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Prove that z=v/c

  1. Nov 19, 2008 #1
    Ok, I feel like an idiot, but can someone remind me how you prove mathematically z = v/c? (z is redshift, v is recessional velocity, c is speed of light.)

    I realise that this equation only works in a non-relativistic Universe, but nevertheless I'd like to see it.
  2. jcsd
  3. Nov 19, 2008 #2


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    You aren't really talking about the cosmological redshift, because that is not the doppler effect of the current recession speed, or the recession speed at the time of emission, or at any other one particular time. the cosmo redshift is determined by the factor by which distances have expanded during the light's travel time. The formula they give you for it, on day one of cosmo class, is 1+z = a(now)/a(then), the ratio of the metric scalefactor now compared to what it was then, when the light was emitted.

    So if you were talking about the cosmo redshift you would have totally the wrong formula. But I think what you are really asking about is the DOPPLER EFFECT shift. If z is defined as the fractional increase in wavelength, and v is actual motion away, of the observer from the source, then you could say z = v/c.

    That would have nothing much to do with universe expansion, but it could apply to some random motions of neighboring galaxies relative to each other, and stars within galaxies, and stuff like that.

    As you point out, it isn't true that z = v/c. But for unrelativistic speeds it is nearly right (if we are clear that it is not cosmo redshift, but some small random motion doppler effect that we are talking about)

    It's the ordinary doppler formula. I have to go out for lunch, but I'll bet they have an explanation in WikiP in the doppler effect article.
    Ill be back. Maybe someone will help in the meantime.

    OK, I'm back. I see nobody stepped in. So let's think about it. Picture it. there's a traintrack running eastwest, and you are standing on a road that runs along the track. And a freighttrain comes along traveling from east to west and every time a break in the cars comes by, you say bingo. The frequency of your saying bingo is the frequency of the train, we'll say.

    then you get on your bicycle and go east at 10 percent of the speed of the train, so you are meeting the cars 10 percent more frequently, and you say bingo 10 percent oftener.

    or if you ride west (same direction as train) at that same speed then it takes 10 percent longer for the next car to catch up with you and you say bingo 10 percent less often.

    You asked for a nonrelativistic picture. So we can interpret the formula for sound. It's more intuitive, quicker to understand, thinking of frequency. So let c be the speed of sound. Let v be your speed, towards. The frequency you hear as you go towards will be the emitted frequency increased by a factor of (1 + v/c)
    You will be meeting the peaks of the waves that much faster, because you are going towards the source.

    Frequency higher by a factor of ((c+v)/c) means wavelength shorter by a factor (c/(c+v))
    But for small velocities (a small percentage of c) that number is about the same as 1 - v/c.

    You know, 1/(1+x) is about the same as 1 - x, for small x.

    So for example 5% higher frequency corresponds approximately to 5% shorter wavelength. The reciprocal of 1.05 is not exactly the same as 0.95, but pretty close.

    What I've described is why the nonrelativistic doppler shift is v/c, where you the receiver are moving towards the source. And v/c applies both to the fractional increase in frequency and the fractional decrease in wavelength (approximately.)

    The story is the same when you are moving away from the source---I just happened to imagine it going towards.

    Given that intuitive framework can you attach algebraic symbols to the various key quantities and construct a proof with equations that you are happy with? If not, let us know. I or someone will help translate into equations.

    I guess you know that the real formula, for the relativistic doppler, is 1+z = sqrt( (c+v)/(c-v))
    To tie up loose ends I guess one should notice that for small v/c that is almost the same number as 1 + v/c
    Last edited: Nov 19, 2008
  4. Nov 20, 2008 #3
    ^Thanks for the explanation, and yeah, I ment the Doppler Effect.

    That was pretty much confirming what I had in my head, which is good to know.
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