Acceleration in special relativity

[Markus Kahn]

Homework Statement

Consider a spaceship that is accelerating with a constant acceleration $a'$ (in its rest frame) along the $x$-direction. Assume that it launches at $t = \tau = 0$ and determine the proper time of the spaceship as a function of coordinate time $\tau (t)$ and vice versa.

Relevant Equations

We weren't given any in particular...

I'm struggling in the details of this exercise. Let $S'$ be the reference frame where the acceleration of the spaceship is constant, in which case we have $u'(t')= a' t'$ (since we assume no acceleration at the beginning). The rest frame of the rocket $S$ is connected to $S'$ via a Lorentz transformation $\Lambda$.

1. My first naive attempt to solve this exercise was to use the definition of proper time and just integrate it $$ \tau (t') = \int_0^{t'} \sqrt{1- \left(\frac{u'(\lambda)}{c}\right)^2} d\lambda.$$Turns out I have no clue how to do this and even if, I would still need to substitute $t'(t,x)$ and hope to obtain a reasonable result only depending on $t$ and $x$, which seems rather unlikely.
2. The only other possibility I could think of to approach this exercise was to start looking at an acceleration $a$ and see how it transforms under a Lorentz-transfomration. Let $v$ be a constant velocity between two reference frames $S$ and $S'$, with velocities $u$ and $u'$ respectively. We then have $$a' = \frac{du'}{dt'} = \frac{d}{dt'} \frac{u-v}{1- \frac{uv}{c^2}} = \frac{dt}{dt'}\frac{d}{dt}\frac{u-v}{1- \frac{uv}{c^2}} = \frac{1}{\gamma}\frac{d}{dt}\frac{u-v}{1- \frac{uv}{c^2}}=\frac{1}{\gamma}\left(\frac{c^2 \left(c^2-v^2\right) }{\left(c^2-v u(t)\right)^2}\right)a,$$ where I just wrote $a\equiv d u /dt$. The problem from here on is that I don't now what values I need to chose to reproduce the situation in the exercise...

I obviously tried googling the solution and found some texts addressing it (see bottom of text), but none of them explained it in a way that I could understand. Section 1.1 of this text uses an approach that seems to resemble my first one, but there signs are different and in general all expression seem to have a slightly different form... (eq. 4 onwards )

Can someone maybe help me on how to approach this exercise?

Some text addressing this exercise:

• https://www.dpmms.cam.ac.uk/~stcs/courses/dynamics/lecturenotes/section6.pdf
• Section 1.1

 

[PeroK]

1. The only other possibility I could think of to approach this exercise was to start looking at an acceleration $a$ and see how it transforms under a Lorentz-transfomration. Let $v$ be a constant velocity between two reference frames $S$ and $S'$, with velocities $u$ and $u'$ respectively. We then have $$a' = \frac{du'}{dt'} = \frac{d}{dt'} \frac{u-v}{1- \frac{uv}{c^2}} = \frac{dt}{dt'}\frac{d}{dt}\frac{u-v}{1- \frac{uv}{c^2}} = \frac{1}{\gamma}\frac{d}{dt}\frac{u-v}{1- \frac{uv}{c^2}}=\frac{1}{\gamma}\left(\frac{c^2 \left(c^2-v^2\right) }{\left(c^2-v u(t)\right)^2}\right)a,$$ where I just wrote $a\equiv d u /dt$.

This is a good start. In this case, we have a constant $a'$ in a continuous sequence of frames where the ship is instantaneously at rest. As the speed of the ship is changing, we have a coordinate acceleration that depends on $t$.

Note that $u' =0$ in this case and $u = v$, which should help simplify things.

