1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: General relativity - gravitational wave/geodesic problem?

  1. May 29, 2010 #1
    Imagine a + polarized gravitational wave propagating in the z direction. A particle in the field of this wave has the following lagrangian:
    [tex] L = -c^2(\frac{dt}{d\tau})^2 + (1 + hcos(wt))(\frac{dx}{d\tau})^2 + (1 - hcos(wt))(\frac{dy}{d\tau})^2 + (\frac{dz}{d\tau})^2[/tex]

    where tau is the proper time in the particle's frame, x,y,z & t are the spacetime coordinates used to describe the particle's trajectory, h is a number far smaller than 1 and w is the wave's angular frequency.

    the problem starts off by asking me to show that there are 3 constants of the motion. The Euler-Lagrange equation yields the constants A, B and C:

    [tex]A = \frac{dz}{d\tau}[/tex],
    [tex]B = (1 + hcos(wt))\frac{dx}{d\tau}[/tex],
    [tex]C = (1 - hcos(wt))\frac{dy}{d\tau}[/tex].

    The problem then gets me to prove that [tex] \frac{dy}{dx} = M(1 + 2hcos(wt))[/tex], which it turns out is done with the combination of the [tex] \frac{dy}{d\tau} [/tex] and [tex] \frac{dy}{d\tau} [/tex] expressions above, and the use of 1st order power series expansions. (It turns out that the constant M = C/B.)

    Now i've reached the last part where I'm not sure how to proceed - this is surely the whole point of doing all the previous calculations and is the bit I am actually supposed to learn something from.

    I am asked "Is a straight line trajectory in the x direction a geodesic here? Is a straight line trajectory along the line x=y a geodesic?" If the answer to either of these is no, I am also asked to sketch a geodesic as h tends to zero.

    What I'm thinking is, I'm looking at the dy/dx expression and seeing that it's non-zero. Is that telling us that you cannot move in the x-direction without moving in the y-direction, so that you cannot move in a straight line trajectory in the x-direction?
    Also, the question is saying that the x-direction is perpendicular to the wave propagation direction (the z-direction). I'm not sure what bearing this has, because if space is distorted, surely that could affect all directions, right? I also have no idea what I would sketch either.

    I really appreciate any answers, getting to this stage has taken me ages of mindless mathematics but I don't feel like I've learned any physics yet.
    Can anyone help out?

  2. jcsd
  3. Jun 3, 2010 #2
    I may be confused here, but what is a straight line in a curved space? From my understanding, there are no more straight lines, but geodesics. So the way I see it, the problem is asking you: is a geodesic a geodesic?

    Now, if we "define" a straight line to be something parametrized as:

    x = at+b \;
    y = ct + d

    I would say the answer is a blatant no...

    I'll get some new ideas when I go outside! I'll be back soon!
    Last edited: Jun 3, 2010
  4. Jun 3, 2010 #3
    Is there any hope of solving the geodesic equation of motion? If so, that could prove your point and allow you to plot geodesics.

    I have another idea, try plotting an x-y, we see that geodesics for a constant time "t0" have the derivative:

    \frac{dy}{dx} = M \left[ 1 + 2h cos\left(\omega t_{o}\right) \right]

    You can at least draw the "geodesics at constant time," though a moving particle will change the "constant" t0 as time goes by.

    As for the straight line geodesics, for an x directed one dy = 0 so it is a geodesic if M is also 0. The y=x one is not a geodesic since assuming it is will give you:

    1 = M \left[ 1 + 2h cos\left(\omega t\right) \right]

    which is just not true, since M is constant.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook