A study of the motion of a relativistic continuous medium

In summary: the observers at the front and back of the rocket would see each other as contracted along the direction of the vector between them.
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  • #3
J.F. said:

That paper would appear to be junk. It claims that the string length in Bell's rocket paradox has been incorrectly calculated in all previous treatments. This is unlikely, because Bell's paradox has the rockets moving with Born rigid acceleration, which has the property that the proper length of the string remains constant. The claim of the paper is tantamount to saying that Born rigid motion does not have the property of constant proper length. The paper also claims that the accepted bunch length of charged particles in an accelerator is not correct. This is also unlikely as accelerators are operated every day somewhere in the world and knowing the bunch length of accelerated particles is probably essential for the routine operation of these machines.
 
  • #4
yuiop said:
That paper would appear to be junk. It claims that the string length in Bell's rocket paradox has been incorrectly calculated in all previous treatments. This is unlikely, because Bell's paradox has the rockets moving with Born rigid acceleration, which has the property that the proper length of the string remains constant. The claim of the paper is tantamount to saying that Born rigid motion does not have the property of constant proper length. The paper also claims that the accepted bunch length of charged particles in an accelerator is not correct. This is also unlikely as accelerators are operated every day somewhere in the world and knowing the bunch length of accelerated particles is probably essential for the routine operation of these machines.
Right. In section 2, the paper says,

Let us consider, for instance, Bell’s well-known problem where two identical pointlike rockets simultaneously (by IRF clocks) begin to move in the same direction, one following the other, with constant and equal accelerations (in the astronauts’ reference frame). Suppose that these rockets are connected by a rubber (a thread) which does not affect their motion.
Now, if there is to be Born rigid acceleration along the lengths of both rockets, such that the thread does not break, then there must be a continuous acceleration gradient of the constant proper accelerations between the front of the first rocket and the back of the second rocket of 1/ aF - 1 / aB = d / c^2, where d is the proper length along the entire length of both rockets and the thread. Since these are proper accelerations, it is what the astronauts will actually measure with accelerometers, but this is not what was stated. Rather, the paper states that the rockets have equal accelerations. So if the acceleration profile of each rocket is the same, yet each rocket also contracts with Born rigid motion, then of course the thread must break.
 
  • #5
Actually, if we just fire a thruster at the back of each ship that applies a constant proper acceleration at those two points and the rest of each ship contracts naturally according to Born rigid motion, then we could simply make d the distance between the thrusters at the back of the ships, or the length of the back ship plus the length of thread. So if the thruster at the back of the back ship applies 1 G of constant proper acceleration and the thruster at the back of the front ship (it doesn't even matter how long the front ship is) applies a constant proper acceleration of (1 G) / [1 + (1 G) d / c^2], then the thread will not break.
 
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  • #6
Oh, I just caught that the setup I quoted from the paper says that the rockets are point-like. So as not to confuse anybody, d would just be the length of the thread itself in both of my posts in that case, then. If the rockets have equal constant proper accelerations, then the thread will break. If the back rocket accelerates faster than the front rocket such that 1 / aF - 1 / aB = d / c^2, then the thread will contract with Born rigid motion at the same rate that the distance between the rockets decreases according to the initial rest frame, and the thread will not break.
 
  • #7
Sorry for the multiple posts, but now I'm wondering about what Rigid born observers at the front and back of a rocket will actually view of each other during acceleration. That is, the equation I posted, 1 / aF - aB = d / c^2, are really for the accelerations required at the front and back such that after shutting off the thrusters and both the front and back becoming inertial again in a final frame so that they are both now traveling at the same speed v, they will be length contracted to that new frame accordingly with sqrt(1 - (v / c)^2) d as viewed from the initial rest frame. But that does not necessarily mean that both will become inertial again simultaneously as they view each other, for instance. Indeed, the proper time of clocks of observers placed at the front and back will read different times as they view each other's clocks in the new frame, with the back clock reading a lesser time than the front clock by arctanh(v / c), and according to the rest frame as well with d v / c^2 - arctanh(v / c).

So it may be that the length does not remain constant during acceleration either, so not truly rigid as the rocket observers view it, but only becomes the same again after reaching the final frame. A ruler placed between the observers during acceleration will continue to lie between the observers during acceleration as it accelerates with rigid Born motion as the rocket does, however, so it will always measure the same distance between them, but I'm not sure if a ruler can be considered a trusted measure of distance during acceleration if the observers do not even measure each other simultaneously according to what their own clocks read, and they will also not directly see things in the same way as light pulses struggle to catch up as the rocket is accelerating. This of course does not change the fact that the thread breaks when both rockets apply equal accelerations, but questions what might really required for the string not to break during the acceleration itself, in a way that is more than just being correctly contracted upon reaching the final frame.
 
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  • #8
grav-universe said:
Now, if there is to be Born rigid acceleration along the lengths of both rockets, such that the thread does not break, then there must be a continuous acceleration gradient of the constant proper accelerations between the front of the first rocket and the back of the second rocket of 1/ aF - 1 / aB = d / c^2, where d is the proper length along the entire length of both rockets and the thread. Since these are proper accelerations, it is what the astronauts will actually measure with accelerometers, but this is not what was stated. Rather, the paper states that the rockets have equal accelerations. So if the acceleration profile of each rocket is the same, yet each rocket also contracts with Born rigid motion, then of course the thread must break.
O.K. I concede you have a point and I may have skimmed the paper too quickly. We are at least agreed that the string breaks in the equal acceleration case. In order not to break the string, the distance between the rockets needs to length contract, as per Born rigid motion, in the inertial reference frame. In the equal acceleration case the separation between the rockets remains constant so the string must be under increasing strain and eventually break.

grav-universe said:
So it may be that the length does not remain constant during acceleration either, so not truly rigid as the rocket observers view it, but only becomes the same again after reaching the final frame.
Now it is my turn to correct you. While the rockets are undergoing Born rigid motion, the proper length between them as measured by rulers (and radar) remains constant. There is no "final frame" if you are implying that the rocket ends up with inertial motion. Under Born rigid motion the rockets remain accelerating forever and always have been accelerating, with each rocket maintaining constant proper acceleration indefinitely. However ...

grav-universe said:
A ruler placed between the observers during acceleration will continue to lie between the observers during acceleration as it accelerates with rigid Born motion as the rocket does, however, so it will always measure the same distance between them, but I'm not sure if a ruler can be considered a trusted measure of distance during acceleration if the observers do not even measure each other simultaneously according to what their own clocks read, and they will also not directly see things in the same way as light pulses struggle to catch up as the rocket is accelerating.

... the radar distance will differ from the ruler distance in the accelerated frame over extended distances just as it does when making radial distance measurements in a gravitational field. I will try and figure out exactly what that difference is in quantifiable terms in a following post.

I think you also might also correct that if the rockets are undergoing constant Born rigid motion and the proper separation is x as measured by rulers (in either the accelerating RF or the ICIRF) then if the rockets were to revert to inertial motion the distance between them would be greater than x. In other words the proper length of an elastic string connecting the rockets when they are accelerating (which is the length measured in the ICIRF) is different from the proper length between the rockets when they have inertial motion and maybe this is what the paper is getting at. Again, it should not be too difficult to quantify this difference.

