# General relativity - gravitational wave/geodesic problem?

1. May 29, 2010

### jeebs

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

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:

$$A = \frac{dz}{d\tau}$$,
$$B = (1 + hcos(wt))\frac{dx}{d\tau}$$,
and
$$C = (1 - hcos(wt))\frac{dy}{d\tau}$$.

The problem then gets me to prove that $$\frac{dy}{dx} = M(1 + 2hcos(wt))$$, which it turns out is done with the combination of the $$\frac{dy}{d\tau}$$ and $$\frac{dy}{d\tau}$$ 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?

thanks.

2. Jun 3, 2010

### rhoparkour

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
3. Jun 3, 2010

### rhoparkour

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.