Can We Construct a Coordinate Chart Where Bell Observers Are At Rest?

  • #1
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I am wondering if the coordinate chart I derive here has appeared anywhere in the literature on the Bell Spaceship Paradox. I have been unable to find it.
A question that might occur to anyone reading about the Bell Spaceship Paradox is, can we construct a coordinate chart in which all of the Bell observers (i.e., observers following worldlines like those of the spaceships in the "paradox" scenario) are "at rest"? I put "at rest" in scare-quotes because, since the expansion scalar of the Bell congruence is positive, these "at rest" observers do not maintain constant proper distance from each other (which is, of course, the point of the "paradox"). However, with that caveat, it seems that we can indeed construct such a chart by a simple series of transformations from the standard Minkowski chart. I'll describe that construction below, but I have been unable to find any reference in the literature that gives it, and I am wondering if any other PF members have.

The construction starts from the specification of a "Bell observer" worldline in the Minkowski chart. We'll restrict to 1+1 spacetime dimensions to keep things simple. Using lower case ##t## and ##x## for the Minkowski chart coordinates, we have each Bell observer worldline specified as follows:

$$
\left( x - X \right)^2 - t^2 = \frac{1}{a^2}
$$

where ##X## is a parameter that is constant along each worldline and so serves to label the worldlines (the reason for the notation ##X## will become evident below), and we are using units in which ##c = 1##.

We now look for a coordinate transformation that will make each of these worldlines a curve of constant spatial coordinate in the new chart. The obvious ansatz to try given the worldline equation above is:

$$
t = \tilde{T}
$$

$$
x = X + \sqrt{\tilde{T}^2 + \frac{1}{a^2}}
$$

where ##\tilde{T}## and ##X## are the new coordinates. (The reason for the notation ##\tilde{T}## instead of just ##T## will be seen shortly.) This gives

$$
dx = dX + \frac{\tilde{T} d\tilde{T}}{\sqrt{\tilde{T}^2 + \frac{1}{a^2}}}
$$

and the line element then becomes

$$
ds^2 = - d\tilde{T}^2 + \left( dX + \frac{\tilde{T} d\tilde{T}}{\sqrt{\tilde{T}^2 + \frac{1}{a^2}}} \right)^2
$$

which simplifies to

$$
ds^2 = - \frac{1}{1 + a^2 \tilde{T}^2} d\tilde{T}^2 + \frac{2 a \tilde{T}}{\sqrt{1 + a^2 \tilde{T}^2}} d\tilde{T} dX + dX^2
$$

We can simplify this further with an obvious rescaling of the time coordinate:

$$
dT = d\tilde{T} \frac{1}{\sqrt{1 + a^2 \tilde{T}^2}}
$$

This gives ##a\tilde{T} = \sinh a T##, and the line element now becomes

$$
ds^2 = - dT^2 + 2 \sinh aT dT dX + dX^2
$$

It is probably helpful to clarify some points. First, the surfaces of constant ##T## in this chart are not orthogonal to the Bell observer worldlines (except at ##T = 0##, as is obvious from the above line element). Instead they are the same surfaces as the surfaces of constant Minkowski coordinate time. One of the consequences of the Bell spaceship paradox is that it is not possible to construct a chart in which all of the surfaces of constant time are orthogonal to the Bell observer worldlines. (You can construct a Rindler chart centered on one Bell observer worldline, and the surfaces of constant coordinate time in that chart will be everywhere orthogonal to that Bell observer worldline, but they won't be orthogonal to any other Bell observer worldline, except at ##t = 0##, and of course the chart will only cover one "wedge" of the spacetime.)

Also, after the rescaling of the time coordinate, ##T## is the same as proper time for all the Bell observers, and the surfaces of constant ##T## are flat. Distance in those surfaces, however, is not the same as proper distance between the Bell observers, because of the non-orthogonality described above. Note that, since the coefficient of the ##dT dX## term in the line element increases with ##T## (and does so exponentially), the non-orthogonality, and hence the mismatch between ##X## distance in the chart and proper distance between Bell observers, gets more and more severe with time.

Again, my question is: has anyone seen a reference to this chart, or a similar one, anywhere in the literature?
 
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  • #2
PeterDonis said:
$$ds^2 = - dT^2 + 2 \sinh aT dT dX + dX^2$$...
Also, after the rescaling of the time coordinate, ##T## is the same as proper time for all the Bell observers
I am missing pseudo-gravitational (##X##-dependent) time dilation of a clock at constant ##X## location (##dX=0##).
 