You agree that the paper does not change the outcome of Bell's paradox and I think you will also agree that scientists operating particle accelerators have not been using the wrong calculation for bunch lengths of charged particles in their accelerators all these years.
 
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  • #9
yuiop said:
... the radar distance will differ from the ruler distance in the accelerated frame over extended distances just as it does when making radial distance measurements in a gravitational field. I will try and figure out exactly what that difference is in quantifiable terms in a following post.
O.K. this is what I get:

If the proper ruler distance in the accelerating or ICIRF is [tex]L = (x_1-x_2)[/tex] then the radar distance [tex]L_R[/tex] measured by an accelerating observer at [tex]x_1[/tex] with constant proper acceleration [tex]a = c^2/x_1[/tex] is:

[tex]L_{R} = L \, \frac{(c^2-aL/2)}{(c^2-aL)} [/tex]

It can be seen that when the acceleration is zero that [tex]L_{R} = L [/tex] and for very small L that [tex]L_{R} \approx L [/tex]

Does that seem reasonable?
 
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  • #10
yuiop said:
O.K. I concede you have a point and I may have skimmed the paper too quickly.
Oh no, I was agreeing with you. :smile: Although I smimmed the paper also, the paper appears to claim that we do not really know what happens to the thread when two rockets equally properly accelerate, that there are at least three possible solutions, but of course we know that is not the case, there is only one, the thread breaks.

We are at least agreed that the string breaks in the equal acceleration case. In order not to break the string, the distance between the rockets needs to length contract, as per Born rigid motion, in the inertial reference frame. In the equal acceleration case the separation between the rockets remains constant so the string must be under increasing strain and eventually break.
Right. By the way, I'm wondering what the name of the paper has to do with its content. Its name gives the impression it has something to do with LET. Bell's paradox doesn't even come close to springing to mind before reading it.

Now it is my turn to correct you. While the rockets are undergoing Born rigid motion, the proper length between them as measured by rulers (and radar) remains constant. There is no "final frame" if you are implying that the rocket ends up with inertial motion. Under Born rigid motion the rockets remain accelerating forever and always have been accelerating, with each rocket maintaining constant proper acceleration indefinitely. However ...
A ruler placed between the two ends of a single rocket will contract at the same rate as the rocket, so the ruler always measures the same length, yes.

... the radar distance will differ from the ruler distance in the accelerated frame over extended distances just as it does when making radial distance measurements in a gravitational field. I will try and figure out exactly what that difference is in quantifiable terms in a following post.
Yes, that would be true under the equivalence principle, wouldn't it? Good call, thanks. :smile: If the distance between two observers at different points in a gravitational field is constant as measured with a ruler, the observer at a greater gravitational potential in the field (the back rocket observer) will observe a constant blueshift of the other while the one at a lesser potential (the front rocket observer) will observe a constant redshift. So working through the equations of acceleration should bear that out. Sounds like fun. :smile: I will have to work on that also.

I think you also might also correct that if the rockets are undergoing constant Born rigid motion and the proper separation is x as measured by rulers (in either the accelerating RF or the ICIRF) then if the rockets were to revert to inertial motion the distance between them would be greater than x. In other words the proper length of an elastic string connecting the rockets when they are accelerating (which is the length measured in the ICIRF) is different from the proper length between the rockets when they have inertial motion and maybe this is what the paper is getting at. Again, it should not be too difficult to quantify this difference.
But that's just it. To visualize better, rather than a rocket, let's say we have two observers stationary within the rest frame with a distance d between them, lined up along the x axis. They then accelerate simultaneously along the x-axis with the relation 1 / aF - 1 / aB = d / c^2 between their constant proper accelerations. Both accelerate simultaneously from the rest frame with these accelerations until each reaches a relative speed to the rest frame of v, then become inertial at v. The difference in proper readings upon their clocks when they reach the final frame will be arctanh(v / c), as according to Born rigid motion. The distance that will be measured between the observers from the rest frame will be sqrt(1 - (v / c)^2) d, so they will also measure the same proper distance d between themselves in the final frame as they did in the rest frame. But since they don't reach the final frame simultaneously according to their own clocks, then the distance between themselves should perhaps have been measured as something else while accelerating, but I'm not sure how that measurement would be performed since a ruler placed between them, for instance, should contract with Born rigid motion to the same degree that the observers accelerate. Since distance is not absolute anyway, maybe there is no ideal way to measure while accelerating, but we can only measure effects. Measuring in terms of the gravitational potential between them while considering the distance constant as you mentioned might be the best way to go, though.

You agree that the paper does not change the outcome of Bell's paradox and I think you will also agree that scientists operating particle accelerators have not been using the wrong calculation for bunch lengths of charged particles in their accelerators all these years.
Right.
 
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  • #11
yuiop said:
O.K. this is what I get:

If the proper ruler distance in the accelerating or ICIRF is [tex]L = (x_1-x_2)[/tex] then the radar distance [tex]L_R[/tex] measured by an accelerating observer at [tex]x_1[/tex] with constant proper acceleration [tex]a = c^2/x_1[/tex] is:

[tex]L_{R} = L \, \frac{(c^2-aL/2)}{(c^2-aL)} [/tex]

It can be seen that when the acceleration is zero that [tex]L_{R} = L [/tex] and for very small L that [tex]L_{R} \approx L [/tex]

Does that seem reasonable?
Well, from the perspective of the initial rest frame, in order for the rocket to be contracting with Born rigid motion, then the front and back must have different proper accelerations, so that must be taken into account for the redshift and blueshift between them. I'm still not sure how the rocket observers should measure the distance between themselves while accelerating, though, but I have an idea based upon what you stated earlier. The redshift and blueshift observed between themselves can also be found from the rest frame, so I will work it out from the rest frame's perspective. Then I will work it out from the perspective of the rocket observers by treating what each measures for their proper accelerations as gravitational potentials instead, to find what the distance between them should be in terms of a gravitational redshift and blueshift using the equivalence principle. This may take a while. :smile:

EDIT - Oh, wait. In a gravitational field, the distance between them can still be anything, depending upon the mass involved, so I cannot go that route, but I can still find the Doppler shifts for now.
 
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  • #12
grav-universe said:
Right. By the way, I'm wondering what the name of the paper has to do with its content. Its name gives the impression it has something to do with LET. Bell's paradox doesn't even come close to springing to mind before reading it.
I was wondering that too. The title alone put me into a "this must be a crackpot etherist thing" alert :-p. I now assume they mean the length of an extended continuous object.
grav-universe said:
Yes, that would be true under the equivalence principle, wouldn't it? Good call, thanks. :smile: If the distance between two observers at different points in a gravitational field is constant as measured with a ruler, the observer at a greater gravitational potential in the field (the back rocket observer) will observe a constant blueshift of the other while the one at a lesser potential (the front rocket observer) will observe a constant redshift. So working through the equations of acceleration should bear that out. Sounds like fun. :smile: I will have to work on that also.
I'll add it to my"to do" list too. :wink:
grav-universe said:
Since distance is not absolute anyway, maybe there is no ideal way to measure while accelerating, but we can only measure effects. Measuring in terms of the gravitational potential between them while considering the distance constant as you mentioned might be the best way to go, though.
We have the same problem in a gravitational field where definitions of distance over extended distances depend on who is doing the measuring and what method is used. Ruler distances at least have the advantage that there is only one answer to the distance between two points while radar distances have the disadvantage that they are dependent upon which end the radar measurement is made from. While measurements of distance in a accelerating reference frame might seem complicated it is easy to demonstrate that can remain constant over time by measuring the height of the room you in. The floor has a different absolute acceleration relative to the ceiling but the height as measured by a ruler does not change over time. The radar distance measured from the ceiling will be different from the radar measurement made from the floor and both will be different from the ruler measurement but they all remain constant over time (if your house does not have subsidence :-p).

grav-universe said:
The redshift and blueshift observed between themselves can also be found from the rest frame, so I will work it out from the rest frame's perspective. Then I will work it out from the perspective of the rocket observers by treating what each measures for their proper accelerations as gravitational potentials instead, to find what the distance between them should be in terms of a gravitational redshift and blueshift using the equivalence principle. This may take a while. :smile:
You have to be careful here. The equivalence principle fails over extended distances and tidal effects in a real gravitational field will cause the measurements from a gravitational field to differ from those made in the artificially accelerating case. An obvious difference is that the proper force in the Born rigid case is proportional to 1/d while in the gravitational case it is more like 1/d^2 modified by a gravitational gamma factor.
 
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  • #13
Relativistic length expansion in general accelerated system revisited

Classical solution of the relativistic length expansion in general accelerated system completely revisited.Instant proper length measurement between J.S.Bell's rockets also is considered successfully.

http://arxiv.org/abs/0910.2298
 
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  • #14
Finally! I tried half a dozen different ways that seemed to just refuse to reduce, but I finally hit upon a way to do it using the relative speeds. Okay, so we're finding the Doppler shift for light pulses emitted at regular intervals from the back of a ship to the front. The ship is initially stationary in the rest frame with a proper length of d. The back and front of the ship then simultaneously begin to accelerate with Born rigid motion, with constant proper accelerations of the back and front of aB and aF, with the relation 1 / aF - 1 / aB = d / c^2. According to the rest frame, the back emits a first pulse to the front at TB1 and a second pulse at TB2, when a clock at the back of the ship reads TB1' and TB2'. The pulses are received at the front of the ship at TF1 and TF2, when a clock at the front reads TF1' and TF2'. The Doppler shift is the ratio of the observed frequency of the pulses to the frequency at which they were emitted, so

D = f_obs / f_emit = [1 / (TF2' - TF1')] / [1 / (TB2' - TB1')] = (TB2' - TB1') / (TF2' - TF1')

Now, according to the rest frame, the first pulse of light will travel from the back to the front in the time TF1 - TB1, traversing the difference in distances that the front and back have traveled during those times according to the relativistic acceleration formula for distance, so we have

c (TF1 - TB1) = d_TF1 - d_TB1

= [d + (c^2 / aF) (sqrt(1 + (aF TF1 / c)^2) - 1)] - [(c^2 / aB) (sqrt(1 + aB TB1 / c)^2) - 1)]

TF1 - TB1 = d / c - c (1 / aF - 1 / aB) + (c / aF) sqrt(1 + (aF TF1 / c)^2) - (c / aB) sqrt(1 + aB TB1 / c)^2), where 1 / aF - 1 / aB = d / c^2, so

TF1 - TB1 = (c / aF) sqrt(1 + (aF TF1 / c)^2) - (c / aB) sqrt(1 + aB TB1 / c)^2)

The relativistic formula for the instantaneous speed at the back of the ship when the photon is emitted is

(vB1 / c) = (aB TB1 / c) / sqrt(1 + (aB TB1 / c)^2)

sqrt(1 + (aB TB1 / c)^2) = aB TB1 / vB1

(c / aB) sqrt(1 + (aB TB1 / c)^2) = TB1 c / vB1

and likewise for TF1, so for TF1 - TB1 that we found before, we now have

TF1 - TB1 = TF1 c / vF1 - TB1 c / vB1

TF1 ((c / aF) - 1) = TB1 ((c / aB) - 1)

The relativistic formula for speed can also be re-arranged further to give

(aB TB1 / c) = (vB1 / c) / sqrt(1 - (vB1 / c)^2)

TB1 = (vB1 / aB) / sqrt(1 - (vB1 / c)^2)

and likewise for TF1. Applying that to the previous equation gives

[(vF1 / aF) / sqrt(1 - (vF1 / c)^2)] ((c / aF) - 1) = [(vB1 / aB) / sqrt(1 - (vB1 / c)^2)] ((c / aB) - 1)

(vF1 / aF) (c / vF1) (1 - vF1 / c) / sqrt(1 - (vF1 / c)^2) = (vB1 / aB) (c / vB1) (1 - vB1 / c) / sqrt(1 - (vB1 / c)^2)

(c / aF) sqrt((1 - vF1 / c) / (1 + vF1 / c)) = (c / aB) sqrt((1 - vB1 / c) / (1 + vB1 / c))

Now we will apply the relativistic formula for proper time with

TB1' = (c / aB) ln[(1 + vB1 / c) / (1 - vB1 / c)] / 2

aB TB1' / c = ln[sqrt((1 + vB1 / c) / (1 - vB1 / c))]

e ^ (aB TB1' / c) = sqrt((1 + vB1 / c) / (1 - vB1 / c))

and likewise for TF1, so the equation now becomes

(c / aF) / (e ^ (aB TB1' / c)) = (c / aB) / (e ^ (aF TF1' / c))

(e ^ (aF TF1' / c)) / (e ^ (aB TB1' / c)) = aF / aB

aF TF1' / c - aB TB1' / c = ln(aF / aB)

and finding the same thing similarly for the second photon emitted from the back to the front also gives us

aF TF2' / c - aB TB2' / c = ln(aF / aB)

and subtracting these last two equations from each other, we get

(aF TF2' / c - aF TF1' / c) - (aB TB2' / c - aB TB2' / c) = 0

(aF / c) (TF2' - TF1') = (aB / c) (TB2' - TB1')

D = (TB2' - TB1') / (TF2' - TF1) = aF / aB

So the observer at the front of the ship will measure a constant redshift of the light coming from the back of the ship of D = aF / aB. Since 1 / aF - 1 / aB = d / c^2, that can also be stated as D = 1 - aF d / c^2 or 1 / (1 + aB d / c^2). If we consider the difference in accelerations between the front and back to be extremely small, then we can say that the acceleration is approximately a = aB = aF, giving an approximate redshift of D = 1 - a d / c^2 or D = 1 / (1 + a d / c^2). Knowing the redshift, we can also now find the Born horizon. If the back of the ship lies precisely upon the Born horizon as viewed from the front, then no pulses will reach the front and the redshift will be zero, whereas since D = aF / aB, then aB would have to be infinite at the limit. So we have

1 / aF - 1 / aB = d / c^2, 1 / aB = 0

1 / aF = d / c^2

d = c^2 / aF
 
  • #15
Your calculations of the Doppler shift - just awesome :wink:

yuiop said:
I think you also might also correct that if the rockets are undergoing constant Born rigid motion and the proper separation is x as measured by rulers (in either the accelerating RF or the ICIRF) then if the rockets were to revert to inertial motion the distance between them would be greater than x. In other words the proper length of an elastic string connecting the rockets when they are accelerating (which is the length measured in the ICIRF) is different from the proper length between the rockets when they have inertial motion and maybe this is what the paper is getting at. Again, it should not be too difficult to quantify this difference.
My guess here was wrong. Now that I have analysed it carefully, the proper length of the string (as measured by a ruler) when it is accelerating is exactly the same as the proper length of the string when it reverts to inertial motion. In other words the length of the string measured in the ICIRF is the proper length of the string, whether it accelerating (with Born rigid motion) or moving inertially. The length of the string as measured by radar distances will however change when the rockets change from accelerating to inertial motion as will (obviously) the redshift between the clocks in the rocket reference frame.

Now that we have some independently derived "facts" for the Bell rocket case and Born rigid motion I will look at the paper again and try to come to a better informed opinion of it.
 
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  • #16
yuiop said:
You have to be careful here. The equivalence principle fails over extended distances and tidal effects in a real gravitational field will cause the measurements from a gravitational field to differ from those made in the artificially accelerating case. An obvious difference is that the proper force in the Born rigid case is proportional to 1/d while in the gravitational case it is more like 1/d^2 modified by a gravitational gamma factor.
Ah yes, that's true. The acceleration gradients are different, so that would make a difference for how the photons travel through the ship as compared to a gravitational field.
 
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  • #17
yuiop said:
Your calculations of the Doppler shift - just awesome :wink:
Thanks. :smile: Working through the equations in the same way for photons that travel from the front to the back gives just the inverse, a blueshift of aB / aF. I have also come to realize that since the acceleration at the back of the ship would have to be infinite at the limit of the Born horizon of d = c^2 / aF, then that also places a limit on the greatest possible constant proper acceleration that can be maintained at the front of the ship. If the ship while initially at rest is d, then in order for the ship to accelerate with Born rigidity without ripping itself apart, the limit of the proper acceleration that can be applied at the front without the ship breaking up would be aF = c^2 / d. Its just a good thing that the value of c is extremely large. :smile:
 
  • #18
yuiop said:
Now that we have some independently derived "facts" for the Bell rocket case and Born rigid motion I will look at the paper again and try to come to a better informed opinion of it.
I put "facts" in quotation marks because there is always the significant possibility that I have made some errors and in fact after checking through again I made an error in my earlier calculation of the radar distance in the accelerating frame.
yuiop said:
O.K. this is what I get:

If the proper ruler distance in the accelerating or ICIRF is [tex]L = (x_1-x_2)[/tex] then the radar distance [tex]L_R[/tex] measured by an accelerating observer at [tex]x_1[/tex] with constant proper acceleration [tex]a = c^2/x_1[/tex] is:

[tex]L_{R} = L \, \frac{(c^2-aL/2)}{(c^2-aL)} [/tex]

It can be seen that when the acceleration is zero that [tex]L_{R} = L [/tex] and for very small L that [tex]L_{R} \approx L [/tex]

Does that seem reasonable?
What I had calculated was ct where t is the time measured in the initial inertial RF where as I should have calculated cT where T is the proper time of the observer measuring the radar distance in the accelerating frame. The corrected equation should be:

[tex]ct = L_0 \, \frac{(c^2-aL_0/2)}{(c^2-aL_0)} [/tex]

[tex]\rightarrow \frac{c^2}{a} \sinh (aT/c) = L_0 \, \frac{(c^2-aL_0/2)}{(c^2-aL_0)} [/tex]

[tex]\rightarrow cT = \frac{c^2}{a}\sinh^{-1}\left( \frac{ac^2L_0-(aL_0)^2/2}{c^4-ac^2L_0} \right) [/tex]

This is the radar distance for the Born rigid case.

The ruler measurement of the proper distance for the equal acceleration case, as a function of velocity or time, is very simply deduced to be:

[tex]L(v) = \frac{L_0}{\sqrt{1-\beta^2}} [/tex]

[tex]L(t) = L_0 \sqrt{1+(at/c)^2} [/tex]

Now the paper gives the proper distance (Equation 23) as a function of time for the equal acceleration case as:

[tex]L(t) = \frac{c^2}{a} \LN \left( \cosh \left( \frac{aL_0}{c^2} \right) + \sinh \left( \frac{aL_0}{c^2} \right) \sqrt{1+\beta^2} \right)[/tex]

The paper does not make it very clear if they mean the radar or the ruler distance. However it can be noted that that it does not make sense to talk of radar distances as a function of velocity when the velocity is changing during the measurement, or as a function of time when the time when the radar signal was sent differs from the time when the radar signal was received, I will assume they mean the ruler distance. Later on the authors say "The use of the radar method for particles at finite distances and for continuum motions which are not rigid in Born’s sense leads to errors" implying the authors do not like radar measurements which backs up the conclusion I made above. Either way, the equation given by the authors does not agree with the equations I gave for either the ruler distance or the radar distance when compared numerically.

The equation I gave for radar distance is only strictly valid for Born acceleration, but can be used for the equal acceleration case when the average velocity of the rockets relative to the ICIRF is zero (and the rockets are monetarily aligned with the x-axis of the ICIRF) and compared with the paper's calculation for L(t) using [tex]\beta =0 [/tex] and the predictions still differ significantly with the paper's calculation predicting that the radar length is equal to the ICIRF measured length which is plainly wrong.

Finally in the conclusion of the paper that authors state "Thus, during integration, the authors leave the physical space comoving with the medium" whatever that means and seemingly hinting at an ether. Having made several errors myself in analysing the paper I won't be rash and call it "junk" again and have now promoted it the category of "highly suspect" or "dodgy" :-p .
 
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  • #19
grav-universe said:
Thanks. :smile: Working through the equations in the same way for photons that travel from the front to the back gives just the inverse, a blueshift of aB / aF. I have also come to realize that since the acceleration at the back of the ship would have to be infinite at the limit of the Born horizon of d = c^2 / aF, then that also places a limit on the greatest possible constant proper acceleration that can be maintained at the front of the ship. If the ship while initially at rest is d, then in order for the ship to accelerate with Born rigidity without ripping itself apart, the limit of the proper acceleration that can be applied at the front without the ship breaking up would be aF = c^2 / d. Its just a good thing that the value of c is extremely large. :smile:

Yep, I think that is about right. I noticed another interesting effect relating to something you said in post #14. If the front and back ships have identical constant proper accelerations (say 1g) so that their speeds are matched at all times in the initial reference frame, then signals from the rear ship are unable to catch up with the front ship after a finite time into the journey. The rear part is a you say behind a Rindler horizon, but it seems passengers in the front of the ship will be able to walk down to the rear of the ship and see it is still there, but will be unable to walk back to the front again! This seems very unintuitive.
 
  • #20
hi yuiop. Radar method does not work if HORIZON exist!

see for example subsection 3 in paper below

Comment on ‘Note on Dewan–Beran–
Bell’s spaceship problem’
doi:10.1088/0143-0807/30/1/L02
D V Peregoudov
http://ivanik3.narod.ru/EMagnitizm/JornalPape/ParadocsCullwick/RedzicPeregudov/ejp9_1_l02Peregoudov.pdf

yuiop:The ruler measurement of the proper distance for the equal acceleration case, as a function of velocity or time, is very simply deduced to be:

[tex]L(v) = \frac{L_0}{\sqrt{1-\beta^2}} [/tex]

[tex]L(t) = L_0 \sqrt{1+(at/c)^2} [/tex] Eq.1
-----------------------------------------------------------------------------------
Sorry but standard procedure of the ruler measurement and Eq.1 valid only if at/c<1
see subsections (1) and (4) in paper below
http://arxiv.org/ftp/arxiv/papers/0910/0910.2298.pdf
 
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  • #21
It seems to me that the expansion tensor of a suitable congruence will tell us about the stretching of an accelerated object. For the unequal acceleration case, where

[tex]ds^2= -g^2x^2dt^2+dx^2+dy^2+dz^2[/tex]

this Wiki article claims to have the answer

http://en.wikipedia.org/wiki/Rindler_coordinates

In this case, the expansion and vorticity of the congruence of Rindler observers vanish. The vanishing of the expansion tensor implies that each of our observers maintains constant distance to his neighbors.

These observers have a proper acceleration 1/x in the x-direction. This seems to be saying that the thread will never break.

The case of equal acceleration hasn't been treated to a kinematic decomposition and my attempt, using the coframe cobasis below seems to be giving dud results - but hinting at a non-zero expansion tensor. Maybe someone with Maple or Mathematica can do the calculation ?
(My method does give the same results as the Wiki article for the unequal acceleration, so it can't be all bad :wink:).

[tex]
\begin{align*}
\vec{\sigma}_t= -\gamma dt + \beta\gamma dx\\
\vec{\sigma}_x= \gamma dx - \beta\gamma dt\\
\vec{\sigma}_y=dy,\ \ \ \vec{\sigma}_z=dz
\end{align}
[/tex]

where [itex]\beta=at[/itex] and [itex]\gamma^{-2}=1-a^2t^2[/itex].

My calculation gives a proper acceleration of [itex]\gamma^3 a[/itex] in the x-direction which seems off the mark.
 
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  • #22
Hi JF, I have only just realized that you are probably one of the authors of the paper and if that is the case I apologise for being rude about your paper :redface: .. it was not my intention to offend an individual and my respect for you has gone up enormously if you are the author of a paper who is prepared to discuss their paper with us commoners. Still, that does not mean I agree with everything you say, but I will try and be more polite. :wink:

J.F. said:
hi yuiop. Radar method does not work if HORIZON exist!
Obviously I agree, because earlier I said "... signals from the rear ship are unable to catch up with the front ship after a finite time into the journey. The rear part is ... behind a Rindler horizon ..." and even earlier I said it does not make sense to use radar measurements over extended distances in an accelerating frame.

J.F. said:
yuiop:The ruler measurement of the proper distance for the equal acceleration case, as a function of velocity or time, is very simply deduced to be:

[tex]L(v) = \frac{L_0}{\sqrt{1-\beta^2}} [/tex]

[tex]L(t) = L_0 \sqrt{1+(at/c)^2} [/tex] Eq.1
-----------------------------------------------------------------------------------
Sorry but standard procedure of the ruler measurement and Eq.1 valid only if at/c<1
see subsections (1) and (4) in paper below
http://arxiv.org/ftp/arxiv/papers/0910/0910.2298.pdf
I am not at all clear why you put an upper limit of at/c<1. There is nothing especially physically significant about that value, it is not even related to when the rear rocket falls behind the Rindler horizon. It also puts an upper limit to the ratio of the stretched length to the rest length of the elastic string at [tex]L(t)/L_0 = \sqrt{1+(at/c)^2} = 1.414[/tex].It is quite easy to stretch an elastic string to much more than twice its rest length even in none relativistic circumstances.

Here is my argument for why the length of the stretched string is given by the equations I wrote above. In the diagram below, the red lines in frame S represent inertial rockets moving with a constant relative velocity of 0.92c. The proper length of a string connecting the front red rocket to the rear red rocket is represented by the length J'-K' in frame S'. (The at/c value in this example is 2.4 btw.) The green curves represent the the spacetime paths of rockets all with a constant proper acceleration of 0.1465 c/s. At events A, B and C the string connecting the red inertial rockets has exactly the same length as a string connecting the green accelerating rockets as measured in frame S when they are momentarily alongside each other. All parts of the string connecting the accelerating rockets have exactly the same velocity as the string connecting the inertial rockets at that instant in frame S and the only difference between the two threads is their prior histories (and futures). We could even design by prior arrangement for the thread connecting the inertial rockets to have the same tension as the string connecting the accelerating rockets at the time they are alongside each other. So if all parts of the two strings have the same velocities and tension simultaneously in frame S when they are measured to have the same length simultaneously in the same frame then there is absolutely no reason to conclude that the proper length of the string connecting the accelerating rockets is different from the string connecting the inertial rockets.
BellLength75.gif


It can be seen that at event B' when the centre of the string connecting the accelerating rockets is at rest in frame S' (the rest frame of the red inertial rockets), the simultaneous measurement of the string's length in that frame is represented by D'-E' and this length is identical to the rest length of the string connecting the outer inertial rockets J'-K'. It can also be seen that if we break the string up into segments and measure each segment when the segment's centre is at rest in frame S', then the sum of the segments (F'-G')+(H'-I') is exactly equal to the rest length of the string connecting the inertial rockets. This is a property of the time symmetry of the accelerating curves about the turn point when the accelerating rocket is momentarily at rest in that frame. It should be obvious if we consider an infinitesimally small segment that the difference in velocities of the two ends of the segment becomes insignificant and certainly eliminates relativistic effects. Since the sum of the segments is exactly equal to the simultaneous measurement of the total length of the accelerating thread in frame S' there is no loss of accuracy over extended distances. It can also be seen that if take the simultaneous measurement of the accelerating string when the front rocket is at rest at event C' this will be shorter than J'-K' and the similar measurement at event A' will be longer than J'-K', but the averaged length will be equal to J'-K' (the rest length of the string connecting the inertial rockets).

Now in the second document linked in your last post, it is stated that
But the key feature of Bell’s problem is that there exists no comoving frame that is common to both spaceships (except for the moment t = 0). In any inertial frame at any time at least one of the spaceships has non-zero velocity. Consequently there exists no comoving frame common to all points of the thread..
This is a valid point and while true, it does not change the physical conclusions. Normally we would define the proper length of an object as the length measured in a reference frame in which all parts of the rod are at rest. As you have pointed out, it is impossible to meet this condition in Bell's problem. There is however, a way around this dilemma. We can use an alternative definition of the proper length of an object as the measured length multiplied by the gamma factor, in a reference frame in which all parts of the object have the same (but not necessarily zero) velocity. This would seem to be the only sensible way to define the proper length of a Bell accelerating object and it does not violate SR, so I stand by my original equations.

Finally we can note that the "clock hypothesis" states that acceleration has no direct effect on time dilation (and by implication no effect on length contraction) and that those relativistic effects are entirely a function of instantaneous relative velocity only removing any objection that the length of the string connecting the inertial rockets is identical to the length of the string connecting the accelerating rockets when they are alongside each other and momentarily comoving in an inertial reference frame where all parts of the strings have identical velocities simultaneously.
 
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  • #23
Mentz114 said:
where [itex]\beta=at[/itex] and [itex]\gamma^{-2}=1-a^2t^2[/itex].

My calculation gives a proper acceleration of [itex]\gamma^3 a[/itex] in the x-direction which seems off the mark.

