# Two particles, constant acceleration in their own frame

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1. May 4, 2014

### HJ Farnsworth

Greetings,

I'm having some issues with a problem that I is related, though not identical, to Bell's spaceship paradox (this is not for a homework or anything, it just sort of occurred to me when I was thinking about something else).

Consider two particles that are initially at rest in some frame $O$, that are separated by a distance $L$. We will say that particle 1 is initially at $x=0$ and particle 2 is initially at $x=L$. At time $t=0$, they both start accelerating in the same direction, which is parallel to their separation, in such a way as to always maintain a constant separation, as viewed in their own constantly changing MCRF, which I will refer to as $O^{\prime}$. (This is the reason I am saying it is related to Bell's spaceship paradox - their separation is constant as viewed in in $O^{\prime}$, rather than as viewed in $O$). My goal is to get equations for the movement of each particle as viewed in $O$.

For the simpler problem of one particle accelerating at a constant acceleration $a^{\prime}$, I have the following equations:
(1): $\beta =\beta (a^{\prime},t^{\prime})=\tanh (\frac{a^\prime}{c}t^{\prime})$
(2): $x=x(a^{\prime},t^{\prime})=\frac{c^{2}}{a^{\prime}}[\cosh (\frac{a^\prime}{c}t^{\prime})-1]$
(3): $t=t(a^{\prime},t^{\prime})=\frac{c}{a^{\prime}}\sinh (\frac{a^{\prime}}{c}t^{\prime})$

Now, one way I could solve this is to simply apply equation (2) to either the front particle or the back particle, and then use my knowledge that Lorentz contraction will occur to determine the equation for the other particle. But, I want to solve this without directly applying Lorentz contraction (although admittedly I used Lorentz contraction to derive equations (1)-(3)).

The issue I am having is that both particles by assumption have constant acceleration $a^{\prime}$ as viewed in $O^{\prime}$, so as far as I can tell, equation (1) should apply to both particles. Then, inverting equation (3) to find $t^{\prime}$ as a function of $a^{\prime}$ and $t$ and substituting into equation (1), we have $\beta = \beta (a^{\prime},t)$ is the same for both particles, and differentiating with respect to $t$, $a=a(a^{\prime},t)$ is the same for both particles. If this is the case, how could Lorentz contraction ever occur? I.e., for $O$ to see the distance between the two particles contracting as the particles speed up, it must see that the speed of the back particle is increasing faster than that of the front particle, but my analysis here has that the acceleration of both particles is always the same as viewed in $O$ (which is clearly wrong, since they are the same as viewed in $O^{\prime}$). So, I have stumbled onto a contradiction.

Does anyone see my mistake? If so, what is it, and how should the analysis in the above paragraph proceed?

Again, I don't want to just apply Lorentz contraction to solve the above problem - instead, I want to analyze it using equations (1)-(3), and hopefully see Lorentz contraction re-emerge as a result. Basically, this is one of those issues where I know two ways to solve a problem that both seem correct to me, but that contradict each other. It is not enough for me to know that one is right - I want to also know why the other is wrong.

Thanks very much for any help that you can give.

-HJ Farnsworth

Last edited: May 4, 2014
2. May 4, 2014

### WannabeNewton

This is almost identical to the Rindler congruence: http://en.wikipedia.org/wiki/Rindler_coordinates

The reason I use the qualifier "almost" is, if I understood your post correctly, you have assumed that both particles have the same uniform proper acceleration $a$ (this is a frame-invariant quantity) and that the distance between them remains constant in either of their instantaneous rest frames. These are contradictory assumptions. In order for the distance between them to remain constant in either instantaneous rest frame, the respective uniform proper accelerations of the particles must be different. If, on the other hand, the uniform proper accelerations are the same for both particles then the distance between them, in either instantaneous rest frame, will be increasing in the proper time of the clock attached to this frame.

You might also be making the calculation more complicated than it needs to be. From the above link, each particle's world-line is a level curve of the implicit equation $x^2 - t^2 = \frac{1}{a^2}$ for different values of a, where we are working in the global inertial frame $O$ defined in your post. Then, in $O$, the trajectories have 3-velocities given by $\frac{\mathrm{d} x}{\mathrm{d} t} = \frac{t}{x} = \frac{t}{\sqrt{t^2 + \frac{1}{a^2}}}$ from which it follows that the particles have different 3-velocities relative to $O$ because they have different values of $a$.

3. May 4, 2014

### HJ Farnsworth

Hi WannabeNewton, thanks for replying so quickly - this is the second time you've helped me out in two days!

Regarding your analysis - that does look quite simple, and seems to do the trick.

Regarding what you mentioned in the paragraph before that - you read my post correctly, I have indeed made both of those assumptions. Could you elaborate on why these are contradictory? This is not obvious to me - for instance, for someone in an accelerating rocketship of length L, I would roughly expect their situation to be analogous to the one I described above - they feel the rocket accelerating as a whole, and see the length of the rocket remain constant.

Thanks very much again.

-HJ Farnsworth

4. May 5, 2014

### WannabeNewton

You're using the Newtonian notion of rigidity to make that judgement but Newtonian rigidity is untenable in special relativity.
If you want the rocket to undergo rigid motion along a given fixed direction then each of its parts has to accelerate at a different rate depending on its radar distance from the back of the rocket. This is called Born rigid motion. If you accelerate all parts of the rocket with the same magnitude then the motion will not be Born rigid and in fact the parts will try to separate from one another in their instantaneous rest frames, eventually exerting enough internal stresses on the rocket to break it apart.

It's clear that if we have two particles accelerated simultaneously and with identical proper acceleration in a global inertial frame, the distance between them as measured in this frame must be constant. If we assume the distance between them is constant in the instantaneous rest frame of either particle and perform a Lorentz boost back to the global inertial frame then the length contraction formula implies that the distance between the particles is changing in time in the global inertial frame which is a contradiction.

5. May 10, 2014

### HJ Farnsworth

Hi WannabeNewton,

Sorry I didn't respond to this, the weekend ended and I've been swamped since. Anyway, thanks again for the help!