Time-reverse symmetry of the principle of relativity

In summary, the conversation discusses the asymmetry in dynamics resulting from the relative position of rest in a two-body collision. The observer's position affects the time-reverse symmetry of the event and raises questions about the principles of relativity. The conversation also delves into time-reverse symmetry in kinematics and dynamics, and how they may not always align. This leads to a discussion about the thermodynamic arrow of time and the laws of mechanics.
  • #36
Hi JesseM,

I take it from our last two posts that we are in agreement that what Chrisc is missing is that linear kinetic energy is converted into heat energy in the forward direction and that it is extremely unlikely that random heat energy would be spontaneously converted to linear motion in the reverse time direction.

A much simpler example is dropping a ball of putty to the floor from 2 meters. Let's assume the putty does not bounce and just comes to stop. Some of the initial kinetic energy is used up in deforming the putty and the remainder isdisperced as heat and vibrations in the floor and the flattened putty. The time reversal of random heat vibrations spontaneously focusing to eject the putty 2 metres into the air (and reform it into a perfect ball) is unlikely.
 
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  • #37
Kev and JesseM, you've both taken this out of its original context.
I had qualified this as a system of rigid, non-composite bodies in order to simplify and focus the
dynamics by removing molecular motion (heat). The idea is to consider the principle of the laws
by examining the exchange and/or conversion of the energies involved (mass, inertia and momentum) through time.
It would be "a miracle" for the constituent particles of a body to align their momentum in such a way
as to "kick" an adjacent body (as a whole) into motion. But this "miracle" would happen in the time-reverse scenario.
It must happen if the laws are correct in the time forward scenario. This miracle is just a collective
molecular example of the simpler version I presented which is a smaller mass brings a larger mass to rest.

You have both described the mechanics that justify our concept of a thermodynamic "direction" of time.
The crux of what I'm trying to figure out is whether the principle of relativity posses enough symmetry
to survive a translation in time, i.e.: is the principle of relativity time-reverse symmetric?
In the observable mechanics of motion (relative velocity) it seems obvious it would. Velocity is kinematical,
as such there is no need to consider dynamics in a causal sense. So the time-reverse symmetry of the
kinematics of Special relativity seems pretty straight forward - it holds.
But, when mass is part of the consideration of the time-reverse symmetry of relativity, its measure and
the laws governing its interactions seem to break the relativistic principle. Holding the dynamical laws
to time-reverse symmetrical appears to reveal a break in the symmetry of the principle of relativity.
It seems there is a "special" case where one can claim a preferred frame of reference.
It is the frame where the dynamics of a single event do NOT break the time-reverse symmetry of the dynamical laws.
I hope that makes sense.
If you consider the collisions as a complete transfer of momentum and inertial energy without dissipating
any energy to heat, you will see the mechanics conflict with the laws. Once the conflict is apparent you
can put the heat energy back in and consider the same event in terms of the direction of heat transfer.
It does not solve the conflict it just changes the mechanics to a more complex collection of smaller masses
where the same issue arises in any collision through the center of mass.
 
  • #38
Chrisc said:
Kev and JesseM, you've both taken this out of its original context.
I had qualified this as a system of rigid, non-composite bodies in order to simplify and focus the
dynamics by removing molecular motion (heat).
But you can't have inelastic collisions that don't involve heat (or radiation I suppose). Energy must be conserved; if the total linear kinetic energy of the two bodies decreases after the collision, where does that energy go, if not to increasing the energy of the random vibrations in the molecules of each body (heat)? And the collision you described in your example was clearly an inelastic one where linear kinetic energy was not conserved. If you have some other physically realistic possibility in mind for where that energy went, please explain.
ChrisC said:
It would be "a miracle" for the constituent particles of a body to align their momentum in such a way
as to "kick" an adjacent body (as a whole) into motion. But this "miracle" would happen in the time-reverse scenario.
It must happen if the laws are correct in the time forward scenario. This miracle is just a collective
molecular example of the simpler version I presented which is a smaller mass brings a larger mass to rest.
What do you mean "it must happen"? Do you just mean it must be possible according to the laws of mechanics (which I would agree with), or are you saying it must be just as likely as the time forward scenario (which I would disagree with)?
ChrisC said:
The crux of what I'm trying to figure out is whether the principle of relativity posses enough symmetry
to survive a translation in time, i.e.: is the principle of relativity time-reverse symmetric?
The "principle of relativity" as it's normally understood is not intended to cover time-reversed coordinate systems. However, it is true that the laws of classical physics are T-symmetric, so that they should work the same in time-reversed coordinate systems, and in quantum field theory the laws are CPT-cymmetric, which means they work the same in coordinate systems where the labels of "forward in time" and "backward in time" have been reversed and where the positive and negative direction on all the spatial axes have been reversed (a change in 'parity', the P in CPT) and where the labels for positive and negative charge have been reversed (the C in CPT).
ChrisC said:
But, when mass is part of the consideration of the time-reverse symmetry of relativity, its measure and
the laws governing its interactions seem to break the relativistic principle. Holding the dynamical laws
to time-reverse symmetrical appears to reveal a break in the symmetry of the principle of relativity.
Only in a statistical sense that the time-reversed scenario is less likely, the time-reversed scenario doesn't violate any of the fundamental laws so it's certainly not forbidden. And this difference in likelihood has to do with cosmology and the fact that our universe apparently started off in a very low-entropy state (for reasons that aren't really understood at present), so that entropy has increasing ever since; if the universe were already at equilibrium, then statistically we'd expect that random fluctuations taking systems from higher entropy to lower would be just as common as fluctuations taking systems from lower to higher (though the vast majority of these fluctuations would be small and not noticeable on macroscopic scales).
Chrisc said:
It seems there is a "special" case where one can claim a preferred frame of reference.
You're misusing the language here. "Frame of reference" only refers to the different coordinate systems related by the Lorentz transformation, which all agree on the order of causally-related events; a coordinate system whose definition of forward in time and backwards in time is reversed is not a valid "frame of reference" in SR.
Chrisc said:
If you consider the collisions as a complete transfer of momentum and inertial energy without dissipating
any energy to heat, you will see the mechanics conflict with the laws.
But in your example there clearly was not a complete transfer of "inertial energy", since the combined linear kinetic energies of the two balls after the collision was different than their combined kinetic energy before the collision. If you consider an elastic collision where combined kinetic energy is conserved, then you will see that this kind of collision is just as likely in the time-reversed version.
 
  • #39
Chrisc, maybe the point you haven't realized is that conservation of momentum is not sufficient to solve the problem of what happens after the collision in post #33. Your solution is one of many that conserve momentum; you need to consider energy, too, to find a unique solution.

If we restrict ourselves to Newtonian mechanics instead of relativistic (i.e. assume small velocities), then to conserve momentum, if I've done the maths right, the post-velocity u of (M = 2m) can take any value between v/3 and v, and the post-velocity of (m) is then 2(v - u).

For an elastic collision, u = v/3 and the post-velocity of (m) is 4v/3 -- this is symmetric as the difference of velocities is v before and after.
 
  • #40
I'm sorry. I see why this makes no sense. I've been misusing the term "inelastic" collision.
I did not mean to use it in the conventional sense.
To be clear, below is the definitions as I understand them, for elastic and inelastic collisions.
An inelastic collision is used to describe the collision of bodies that, due to their inability
to deflect incident kinetic energy of the collision to the internal motion of their constituent parts,
must disperse (radiate)the energy in excess of the total energy of the system in some other form such as heat.
An elastic collision describes the collisions between bodies that are capable of such deflection,
resulting in the total energy being conserved in the sum of the main body momentums and
the motion of their constituent parts (or molecules).

Again, I am sorry for not checking the proper use.
I meant neither, but inelastic in the sense of a single, non-composite, rigid body that does not flex
and has no constituent parts onto which it can defer the energy internally.
A body that moves according to the exchange of the inertial and kinetic energy of the collision.
This is thought experiment centered on the principles of the laws, not a real description of collisions.

The total energy of the system is symmetric in all cases when the collisions are considered as mentioned above.(conserved after collision)
The lack of symmetry is not in the total energy before and after the collision, it is in the physical dynamics before and after the collision.

I had mentioned earlier, and forgot to follow up on the objections, that the time-reverse symmetry of
the laws must be distinguished from the time-reverse symmetry of the mechanics.
If time-reversal is taken as just the reversal of "clock-ticks", then the same laws must define
the observed reverse kinematics.
If time-reversal is taken to include the reversal of physical dynamics, then the same laws must be
considered in their full time-reverse symmetry.
For example: when we consider the broken glass collecting its pieces together and jumping back up onto the table,
we are considering the probability of all the momentum of the pieces being properly directed such that the
glass becomes unbroken and the continued momentum of the glass launches itself back up on the table.
It is extremely improbable, but not impossible according to the laws. (assuming there is nothing more to know about the nature of time)
This analogy considers the reverse kinematics (clock-ticks) driven by the "same" laws not the reversal of the laws.

When this analogy is considered under the time-reversal of the laws, the physical dynamics of the laws are reversed.
Gravity is a repelling force. The table will not be waiting for the glass to leap back up. It is impossible to consider
the kinematics being exactly reversed in this type of time-reversal. The dynamics giving rise to the kinematics may express
a symmetry of reversal (the axis being a point in time where the reversal is considered) but the kinematics from that point on
are exactly reversed of what the laws predict in forward time. Small masses bring large masses to rest.
Masses repel others proportional to the sqrt of their distance.

This being the case, the example I gave represents both types of time-reversal.
In (2.) the same laws accurately predict the dynamics necessary to give rise to the kinematics observed.
There is nothing unusual in the time-reverse of (2) If you were to look at a film of (1) running backward you would find no conflict with the laws.
In (4) although the total energy of the system is conserved as in (1,2, and 3), it is at the expense of contradicting the laws.
If you watched a film of (3) running backward you would see (4) and claim "the film is running backward".
Because a small mass cannot bring a larger mass to rest.
So if (2) and (4) are distinctly different in the dynamics required to give rise to such kinematics,
they are a result of the frame of the observer being initially at rest with respect M or m.
 
  • #41
Chrisc said:
This being the case, the example I gave represents both types of time-reversal.
In (2.) the same laws accurately predict the dynamics necessary to give rise to the kinematics observed.
There is nothing unusual in the time-reverse of (2) If you were to look at a film of (1) running backward you would find no conflict with the laws.
In (4) although the total energy of the system is conserved as in (1,2, and 3), it is at the expense of contradicting the laws.
If you watched a film of (3) running backward you would see (4) and claim "the film is running backward".
Because a small mass cannot bring a larger mass to rest.
So if (2) and (4) are distinctly different in the dynamics required to give rise to such kinematics,
they are a result of the frame of the observer being initially at rest with respect M or m.

Back in post#35 JesseM showed you that the total kinetic energy of the experiment you described in post#33 is not conserved. If you do not want to consider energy dissapated as thermal energy then you can not say the total energy of the system is conserved. You need to consider a perfectly elastic collison to be able to justify the conservation of energy in a system that does not include thermal energy or thermodynamic miracles. Back in post#17 I showed that with initial conditions as described in your first post:

Ball B1: [tex]m_1=2m, v_1=2v[/tex]
Ball B2: [tex]m_2=1m, v_2=0v[/tex]

the final velocities would be 2/3v and 8/3v for B1 and B2 respectively in a perfectly elastic collision. All positive velocities are to the right with respect to the inertial observer in the reference frame that ball B2 is initially at rest in.

Now if you considerthe point of view of a different observer who has velocity 2/3v relative to the first observer, the final velocities of the balls according to him are 0v for B1 and 2v for B2. So in this perfectly elastic collison the second observer sees the large ball (B1) as being brought to a stop by the smaller ball. This can be observed in any collision simply by choosing a reference frame for the observer that coincides with the final velocity of the larger mass. There is no law of physics that says a large ball can not be brought to rest by a smaller ball but there are laws about conservation of momentum and total energy. You have to account for all the energy.
 