Shouldn't that be [itex]\beta = at(1+a^2t^2)^{-1/2} [/itex]
and [itex]\gamma^{2}= (1-v^2)^{-1} = 1+a^2t^2[/itex] ?

http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html
 
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  • #24
yuiop said:
Shouldn't that be [itex]\beta = at(1+a^2t^2)^{-1/2} [/itex]
and [itex]\gamma^{2}= (1-v^2)^{-1} = 1+a^2t^2[/itex] ?

http://www.xs4all.nl/~johanw/PhysFAQ/Relativity/SR/rocket.html

Thanks, I'll check out the reference. The coframe I've quoted is got by boosting the static frame, but it doesn't appear anywhere so there could be something wrong with it.

I've been thinking about this and I've concluded that the expansion tensor of a congruence is not telling us what we want to know. It only expresses the change in the volume element. For instance in curved space we can get zero expansion but a change in 3D shape.
 
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  • #25
This problem is still bugging me. My convoluted arguments in #22 are probably more of a counter argument to my own position :-p. A very simple argument is that in frame S (the initial inertial frame) the proper time of every single infinitesimal element of the moving string are equal simultaneously in that frame, so the length adjusted by the gamma factor for that velocity gives the proper length of the string at equal proper times of all parts of the string. <end of argument>

In frame S' in #22 the measured length can be argued to be slightly shorter than the proper length of the string because not all parts of the string are ever exactly at rest simultaneously in that approximately comoving frame and there is no good way to identify a time for the string in frame S' with a time for the string in S due to simultaneity issues so analysis in the "nearly comoving" frame is probably not a good way to go.

To try and clear things up I did a numerical check of the various formulas given by the papers compared to the formulas I gave.

The classical relativistic proper length L(T) of the string in terms of proper time (T) that the string has accelerated for is:

[tex]L(T) = L_0 \sqrt{1+(sinh(aT))^2} [/tex]

For a string of initial rest length [tex]L_0 = 1[/tex] and a terminal velocity of 0.999999999 and an acceleration of 5 this equation predicts the stretched length of the string to be 22,360.68 times longer than its original length.

Equation (23) of the first paper http://www.scribd.com/doc/38943606/Gravitation-and-Cosmology-BELL mentioned in this thread predicts the stretched length of the string (for the same parameters) to be only 1.04 times longer than the initial length.

Equation (34) of the the third paper mentioned in this thread http://arxiv.org/ftp/arxiv/papers/0910/0910.2298.pdf (by the same authors) predicts a stretched length of only 2.86 times the initial length of the string.

The predictions in the papers differ hugely from the classical relativistic ones and tend to little or no stretching for large accelerations (a>100). It is not clear why the equation for the physical length of the string differs in the two papers when they are by the same authors.
 
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  • #26
Mentz114 said:
this Wiki article claims to have the answer

http://en.wikipedia.org/wiki/Rindler_coordinates



These observers have a proper acceleration 1/x in the x-direction. This seems to be saying that the thread will never break.
I don't see that. I had to read through it a few times, but the author seems to be placing Rindler observers all along the x-axis in both directions originally in the rest frame, and then starting each of the Rindler observers off originally from the rest frame with a constant proper acceleration that will place a Rindler observer that starts off from the origin at the Rindler horizon to each of the others. In that case, the acceleration of each Rindler observer, depending upon the starting position X from the origin, will have an acceleration g_i = c^2 / X for any particular Rindler observer i, and the closer to the origin each Rindler observer starts off, the greater the proper acceleration must be. In the next section, it talks about Bell's paradox and says

But in relativistic physics, we see that the trailing endpoint of a rod which is accelerated by some external force (parallel to its symmetry axis) must accelerate a bit harder than the leading endpoint, or else it must ultimately break.

Of course, this means that the thread must ultimately break, since two rockets that have identical acceleration profiles will always maintain a constant distance according to the rest frame between any two matching points upon each of the rockets, but also while each of the rockets is contracting as it accelerates, which means that the distance between the back of the front ship and the front of the back ship is expanding according to the rest frame also, just the opposite of what would be necessary to maintain rigidity between the rockets, and it is expanding at an even greater rate according to the rocket observers than it is to the rest frame. This also reverses the acceleration profile for the thread from that of the rockets, whereas the back of the thread accelerates at a rate equal to the constant proper acceleration at the front of a rocket and vice versa, so the front of the thread accelerates harder than the back and the thread breaks.
 
  • #27
One interesting thing I've noticed when working through some of this that I thought I would mention is that the midpoint of the rocket when stationary in the initial inertial frame and the final inertial frame is not the midpoint while accelerating. While accelerating, the midpoint tends to fall back closer to the back of the rocket than the front. The midpoint of the rocket only regains its original position upon the rocket becoming inertial again, due to the differences in the times of acceleration of different points upon the rocket when accelerating to the same speed v. Of course, this also follows for other points besides the midpoint.

One can visualize the effect by imagining two stationary rockets of equal length in the rest frame that are lined up nose to tail. The point where the two rockets meet is the midpoint of the entire length of both rockets. The rockets then begin to accelerate such that they remain nose to tail. Let's say that the front of the forward rocket has some constant proper acceleration a1. In order to retain Born rigidity, the back of the forward rocket, then, must have a greater acceleration a2 than the front. In order to remain nose to tail while accelerating, the front of the rear rocket must have the same acceleration as the back of the forward rocket, so a2 also. The back of the rear rocket, in order to retain Born rigidity as well, must have a greater acceleration a3 than the front of the rear rocket. So we have a3> a2 > a1. We can see that the rear rocket, with points upon that rocket having accelerations between a2 and a3, must accelerate faster than the forward rocket in general, with accelerations between a1 and a2. So the rear rocket, having a greater instantaneous speed at any given time according to the rest frame, will contract quicker than the forward rocket, causing the original midpoint between them to now fall more toward the back of the entire length of both rockets, and since both rockets continue to travel nose to tail as one entity, the same thing will occur upon a single rocket. I don't know if this would have any effect upon distance measurements or not, probably not, but I thought it was interesting.
 
  • #28
Oops. It appears I made an error in the math of post #14. I had it worked out right on paper, but somehow typed it in wrong. The bolded part is where the error lies in the other post. I had ln(aF / aB) when it should have been ln(aB / aF). Thankfully, the next part subtracted the two results in the same way, so it didn't make a difference for the result of the Doppler shift, but it will for my next post. The correct math should have been

D = f_obs / f_emit = [1 / (TF2' - TF1')] / [1 / (TB2' - TB1')] = (TB2' - TB1') / (TF2' - TF1')

Now, according to the rest frame, the first pulse of light will travel from the back to the front in the time TF1 - TB1, traversing the difference in distances that the front and back have traveled during those times according to the relativistic acceleration formula for distance, so we have

c (TF1 - TB1) = d_TF1 - d_TB1

= [d + (c^2 / aF) (sqrt(1 + (aF TF1 / c)^2) - 1)] - [(c^2 / aB) (sqrt(1 + aB TB1 / c)^2) - 1)]

TF1 - TB1 = d / c - c (1 / aF - 1 / aB) + (c / aF) sqrt(1 + (aF TF1 / c)^2) - (c / aB) sqrt(1 + aB TB1 / c)^2), where 1 / aF - 1 / aB = d / c^2, so

