Equation of accelerated motion in GR

[Decibit]

Hello forum members,
I decided to post in the homework section because my question seems very basic to me. Still I'm getting stuck with it and would appreciate any help.

Homework Statement

I am teaching myself foundations of GR with the goal of simulating numerically some motion in flat and curved space-time. So I start with the geodesic equation
\frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau} = 0
According to the forum thread: https://www.physicsforums.com/threads/accelerated-motion.118435/ one can use for constant acceleration the equation like this:
\frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau} = a^\mu
where a^\mu is a constant vector with the timelike being zero.
Let us first consider flat Minkowski space so \Gamma^\mu_{\alpha\beta}=0 and the equation of motion takes the form:
\frac{d^2x^\mu}{d\tau^2} = a^\mu
At this point it almost looks like the Newton's equation. But the problem remains: how can I integrate it?

Homework Equations

\eta=diag(-1,1,1,1)
d\tau=-\eta_{\alpha\beta}dx^\alpha dx^\beta
Greek indices run from 0 to 3, latin indices are from 1 to 3.

The Attempt at a Solution

If I use $\tau$ as an independent variable the result doesn't make any sense
\begin{array}{l}<br /> x^i=C^i_0+C^i_1 \tau+a^i\frac{\tau^2}{2}\\<br /> x^0=C^0_0+C^0_1 \tau\end{array}
Where $C^\mu_0$ and $C^\mu_1$ are some constants. I suppose that we can fix these constants at $C^\mu_0 = 0$ , $C^i_1=0$ and $C^0_1=1$ if we start at MCRF. So the equations are further simplified to
\begin{array}{l}<br /> x^i=a^i\frac{\tau^2}{2}\\<br /> x^0=\tau\end{array}
Quick consistency check with $a^i=(1,0,0)$.
\begin{array}{l}<br /> x^1=\frac{\tau^2}{2}\\<br /> x^0=\tau\end{array}
At $\tau = 4$ the particle has traveled 8 units of lengths and 4 units of time. As FTL travel is not possible I'm sure that I'm making a mistake. So, what is the proper way to integrate \frac{d^2x^\mu}{d\tau^2} = a^\mu (with a possibility to extend it to non-flat space-time)?
Thanks!

 

[stevendaryl]

If a^\mu is a constant, then you can integrate immediately:

x^\mu = \frac{1}{2} a^\mu \tau^2

However, it is not actually possible for an object to maintain a constant proper acceleration. You can have a constant magnitude for the acceleration.

For a slower-than-light object, the 4-velocity satisfies: U^\mu U_\mu = c^2 (or -c^2, depending on the convention). If you take a derivative with respect to proper time, you find that: U_\mu \frac{dU^\mu}{d\tau} = 0. So the 4-acceleration A^\mu = \frac{dU^\mu}{d\tau} is always "perpendicular" to the 4-velocity. That relation can't be maintained if A^\mu is constant and U^\mu is not.

What you can do is to let the squared magnitude, A_\mu A^\mu, be constant. That means (taking a derivative with respect to proper time) that A_\mu \frac{dA^\mu}{d\tau} = 0.

 

[Decibit]

Great answer! Thanks!

However, it is not actually possible for an object to maintain a constant proper acceleration. You can have a constant magnitude for the acceleration.

This was probably the whole point I was missing! So, my initial equation was wrong. No wonder that the answer didn't make sense.

Another statement you've said, namely

So the 4-acceleration A^\mu = \frac{dU^\mu}{d\tau} is always "perpendicular" to the 4-velocity.
...
What you can do is to let the squared magnitude, A_\mu A^\mu, be constant.

reminded me of something I've seen in a textbook:

§ 2.6: \vec{a}=\frac{d\vec{U}}{d\tau}\vec{U} \cdot \vec{a} = 0
§ 2.9 Exer. 19 A body is said to be uniformly accelerated if its acceleration four-vector \vec{a} has constant spatial direction and magnitude, say \vec{a} \cdot \vec{a} = \alpha^2 \geqslant 0

Working out through the exercise we start with
\begin{array}{l}<br /> U^\mu U_\mu = 1 \\<br /> A^\mu U_\mu = 0 \\<br /> A^\mu A_\mu = \alpha^2<br /> \end{array}
As we consider only one direction we can assume A^2=A^3=U^2=U^3=0 This allows us to express
\begin{array}{l}<br /> A^0=\frac{dU^0}{d\tau}=\alpha U^1\\<br /> A^1=\frac{dU^1}{d\tau}=\alpha U^0<br /> \end{array}
Which is a system of linear DE with constant coefficients. The solution is straightforward
\begin{array}{l}<br /> U^0=C_1 e^{\alpha \tau} + C_2 e^{-\alpha \tau}\\<br /> U^1=C_1 e^{\alpha \tau} - C_2 e^{-\alpha \tau}<br /> \end{array}
Using the MCRF initial conditions U^0(\tau=0)=1,U^1(\tau=0)=0 we get
\begin{array}{l}<br /> U^0=cosh(\alpha \tau)\\<br /> U^1=sinh(\alpha \tau)<br /> \end{array}
and after integrating these we get the solution:
\begin{array}{l}<br /> x^0=\frac{1}{\alpha}sinh(\alpha \tau)\\<br /> x^1=\frac{1}{\alpha}cosh(\alpha \tau)<br /> \end{array}

Using cosh(arcsinh(x))=\sqrt{1+x^2} it is easy to show that
x^1 = \frac{1}{\alpha} \sqrt{1+\alpha^2 {x^0}^2}
This is consistent with the derivation here http://www.physicspages.com/2011/05/25/acceleration-in-special-relativity/ So I'm quite happy now.
I think that I can figure out arbitrary acceleration direction using matrix transformations.
Thanks again for your excellent counseling!