# Looking for clarification regarding equations describing uniformly accelerated motion in SR

I am looking for clarification regarding Problem 1.17. In "Problem Book In Relativity and Gravitation" because I am probably misunderstanding some details about it, which lead me to arrive at a different solution than the one included in the book. The problem statement is as follows:

An observer experiences a uniform acceleration in the $$x$$ direction, of magnitude $$g$$. Define a coordinate system $$(\bar{t},\bar{x},\bar{y},\bar{z})$$ for him in the following way:

$$\$$(i) Let the observer be at $$\bar{t}=\bar{x}=\bar{y}=\bar{z}=0$$ and let $$\bar{t}$$ be his proper time.

$$\$$(ii) Let his hyperplanes of simultaneity agree with the hyperplanes of simultaneity of an instanteneously comoving inertial frame.

$$\$$(iii) Let the other "coordinate stationary observers" (for whom $$\bar{x},\bar{y},\bar{z}$$ are constant) move in such a way that they are always at rest with respect to the observer on the hyperplanes of simultaneity. At $$t=0$$ label all spatial points with the same labels as the momentarily comoving inertial system $$t=0, x, y, z$$.

Give the coordinate transformation between $$t,x,y,z$$ and $$\bar{t},\bar{x},\bar{y},\bar{z}$$. Show that coordinate stationary clocks cannot remained synchronized.

While I recognized that I don't fully understand the meaning of condition (ii) and (iii), it did seem to me that this problem is generally asking for the equations relating a uniformly accelerating reference frame, whose local coordinates are $$\bar{x},\bar{y},\bar{z},\bar{t}$$, with an inertial reference frame whose coordinates are $$x,y,z,t$$. Assuming this is the case, the solution is well known to be the hyperbolic motion solution, in this case it would look as follows: $$ct = \frac{c^2}{g}\sinh {\dfrac{g}{c}\bar{t}}$$ $$x = \frac{c^2}{g}\left(\cosh \dfrac{g}{c}\bar{t}-1\right)$$ Which further assumes the initial condition $$x(0)=0$$.

However, the solution provided in the book is: $$ct = \left(\frac{c^2}{g}+\bar{x}\right)\sinh {\dfrac{g}{c}\bar{t}}$$ $$x = \left(\frac{c^2}{g}+\bar{x}\right)\cosh \dfrac{g}{c}\bar{t}-\frac{c^2}{g}$$

So as mentioned, I am assuming that the difference somehow stems from me failing to understand conditions (ii) and (iii) as they are stated in the problem, and I would be grateful if someone can clarify them and how they lead to this additional dependence on $$\bar{x}$$.

There is a discrepancy between your answer and the solution provided, because you answered a different question than was asked. The question was:

Give the coordinate transformation between $$t, x, y, z$$ and $$\overline{t}, \overline{x}, \overline{y}, \overline{z}$$. Show that coordinate stationary clocks cannot remained synchronized.

So you're being asked to provide a coordinate transformation. That means specifying four separate functions $$t = \varphi_{t}(\overline{t}, \overline{x}, \overline{y}, \overline{z})\ ,\quad x = \varphi_{x}(\overline{t}, \overline{x}, \overline{y}, \overline{z})\ ,\quad y = \varphi_{y}(\overline{t}, \overline{x}, \overline{y}, \overline{z})\ ,\quad z = \varphi_{z}(\overline{t}, \overline{x}, \overline{y}, \overline{z})\ .$$ Each of these functions is defined over the full 4D spacetime, in overlined coordinates. (Actually, it turns out that it's only defined over a region of the full Minkowski spacetime, but the point is you're looking for four functions defined over 4D.)

The answer you suggested gives parametric equations for the worldline of a single particle. So your functions are defined over just one parameter, the proper time of the uniformly accelerating observer. Conditions (ii) and (iii) simply don't apply to your answer, because these conditions are about setting up an entire coordinate system, not giving the coordinates of a single worldline. (By the way, your equations for the worldline are entirely correct, given condition (i). But you should add $$y = z = 0$$.)

The answers are not the same, but they are related. Given your answer, all that remains to do is to extend the equations for the worldline of the accelerating observer to a full coordinate system. Conditions (ii) and (iii) tell you how to do that uniquely.

