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Horizontal photon in a vertical lift

  1. Oct 18, 2014 #1
    supposing a constant acceleration Vertically a horizontal incoming photon is considered. We know that in the frame of the lift the vertical speed of the photon will increase, hence its horizontal one will decrease and then the photon trajectory has a vertical asymptote ?
     
  2. jcsd
  3. Oct 18, 2014 #2

    Dale

    Staff: Mentor

    This is not that hard to calculate directly. It is a good exercise to go through. Start with the world line of a pulse of light in the y direction. Then look up the transformation to Rindler coordinates. Apply the equations, and what is the result?
     
  4. Oct 18, 2014 #3
    I got [tex]y(x)=c^2/a*cosh(ax/c^2)[/tex] is this correct ?
     
  5. Oct 19, 2014 #4
    I found this by putting a the acceleration, finding the speed through the formula d(gamma v)/dt=a, then I have the orthogonal speed by the relativistic postulate that the speed of light is constant, hence vx=sqrt(c^2-vy^2). By integration I find the x and y coordinate in function of time, then it's just to express y as a function of x.

    Well, this means the relativistic photon "falls" quicker than a parabola. My idea is now to compare this result with the general relativistic approach.
     
  6. Oct 19, 2014 #5

    Dale

    Staff: Mentor

    I don't know. I got a different formula, but they may be equivalent.

    In Minkowski coordinates in units where c=1 ##(T,X,Y,Z)=(T,X_0,T,Z_0)##. Transforming to the Rindler frame http://en.wikipedia.org/wiki/Rindler_coordinates we get
    ##(x \sinh(gt), x \cosh(gt),y,z)=(x \sinh(gt),X_0,x \sinh(gt), Z_0)##
    which simplifies to
    ##(t,x,y,z)=(t,X_0 \text{sech}(gt), X_0 \tanh(gt),Z_0)##
     
  7. Oct 21, 2014 #6
    The result i got is probably wrong since this one is ##y(x)=\sqrt(x_0^2-x^2)##. It is weird that the curve does not depend on the acceleration
     
  8. Jan 24, 2015 #7
    I tried a general relativistic approach and started with the metric (A(y),-1,-B(y)) for dt,dx,dy ( I suppose from symmetry that there can be only contraction depending in y where the acceleration is.

    I got in vaccum a single equation . R_mu,nu=0 => B'/B=(A'^2-2AA'')/(AA')

    hence there apparently exist an infinite number of solution, it suffices to choose A(y) a function and we can find B(y) such that it is a solution of the field equation ??
     
  9. Jan 24, 2015 #8

    Nugatory

    User Avatar

    Staff: Mentor

    You don't have to go back to the EFE; this is a flat space-time, so it is a vacuum solution of the EFE no matter what coordinate transformations you make. All you need to do is apply the tensor coordinate transformation to the metric written in Minkowski coordinates and you'll have the metric in your new coordinates.
     
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