  • #42
Chrisc said:
Again, I am sorry for not checking the proper use.
I meant neither, but inelastic in the sense of a single, non-composite, rigid body that does not flex
and has no constituent parts onto which it can defer the energy internally.
A body that moves according to the exchange of the inertial and kinetic energy of the collision.
This is thought experiment centered on the principles of the laws, not a real description of collisions.
But even in a thought experiment, we must respect principles of physics like conservation of energy. Do you not see that in your example above, if we calculate the kinetic energy for each body according to the standard formula (1/2)*mv^2, then the total kinetic energy after the collision was different from the total kinetic energy before? If you want a thought experiment involving bodies that move "according to the exchange of the inertial and kinetic energy" (although I don't know what you mean by 'inertial energy', are you talking about momentum?), then you need one where energy is conserved. Do you wish to provide a new example where energy is conserved, or do you not understand that in your example energy was not conserved, if none of the kinetic energy was transformed into heat?
Chrisc said:
The total energy of the system is symmetric in all cases when the collisions are considered as mentioned above.(conserved after collision)
The lack of symmetry is not in the total energy before and after the collision, it is in the physical dynamics before and after the collision.
If you pick an example where both momentum and kinetic energy are conserved--which, by definition, is an "elastic collision"--then there will be no "lack of symmetry", the collision will be just as consistent with all laws (and just as probably thermodynamically) in the time-reversed version as in the forward-time version.
Chrisc said:
I had mentioned earlier, and forgot to follow up on the objections, that the time-reverse symmetry of
the laws must be distinguished from the time-reverse symmetry of the mechanics.
Could you explain what you mean by this distinction? Time-reverse symmetry of the laws logically implies that for any physical scenario consistent with the laws (like a collision), if you look at a backwards movie of that system's behavior, you could in principle set up a different physical system with different initial conditions such that its behavior in the forward-time direction looks precisely like the backwards movie of the first system. Is this different than what you meant by "time-reverse symmetry of the mechanics"?
Chrisc said:
If time-reversal is taken as just the reversal of "clock-ticks", then the same laws must define
the observed reverse kinematics.
Huh? Why? It is perfectly possible to imagine laws of physics (different from the actual laws of physics in our universe) where the reverse kinematics are not governed by the same laws. For example, imagine that objects in a gravitational field could only move down as time moves forward, never up (in reality of course an object can bounce on the ground and move up). If you switch which direction in time you label "forward", objects in a gravitational field could only move up as time moves forward, never down. This would require a different set of equations to describe the object's motion as a function of time.
Chrisc said:
For example: when we consider the broken glass collecting its pieces together and jumping back up onto the table,
we are considering the probability of all the momentum of the pieces being properly directed such that the
glass becomes unbroken and the continued momentum of the glass launches itself back up on the table.
It is extremely improbable, but not impossible according to the laws. (assuming there is nothing more to know about the nature of time)
This analogy considers the reverse kinematics (clock-ticks) driven by the "same" laws not the reversal of the laws.
I don't know why you consider kinematics as "clock ticks", and I don't understand what you mean by "reversal of the laws" as opposed to "reverse kinematics driven by the same laws". Could you explain specifically how these terms apply to your example?
Chrisc said:
When this analogy is considered under the time-reversal of the laws, the physical dynamics of the laws are reversed.
Gravity is a repelling force.
No no no, you're totally misunderstanding the idea of time-reversal symmetry here! The symmetry means that the laws do not change in any way in the time-reversed version, the scenario where the broken glasses are shot upwards can still be understood in terms of the same old attractive gravitational force. The point is that when you reverse all the molecular dynamics, the random motions of molecules due to heat suddenly synchronize and give the pieces of glass an upwards "kick" which shoots it into the air, just like you can send a soccer ball up into the air by kicking it with your foot--this certainly doesn't require gravity to be repulsive! As I said before, time-reversal symmetry means that by setting up precisely the right initial conditions (and assuming deterministic laws), you can create a physical situation whose behavior in the forward time direction looks precisely the same as the backwards movie of the first situation. In classical terms, if you take a snapshot of the positions and velocities of every particle at some time after the glass has fallen to the floor, and then you create a new set of initial conditions where all the positions are the same but all the velocities are reversed in direction, then when you evolve this new set of initial conditions forward, it will behave just like the backwards version of the original system.
Chrisc said:
Small masses bring large masses to rest.
Masses repel others proportional to the sqrt of their distance.
Nope, if the laws of physics are time-symmetric, the laws are precisely the same in the backwards version as the forwards version. And of course your "small masses bring large masses to rest" is very vague--it is quite possible to come up with collisions where large masses are brought to rest by small ones in the forward-time direction, it just depends on the details of each object's speed and mass.
Chrisc said:
This being the case, the example I gave represents both types of time-reversal.
The example you gave is simply impossible unless some kinetic energy is dispersed as heat, because the combined kinetic energy of both masses before the collision was different than the combined kinetic energy of both masses afterwards. Do you disagree?
Chrisc said:
In (2.) the same laws accurately predict the dynamics necessary to give rise to the kinematics observed.
There is nothing unusual in the time-reverse of (2) If you were to look at a film of (1) running backward you would find no conflict with the laws.
Again, the only possible way (2) could happen is if heat suddenly becomes kinetic energy, since in (2) the kinetic energy of both masses after the collision is larger than the kinetic energy of both masses before. I suspect the problem here is that you are just thinking of the laws in terms of vague qualitative terms, you think that (2) is OK because it doesn't involve a "large mass being stopped by a small mass" like in (4), but it really is necessary to make a quantitative calculation of the energy before and after the collision in order to have a sensible discussion of whether this is consistent with the laws of physics.
Chrisc said:
In (4) although the total energy of the system is conserved as in (1,2, and 3), it is at the expense of contradicting the laws.
Total energy is conserved?? Where are you getting this? I already gave you a detailed analysis in post #35 showing that the combined kinetic energy is not conserved in any of your scenarios, did you even read that? Do you understand that kinetic energy is given by the formula (1/2)*mv^2, and that in order for momentum to be conserved in your scenario, the larger object must have twice the mass of the smaller one, so if the larger object has mass M and the smaller has mass m, then m = M/2? If you agree with that, then once again, in scenario (1) you said:
large mass M is moving with constant velocity v toward a small mass m as measured by an observer R
initially at rest with respect to m.
After a collision through the center of their masses, R measures the velocity of M to be 1/2v and the velocity of m to be v.
So before the collision the first object has mass M and velocity v, so its kinetic energy is (1/2)*Mv^2, while the second object has velocity v=0 and mass m = M/2, so its kinetic energy is (1/2)*(M/2)*(0)^2 = 0. So, the total kinetic energy before the collision is (1/2)*Mv^2. Then after the collision, the first object still has mass M but now has velocity (v/2), so its kinetic energy is (1/2)*(M)*(v/2)^2 = (1/8)*Mv^2, while the second object has mass m = M/2 and velocity v so its kinetic energy is (1/2)*(M/2)*v^2 = (1/4)*Mv^2. So, the combined kinetic energy after the collision is (1/8)*Mv^2 + (1/4)*Mv^2 = (3/8)*Mv^2. So, the total kinetic energy of both objects has decreased from (1/2)*Mv^2 to (3/8)*Mv^2, meaning that unless some of that kinetic energy is dispersed as heat or electromagnetic waves or some other form, the scenario would be physically impossible because energy is not conserved. If you disagree, what part of my analysis do you object to?
 
  • #43
Chrisc said:
When this analogy is considered under the time-reversal of the laws, the physical dynamics of the laws are reversed.
Gravity is a repelling force. ... Masses repel others proportional to the sqrt of their distance.
Forces are invariant under time reversal. Remember: a = d²x/dt². So the power of 2 on the dt means acceleration is "even" under time reversal. Since acceleration is even and force is proportional to acceleration then force is also even under time reversal. Time reversed Newtonian gravity is still an attractive force.

You may want to do a little study before coming back to this. I don't really see why you care about time symmetry, but if it is important to you then you really should approach it in a little more organized manner.

Again, you don't need to focus on a specific example, you need only consider the laws themselves. If the laws exhibit some symmetry then any example you provide must either exhibit the same symmetry or it must not follow the laws.
 
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  • #44
So, the total kinetic energy of both objects has decreased from (1/2)*Mv^2 to (3/8)*Mv^2, meaning that unless some of that kinetic energy is dispersed as heat or electromagnetic waves or some other form, the scenario would be physically impossible because energy is not conserved. If you disagree, what part of my analysis do you object to?
The energy is conserved in the momentum of the system measured in the frame of the observer.
The 1/8*Mv^2 that you claim is missing is the inertia of m.
The force required to set a body in motion must equal or exceed its inertia.
You cannot move a body from rest to a velocity less than that of the body
imparting the force without also slowing the latter.
This is why M continues after the collision at 1/2v not v, and why m is set
in motion with velocity v not 1/2v.
The missing energy of motion is not missing in the total energy of the system
it is converted to momentum.
This is a thought experiment, the bodies are non-composite, so consider the
heat energy if you must, as the motion of the whole body instead of the motion
of its constituents, as it has no constituents.
Momentum and kinetic energy are frame dependent quantities.
The issue here is that the inertia of M is greater than the inertia of m therefore
as a frame dependent quantity, the total momentum of the system as measured
by the observer when at rest with M, is less than the total momentum of the
system when they are at rest with m.
So although the velocity of M and m is the same in both cases (relative), the total energy is not.
If we simply consider a change in the observers position of rest with respect to each
mass before and after collision, the dynamics are valid with respect to the frame
in each case.
But, in the time-reverse version this symmetry of the dynamics holds in the case
where the observer is initially at rest with respect to m, but fails when initially
at rest with respect to M.
 
  • #45
Chrisc said:
The energy is conserved in the momentum of the system measured in the frame of the observer.
That makes no sense at all. Do you understand that "momentum" and "energy" are entirely separate quantities, with the momentum of an object with mass m and velocity v being given by the formula m*v, while the kinetic energy of the same object is given by the formula (1/2)*m*v^2? There are of course other forms of energy, like potential energy and heat (which is really just the combined kinetic energy of all the molecules of the object moving relative to the object's center of mass). But momentum is definitely not a form of energy--in the MKS system of units, momentum has units of kilograms*meters/second, while all forms of energy have units of kilograms*meters^2/second^2.
Chrisc said:
The 1/8*Mv^2 that you claim is missing is the inertia of m.
The force required to set a body in motion must equal or exceed its inertia.
I'm sorry, but you are obviously using terminology without having bothered to read the accepted definitions here (as you were in the case of inelastic vs. elastic collisions). "Inertia" is understood as the tendency of objects to resist acceleration, and can be understood in terms of the Newtonian formula F = m*a, where F is the force on an object, m is its inertial mass, and a is the amount of acceleration it experiences when that force is applied. You can see that for a given applied force F, then the greater the inertial mass m, the smaller the acceleration a. But if F is the only force on the object--if there is no counter-force in the opposite direction to balance it out, like friction--then an object will always accelerate a little bit when force is applied, so your statement "the force required to set a body in motion must equal or exceed its inertia" is totally wrong. It is also totally wrong to treat "inertia" as a form of energy.
Chrisc said:
You cannot move a body from rest to a velocity less than that of the body
imparting the force without also slowing the latter.
Sure, that's obvious just based on conservation of momentum. If the body started at rest then its momentum m*v was zero, so after the collision its momentum increases, which means the momentum of the other object must decrease by the same amount in order for momentum to be conserved. However, just because momentum is conserved, that alone is not enough to guarantee that the collision is a physically realistic one; you must also check whether energy is conserved, if it's not then your scenario is every bit as impossible as the scenario where both objects' velocities increase.
Chrisc said:
The missing energy of motion is not missing in the total energy of the system
it is converted to momentum.
Again, this is nonsense. Momentum is not a form of energy, and energy can only be converted to another form of energy, like linear kinetic energy being converted to heat or to potential energy (like a ball thrown upward in a gravitational field).
Chrisc said:
This is a thought experiment, the bodies are non-composite, so consider the
heat energy if you must, as the motion of the whole body instead of the motion
of its constituents, as it has no constituents.
Heat is always just the combined kinetic energy of all the different parts of the object moving relative to one another, so this doesn't really make much sense. I suppose you could imagine a continuous jello-like object that wasn't made out of discrete components like atoms, but which was nonrigid so that different points in the object could be in motion relative to one another, but it definitely wouldn't make sense to talk about the heat of a rigid, non-composite object.
Chrisc said:
Momentum and kinetic energy are frame dependent quantities.
The issue here is that the inertia of M is greater than the inertia of m therefore
as a frame dependent quantity, the total momentum of the system as measured
by the observer when at rest with M, is less than the total momentum of the
system when they are at rest with m.
So although the velocity of M and m is the same in both cases (relative), the total energy is not.
It's true that momentum and energy are frame-dependent. However, within any single frame, it is required by the laws of physics that the total momentum of a system as measured by that frame must be constant as long as there are no external forces acting on the system (just imagine a collision in deep space), and that the total energy of the system as measured by that frame must be constant as well. It works out so that if the momentum and energy are conserved in one inertial frame, that guarantees that the momentum and energy will also be conserved in every other inertial frame, even though different frames have different numbers for the total momentum and energy.

When I pointed out that in your scenario, total linear kinetic energy went from (1/2)*M*v^2 before the collision to (3/8)*M*v^2 afterwards, this was from the perspective of a single frame, not two different frames. So again, unless you can provide a physically reasonable explanation as to what other form of energy that missing kinetic energy was converted to, your scenario is violating the basic laws of physics of our universe, and so is completely impossible.
 
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  • #46
JesseM said:
That makes no sense at all. Do you understand that "momentum" and "energy" are entirely separate quantities ... But momentum is definitely not a form of energy--in the MKS system of units, momentum has units of kilograms*meters/second, while all forms of energy have units of kilograms*meters^2/second^2.
Not only that, but momentum is a vector and energy is a scalar. Even if they had the same units they would still not be the same thing.
 
  • #47
JesseM, I have attempted to explain this from a number of different perspectives with the intent of helping you
to see the core concepts in question.
I can only assume your fixation with the literal interpretation of the equations has provided you with
a level of understanding beyond my comprehension.
Perhaps in another thread when I have more time to properly consider it, I will ask you to explain
your justifications for arguing the proper interpretation of energy.

Until then, and in order to see if you have an answer to my question, consider the collisions as having
any real properties of matter you see fit. (i.e. the production of heat, radiation, conservation of
kinetic energy, momentum, and frame dependency of all throughout the collisions)

From your previous posts as well as those of others, I will assume, and please correct me if I'm wrong,
that your explanation for the "time-reverse" mechanics that violate the laws of dynamics is as follows:
That all the kinetic energy of the constituent particles of the larger mass might be completely directed
to the acceleration of the smaller mass, thereby bringing the larger to rest, is so extremely improbable
that it has never been observed, but is not impossible and therefore does not violate the laws.
If this is correct, I can present the core concept in these terms.
If not, please explain how the smaller mass brings the larger to rest in the "time-reverse" version
without violating the laws.
 
  • #48
Chrisc said:
JesseM, I have attempted to explain this from a number of different perspectives with the intent of helping you
to see the core concepts in question.
I can only assume your fixation with the literal interpretation of the equations has provided you with
a level of understanding beyond my comprehension.
Perhaps in another thread when I have more time to properly consider it, I will ask you to explain
your justifications for arguing the proper interpretation of energy.
It's very basic physics that momentum and inertia are not forms of energy, that energy always has units of mass*distance^2/time^2, and that energy is always conserved. Feel free to ask any of the mentors on this forum and I'm sure they'll tell you the same thing.
Chrisc said:
Until then, and in order to see if you have an answer to my question, consider the collisions as having
any real properties of matter you see fit. (i.e. the production of heat, radiation, conservation of
kinetic energy, momentum, and frame dependency of all throughout the collisions)
If the linear kinetic energy is constant (an elastic collision), then the collision is just as possible in the time-reversed version as it is in the forward version. If the linear kinetic energy decreases, it must be because some of it was converted to light or heat, in which case thermodynamics explains why the time-reversed version is much less likely in our universe than the forward version, although there will always be some probability larger than zero of seeing it happen in reverse.
Chrisc said:
From your previous posts as well as those of others, I will assume, and please correct me if I'm wrong,
that your explanation for the "time-reverse" mechanics that violate the laws of dynamics is as follows:
That all the kinetic energy of the constituent particles of the larger mass might be completely directed
to the acceleration of the smaller mass, thereby bringing the larger to rest, is so extremely improbable
that it has never been observed, but is not impossible and therefore does not violate the laws.
Your comment about "bringing to rest" is much too vague. It is perfectly possible to have an elastic collision where a smaller mass brings a larger one to rest, and there would be no time asymmetry in this situation. It is only in the case of an inelastic collision, where linear kinetic energy changes before vs. after the collision, that thermodynamics becomes relevant, and the time-reversed version is more or less likely than the time-forward version.

For an example of an elastic collision where a larger mass is brought to rest, suppose we have two masses, the smaller with mass m and the larger with mass 2m, and the smaller is moving to the right at speed (1/3)*v, while the larger is moving to the left at speed (2/3)*v. After the collision, the smaller will be moving to the left at speed v, while the larger will have come to rest. Both before and after the collision, the total momentum of the system is mv to the left. And both before and after the collision, the total kinetic energy is (1/2)*mv^2. This is perfectly reasonable in the forward direction, and there would also be absolutely nothing about a reversed movie of this collision that would give away to a physicist that the movie was being played backwards.
 
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  • #49
Chrisc, here is one more attempt to get you to see that there is no asymmetry in the laws of dynamics in general.

Lets talk about two arbitrary point masses with arbitrary initial velocity that interact with each other through some force f.
[tex]
\begin{array}{111}
\text{1)} & \text{equation:} & f = m a = m x''\\
\text{2)} & \text{forces:} & f \text{ and } -f\\
\text{3)} & \text{masses:} & m1 \text{ and } m2\\
\text{4)} & \text{initial position:} & x1_0 \text{ and } x2_0\\
\text{5)} & \text{initial velocity:} & v1_0 \text{ and } v2_0\\
\end{array}
[/tex]

So 1) is Newton's 2nd law and 2) is Newton's 3rd law. Together they comprise "the laws of dynamics" as you keep saying. And 3)-5) are the initial conditions. So, solving we get the equations of motion:

[tex]
\begin{array}{11}
\text{6)} & x1(t) = x1_0 + v1_0 t + \frac{f}{2m} t^2\\
\text{7)} & x2(t) = x2_0 + v2_0 t - \frac{f}{2m} t^2\\
\end{array}
[/tex]

Now, let's see what happens under time reversal

[tex]
\begin{array}{11}
\text{8)} & \tau = -t\\
\text{9)} & x1(\tau) = x1_0 - v1_0 \tau + \frac{f}{2m} \tau^2\\
\text{10)} & x1'' = f/m\\
\text{11)} & x2(\tau) = x2_0 - v2_0 \tau - \frac{f}{2m} \tau^2\\
\text{12)} & x2'' = -f/m\\
\end{array}
[/tex]

Note that 10) and 12) together are the same as 1) and 2), the laws of dynamics. Therefore the laws of dynamics are time-reverse symmetric. The masses and initial velocities were left arbitrary, as were the forces, so this applies for any interaction of any masses in any reference frame.
 
  • #50
DaleSpam said:
Now, let's see what happens under time reversal

[tex]
\begin{array}{11}
\text{8)} & \tau = -t\\
\text{9)} & x1(\tau) = x1_0 - v1_0 \tau + \frac{f}{2m} \tau^2\\
\text{10)} & x1'' = f/m\\
\text{11)} & x2(\tau) = x2_0 - v2_0 \tau - \frac{f}{2m} \tau^2\\
\text{12)} & x2'' = -f/m\\
\end{array}
[/tex]

Note that 10) and 12) together are the same as 1) and 2), the laws of dynamics. Therefore the laws of dynamics are time-reverse symmetric. The masses and initial velocities were left arbitrary, as were the forces, so this applies for any interaction of any masses in any reference frame.

Shouldn't [itex]f[/itex] change sign under time reversal?

Regards,

Bill
 
  • #51
Antenna Guy said:
Shouldn't [itex]f[/itex] change sign under time reversal?
No, forces and accelerations are invariant under time reversal (which is why Newton's laws are symmetric under time reversal). If forces changed sign under time reversal then Newton's laws would be anti-symmetric.

I am not asking you to take my word for it. Look at the second derivative of 9) and 10) and see for yourself.
 
  • #52
DaleSpam said:
Look at the second derivative of 9) and 10) and see for yourself.

I assume you mean 9) and 11), and I thought I did.

Within a Newtonian context, are you not saying that gravitational forces attract irrespective of which direction time flows?

Regards,

Bill
 
  • #53
Antenna Guy said:
I assume you mean 9) and 11), and I thought I did.
Yes, 9) and 11).
Antenna Guy said:
Within a Newtonian context, are you not saying that gravitational forces attract irrespective of which direction time flows?
That is correct. Time-reversed gravity is attractive, time-reversed ropes pull in tension, time-reversed bars of steel push in compression, etc. If any of those were not so then you could take a movie of some Newtonian scenario and immediately know if it were running forwards or backwards. Then Newton's laws would not be time reverse symmetric.
 
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  • #54
DaleSpam said:
If any of those were not so then you could take a movie of some Newtonian scenario and immediately know if it were running forwards or backwards. Then Newton's laws would not be time reverse symmetric.

I don't know - a water wheel feeding a sluice might be a dead giveaway.:smile:

Regards,

Bill
 
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  • #55
Antenna Guy said:
I don't know - a water wheel feeding a sluce might be a dead giveaway.:smile:
The only reason it is a dead giveaway is because of the laws of thermodynamics, which are not time-reverse symmetric. The entropy of the water in the end is greater than at the beginning. If you look at any scenario where the entropy is equal before and after then you cannot tell which way time is running.
 
  • #56
DaleSpam said:
The only reason it is a dead giveaway is because of the laws of thermodynamics, which are not time-reverse symmetric. The entropy of the water in the end is greater than at the beginning. If you look at any scenario where the entropy is equal before and after then you cannot tell which way time is running.

I must say that I'm not comfortable with the qualifications that you are adding.

I'd like to think that if I see a log pulling a horse (or an apple falling into a tree:rolleyes:), I have a pretty good idea which direction time is flowing.

Regards,

Bill
 
  • #57
Antenna Guy said:
I must say that I'm not comfortable with the qualifications that you are adding.
I'm not adding any qualifications. I never said that all laws of physics, were time-reverse symmetric. I only said that Newton's laws were time-reverse symmetric.

Antenna Guy said:
I'd like to think that if I see a log pulling a horse (or an apple falling into a tree:rolleyes:), I have a pretty good idea which direction time is flowing.
I would hope you would figure that out too. In both cases entropy increases in only one direction. This follows the second law of thermodynamics, which is not time-reverse symmetric. I have clearly pointed that out already starting in post 16 of this thread.
 
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  • #58
Hi Chrisc,

Here is a proof of a smaller mass bringing a larger mass to a stop without even requiring time to be reversed.

A large mass of 20m moving at 10v collides with a stationary mass of 1m.
After the collision the large mass comes to rest and the smaller mass moves off with a velocity of 200v.

Momentum is conserved because the initial momentum 20m*10v is equal to the total final momentum 1m*200v.

The problem with this "proof" is that I have chosen to ignore the conservation of energy law. In fact I can proove just about any fancyful conjecture I like if I choose to ignore the conservation of energy law and it does not make sense to ask why the consequences of a thought experiment do not seem to adhere to the laws of physics if we choose to ignore the laws of physics in setting the thought experiment up. In reality the smaller mass will not depart with a velocity of 200v and the larger mass will not come to rest as I suggested because even though it seems to conserve momentum it does not conserve energy. To make a sensible thought experiment you must ensure both energy and momentum are conserved in the first place if you want it to relate to anything real.
 
  • #59
DaleSpam said:
In both cases entropy increases in only one direction. This follows the second law of thermodynamics, which is not time-reverse symmetric. I have clearly pointed that out already starting in post 16 of this thread.

Point taken regarding what you have said previously.

Not to belabor this, but if a mass loses velocity (decelerates, without changing direction) in a frictionless environment due to gravity alone, in which direction must the source of gravitational force be with respect to the mass' velocity vector - and is that direction the same for both the forward and reverse time scenarios?

I guess the point I'm trying to make here is that the first derivatives with respect to [itex]\tau[/itex] of 9) and 11) (velocity at time [itex]\tau[/itex]) do not equal the first derivatives with respect to time of 6) and 7) (velocity at time t).

Regards,

Bill
 
  • #60
Antenna Guy said:
Not to belabor this, but if a mass loses velocity (decelerates, without changing direction) in a frictionless environment due to gravity alone, in which direction must the source of gravitational force be with respect to the mass' velocity vector - and is that direction the same for both the forward and reverse time scenarios?
OK, let's think it through here.

What is the equation of an orbit? What does that equation imply about the form of the Newtonian law of gravity? If you knew the masses in addition to the equation of an orbit could you determine G?

If you time-reverse the equation of an orbit, what do you get? What does that imply about the form of the time-reversed Newtonian law of gravity? If you knew the masses in addition to the time-reversed equation could you determine G?

Based on the above, if you made a movie of an orbit could you tell if it was running forwards or backwards simply by looking at the movie?

Antenna Guy said:
I guess the point I'm trying to make here is that the first derivatives with respect to [itex]\tau[/itex] of 9) and 11) (velocity at time [itex]\tau[/itex]) do not equal the first derivatives with respect to time of 6) and 7) (velocity at time t).
This is true. All odd derivatives are anti-symmetric wrt reversal, and all even derivatives are symmetric. Newton's laws describe the behavior of the second time derivative, so the laws are even wrt time reversal. The reversal of the first derivatives changes the initial conditions of a given scenario, but not the laws that govern the motion.
 
  • #61
Antenna Guy said:
Not to belabor this, but if a mass loses velocity (decelerates, without changing direction) in a frictionless environment due to gravity alone, in which direction must the source of gravitational force be with respect to the mass' velocity vector - and is that direction the same for both the forward and reverse time scenarios?
One simple example would be a ball tossed upwards on the moon's surface--as its height increases, its velocity decreases, at least until it turns around (if the ball's velocity is greater than or equal to the escape velocity it'll never turn around, of course). What would be the time-reverse of this? Well, it would just be a ball moving downwards towards the moon's surface, its velocity increasing as it falls. Both imply that the gravitational force is pointing in the same downward direction.
 
  • #62
Kev, that is not my point.
I have offered no proof of the mechanics. They follow from the time forward mechanics.
When a small mass is brought to rest by a larger, the time-reverse is a large mass brought
to rest by a smaller, even when the conservation of kinetic energy is considered.
I have not mentioned the kinetic energy as it does not change the principle of the mechanics.
Go ahead and conserver the kinetic energy of the collision forward in time so as to reduce
the velocity of the larger mass after collision from what I've shown.
It's total kinetic energy is still such that it is too great to be assumed by the smaller mass
when the collision is reversed.
But as such a collision occurs forward in time, it's reverse must follow the same(inverse)
kinematics. If not the second law would predict increased entropy backward in time.

Let me make it clear I am not disputing the time-reverse symmetry of the laws of dynamics,
the time-reverse asymmetry of the second law of thermodynamics,
the time-reverse symmetry of SR or the conservation of energy.

I am pointing out and questioning the conflict or discrepancy that arises
when one attempts to hold each true under the time-reverse mechanics
of a particular event observed from both inertial frames of the bodies in question.

You can't have your cake and eat it too.

You must in principle hold to one of the following:
One can claim the second law does not statistically forbid
such dynamics but qualifies them as so astronomically improbable
that they have never (and probably never will) been observed.
But in taking this position, one must then explain how and why
the time reverse version of the same event when observed
at rest with respect to the smaller mass, is not ALSO as astronomically
improbable but it is in fact the ONLY probable dynamics.
This is not a trivial exception to the law, but the necessary norm of the law
under time-reverse symmetry of the dynamical laws.
This fails the principle of the second law.
-OR-
One can take the position that time-reversal requires time defines a
continuous symmetry of dynamics whereby time is not a simple
reversal of the order of kinematics but a physical dynamic that
will produce, indeed demand the astronomically improbable events
of decreasing entropy.
This fails the principle of time-reverse symmetry of the laws.

In taking either position, one realizes that the observed(considered)
mechanics present, in principle, a discrepancy between frames.
A discrepancy that implies, depending on which laws one chooses
to uphold and which one chooses to question, a unique or
privileged inertial frame of reference in that it distinguishes
relative motion from absolute motion with respect to these laws of dynamics.

DaleSpam, as I mentioned above I am not disputing the time-reverse symmetry
of the laws of dynamics in what they express as you've written them above.
When in previous posts I've said, the laws are violated
I am simply pointing out that they must be, or the second law must be.
If you prefer to hold the laws symmetric through time-reversal, I won't argue
that choice, but I will ask you to explain how you rationalize that choice against
its contradiction of the second law.
As for the time-reverse symmetry you offer in the orbital example, it is not
an example in which the principle of time-reverse symmetry of dynamics is tested.
You've given an example of the quantitative symmetry of kinematics through
time reversal. Until your example includes a change in dynamics
in the forward time scenario, the time-reverse symmetry of the "dynamical" laws
governing that change are not tested.