Condition (ii) is telling you that $$\overline{t} = const.$$ surfaces are hyperplanes Minkowski-orthogonal to the observer's four-velocity. As a function of observer proper time $$\overline{t}$$, this four-velocity is

$$U^\mu = \left(\frac{dt}{d\overline{t}}, \frac{dx}{d\overline{t}}, \frac{dy}{d\overline{t}}, \frac{dz}{d\overline{t}}\right) = \left(\cosh\frac{g}{c}\overline{t}, c\sinh\frac{g}{c}\overline{t}, 0, 0\right)$$

Edited to add some detail here: We're looking for a hyperplane Minkowski-orthogonal to this vector. It is given by the constraint

$$ct \cosh\frac{g}{c}\overline{t} - x \sinh\frac{g}{c}\overline{t} = \kappa$$ for some constant $$\kappa$$ to be determined. We determined it by stipulating that the plane passes through the point $$(t, x, y, z) = \left(\frac{c}{g}\sinh\frac{g}{c}\overline{t}, \frac{c^2}{g}\left(\cosh\frac{g}{c}\overline{t} - 1\right), 0, 0\right)$$, so the constraint is satisfied by this point. So:

$$\frac{c^2}{g}\sinh\frac{g}{c}\overline{t}\cosh\frac{g}{c}\overline{t} - \frac{c^2}{g}\left(\cosh\frac{g}{c}\overline{t} - 1\right)\sinh\frac{g}{c}\overline{t} = \kappa\ ,$$ giving $$\kappa = \frac{c^2}{g}\sinh\frac{g}{c}\overline{t}$$. After some tidying, the hyperplane is finally given by the constraint

$$\mbox{for } \overline{t} = const.,\qquad ct\cosh\frac{g}{c}\overline{t} - \left(x + \frac{c^2}{g}\right)\sinh\frac{g}{c}\overline{t} = 0\ , \quad \mbox{or} \quad \frac{ct}{x + \frac{c^2}{g}} = \tanh\frac{g}{c}\overline{t}\ .$$

Condition (iii) is telling you that, on each $$\overline{t} = const.$$ hyperplane, the $$(\overline{x}, \overline{y}, \overline{z}) = const.$$ worldlines have four-velocities parallel to the accelerating observer. It follows that these four-velocities satisfy:

$$\left(\frac{dt}{d\overline{t}}, \frac{dx}{d\overline{t}}, \frac{dy}{d\overline{t}}, \frac{dz}{d\overline{t}}\right) = \alpha\left(\cosh\frac{g}{c}\overline{t}, c\sinh\frac{g}{c}\overline{t}, 0, 0\right)\ ,$$ where $$\alpha$$ is some function of $$\overline{x}, \overline{y}, \overline{z}$$ and $$\alpha >0$$. (Warning: $$\overline{t}$$ is not in general a proper time parameter for these worldlines!) Integrating, one obtains the equations

$$\left(t, x, y, z\right) = \left(\alpha\frac{c}{g}\sinh\frac{g}{c}\overline{t} + t_0, \alpha\frac{c^2}{g}\left(\cosh\frac{g}{c}\overline{t} - 1\right) + x_0, y_0, z_0\right)\ .$$

Now, condition (ii) implies that the $$\overline{t} = 0$$ hyperplane, which coincides with the $$t=0$$ hyperplane, is such that $$(t, x, y, z) = (0, \overline{x}, \overline{y}, \overline{z})$$. This yields $$(t_0, x_0, y_0, z_0) = (0, \overline{x}, \overline{y}, \overline{z})$$. Moreover, each of these worldlines must satisfy the $$\overline{t} = const.$$ hyperplane constraint above. So

$$\frac{\frac{\alpha c^2}{g}\sinh\frac{g}{c}\overline{t}}{\frac{\alpha c^2}{g}\left(\cosh\frac{g}{c}\overline{t} - 1\right) + \overline{x} + \frac{c^2}{g}} = \tanh\frac{g}{c}\overline{t}\ .$$

It follows that $$\alpha$$ must satisfy

$$-\frac{\alpha c^2}{g} + \overline{x} + \frac{c^2}{g} = 0\ ,$$ which after a bit of algebra gets you $$\alpha = 1 + \frac{g\overline{x}}{c^2}\ .$$

Putting all this together, you obtain

$$\left(ct, x, y, z\right) = \left(\left(\frac{c^2}{g} + \overline{x}\right)\sinh\frac{g}{c}\overline{t},\ \left(\frac{c^2}{g}+\overline{x}\right)\cosh\frac{g}{c}\overline{t} - \frac{c^2}{g},\ \overline{y},\ \overline{z}\right)\ .$$

Edit: I forgot the second part of the question. You are asked:

Show that coordinate stationary clocks cannot remained synchronized.

