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45 dergree mirrors in Special Relativity

  1. Nov 6, 2009 #1
    45 degree mirrors in Special Relativity

    If a moving 45 degree mirror is length contracted doesn't that mean a ray hitting it from the front will appear (from the stationary observer) to contradict the laws of reflection. ie the angle of incidence will not be equal to the angle of reflection. That is, the mirror will appear to be sitting at say something like 30 degrees from the vertical (depending on speed) but still deflect a horizontral ray straight up !

    Likewise if the moving mirror DOES appear to be at 45 degrees after length contraction, the ray will NOT go straight up but rather deflect slightly behind the mirror :-)

    Is this true ? And does SR have any problems with this ?
    Or is it just an example of relativistic aberration ?
    Last edited: Nov 6, 2009
  2. jcsd
  3. Nov 7, 2009 #2
    Actually it worse than that :P

    In the rest frame of the 45 degree mirror the ray is reflected directly upwards, but to an observer that sees the mirror moving to the right relative to him, the mirror is at 30 degrees from the vertical and the ray is deflected forward (to the right) but still hits the same target above the mirror that is comoving the mirror. So yes, the laws of reflection in relativity are different from the classical ones.

    SR does not have a problem with this as long as the all observers measure the speed of light to be constant and observe events to happen in the same sequence and the laws of physics are the same in all reference frames. If a "law of physics" is not the same in all reference frames, then it not a law of physics and we have to modify it, until it is!
    Last edited: Nov 7, 2009
  4. Nov 7, 2009 #3


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    So the angle of incidence is still equal to the angle of reflection, isn't it?
  5. Nov 7, 2009 #4
    No, to the observer that sees the length contracted mirror at 30 degrees from the vertical, the angle of incidence of the horizontal ray is 30 degrees and the angle of reflection is greater than 30 degrees, possibly greater than 90 degrees but I have not calculated it.

    Also, I imagine that if you throw a ball horizontally at a 45 degree surface that is moving away from you, that it will bounce off at angle that is greater than 45 degrees.
  6. Nov 7, 2009 #5
    [EDIT] I tried to find a formula on the internet but had luck. I am pretty sure if an angled mirror is moving away from the light source, the angle of reflection will be greater than the angle of incidence and if the mirror is moving towards the source, the angle of reflection will be less than the angle of incidence.
  7. Nov 7, 2009 #6


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    Yes that makes sense, since it has to hit something moving with the mirror above it.

    Yet when you think about it, even in classical mechanics (ball & 45°-wall) the rule: incidence angle = reflection angle applies only the rest frame of the wall. So, are the rules of reflection really that different in SR?
    Last edited: Nov 7, 2009
  8. Nov 7, 2009 #7

    Hans de Vries

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    - The angle of reflection is also a function of the velocity.
    - Non simultaneity changes the direction of the wave front.

    I did work this out with an example which involves reflections at all
    possible angles, The moving spherical mirror clock:

    A light flash at the center of a hollow sphere with a reflective inner
    surface is reflected and refocused to the center periodically and
    constitutes an elementary clock.

    See section 4.13 here:

    Figure 4.22 shows how the angle of reflection is changed by the velocity.

    Sections 4.16 & 4.17 discuss how the light wave front rotates
    as a result of non-simultaneity.

    Regards, Hans
  9. Nov 7, 2009 #8
    A beautifully crafted and presented document Hans!

    I thought I had better quantify the above statement.

    The aberration formula for a ray of light emitted at an angle [itex]\theta '[/itex] as measured in the moving frame, by a source moving with velocity v, the angle of the ray [itex]\theta[/itex] in the stationary frame is:

    [tex] \theta = - 2*\tan^{-1}\left(\frac{tan(\pi-\theta ')}{2}\right) * \sqrt{\frac{(1-v)}{(1+v)}}+\pi[/tex] (Adapted from http://www.mathpages.com/rr/s2-05/2-05.htm)

    where all angles are in radians and are measured clockwise from the negative x axis and v is the is in the positive x direction.

    It is farly easy to work out that the horizontal velocity required for the angle of the mirror to change from 45 degrees (Pi/4 radians) to 30 degrees from the vertical is 0.8165c. Plugging these numbers in gives:

    [tex] \theta = - 2*\tan^{-1}\left(\frac{tan(\pi-\pi/4)}{2}\right) * \sqrt{\frac{(1-0.8165)}{(1+0.8165)}}+\pi [/tex]

    = 2.52611 radians = aprox 144.7 degrees.

    The normal of the moving mirror is 30 degrees so the angle of reflection with the mirror moving away from the source is 144.7-30 = 114.7 degrees which is much greater than the angle of incidence (30 degrees).

