not going to comment on scenarios you've suggested that appear to me to violate this rule. For example, this:

Given that the magnitude of force is within the materials ability to transmit it fast enough, applied energy, momentum, propagates through the system, not as motion, but as a reciprocal oscillation.

doesn't really capture what's going on. To see how the oscillation comes about, you have to model the object as a bunch of individual segments that can move independently (but may exert forces on adjacent segments), as we did with the ship; then you can see that *some* segment is always moving in response to the applied force, and the sum of the motions of all the segments always equals the total momentum gained from the applied force, so that conservation of momentum is satisfied. If this is what you mean by "reciprocal oscillation", that's fine, but it's not something that happens instead of the system moving as a whole; the system *is* moving as a whole at the same time that the oscillation is propagating through it.

One point I am unclear on : given this equivalence of methods and balance of forces does this mean that the resultant velocity would then be the same at both ends??

Let's go back to the full model we considered in my last post, where we have three segments of the ship, with thrusts applied at the front and rear segments and no thrust in the middle segment. For your question quoted just above to be meaningful, the ship must reach some sort of steady-state equilibrium, in which there is no relative motion between the parts of the ship--otherwise, as we found last time, there will still be oscillations that haven't been damped out. (This does make a fairly general assumption about the material structure of the ship--that it is adequately modeled by the general scheme of tension and compression forces we used last time.)

But remember the definition of Born rigid acceleration: the whole object must accelerate such that the distance between individual parts of the object, in each part's rest frame, remains constant. That means that any steady-state equilibrium of an accelerating object *must* be a state of Born rigid acceleration! In other words, Born rigid acceleration, while extremely unlikely to happen at the *start* of an object's acceleration (because the acceleration of each part of the object would have to be controlled so precisely), is a likely *end state* of an object's acceleration in response to constant continuous applied forces.

This is very helpful because it tells us how the net forces on each segment of the ship must be related (since the acceleration of each segment is determined by the net force on that segment). The net force must be highest at the rear end of the ship, and must decrease linearly with distance towards the front of the ship ("distance" as measured in the rest frame of the ship). That means that, if the applied thrusts at each segment are different than that relationship, then internal forces will develop within the ship to "correct" the total net force on each segment to what it "should" be to meet the Born rigid condition. (The forces may oscillate for a while until they settle into this equilibrium state, but assuming that there is at least some damping, they will eventually settle in.)

This also answers your question above: no matter *what* the starting thrusts are, *if* there is a steady-state equilibrium reached, then, since that equilibrium is a state of Born rigid acceleration, the acceleration at the front end of the ship must be *less* than at the rear end. The relative *velocities* of the end will depend on what frame we measure them in; but in the original rest frame of the ship, the velocity of the front end will be *less* than that of the rear end, while the ship continues to accelerate.

Let's now consider some examples to see what the final states look like; this will get back to your question about whether the ship will be stretched or compressed. Because the final state is one of Born rigid acceleration, in all cases, we must have

$$F_R > F_M > F_F$$

for the relationship between the forces on each segment (for a more exact relationship we would need to know the length of each segment in the ship's rest frame). Also, the net force on each segment must be positive, since the ship as a whole is moving in the positive x-direction.

(Case 1) The only thrust is on the rear segment. Then the final equilibrium state will look something like this:

$$F_R = T - C_R$$

$$F_M = C_R - C_F$$

$$F_F = C_F$$

where $C_R$ is the pressure force between the rear and middle segments, and $C_F$ is the pressure force between the middle and front segments. Note the signs of the forces; they *must* be as they are for the inequality $F_R > F_M > F_F$ to hold. Thus, in this case, the ship will end up compressed.

(Case 2) There are equal thrusts on the front and rear segments. Then the final equilibrium state will look something like this:

$$F_R = T + S_R$$

$$F_M = S_F - S_R$$

$$F_F = T - S_F$$

where now we have stretching forces between the segments. Note, once again, the signs of the forces; if the thrusts on the front and rear segments are equal, then the forces *must* be stretching forces for the inequality $F_R > F_M > F_F$ to hold. Thus, in this case, the ship will end up stretched.

Obviously, if the thrust on the front segment is *larger* than that on the rear, the ship will still end up stretched, just by a larger amount. If the thrust on the front segment is *smaller* than that on the rear, there should be a point at which the ship, in its final equilibrium state, will be neither stretched nor compressed (since if there is no force on the front end, the ship will end up compressed). However, that does *not* mean that the individual segments will be equidistant! In fact, for the overall length of the ship to remain the same even though there are net forces on each segment (which there must be for the segments to be accelerating), the forces must look like this:

$$F_R = T_R - C_R$$

$$F_M = C_R + S_F$$

$$F_F = T_F - S_F$$

In other words, there must be compression between the rear and middle segments, and stretching between the middle and front segments, with the amounts just canceling each other out, so that the overall length of the ship is the same as its original rest length.

A) [Local] The rod is stretched locally from the ends. The cross section decreasing and the length increasing at both ends with the middle retaining its original cross section or at least greater than the ends. Eventually breaking at one end or the other.

B) [Global] The rod stretches initially equally throughout its length, but because the force is applied to the ends they would tend to retain the cross section, so the middle of the rod would be the area of maximal stretching and eventual breakdown.

A or B ?

There are three issues with the above:

First, I'm not sure how you're deriving these two different predictions from a local vs. a global picture. The local picture is the one I'm going to expound below; the global picture would be something pretty simple, like this: the rod stretches until it breaks. *Where* it breaks is a local question, not a global one. :-) (I put the smiley face there, but I'm actually serious: as soon as you ask a question like "where does the rod break?", you're implicitly modeling the rod as an extended system with internal parts, not a single object, which means that you need a local model that looks at the internal parts of the rod and their interactions.)

Second, if you want to try and model how the cross section of the rod will respond to applied stresses, you will need to introduce additional assumptions about the material properties of the rod. I don't want to introduce that additional complication, so I'm just going to ignore the rod's cross section and talk about the internal energy stored in each segment, without discussing specifics about how it's stored.

Third, the models we've discussed so far don't include any internal energy stored in any of the segments, so they aren't adequate to model something like a rod breaking, since the general condition for a material to break under stress is that the stored internal energy in the material due to the stress exceeds the material's breaking strength. In order to add internal energy to our model, we would need, once again, additional assumptions about the material properties of the rod, to calculate how much internal energy is stored as a result of a given set of stresses. I'll give a very simple assumption of this sort below, but it will be very simple and crude, just to see in general how such a scheme might work; any real material would require more complicated models that, once again, I don't want to discuss here.

Let's model the rod the same way we did the ship above, with three segments, but now we have the force on the "rear" segment in the opposite direction from that on the front. The forces on the segments will then look like this:

$$F_R = S_R - T$$

$$F_M = S_F - S_R$$

$$F_F = T - S_F$$

where we have assumed that the force (T) on the front and rear ends is the same magnitude (but in opposite directions). If we further assume that the response of the material to stretching is uniform, then we should have [itex]S_F = S_R[/tex], and we have

$$F_R = S - T$$

$$F_M = S - S = 0$$

$$F_F = T - S$$

so that the middle of the rod will not move at all (but it will still be pulled on from each side). In other words, the rod will gain *no* net momentum in its original rest frame (its center of mass remains motionless). This is pretty much the simplest situation we can model.

Now, as I noted above, in order to determine where the rod will break, we need to make some assumption about how the rod will store internal energy due to applied stress. Here's a simple assumption of that sort: each segment stores an amount of internal energy equal to

$$U = K S$$

where K is a constant that we assume to be an innate property of the material, and S is the "balanced" stress on the segment--in other words, the amount of stress that is "equalized" by forces in opposite directions, so that it doesn't actually result in any motion of the segment. For each segment, this will be the stretching force S between the segments, since that is the amount of stress that is "balanced" on both sides--for the middle segment, this is obvious, and for the rear and front segments, we can see that, for the rod to stretch at all, we must have $T > S$, so that S will be the stress that is "balanced" by equal forces in opposite directions, and $+/- \left( T - S \right)$ will be the stress that is "unbalanced" and therefore results in motion of the segment.

With this simple assumption about internal energy stored, we can see that the internal energy will be the same for each segment, since the balanced stress on each segment is the same. Therefore, under this model, the rod is equally likely to break *anywhere* along its length. If we had a material with a more complicated relation of internal energy to stress, we might find that the internal energy was greater in one segment than the others, so that one would be the one most likely to break. But again, I don't want to get into those complications here. Someone who has, in considerably more detail than I do here, is Greg Egan, at http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/SimpleElasticity.html" .

Agreed some of the heat will result from the propulsion mechanism itself and the impossibility of 100% efficiency. I assume we agree this is not relevant and simply assumes some infrared radiative dissapation or the ship will eventually cook.

Yes. In any real problem we'd have to include heat dissipation, but that's a complication I didn't want to introduce for these simple models.

Can we agree that the 2nd law of TD will still operate but that the results would be negligable and irrelevant to the real questions we are addressing??

I think we don't need to worry about the *details* of how dissipation happens, but we need to recognize that it has to happen somehow in order for the oscillations to damp out and the ship to reach an equilibrium. An idealized system with zero dissipation would continue to oscillate forever.

I think in actuality I am trying to integrate momentum and kinetic energy.

"Kinetic energy" isn't really a useful concept in relativistic problems. The 4-momentum and its components, total energy and total momentum, are the useful concepts. I've never seen a relativistic problem where separating out the kinetic energy helped in the solution.

That there are three fundamental principles involved here.
Conservation of mass and energy.
Conservation of momentum.
2nd law of thermodynamics.

The first two are aspects of a single law in relativity, conservation of energy-momentum. Or, when you get into more complicated problems that require using a stress-energy tensor instead of just a 4-momentum vector, the single law is conservation of the stress-energy tensor, in the form:

$$\frac{D T^{ab}}{D x^b} = 0$$

which says that the covariant divergence of the stress-energy tensor is zero.

I trust we agree that the Lorentz mass increase is not relevant to our discussion.

Yes, if by this you mean mass increase solely due to the effects of changing reference frames.

The 4 momentum differs in its coordinate time component but still defines instantaneous change.

"Instantaneous" in a relative sense, yes. Relativity includes the concept of "events", which are taken to be points in spacetime, with no extension in any dimension. We've been modeling the continuous application of forces, such as the thrust of a rocket engine on a segment of a ship, as a continuous succession of events along the worldline of the ship segment. Since the events have no extension, what happens at each event happens instantaneously, to the level of precision of our model; essentially we are modeling each ship segment as a spatial point. If we wanted a more accurate model, we could increase the resolution so that we were looking at individual atoms instead of macroscopic segments of the ship; then we would be able to see the "travel time" of forces internal to each ship segment, but the forces applied to individual atoms would be modeled as a succession of instantaneous events.

Mass/energy is included in the concept of rest mass, invariant mass etc. but a possible change in inertial mass through applied force is not included in the basic formulations.

Not in the ones in my previous posts, but it is (in a very simple fashion) in the model earlier in this post. Such models do require assumptions beyond the basic postulates of relativity, about the properties of materials, as I've said, and we may not want to get into those complexities. For many problems it isn't necessary to do so.

 

[Austin0]

=PeterDonis;2389087]Austin0: Let me start with this from your post because I think it's the most basic point that needs to be clarified:

Originally Posted by Austin0
I am in complete agreement with this . With one small caveat.
It is agreed that initially the front and rear will unequivocally gain momentum before the middle within the context of momentum as energy. IMHO --Whether that momentum results in actual coordinate translation or not depends on magnitude.

This is false. *Any* gain in momentum must result in *some* coordinate translation. More precisely: if, viewed in a given frame, an object gains some 4-momentum, and the 4-momentum gained has a spatial component (i.e., momentum as well as energy) in that frame, then the object must gain *some* coordinate velocity in that frame.

We may be having a semantic problem here. The context of my quote was your description of the stage where applied momentum had not yet reached the middle of the system. I was making no statement here as to whether or not it would result in coordinate translation of the system when it had propagated throughout that system.
I think we both agree that no system motion can occur until the momentum has propagated to that point. Yes??

I think we both agree that the energy component of 4-momentum , the scalar product of mass and speed, is both instantaneous and absolute upon application to a system. ?

SO the question is regarding the time it is internally propagating and whether that results in instantaneous motion of the back of the system,, not whether it results in internal motion at some internal point

So a question; assuming a purely x direction of the initial vector do you think that this component of the 4-vector is quantitatively conserved?? That there is zero propagated motion in y or z ??

Assume a thin slice ,one molecule thick in a 1000 x 1000 atom matrix.
Assume a single particle impacting one atom squarely in the x direction.
This is then propagated generally in the x dir. but inevitably there will be a yand z component also , yes?

Initially the impacted atom will translate in the x dir. but by the time the momentum has propagated to the middle, that original atom will have rebounded back into its original position in the matrix and will be oscillating at a decreasing amplitude and probably in additional directions due to the tensile forces from neighboring atoms.

We agree, no motion at this point in the front. So do you think there is net motion at the back??
For there to be would seem to neccessitate a coordinated pressure sustained all along the back. For system coordinate motion at the back at this point could only be a result of overall pressure and contraction along the whole face ,not simply the motion initially imparted to the first atom which was exceedingly localized and is now an oscillation..

Now at the point when the momentum reaches the front, what is the state of the back of the system?

Isn't it essentially in an inertial state with a slightly increased kinetic energy distributed as vibration??

Wouldn't the same consideration that applied at the beginning. I.e that the system couldn't move at the front until the energy reached there , also apply in this case?

That reaching the front would effect some degree of actual local motion of atoms but the inertia of the rest of the system would constrain that to an oscillation, which would then propagate backward through the system.

You have agreed that internal disappation can result in increased inertial mass with no necessary increase in coordinate velocity.

However, an object may have internal degrees of freedom (for example, its individual molecules may vibrate). Energy that is in these internal degrees of freedom--for example, heat--is included, relativistically, in the rest mass of the object. In the example above of the ship oscillating in response to applied thrust, and part of the energy of the oscillation being converted into heat, the effect would be an increase in the ship's rest mass. This is still an increase in 4-momentum, but it would not appear as an increase in velocity; it would appear as an increase in inertial mass, but due to the object's rest mass increasing, not due to its 4-velocity increasing.

Doesn't it follow that, in principle, given a large enough system ,that a single particles momentum could be totally internally converted without ever reaching the opposite boundary??

Resulting in increased heat and inertial mass but no velocity ?

Conforming to the overall conservation of energy and mass and the conservation of energy-momentum also??
The math for the application of Force treats a complex system as a black box labled M
, no explicit consideration of internal structure etc.

What you put into one side of the box is what you get coming out of the box divided by overall mass, instantly.
I have agreed that within a certain macroscopic range this is operationally sound. With two cannonballs any other considerations are so neglible as to make the effort calculating them pointless.
But this discussion has taken us into the realm of absolutes, the extremes of range, fundamental questions of the mechanism of the propagation of momentum.
Part of me would just like to agree to anything you say and get on to other aspects of the original enquiry but some of these questions are central to the larger issue so I can only hope we can reach some grounds of agreement sufficient to carry on.

This observation affects a number of the scenarios you've suggested: basically, *any* scenario where an application of a force in a particular direction does not immediately result in the object acquiring a momentum in that direction *cannot* be correct. The momentum may reside in only one part of the object (as above, only the rear segment has a momentum on the initial application of the force), but *some* part of the object must move immediately on application of a force. Given that observation, I'm not going to comment on scenarios you've suggested that appear to me to violate this rule. For example, this:

Originally Posted by Austin0
Given that the magnitude of force is within the materials ability to transmit it fast enough, applied energy, momentum, propagates through the system, not as motion, but as a reciprocal oscillation.

doesn't really capture what's going on. To see how the oscillation comes about, you have to model the object as a bunch of individual segments that can move independently (but may exert forces on adjacent segments), as we did with the ship; then you can see that *some* segment is always moving in response to the applied force, and the sum of the motions of all the segments always equals the total momentum gained from the applied force, so that conservation of momentum is satisfied. If this is what you mean by "reciprocal oscillation", that's fine, but it's not something that happens instead of the system moving as a whole; the system *is* moving as a whole at the same time that the oscillation is propagating through it.

We have agreed that no motion can take place at the front until momentum has reached it.
If we take a single vanishingly short pulse , then yes there is actual motion locally at the leading edge of the resultant wave as it propagates. But as it moves it does so as a small displacement within the tensile elasticity of the matrix, at a velocity totally unrelated to the velocity vector of the original impulse .
Behind it how can there be general displacement?
Atomic reactions are exceedingly fast compared to the velocity of sound right??

If the energy of momentum is located at this internal front, conservation would seem to mean that it could not reside in actual increased momentum as velocity ,of all the particles behind it wouldn't it?

If this applies then wouldn't it seem to follow that a continuous application would operate on the same principle. Overall system velocity increase results from propagation throughout the system not locally except as an initial pressure adjustment?

We already agreed earlier that in a sound wave there is always a local motion someplace but no overall displacement of the medium. You suggested that sustained applied momentum was different
I thought that had been addressed but maybe not.
DO you think there is a fundamental difference in propagation between coherent sound and white sound??

DO you think there is any significant difference between white sound and the applied momentum of a large number of random particles??

Isn't each individual collision equivalent to microscopically ringing a tiny atomic resonant bell?

Hi PeterDonis I am out of time so will get back for the rest of your post.
Hopefully we can resolve these issues soon, Thanks

 

[Austin0]

=PeterDonis;2389087]
This also answers your question above: no matter *what* the starting thrusts are, *if* there is a steady-state equilibrium reached, then, since that equilibrium is a state of Born rigid acceleration, the acceleration at the front end of the ship must be *less* than at the rear end. The relative *velocities* of the end will depend on what frame we measure them in; but in the original rest frame of the ship, the velocity of the front end will be *less* than that of the rear end, while the ship continues to accelerate.

This appears to be entering a desired end result , the conclusion , directly into the deductive chain.
ANd then simply determining the physics and intermediate arguments in retrospect to conform to this conclusion.

Even going with the premise that unequal force applied at different locations would result in different stresses after equilibrium is established; If the force applied is constant
and the slight difference due to contraction or expansion is stabalized........ what would lead to continued relative motion between different parts of the ship?
Are you talking about a differing dynamic thrust increase at different parts of the ship?

(Case 1) The only thrust is on the rear segment. Then the final equilibrium state will look something like this:

$F_R > F_M > F_F$ to hold. Thus, in this case, the ship will end up compressed.

(Case 2) There are equal thrusts on the front and rear segments. Then the final equilibrium state will look something like this:

$$F_R = T + S_R$$

$$F_M = S_F - S_R$$

$$F_F = T - S_F$$

where now we have stretching forces between the segments. Note, once again, the signs of the forces; if the thrusts on the front and rear segments are equal, then the forces *must* be stretching forces for the inequality $F_R > F_M > F_F$ to hold. Thus, in this case, the ship will end up stretched.

I don't understand where any net stretching would occur if the forces are balanced at each end?
Or why you think that the inequality would hold??

Obviously, if the thrust on the front segment is *larger* than that on the rear, the ship will still end up stretched, just by a larger amount.

How does this equate. If the force is greater at the front then the force at the rear in the same direction would seem to have to decrease the inertial resistence from the rear and thus lessen the stretching forces from the front or am I missing something here?

If the thrust on the front segment is *smaller* than that on the rear, there should be a point at which the ship, in its final equilibrium state, will be neither stretched nor compressed (since if there is no force on the front end, the ship will end up compressed). However, that does *not* mean that the individual segments will be equidistant! In fact, for the overall length of the ship to remain the same even though there are net forces on each segment (which there must be for the segments to be accelerating), the forces must look like this:

$$F_R = T_R - C_R$$

$$F_M = C_R + S_F$$

$$F_F = T_F - S_F$$

In other words, there must be compression between the rear and middle segments, and stretching between the middle and front segments, with the amounts just canceling each other out, so that the overall length of the ship is the same as its original rest length.

Would you agree that placing a weight on top of a vertical spring would be equivalent to applying a constant acceleration??
Initially compression would start at the locale of the force and propagate through the spring and then recoil. After equilibrium was regained we agree there would be a net compression.
Do you think there would then be a gradient of compression, greater at the top and decreasing toward the bottom??
Or uniform throughtout the spring , given that it is uniform in construction??


Let's model the rod the same way we did the ship above, with three segments, but now we have the force on the "rear" segment in the opposite direction from that on the front. The forces on the segments will then look like this:

$$F_R = S_R - T$$

$$F_M = S_F - S_R$$

$$F_F = T - S_F$$

where we have assumed that the force (T) on the front and rear ends is the same magnitude (but in opposite directions). If we further assume that the response of the material to stretching is uniform, then we should have [itex]S_F = S_R[/tex], and we have

$$F_R = S - T$$

$$F_M = S - S = 0$$

$$F_F = T - S$$

so that the middle of the rod will not move at all (but it will still be pulled on from each side). In other words, the rod will gain *no* net momentum in its original rest frame (its center of mass remains motionless). This is pretty much the simplest situation we can model.

Now, as I noted above, in order to determine where the rod will break, we need to make some assumption about how the rod will store internal energy due to applied stress. Here's a simple assumption of that sort: each segment stores an amount of internal energy equal to

$$U = K S$$

where K is a constant that we assume to be an innate property of the material, and S is the "balanced" stress on the segment--in other words, the amount of stress that is "equalized" by forces in opposite directions, so that it doesn't actually result in any motion of the segment. For each segment, this will be the stretching force S between the segments, since that is the amount of stress that is "balanced" on both sides--for the middle segment, this is obvious, and for the rear and front segments, we can see that, for the rod to stretch at all, we must have $T > S$, so that S will be the stress that is "balanced" by equal forces in opposite directions, and $+/- \left( T - S \right)$ will be the stress that is "unbalanced" and therefore results in motion of the segment.

With this simple assumption about internal energy stored, we can see that the internal energy will be the same for each segment, since the balanced stress on each segment is the same. Therefore, under this model, the rod is equally likely to break *anywhere* along its length. If we had a material with a more complicated relation of internal energy to stress, we might find that the internal energy was greater in one segment than the others, so that one would be the one most likely to break. But again, I don't want to get into those complications here. Someone who has, in considerably more detail than I do here, is Greg Egan, at http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/SimpleElasticity.html" .

"Kinetic energy" isn't really a useful concept in relativistic problems. The 4-momentum and its components, total energy and total momentum, are the useful concepts. I've never seen a relativistic problem where separating out the kinetic energy helped in the solution.

But it certainly is a component of total energy yes??So if you are going to actually consider total energy, either quantitatively in an actual problem or situation , or simply as a matter of principle without detailed computation as we are doing ,,it still must be factored in.No?
Can you apply OHM's law, in principle, across its range of applicability without considering temperature at some levels of that range?
Thanks


.

"Instantaneous" in a relative sense, yes. Relativity includes the concept of "events", which are taken to be points in spacetime, with no extension in any dimension. We've been modeling the continuous application of forces, such as the thrust of a rocket engine on a segment of a ship, as a continuous succession of events along the worldline of the ship segment. Since the events have no extension, what happens at each event happens instantaneously, to the level of precision of our model; essentially we are modeling each ship segment as a spatial point. If we wanted a more accurate model, we could increase the resolution so that we were looking at individual atoms instead of macroscopic segments of the ship; then we would be able to see the "travel time" of forces internal to each ship segment, but the forces applied to individual atoms would be modeled as a succession of instantaneous events.[/QUOTE]
Agreed entirely. Everything you have said. In fact we have both agreed that any input of momentum would result in immediate motion at the local atomic level.
But the math does not make this distinction. A complex system appears in the equations as a discrete entity. Simply Mass

Not in the ones in my previous posts, but it is (in a very simple fashion) in the model earlier in this post. Such models do require assumptions beyond the basic postulates of relativity, about the properties of materials, as I've said, and we may not want to get into those complexities. For many problems it isn't necessary to do so

Agreed we don't want to get sidetracked into specific consideration of materials etc.

One last quickie. If there was a long line of spheres as before with an impacting sphere:
A) At some finite number of spheres the last sphere would not accelerate. No coordinate translation.
or
B) No matter how many spheres were added to the line the last sphere would undergo "some" translation.
So A) or B) ?

 

[PeterDonis]

Austin0: First, a general comment. You keep on throwing different, more complicated scenarios at me when we apparently haven't reached complete agreement on the simple one I've been discussing. Let's please hold off on more complicated scenarios (throwing in thousands of atoms, y and z dimensions of space, requiring assumptions about the tensile strength of materials and how they respond to stresses, etc.) until we've got complete agreement on the simple one.

We may be having a semantic problem here. The context of my quote was your description of the stage where applied momentum had not yet reached the middle of the system. I was making no statement here as to whether or not it would result in coordinate translation of the system when it had propagated throughout that system.
I think we both agree that no system motion can occur until the momentum has propagated to that point. Yes??

No, according to the definition of system motion that corresponds to the global version of conservation of momentum I gave several posts ago. According to that definition, the "system motion" is just the sum of the motions of all the individual parts of the system. If that sum results in a net motion, then the system as a whole is moving. So, for example, if we have a rocket just at the rear segment of a ship, and we turn on its engine, initially only the rear segment is moving, but since that equates to net motion when all the segments' motions are summed (since no other segment is moving), the system as a whole is moving--it has net momentum in a particular direction.

If this definition bothers you, then forget the term "system motion" and let's just talk about the individual motions of each part of the system. We'll get the same answers anyway when we try to predict actual measurements.

You have agreed that internal disappation can result in increased inertial mass with no necessary increase in coordinate velocity.

I have agreed that dissipation can result in increased inertial mass (rest mass). I have *not* agreed that there is no necessary increase in coordinate velocity if the external applied forces to the system are in a particular direction. The only way to increase a system's mass (e.g., by heating it) without changing its coordinate velocity is to add energy to it in such a way that the net momentum cancels out--for example, by heating it with two laser beams from opposite directions, with the beam intensities exactly equal, so that their net momentum cancels out. Internal dissipation in response to an applied force in a particular direction won't do that.

Remember the rule I gave last time: *any* applied force in a particular direction *must* result in *some* coordinate velocity in that direction.

What you put into one side of the box is what you get coming out of the box divided by overall mass, instantly.

At the level of accuracy we've been modeling, yes. You're correct that how we see the momentum propagate (this isn't quite the term I would use) depends on the level of accuracy of our model. But that means that this:

But this discussion has taken us into the realm of absolutes, the extremes of range, fundamental questions of the mechanism of the propagation of momentum.

is not quite accurate. The mechanisms in the models we've been discussing for propagating momentum are anything but fundamental. We've been modeling what are basically contact forces between macroscopic segments of objects; in this type of model, momentum transfer is basically instantaneous between adjacent objects. But those contact forces aren't fundamental; they're due to the underlying physics of atoms and the internals of atoms, which are ultimately due to quantum mechanics and mechanisms of momentum transfer that are quite different.

We've been discussing how we would deal relativistically with the motion of macroscopic objects that might have internal parts, but for which all the forces we're interested in can be modeled in the simple ways we've been modeling them. That let's us concentrate on the aspects that are due to relativity itself, rather than to other things like the material properties of objects. But that means we won't be able to answer some questions that are natural to ask, and still remain within those boundaries.

If we take a single vanishingly short pulse , then yes there is actual motion locally at the leading edge of the resultant wave as it propagates. But as it moves it does so as a small displacement within the tensile elasticity of the matrix, at a velocity totally unrelated to the velocity vector of the original impulse .

Once again, let's keep to the simple model before introducing complexities. Let's take the ship with three segments, but now, instead of applying a steady thrust at the rear, we'll apply the thrust only for a very short time--essentially, long enough to get the rear segment to move, but not long enough for its movement to develop any internal forces with other segments. Then, at time t = 0 in the original rest frame, when we apply the force at the rear, we have:

$$F_R = T$$

$$F_M = 0$$

$$F_F = 0$$

After a very short time dt, the rear segment will have moved slightly towards the middle one, and it will have momentum equal to $T dt$. But by then the thrust will be removed, so the forces will be:

$$F_R = - C$$

$$F_M = C$$

$$F_F = 0$$

Now let's suppose, just to make things simple, that the compressive force C just happens to be equal to the original thrust T. Then, after another short time dt, the rear segment will have stopped moving, and the middle one will be moving forward. Thus, the rear segment, in the original rest frame, will have zero momentum, and the middle one will now have momentum $T dt$, so that momentum is conserved. Then the forces will be:

$$F_F = 0$$

$$F_M = - T$$

$$F_F = T$$

Now wait one more short interval of time dt. Then the middle segment will have stopped moving, and the front segment will have moved forward, and will have momentum $T dt$, so that momentum is still conserved. At that point, the forces will be zero on all segments. But has the ship stopped moving? No--the front segment is still moving forward. That means that, if we wait another short interval of time dt, we will have:

$$F_F = 0$$

$$F_M = T$$

$$F_F = - T$$

as the front segment now moves further forward and stretches the ship. So the ship will keep moving forward. If we add some damping, to be more realistic, the oscillations will eventually damp out, and when they do, the final forward momentum of the ship will still be $T dt$, but of course in a steady state it will be equally distributed among the three segments, so the final velocity of the ship as a whole, as seen from its original rest frame, will be (assuming each segment has equal mass, and that the heating from the damping is negligible) one-third of the velocity that the original impulse force imparted to the rear segment.

This general picture would be the same if we took a bulk material and looked at the motion of its atoms; the final velocity of the object as a whole would still be the original momentum imparted by the impulse force, divided by the total mass of the object. If the original impulse force were applied to a single atom, then of course the initial velocity of that atom would be much higher than the final velocity--but the initial *momentum* would be the same as the final momentum.

We already agreed earlier that in a sound wave there is always a local motion someplace but no overall displacement of the medium. You suggested that sustained applied momentum was different.

The question whether sound waves are propagating in an object is independent of the question whether the object is undergoing overall motion. The phenomena may be related in a specific case (for example, the sound waves were caused by a very short impulse force applied to the object, which also induced overall motion because it was applied in a specific direction), but they are independent. The easiest way to separate them is as follows: overall motion of the object is motion of its center of mass, relative to some reference position we're interested in. Sound waves are waves of compression and expansion, viewed from the center of mass frame of the object (in other words, the frame in which its center of mass is at rest).

This appears to be entering a desired end result , the conclusion , directly into the deductive chain.

Then you haven't followed the argument correctly. Here's the deductive chain:

(1) For an object to undergo steady-state, equilibrium motion, in which all internal oscillations have been damped out, there must be no relative motion between the internal parts of the object, as viewed from the MCIF of each part at any event along the part's worldline.

(2) For an object under acceleration, there is only one state of motion which corresponds to no relative motion between internal parts of the object as specified above: Born rigid acceleration.

(3) Therefore, any state of motion of an accelerated object which is a steady-state equilibrium, with no internal oscillations, as described above, must be a state of Born rigid acceleration.

The reason I introduced the condition of steady-state equilibrium is that that was the kind of state of motion we were interested in, and it's only that condition that leads to Born rigid acceleration as the final state. That doesn't mean that *every* state of motion of an accelerated object must be a state of Born rigid acceleration; there are at least two categories of states for which that isn't true:

(i) States in which the object is still undergoing internal oscillations that haven't yet been damped out--or, of course, oscillations that will never damp out, because we've modeled a perfectly elastic object with no damping;

(ii) States in which the object is not tending towards a steady-state equilibrium--for example, the scenario I discussed a number of posts ago, in which the acceleration of the front of the ship is such that the ship keeps stretching until it breaks. I'm sure we could come up with other such scenarios.

How does this equate. If the force is greater at the front then the force at the rear in the same direction would seem to have to decrease the inertial resistence from the rear and thus lessen the stretching forces from the front or am I missing something here?

You're missing what I was comparing it to. If the thrust on the front is greater than the thrust on the rear, the ship will be stretched by a greater amount than it would be if the thrust on the front was *equal* to the thrust on the rear.

Would you agree that placing a weight on top of a vertical spring would be equivalent to applying a constant acceleration??
Initially compression would start at the locale of the force and propagate through the spring and then recoil. After equilibrium was regained we agree there would be a net compression.
Do you think there would then be a gradient of compression, greater at the top and decreasing toward the bottom??
Or uniform throughtout the spring , given that it is uniform in construction??

There would be a gradient of compression; it would be greater at the bottom of the spring and less at the top, just as the water pressure in the ocean is greater the deeper you go, or air pressure at sea level is greater than at a high altitude.

But it certainly is a component of total energy yes??So if you are going to actually consider total energy, either quantitatively in an actual problem or situation , or simply as a matter of principle without detailed computation as we are doing ,,it still must be factored in.No?

My point is just that in relativity, when you calculate total energy, you're already including kinetic energy; you don't have to worry about it separately. And in virtually all problems where relativistic effects are significant, total energy is both easier and more useful to calculate.

But the math does not make this distinction. A complex system appears in the equations as a discrete entity. Simply Mass

At the level of accuracy we've been modeling, yes. You can always increase the accuracy of your model, at the expense of greater complexity and difficulty in calculation.

One last quickie. If there was a long line of spheres as before with an impacting sphere:
A) At some finite number of spheres the last sphere would not accelerate. No coordinate translation.
or
B) No matter how many spheres were added to the line the last sphere would undergo "some" translation.
So A) or B) ?

Good, we've gotten to the spheres, which was what I wanted to discuss next. :-)

The quick answer is B. Can you see why from what I've posted already?

 

[Austin0]

Austin0: First, a general comment. You keep on throwing different, more complicated scenarios at me when we apparently haven't reached complete agreement on the simple one I've been discussing. Let's please hold off on more complicated scenarios (throwing in thousands of atoms, y and z dimensions of space, requiring assumptions about the tensile strength of materials and how they respond to stresses, etc.) until we've got complete agreement on the simple one.

I am sorry you think my matrix scenario was unduly complex. We have agreed there is no point in complicating things with specific tensile strengths etc. I was not suggesting we do so. But within the context of basic principles I fail to see how it is essentially more complex that any of the scenarios we are considering. As it relates directly to the current question I am curious as to why it is simply not addressed.
It seems to me that part of the problem we are having is a basic difference in mindset.
I am viewing the question of applied momentum as a complete continuum, starting from zero and progressing through the range of magnitudes and conditions with different responses varying through that range.
You view it as an absolute linear response regardless of magnitude or internal considerations. ALthough I would agree that the conservation of mass and energy can be viewed in this way, I do not see how this could possibly apply to momentum unless the 2nd law of thermodynamics was not valid.
Or the fundamental conservation of mass and energy for that matter.
For the math , and your interpretation from the math, posits that a specific amount of energy could enter a system and increase internal kinetic energy and at the same time be 100% translated into system translational momentum without consideration of the internal characteristics and composition of that system.
I just don't see how that could work.

Originally Posted by Austin0
We may be having a semantic problem here. The context of my quote was your description of the stage where applied momentum had not yet reached the middle of the system. I was making no statement here as to whether or not it would result in coordinate translation of the system when it had propagated throughout that system.
I think we both agree that no system motion can occur until the momentum has propagated to that point. Yes??

No, according to the definition of system motion that corresponds to the global version of conservation of momentum I gave several posts ago. According to that definition, the "system motion" is just the sum of the motions of all the individual parts of the system. If that sum results in a net motion, then the system as a whole is moving. So, for example, if we have a rocket just at the rear segment of a ship, and we turn on its engine, initially only the rear segment is moving, but since that equates to net motion when all the segments' motions are summed (since no other segment is moving), the system as a whole is moving--it has net momentum in a particular direction.

This quote does not relate to my post at all but is addressed to a different stage where the momentum has progressed farther. See your quote following.

overall motion of the object is motion of its center of mass, relative to some reference position we're interested in.

Originally Posted by Austin0
You have agreed that internal disappation can result in increased inertial mass with no necessary increase in coordinate velocity.

I

have agreed that dissipation can result in increased inertial mass (rest mass). I have *not* agreed that there is no necessary increase in coordinate velocity if the external applied forces to the system are in a particular direction. The only way to increase a system's mass (e.g., by heating it) without changing its coordinate velocity is to add energy to it in such a way that the net momentum cancels out--for example, by heating it with two laser beams from opposite directions, with the beam intensities exactly equal, so that their net momentum cancels out. Internal dissipation in response to an applied force in a particular direction won't do that.

Are you now saying that directed momentum entering a system cannot result in internal dissipation?


Originally Posted by Austin0
But this discussion has taken us into the realm of absolutes, the extremes of range, fundamental questions of the mechanism of the propagation of momentum.

is not quite accurate. The mechanisms in the models we've been discussing for propagating momentum are anything but fundamental. We've been modeling what are basically contact forces between macroscopic segments of objects; in this type of model, momentum transfer is basically instantaneous between adjacent objects. But those contact forces aren't fundamental; they're due to the underlying physics of atoms and the internals of atoms, which are ultimately due to quantum mechanics and mechanisms of momentum transfer that are quite different.

Would you agree that the only actual instantaneous transfer of momentum does happen on the level of atoms not between macroscopic segments which each require a finite propagation time interval ?
Doesn't any discussion of the propagation of momentum neccessitate going down to the phonon level? Not in specifics but in principles.

We've been discussing how we would deal relativistically with the motion of macroscopic objects that might have internal parts, but for which all the forces we're interested in can be modeled in the simple ways we've been modeling them. That let's us concentrate on the aspects that are due to relativity itself, rather than to other things like the material properties of objects. But that means we won't be able to answer some questions that are natural to ask, and still remain within those boundaries.

We really have several different questions going which itself leads to confusion.
1) The fundamental propagation and how it relates to the initial stage of application and attaining equilibrium.
2) the actually more important question of the sustained period after the intial stage and how they evolve. For this question I agree that simpler models will suffice.
3) A peripheral question of the absolute application of the conservation of momentum which doesn't neccessarily add anything significant to the central enquiry but has come and up begs resolution.

Originally Posted by Austin0
If we take a single vanishingly short pulse , then yes there is actual motion locally at the leading edge of the resultant wave as it propagates. But as it moves it does so as a small displacement within the tensile elasticity of the matrix, at a velocity totally unrelated to the velocity vector of the original impulse .


This general picture would be the same if we took a bulk material and looked at the motion of its atoms; the final velocity of the object as a whole would still be the original momentum imparted by the impulse force, divided by the total mass of the object. If the original impulse force were applied to a single atom, then of course the initial velocity of that atom would be much higher than the final velocity--but the initial *momentum* would be the same as the final momentum.



Originally Posted by Austin0
This appears to be entering a desired end result , the conclusion , directly into the deductive chain.

Then you haven't followed the argument correctly. Here's the deductive chain:

(1) For an object to undergo steady-state, equilibrium motion, in which all internal oscillations have been damped out, there must be no relative motion between the internal parts of the object, as viewed from the MCIF of each part at any event along the part's worldline.

(2) For an object under acceleration, there is only one state of motion which corresponds to no relative motion between internal parts of the object as specified above: Born rigid acceleration.

(3) Therefore, any state of motion of an accelerated object which is a steady-state equilibrium, with no internal oscillations, as described above, must be a state of Born rigid acceleration.

Right here in number 2 is the conclusion stated as an argument.

The reason I introduced the condition of steady-state equilibrium is that that was the kind of state of motion we were interested in, and it's only that condition that leads to Born rigid acceleration as the final state. That doesn't mean that *every* state of motion of an accelerated object must be a state of Born rigid acceleration; there are at least two categories of states for which that isn't true:

Here again

(

ii) States in which the object is not tending towards a steady-state equilibrium--for example, the scenario I discussed a number of posts ago, in which the acceleration of the front of the ship is such that the ship keeps stretching until it breaks. I'm sure we could come up with other such scenarios

.

Here is another conclusion. You have not presented any physical principle that I am aware of to support this stretching that continues until disruption.



Originally Posted by Austin0
Would you agree that placing a weight on top of a vertical spring would be equivalent to applying a constant acceleration??
Initially compression would start at the locale of the force and propagate through the spring and then recoil. After equilibrium was regained we agree there would be a net compression.
Do you think there would then be a gradient of compression, greater at the top and decreasing toward the bottom??
Or uniform throughtout the spring , given that it is uniform in construction??

There would be a gradient of compression; it would be greater at the bottom of the spring and less at the top, just as the water pressure in the ocean is greater the deeper you go, or air pressure at sea level is greater than at a high altitude.

You are quite right. I was failing to consider the weight of the spring itself. There would neccessarily be some gradient even in a vertical spring here on Earth without any weight.

Originally Posted by Austin0
But it certainly is a component of total energy yes??So if you are going to actually consider total energy, either quantitatively in an actual problem or situation , or simply as a matter of principle without detailed computation as we are doing ,,it still must be factored in.No?

My point is just that in relativity, when you calculate total energy, you're already including kinetic energy; you don't have to worry about it separately. And in virtually all problems where relativistic effects are significant, total energy is both easier and more useful to calculate.

I don't understand what you are saying here. It seems like you are saying ,,,when we calculate net profit we of course include loss in the calculation so you don't need to think about loss separately. This seems to say there is some method of calculating total energy without having to calculate kinetic energy.


Originally Posted by Austin0
One last quickie. If there was a long line of spheres as before with an impacting sphere:
A) At some finite number of spheres the last sphere would not accelerate. No coordinate translation.
or
B) No matter how many spheres were added to the line the last sphere would undergo "some" translation.
So A) or B) ?

Good, we've gotten to the spheres, which was what I wanted to discuss next. :-)

The quick answer is B. Can you see why from what I've posted already?

Only if 2nd law of TD is inoperative for some reason.
Only if energy conservation means that a discrete quantity of energy can be spread out within a system without limit by some process of infinite division.

Getting on to the previous spheres would be fantastic. :-) Thanks

 

[PeterDonis]

Austin0:

I am viewing the question of applied momentum as a complete continuum, starting from zero and progressing through the range of magnitudes and conditions with different responses varying through that range.
You view it as an absolute linear response regardless of magnitude or internal considerations. ALthough I would agree that the conservation of mass and energy can be viewed in this way, I do not see how this could possibly apply to momentum unless the 2nd law of thermodynamics was not valid.
Or the fundamental conservation of mass and energy for that matter.
For the math , and your interpretation from the math, posits that a specific amount of energy could enter a system and increase internal kinetic energy and at the same time be 100% translated into system translational momentum without consideration of the internal characteristics and composition of that system.
I just don't see how that could work.

I think the above is really asking the same question as the following:

Are you now saying that directed momentum entering a system cannot result in internal dissipation?

No; I'm saying that directed momentum must result in directed momentum. :-)

Suppose I start with an object of rest mass M, at rest. I add some 4-momentum P to it; to keep it simple, say that in the object's original rest frame, P has only a spatial component in the x-direction, p_x. Then there are two possible results:

(1) No dissipation; then the object's final rest mass will be the same as its initial rest mass, M, and its final velocity will be (if the momentum p_x is small enough that the non-relativistic approximation is valid) v = p_x / M.

(2) With dissipation; then the object's final rest mass will be M' > M, and its final velocity will be (in the non-relativistic approximation) v' = p_x / M' < v. In other words, it will have the same final momentum, but its final velocity will be smaller because some of the applied energy went into its rest mass instead.

In either case, the object will have *some* coordinate velocity in the x-direction. There is no way to add a directed momentum p_x to the object and *not* have it end up with *some* coordinate velocity in the x-direction. Of course the "no dissipation" case is an idealization; no real object will have zero dissipation. But there are real cases where the dissipation is small enough that it can be ignored for the accuracy required for the specific problem. Also, of course, I'm ignoring the fact that a real object that was heated up by dissipation would then radiate that heat, which would further change its 4-momentum (since it would lose rest mass)--but then we would also have to account for the 4-momentum carried away by the radiation, in order to balance the books properly.

Does this address what you were thinking in the longer quote above? If not, please say what else isn't clear.

This quote does not relate to my post at all but is addressed to a different stage where the momentum has progressed farther. See your quote following.

The quote following was my comment that overall motion of a system is motion of its center of mass. But if any part of a system is moving, then its center of mass must be moving. The center of mass is not a physical object; it's an abstraction which is determined by the average location of all parts of the system. Saying that the center of mass is moving is *not* the same thing as saying that the particular physical segment of the object which happened to be located at the center of mass before it started moving, is moving. The latter seems to be what you are talking about, but it's not what I was talking about. Once again, though, if you're not comfortable with my definitions of "global" terms, we can just leave them out and talk about the motions of each individual part of the system. We'll end up with the same answers.

Would you agree that the only actual instantaneous transfer of momentum does happen on the level of atoms not between macroscopic segments which each require a finite propagation time interval ?

No; I would say the opposite. If you have a model that includes instantaneous transfer of momentum, it's because your model is not fundamental; it's at a level of detail that's not sufficient to capture the finite propagation times that are always present at the level of fundamental particles. It's at the level of fundamental particles and interactions that propagation times will always appear in one form or another.

Doesn't any discussion of the propagation of momentum neccessitate going down to the phonon level? Not in specifics but in principles.

In principle, if you want to understand how momentum propagates at a fundamental level, you have to look at quantum fields and interactions. (Phonons are quanta of vibration, but they're not really fundamental the way that, say, electrons and photons are.) At that level, momentum is transferred between real particles by the exchange of virtual particles; for example, the nucleus of an atom "holds" the electrons in certain orbitals by exchanging virtual photons with them in order to keep their momentum within certain limits. (Even this picture is arguably not fully fundamental, since it depends on perturbation theory, which is only an approximate solution to the equations of quantum field theory--but I don't want to get into that here.) This virtual photon exchange looks to us, at a macroscopic level, like an electromagnetic force between the nucleus and the electrons. Virtual particle exchanges take some amount of time--roughly speaking, that time is determined by the uncertainty principle combined with the magnitude of the momentum and energy being exchanged. But that time is so short that it can be ignored for many problems, and the force between the nucleus and the electrons can be treated as instantaneous, so that the atom acts like a single object.

Similarly, at the quantum level, interactions between atoms involve exchanges of virtual photons between the electrons in the atoms' outer shells, which take time. It's only in macroscopic models, when we don't care about the details of the interatomic forces, and we're working on time scales much larger than the interaction time for virtual particle exchanges, that we idealize the interatomic forces as instantaneous "contact forces" between atoms that have a definite size and collide like little billiard balls.

Right here in number 2 is the conclusion stated as an argument.

My premise 2 is a proposition that's already been proved by other means; it's not an assumption. You're correct that it's the crucial premise that leads to my conclusion, but that doesn't mean I'm arguing in a circle; it means my conclusion is correct.

Here is another conclusion. You have not presented any physical principle that I am aware of to support this stretching that continues until disruption.

Well, I specified exactly the scenario that would lead to it--that the *acceleration* experienced at the front and rear ends of the ship was the same. I explained how that was a *different* specification than saying that the *rocket thrust* at each end of the ship was constant and equal; the acceleration experienced at either end of the ship depends on the *net* force at that end, which is the resultant of the rocket thrust and any internal forces exerted on the end by other parts of the ship. I admitted that the specification of constant acceleration, rather than constant rocket thrust, was extremely unlikely to be realized by any real rocket (since it would require continually increasing rocket thrust at the front end of the ship), but unlikely is not the same as physically impossible.

It is true that I didn't make any specification of *how much* the ship would have to stretch before it broke, but I don't have to to know that it would eventually have to break with the specification I gave (constant acceleration of each end). For the ship not to break eventually, it would have to have infinite breaking strength, and that *is* physically impossible--not because values tending to infinity are impossible (after all, the scenario posits rocket thrust at the front end of the ship that must tend to infinity) but because relativity places finite limits on the breaking strength of materials. This condition is called the "weak energy condition", and it says that the stress in a material can't exceed its rest energy density (its rest energy divided by its volume in its rest frame). Maybe this is the additional physical principle you were looking for.

In any case, that specific scenario is not very important for our discussion; I brought it in only to show that it is possible to construct scenarios that do not tend to a steady-state equilibrium. Such scenarios may be extremely unlikely; that's fine.

I don't understand what you are saying here. It seems like you are saying ,,,when we calculate net profit we of course include loss in the calculation so you don't need to think about loss separately. This seems to say there is some method of calculating total energy without having to calculate kinetic energy.

Exactly; there is. Simple example: an object of rest mass m moving with velocity $\beta$ (in units such that the speed of light = 1). The total energy of the object is $E = \gamma m$, where $\gamma = 1 / \sqrt{1 - \beta^2}$. Simple and direct. There is no simple and direct formula for the relativistic kinetic energy of the object; to obtain that, I have to subtract its rest mass m from its total energy E and call what's left over "kinetic energy". Take any other relativistic problem and you'll find the same thing: there will be a simple and direct formula for the total energy, but to get the kinetic energy, you'll have to subtract the rest energy from the total energy; there won't be a simple and direct formula for the kinetic energy alone.

Only if 2nd law of TD is inoperative for some reason.
Only if energy conservation means that a discrete quantity of energy can be spread out within a system without limit by some process of infinite division.

I'm not sure either of these is relevant to the specific example where I gave the quick answer B. The reason B was the obvious quick answer is simple: as I said above, directed momentum must result in directed momentum. Let there be a row of spheres extending along the x-direction. Now hit the sphere on the left with an impulse to the right (i.e., in the positive x-direction). Call that sphere sphere #1. Sphere #1 now moves to the right and hits sphere #2. Sphere #2 *must* acquire *some* momentum to the right in this collision. If this isn't obvious to you, consider the possibilities:

(1) Sphere #1 stops moving to the right after the collision--either it is at rest, or it is now moving to the left. In this case sphere #2 must move to the right to conserve momentum.

(2) Sphere #1 continues moving to the right after the collision. In this case, sphere #2 must move to the right at least as fast as sphere #1, because the collision brought them into contact.

Either way, sphere #2 must be moving to the right after the collision. The same argument then carries through to sphere #3, #4, etc., up to any number of spheres.

Getting on to the previous spheres would be fantastic. :-)

OK, let's suppose now that spheres #1 through #N are in contact, each with its adjacent neighbors, and they all lie along the x-direction. We then hit sphere #1, from the left, with another sphere, sphere #0, which starts out with a certain momentum directed to the right--i.e., in the positive x-direction. (I'm leaving out the part about the spheres being suspended, which was in your original specification--I'm assuming the spheres to start out at rest, far out in empty space, with no other objects near enough to affect their motion.) Macroscopically, it looks like the impulse passes all the way through the row of spheres with no motion, until it finally "comes out" as motion at the other end, with sphere #N taking off while the rest of the spheres stay put. However, if we were to photograph the process with high-speed cameras, with enough spatial resolution to see deformations in the individual spheres, we would see this:

(1) Sphere #0 hits the left side of sphere #1. As a result, sphere #1 deforms--its left side is pushed inward, so there is now a net motion to the right of sphere #1. We'll assume, for simplicity, that sphere #0's motion stops completely as a result of its hitting sphere #1. Thus, sphere #1 now contains the momentum that was contained before in sphere #0. This will cause the center of mass of sphere #1 to shift to the right.

(2) Since sphere #1 is in contact with sphere #2, as soon as the wave of deformation has time to pass across sphere #1, and hit sphere #2, sphere #2 deforms, with its left side being pushed inward. If everything is tuned just right, sphere #2 deforming will coincide with sphere #1 "un-deforming"--the right side of sphere #1 will now push back on the rest of sphere #1, and sphere #1 will restore itself to its original shape, and its center of mass will shift back to the left, to where it was at the start of the process. Sphere #2 then contains the momentum transferred by the initial impulse.

(N) This process continues down the line until sphere #N is reached. Since this sphere has no neighbor to the right, once the wave of deformation starts at its left side (and sphere #(N-1) is pushed back into its original shape and position), there is nothing to keep all of sphere #N from continuing the motion. So now we see, macroscopically, that all the momentum of the initial impulse appears as motion of sphere #N to the right.

In other words, each sphere in succession does move to the right, but all of the motions except for the last one are reversed in the course of propagating the wave of deformation down the row of spheres. In a real case, things wouldn't be as "clean" as I've presented them, meaning that the momentum of the intial impulse would probably be distributed among two or more of the spheres during the process, though it would all come out contained in sphere #N at the end (if everything is tuned just right). But at any instant of time, viewed from the original rest frame of the spheres, the total momentum of the system as a whole--the sum of the net momentum contained in all the spheres, which must be directed to the right--will be equal to the initial impulse.

 

[Austin0]

The quote following was my comment that overall motion of a system is motion of its center of mass. But if any part of a system is moving, then its center of mass must be moving. The center of mass is not a physical object; it's an abstraction which is determined by the average location of all parts of the system. Saying that the center of mass is moving is *not* the same thing as saying that the particular physical segment of the object which happened to be located at the center of mass before it started moving, is moving. The latter seems to be what you are talking about, but it's not what I was talking about. Once again, though, if you're not comfortable with my definitions of "global" terms, we can just leave them out and talk about the motions of each individual part of the system. We'll end up with the same answers.

Of course the center of mass is an abstraction. A coordinate location.
But it is frame independant yes?? ANd it does correspond to a specific physical location within a system.
So saying the center of mass is moving, has a specific meaning of motion wrt an outside inertial frame. The center does not move relative to a comoving frame true?

So you have agreed that there can be no motion at a point in a system until momentum has reached it. YOu have proposed that there is actual instantaneous coordinate motion at the origin of applied momentum.
Now you are saying "But if any part of a system is moving, then its center of mass must be moving."
How is this consistent?

(2) For an object under acceleration, there is only one state of motion which corresponds to no relative motion between internal parts of the object as specified above: Born rigid acceleration.

My premise 2 is a proposition that's already been proved by other means; it's not an assumption. You're correct that it's the crucial premise that leads to my conclusion, but that doesn't mean I'm arguing in a circle; it means my conclusion is correct.

Certainly it has not been mentioned in this discussion let alone proved.
At this point we are talking about the basic physics of acceleration as applied to some models of propulsion.

Originally Posted by Austin0
Here is another conclusion. You have not presented any physical principle that I am aware of to support this stretching that continues until disruption.

Well, I specified exactly the scenario that would lead to it--that the *acceleration* experienced at the front and rear ends of the ship was the same.

equal acceleration does not, in itself, constitute either an explanation or a proof of expansion to the point of disruption.

I admitted that the specification of constant acceleration, rather than constant rocket thrust, was extremely unlikely to be realized by any real rocket (since it would require continually increasing rocket thrust at the front end of the ship)

The requirement of increasing thrust at the front to maintain constant acceleration is also not demonstrated at this point.
ANd you never did answer when I asked if Born acceleration requires increasing thrust at all points.

It is true that I didn't make any specification of *how much* the ship would have to stretch before it broke

,
We have agreed that specific quantitative details may be disregarded for the purposes of our inquiry.

Originally Posted by Austin0
I don't understand what you are saying here. It seems like you are saying ,,,when we calculate net profit we of course include loss in the calculation so you don't need to think about loss separately. This seems to say there is some method of calculating total energy without having to calculate kinetic energy.

Exactly; there is. Simple example: an object of rest mass m moving with velocity $\beta$ (in units such that the speed of light = 1). The total energy of the object is $E = \gamma m$, where $\gamma = 1 / \sqrt{1 - \beta^2}$. Simple and direct. There is no simple and direct formula for the relativistic kinetic energy of the object; to obtain that, I have to subtract its rest mass m from its total energy E and call what's left over "kinetic energy". Take any other relativistic problem and you'll find the same thing: there will be a simple and direct formula for the total energy, but to get the kinetic energy, you'll have to subtract the rest energy from the total energy; there won't be a simple and direct formula for the kinetic energy alone.

As far as I can see using this formula there will always be zero kinetic energy outside of the acquired momentum from acceleration. Of course there is no simple way of calculating internal kinetic energy. But we have agreed that for our purposes there is no need for actual quantitative values.
But an actual calculation of either rest mass or E in the foregoing equation would seem to neccessitate some non simple and direct calculation in order to include internal kinetic energy from dissipation.

I'm not sure either of these is relevant to the specific example where I gave the quick answer B. The reason B was the obvious quick answer is simple: as I said above, directed momentum must result in directed momentum. Let there be a row of spheres extending along the x-direction. Now hit the sphere on the left with an impulse to the right (i.e., in the positive x-direction). Call that sphere sphere #1. Sphere #1 now moves to the right and hits sphere #2. Sphere #2 *must* acquire *some* momentum to the right in this collision. If this isn't obvious to you, consider the possibilities:

(1) Sphere #1 stops moving to the right after the collision--either it is at rest, or it is now moving to the left. In this case sphere #2 must move to the right to conserve momentum.

(2) Sphere #1 continues moving to the right after the collision. In this case, sphere #2 must move to the right at least as fast as sphere #1, because the collision brought them into contact.

Either way, sphere #2 must be moving to the right after the collision. The same argument then carries through to sphere #3, #4, etc., up to any number of spheres.

On collision between sp/#1 and #2 there is an internal sound. A literal coherent sound wave the propagates both ways into #1 and #2 We know that these waves will resonate and reflect throughout the spheres. We know that some infintesimal amount of the propagated momentum itself will also be diisipated internally. SO (sph #1) p > (#2) p or conversely
(n+1) p < (n) p So unless you propose some kind of Zenoesque infinite divisibility there will be an infintesimal but steady attenuation of the magnitude of momentum until eventual total diminishment yes?

OK, let's suppose now that spheres #1 through #N are in contact, each with its adjacent neighbors, and they all lie along the x-direction. We then hit sphere #1, from the left, with another sphere, sphere #0, which starts out with a certain momentum directed to the right--i.e., in the positive x-direction. (I'm leaving out the part about the spheres being suspended, which was in your original specification--I'm assuming the spheres to start out at rest, far out in empty space, with no other objects near enough to affect their motion.) Macroscopically, it looks like the impulse passes all the way through the row of spheres with no motion, until it finally "comes out" as motion at the other end, with sphere #N taking off while the rest of the spheres stay put. However, if we were to photograph the process with high-speed cameras, with enough spatial resolution to see deformations in the individual spheres, we would see this:

(1) Sphere #0 hits the left side of sphere #1. As a result, sphere #1 deforms--its left side is pushed inward, so there is now a net motion to the right of sphere #1. We'll assume, for simplicity, that sphere #0's motion stops completely as a result of its hitting sphere #1. Thus, sphere #1 now contains the momentum that was contained before in sphere #0. This will cause the center of mass of sphere #1 to shift to the right.

(2) Since sphere #1 is in contact with sphere #2, as soon as the wave of deformation has time to pass across sphere #1, and hit sphere #2, sphere #2 deforms, with its left side being pushed inward. If everything is tuned just right, sphere #2 deforming will coincide with sphere #1 "un-deforming"--the right side of sphere #1 will now push back on the rest of sphere #1, and sphere #1 will restore itself to its original shape, and its center of mass will shift back to the left, to where it was at the start of the process. Sphere #2 then contains the momentum transferred by the initial impulse.

(N) This process continues down the line until sphere #N is reached. Since this sphere has no neighbor to the right, once the wave of deformation starts at its left side (and sphere #(N-1) is pushed back into its original shape and position), there is nothing to keep all of sphere #N from continuing the motion. So now we see, macroscopically, that all the momentum of the initial impulse appears as motion of sphere #N to the right.

In other words, each sphere in succession does move to the right, but all of the motions except for the last one are reversed in the course of propagating the wave of deformation down the row of spheres. In a real case, things wouldn't be as "clean" as I've presented them, meaning that the momentum of the intial impulse would probably be distributed among two or more of the spheres during the process, though it would all come out contained in sphere #N at the end (if everything is tuned just right). But at any instant of time, viewed from the original rest frame of the spheres, the total momentum of the system as a whole--the sum of the net momentum contained in all the spheres, which must be directed to the right--will be equal to the initial impulse.[/QUOTE]

How could the energy be spread out through two or more of the spheres when it is clear that at the instant it reaches the end of the final sphere, that sphere will definitely move and break contact ending the transfer of momentum??
SO do you think that if we assume ideal crystal spheres at 0 deg. K , as close to rigid as theoretically possible that there would be deformation or that anything significant would be different as far as the propagation of momentum?
YOu still seem to be proposing a universe where a sound wave in a rod will travel forever without losing either coherence or energy. Without attenuation or decay. Without the 2nd law of TD

 

[PeterDonis]

Of course the center of mass is an abstraction. A coordinate location.
But it is frame independant yes?? ANd it does correspond to a specific physical location within a system.
So saying the center of mass is moving, has a specific meaning of motion wrt an outside inertial frame. The center does not move relative to a comoving frame true?

The center of mass (CoM, to save typing) corresponds to a *coordinate* location within a system. Which particular physical part of the system, if any, is located at that coordinate location can change. The CoM is not "anchored" to any particular physical part of the system. It's just wherever the "average" location of all the parts of the object happens to be.

Whether or not the CoM is moving is frame dependent. You're correct that the CoM does not move in a comoving frame, if "comoving" is defined properly; but a "comoving" frame in this sense will not be an inertial frame, if there are external forces acting on the object.

The Wikipedia pages on http://en.wikipedia.org/wiki/Center_of_mass" have good (if brief) discussions of this stuff. However, they don't discuss what happens if the object is being acted on by an external force and so is being accelerated.

So you have agreed that there can be no motion at a point in a system until momentum has reached it. YOu have proposed that there is actual instantaneous coordinate motion at the origin of applied momentum.
Now you are saying "But if any part of a system is moving, then its center of mass must be moving."
How is this consistent?

As above, the CoM isn't "anchored" to any particular part of the object. It's just the "average" location of all the parts. It's not necessary that all the parts of the system are moving for the "average" location of the parts to be moving; just one part moving is enough.

Certainly it has not been mentioned in this discussion let alone proved.

The fact that Born rigid acceleration is the only state of motion of an accelerated object for which the internal distances between the parts remain constant has been mentioned several times in this thread, not just by me. We haven't proven it here, but there are plenty of discussions of this fact and how it comes about online and in textbooks. It's not a new idea; Born came up with the concept in, I believe, the 1920's. For a good quick reference to start with, try these online discussions of Rindler coordinates: http://en.wikipedia.org/wiki/Rindler_coordinates" (I think I've already posted these links earlier in this thread).

equal acceleration does not, in itself, constitute either an explanation or a proof of expansion to the point of disruption.

Once again, this has already been discussed in this thread. We discussed the motion of two separate spaceships with equal, constant acceleration, as in the "[URL spaceship paradox[/URL], and confirmed that their separation would continuously increase. Then I specified the motions of two ends of a single rocket ship in such a way that the worldlines of the two ends would be identical to those of the two separate spaceships. I admitted that my specification was extremely unlikely to occur in practice, but that doesn't make it physically impossible.

ANd you never did answer when I asked if Born acceleration requires increasing thrust at all points.

Not directly, but I did implicitly when I described the scenario that would produce Born rigid acceleration. I'll describe it again: Let there be a rocket ship that starts at rest along the x-axis of an inertial frame. At time t = 0 in that frame, let each individual segment of the ship start accelerating, with an acceleration equal to $c^2 / x_s$, where x_s is the x-coordinate of that segment in the original inertial frame. Then the ship as a whole will undergo Born rigid acceleration.

This specification, in itself, doesn't tell exactly what thrust will be required at each individual segment to produce the specified acceleration. However, if we assume that there are no internal stresses to start with in the ship, then, since Born rigid acceleration preserves the distances between adjacent segments, it should not produce any internal stresses. So the thrust required for each segment should simply be the rest mass of the segment times its specified acceleration. This does depend, though, on our assumption about the material properties of the ship being correct.

Also, the above specification is what would be required for an object to *start out* with Born rigid acceleration, and of course, as has been said before in this thread, it's extremely unlikely that it would happen that way in any real case. However, suppose an object starts out with some other "force profile" (for example, thrust only at the rear end), but then, as a result of the development of internal forces, reaches a state such that the net force on each segment of the object is exactly what it needs to be to cause all the segments to accelerate according to the same profile I described above. At that point, the object as a whole would have settled into a state of Born rigid acceleration; it's this kind of scenario I was talking about in previous posts. Of course in this case the internal stresses in the object would be non-zero, since the internal forces would be supplying part of the net force on each segment to make the force profile just right. But the internal stresses would not *change* any further once this state was reached; it would still be a stable steady-state equilibrium, just as an object resting on the surface of the Earth is in a stable steady-state equilibrium even though the stresses inside it are non-zero because of gravity.

We know that some infintesimal amount of the propagated momentum itself will also be diisipated internally.

No, we do *not* know this, at least I don't. One more time: **this would violate conservation of momentum**. Conservation of momentum requires that the directed momentum before the collision must be equal to the directed momentum after the collision. So we must have (for the case where only sphere #1 is moving before the collision but both spheres #1 and #2 may be moving after the collision):

$$p_{1-before} = p_{1-after} + p_{2-after}$$

From this, the rest of what I said follows. Directed momentum simply cannot be "dissipated internally" like you're saying.

How could the energy be spread out through two or more of the spheres when it is clear that at the instant it reaches the end of the final sphere, that sphere will definitely move and break contact ending the transfer of momentum??

It could be spread between two or more spheres *during the process*; I'm just allowing for this possibility since I haven't done a detailed calculation to show microscopically how the momentum propagates. After the process ends, it will all be in the final sphere.

SO do you think that if we assume ideal crystal spheres at 0 deg. K , as close to rigid as theoretically possible that there would be deformation or that anything significant would be different as far as the propagation of momentum?

Assuming that the material properties of the spheres remain intact during the process (i.e., none of the spheres shatter, melt, fragment, etc.), and that those properties support elastic deformation (i.e., spheres made of clay or something like that wouldn't work), then there would be deformation as I've described it. The temperature of the spheres doesn't matter (as long as it's below the melting point).

YOu still seem to be proposing a universe where a sound wave in a rod will travel forever without losing either coherence or energy. Without attenuation or decay. Without the 2nd law of TD

I'm proposing no such thing. I've said several times that any real object would have dissipation, and dissipation means that some of the *energy* added to the object gets turned into heat--i.e., it appears as an increase in the object's rest mass instead of an increase in the object's velocity. That is the sort of thing the 2nd law of TD deals with. But none of that changes the fact that *directed momentum* is conserved. I've already shown how that can happen, but one more time: consider an object of mass M, initially at rest, which then has some linear momentum p_x added to it:

(1) In the idealized case where there's no dissipation, the object's rest mass is unchanged, its linear momentum is p_x, and its final velocity (in the non-relativistic limit) will be v = p_x / M.

(2) In the realistic case where there is dissipation, the object's rest mass will increase to M' > M, so its final velocity will be v' = p_x / M', which is less than v. But its final linear momentum will still be p_x.

 

[PeterDonis]

Oops, forgot to include a comment on this:

As far as I can see using this formula there will always be zero kinetic energy outside of the acquired momentum from acceleration. Of course there is no simple way of calculating internal kinetic energy. But we have agreed that for our purposes there is no need for actual quantitative values.
But an actual calculation of either rest mass or E in the foregoing equation would seem to neccessitate some non simple and direct calculation in order to include internal kinetic energy from dissipation.

You call it "internal kinetic energy" but if it appears in the object's rest mass, it's not "kinetic energy" as far as that level of modeling is concerned. Of course all sorts of things that contribute to rest mass in a macroscopic model, like internal vibrations of molecules, will be treated as kinetic energy in a microscopic model. But if molecular vibrations, for example, are treated as kinetic energy in the microscopic model, they don't appear in the rest mass of the molecules or atoms; the model will explicitly include the motion of the molecules or atoms, and will track the explicit coordinates and velocities of those molecules or atoms, so that the energy in the vibrations appears as "external" kinetic energy, just like in the formula I gave.

For a macroscopic model, none of this is necessary; all you need is some kind of coefficient to tell you how much dissipation there is--i.e., how much of the energy applied to the object goes into increasing its rest mass. The viscosity of a fluid is an example of such a coefficient. In macroscopic models these coefficients are just intrinsic properties of materials, measured empirically; the underlying microscopic details are simply ignored.

 

[Austin0]

=PeterDonis;2395420]The center of mass (CoM, to save typing) corresponds to a *coordinate* location within a system. Which particular physical part of the system, if any, is located at that coordinate location can change. The CoM is not "anchored" to any particular physical part of the system. It's just wherever the "average" location of all the parts of the object happens to be.

If we are considering a system in motion then, practically speaking, couldn't we consider a line orthagonal to the vector of motion at some point in the system , whereby the mass to the rear of this line is equal to the mass in front of this line.
That any spatial redistribution or relative translation within the system is irrelevant unless it moves mass from one side of this line to the other?

As above, the CoM isn't "anchored" to any particular part of the object. It's just the "average" location of all the parts. It's not necessary that all the parts of the system are moving for the "average" location of the parts to be moving; just one part moving is enough.

As above

The fact that Born rigid acceleration is the only state of motion of an accelerated object for which the internal distances between the parts remain constant has been mentioned several times in this thread, not just by me. We haven't proven it here, but there are plenty of discussions of this fact and how it comes about online and in textbooks.

I am somewhat familiar with the available info on the web regarding both Born and Rindler.
I have encountered descriptions in both cases but nothing in the way of explicit proof. I am also unaware of any empirical tests validating either ,,, have I missed something?

Once again, this has already been discussed in this thread. We discussed the motion of two separate spaceships with equal, constant acceleration, as in the "[URL spaceship paradox[/URL], and confirmed that their separation would continuously increase. Then I specified the motions of two ends of a single rocket ship in such a way that the worldlines of the two ends would be identical to those of the two separate spaceships

.

You specified. Does this mean that the physics neccessarily conforms to your specifications?
That a single ship would neccessarily conform to the identical world lines??
That this justifies an automatic assumption that a single extended ship is absolutely the same as two separate ships and there is no consequence derived from the matter connecting the ends?

Not directly, but I did implicitly when I described the scenario that would produce Born rigid acceleration. I'll describe it again: Let there be a rocket ship that starts at rest along the x-axis of an inertial frame. At time t = 0 in that frame, let each individual segment of the ship start accelerating, with an acceleration equal to $c^2 / x_s$, where x_s is the x-coordinate of that segment in the original inertial frame. Then the ship as a whole will undergo Born rigid acceleration.

This specification, in itself, doesn't tell exactly what thrust will be required at each individual segment to produce the specified acceleration. However, if we assume that there are no internal stresses to start with in the ship, then, since Born rigid acceleration preserves the distances between adjacent segments, it should not produce any internal stresses. So the thrust required for each segment should simply be the rest mass of the segment times its specified acceleration. This does depend, though, on our assumption about the material properties of the ship being correct.

Also, the above specification is what would be required for an object to *start out* with Born rigid acceleration, and of course, as has been said before in this thread, it's extremely unlikely that it would happen that way in any real case. However, suppose an object starts out with some other "force profile" (for example, thrust only at the rear end), but then, as a result of the development of internal forces, reaches a state such that the net force on each segment of the object is exactly what it needs to be to cause all the segments to accelerate according to the same profile I described above. At that point, the object as a whole would have settled into a state of Born rigid acceleration; it's this kind of scenario I was talking about in previous posts. Of course in this case the internal stresses in the object would be non-zero, since the internal forces would be supplying part of the net force on each segment to make the force profile just right. But the internal stresses would not *change* any further once this state was reached; it would still be a stable steady-state equilibrium, just as an object resting on the surface of the Earth is in a stable steady-state equilibrium even though the stresses inside it are non-zero because of gravity.

From this I understand that you are saying the the thrust would be different at different locales but would remain constant and not increase over time with Born acceleration. Correct?
It also appears you are saying it is not impossible that simple rear thrust could result in essentially Born acceleration or am I misreading you here?

: as I said above, directed momentum must result in directed momentum. Let there be a row of spheres extending along the x-direction. Now hit the sphere on the left with an impulse to the right (i.e., in the positive x-direction). Call that sphere sphere #1. Sphere #1 now moves to the right and hits sphere #2. Sphere #2 *must* acquire *some* momentum to the right in this collision. If this isn't obvious to you, consider the possibilities:

(1) Sphere #1 stops moving to the right after the collision--either it is at rest, or it is now moving to the left. In this case sphere #2 must move to the right to conserve momentum.

(2) Sphere #1 continues moving to the right after the collision. In this case, sphere #2 must move to the right at least as fast as sphere #1, because the collision brought them into contact.

Either way, sphere #2 must be moving to the right after the collision. The same argument then carries through to sphere #3, #4, etc., up to any number of spheres.

Originally Posted by Austin0
Only if 2nd law of TD is inoperative for some reason.
Only if energy conservation means that a discrete quantity of energy can be spread out within a system without limit by some process of infinite division.



Originally Posted by Austin0
SO (sph #1) p > (#2) p or conversely
(n+1) p < (n) p So unless you propose some kind of Zenoesque infinite divisibility there will be an infintesimal but steady attenuation of the magnitude of momentum until eventual total diminishment yes?



Originally Posted by Austin0
We know that some infintesimal amount of the propagated momentum itself will also be diisipated internally.

No, we do *not* know this, at least I don't. One more time: **this would violate conservation of momentum**. Conservation of momentum requires that the directed momentum before the collision must be equal to the directed momentum after the collision. So we must have (for the case where only sphere #1 is moving before the collision but both spheres #1 and #2 may be moving after the collision):

$$p_{1-before} = p_{1-after} + p_{2-after}$$-

From this, the rest of what I said follows. Directed momentum simply cannot be "dissipated internally" like you're saying.

Total agreement here- $$p_{1-before} = p_{1-after} + p_{2-after}$$-as I have explicitly stated before. ANd this continues to hold for

$$p_{1-before} = p_{1-after} + p_{2-after} + p_{3-after}+,,,,,,,,,,+ p_{N-after}$$

I have a couple of responses to this.

#1

Original PeterDonis (2) In the realistic case where there is dissipation, the object's rest mass will increase to M' > M, so its final velocity will be v' = p_x / M', which is less than v. But its final linear momentum will still be p_x

From this it follows that: $$v_{1-before} > v_{2-after}$$-

ANd through extrapolation $$v_{1-before} > v_{2-after}$$> $$v_{3-after} > v_{4-after}$$> $$v_{5-after} > v_{6-after}$$....>$$v_{N-after}= 0$$

Unless there is some kind of infinite divisibility

COnsidering Sphere N with a v = 0 $$p_{1-before} = p_{N-after}$$

only if the increase in inertial mass of N was equivalent to the energy of $$v_{1-before}$$

or equivalent to the total increase in inertial mass distributed among all the previous spheres.

Clearly this could not be true in either case so logically
so it would seem to follow that:

$$p_{1-before} > p_{N-after}$$ And as there is an infintesimal incremental continuum between $$p_{1-before} and p_{N-after}$$

it follows that :

$$p_{1-before} > p_{2-after} > p_{3-after}>,,,,,,,,,,> p_{N-after}$$

ANd that there is no conflict between A) and B)

A) $$p_{1-before} = p_{1-after} + p_{2-after} + p_{3-after}+,,,,,,,,,,+ p_{N-after}$$

B) $$p_{1-before} > p_{2-after} > p_{3-after}>,,,,,,,,,,> p_{N-after}$$

That in both descriptions the total mass/energy content of the initial momentum is simply distributed among a large number of spheres with a steady decrease in velocity.





Originally Posted by Austin0
YOu still seem to be proposing a universe where a sound wave in a rod will travel forever without losing either coherence or energy. Without attenuation or decay. Without the 2nd law of TD

I'm proposing no such thing. I've said several times that any real object would have dissipation, and dissipation means that some of the *energy* added to the object gets turned into heat--i.e., it appears as an increase in the object's rest mass instead of an increase in the object's velocity. That is the sort of thing the 2nd law of TD deals with. But none of that changes the fact that *directed momentum* is conserved. I've already shown how that can happen, but one more time: consider an object of mass M, initially at rest, which then has some linear momentum p_x added to it:

(1) In the idealized case where there's no dissipation, the object's rest mass is unchanged, its linear momentum is p_x, and its final velocity (in the non-relativistic limit) will be v = p_x / M.

(2) In the realistic case where there is dissipation, the object's rest mass will increase to M' > M, so its final velocity will be v' = p_x / M', which is less than v. But its final linear momentum will still be p_x.

Is there really any meaningful difference between the cases we are considering?

Propagation between a series of spatially separated interactions,
Propagation between a series of contact interactions
Propagation between atoms in a matrix
Or propagation of sound , which in the final analysis is just momentum.
SO if you are not proposing that a sound wave would propagate without limit why would you think that any of these other interactions or mediums etc could propagate without limit??

Thanks

 

[PeterDonis]

Austin0:

If we are considering a system in motion then, practically speaking, couldn't we consider a line orthagonal to the vector of motion at some point in the system , whereby the mass to the rear of this line is equal to the mass in front of this line.
That any spatial redistribution or relative translation within the system is irrelevant unless it moves mass from one side of this line to the other?

Why would motion be irrelevant unless it moved mass from one side of the line to the other? That condition doesn't correspond to any meaningful physical quantity that I'm aware of. For example, I can move parts of the object around without violating your condition and still change the location of the object's CoM, or the object's moment of inertia about an axis, etc. What meaningful physical invariant corresponds to the condition you're specifying?

I am somewhat familiar with the available info on the web regarding both Born and Rindler.
I have encountered descriptions in both cases but nothing in the way of explicit proof. I am also unaware of any empirical tests validating either ,,, have I missed something?

The specifications for Born rigid acceleration and Rindler coordinates are constructed in such a way that the distances between internal parts of the object remain constant *by construction*. If the references you've read don't explicitly prove, from the construction, that the distances remain constant, that's probably because the writers considered it sufficiently obvious that they could leave it to the reader to fill in the details. But if you want me to explicate the proof, here it is:

Let there be a ship which starts out at rest in an inertial frame (x, t), lying along the x-axis; each segment of the ship starts out at a given x-coordinate x_s. At time t = 0 in this original inertial frame, each segment starts accelerating as I specified previously for Born rigid acceleration. Then the worldline of each segment of the ship, in the coordinates of the original rest frame, will be

$$x^2 - t^2 = x_s^2$$

where I'm using units such that the speed of light is 1. (Rindler coordinates are constructed so that these worldlines, which look like hyperbolas in ordinary Minkowski coordinates, look like straight lines, but the invariant properties of the worldlines themselves are the same either way.) Let's re-write this equation in terms of x as a function of t:

$$x = \sqrt{t^2 + x_s^2}$$

Then the coordinate velocity of the segment, at a given event (t, x), will be

$$v = \frac{dx}{dt} = \frac{t}{\sqrt{t^2 + x_s^2}} = \frac{t}{x}$$

And the relativistic gamma factor will be

$$\gamma = \frac{1}{\sqrt{1 - v^2}} = \frac{1}{\sqrt{1 - \frac{t^2}{x^2}}} = \sqrt{\frac{x^2}{x^2 - t^2}} = \frac{x}{x_s}$$

Now, the equation for coordinate velocity tells us that any line through the origin of the (x, t) frame, x = 0, t = 0 (these lines are the lines of constant q in Greg Egan's diagram), will be a line of simultaneity for any segment, at the point where the line intersects that segment's worldline. Thus, spacelike intervals along such a line will correspond to distances as measured by observers moving with each segment. In particular, given a point (x, t) along a segment's worldline, the proper distance from that point to the point x = 0, t = 0 is given by the Lorentz transformation:

$$x' = \gamma \left( x - v t \right)$$

Substituting for v and gamma, we have

$$x' = \frac{x}{x_s} \left( x - \frac{t^2}{x} \right) = \frac{1}{x_s} \left( x^2 - t^2 \right) = \frac{x_s^2}{x_s} = x_s$$

In other words, as Greg Egan notes in his discussion, each segment maintains a constant proper distance from the event x = 0, t = 0. That means that the proper distance between any two segments must also remain constant, since each maintains a constant proper distance from a fixed point.

I don't know that anyone has done a direct experiment where an object was put under Born rigid acceleration; such an experiment would be extremely difficult in practice. However, since everything I've said about Born rigid acceleration is a simple consequence of the laws of SR, and the laws of SR have been experimentally verified, I'm not too worried about my statements about Born rigid acceleration being correct.

You specified. Does this mean that the physics neccessarily conforms to your specifications?
That a single ship would neccessarily conform to the identical world lines??
That this justifies an automatic assumption that a single extended ship is absolutely the same as two separate ships and there is no consequence derived from the matter connecting the ends?

My specification is physically possible, as I've said several times. That's all that's required to consider its implications. In other words: there's nothing in the laws of physics that *prevents* the front and rear ends of a ship from following the worldlines I've specified, at least until the ship breaks from the stretching. It might be highly unlikely in practice for the front and rear ends of a ship to follow the worldlines I've specified; I've already said several times that for it to happen in practice, the rocket thrust at the front end would have to continually increase, to compensate for the increasing internal force of the rest of the ship pulling back on the front end because of the stretching. (Conversely, the rocket thrust at the rear end would have to continually decrease, because the increasing internal force from the rest of the ship due to the stretching would be pulling the rear end forward.) But the laws of physics permit the rocket thrust to behave this way, so my scenario is physically possible.

If it makes you feel any better, consider my scenario as a hypothetical. In other words, all I'm saying is that *if* the front and rear ends of the ship followed the worldlines I specified, the ship would continually stretch until it broke. It's physically possible for the front and rear ends to follow those worldlines, so my hypothetical is not ruled out by the laws of physics; but it's still only a hypothetical. I'm certainly not saying that *any* rocket ship with engines at the front and rear ends *has* to move this way; I'm only exploring the consequences *if* one did so.

From this I understand that you are saying the the thrust would be different at different locales but would remain constant and not increase over time with Born acceleration. Correct?

Correct.

It also appears you are saying it is not impossible that simple rear thrust could result in essentially Born acceleration or am I misreading you here?

You're correct here too, in the sense that simple rear thrust could lead to a steady-state equilibrium in which the object as a whole was undergoing Born rigid acceleration; but in that equilibrium state, there would be internal stresses in all parts of the ship, in addition to the rear thrust.

From this it follows that: $v_{1-before} > v_{2-after}$

ANd through extrapolation $v_{1-before} > ... > v_{N-after} = 0$

Unless there is some kind of infinite divisibility

The last "= 0" here is not correct. The velocity will never go to zero; it can't, because that would imply zero momentum in that direction. This does not imply infinite divisibility; it implies that there are a finite number of objects among which the momentum can be shared. In this example, there were N spheres, a finite number; in the limit, we would be counting atoms, or subatomic particles. There are a finite number of those too, and even when the momentum has been shared among every last one of them, the velocity will still be nonzero.

ANd that there is no conflict between A) and B)

A) $$p_{1-before} = p_{1-after} + p_{2-after} + p_{3-after}+,,,,,,,,,,+ p_{N-after}$$

B) $$p_{1-before} > p_{2-after} > p_{3-after}>,,,,,,,,,,> p_{N-after}$$

That in both descriptions the total mass/energy content of the initial momentum is simply distributed among a large number of spheres with a steady decrease in velocity.

I have no problem with A), since it's just conservation of momentum. I'm not entirely sure about B). I would agree that, in general, $p_{1-before} > p_{N-after}$, since some of the original momentum is likely to be distributed among other spheres besides sphere #N. But I'm not sure that this implies B). There's no reason why each of the momenta of the intermediate spheres *has* to fit into the "infinitesimal incremental continuum" between $p_{1-before}$ and $p_{N-after}$; they could equally well be less than $p_{N-after}$.

Is there really any meaningful difference between the cases we are considering?

Propagation between a series of spatially separated interactions,
Propagation between a series of contact interactions
Propagation between atoms in a matrix
Or propagation of sound , which in the final analysis is just momentum.
SO if you are not proposing that a sound wave would propagate without limit why would you think that any of these other interactions or mediums etc could propagate without limit??

I'm not sure what you mean here by "propagate without limit", but certainly all of the interactions you've described must obey the laws of conservation of momentum and energy. In that sense there's no meaningful difference between them. But the specifics of how the interactions are modeled in physics can vary quite a bit.

 

[Austin0]

Austin0;2397188]If we are considering a system in motion then, practically speaking, couldn't we consider a line orthagonal to the vector of motion at some point in the system , whereby the mass to the rear of this line is equal to the mass in front of this line.
That any spatial redistribution or relative translation within the system is irrelevant unless it moves mass from one side of this line to the other?

=PeterDonis;2398074]Austin0:
Why would motion be irrelevant unless it moved mass from one side of the line to the other? That condition doesn't correspond to any meaningful physical quantity that I'm aware of. For example, I can move parts of the object around without violating your condition and still change the location of the object's CoM, or the object's moment of inertia about an axis, etc. What meaningful physical invariant corresponds to the condition you're specifying?

We are talking about systems accelerating along a linear path. No rotation. No systems moving at angles relative to the direction of motion.
Limited to this context do you think moving mass around before or behind the line would shift the CoM relative to the direction of travel?

The specifications for Born rigid acceleration and Rindler coordinates are constructed in such a way that the distances between internal parts of the object remain constant *by construction*. If the references you've read don't explicitly prove, from the construction, that the distances remain constant, that's probably because the writers considered it sufficiently obvious that they could leave it to the reader to fill in the details. But if you want me to explicate the proof, here it is:

Let there be a ship which starts out at rest in an inertial frame (x, t), lying along the x-axis; each segment of the ship starts out at a given x-coordinate x_s. At time t = 0 in this original inertial frame, each segment starts accelerating as I specified previously for Born rigid acceleration. Then the worldline of each segment of the ship, in the coordinates of the original rest frame, will be

$$x^2 - t^2 = x_s^2$$

where I'm using units such that the speed of light is 1. (Rindler coordinates are constructed so that these worldlines, which look like hyperbolas in ordinary Minkowski coordinates, look like straight lines, but the invariant properties of the worldlines themselves are the same either way.) Let's re-write this equation in terms of x as a function of t:

$$x = \sqrt{t^2 + x_s^2}$$

Then the coordinate velocity of the segment, at a given event (t, x), will be

$$v = \frac{dx}{dt} = \frac{t}{\sqrt{t^2 + x_s^2}} = \frac{t}{x}$$

And the relativistic gamma factor will be

$$\gamma = \frac{1}{\sqrt{1 - v^2}} = \frac{1}{\sqrt{1 - \frac{t^2}{x^2}}} = \sqrt{\frac{x^2}{x^2 - t^2}} = \frac{x}{x_s}$$

Now, the equation for coordinate velocity tells us that any line through the origin of the (x, t) frame, x = 0, t = 0 (these lines are the lines of constant q in Greg Egan's diagram), will be a line of simultaneity for any segment, at the point where the line intersects that segment's worldline. Thus, spacelike intervals along such a line will correspond to distances as measured by observers moving with each segment. In particular, given a point (x, t) along a segment's worldline, the proper distance from that point to the point x = 0, t = 0 is given by the Lorentz transformation:

$$x' = \gamma \left( x - v t \right)$$

Substituting for v and gamma, we have

$$x' = \frac{x}{x_s} \left( x - \frac{t^2}{x} \right) = \frac{1}{x_s} \left( x^2 - t^2 \right) = \frac{x_s^2}{x_s} = x_s$$

In other words, as Greg Egan notes in his discussion, each segment maintains a constant proper distance from the event x = 0, t = 0. That means that the proper distance between any two segments must also remain constant, since each maintains a constant proper distance from a fixed point.

I don't know that anyone has done a direct experiment where an object was put under Born rigid acceleration; such an experiment would be extremely difficult in practice. However, since everything I've said about Born rigid acceleration is a simple consequence of the laws of SR, and the laws of SR have been experimentally verified, I'm not too worried about my statements about Born rigid acceleration being correct.

You have just given a description of Born acceleration. WHich is itself basically a definition of coordinate events.
The events described are exactly those of the essential Lorentz contraction and lead to exactly the same world lines.
SPECIFICALLY: the fundamental postulates propose that internal proper distances remain the same in inertial frames. ANd that the Lorentz contraction is regarded as a purely kinematic effect.

SO on the basis of this and the clock hypothesis, you could plot the worldline of the back of a normally propulsed ship, simply on the basis of constant proper acceleration.
You could then plot the front of the ship on the basis of instantaneous Lorentz contraction
and would then have a pair of worldlines that were indistinguishable from Born worldlines
Correct?

Based on these worldlines it would inevitably appear as if there had to be unequal acceleration at the front and the back. But this does not neccessarily have any physical meaning. It is simply a result of the conventions of Minkowski space and will have to be this way even if there is NO PHYSICAL cause of the contraction at all, but it is simply a kinematic effect taking place purely in the relative observation frame.
So aside from the description, the Born hypothesis is making a statement about the nature of the Lorentz contraction and is proposing that it will not take place through equal acceleration throughout a system , which will result in the opposite result . Catastrophic expansion.
It also seems to place a meaning to lines of simultaneity that I don't quite understand.

My specification is physically possible, as I've said several times. That's all that's required to consider its implications. In other words: there's nothing in the laws of physics that *prevents* the front and rear ends of a ship from following the worldlines I've specified, at least until the ship breaks from the stretching. It might be highly unlikely in practice for the front and rear ends of a ship to follow the worldlines I've specified; I've already said several times that for it to happen in practice, the rocket thrust at the front end would have to continually increase, to compensate for the increasing internal force of the rest of the ship pulling back on the front end because of the stretching. (Conversely, the rocket thrust at the rear end would have to continually decrease, because the increasing internal force from the rest of the ship due to the stretching would be pulling the rear end forward.) But the laws of physics permit the rocket thrust to behave this way, so my scenario is physically possible.

If it makes you feel any better, consider my scenario as a hypothetical. In other words, all I'm saying is that *if* the front and rear ends of the ship followed the worldlines I specified, the ship would continually stretch until it broke. It's physically possible for the front and rear ends to follow those worldlines, so my hypothetical is not ruled out by the laws of physics; but it's still only a hypothetical. I'm certainly not saying that *any* rocket ship with engines at the front and rear ends *has* to move this way; I'm only exploring the consequences *if* one did so.

WE need to go into this a little more perhaps. We talked about it in regard to two separate ships but haven't actually covered it at all with a single extended ship.


.


$$v_{1-before} > v_{2-after}$$> $$v_{3-after} > v_{4-after}$$> $$v_{5-after} > v_{6-after}$$....>$$v_{N-after}= 0$$

The last "= 0" here is not correct. The velocity will never go to zero; it can't, because that would imply zero momentum in that direction. This does not imply infinite divisibility; it implies that there are a finite number of objects among which the momentum can be shared. In this example, there were N spheres, a finite number; in the limit, we would be counting atoms, or subatomic particles. There are a finite number of those too, and even when the momentum has been shared among every last one of them, the velocity will still be nonzero.