I.e. we have:

$a(t) =$ some function of $a'$ and $v(t)$ (the instantaneous speed of the ship)

Where $a(t) = \frac{dv}{dt}$

 

[Markus Kahn]

If we assume $u=v$ we end up with
$$a' = \frac{1}{\gamma} \frac{c^2}{(c^2-v^2)} a(t) \Longleftrightarrow a(t)=\frac{d v}{dt}=\frac{c^2-v^2}{c^2}\gamma a' .$$

I'm sorry, but I have absolutely no idea what I'm supposed to do now... For me it looks like we are just manipulating some equations that have vaguely something to do with what I would like to figure out... What are the logical steps here and what am I trying to find in order to obtain $\tau(t)$ and vice versa?

If I had to guess I'd probably try the following: we solve the last equation for $dt$ and integrate then.
$$ \int \frac{c^2}{a'} \sqrt{1-\frac{v^2}{c^2}} dv = \int dt = t.$$
This seems kind of weird, especially since I don't really think there is an analytical solution for the lhs, or at the very least I don't know how to calculate it...

Sorry that I can't contribute here... Could you maybe help me a little bit more?

 

[PeroK]

I'm not sure what the problem is as this is all standard mathematical physics.

If you find $v(t)$ you have $\gamma(t)$ which relates $dt$ and $d\tau$.

 

[PeroK]

Ps check your working. You should have a factor of $\gamma^3$ relating $a$ and $a'$.

 

[Markus Kahn]

Alright, I'll go trough everything again tomorrow. Thanks for the help!

 

[Markus Kahn]

@PeroK
Alright, so another try. Let $u(t)$ and $u'(t')$ be the velocities of the rocket in the observers and the rest frame of the rocket respectively and let $v$ be the constant velocity between the the two frames. We then have
$$\begin{align*}a'
&= \frac{du'}{dt'} = \frac{dt}{dt'} \frac{du'}{dt} = \frac{dt}{dt'} \frac{d}{dt} \frac{u-v}{1-\frac{uv}{c^2}}.
\end{align*}$$

• Since $t' = \gamma(t-\frac{v}{c^2}x)$ we have $$\frac{dt}{dt'}=\frac{d}{dt'} \left(\frac{t'}{\gamma} + \frac{v}{c^2}x\right) = \frac{1}{\gamma}$$
• and here we have with the product rule $$\begin{align*} \frac{d}{dt} \frac{u-v}{1-\frac{uv}{c^2}} &= \frac{u' (1-\frac{vu}{c^2}) - (u-v) (-\frac{u'v}{c^2}) }{(1-\frac{uv}{c^2})^2} =\frac{(1-\frac{vu}{c^2}) + (\frac{uv}{c^2}-\frac{v^2}{c^2}) }{(1-\frac{uv}{c^2})^2}u' \\&= \frac{1-\frac{v^2}{c^2} }{(1-\frac{uv}{c^2})^2}u' =\frac{1}{(1-\frac{uv}{c^2})^2} \frac{u'}{\gamma^2}.\end{align*}$$

So everything together gives
$$a' =\frac{1}{(1-\frac{uv}{c^2})^2} \frac{a}{\gamma^3} $$

The thing that confuses me now is that if we set $u=v$ we will find
$$a' = \frac{a}{\gamma^3} \frac{1}{(1-\frac{v^2}{c^2})^2} = \frac{a}{\gamma^3} \gamma^4 = a\gamma \quad\Longleftrightarrow\quad a =\frac{a'}{ \gamma}.$$
Is this correct? It almost seems to simple to be true...

Assuming this to be correct, we have
$$ a=\frac{dv}{dt} = a'/\gamma \quad \Longleftrightarrow \quad \int_{v_0}^v \sqrt{1-\frac{v^2}{c^2}}dv = a'(t-t_0).$$
The result of this integral is not really nice and doesn't seem to have anything to do with the solution that I'm expecting (the solution presented in my opening post in the links).. So once again, I'm doing something wrong... Can you maybe point me to my mistake?

 

[PeroK]

Here's the mistake:

• Since $t' = \gamma(t-\frac{v}{c^2}x)$ we have $$\frac{dt}{dt'}=\frac{d}{dt'} \left(\frac{t'}{\gamma} + \frac{v}{c^2}x\right) = \frac{1}{\gamma}$$

The fundamental problem there is that you took $\frac{dx}{dt'} = 0$.