    For the mirror moving in the negative x direction towards the source:

    [tex] \theta = - 2*\tan^{-1}\left(\frac{tan(\pi-\pi/4)}{2}\right) * \sqrt{\frac{(1+0.8165)}{(1-0.8165)}}+\pi[/tex]

    = 0.61548 radians = aprox 35.3 degrees,

    so the angle of reflection in this case is 35.3-30 = 5.3 degrees from the normal of the mirror, which is much less than angle on incidence (30 degrees).

    Pretty much... as the worked example above shows.
    Last edited: Nov 8, 2009
  10. Nov 8, 2009 #9
    Re: 45 degree mirrors in Special Relativity

    Shouldn't it look like exactly 60 degrees ?

    - So that it hits the ceiling vertically. Otherwise if a light detector on the ceiling triggers a bomb, the two observes will disagree about whether the bomb went off.
  11. Nov 8, 2009 #10
    Re: 45 degree mirrors in Special Relativity

    If I have done the calculations correctly, then the reflected ray should still hit the target/bomb on the ceiling in both frames. The increased complexity over a simple application of angle of reflection is equal to angle of incidence is required to do that.

    Bernhard Rothenstein who is a member of physicsforums has gave a formula (Eq 11) in this document here http://arxiv.org/ftp/physics/papers/0508/0508084.pdf but his formula causes a division by zero error when the angle of incidence is parallel to the x axis which is exactly the case we are considering. I am trying to put together a more flexible formula.
    Last edited: Nov 8, 2009
  12. Nov 8, 2009 #11
    O.K. I think I have worked it out now.

    Given the angle [itex] \theta '[/itex], in the rest frame of a mirror, of a light ray from the normal of the mirror, the following formula gives the angle [itex]\theta[/itex] of the same ray as measured in a frame with relative velocity (v).

    [tex] \theta = \pi - 2*\tan^{-1}\left(\frac{tan(\pi-\theta ')}{2}\right) * \sqrt{\frac{(1-v/c)}{(1+v/c)}} - \tan^{-1} \left( \frac{ \sqrt{1-v^2/c^2}}{\tan(\theta_{M} ')} \right) [/tex]

    where [itex]\theta_{M} '[/itex] is the angle of the angle of the mirror from the y axis as measured in the rest frame of the mirror and v is parallel to the x axis.

    More explicitly the transformed angle of reflection is given by:

    [tex] \theta_{R} = \pi - 2*\tan^{-1}\left(\frac{tan(\pi-\theta_{R} ')}{2}\right) * \sqrt{\frac{(1-v/c)}{(1+v/c)}} - \tan^{-1} \left( \frac{ \sqrt{1-v^2/c^2}}{\tan(\theta_{M} ')} \right) [/tex]

    and the transformed angle of incidence is given by:

    [tex] \theta_{I} = \pi - 2*\tan^{-1}\left(\frac{tan(\pi-\theta_{I} ')}{2}\right) * \sqrt{\frac{(1-v/c)}{(1+v/c)}} - \tan^{-1} \left( \frac{ \sqrt{1-v^2/c^2}}{\tan(\theta_{M} ')} \right) [/tex]

    All angles are positive in the clockwise direction so although the formulas look almost identical, it must be remembered that if a positive value of [itex]\theta_{R} '[/itex] is used for the angle of reflection, then a negative value of [itex] \theta_{I} '[/itex] must be used for the incident angle and vice versa.

    For v=0 the expected result of [itex]\theta_{R} '[/itex] = [itex] -\theta_{I} '[/itex] is obtained.

    The last term in the above equations allows for the change in angle of the moving mirror due to length contraction and the first two terms allow for the relativistic aberration of the light rays.

    It is also worth noting that a physical rod is welded at right angles to the mirror in its rest frame, then when the mirror is moving relative to an observer, the rod no longer appears to be at right angles to the moving mirror and is not parallel to the assumed normal of the moving mirror.

    Maybe someone could check the equations, as there is plenty of room for error.
  13. Nov 8, 2009 #12
    Re: 45 degree mirrors in Special Relativity

    The attached diagram gives an idea of what is happening. The mirror is red, the normal of the mirror is in solid black and the light rays are blue. The reflected light ray in the transformed diagram appears to go through the mirror but that does not actually happen because the mirror is moving.

    Attached Files:

  14. Nov 8, 2009 #13
    Thanks for taking an interest and making the effort kev.

    For some reason I imagined everything the exact other way around to you. It seems that great minds actually don't think alike.

    For simplicity I visualised this :
    Length contracted mirror moving to the right (looks like it's at 30'), light ray comes in from from the right (parallel to the line of motion) and gets deflected vertically upward. Simple as that...

    I would check the equations, but unfortunately I can't even see them on my computer, just a bunch of squiggles on a black background.. I'll see if I can log in to physicsforums from a friends comp. Cheers
  15. Nov 8, 2009 #14
    Here is another drawing, but this time with mirror moving towards the light source, which might be more in line with what you had in mind (except the mirror is moving to the left).

    Attached Files:

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