Ehrenfest / rod thought experiment.

[Passionflower]

Perhaps you should start with a definition of such a reference frame (CMIRF). In an accelerating system we can have an instantaneously co-moving observer. Or is it specifically a frame what you mean? Such a frame obviously applies to one instant of the acceleration, the next instant there will be a completely different frame. Or are you thinking about Fermi-Walker transporting such a frame?

 

[Austin0]

Perhaps you should start with a definition of such a reference frame (CMIRF). In an accelerating system we can have an instantaneously co-moving observer. Or is it specifically a frame what you mean? Such a frame obviously applies to one instant of the acceleration, the next instant there will be a completely different frame. Or are you thinking about Fermi-Walker transporting such a frame?

As I understand the problem, it is in effect an attempt to duplicate a dynamic series of CMIRFs using the clocks and rulers of the accelerating frame.
Using accelerometers, calculators. synch signals or whatever to artificially calibrate the system to simulate and update a quasi -inertial frame [dynamic]
A variation on your idea of using such means to artificially keep a home clock going and updated based on computations derived from acceleration/time calculations.
I will have to look up Fermi-Walker transport, I seem to remember their names from my seach of the Ehrenfest question but never found the papers mentioned.

 

[Passionflower]

As I understand the problem, it is in effect an attempt to duplicate a dynamic series of CMIRFs using the clocks and rulers of the accelerating frame.
Using accelerometers, calculators. synch signals or whatever to artificially calibrate the system to simulate and update a quasi -inertial frame [dynamic]
A variation on your idea of using such means to artificially keep a home clock going and updated based on computations derived from acceleration/time calculations.

Well but in which way, what is the artificial clock supposed to measure?

I already mentioned it is possible to measure coordinate time, or the longest possible time between the start and clock measuring event. So what third option are you looking for?

Personally I would be very interested in comparing a "straight line" clock on a spatial path from A to B with a helical path but this seem very complicated to calculate. In case we have someone here make the claim it is easy then please provide the formula.

 

[yuiop]

The problem seems to be how to adjust clock B. If we adjust B so that A-B works out we seem to have a problem with B-C and vice versa. No?

Yes, I think you are right. There appears to be no single adjustment that can be made to clock B that can get the radar distance BAB to agree with BCB if the ruler lengths AB and BC are equal. So it seems my suggested scheme is doomed and there is no way to come up with a synchronisation scheme that can make the speed of light isotropic over extended distances in a reference frame undergoing Born rigid acceleration. Basically radar signals sent out simultaneously from B in both directions to A and C, do not return to B simultaneously and there is no calibration procedure that can make those signals from different directions arrive simultaneously. Looks like your shooting from the hip hit the target

I may have to rescind my possibly premature capitulation here, as I have thought of way around the objection you have raised. I have up to now been assuming that that the distances between clocks have to be regularly spaced in the instantaneously co-moving frame. Here is a method htat might work. We have a multitude of clocks, A,B,C etc with A at the nose. We place B at location further back in the rocket so that radar distance A,B,A as measured from A is say 1 light second. We adjust the rate of clock B so that B agrees that the radar distance B,A,B is also one light second. Now we do the same with clock C. Clock C is placed at a distance further back where the radar distance of C measured by B is also 1 light second. Clock C is adjusted so that observer C also agrees that the radar distance is also 1 light second using the adjusted clock. We can continue adding clocks like this so that they all the agree that the coordinate radar distance between any two adjacent clocks is one light second and that the the coordinate distance between any N number of clocks is N-1 light seconds, measured in any direction. Now, while the observers on the accelerating rocket consider the clocks to placed at uniform distance intervals as measured by coordinate radar distances, any instantaneous co-moving frame will see the clocks towards the back as getting more bunched together. Obviously I will need to check these conclusions and provide some equations, but what do you think of the outline of the method so far. Any immediate objections?

 

[Passionflower]

I may have to rescind my possibly premature capitulation here, as I have thought of way around the objection you have raised. I have up to now been assuming that that the distances between clocks have to be regularly spaced in the instantaneously co-moving frame. Here is a method htat might work. We have a multitude of clocks, A,B,C etc with A at the nose. We place B at location further back in the rocket so that radar distance A,B,A as measured from A is say 1 light second. We adjust the rate of clock B so that B agrees that the radar distance B,A,B is also one light second. Now we do the same with clock C. Clock C is placed at a distance further back where the radar distance of C measured by B is also 1 light second. Clock C is adjusted so that observer C also agrees that the radar distance is also 1 light second using the adjusted clock. We can continue adding clocks like this so that they all the agree that the coordinate radar distance between any two adjacent clocks is one light second and that the the coordinate distance between any N number of clocks is N-1 light seconds, measured in any direction. Now, while the observers on the accelerating rocket consider the clocks to placed at uniform distance intervals as measured by coordinate radar distances, any instantaneous co-moving frame will see the clocks towards the back as getting more bunched together. Obviously I will need to check these conclusions and provide some equations, but what do you think of the outline of the method so far. Any immediate objections?

Hmmm.

The relationship between the clocks at different locations A, B, C etc is:

\frac{\Delta \tau_A}{\alpha_A} = \frac{\Delta \tau_B}{\alpha_B} = \frac{\Delta \tau_C}{\alpha_C}

While the accelerations are:

\alpha = \frac{c^2}{x_A + h}

The sum of the roundtrip time AB (e.g A's clock + B's clock) is larger than BC, which is larger than CD etc. So how are you going to make that fit?

Note that this is not the case in a homogenic gravitational field where the sum is always twice the inertial radar distance. So there it seems easier

 

[bcrowell]

As far as I can tell, the last ~30 posts have been about whether Einstein synchronization is possible aboard an accelerating rocket. There is a good discussion of Einstein synchronization here: http://en.wikipedia.org/wiki/Einstein_synchronization

Or is the question really whether or not the observers in the rocket are static observers in a static spacetime?

Einstein synchronization is not possible because the no-redshift criterion given in the WP article is not satisfied.

