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Doppler, mirror vs lamp

  1. Jul 25, 2015 #1
    Dear PF Forum,
    I have a question again about Doppler effect in relativity.
    Before I ask about mirror and lamp, I want to refresh the discussion again
    as in my previous thread: https://www.physicsforums.com/threads/doppler-for-light.823942/
    The formula for Doppler factor for light is.
    ##f = \sqrt{\frac{1+v}{1-v}}f0##
    Is this for the frequency of light or for the frequency of receiving?
    Can anyone give me the answer, so I'll never misunderstood it again.
    B moves toward A at 0.3c
    Doppler factor is ##\sqrt{\frac{1.3}{0.7}}=1.36##
    B shines a red light 400 THz: https://en.wikipedia.org/wiki/Visible_spectrum
    Will A see the light as somewhat violet, 742 THz?
     
  2. jcsd
  3. Jul 25, 2015 #2

    Mentz114

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    In the formula you wrote ##f## is the received frequency and ##f0## is the sending frequency if the receiver and sender are approaching ( moving towards each other ).

    Yes, calculation looks OK. The light has been 'blue-shifted'.

    No mirror is involved - the light is being emitted from the source and received without any reflection.
     
    Last edited: Jul 25, 2015
  4. Jul 25, 2015 #3
    Thanks Mentz114
    Ok, now this.
    Let ##k = \sqrt{\frac{1+V}{1-V}}##

    A see the light blue-shifted, and re emitted the blue-shifted light to B.
    Let's say this.
    B sends an F frequency light to A. A see the light frequency as F*k.
    Now A sends the light with ##F*k## frequency to B again.
    B will see the light as ##F*k^2## is that right?
    No wait,... if A moves, then every atom in A also dilated.
    So actually A sends ##\frac{F*k}{\gamma}##
    So B will see the light as ##\frac{F*k^2}{\gamma}##
    No wait, ... the situation is similar. We should use the formula as it is, because A moves toward B and B moves toward A should use the same formula.
    1. Is this right that B will see ##F*k^2## if A bounce the signal back to B as it is.
    2. What if A bounces the signal not by lamp, but by mirror, what will B see?

    Thanks for any help.
     
  5. Jul 25, 2015 #4

    PAllen

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    An ideal mirror will act like your re-emitter, and the light will be more and more blue shifted ... but the A and B relative speed will be decreasing as energy and momentum are transferred to the light at each reflection.
     
  6. Jul 25, 2015 #5
    Yes, that makes sense.
    Light has momentum? Hmhh, reminds me to previous thread about photon with mass.
    Okay, confirmation: So it's F*k, F*k2, F*k3, etc...? Please confirm.
    I already have all the answers that I need.
    But one last question, out of curiousity.
    What if the mirror getting closer? The frequency increases, right. Can it go beyond gamma ray frequency?
    What if the mirror getting closer and closer, the frequency increases and the wavelength decreases. Can it go below planck length?
     
  7. Jul 25, 2015 #6

    PAllen

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    Each k in your series is smaller each time due to decreasing relative velocity. If you look at this in a COM frame, you would (in an idealized case) see all of the body's momentum transferred to the bouncing light at some point, then the bodies would reverse direction, propelled by the bouncing light, asymptotically approaching their relative speed before the light was emitted (because now, each bounce is more redshifted). Of course, this is all nonsense - no mirror is anywhere near that perfect. Also, of course, I do ignore quantum effects because by far the dominant issue is impossibility of 100% reflectance. [edit: this would 'work' better if each body emitted a pulse of light at the same time, and then they both bounced back and forth between the mirrors. Then the COM frame analysis becomes particularly simple.]
     
    Last edited: Jul 25, 2015
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