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  Laser interferometric gravitational wave detectors

+ 1 like - 0 dislike
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so I was trying to solve this excercise:

Ex

Now I was able to find the eq. of geodetics (or directly by Christoffel formulas calculation or by the Lagrangian for a point particle). And I verified that such space constant coordinate point is a geodetic.

Now, for the second point I considered

$$ds^2=0$$

to separate the

$$dt=0$$

and find the separation time. But I don't know how to solve for a generic path of a light ray. So I considered that maybe the text wants a light ray travelling along x axis and the second along y axis.

I checked in other sources and all people make the same, by considering a light ray along x-axis and then setting

$$dy=dz=0$$

.

But when I substitute these in my geodesic equations it turns out that they are not true even at first order in A! So these people that consider a light ray travelling along x-axis, such as in an interferometer, are not considering a light geodesic. All of this if and only if my calculations are true.

So I know that if

$$ds^2=0$$

I have a light geodesic. And so it should solve my eq. of geodesics. But if I restrain my motion on x axis what I can say is that the

$$ds^2=0$$

condition now is on a submanifold of my manifold. So, the light wave that I consider doesn't not move on a geodesic of the original manifold but on one of the x axis. This is the only thing that came in my mind.

Is there any way to say that I can set

$$dy=dz=0$$

without worring? And if I can't set it how can I solve the second point?

asked Sep 1, 2016 in Theoretical Physics by anonymous [ no revision ]
recategorized Sep 1, 2016 by Dilaton

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