The spacetime is static because staticity is coordinate-independent, and the observers aboard the rocket are just seeing flat spacetime in a different set of coordinates. I believe they are also static observers, because they don't observe any time-variation of the geometry (the geometry is always flat), and they aren't rotating (they see a vanishing Sagnac effect).

Actually, is there a standard definition of a static observer? The books I have seem to define a clear notion of a static spacetime and of static coordinates used to describe a static spacetime (i.e., coordinates in which the metric is diagonal and time-independent). But defining a property of coordinates isn't quite the same as defining a property of an observer, since observers are local and coordinates are global.

 

[yuiop]

Hmmm.

The relationship between the clocks at different locations A, B, C etc is:

\frac{\Delta \tau_A}{\alpha_A} = \frac{\Delta \tau_B}{\alpha_B} = \frac{\Delta \tau_C}{\alpha_C}

While the accelerations are:

\alpha = \frac{c^2}{x_A + h}

The sum of the roundtrip time AB (e.g A's clock + B's clock) is larger than BC, which is larger than CD etc. So how are you going to make that fit?

Note that this is not the case in a homogenic gravitational field where the sum is always twice the inertial radar distance. So there it seems easier

Not quite sure what you are doing here. The roundtrip time (eg A to B and back to A) is measured by a single clock at A.

I have uploaded some diagrams of the method of synchronising accelerating clocks that I described earlier. https://www.physicsforums.com/showpost.php?p=2882964&postcount=64 The first diagram is the radar distances as measured by unsynchronised clocks moving with Born rigid motion and the second diagram is with clocks synchronised so that the radar distance between any 2 clocks is 1 light second and the distance between any N accelerating clocks is N-1 light seconds. This method of synchronisation has most of the properties required of it that bcrowell mentioned:

The problem is whether this synchronisation does really succeed in assigning a time label to any event in a consistent way. To that end one should find conditions under which
(a) clocks once synchronized remain synchronized,
(b1) the synchronisation is reflexive, that is any clock is synchronized with itself (automatically satisfied),
(b2) the synchronisation is symmetric, that is if clock A is synchronized with clock B then clock B is synchronized with clock A,
(b3) the synchronisation is transitive, that is if clock A is synchronized with clock B and clock B is synchronized with clock C then clock A is synchronized with clock C.

Condition (a) is met, but it requires that the raw output of the proper clocks has a constant correction factor of a_i/a_m where a_i is the proper acceleration of the i'th clock under consideration and a_m is the proper acceleration of the "master clock". For constant proper acceleration and Born rigid motion, this correction factor does not change over time.

Conditions (b1), (b2) and (b3) are also met, without any difficulty. Radar distances are the same measured in either direction and over long distances on the accelerating rocket with this method. Constant distant intervals in the accelerating frame appear progressively more length contracted towards the rear of the rocket in the inertial momentarily co-moving frame and this is perhaps the "differential length contraction" that austin0 was intuitively referring to.

 

[Passionflower]

Not quite sure what you are doing here. The roundtrip time (eg A to B and back to A) is measured by a single clock at A.

Of course it is.

I was merely giving additional information that if we add the roundtrip times measured by both clock A (ABA) and clock B (BAB) then also this is not constant, unlike in a homogenic gravitational field where the total time remains constant (e.g. double the inertial rountrip time).

I have uploaded some diagrams of the method of synchronising accelerating clocks that I described earlier.

Nice, but I prefer to see the formulas or the raw data. I like to verify it for myself.

 

[yuiop]

Hmmm.

The relationship between the clocks at different locations A, B, C etc is:

\frac{\Delta \tau_A}{\alpha_A} = \frac{\Delta \tau_B}{\alpha_B} = \frac{\Delta \tau_C}{\alpha_C}

While the accelerations are:

\alpha = \frac{c^2}{x_A + h}

The sum of the roundtrip time AB (e.g A's clock + B's clock) is larger than BC, which is larger than CD etc. So how are you going to make that fit?

Note that this is not the case in a homogenic gravitational field where the sum is always twice the inertial radar distance. So there it seems easier

We have the equation for the instantaneous velocity v of an accelerating clock with constant proper acceleration \alpha given as:

v = tanh(\alpha \Delta T)

Since v is the same for all clocks on the "line of simultaneity" described in mathpages that pivots about the origin, we can say:

tanh(\alpha_A \Delta T_A) = tanh(\alpha_B \Delta T_B) = tanh(\alpha_C \Delta T_C)

\alpha_A \Delta T_A = \alpha_B \Delta T_B = \alpha_C \Delta T_C

\Delta T_A = \frac{\alpha_B}{\alpha_A} \Delta T_B = \frac{\alpha_C}{\alpha_A} \Delta T_C

So as long as we apply a correction factor of {\alpha_B}/{\alpha_A} to the raw output of clock B and {\alpha_C}/{\alpha_A} to the raw output of clock C and so on, the processed outputs of the clocks are guaranteed to remain in sync with master clock A that is arbitrarily chosen. By suitable positioning of the clocks, the upstream radar signal returns simultaneously with downstream radar signal for any given clock. This coordinate system/scheme gives a speed of light and radar distances that are constant/isotropic/symmetric/transitive over long distances (and consistent with ruler distances) in an accelerating reference frame with Born rigid acceleration.

 

[Passionflower]

So as long as we apply a correction factor of {\alpha_B}/{\alpha_A} to the raw output of clock B and {\alpha_C}/{\alpha_A} to the raw output of clock C and so on, the processed outputs of the clocks are guaranteed to remain in sync.

I just tested this and while the numbers are very close they are not exactly right!

For instance for:

\alpha = 1

\tau_{h_0} = 1

h_0= 1/0.1

h_A= 1/0.3

h_B= 1/0.2

We get:

\alpha_A = 0.83333

\tau_A = 0.83333

\alpha_B = 0.90909

\tau_B = 0.90909

d_{ABA} = 0.20833

d_{BAB} = 0.19091

Then if we take:

d_{BAB} * \frac{\alpha_B}{\alpha_A}

We get: 0.20826 which is close but not equal to 0.20833 (and it is certainly not due to rounding errors as my precision is much higher).