Perhaps you could explain this . You say there are a finite number of objects among which the momentum can be shared but the momentum can never reach zero in that direction. Well that would seem to neccessitate an infinite number of objects or infinite division of the quantity of momentum . I.e. smaller and smaller momenta for each succeeding object but never reaching 0 ,,exactly like Achilles' smaller and smaller divisions of the distance from the tortoise which he never reaches.

I have no problem with A), since it's just conservation of momentum. I'm not entirely sure about B). I would agree that, in general, $p_{1-before} > p_{N-after}$, since some of the original momentum is likely to be distributed among other spheres besides sphere #N. But I'm not sure that this implies B). There's no reason why each of the momenta of the intermediate spheres *has* to fit into the "infinitesimal incremental continuum" between $p_{1-before}$ and $p_{N-after}$; they could equally well be less than [itex]p_{N-after}[/itex].

Well if any of them were less than a later sphere then there were seem to be a definite violation of conservation of momentum. Energy out of nowhere

____________________________________________________________________

Originally Posted by Austin0
YOu still seem to be proposing a universe where a sound wave in a rod will travel forever without losing either coherence or energy. Without attenuation or decay. Without the 2nd law of TD



Originally Posted by Austin0
Is there really any meaningful difference between the cases we are considering?

Propagation between a series of spatially separated interactions,
Propagation between a series of contact interactions
Propagation between atoms in a matrix
Or propagation of sound , which in the final analysis is just momentum.
SO if you are not proposing that a sound wave would propagate without limit why would you think that any of these other interactions or mediums etc could propagate without limit??

I'm not sure what you mean here by "propagate without limit", but certainly all of the interactions you've described must obey the laws of conservation of momentum and energy. In that sense there's no meaningful difference between them. But the specifics of how the interactions are modeled in physics can vary quite a bit

Without limit. Velocity and momentum never reaching zero.

Had a bit of flu ,, more later. Thanks

 

[PeterDonis]

Austin0:

We are talking about systems accelerating along a linear path. No rotation. No systems moving at angles relative to the direction of motion.
Limited to this context do you think moving mass around before or behind the line would shift the CoM relative to the direction of travel?

Yes. Suppose, for example, that our system is composed of two subsystems, a front segment and a rear segment. The segments are initially at rest, and are identical in length and mass in the original rest frame. That means your imaginary line would be drawn precisely at the boundary between the two segments.

Now apply a forward impulse to the rear edge of the rear segment. The rear segment will compress, but until a time equal to the length of the rear segment divided by the speed of light has elapsed, the front edge of the rear segment will not move. So we have the rear segment compressed, meaning the CoM of the rear segment (and hence of the whole system) moves forward, but no mass crosses your imaginary line.

You have just given a description of Born acceleration. WHich is itself basically a definition of coordinate events.
The events described are exactly those of the essential Lorentz contraction and lead to exactly the same world lines.
SPECIFICALLY: the fundamental postulates propose that internal proper distances remain the same in inertial frames. ANd that the Lorentz contraction is regarded as a purely kinematic effect.

No. The worldlines of each segment of the ship are accelerated worldlines, not inertial worldlines. The MCIF at each event along those worldlines is an inertial frame, but the ship segments are at rest in each MCIF only momentarily. So the fact that "internal proper distances remain the same in inertial frames" is irrelevant, since that only applies to objects moving on inertial worldlines.

You are correct that the distance between any two adjacent segments (i.e., between appropriate events on their worldlines), which remains constant in each MCIF along the worldlines, will appear to be increasingly Lorentz contracted in the "initial" frame, the one in which Born rigid acceleration begins at time t = 0. This is indeed a "kinematic" effect. However, the physics that determines which worldlines each segment follows is *not* just "kinematics"; it requires knowledge of forces and accelerations, which are genuine physical quantities. See further comments below.

SO on the basis of this and the clock hypothesis, you could plot the worldline of the back of a normally propulsed ship, simply on the basis of constant proper acceleration.
You could then plot the front of the ship on the basis of instantaneous Lorentz contraction
and would then have a pair of worldlines that were indistinguishable from Born worldlines
Correct?

Not really. You are ignoring the internal forces between the parts of the ship. Remember how I answered your question about simple rear thrust and Born acceleration:

You're correct here too, in the sense that simple rear thrust could lead to a steady-state equilibrium in which the object as a whole was undergoing Born rigid acceleration; but in that equilibrium state, there would be internal stresses in all parts of the ship, in addition to the rear thrust.

So the object would only be undergoing Born rigid acceleration *after* all the oscillations of internal forces had damped out, and the ship had settled into a steady-state equilibrium. In that equilibrium state, the *net* force on the rear end of the ship would be *less* than the rocket thrust applied there, because there would also be a compression force from the next segment forward that would act towards the rear, in the opposite direction from the thrust. So the *final* proper acceleration of the rear segment would be *less* that the proper acceleration you would calculate just from the rocket thrust alone.

When I gave the proof that Born rigid acceleration preserves internal distances, I ignored the complications above. To make my specification consistent with those complications, we would simply define the "initial" reference frame, the one in which each segment starts accelerating at time t = 0 in accordance with the precise profile required for Born rigid acceleration, to be the MCIF of the ship at the instant when all the internal oscillations are gone and the steady-state equilibrium has just been reached. The ship as a whole would be momentarily at rest in that frame, and the accelerations on each segment of the ship (due to internal forces, plus the rocket thrust at the rear end only) would be in just the right profile to produce Born rigid acceleration from then on. In essence, we would be ignoring the details of how the ship got to the steady-state equilibrium and just focusing on the equilbrium state itself.

If we do focus just on the equilibrium state, then you're basically correct that we can calculate what all the worldlines have to look like for Born rigid acceleration based on relativistic kinematics. However, doing that can lead to confusion if we forget that we were originally led to calculate those worldlines by specific *physical* assumptions, so our calculations are only relevant in specific physical situations where those assumptions are applicable.

Based on these worldlines it would inevitably appear as if there had to be unequal acceleration at the front and the back. But this does not neccessarily have any physical meaning.

Wrong. The acceleration of a worldline is simply the derivative of its 4-velocity with respect to proper time:

$$A = \frac{dU}{d \tau}$$

This is an invariant, the same in all reference frames, so it's a bona fide physical quantity. (An "inertial" worldline is just one for which A = 0.) For the specific worldlines I gave for Born rigid acceleration for each segment, you can compute this acceleration, and verify that it is equal to x_s for each segment (i.e., the original x-coordinate of the segment in the "initial" frame).

So aside from the description, the Born hypothesis is making a statement about the nature of the Lorentz contraction and is proposing that it will not take place through equal acceleration throughout a system , which will result in the opposite result . Catastrophic expansion.

You're correct that equal *proper* acceleration of all parts of a system will result in expansion, since equal *proper* acceleration of all parts is enough to specify the worldline of each part, and from there it's all kinematics, as I've said before. However, this doesn't mean that the Born hypothesis is just a statement about kinematics. I've already addressed this somewhat in my comments above, but let me restate them in more detail:

-- The original motivation for working out the specification of Born rigid acceleration was to determine what kind of accelerated motion would result in unchanging proper distances between the parts of a system.

-- The reason for wanting to know *that* was the *physical* assumption that internal stresses between the parts of a system depend on the proper distances between those parts. If that weren't the case, nobody would care about Born rigid acceleration; we'd be discussing some other specification (whatever one turned out to keep internal stresses constant).

So the original motivation for studying Born rigid acceleration was physical, not just kinematic. Also:

-- The proper acceleration of each part of a system depends on the *net* force on that part, *not* just the externally applied force. You must also take into account the internal forces between the parts, as we've seen already.

So, as we've seen before, equal proper acceleration does *not* necessarily mean equal rocket thrust. You need physical assumptions to determine *how* equal proper acceleration would be realized (or any other acceleration profile, for that matter).

It also seems to place a meaning to lines of simultaneity that I don't quite understand.

The meaning of each of the lines of simultaneity (lines of constant q in Greg Egan's diagram) is just what it seems: each line of constant q is the line of simultaneity for the MCIF of each worldline it crosses, at the point where it crosses that worldline. In other words, all events on a line of constant q will be seen to be simultaneous by observers traveling on any of the accelerated worldlines we've specified (the lines of constant s in Greg Egan's diagram).

What may be confusing you is the fact that all of those lines of simultaneity intersect at the origin of the "initial" frame, x = 0, t = 0. That may seem paradoxical, but it's not: it's just an unusual fact about accelerated worldlines. It does, however, mean that the "pivot point" (x = 0, t = 0 in this case), and the past and future light cone at that point (the lines t = x and t = -x; part of the t = x line appears as a 45-degree dotted line on Greg Egan's diagram) have a special meaning to observers traveling on those accelerated worldlines. Look up "Rindler horizon" for more information (Greg Egan's page has some discussion of this).

$$v_{1-before} > v_{2-after} > v_{3-after} > v_{4-after} > v_{5-after} > v_{6-after} .... > v_{N-after} = 0$$

...

Perhaps you could explain this . You say there are a finite number of objects among which the momentum can be shared but the momentum can never reach zero in that direction. Well that would seem to neccessitate an infinite number of objects or infinite division of the quantity of momentum .

I don't understand how you're coming up with these conclusions. Let me re-state once more what I've been saying:

-- If the total momentum of a system is p_x before some process, the total momentum of the same system (provided there are no external forces on the system) after the process will also be p_x (we're assuming here that all motion is in the x-direction only).

For example, in the case of N spheres, we have (when sphere #1 is the only one with any momentum before the process):

$$p_{1-before} = p_{1-after} + p_{2-after} + ... + p_{N-after}$$

We appear to have agreement on this; but you may not be realizing that when I say "the momentum can never reach zero in that direction", the above, conservation of momentum, is *all* that I'm saying. I'm *not* saying that the momentum of one particular sphere could not be zero; only that the *total* momentum, the sum of the momenta of all the spheres, must remain constant, so if it starts out non-zero, it must end up non-zero (meaning that at least one sphere must have a non-zero momentum after the process).

-- If the original momentum, which was carried by only one sphere before the process, is shared between two or more spheres *after* the process, we will have

$$p_{1-before} > p_{i-after}$$

for any value of i from 1 to N (i being the number of any particular sphere). But this does *not* imply that the momenta are strictly decreasing as we go from left to right (i.e., as i increases from 1 to N). Nothing I've said implies that, and in general it won't be true. See below.

-- In the presence of dissipation, the rest mass of any given sphere after the process will be greater than it was before the process. Thus, the velocity of any given sphere after the process will be *less* than it would have been without dissipation (because some of the energy of the process goes into increasing the sphere's rest mass instead of its velocity). So even in a limiting case where virtually all of the original momentum ends up in a single sphere after the process, we will still have

$$v_{1-before} > v_{i-after}$$

for any value of i from 1 to N. But this, again, does *not* imply that the velocities will be strictly decreasing as we go from left to right (i.e., as i increases from 1 to N). In fact, since the original momentum was to the right (increasing x-direction), it is easy to see that, since sphere i + 1 is to the right of sphere i, and since one sphere can't move through another, we must have

$$v_{i-after} <= v_{i+1-after}$$

for any value of i from 1 to N - 1. If this were not true, we would have, for some i, sphere i moving to the right *faster* than sphere i + 1, which is to its right, *after* spheres i and i + 1 have collided. That's obviously impossible.

Does this help any?

Well if any of them were less than a later sphere then there were seem to be a definite violation of conservation of momentum. Energy out of nowhere

Once again, I don't know how you're coming up with this conclusion. Let's consider the simplest interesting case, with three spheres. Before the collisions, sphere #1 has velocity v_0 in the positive x-direction, and spheres #2 and #3 are at rest. After the collisions, sphere #1 has velocity v_1, sphere #2 has velocity v_2, and sphere #3 has velocity v_3 (all velocities are in the x-direction, but we don't yet know anything about their signs). Also, before the collisions, all spheres have the same initial rest mass M; after the collisions, dissipation has increased the rest masses of all the spheres by the same factor k > 1.

The equations for conservation of energy and momentum then look like this, using the fully relativistic formulas (I've eliminated the common factor M in both equations, since it appears in every term):

$$\gamma_0 + 2 = k \left( \gamma_1 + \gamma_2 + \gamma_3 \right)$$

$$\gamma_0 v_0 = k \left( \gamma_1 v_1 + \gamma_2 v_2 + \gamma_3 v_3 \right)$$

where we have written the relativistic $\gamma$ factors with the same subscripts as their corresponding velocities. Unless I'm misunderstanding you, you're claming that it must be the case that p_{2-after} > p_{3-after}, or

$$\gamma_2 v_2 > \gamma_3 v_3$$

or the above conservation equations will be violated. That's not correct; there are plenty of solutions of the above equations for which $\gamma_2 v_2 < \gamma_3 v_3$. (In fact, if we consider the additional constraint I gave above, that the spheres can't move through each other, we see that we must have $v_2 <= v_3$, meaning that

$$\gamma_2 v_2 <= \gamma_3 v_3$$

must hold.)