It was simpler just to do:

• Since $t' = \gamma(t-\frac{v}{c^2}x)$ we have $$\frac{dt'}{dt}=\gamma(1 - \frac{vu}{c^2})$$

Also, it seems better to me to aim for an expression for $a$ in terms of $a'$. It all comes out the same, of course, but it seemed more logical to me to start with:

$a = \frac{du}{dt} = \dots$

 

[PeroK]

• and here we have with the product rule $$\begin{align*} \frac{d}{dt} \frac{u-v}{1-\frac{uv}{c^2}} &= \frac{u' (1-\frac{vu}{c^2}) - (u-v) (-\frac{u'v}{c^2}) }{(1-\frac{uv}{c^2})^2} =\frac{(1-\frac{vu}{c^2}) + (\frac{uv}{c^2}-\frac{v^2}{c^2}) }{(1-\frac{uv}{c^2})^2}u' \\&= \frac{1-\frac{v^2}{c^2} }{(1-\frac{uv}{c^2})^2}u' =\frac{1}{(1-\frac{uv}{c^2})^2} \frac{u'}{\gamma^2}.\end{align*}$$

So everything together gives
$$a' =\frac{1}{(1-\frac{uv}{c^2})^2} \frac{a}{\gamma^3} $$

PS all those $u'$ should be $a'$.

 

[PeroK]

Here's my working:

Spoiler

$a = \frac{du}{dt} = \frac{du/dt'}{dt/dt'}$

$t = \gamma(t' + vx'/c^2) \ \Rightarrow \ \frac{dt}{dt'} = \gamma(1 + vu'/c^2)$

$\frac{du}{dt'} = \frac{d}{dt'}(\frac{v + u'}{1 + vu'/c^2}) = \frac{a'}{1 + vu'/c^2} - \frac{(v + u')}{(1 + vu'/c^2)^2}(\frac{va'}{c^2}) = \frac{1}{(1 + vu'/c^2)^2}(1 - v^2/c^2)a' = \frac{a'}{\gamma^2(1 + vu'/c^2)^2}$

Putting this together gives:

$a = \frac{a'}{\gamma^3(1 + vu'/c^2)^3}$

In this case we set $u' = 0$ to get $a = \frac{a'}{\gamma^3}$

 

[Markus Kahn]

Alright, this took me way longer than it should have, but I think I've got it now... Again, I did a slightly different way $\frac{du}{dt} = \frac{dt'}{dt}\frac{du}{dt'}$ instead of $\frac{du}{dt} = \frac{du/dt'}{dt/dt'}$, but I arrive at the same result. Thank you for your patience!

But just to make sure that I understand what we are doing here: What we have found now is the acceleration $a$ that an external observer in a reference frame $S$ would measure if the rocket was accelerating in a reference frame $S'$ with constant acceleration $a'$. Is this correct?

Our goal is now to obtain $v(t)$ right? For this I would need to solve the Differential equation from above.

I don't really know how to solve
$$\frac{dv}{dt} = \frac{a'}{(1-\frac{v^2}{c^2})^3}$$
but by fixing $v(0)=0$ as boundry condition, wolfram alpha gives
$$a c^2 t = \frac{3}{8} c^3 \sin^{-1}(v(t)/c) + \frac{1}{8} v(t) (5 c^2 - 2 v(t)^2) \sqrt{1 - v(t)^2/c^2},$$
which indicates that there is no closed form for $v(t)$. So how does one proceed to find the solution for this DE?

 

[PeroK]

Alright, this took me way longer than it should have, but I think I've got it now... Again, I did a slightly different way $\frac{du}{dt} = \frac{dt'}{dt}\frac{du}{dt'}$ instead of $\frac{du}{dt} = \frac{du/dt'}{dt/dt'}$, but I arrive at the same result. Thank you for your patience!