However, as I suspected a little bit, if we do the same calculation in a homogeneous gravitational field your adjustment seems to work. I think this is due to the fact the the roundtrip time between A and B from A plus the roundtrip time between A and B from B is always double the inertial roundtrip time in a homogeneous gravitational field but not in the case of a Born rigid system having a constant proper acceleration.

 

[yuiop]

I just tested this and while the numbers are very close they are not exactly right!

For instance for:

\alpha = 1

h_0= 1/0.1

h_A= 1/0.3

h_B= 1/0.2

\tau_{h_0} = 1

We get:

\alpha_A = 0.83333

\tau_A = 0.83333

\alpha_B = 0.90909

\tau_B = 0.90909

d_{ABA} = 0.20833

d_{BAB} = 0.19091

Then if we take:

d_{BAB} * \frac{\alpha_B}{\alpha_A}

We get: 0.20826

However, as I suspected a little bit, if we do the same calculation in a homogeneous gravitational field your adjustment seems to work. I think this is due to the fact the the roundtrip time between A and B from A plus the roundtrip time between A and B from B is always double the inertial roundtrip time in a homogeneous gravitational field.

In Born rigid linear acceleration the proper acceleration is proportional to the inverse of the distance (your h I assume) from the pivot point. This distance h is measured in the momentarily Co-Moving Inertial Reference Frame. When the clocks are synchronised the way I specified, the clocks are no longer regularly spaced at constant intervals apart in the CMIRF. You are using constant intervals of 0.1 between h0, hA and hC, and I think that is why you are not getting the correct numbers.

 

[Passionflower]

In Born rigid linear acceleration the proper acceleration is proportional to the inverse of the distance (your h I assume) from the pivot point. This distance h is measured in the momentarily Co-Moving Inertial Reference Frame. When the clocks are synchronised the way I specified, the clocks are no longer regularly spaced at constant intervals apart in the CMIRF. You are using constant intervals of 0.1 between h0, hA and hC, and I think that is why you are not getting the correct numbers.

Hmmm, I am not convinced. What are you saying, that it works only for special locations?

Here is what we can do, you give me two clock locations that you think will work and I can verify your results. Ok?

Did you actually calculate the radar distances to verify your calculation is actually correct?

 

[yuiop]

Hmmm, I am not convinced.

Here is what we can do, you give me two clock locations that you think will work and I can verify your results. Ok?

Did you actually calculate the radar distances to verify your calculation is actually correct?

Actually, if all you want to do is verify that d(ABA) = d(BAB) then the spatial separations in the CMIRF are not important. So keep it simple and say the pivot point is at the origin, clock A is initially at (x'A,t'A) = (3,0) and clock B is at (x'B,t'B) = (2,0) in the initial CMIRF (S'). The proper accelerations of A and B are 1/3 and 1/2 respectively.

 

[Passionflower]

Actually, if all you want to do is verify that d(ABA) = d(BAB) then the spatial separations in the CMIRF are not important. So keep it simple and say the pivot point is at the origin, clock A is initially at (x'A,t'A) = (3,0) and clock B is at (x'B,t'B) = (2,0) in the initial CMIRF (S'). The proper accelerations of A and B are 1/3 and 1/2 respectively.

Well that gives an even bigger difference, in fact the error increases the higher the acceleration differential.

Radar roundtrip AB_A = 2.33333
Radar roundtrip AB_B = 1.5

1.5 * ( 0.5 /0.33333) = 2.25

2.25 is not equal to 2.33333

 

[yuiop]

Well that gives an even bigger difference, in fact the error increases the higher the acceleration differential.

Radar roundtrip AB_A = 2.33333
Radar roundtrip AB_B = 1.5

1.5 * ( 0.5 /0.33333) = 2.25

2.25 is not equal to 2.33333

What exactly does AB_B mean? I would assume it means the distance (A to B) as measured by clock B, but I would not call that a round trip, but a one way trip or one leg.

Any chance of showing more detail of how you are obtaining the radar distances 2.3333 and 1.5?

 

[Passionflower]

What exactly does AB_B mean? I would assume it means the distance (A to B) as measured by clock B, but I would not call that a round trip, but a one way trip or one leg.

No, no, no, it is the roundtrip.

Any chance of showing more detail of how you are obtaining the radar distances 2.3333 and 1.5?

Well did you get the right answer, perhaps my calculations are wrong?

 

[pervect]

By far the easiest operational way to syncrhronize clocks in an accelerated frame is to use the Einstein method (the midpoint method) and make sure that the clocks are not "too far apart". If they are too far apart, you need to put in a bunch of intermediate clocks, and synchronize them all using the midpoint method.

The adjusted clocks should then keep "Rindler" time. Similarly , one can set up a lattice of rulers - those will measure "Rindler" distance.

the transformation from the Rindler coordinates (X,T) to inertial coordinates (x,t) with the simplification that units where c=1are used is (see MTW pg 173)

x = (1/g + X)*cosh(g*T)
t = (1/g + X)*sinh(g*T)I'm not aware of any simple closed form solution for (X,T) in terms of (x,t)

Now, because we've used units where c=1, the invariant lorentz interval is just

dt^2 - dx^2

We can write dt = (dt/dX)*dX + (dt/dT)*dT, and also dx = (dx/dX)*dX + (dx/dT)*dT

It would be messy to do by hand, but computer algebra quickly caluclates the well known result for the Rindler metric which can also be found in MTW:

dt^2 - dx^2 = (1+gX)^2*dT^2 - dX^2

Using this result reveals why the radar distance is acting funny. One is used to inertial frames, where the metric is always unity, and the coordinate speed of light is therefore always c (or 1, because of our unit simplifications). We can demonstrate for this for solving for the equation of motion for light, knowing that the lorentz interval is always zero. We get dt^2 - dx^2 = 0, giving the two solutions dx = dt or dx= -dt for the equations of light propagation in an inertial frame.

But, for the Rindler metric, we have (1+gX)^2 dT^2 - dx^2 = 0

We proceed in the same manner to find the equation for the path that light takes - we set the Lorentz interval to zero. Our solution is (1+gX)*dT = dX or (1+gX)*dT = -dX

i.e. the coordinate speed of light dX/dT is not a constant, rather it is equal to (1+gX). Note the pathology in the coordinate speed of light at X = -1/g. It goes to zero. This
is due to the ill-behavior of the Rindler coordinates at the "Rindler horizon", which is somewhat analogous to the bad behavior of Schwarzschild coordinates at the event horizon of a black hole. Just as outgoing light appears to "hang forever" at the event horizon of a black hole, light emitted from the "Rindler horizon" never catches up with our accerleating observer, but seems to keep a constant "Rindler distance" coordinate behind the accelerating observer.

 

[Austin0]

If B=back M=middle and F= front with D being proper distance wouldn't:

Radar time M-->F-->M be dt=(dx_MF+dx_FM/c)
M-->F=
dx_MF= D+dt_MF*v+0.5 a_F dt_MF²
F-->M=
dx_FM=D-(dt_FM*v +0.5 a_M dt_FM ²)

Radar time F-->M-->F then being
dt=(dx_MF+dx_FM/c)

F-->M dx_FM=
D-(dt_FM*v +0.5 a_M dt_FM ²)

M-->F=
dx_MF=
D+dt_MF*v+0.5 a_F dt_MF²

It looks like the round trip, distance and time for F-M-F and M-F-M would be equivalent as far as the acceleration is concerned doesn't it??
This assuming the time dilation differential is compensated for.
It would seem to follow that M-F-M would be equivalent to M-B-M also as far as differential acceleration is concerned so the only difference would seem to be differential contraction between the front and the back which should be very negligable, Am I far astray here?
It does seem possible to make it possible for one way isotropic measurements of c
Just some thoughts anyway

I have uploaded some diagrams of the method of synchronising accelerating clocks that I described earlier. https://www.physicsforums.com/showpost.php?p=2882964&postcount=64 The first diagram is the radar distances as measured by unsynchronised clocks moving with Born rigid motion and the second diagram is with clocks synchronised so that the radar distance between any 2 clocks is 1 light second and the distance between any N accelerating clocks is N-1 light seconds. This method of synchronisation has most of the properties required of it that bcrowell mentioned:



Condition (a) is met, but it requires that the raw output of the proper clocks has a constant correction factor of a_i/a_m where a_i is the proper acceleration of the i'th clock under consideration and a_m is the proper acceleration of the "master clock". For constant proper acceleration and Born rigid motion, this correction factor does not change over time.

Conditions (b1), (b2) and (b3) are also met, without any difficulty. Radar distances are the same measured in either direction and over long distances on the accelerating rocket with this method. Constant distant intervals in the accelerating frame appear progressively more length contracted towards the rear of the rocket in the inertial momentarily co-moving frame and this is perhaps the "differential length contraction" that austin0 was intuitively referring to.

Hi yuiop Just a little clarification here:
Was I able to make the math and statement above clear?
DO you agree that light paths from (Front to Back to Front) and (Back to Front to Back) would be equal in the same way that would apply with inertial frames?? This would seem to be problematic as measured from inertial frames dus to simultanieity issues.
This raises the question of where the midpoint of the system would be located as determined by simultaneous reception of signals sent in both directions yes?
Assuming that the midpoint was so determined then if the signals are sent from the midpoint and relfected from the front, back to the midpoint and from the midpoint to the back and return it seems like observing inertial frames would agree on this equivalence. So am I incorrect in this conclusion??

If this is the case then the difference in radar times in comparison of these cases could only be a consequence of the differential dilation between the clocks at those locations . Agreed??
If this is so there are some inferences to be made:
a) If the difference in dilation is compensated for by artificial calibration then the times should be the same, no?
b) If the spatial distances of light paths are equivalent in both cases it would seem to indicate that the Minkowski diagrams may be misleading and not accurate descriptions of reality in this regard perhaps?
That unless there are truly huge distances between the front and back and large differential accelerations involved , the dilation factor between front and back should be very small. So as spacetime intervals, even with naturally dilated clocks , the difference between F-B-F abd B-F-B should be small. Or no? The diagrams appear to indicate very large differences in both time and space in all cases. DO you have any explantion or thoughts regarding this?

If reflected measurements are made from the central point in both directions then distances from that point as derived by (0.5*dt*c) should be equivalent in both directions and one way synchronizations from a continuous synch signal based on those distances should then work throughout the system. The intervals should also be uniform through the system and no special spacing required Yes??
Or am I off here?
If this is at alll accurate this setup would then be equivalent to the CMIRF centered on the system.
WIth continued agreement between this system and successive CMIRFs if calibration is based on a continuous synch signal or equivalent computaitonal clock adjustment.
But there would not be agreement with local natural rulers.
If there was a long ruler running the length of the system and this ruler was naturqally Born accelerated
then indicated intervals would be shorter at the back and longer at the front, relative to the artificially calibrated ruler lengths derived from the light synch and calibration of the quasi-inertial coordinate frame. Ruler lengths derived from naturally dilated clocks at the back would also seem to mean a shorter length in comparison with the artificailly derived length , would you agree?
This is the differential contraction I was "intuitively" referring to.
So any place I am astray here please let me know .

 

[starthaus]

But, for the Rindler metric, we have (1+gX)^2 dT^2 - dx^2 = 0

We proceed in the same manner to find the equation for the path that light takes - we set the Lorentz interval to zero. Our solution is (1+gX)*dT = dX or (1+gX)*dT = -dX

i.e. the coordinate speed of light dX/dT is not a constant, rather it is equal to (1+gX). Note the pathology in the coordinate speed of light at X = -1/g. It goes to zero. This
is due to the ill-behavior of the Rindler coordinates at the "Rindler horizon", which is somewhat analogous to the bad behavior of Schwarzschild coordinates at the event horizon of a black hole. Just as outgoing light appears to "hang forever" at the event horizon of a black hole, light emitted from the "Rindler horizon" never catches up with our accerleating observer, but seems to keep a constant "Rindler distance" coordinate behind the accelerating observer.

Very nice. The above is also true for any type of general linear transform that is not Lorentz (an example of such transform is the Galilei transform) . Indeed:

x=aX+bT
t=eT+gX

In the preferential frame s(x,t), light speed is isotropic and equal to c (since (cdt)^2-dx^2=0).
Yet, in any other frame S(X,T), light speed is not isotropic:

0=(cdt)^2-dx^2=c^2(e*dT+g*dX)^2-(a*dX+b*dT)^2

Solving the above for dX/dT gives the (anisotropic) speed of light in S(X,T)

 

[yuiop]

Well that gives an even bigger difference, in fact the error increases the higher the acceleration differential.

Radar roundtrip AB_A = 2.33333
Radar roundtrip AB_B = 1.5

1.5 * ( 0.5 /0.33333) = 2.25

2.25 is not equal to 2.33333

I get the radar round trip time for ABA as measured by the proper time of clock A to be 2.432791 seconds (or 2.708333 seconds coordinate time in the initial CMIRF). I noticed you got your solution very quickly and unless you have a a formula ready to hand, I think you may have done the calculation too quickly, because although it is fairly tractable, there are several steps/ stages to the calculation. Could you check your result for Radar roundtrip (proper) time AB_A against mine again?

My method was:

Calculate the CMIRF coordinate one way time for t(AB).
Calculate the CMIRF coordinate roundtrip time for t(ABA).
Transform the coordinate time t(ABA) to proper time tau_A(ABA) as measured by clock A.

I am fairly sure my result is correct, because it is in agreement with measurements directly off the plotted graphs.

 

[Passionflower]

I get the radar round trip time for ABA as measured by the proper time of clock A to be 2.432791 seconds (or 2.708333 seconds coordinate time in the initial CMIRF). I noticed you got your solution very quickly and unless you have a a formula ready to hand, I think you may have done the calculation too quickly, because although it is fairly tractable, there are several steps/ stages to the calculation. Could you check your result for Radar roundtrip (proper) time AB_A against mine again?

Interesting.

I think the literature is rather vague on this as they often quote approximations or no formulas at all (but it may be due to the books I read, although I have a rather large collection). Folks who disagree, please provide a few references that have the actual formulas, would be tremendously helpful!

I am getting doubts now I do not get the 2.432791 seconds you gave.

If I am using the formula:

\Delta \tau_{ABA} = \frac{2h}{c\left(1-\frac{ah}{2c^2}\right)}

It gives me: 2.4 seconds

And this formula, which seems an approximation:

\Delta \tau_{ABA} = \frac{2h}{c} \left(1+\frac{ah}{2c^2}\right)

gives me: 2.33333 (2 1/3) seconds

Perhaps the first formula is also an approximation and the difference is due to some higher terms?

How many seconds do you get for BAB? I get resp. 1.5 and 1.6 seconds.

For folks to whom all this is simple please provide the right and exact formula and how many seconds you get, so at least I do not have to flip around different formulas and keep having doubts.

 

[Passionflower]

Update:

Looks like you are correct, my formula is only an approximation. If we take Rindler - "Relativity: Special, General and Cosmological": Example 3.24 we see an exact formula (give or take a typo in the book).

Based on this formula we get for the roundtrip time:

|\Delta \tau_{ABA}| = 2 \left( \frac{1}{\alpha_A} sinh^{-1} \left[ 2 \alpha_A \alpha_B (\frac{1}{\alpha_B^2} - \frac{1}{\alpha_A^2}} )\right] \right)

Where \alpha_B = \frac{1}{\frac{1}{\alpha_A} - h}

Which is indeed: 2.432790649

Given:
\alpha_B = \frac{1}{\frac{1}{\alpha_A} - h}

I have a feeling that the following part can be simplified, anybody readily see it?

2 \alpha_A \alpha_B ( \frac{1}{\alpha_B^2} - \frac{1}{\alpha_A^2}} )

By the way while we are at it, let's compare this to the radar roundtrip time in a Schwarzschild solution (d\theta = 0, \, d\phi = 0), here we have (if I am not mistaken):

Coordinate radar roundtrip distance between two radial coordinates r₁ and r₂:

\Delta t = 2 \left(r_2 - r_1 + r_s \log \left[ \frac{r_s + r_2}{r_1 - r_s}\right]\right)

Conversion factor from coordinate to proper time:

\sqrt{1-r_s/r}

Where:

r_s = 2m

For stationary observers the r values are related to proper acceleration the following way:

\alpha_r = \frac{m}{r^2} \frac{1}{\sqrt{1-r_s/r}}

 

[yuiop]

Update:

Looks like you are correct, my formula is only an approximation. If we take Rindler - "Relativity: Special, General and Cosmological": Example 3.24 we see an exact formula (give or take a typo in the book).

Based on this formula we get for the roundtrip time:

|\Delta \tau_{ABA}| = 2 \left( \frac{1}{\alpha_A} sinh^{-1} \left[ 2 \alpha_A \alpha_B (\frac{1}{\alpha_B^2} - \frac{1}{\alpha_A^2}} )\right] \right)

Where \alpha_B = \frac{1}{\frac{1}{\alpha_A} - h}

Which is indeed: 2.432790649

I am glad you obtained 2.432790649 in agreement with what I got earlier, but for some reason I can not reproduce your result given the Rindler equation you have quoted using \alpha_A = 1/3, \alpha_B = 1/2 and h=1. Any ideas? I keep getting |\Delta \tau_{ABA}| = 7.70277 using your Rindler formula.

If |\Delta \tau_{ABA}| = 2.432790649 is correct, then my method predicts |\Delta \tau_{BAB}| = 2.432790649*2/3 = 1.62186043

 

[Passionflower]

I am glad you obtained 2.432790649 in agreement with what I got earlier, but for some reason I can not reproduce your result given the Rindler equation you have quoted using \alpha_A = 1/3, \alpha_B = 1/2 and h=1. Any ideas?

Aaargh...forgot to move a factor 2 to the other side.

