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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 $$dt$$ 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?

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I think one could use the metric alone to solve the second part.

One of the light signals is emitted at the origin and travels to a mirror located at (L,0,0). so there is no change in $y$ or $z$. Hence setting $dx=dy=0$ and $ds^2=0$ for light ray: $$0=-dt^2 +(1+A\cos kt)dx^2$$ Taking the square root and integrating both sides $\int \frac{dt}{1+A \cos kt}=\int dx$ using the boundary conditions for the $x$ values one should be able to find the travelling time in the reference frame. Similar computation for the other light ray.

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  • $\begingroup$ I did this and calculated the time. What I don't understand is that why I can put $$dy=dz=0$$ without any problem. And why this light signal with $$dy=dz=0$$ is not a geodesic $\endgroup$ – Saladino Sep 1 '16 at 18:59
  • $\begingroup$ You set them zero because you assume $y$ and $z$ are not changing along the path you consider for the calculation. $\endgroup$ – Statics Sep 1 '16 at 21:18
  • $\begingroup$ For null geodesics it is not quite straightforward, see here $\endgroup$ – Statics Sep 1 '16 at 21:23
  • $\begingroup$ Where it is explained that I can set dy=dz=0? $\endgroup$ – Saladino Sep 1 '16 at 22:49
  • $\begingroup$ Just look at your movement. It is restricted to the x-axis in the first case. Why would you need to consider the change in $y$ or $z$ for that? $\endgroup$ – Statics Sep 2 '16 at 10:31

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