Equation of accelerated motion in GR

  • Thread starter Decibit
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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
[tex]\frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau} = 0[/tex]
According to the forum thread: https://www.physicsforums.com/threads/accelerated-motion.118435/ one can use for constant acceleration the equation like this:
[tex]\frac{d^2x^\mu}{d\tau^2} + \Gamma^\mu_{\alpha\beta}\frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau} = a^\mu[/tex]
where [itex]a^\mu[/itex] is a constant vector with the timelike being zero.
Let us first consider flat Minkowski space so [itex]\Gamma^\mu_{\alpha\beta}=0[/itex] and the equation of motion takes the form:
[tex]\frac{d^2x^\mu}{d\tau^2} = a^\mu[/tex]
At this point it almost looks like the Newton's equation. But the problem remains: how can I integrate it?

Homework Equations


[tex]\eta=diag(-1,1,1,1)[/tex]
[tex]d\tau=-\eta_{\alpha\beta}dx^\alpha dx^\beta[/tex]
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
[tex]\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}[/tex]
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
[tex]\begin{array}{l}
x^i=a^i\frac{\tau^2}{2}\\
x^0=\tau\end{array}[/tex]
Quick consistency check with ##a^i=(1,0,0)##.
[tex]\begin{array}{l}
x^1=\frac{\tau^2}{2}\\
x^0=\tau\end{array}[/tex]
At ##\tau = 4## the particle has travelled 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 [itex]\frac{d^2x^\mu}{d\tau^2} = a^\mu[/itex] (with a possibility to extend it to non-flat space-time)?
Thanks!
 

Answers and Replies

  • #2
stevendaryl
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If [itex]a^\mu[/itex] is a constant, then you can integrate immediately:

[itex]x^\mu = \frac{1}{2} a^\mu \tau^2[/itex]

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

What you can do is to let the squared magnitude, [itex]A_\mu A^\mu[/itex], be constant. That means (taking a derivative with respect to proper time) that [itex]A_\mu \frac{dA^\mu}{d\tau} = 0[/itex].
 
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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 [itex]A^\mu = \frac{dU^\mu}{d\tau}[/itex] is always "perpendicular" to the 4-velocity.
...
What you can do is to let the squared magnitude, [itex]A_\mu A^\mu[/itex], be constant.
reminded me of something I've seen in a text book:
Bernard Schutz: A first Course in General Relativity said:
§ 2.6: [itex] \vec{a}=\frac{d\vec{U}}{d\tau}\vec{U} \cdot \vec{a} = 0[/itex]
§ 2.9 Exer. 19 A body is said to be uniformly accelerated if its acceleration four-vector [itex]\vec{a}[/itex] has constant spatial direction and magnitude, say [itex]\vec{a} \cdot \vec{a} = \alpha^2 \geqslant 0[/itex]
Working out through the exercise we start with
[tex]\begin{array}{l}
U^\mu U_\mu = 1 \\
A^\mu U_\mu = 0 \\
A^\mu A_\mu = \alpha^2
\end{array}[/tex]
As we consider only one direction we can assume [itex]A^2=A^3=U^2=U^3=0[/itex] This allows us to express
[tex]\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}[/tex]
Which is a system of linear DE with constant coefficients. The solution is straightforward
[tex]\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}[/tex]
Using the MCRF initial conditions [itex]U^0(\tau=0)=1,U^1(\tau=0)=0[/itex] we get
[tex]\begin{array}{l}
U^0=cosh(\alpha \tau)\\
U^1=sinh(\alpha \tau)
\end{array}[/tex]
and after integrating these we get the solution:
[tex]\begin{array}{l}
x^0=\frac{1}{\alpha}sinh(\alpha \tau)\\
x^1=\frac{1}{\alpha}cosh(\alpha \tau)
\end{array}[/tex]

Using [itex]cosh(arcsinh(x))=\sqrt{1+x^2}[/itex] it is easy to show that
[tex] x^1 = \frac{1}{\alpha} \sqrt{1+\alpha^2 {x^0}^2} [/tex]
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!
 
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