Hopefully it is correct now:

|\Delta \tau_{ABA}| = 2 \left( \frac{1}{\alpha_A} sinh^{-1} \left[ \frac{1}{2} \alpha_A \alpha_B (\frac{1}{\alpha_B^2} - \frac{1}{\alpha_A^2}} )\right] \right)

If |\Delta \tau_{ABA}| = 2.432790649 is correct, then my method predicts |\Delta \tau_{BAB}| = 2.432790649*2/3 = 1.62186043

Yes that is correct, the above, corrected, formula gives that too.

 

[yuiop]

Aaargh...forgot to move a factor 2 to the other side.

Hopefully it is correct now:

|\Delta \tau_{ABA}| = 2 \left( \frac{1}{\alpha_A} sinh^{-1} \left[ \frac{1}{2} \alpha_A \alpha_B (\frac{1}{\alpha_B^2} - \frac{1}{\alpha_A^2}} )\right] \right)Yes that is correct, the above, corrected, formula gives that too.

Cheers, now if we apply a "correction factor" to the proper time of each clock, equal to the proper acceleration of each clock, then this gives a synchronised coordinate time that can be used to obtain coordinate radar distances that are isotropic and transitive over long distances in the Born rigid accelerating reference frame, except at the singularity, where no physical clock can actually be located because it would have to be moving at the speed of light. I think Pervect has confirmed my earlier assumption, that by applying such a scheme we would end up with Rindler coordinates, is correct. It is nice to have arrived at the Rindler coordinate scheme from scratch, as it gives an intuitive physical interpretation to those coordinates.

In the example we were working on,

|\Delta \T_{ABA}| = |\Delta \tau_{ABA}|*\alpha_A = 2.432790649 *1/3 = 0.81093

and,

|\Delta \tau_{BAB}| = |\Delta \tau_{BAB}|*\alpha_B = 1.62186043 *1/2 = 0.81093

where \Delta T = 0.81093 is the synchronised coordinate time measured in the accelerating reference frame.

 

[Passionflower]

Simplifying the formula a little gives:

<br /> |\Delta \tau_{ABA}| = 2 \left( \frac{1}{\alpha_A} sinh^{-1} \left[ \frac{1}{2} (\frac{\alpha_A}{\alpha_B} - \frac{\alpha_B}{\alpha_A}} )\right] \right) <br />

 

[yuiop]

Simplifying the formula a little gives:

<br /> |\Delta \tau_{ABA}| = 2 \left( \frac{1}{\alpha_A} sinh^{-1} \left[ \frac{1}{2} (\frac{\alpha_A}{\alpha_B} - \frac{\alpha_B}{\alpha_A}} )\right] \right) <br />

Nice work.

The proper radar times in the two directions (ABA and BAB), using the notation \alpha_{AB} = \frac{1}{2} \left( \frac{\alpha_A}{\alpha_B} - \frac{\alpha_B}{\alpha_A} \right) can now be stated as:

|\Delta \tau_{ABA}| = \frac{2}{\alpha_A}\, sinh^{-1} (\alpha_{AB})

|\Delta \tau_{BAB}| = \frac{2}{\alpha_B}\, sinh^{-1} (\alpha_{AB})


and the Rindler coordinate radar times are:


|\Delta T_{ABA}| = 2 \, sinh^{-1} (\alpha_{AB})

|\Delta T_{BAB}| = 2 \, sinh^{-1} (\alpha_{AB})

 

[yuiop]

If B=back M=middle and F= front with D being proper distance wouldn't:

Radar time M-->F-->M be dt=(dx_MF+dx_FM/c)
M-->F=
dx_MF= D+dt_MF*v+0.5 a_F dt_MF²
F-->M=
dx_FM=D-(dt_FM*v +0.5 a_M dt_FM ²)

Radar time F-->M-->F then being
dt=(dx_MF+dx_FM/c)

F-->M dx_FM=
D-(dt_FM*v +0.5 a_M dt_FM ²)

M-->F=
dx_MF=
D+dt_MF*v+0.5 a_F dt_MF²

It looks like the round trip, distance and time for F-M-F and M-F-M would be equivalent as far as the acceleration is concerned doesn't it??
This assuming the time dilation differential is compensated for.
It would seem to follow that M-F-M would be equivalent to M-B-M also as far as differential acceleration is concerned so the only difference would seem to be differential contraction between the front and the back which should be very negligable, Am I far astray here?
It does seem possible to make it possible for one way isotropic measurements of c
Just some thoughts anyway

Hi yuiop Just a little clarification here:
Was I able to make the math and statement above clear?

Hi Austin, sorry for the delay getting back to you on your questions here. This was due to real life commitments and the requirement to come to an agreement in this thread on the equations that applied here, before we could sensibly talk about the issues you have raised.
Your equations seem to suggest that the accelerating clocks would agree on on the distances MFM and FMF without any adjustment of the clocks and in fact you obtain MFM=FMF=2D. I see you have used the equation x = (1/2)aT^2, because the time used in that expression is not the elapsed proper time of the accelerating clocks and nor is it the elapsed time of a clock in the CMIRF. It is the time measured by a third coordinate system of an inertial system moving at the average speed of the accelerating clock between two events and in this case, T = sqrt(x^2+t^2) where t is the time measured in the CMIRF that is moving at the instantaneous speed of the accelerating clock. If you have still been following this thread, I hope you will now agree that we have demonstrated that we can come up with a scheme that makes the one way speed of light isotropic.

This raises the question of where the midpoint of the system would be located as determined by simultaneous reception of signals sent in both directions yes?
Assuming that the midpoint was so determined then if the signals are sent from the midpoint and relfected from the front, back to the midpoint and from the midpoint to the back and return it seems like observing inertial frames would agree on this equivalence. So am I incorrect in this conclusion??

This is easy enough. First we place a clock at an assumed point M. We then place a mirror upstream at a suitable point (F) upstream where the radar signal returns in time t as measured by the clock at M. We place another mirror downstream which is located at a point (B) where the radar signal also returns simultaneously in time t. We can now agree that M is the midpoint between F and B in the accelerating system. This midpoint will not look like it is located midway in the CMIRF and in the CMIRF the distance MB looks shorter than MF and in effect, both distances look length contracted in the CMIRF with the rear section looking more length contracted than the front section.

a) If the difference in dilation is compensated for by artificial calibration then the times should be the same, no?
b) If the spatial distances of light paths are equivalent in both cases it would seem to indicate that the Minkowski diagrams may be misleading and not accurate descriptions of reality in this regard perhaps?

