# Motion of uniform accelerating rocket in special relativity

I have been trying to learn more about acceleration in special relativity. For the purpose I consulted the great Lev Landau. He has an example about this in his book but one of his equations puzzles me. Consider a rocket accelerating in the positive $x$ direction.The four acceleration for an observer that sees the rocket moving at speed $v$ is given by

$$a^{\mu}=\gamma \frac{d}{dt}(c\gamma,v\gamma)$$

In the instantaneous rest frame of the rocket where $v=0$, we have $$a^{\mu}=(0,a_0,0,0).$$ This instantaneous frame I picture as the passengers of the rocket being pressed to their seats.

Then Landau claims that you could square the expressions for the two rest frames and obtain

$$\frac{d}{dt}\frac{v}{\sqrt{1-\frac{v^2}{c^2}}} = \frac{d}{dt}(\gamma v) = a$$

The right hand side of this equation is clear, obviously it is the square of the four acceleration in the instantaneous rest frame. But how to get the right hand side from $a^{\mu}a_{\mu}$ in the system in which the rocket moves? I would appreciate your help.

EDIT: Some Remarks Written in an explicit form $$a^{\mu}=\gamma^2\left(\frac{\textbf{v}\cdot \textbf{ a}}{c}\gamma ,\frac{\textbf{v}\cdot \textbf{a}}{c^2}\gamma v +\textbf{a } \right)$$

Where $$\frac{d\gamma}{dt}=\frac{\textbf{v}\cdot \textbf{ a}}{c^2}\gamma^3$$

Along these lines

$$a_{\mu}a^{\mu}=\frac{\left(\textbf{v}\cdot \textbf{ a}\right)^2}{c^2}\gamma^6 +\gamma^4a^2 = a_0^2$$

with some vector algebra: $$a_{\mu}a^{\mu}=\frac{\left(\textbf{v}\cdot \textbf{ a}\right)^2}{c^2}\gamma^6 +\gamma^4a^2 = \gamma^6\left(\frac{a^2v^2-(\textbf{a}\times \textbf{v})^2}{c^2} \right)+\gamma^4a^2=a_0^2$$ If $\textbf{a} \parallel \textbf{v}$ the above reduces to $$a_{\mu}a^{\mu}= \gamma^6\left(\frac{a^2v^2}{c^2} \right)+\gamma^4a^2=a_0^2,$$ which is different from what Landau has. I should be getting instead: $$**a_0=\gamma^3a**$$

Answer: $$\gamma^6\left(\frac{a^2v^2}{c^2} \right)+\gamma^4a^2=\gamma^4a^2\left(1+\gamma^2\frac{v^2}{c^2}\right)=\gamma^4a^2\left(\frac{1}{1-\frac{v^2}{c^1}}\right)=\gamma^6a^2$$

Remark: What is an instantaneous rest frame?

Consider a rocket which undergoes constant acceleration along $\hat{x}$. Although the velocity is changing, we consider a frame in which the rocket is instantaneously at rest. In such a frame, the rocket’s 4-acceleration is $(0, a_0 )$. Which is a key concept for this problem.

• If you are considering so-called "instantaneous rest frame", i.e. v=0, then there is no motion whatsoever, so what is the role of SR here? SR applies to situations, where there is not only a difference of speeds between two objects, but also this difference is ... well ... huge ... Aug 7 '16 at 10:28
• @brightmagus I don't really understand what you are trying to say? This is a standard problem with two frames of reference. Aug 7 '16 at 11:56
• I'm not trying anything. I said what I said: if the frames are not moving wrt. to each other (v=0), SR is not applicable. That's "standard" conclusion. Aug 7 '16 at 13:26
• @brightmagus I think that you are getting confused how acceleration is treated in SR and what is instantaneous rest frame. Aug 7 '16 at 15:45
• $\frac{v^2}{c^2}$ may be rewritten as $1-\frac{1}{\gamma^2}$. Try plugging that in. Aug 7 '16 at 16:30

Your confusion is understandable, because Landau used the same symbol for the proper acceleration and the coordinate acceleration. What Landau actually claims is that in the observer frame $$\frac{d}{dt}(\gamma v) = a_0,$$ where I used your notation $a_0 = \sqrt{a^\mu a_\mu}$. This quantity is a four-scalar (which means it is Lorentz invariant) and is called the proper acceleration. You correctly deduced that $a_0 = \gamma^3 a$ (if the acceleration is parallel to the velocity), where $a = dv/dt$ is the ordinary coordinate acceleration (which is not a four-scalar, and therefore depends on the reference frame). Landau then remarks that $a_0$ coincides with $a$ in the instantaneous rest-frame, since in that frame $v=0$ and $\gamma = 1$.

EDIT

\begin{align} \frac{d}{dt}(\gamma v) &= \frac{d\gamma}{dt}v + \gamma\frac{dv}{dt}\\ &= \frac{v/c^2}{(1 - v^2/c^2)^{3/2}}\frac{dv}{dt}v + \frac{1}{(1 - v^2/c^2)^{1/2}}\frac{dv}{dt}\\ &= \frac{v^2/c^2}{(1 - v^2/c^2)^{3/2}}a + \frac{1 - v^2/c^2}{(1 - v^2/c^2)^{3/2}}a\\ &= \frac{1}{(1 - v^2/c^2)^{3/2}}a = \gamma^3 a = a_0. \end{align}

• I don't really understand how to go from $a_0=\gamma^3a \to \frac{d}{dt}(v\gamma)=a_0$. I used $\gamma^3=\frac{d}{dv}(\gamma v)$, but still I have derivative with respect to $v$ and not $t$ as Landau has it. i.e $a\frac{d}{dv}(\gamma v)=a_0$ Aug 8 '16 at 9:08
• @AlexanderCska I've added the calculation. Aug 8 '16 at 12:52

The product $a^\mu a_\mu$ can be written as $a^\mu a^\nu \eta_{\mu\nu}$. This metric (the Minkowski metric) usually carries the signature $(-,+,+,+)$ or $(+,-,-,-)$. Either way, the time coordinate has to have the opposite sign. The product here gives you some nice cancellations when you take the product of

$a^{\mu}=\gamma \frac{d}{dt}(c\gamma,u\gamma)$ --- (here $u$ is the four-velocity)

with itself, calculating the time derivative first, then doing very simple multiplications.

You'll get to a point where you have the dot product equal to the following:

$a^\mu a_\mu = a^2 \gamma_u^6$

Which when expanded using Taylor's Theorem to first order in $u$ gives simply

$a^\mu a_\mu \approx a^2$, as $\gamma_u^6$ is very very small for even relativistic values of $u$.

• @hebetudionus no this is wrong and there are no expansions needed. The expression in Landau-Lifschitz is a precise one! Aug 7 '16 at 10:04
• You're right, if the value you're trying to find is $a_0 = a\gamma^3$. In my notebook it was from a problem where I had to show that $a_\mu a^\mu = a^2$. My mistake. Aug 7 '16 at 16:32