# A Gravitational Interferometer and geodesics

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1. Sep 7, 2016

### Salah93

I was trying to solve this excercise:

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 isolate the

$$dt$$ and find the time difference between the two routes. 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?

I want also to ask is there other geodesics that go from the 3d point (0,0,0) to (L,0,0)?

2. Sep 7, 2016

### GeorgeDishman

I'm working beyond my knowledge here (very much a beginner), but if you put $dy=0$ and $dz=0$ into the above, you still have $ds^2 = - dt^2 + (1+ A cos( k(z+t))dx^2$ and if $ds^2 = 0$ for light, you get a relationship between $dt$ and $dx$.

To the admins, I hope that doesn't constitute too much of an answer, I'm hoping to test my own rudimentary knowledge too. I've deliberately left out the final step I think.

3. Sep 7, 2016

### Salah93

I solved the excercise and made all the necessary calculations and obtained the correct results. What I asking is a more theoretical question, to justify what I did. If the question is not formulated well please tell me.

4. Sep 7, 2016

### Staff: Mentor

You will need to show more of your work. In particular you should show explicitly what you did here:

5. Sep 7, 2016

### Salah93

Ok, I write the equations that I obtain(one can use action variation with an affine parameter, or EL eq. with affine parameter or use directly geodesic eqs with affine parameter by first calculating Christoffel symbols):

$$\ddot{t}=\frac{Ak}{2}sin(k(z+t)) (\dot{x}^2-\dot{y}^2)$$

$$\ddot{t}=\frac{Ak \sin(k(z+t))}{1+Acos(k(z+t))} (\dot{z}+\dot{t})\dot{x}$$

$$\ddot{y}=-\frac{Ak \sin(k(z+t))}{1-Acos(k(z+t))} (\dot{z}+\dot{t})\dot{y}$$

$$\ddot{z}=-\frac{Ak}{2}sin(k(z+t)) (\dot{x}^2-\dot{y}^2)$$

Now these eqs define a geodesic.

I know that taking $$ds^2=0$$ this defines a light geodesic(and I can use any monotone function to parametrize it). Now I can set freely $$dy=dz=0$$ and take a light ray that travels along x axis form (0,0,0) to (L,0,0) and then obtain the result of the second point for this particular light ray.

My questions are:
1) if I substitute $$dy=dz=0$$ I correctly have the eq. for $$\ddot{y}$$ zero. But the ones for $$\ddot{t}$$, $$\ddot{x}$$ , $$\ddot{z}$$ are not zero(if I made calculations well). So or I made wrong calculations or I can't use these equations for the light ray.

2)I saw that all the books ,ex. Schultz for RG, that treat interference for light rays take directly I ray that travels along x or y axis (with a gravitational wave propagating in the z direction). Now are there other light rays that can go from (0,0,0) to (L,0,0)?

I hope I posted better the question. Thank you