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Law of reflection in SR

  1. Nov 12, 2009 #1

    I'm working through an Excercise in Sean Carrol's spacetime and geometry book. The question asks you to consider an inertial frame S with coordinates [tex] x^\mu=(t,x,y,z) [/tex] and a frame S' with primed coordinates. Which is related to S by a boost v in the y direction. Imagine a wall( or mirror) lying along the line x'=-y'. From the point of view of S, what is the relationship between the angle of incidence (assuming ball travels in x-y plane only) and angle of reflection? Also what is velocity before and after?

    Unfortunately there are not even one word answers let alone solutions in this book, so I don't even know if what I have is correct.

    The way I proceeded, was firstly to consider what angle the mirror would appear at in S. Since the y-direction will be Lorentz contracted, but the x -direction will remain unchanged, the mirror should appear to be at a greater angle than 45 to someone in the S frame. This angle would precisely be:
    [tex] \theta_M=arctan(\gamma \frac{\Delta x'}{\Delta y'})=arctan(\gamma) [/tex], the last equality following from the angle being 45 degrees in the primed frame meaning [tex] \Delta x'=\Delta y' [/tex]

    OK so that is my step 1. Now we need to work out how the incident velocity looks in S.

    e.g. [tex] u_x(init)=\frac{\Delta x}{\Delta y}=\frac{\Delta x'}{\gamma (\Delta t'+v\Delta y')}= \frac{u_x'(init)}{\gamma(1+vu_y'(init))} [/tex] Similarly, [tex] u_y(init)=\frac{u_y'(init)+v}{(1+vu_y'(init))} [/tex].

    From this we can work out the angle of the incident ray (wrt -x-axis) in the S frame, this is therefore [tex] \alpha=\frac{u_y(init)}{u_x(init)} [/tex] (which you can sub into from above). I work out then the actual incident angle, [tex] \theta_I=\alpha+\theta_M-90 [/tex]. (not 100% sure if this is correct)

    Does this look like I'm going down the right path?
  2. jcsd
  3. Nov 12, 2009 #2
    I come out with this monster finally anyway:

    [tex] \theta_r-\theta_i=-2tan(\gamma)+arctan(\frac{u_(y')(i)+vu_(x')(i)u_(y')(i)}{\gamma(u_(x')(i)+v+v[u_(x')(i)]^2+v^2u_(x')(i)})+arctan(\frac{u_(x')(i)+vu_(x')(i)u_(y')(i)}{\gamma(u_(y')(i)+v+v[u_(y')(i)]^2+v^2u_(y')(i)}) [/tex]

    where the [tex]u_(y')(i) [/tex] etc, represents the component of the incident velocity in the S', wrt to y' axis. [tex]\gamma[/tex] is just the Lorentz factor. v is the boost between S and S'.

    Not sure if anyone knows what the expression shouled be for such a shift, and if this looks on the right track?
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