"Stationary" clocks are those for which $$(\overline{x},\overline{y}, \overline{z}) = const.$$ So, holding $$\overline{x}, \overline{y}, \overline{z}$$ all fixed, vary the coordinate transformations above with respect to $$\overline{t}$$:

$$\frac{\partial t}{\partial \overline{t}} = \left(1 + \frac{g\overline{x}}{c^2}\right)\cosh\frac{g}{c}\overline{t}\ , \quad \frac{\partial x}{\partial \overline{t}} = c\left(1 + \frac{g\overline{x}}{c^2}\right)\sinh\frac{g}{c}\overline{t}\ , \quad \frac{\partial y}{\partial \overline{t}} = \frac{\partial z}{\partial \overline{t}} = 0 \ .$$

The rate at which proper time varies with $$\overline{t}$$ under the same conditions -- this is the inverse time-dilation factor for these clocks -- is then

$$\frac{\partial \tau}{\partial \overline{t}} = \sqrt{\left(\frac{\partial t}{\partial \overline{t}}\right)^2 - \frac{1}{c^2}\left(\left(\frac{\partial x}{\partial \overline{t}}\right)^2 + \left(\frac{\partial y}{\partial \overline{t}}\right)^2 + \left(\frac{\partial z}{\partial \overline{t}}\right)^2\right)} = {1 + \frac{g\overline{x}}{c^2}} \ .$$

For the clocks to remain synchronized, we must have $$\frac{\partial \tau}{\partial \overline{t}}$$ not depend on $$\overline{x}, \overline{y}$$ or $$\overline{z}$$. But we have found a dependency on $$\overline{x}$$. So the clocks do not remain synchronized.

• Thanks a lot! I would be glad to clarify just two issues: 1. When finding the constraint for the simultaneous hyperplane I don't exactly see the step where you demand it to pass through the point $(t, x, y, z) = \left(\frac{c}{g}\sinh\frac{g}{c}\overline{t}, \frac{c^2}{g}\left(\cosh\frac{g}{c}\overline{t} - 1\right), 0, 0\right)$, I can see it does but I don't see how you used it to get the following equation. 2. After reading a bit more about accelerated frames in SR, it seems that conditions (ii) and (iii) are really ways to ask us to construct the Rindler coordinates. Is that correct?
– Amit
Commented Jul 24 at 16:34
• No problem! 1. Any plane perpendicular to the vector $(v_1, \ldots, v_d)$ is a set of points $(x_1, \ldots, x_d)$ satisfying $v_1x_1 + \ldots + v_dx_d = c$. Proof: take any two points on the plane $(a_1, \ldots, a_d)$ and $(b_1, \ldots, b_d)$. These define the vector $(a_1 - b_1, \ldots, a_d - b_d)$ parallel to the plane, and it satisfies $v_1(a_1 - b_1) + \ldots + v_d(a_d - b_d) = 0$. $c$ is then determined by plugging in any point known to lie on the plane. Commented Jul 24 at 17:34
• I edited my answer to go through it a bit more slowly. Commented Jul 24 at 19:27
• Thanks! That's clear now. Just a nitpick to let you know -- I think you got the wrong sign for $\kappa$.
– Amit
Commented Jul 24 at 19:36
• Good catch, thanks! Commented Jul 24 at 19:54

Here is the short and sweet version. You are correct that conditions (ii) and (iii) are basically requesting you to construct the Rindler coordinate system.

Show that coordinate stationary clocks cannot remained synchronized.

In Rindler coordinates the proper acceleration of the Rindler observers is proportional to 1/X where X is the coordinate distance from the origin at coordinate time 0. For two different Rindler observers with different X coordinates, their velocities at any given coordinate time T are different because they have different accelerations. Since they are moving in flat spacetime, their clocks are time dilated according to the SR time dilation factor $$T = t' \sqrt{1-v^2/c^2}$$ and are therefore running at different rates at any instant $$(T \ne 0 )$$, so they obviously cannot remain synchronised. However they measure their proper spatial separation to remain constant and this is called the "comoving distance".

• That's a very nice summary indeed. I think you meant to write conditions (ii) and (iii) -- condition (i) seems to be just the initial condition of the accelerated frame.
– Amit
Commented Jul 24 at 20:32
• Yes, my bad. I'll fix that typo.
– KDP
Commented Jul 25 at 1:19