That's just the same thing, using $dt'/dt = \frac{1}{dt/dt'}$

But just to make sure that I understand what we are doing here: What we have found now is the acceleration $a$ that an external observer in a reference frame $S$ would measure if the rocket was accelerating in a reference frame $S'$ with constant acceleration $a'$. Is this correct?

It's more general than that. $a'$ does not have to be constant. The formula relates the acceleration (for 1D motion) between reference frames. Note that you can do the same thing for the other components of velocity and acceleration to get a more general acceleration transformation for 3D motion (under a Lorentz boost).

If the particle is instantaneously at rest in S', then we have the simplification using $u' =0$. Note that these formulas relate instantaneous, time-dependent quantities in general.

The special case we have here is that $a'$ is constant and S' is the instantaneous rest frame of the ship.

I don't really know how to solve
$$\frac{dv}{dt} = \frac{a'}{(1-\frac{v^2}{c^2})^3}$$
but by fixing $v(0)=0$ as boundry condition, wolfram alpha gives
$$a c^2 t = \frac{3}{8} c^3 \sin^{-1}(v(t)/c) + \frac{1}{8} v(t) (5 c^2 - 2 v(t)^2) \sqrt{1 - v(t)^2/c^2},$$
which indicates that there is no closed form for $v(t)$. So how does one proceed to find the solution for this DE?

Okay, so there might be a trick! What about looking at $\frac{d}{dt}({\gamma v})$?

There's a general rule of thumb in SR. If it doesn't work out with velocity, try energy-momentum. In this case the momentum (per unit mass) is $\gamma v$. So, it's worth trying the time derivative of momentum.

 

[Markus Kahn]

Thanks for the hint, but this time I didn't work because I again didn't pay enough attention to the details... The differential equation, with the same boundary condition, is actually
$$\frac{dv}{dt} = {\left(1-\frac{v^2}{c^2}\right)^{3/2}}a'\quad\Longrightarrow \quad \frac{v}{\sqrt{1-\frac{v^2}{c^2}}}= a't.$$

From this we find
$$v(t)= \frac{a' c t}{\sqrt{{a'}^2 t^2+c^2}}\quad \Longrightarrow \sqrt{1-v^2/c^2} = \frac{1}{\sqrt{1+\frac{{a'}^2t^2}{c^2}}}$$
We therefore find
$$\tau(t) = \int \frac{dt}{\sqrt{1+\frac{{a'}^2t^2}{c^2}}} = \frac{c}{a'}\sinh^{-1}\left(\frac{a't}{c}\right).$$

Thank you so much for the help, I really appreciate it! Just out of curiosity, what exactly was your idea with the $\gamma v$? I tried taking the derivative of it, but in the end I just got a way more complicated expression..

 

[PeroK]

Thank you so much for the help, I really appreciate it! Just out of curiosity, what exactly was your idea with the $\gamma v$? I tried taking the derivative of it, but in the end I just got a way more complicated expression..

The derivative of $\gamma$ is quite important in SR kinematics. You should get:

$\frac{d}{dt}\gamma = \frac{d}{dt}(1 - v^2/c^2)^{-1/2} = \gamma^3(v/c^2)\frac{dv}{dt} = \gamma^3(v/c^2)a$

If we now include the mass of the object, we have:

$\frac{dp}{dt} = \frac{d}{dt}(\gamma mv) = \gamma^3(mv^2/c^2)a + \gamma ma = \gamma^3 ma (v^2/c^2 + 1/\gamma^2) = \gamma^3 ma$

And that's quite an important result, because if we use the formula for the accelerations above we get:

$\frac{dp}{dt} = ma'$

If we now define the three-force, $F$, as the rate of change of momentum, then we have:

$F = \frac{dp}{dt} = ma' = F'$

Where $F'$ is measured in the rest frame of the particle.

Also, as $E = \gamma mc^2$, we have $\frac{dE}{dt} = \gamma^3(v/c^2)amc^2 = Fv$

And, in fact, this generalises to $\frac{dE}{dt} = \vec{F}.\vec{v}$

In any case, it's important to remember how to do these derivatives.