# Equation of accelerated motion in GR

• Decibit
In summary, the student attempted to integrate the geodesic equation for a constant proper acceleration, but found that the solution did not make sense. He realized that he had made a mistake and solved the equation using MCRF initial conditions.f

#### 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
$$\begin{array}{l} x^i=C^i_0+C^i_1 \tau+a^i\frac{\tau^2}{2}\\ x^0=C^0_0+C^0_1 \tau\end{array}$$
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
$$\begin{array}{l} x^i=a^i\frac{\tau^2}{2}\\ x^0=\tau\end{array}$$
Quick consistency check with ##a^i=(1,0,0)##.
$$\begin{array}{l} x^1=\frac{\tau^2}{2}\\ x^0=\tau\end{array}$$
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!

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
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:
Bernard Schutz: A first Course in General Relativity said:
§ 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 $\vec{a} \cdot \vec{a} = \alpha^2 \geqslant 0$

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

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!

Last edited:
stevendaryl