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A Gravitational Interferometer and geodesics

  1. Sep 7, 2016 #1
    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


    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


    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

    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

    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

    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. jcsd
  3. Sep 7, 2016 #2
    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.
  4. Sep 7, 2016 #3
    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.
  5. Sep 7, 2016 #4


    Staff: Mentor

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

  6. Sep 7, 2016 #5
    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
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