  • #3
Sagittarius A-Star said:
I am missing pseudo-gravitational (##X##-dependent) time dilation of a clock at constant ##X## location (##dX=0##).
There isn't any. Remember this is not Rindler coordinates. The way the surfaces of constant time are chosen, all Bell observers have the same "time dilation" relative to an inertial observer at rest in the original inertial frame (the one in which all the Bell observers are at rest at ##t = 0##). So relative to each other, in this chart, the Bell observers have no time dilation.
 
  • #4
PeterDonis said:
There isn't any.
In the inertial frame, in which the 2 Bell rockets start at the "same" coordinate-time and have the same velocity at each point in coordinate-time, on can understand best, that the trailing rocket receives light pulses, sent out by the leading rocket, blue-shifted (longitudinal Doppler effect). So there must be time-dilation of the leading rocket in the rest-frame of the trailing rocket.
 
  • #5
Sagittarius A-Star said:
In the inertial frame, in which the 2 Bell rockets start at the "same" coordinate-time and have the same velocity at each point in coordinate-time, on can understand best, that the trailing rocket receives light pulses, sent out by the leading rocket, blue-shifted
No, redshifted.

Sagittarius A-Star said:
So there must be time-dilation of the leading rocket in the rest-frame of the trailing rocket.
The frame I have derived in this thread is not the frame described by the bolded phrase in your quote above.

Please bear in mind that this is an "A" level thread. I gave it that thread level on purpose. If what I said in the sentence above is not obvious to you, then you don't have the requisite background knowledge for an "A" level thread on this topic.
 
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  • #8
So in this chart all of the Bell observers are at rest, but the distance between them increases. In that sense it is a little like cosmological coordinates
 
  • #9
I cannot say that I remember seeing these coordinates explicitly. I will however say that it is an implementation of a more general way of assigning coordinates, namely picking a spatial hypersurface and a set of non intersecting world lines. The spatial coordinates are then defined as some set of coordinates on the hypersurface and the time coordinate as the proper time along the world line passing through the event (from the hypersurface). As such I would probably have gone straight for ##T## as the choice of time coordinate. It seems a more natural choice.

Sagittarius A-Star said:
So there must be time-dilation of the leading rocket in the rest-frame of the trailing rocket.
No, this is a faulty conclusion. Time dilation is contingent upon the choice of simultaneity convention (and here simultaneity is defined simply as the simultaneity in the original rest frame) whereas a frequency shift is something invariant and calculable. Even in standard Minkowski coordinates the Doppler factor is not the same as the gamma factor in the time dilation formula.
 
  • #10
Orodruin said:
I would probably have gone straight for as the choice of time coordinate. It seems a more natural choice.
Yes, one could combine both of my coordinate transformations into one. I split them up only because for me it was easier to see how to rescale the time coordinate after I'd done the other transformation.
 
  • #11
PeterDonis said:
one could combine both of my coordinate transformations into one
Just to round this off, here is the explicit transformation in one step:

$$
t = \frac{1}{a} \sinh a T
$$
$$
x = X + \frac{1}{a} \cosh a T
$$

This gives

$$
dt = \cosh aT dT
$$
$$
dx = dX + \sinh aT dT
$$

and substituting that into the Minkowski line element gives the line element I posted earlier.
 
  • #12
Orodruin said:
No, this is a faulty conclusion. Time dilation is contingent upon the choice of simultaneity convention (and here simultaneity is defined simply as the simultaneity in the original rest frame) whereas a frequency shift is something invariant and calculable.
Correct, I missed to check the simultaneity convention.

Orodruin said:
Even in standard Minkowski coordinates the Doppler factor is not the same as the gamma factor in the time dilation formula.
I think the transversal Doppler factor, as measured by an observer at rest in this inertial frame, validates the TD-gamma factor with respect to the standard Minkowski coordinates, because they use Einstein-synchronization.

Consider a moving particle (optical clock) with consecutive events A and B on it's worldline in the below drawing of a Galilei-telescope.
If the velocity-direction of the particle is transversal to the telescope-view-direction, with respect to the receiver's rest frame, then you get a gamma-redshift.
If the frequency is higher than shown (much more than 2 consecutive events would be shown in the drawing), then you observe a frequency-band, and it's mid-frequency has the gamma-redshift.

Linsenfernrohre2a.gif
Source:
https://de.wikipedia.org/wiki/Fernrohrbrille#Optischer_Hintergrund
 

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