JesseM, a ball tossed upwards on the moon is under the force you impart by throwing
(momentum, kinetic energy)and the gravitational forces of the moon and the ball.
Let the ball free fall from infinity and then reverse the film. Why is the ball being
"pushed" away with less force as the distance increases?
Is the ball's decelerating frame an inertial frame?

I am out of town for the next few days, but I look forward to your answer
to the ball toss question JesseM.


GeorgeJones, this is a struggle, but it is still primarily a question of relativity.
 
  • #63
Chrisc said:
Kev, that is not my point.
I have offered no proof of the mechanics. They follow from the time forward mechanics.
When a small mass is brought to rest by a larger, the time-reverse is a large mass brought
to rest by a smaller, even when the conservation of kinetic energy is considered.
I have not mentioned the kinetic energy as it does not change the principle of the mechanics.
Do you disagree that the example I provided in my last response to you, an elastic collision where a large mass is brought to rest by a smaller one which has a smaller initial velocity, makes perfect sense in the time-forward version? The point, again, is that your description is far too vague, the only situation where a collision would seem strange in reverse is if there is a change in the linear kinetic energy after the collision (an inelastic collision), and in this case all inertial frames will agree on whether the linear kinetic energy increased or decreased when they analyze the same collision.
Chrisc said:
JesseM, a ball tossed upwards on the moon is under the force you impart by throwing
(momentum, kinetic energy)and the gravitational forces of the moon and the ball.
Let the ball free fall from infinity and then reverse the film. Why is the ball being
"pushed" away with less force as the distance increases?
It isn't being pushed in the reverse movie, it's still being pulled. That's what time-symmetry means, that the laws are exactly the same whether you play the movie forward or backward, meaning there's no way to tell whether a given film is actually being played forward or backwards.

In your example of a ball free falling from infinity, the reverse version just looks like a ball shot away from the planet with an initial velocity greater than or equal to the planet's escape velocity. In this case it will continually be moving away from the planet, but its velocity away from the planet is constantly decreasing because the planet is constantly exerting force back in the direction of the planet (though as the distance increases the force decreases because gravitational force follows an inverse-square law). Do you understand that if we toss a ball upwards and its velocity decreases as it rises, this is because the planet is pulling on the ball, not pushing?
Chrisc said:
Is the ball's decelerating frame an inertial frame?
What do you mean by "the ball's decelerating frame"? Do you mean a coordinate system where the ball is at rest while the planet moves away from it, or do you mean a frame where the ball is decelerating as it moves away from the planet? If the former, no, it's not inertial, at least not in Newtonian mechanics which is what we're discussing (in general relativity any freefalling object will be at rest in a local inertial frame which only covers an infinitesimally small region of spacetime). If the latter, that wouldn't be the "ball's frame", because when you talk about an object's frame in physics that's always shorthand for the object's rest frame, if an object is decelerating in a given frame, that isn't the object's own rest frame obviously. You can certainly use an inertial frame to analyze the behavior of an accelerating object though, and that's what's normally done in Newtonian mechanics when we look at the motion of a projectile fired away from a planet, using a frame where the center of mass of the planet/object system is defined to be at rest (if the object's mass is negligible compared to the planet's, then this is just the rest frame of the planet's center).
 
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  • #64
Chrisc said:
When a small mass is brought to rest by a larger, the time-reverse is a large mass brought to rest by a smaller.
No, the time-reverse of "a small mass being brought to rest by a larger" is "a small mass being accelerated from rest by a larger". From this comment and from previous comments about time-reversed gravity being repulsive it is clear that you don't even understand the concept of time-reversal.

Chrisc said:
DaleSpam, as I mentioned above I am not disputing the time-reverse symmetry
of the laws of dynamics in what they express as you've written them above.
When in previous posts I've said, the laws are violated
I am simply pointing out that they must be, or the second law must be.
If you prefer to hold the laws symmetric through time-reversal, I won't argue
that choice, but I will ask you to explain how you rationalize that choice against
its contradiction of the second law. As for the time-reverse symmetry you offer in the orbital example, it is not
an example in which the principle of time-reverse symmetry of dynamics is tested.
You've given an example of the quantitative symmetry of kinematics through
time reversal. Until your example includes a change in dynamics
in the forward time scenario, the time-reverse symmetry of the "dynamical" laws
governing that change are not tested.
Is English a second language for you? Frankly, I am beginning to think that either there must be some severe language barrier or you are being deliberately obtuse. Between kev, JesseM, and myself all of your points have been correctly explained multiple times. Your responses often border on word-salad, and your useage of physics terminology is generally nonsensical and illogical.

I cannot tell if by "second law" you mean "the second law of thermodynamics" or if you mean "Newton's second law". So here is my last try, and I will address both the second law of thermo and Newton's second law.

When we say that a law is symmetric under time reversal we mean that it has the same form wether expressed in terms of t or tau = -t.

Newton's second law:
(1) [tex]f = m \frac{d^2x}{dt^2}[/tex] making the substitution t = -tau we obtain the time-reversed law
(2) [tex]f = m \frac{d^2x}{d(-\tau)^2} = m \frac{d^2x}{d\tau^2} [/tex] which is the same form as (1) so Newton's 2nd law is time-reverse symmetric.

Newton's law of gravitation:
(3) [tex] \frac{d^2x}{dt^2} = -\frac{G m x}{|x|^3}[/tex] making the substitution t = -tau we obtain the time-reversed law
(4) [tex] \frac{d^2x}{d(-\tau)^2} = \frac{d^2x}{d\tau^2} = -\frac{G m x}{|x|^3}[/tex] which is the same form as (3) so Newton's law of gravity is time-reverse symmetric.

Second law of thermodynamics:
(5) [tex]\frac{dS}{dt} \geq 0[/tex] making the substitution t = - tau we obtain the time-reversed law
(6) [tex]\frac{dS}{d(-\tau)} = -\frac{dS}{d\tau} \geq 0[/tex] which is not the same form as (5) so the second law of thermodynamics is not time-reverse symmetric.

You can use the same approach for any law. It is really not any more difficult or complicated than this, and there is no need to conjure up example after example when you can see the symmetry (or asymmetry) in general from the laws themselves.
 
  • #65
JesseM said:
Do you disagree that the example I provided in my last response to you, an elastic collision where a large mass is brought to rest by a smaller one which has a smaller initial velocity, makes perfect sense in the time-forward version? The point, again, is that your description is far too vague,
I don't disagree. Yes it makes perfect sense.
My description seems vague because you are trying to read too much into it.
I am talking about the principles of the laws not the time-reverse symmetry of the equations.
Perhaps it will be more obvious to you if I use your example.

