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Equation of accelerated motion in GR

  1. Apr 25, 2016 #1
    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.
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    [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.

    3. The attempt at a solution
    If I use ##\tau## as an independent variable the result doesn't make any sense
    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
    Quick consistency check with ##a^i=(1,0,0)##.
    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)?
  2. jcsd
  3. Apr 26, 2016 #2


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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].
  4. Apr 30, 2016 #3
    Great answer! Thanks!
    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
    reminded me of something I've seen in a text book:
    Working out through the exercise we start with
    U^\mu U_\mu = 1 \\
    A^\mu U_\mu = 0 \\
    A^\mu A_\mu = \alpha^2
    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
    A^0=\frac{dU^0}{d\tau}=\alpha U^1\\
    A^1=\frac{dU^1}{d\tau}=\alpha U^0
    Which is a system of linear DE with constant coefficients. The solution is straightforward
    U^0=C_1 e^{\alpha \tau} + C_2 e^{-\alpha \tau}\\
    U^1=C_1 e^{\alpha \tau} - C_2 e^{-\alpha \tau}
    Using the MCRF initial conditions [itex]U^0(\tau=0)=1,U^1(\tau=0)=0[/itex] we get
    U^0=cosh(\alpha \tau)\\
    U^1=sinh(\alpha \tau)
    and after integrating these we get the solution:
    x^0=\frac{1}{\alpha}sinh(\alpha \tau)\\
    x^1=\frac{1}{\alpha}cosh(\alpha \tau)

    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!
    Last edited: Apr 30, 2016
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