TF1 - TB1 = (c / aF) sqrt(1 + (aF TF1 / c)^2) - (c / aB) sqrt(1 + aB TB1 / c)^2)

The relativistic formula for the instantaneous speed at the back of the ship when the photon is emitted is

(vB1 / c) = (aB TB1 / c) / sqrt(1 + (aB TB1 / c)^2)

sqrt(1 + (aB TB1 / c)^2) = aB TB1 / vB1

(c / aB) sqrt(1 + (aB TB1 / c)^2) = TB1 c / vB1

and likewise for TF1, so for TF1 - TB1 that we found before, we now have

TF1 - TB1 = TF1 c / vF1 - TB1 c / vB1

TF1 ((c / aF) - 1) = TB1 ((c / aB) - 1)

The relativistic formula for speed can also be re-arranged further to give

(aB TB1 / c) = (vB1 / c) / sqrt(1 - (vB1 / c)^2)

TB1 = (vB1 / aB) / sqrt(1 - (vB1 / c)^2)

and likewise for TF1. Applying that to the previous equation gives

[(vF1 / aF) / sqrt(1 - (vF1 / c)^2)] ((c / aF) - 1) = [(vB1 / aB) / sqrt(1 - (vB1 / c)^2)] ((c / aB) - 1)

(vF1 / aF) (c / vF1) (1 - vF1 / c) / sqrt(1 - (vF1 / c)^2) = (vB1 / aB) (c / vB1) (1 - vB1 / c) / sqrt(1 - (vB1 / c)^2)

(c / aF) sqrt((1 - vF1 / c) / (1 + vF1 / c)) = (c / aB) sqrt((1 - vB1 / c) / (1 + vB1 / c))

Now we will apply the relativistic formula for proper time with

TB1' = (c / aB) ln[(1 + vB1 / c) / (1 - vB1 / c)] / 2

aB TB1' / c = ln[sqrt((1 + vB1 / c) / (1 - vB1 / c))]

e ^ (aB TB1' / c) = sqrt((1 + vB1 / c) / (1 - vB1 / c))

and likewise for TF1, so the equation now becomes

(c / aF) / (e ^ (aF TF1' / c)) = (c / aB) / (e ^ (aB TB1' / c))

(e ^ (aF TF1' / c)) / (e ^ (aB TB1' / c)) = aB / aF

aF TF1' / c - aB TB1' / c = ln(aB / aF)

and finding the same thing similarly for the second photon emitted from the back to the front also gives us

aF TF2' / c - aB TB2' / c = ln(aB / aF)


and subtracting these last two equations from each other, we get

(aF TF2' / c - aF TF1' / c) - (aB TB2' / c - aB TB2' / c) = 0

(aF / c) (TF2' - TF1') = (aB / c) (TB2' - TB1')

D = (TB2' - TB1') / (TF2' - TF1) = aF / aB
 
  • #29
Here is the radar distance. From post #28, in the bolded section :smile:, we found that for a photon emitted from the back of a ship accelerating with Born rigidity to the front, if the proper reading upon a clock at the back when the photon is emitted there is TB1' and a clock at the front when the photon is received reads TF1', then we have the relation

1) aF TF1' / c - aB TB1' / c = ln(aB / aF) {back to front}

TF1' = (c / aF) [ln(aB / aF) + aB TB1' / c], giving a difference in the measured time between the readings of

TF1' - TB1' = (c / aF) [ln(aB / aF) + aB TB1' / c] - TB1'

One can see that the difference in proper readings depends upon the time of emission for the one way transmission from one end of the ship to the other, and we already know that two clocks upon the ship do not remain synchronized to each other anyway, so we couldn't trust that measurement, but let's find out what the two way time will be away and back to the same single clock. If a photon is emitted at the back and then reflected back from the front to the same back clock again at a proper reading of TB2', then working it out by the same method for a photon emitted at the front and received at the back, the relation is

2) aB TB2' / c - aF TF1' / c = ln(aB / aF) {front to back}

and adding equations 1 and 2 gives us

aB TB2' / c - aB TB1' / c = 2 ln(aB / aF)

TB2' - TB1' = 2 (c / aB) ln(aB / aF)

This is the two way time for a photon to travel away and back to the same clock. In this case, one see that all of the variables on the right this time are constant, giving a constant two way time that is independent of the time of emission and synchronization should not be much of an issue with a single clock. Since the time that a photon travels away from some point and is reflected back from some other point upon the ship back to the same clock is a constant, that is another acceptable means by which to measure distance. If observers upon the ship assume the speed of light to be measured at c as it is for inertial frames, then that would give

dB_radar = c (TB2' - TB1') / 2

= (c^2 / aB) ln(aB / aF)

If the photon is now reflected to the front again, then in the same way as with equation 1, the one way time is

aF TF2' / c - aB TB2' / c = ln(aB / aF), so adding that to the equation 2 gives

aF TF2' / c - aF TF1' / c = 2 ln(aB / aF)

TF2' - TF1' = 2 (c / aF) ln(aB / aF)

dF_radar = (c^2 / aF) ln(aB / aF)

One problem with this type of distance measurement, however, is that the front and back observers will disagree about the radar distance between themselves, with the front observer claiming the distance is aB / aF greater than the back observer claims.

Since we also have the relation between the back and front for a ship with Born rigity of

1 / aF - 1 / aB = d / c^2, then

1 / aB = 1 / aF - d / c^2

aB = 1 / (1 / aF - d / c^2), and the radar distances in respect to the front observer become

dB_radar = c^2 (1 / aF - d / c^2) ln[1 / (1 - aF d / c^2)]

= (c^2 / aF) (1 - aF d / c^2) ln[1 / (1 - aF d / c^2)]

dF_radar = (c^2 / aF) ln[1 / (1 - aF d / c^2)]

Making m = aF d / c^2, they become

dB_radar = (d / m) (1 - m) ln[1 / (1 - m)]

dF_radar = (d / m) ln[1 / (1 - m)]

We can see that if d approaches c^2 / aF, where m approaches 1, then the back observer lies very near the Rindler horizon in respect to the front observer, giving dF_radar approaches infinity according to the front observer while dB_radar works toward zero. In the case of the back observer's time, from the point of view of the rest frame, the time for a photon to travel from the back to the front still approaches infinity before it is then reflected back, but apparently the back clock time dilates to nearly zero with aB approaching infinity by a greater factor, so the radar distance the back observer measures to the front and reflected back ends up approaching zero. If m is small, then dB_radar approaches d (1 - m / 2) and dF_radar approaches d (1 + m / 2).
 
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  • #30
yuiop said:
Hi JF, I have only just realized that you are probably one of the authors of the paper and if that is the case I apologise for being rude about your paper :redface: .. it was not my intention to offend an individual and my respect for you has gone up enormously if you are the author of a paper who is prepared to discuss their paper with us commoners.

Yes of course, I agree. Thank you for attention to this paper.

yuiop said:
Still, that does not mean I agree with everything you say, but I will try and be more polite.
Ok.
I am not at all clear why you put an upper limit of at/c<1. There is nothing especially physically significant about that value, it is not even related to when the rear rocket falls behind the Rindler horizon. It also puts an upper limit to the ratio of the stretched length to the rest length of the elastic string at [tex]L(t)/L_0 = \sqrt{1+(at/c)^2} = 1.414[/tex].It is quite easy to stretch an elastic string to much more than twice its rest length even in none relativistic circumstances.

Here is my argument for why the length of the stretched string is given by the equations I wrote above. In the diagram below, the red lines in frame S represent inertial rockets moving with a constant relative velocity of 0.92c. The proper length of a string connecting the front red rocket to the rear red rocket is represented by the length J'-K' in frame S'. (The at/c value in this example is 2.4 btw.) The green curves represent the the spacetime paths of rockets all with a constant proper acceleration of 0.1465 c/s. At events A, B and C the string connecting the red inertial rockets has exactly the same length as a string connecting the green accelerating rockets as measured in frame S when they are momentarily alongside each other. All parts of the string connecting the accelerating rockets have exactly the same velocity as the string connecting the inertial rockets at that instant in frame S and the only difference between the two threads is their prior histories (and futures). We could even design by prior arrangement for the thread connecting the inertial rockets to have the same tension as the string connecting the accelerating rockets at the time they are alongside each other. So if all parts of the two strings have the same velocities and tension simultaneously in frame S when they are measured to have the same length simultaneously in the same frame then there is absolutely no reason to conclude that the proper length of the string connecting the accelerating rockets is different from the string connecting the inertial rockets.
BellLength75.gif
yuiop said:
Since the sum of the segments is exactly equal to the simultaneous measurement of the total length of the accelerating thread in frame S' there is no loss of accuracy over extended distances.

I do not agree with this statement. The summation procedure of the segments You considered above, does not equal to the simultaneous measurement of the total length of the accelerating thread in frame S'.
The correct summation procedure of the segments, such that equal to the simultaneous measurement of the total length of the accelerating thread in frame S' is not trivial. See paper
http://arxiv.org/ftp/arxiv/papers/0910/0910.2298.pdf
subsection 4 in particular Pic.1-2.

The correct condition for a physical (leading to strain) deformation was formulated by M. Born in the early days of relativity: one has to look at the stretch in the comoving frame.

The correct summation procedure is not carried out by means of of the inertial observers but only by means of of the accelerated comoving observers.

yuiop said:
Equation (23) of the first paper http://www.scribd.com/doc/38943606/G...Cosmology-BELL mentioned in this thread predicts the stretched length of the string (for the same parameters) to be only 1.04 times longer than the initial length.

Equation (34) of the the third paper mentioned in this thread http://arxiv.org/ftp/arxiv/papers/0910/0910.2298.pdf (by the same authors) predicts a stretched length of only 2.86 times the initial length of the string.

The predictions in the papers differ hugely from the classical relativistic ones and tend to little or no stretching for large accelerations (a>100). It is not clear why the equation for the physical length of the string differs in the two papers when they are by the same authors.
Agree. But Equation (34) of the the third paper was obtained by using some rough simplifications.
 
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  • #31
I've been finding this paper a bit hard to follow, and been busy to boot, so I've been keeping out of it. But I think I may have some comments. And a couple of questions.

I'll throw some things out, I'd like to see if everyone (esp. one of the original authors, who apparently is posting here) agrees with them.

1) There seem to be about five notions of distance that are mentioned in the paper or in this thread.

1a)There are two inertial frames corresponding to the front of the rocket and the back. These have different notions of simultaneity and hence define different notions of distance. These are mentioned near the start of the paper.

We can conclude that in either of these frames, the string breaks when the front and rear rocketships have the same proper acceleration.

1b) There are 2 non-inertial frames which I didn't see mentioned in the paper (I could have missed them), but think they are important enough to deserve special mention and have been mentioned in this thread. These would be the Rindler frames associated with the front of the rocket, and with the rear. By "Rindler frame", I mean the frame associated with a metric of the form

ds^2 = -(1+gt)^2 + dx^2 + dy^2 + dz^2

In the case of equal proper accelerations of the front and rear rocket, these are different frames, and in both of them the string breaks.

1c) The fifth case is the case analyzed in the paper, where the notion of distance is clearly stated to be measured by the hypersurface orthogonal to the congruence of worldlines of the string. I'm not sure if it was explicitly stated in the paper, but I _assume_ that all particles in the string have a constant proper acceleration (?).

This has a different notion of simultaneity than 1a) or 1b) and hence defines a different notion of distance.

One thing I'd like to make sure everyone agrees with: the string always breaks, there is nothing in the analysis to indicate that it does not break.

Another comment. I believe that the notion of distance in this "medium" frame 1c) is inherently incompatible with the notion of Born rigidity - basically, because the "medium" is stretching, objects at rest with respect to the medium are not maintaining a constant distance from each other as is required by the Born notion of rigidity. The other notions of distance 1a) and 1b) have the property that objects at rest with respect to the frames are in Born-rigid motion.

One last point puzzles me. The authors mention that the hypersurface is spatially curved, and that this is important to measuring the distance. This would seem to me to only be important in the case where the underlying geometry is not one dimensional. I.e. if we consider two rockets moving in some Minkowskian frame where y=z=0 for both rockets, we can describe the rockets by their x coordinates. There will be no spatial curvature in the x frame, because you can't have curvature in one spatial dimension, and therefore we won't need to consider the spatial curvature. Only if we had y or z nonzero and different for the front and rear rockets would we have to worry about spatial curvature.
 
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pervect:"The fifth case is the case analyzed in the paper, where the notion of distance is clearly stated to be measured by the _hypersurfaces_ orthogonal to the congruence of worldlines of the string".

The notion of distance which is considered above in fact was introduced A. Zelmanov:
"They can be also considered as tensors in a space, all elements of which (i.e. threedimensional local spaces) are definitely orthogonal to the time lines ander any given coordinate of time".
See subsection 3, pp.37-38 This isn't new. It just isn't in the standard textbooks.
http://zelmanov.ptep-online.com/papers/zj-2008-05.pdf

https://www.physicsforums.com/showthread.php?t=164305

pervect: "I'm not sure if it was explicitly stated in the paper, but I _assume_ that all particles in the string have a constant proper acceleration (?)".

Yes of course the all particles in the string have a constant proper acceleration

pervect:
"This has a different notion of simultaneity than 1a) or 1b) and hence defines a different notion of distance".
Agree.

pervect: "One thing I'd like to make sure everyone agrees with: the string always breaks, there is nothing in the analysis to indicate that it does not break".
Agree.

pervect:"One last point puzzles me. The authors mention that the hypersurface is spatially curved, and that this is important to measuring the distance. This would seem to me to only be important in the case where the underlying geometry is not one dimensional. I.e. if we consider two rockets moving in some Minkowskian frame where y=z=0 for both rockets, we can describe the rockets by their x coordinates. There will be no spatial curvature in the x frame, because you can't have curvature in one spatial dimension, and therefore we won't need to consider the spatial curvature. Only if we had y or z nonzero and different for the front and rear rockets would we have to worry about spatial curvature".

In one-dimensional case, (i.e. y=z=0) all Zelmanov's _hypersurfaces_ is degenerate in curves orthogonal to the congruence of worldlines of the string.
 
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What is a relativistic continuous medium?

A relativistic continuous medium refers to a material or substance that is in constant motion and is subject to the laws of relativity. This means that the speed of the medium is close to the speed of light and its behavior cannot be described by classical mechanics.

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