A ball in free fall from infinity to the moon.
The gravitational potential of the ball at infinity is theoretically infinite.
The gravitational potential of the same ball upon "approaching" the surface of the moon is
by comparison infinitesimal.
Between infinite and infinitesimal there exists a substantial difference in potential energy.
When your film runs backwards this difference must be reasoned.
How?
There is no one throwing the ball. There is no way to reason the motion of the ball
"decelerating" (negative acceleration) away from the moon except via the
gravitational energy of each.
The gravitational potential must be inverted to account for the motion of the ball.
Do not think to include the collision of the ball as the energy that launches
it back into space. The event is the free-fall of the ball, not the collision.
The free-fall can be explained without the collision, so the inverse of this free-fall must
also be explained without the collision.

When the ball is in free fall to the moon, the ball's frame is an inertial frame.
When the ball is in the reverse of free fall away from the moon under no more
influence than inverted gravitational potential, is it also an inertial frame?
 
  • #66
DaleSpam said:
No, the time-reverse of "a small mass being brought to rest by a larger" is "a small mass being accelerated from rest by a larger". From this comment and from previous comments about time-reversed gravity being repulsive it is clear that you don't even understand the concept of time-reversal.

According to your distinction between the kinematics of time-reversal above,
you must then admit the time-reversal of "a small mass accelerating a larger mass"
is "a larger mass being brought to rest by a smaller".
You have simply chosen a preferred frame as did I. In either case the energy
transfered between the masses when the larger is brought to rest or as you prefer
to state it, the smaller is accelerated, contradicts the second law of thermodynamics.

English is my first language.
I think the language barrier is a matter of preconceptions.
You assume I don't know the meaning of the time-reverse symmetry of the laws.
Instead of trying to understand what I'm saying you keep restating what you
think I don't understand.
There are three considerations to the symmetry of the dynamical laws under time-reversal.
The symmetry of the equations:
the quantitative equality of the statements when time is + or -.

The symmetry of the principle of the law:
the principle is unchanged when time is + or -.

The symmetry of the mechanics governed by the law:
the kinematics "AND" dynamics are unchanged when time is + or -.

You have pointed out the symmetry of the quantitative statements of the equations.
I assume you would then agree that the principle of causality remains unchanged
during time-reversal, even though what we identify as cause and effect in the
forward time event become reversed to uphold the law in the time-reverse event.
Effect becomes cause and cause becomes effect.
If you agree, what is the answer to the question I asked of JesseM regarding the
motion of the ball moving away from the moon in time-reversal?
What is the "cause" of the balls motion with respect to the moon?
 
  • #67
Oh well Chrisc, I really tried to help you understand here. Enjoy your ignorance, it appears deliberate to me.
Chrisc said:
If you agree, what is the answer to the question I asked of JesseM regarding the motion of the ball moving away from the moon in time-reversal? What is the "cause" of the balls motion with respect to the moon?
I'm sure JesseM will point out all of the incredibly obvious and basic mistakes you made, starting with the second sentence of your moon example.
 
  • #68
DaleSpam said:
...all of the incredibly obvious and basic mistakes you made, starting with the second sentence of your moon example.

Upon reading of the fourth sentence of the referenced example, it seems obvious that the word "energy" should be understood to follow the word "potential" in the second and third sentences. If this clarified wording is still an "obvious and basic mistake", I'm curious to hear why you think so.

Regards,

Bill
 
  • #69
Chrisc said:
There is no one throwing the ball. There is no way to reason the motion of the ball
"decelerating" (negative acceleration) away from the moon except via the
gravitational energy of each.
The gravitational potential must be inverted to account for the motion of the ball.
Do not think to include the collision of the ball as the energy that launches
it back into space. The event is the free-fall of the ball, not the collision.
The free-fall can be explained without the collision, so the inverse of this free-fall must
also be explained without the collision.

Here is an experiment you can actually do in your back garden. Film a ball released from a fixed height onto a trampoline and film the rebound (almost) back to its starting point. When the film is played backward it will be hard to tell the difference from the film playing forward. Sure the final height is not as high as the starting height but that is due to friction from the air and the trampoline. If there was no air and the trampoline was frictionless it would not be possible to tell which way the film was playing. The ball is being accelerated downwards as it falls AND as it rebounds. With no energy losses the situation is perfectly symmetrical. With energy losses the second law of thermodynamics comes into play and gives an arrow of time that removes the symmetry.


Chrisc said:
When the ball is in free fall to the moon, the ball's frame is an inertial frame.
When the ball is in the reverse of free fall away from the moon under no more
influence than inverted gravitational potential, is it also an inertial frame?

Yes.
 
  • #70
Chrisc said:
I don't disagree. Yes it makes perfect sense.
My description seems vague because you are trying to read too much into it.
I am talking about the principles of the laws not the time-reverse symmetry of the equations.
Perhaps it will be more obvious to you if I use your example.
Do you agree that for any system governed by time-reversible equations, if you film a situation where there is no change in entropy, there will be absolutely nothing to tell a physicist whether the film is being played backwards or forwards? If you disagree with this, it's apparent that you don't understand what "time-reverse symmetry" actually means, this is logically implied by time-reversible equations.
Chrisc said:
A ball in free fall from infinity to the moon.
The gravitational potential of the ball at infinity is theoretically infinite.
Actually, by convention the gravitational potential is usually treated as being zero at infinity, and being negative at finite distances, the absolute value of the negative potential being greater the closer you approach the object's surface.
Chrisc said:
The gravitational potential of the same ball upon "approaching" the surface of the moon is
by comparison infinitesimal.
No, it'd be some finite negative number.
Chrisc said:
Between infinite and infinitesimal there exists a substantial difference in potential energy.
The difference in potential energy would be finite; zero at infinity, some finite negative number at the surface. Of course where you set the zero is somewhat arbitrary, you could define the potential function to be zero at the surface, but in this case it would have some finite positive value at infinity (the difference between the potential at the surface and the potential at infinity would have to be the same regardless of where you set the zero--the potential is always larger at greater distances from the surface).
Chrisc said:
When your film runs backwards this difference must be reasoned.
How?
There is no one throwing the ball. There is no way to reason the motion of the ball
"decelerating" (negative acceleration) away from the moon except via the
gravitational energy of each.
The gravitational potential must be inverted to account for the motion of the ball.
You're totally confused about basic vector analysis here. If a ball is thrown away from the moon, and the gravitational force vector is pointing back at the center of the moon, then naturally the ball's speed in the "upward" direction will be decreasing. This does not require any change in the potential function (the potential must be greater at bigger distances, since increasing potential energy balances out decreasing kinetic energy as the ball's speed decreases while it's moving away from the moon--the sum of kinetic plus potential energy must always be conserved!) Perhaps you should study some basic Newtonian mechanics before pontificating about how everything physicists believe about time-reversal symmetry is wrong.
Chrisc said:
When the ball is in free fall to the moon, the ball's frame is an inertial frame.
Not in Newtonian mechanics it isn't. In Newtonian mechanics, the ball is being acted on by a force (gravity), so it is not moving inertially. In general relativity a freely-falling object does have a local inertial rest frame, but I think you need to review basic Newtonian physics before trying to tackle general relativity.
 

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