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Is the frequency of light relative?

  1. Apr 10, 2014 #1
    Given that distance between two points can vary depending on your frame of reference wouldn't that mean that the distance between two points on a wave would vary in different frames and therefore the frequency of light would be relative. So as you speed up would this mean there would be a shift in the frequencies of visible light?
     
  2. jcsd
  3. Apr 10, 2014 #2
    Yes, This is the known as the relativistic doppler shift.
    Part of it has nothing to do with relativity, it is just a classical effect of light emitted from closer and closer ( or further and further) distances due to the relative velocity of the source and the observer. a simple calculation shows [itex]\frac{\delta f}{f}=1+\frac{v}{c}[/itex] (where v is positive when the source is moving towards the observer.

    As you say, Relativistic effects add time dilation which gives an additional factor of [itex]\frac{\delta t}{\delta t`}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/itex] (where the tagged frame is the frame where the source is at rest and the untagged is the frane where the observer at rest). Combining this gives [itex]\frac{\delta f}{f}=(1+\frac{v}{c})\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\frac{\sqrt{v+c}}{\sqrt{v-c}}[/itex]

    Note that the dependence on the sign of the velocity only comes from the classical effect.
     
  4. Apr 10, 2014 #3

    ghwellsjr

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    Yes, but not for the reason that you stated. It doesn't matter what frame of reference you use to analyze or depict a scenario. If you see light of one color while stationary in a particular reference frame and then you speed up, assuming that you mean you start moving toward the source of the light, you will see it blue-shifted because you are encountering the waves at a faster rate. This is a Doppler effect. You can analyze this in your original rest frame so that you end up moving in that same frame but the source of the light remains the same or you can analyze it in your final rest frame so that you have accelerated from your original speed and the source of the light is at a different speed. In these two frames, the wavelength of the light is different, but the time dilations for you and of the source of the light are also different so that your observation is independent of any frame.
     
    Last edited: Apr 10, 2014
  5. Apr 10, 2014 #4
    Yes. Let's calculate it:

    Let's say the distance between the start and the end of a wave is one light minute. The wave passes a static observer in one minute and at velocity c. We can conclude that the observer concludes that the length of the wave is one light minute.

    Another observer moves through the same wave at velocity 0.5 c. We know that it takes 40 seconds of our time for him to pass the wave. We can conclude that the observer concludes that the wave is little less than 40 light seconds long, because we know that the clock of the observer is a little bit slow compared to our clock.
     
    Last edited: Apr 10, 2014
  6. Apr 10, 2014 #5
    Thanks very much for all your responses. Much appreciated.
     
  7. Apr 10, 2014 #6

    ghwellsjr

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    Let me illustrate with some spacetime diagrams of different frames of reference for the same scenario. We start with the rest frame of the observer and the light source and then the observer starts moving toward the light source at 0.6c. The observer is shown in blue and the light source is shown as the thick red line with light pulses emitted every nanosecond shown as the thin red lines. The dots represent one-nanosecond intervals of Proper Time. The speed of light is one foot per nanosecond:

    attachment.php?attachmentid=68509&stc=1&d=1397123453.png

    Note that the observer sees one light pulse every nanosecond before he accelerates and two light pulses every nanosecond after he accelerates even though the wavelength between the light pulses is one foot (twelve inches) all the time.

    Now we use the Lorentz Transformation on the coordinates of all the events (dots) to see what the same scenario looks like in the final rest state of the observer:

    attachment.php?attachmentid=68510&stc=1&d=1397123453.png

    Note that the observer continues to see one pulse every nanosecond of his Proper Time before his acceleration and two pulses afterwards but now the wavelength between the pulses is six inches (half what it was before). The Time Dilation of the light source and of the observer adjust so that the observer continues to see the same Doppler shift that he saw in the first diagram.

    We can go to another reference frame in which the observer is moving at the same speed, 0.333c, before and after acceleration, he just changes direction:

    attachment.php?attachmentid=68511&stc=1&d=1397123453.png

    Now the wavelength is eight inches and the Time Dilation of the observer is the same before and after acceleration and it's the same as the source.

    One more randomly picked reference frame transformed from the original frame to -0.3715c:

    attachment.php?attachmentid=68512&stc=1&d=1397123453.png

    This one has a wavelength of 18 inches, and, as always, the Doppler shifts are still the same.

    Does this all make perfect sense to you? Any questions?
     

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