I am not sure I fully understand what you are getting at here.

That unless there are truly huge distances between the front and back and large differential accelerations involved , the dilation factor between front and back should be very small. So as spacetime intervals, even with naturally dilated clocks , the difference between F-B-F abd B-F-B should be small. Or no? The diagrams appear to indicate very large differences in both time and space in all cases. DO you have any explantion or thoughts regarding this?

The difference in acceleration between for example the floor of your room (B) and the ceiling (F) of your room, is very small and it would indeed to difficult to detect any difference between FBF and BFB in practice, although it could be said that the accuracy of the Pound Rebka experiment was equivalent to doing this. In the diagrams I uploaded, the measurements are all close to the fulcrum at the origin and this is equivalent to being in the extreme curved space near the event horizon of a black hole and the effects are much more noticeable there.

If reflected measurements are made from the central point in both directions then distances from that point as derived by (0.5*dt*c) should be equivalent in both directions and one way synchronizations from a continuous synch signal based on those distances should then work throughout the system. The intervals should also be uniform through the system and no special spacing required Yes??
Or am I off here?

Yes and no. It looks like I have applied special spacing but this would come about naturally due to length contraction.

If this is at alll accurate this setup would then be equivalent to the CMIRF centered on the system.
WIth continued agreement between this system and successive CMIRFs if calibration is based on a continuous synch signal or equivalent computaitonal clock adjustment.
But there would not be agreement with local natural rulers.

Actually, there would be agreement with local natural rulers. Let us say we have a long rocket that is moving inertially and has a series of rulers layed end to end that are all the same length and in agreement with the radar distances measured inertially by clocks placed at the end of each ruler. If the rocket now transfers from inertial motion to accelerating motion to Born rigid accelerating motion the rulers will length contract (more so at the back) and have the "special spacing" that I depicted in the second diagram I uploaded earlier. The length contraction of the rulers is natural. THe adjustment to the clocks to get the radar measurements to agree with the ruler lengths isotropically is artificaial, but can be done.

If there was a long ruler running the length of the system and this ruler was naturally Born accelerate, then indicated intervals would be shorter at the back and longer at the front, relative to the artificially calibrated ruler lengths derived from the light synch and calibration of the quasi-inertial coordinate frame. Ruler lengths derived from naturally dilated clocks at the back would also seem to mean a shorter length in comparison with the artificailly derived length , would you agree?

As above, natural ruler lengths would be in agreement with "quasi-inertial coordinate" calibrated rulers. If you try to calibrate rulers in the accelerating system using radar measurements made with naturally dilated clocks (i.e proper time) then then "proper radar distances" would not be in agreement with the natural ruler lengths and they would also disagree about the length of a given ruler depending upon which end of the ruler the proper radar distance was measured from. Natural radar measurements would also not agree that when two rulers of equal proper length (L) when laid along side each other would measure 2L when placed end to end. For these reasons natural radar distances are not very good for constructing a sensible coordinate system.

This is the differential contraction I was "intuitively" referring to.

With the clarifications I have made above, I would indeed agree that there is a notion of differential length contraction of an Born accelerating rocket, when observed from a CMIRF.

Passionflower has kindly provided some equations for us to play with and maybe you could try them out and see if things make more sense now?

 

[Austin0]

Hi Austin, sorry for the delay getting back to you on your questions here. ..

Hi yuiop ,,no problem ,,,me too .

Your equations seem to suggest that the accelerating clocks would agree on on the distances MFM and FMF without any adjustment of the clocks and in fact you obtain MFM=FMF=2D. I see you have used the equation x = (1/2)aT^2, because the time used in that expression is not the elapsed proper time of the accelerating clocks and nor is it the elapsed time of a clock in the CMIRF. It is the time measured by a third coordinate system of an inertial system moving at the average speed of the accelerating clock between two events and in this case, T = sqrt(x^2+t^2) where t is the time measured in the CMIRF that is moving at the instantaneous speed of the accelerating clock. If you have still been following this thread, I hope you will now agree that we have demonstrated that we can come up with a scheme that makes the one way speed of light isotropic.

No actually I thought it was the time measured in the inertial frame where the accelerated system was initially at rest. The frame which is the frame for the diagrams.

This raises the question of where the midpoint of the system would be located as determined by simultaneous reception of signals sent in both directions yes?
Assuming that the midpoint was so determined then if the signals are sent from the midpoint and relfected from the front, back to the midpoint and from the midpoint to the back and return it seems like observing inertial frames would agree on this equivalence. So am I incorrect in this conclusion??

This is easy enough. First we place a clock at an assumed point M. We then place a mirror upstream at a suitable point (F) upstream where the radar signal returns in time t as measured by the clock at M. We place another mirror downstream which is located at a point (B) where the radar signal also returns simultaneously in time t. We can now agree that M is the midpoint between F and B in the accelerating system. This midpoint will not look like it is located midway in the CMIRF and in the CMIRF the distance MB looks shorter than MF and in effect, both distances look length contracted in the CMIRF with the rear section looking more length contracted than the front section.

Yes your easy method was just what I described and the question I posed was based on the understanding that the midpoint so deirived would not be in the middle of the system or the CMIRF.
I think you may be mistaken about MB and MF both appearing contracted to the middle CMIRF.
I thin the back would be contracted and the front would be relatively larger than the local metric or section of the middle CMIRF
If the system is viewed as a composite of 3 CMIRFs the back 1/3 would be contracted and the front 1/3 would be expanded relative to the middle CMIRF yes?

If reflected measurements are made from the central point in both directions then distances from that point as derived by (0.5*dt*c) should be equivalent in both directions and one way synchronizations from a continuous synch signal based on those distances should then work throughout the system. The intervals should also be uniform through the system and no special spacing required Yes??
Or am I off here?

Yes and no. It looks like I have applied special spacing but this would come about naturally due to length contraction.

WHy would length contraction be relevant if using calibrated clocks [all running same rate] and synchronization from a central master clock?
Wouldn't lengths derived from such clocks be equivalent throughout the system?

If this is at alll accurate this setup would then be equivalent to the CMIRF centered on the system.
WIth continued agreement between this system and successive CMIRFs if calibration is based on a continuous synch signal or equivalent computaitonal clock adjustment.
But there would not be agreement with local natural rulers.

Actually, there would be agreement with local natural rulers. Let us say we have a long rocket that is moving inertially and has a series of rulers layed end to end that are all the same length and in agreement with the radar distances measured inertially by clocks placed at the end of each ruler. If the rocket now transfers from inertial motion to accelerating motion to Born rigid accelerating motion the rulers will length contract (more so at the back) and have the "special spacing" that I depicted in the second diagram I uploaded earlier. The length contraction of the rulers is natural. THe adjustment to the clocks to get the radar measurements to agree with the ruler lengths isotropically is artificaial, but can be done.

Using the analogy of the back CMIRF wouldn't the local clock radar measurements agree with the ruler measurements if it is in fact equivalent to a locally instantaneous inertial frame? It would only be with radar measurements over greater distances in the system that the natural dilation of the rear clock would not agree with distance or am I missing again?

If there was a long ruler running the length of the system and this ruler was naturally Born accelerate, then indicated intervals would be shorter at the back and longer at the front, relative to the artificially calibrated ruler lengths derived from the light synch and calibration of the quasi-inertial coordinate frame. Ruler lengths derived from naturally dilated clocks at the back would also seem to mean a shorter length in comparison with the artificailly derived length , would you agree?

As above, natural ruler lengths would be in agreement with "quasi-inertial coordinate" calibrated rulers. If you try to calibrate rulers in the accelerating system using radar measurements made with naturally dilated clocks (i.e proper time) then then "proper radar distances" would not be in agreement with the natural ruler lengths and they would also disagree about the length of a given ruler depending upon which end of the ruler the proper radar distance was measured from. Natural radar measurements would also not agree that when two rulers of equal proper length (L) when laid along side each other would measure 2L when placed end to end. For these reasons natural radar distances are not very good for constructing a sensible coordinate system.

This is the differential contraction I was "intuitively" referring to.

With the clarifications I have made above, I would indeed agree that there is a notion of differential length contraction of an Born accelerating rocket, when observed from a CMIRF.

Passionflower has kindly provided some equations for us to play with and maybe you could try them out and see if things make more sense now?

I have been following with great interest the calculations the two of you have been making but haven't had time to comment.
I have some questions regarding what you and PassionFlower are doing and some areas where I may not correctly understand Born rigidity and Rindler coordinates so I am hoping we can continue this to a point of understanding.
So I would appreciate some clatification on the previous points.

Originally Posted by yuiop
As far as I can tell, the line of simultaneity pivoting about the origin is only true, if the clocks of the accelerating rocket nearer the origin are artificially sped up relative to the clocks nearer the nose.

Originally Posted by DrGreg
True.

austin0...A) I thought the difference in periodicity between the clocks at the front and the back was a result of the difference in proper acceleration and not an artificial contrivance ??
B) That the convergence of L's of S at the pivot event was due to the clocks at the back being naturally dilated relative to the front , not sped up either natrually or artificially . Is this incorrect?

yuiop...If the clocks are left to do their own thing so that they agree with local natural processes, the natural line of simultaneity would be tilted in the opposite direction and would not pass through the origin, no?

Originally Posted by DrGreg
Yes, although now "simultaneity" would bear no relationship to Einstein synchronisation.

Here are some other points I am unsure of:
1) Born rigidity and Dindler coordinates are both based on a natural dilation differential between the front and the back. yes??
This is taken to be equivalent to both a comparable gravitational dilation and a dilation equivalent to a difference in relative velocities??

2) This is assumed to be constant through time with a constant differential factor?

3) A constant acceleration differential would seem to imply a linear increase in relative velocity between the various locations, eg. the front and the back Yes??

4) How does a linear increase in relative velocities work out to be consistent with a constant dilation factor??

5) The coordinate difference in acceleration and resulting relative velocities , as measured in the inertial frame where the accelerated system was initially at rest, must be consistent with the measured Lorentz contraction relative to that inertial frame, correct??

How then could a linear increase in coordinate relative velocities between the front and the back be consistent with the non-linear increase in contraction??
At the begininng stages of acceleration where the contraction is flat and negligable a linear increase would seem to produce too much contraction. At later stages with greater acheived velocities a linear increase in relative velocities would seem to be insufficient to produce the asymptotic increase in contraction, no?
It would seem that an increasing acceleration differential would be needed to be consistent.
On top of this there is the coordinate acceleration dropoff factor of \gamma ³ so at the ponts of the acceleration course, where it would appear to require greater relative acceleration at the rear, the acceleration there would be decreasing by the cubic factor relative to the front, which would be at a lesser relative instantaneous velocity. Yes?

6) A hyperbolic world line for an accelerated system is a graph of the coordinate acceleration of that system as measured in the frame where it was initially at rest correct??

WHile not explicitly an acceleration profile , can't we extract an approximation from this graph by looking at various coordinate time intervals on the vertical axis and the relative increase in the tangent slopes at those points ?

Looking at a graph with this view, it seems to me that the acceleration dropoff, the increase in coordinate time to reach comparable increases in relative velocity are consistent with a single gamma factor. Fairly flat up to high velocities 0.8->0.9+ c where it then becomes increasingly radical.

It does not appear to be consistent with a falloff of \gamma ³ where the increase in coordinate time would become radical at much lower velocities.
Now am I way off here ?? DO you see what I am referring to??

7) You have mentioned that with your calculations for clock calibration you have derived an artificial system that is equivalent to a Rindler system.

COuld you expand upon this as it seems to me that you have achieved a completely different system so I may be completely misinformed regarding Rindler coordinates.
In any case this whole question is fascinating and I hope it progresses. I also apologise for slow responces. Thanks

 

[Passionflower]

Here is a graph showing the radar roundtrip times of a height of 1 for various accelerations.