1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Light bounced off mirror

  1. Jan 27, 2013 #1
    Hi I'm new here. Been interested in physics all my life and understand it quite well. Been reading a book about Einstein which lead to a question i've been pondering. What happens if a mirror, that is traveling near the speed of light, has a beam of light pointed at it? What is the status of the light coming off the mirror? Does it come off at the speed of light and would the wave length be shifted?

    Your input please.

  2. jcsd
  3. Jan 27, 2013 #2
    Hi Ratchet....

    Light always travels at speed c, locally, no matter the observers speed.

    That's convenient for police because it enables the Doppler effect for radar traps:


    Same effect as your posted example.

    Standing as an observer on, say, ground while your mirror moves past you at high velocity means you observe the frequency [not the velocity] of the light differently than if you were moving along with the mirror. The mirror is impinged on by the light beam at a higher apparent [relative] frequency than you as a stationary ground observer. So in an ideal reflection, the frequency, that is wavelength, would be different....

    it gets reflected at a higher frequency, meaning more energy...E = hf....
    where did that extra energy come from do you think??
  4. Jan 27, 2013 #3
    Remember you can always find a reference frame where the mirror is not moving, and the physics have to be the same, so there will be no difference wheather it is moving or not.
  5. Jan 27, 2013 #4
    rudolf...good point...I agree

    I should have included in my prior post that from the frame of the mirror the frequency/wave length appears unchanged....and in that in that frame the frequency/wavelength was never the same as that of the 'ground' observer.
    edit: I am assuming ideal reflection here....whereas Drakkith assumes
    a tiny energy change for the stationary mirror....

    'Observations' in relativity are not so simple!!!
    Last edited: Jan 27, 2013
  6. Jan 27, 2013 #5
    I'm thinking about things that are at the fringe of my knowledge to understand them but that is what draws me. That said, I'll have to give your energy question some thought. My first thought wonders if its from the differance in speed of the mirror compaired to the lights speed. Thinking out loud, if for example the mirror and the light beam originated from the same point and both were traveling at he speed of light, then the energy reflected from the mirror would be the same because the frequency would be the same. So as you slow the mirror down, the reflected light would gain energy, or could it be just the opposit? I'll think about this more and respond again.

    Thanks, this is going to be a fun, stimulating and enlightening forum.

    Last edited: Jan 27, 2013
  7. Jan 27, 2013 #6


    User Avatar
    Staff Emeritus
    Science Advisor

    You cannot have the mirror travel at c, sorry. It must remain at some fraction of c.

    Now, consider the following.
    Let's say we are observing a mirror stationary to an emitting light source.
    When light hits a mirror it contains momentum found by the equation E=pc, with p being momentum, c being the speed of light, and E being the energy of the photon. When reflection occurs the light goes from a momentum of E/c to a momentum of -E/c since it reverses direction. This indicates a change of momentum of 2E/c. This change in momentum is felt on the mirror as a force, known as radiation pressure and it causes the mirror to accelerate away from the incoming radiation. The radiation leaves the mirror with LESS energy than it had before the reflection, with the difference being applied to the mirror as work.

    So when the mirror was initially stationary with respect to the emitting light source, the light hits the mirror, reflects, and leaves with less energy, causing the mirror to accelerate backwards. But what about when the mirror is moving towards the light source?

    Well, in that case the light hits the mirror and again applies a force to the mirror. But this time when the mirror is moving towards the emitting source, instead of the light LOSING energy, it GAINS energy. It must gain energy, as the mirror is losing momentum, and the only way for momentum to be conserved is for the light itself to gain it, increasing its energy. So the light reflects off of the mirror, gains energy, and by that the frequency increases. In things like radar guns this change in frequency is used to calculate the speed of the object.

  8. Jan 27, 2013 #7
    Ok I'm starting to get it. So when I considered the differance in the mirror and light speeds being the source of the energy, I'm thinking I was sort of right and that is why I went to the mirror going the same speed as light in the same direction. I do know that isn't possible. Lets then change it to as the mirror approach infinitely close to the speed of light. The energy then produced would approach zero, wouldn't it? I was not aware of radiation pressure being the source but I did feel that if light is energy then if it bounces off of something, it must push on it and thus transfer energy to it. So the radiation pressure explains the source of the energy. In my case with the mirror going the same direction as the light, the mirror would then gain momentum. Now if the light wave lengths get longer (less energy) as the mirror approaches the speed of light, could the wave length then approach infinitely long? Thanks for the lessons, I'm learning which is something I like do as continously as possible. Continued thanks.

    Last edited: Jan 27, 2013
  9. Jan 27, 2013 #8


    User Avatar
    Staff Emeritus
    Science Advisor

    Yes, as the velocity of the mirror increases with respect to the emitting source, the wavelength of the light gets longer and longer.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook