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Relativistic doppler shift

  1. Sep 11, 2004 #1
    For the relativistic doppler shift:

    change in wavelength = (c - Vs) To / (1 - Vs ^2 /c^2)^1/2

    where Vs is emitter velocity, c is speed of light and To is time.

    Suppose change in wavelength was equal to just 1 / (1 - Vs ^2 / c^2)^1/2

    then (c - Vs) To = 1
    c -Vs = 1 / To

    c = Vs + 1 / To
    c = Vs + frequency of emitted wave

    I now suggest that for M = Mo / (1 - v^2 /c^2) ^ 1/2

    that this relation is actually

    M = Mo x (c - Vs) To / (1 - Vs ^2 /c^2)^1/2

    when (c - Vs) To = 1
    and c = Vs + frequency of emitted wave
    In other words a mass emits a wave as it travels through space at constant velocity.The slower the mass travels, the greater the frequency of the
    emitted wave.A mass at rest would emit the highest frequency.
     
  2. jcsd
  3. Sep 11, 2004 #2

    Integral

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    This is certianly not Quantum Mechanics, it is not clear if it is even related to relativity. I'll let Janus and or Phobos take a look it to make that call.
     
  4. Sep 11, 2004 #3
    I am gonna be honest and say : I don't get this???

    regards
    marlon
     
  5. Sep 11, 2004 #4
    This not even wrong.
     
  6. Sep 11, 2004 #5

    pervect

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    http://scienceworld.wolfram.com/physics/DopplerEffect.html

    gives the correct formula for relativistic doppler shift
    [tex]
    \frac{\lambda}{\lambda_0} = \sqrt{\frac{c+v}{c-v}}
    [/tex]
    where v is the relative velocity.

    I can't even figure out the what the rest of the post is supposed to be computing.
     
  7. Sep 11, 2004 #6

    Nereid

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    Well, a mass with a temperature >0K will emit 'thermal' radiation, and if that mass is travelling away from you, the observer, you will see that thermal radiation as having a lower frequency (redshifted) that your friend, who is sitting on the mass ... however, if the mass is travelling towards you, it will be blueshifted.

    Would you care to say a little more about this idea, Rothiemurchus? For example, how does it relate to the Special Relativity which is taught at university (or as described by Einstein)?
     
  8. Sep 11, 2004 #7
    Neither do I... since when does time enter the calculation for a Doppler shift?
     
  9. Sep 11, 2004 #8
    Hmm; I think I see where you're coming from Rothie. But if you're going to do such a derivation, don't you think you will need to specify more precisely To (in the first eqn.). From what frame is this time measured?
    Let me show you another error that will nullify your conclusion:

    In your statement that:
    "Suppose change in wavelength was equal to just 1 / (1 - Vs ^2 / c^2)^1/2",
    You said,
    "then (c - Vs) To = 1"
    This is not true.
    In reality, if the wavelenth change is 1/(1-Vs^2/c^2)^1/2,
    then (c - Vs) To = 1 meter (Not simply a dimensionless 1).

    So c-Vs = 1METER / To

    and thus c = Vs + 1METER / To
    SO your statement,
    "c = Vs + frequency of emitted wave" must be false since meters/time is NOT frequency! It is simply velocity, and now the resultant eqn. is
    c = Vs + Velocity (of whatever)

    Creator
     
    Last edited: Sep 11, 2004
  10. Sep 11, 2004 #9

    Janus

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    Well, one place where he runs afoul is that old bugaboo, maintaining proper units.

    He starts with

    [tex]\Delta \lambda = \frac{t_{0}(c-v_{s})}{1-\frac{v_{s}^2}{c^2}}[/tex]

    Let's ignore the veracity of this formula for now and analyse its units.

    The right side of the equation reduces to velocity * time, or distance, so the left side must represent a distance also, which is what he must mean by change of wavelength.

    Now his next step is to say that we let :

    [tex]\Delta\lambda = \frac{1}{1-\frac{v_{s}^2}{c^2}}[/tex]

    Okay, right now we have a problem. We already established that the change of wavelength was a measurement of distance, but the right side of this equation is a dimensionless number. To make it correct a unit of distance (or length must be added to the right side. If we use "d" to represent our unit of length, we get.

    [tex]\Delta\lambda = \frac{1d}{1-\frac{v_{s}^2}{c^2}}[/tex]

    Next he reduces to get

    [tex]1 = t_{0}(c-v_{s})[/tex]

    still with the missing unit. He rearranges to get:

    [tex] c = v_{s}+ \frac{1}{t_{0}}[/tex]

    and now the missing unit becomes really apparent.

    A velocity equals a velocity plus a frequency?

    Any conclusion he derives from this relationship must obviously be flawed.
     
    Last edited: Sep 11, 2004
  11. Sep 12, 2004 #10

    Chronos

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    Well, Janus beat me to that one. Background independence wins again.
     
  12. Sep 12, 2004 #11
    JANUS:
    A velocity equals a velocity plus a frequency?

    Any conclusion he derives from this relationship must obviously be flawed.

    ROTHIE M:

    A velocity equals a frequency provided wavelength = 1 metre.
    So conclusion is not flawed.
    1 x M = Mo x (c - Vs) To / (1 - Vs ^2 /c^2)^1/2

    where 1 = (c - Vs) To
    distance x mass = distance x mass

    Because doppler shift is inverted this could explain away Feynman's objection to particles causing gravity by striking Earth in all directions.The greater momentum from a greater number of particles on the side of Earth approaching particle shower, would be compensated for by the lower momentum of the inverted doppler shift for each particle.The Earth would not slow down as Feynman said (could be a justification for Newton's first law too).
     
    Last edited: Sep 12, 2004
  13. Sep 12, 2004 #12

    Nereid

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    Why is 'wavelength' equal to 1 metre?
    Since we have a pretty good handle on the full spectrum of EM incident on the Earth, we could conclude that your idea could only make sense if the 'particles' are not photons. However, your initial post seems to refer to only EM ... does your idea thus contain a fatal inconsistency?
     
  14. Sep 12, 2004 #13
    You are right because there is no way to explain how waves of any kind catching up with the Earth could increase their frequency.Although a medium
    of particles that behaved according to my equation, and allowed waves to form in it, would allow a mass to move at constant speed through it (in keeping with Newton's first law).
    1 metre was chosen so that 1/To x 1 metre = frequency x wavelength = a velocity, and the equation would remain valid.
    Feynman's objection could be overcome by particles hitting the Earth and being re-emitted in such a way as to oppose the momentum of the incoming particles .The smaller number of incoming particles at the back of the Earth (the back means pointing away from the direction in which the Earth is travelling) would not be opposed by as much momentum from re-emitted particles (some momentum would be taken away from the re-emitted particles by the forward motion of the Earth)and so the net force on the front and back of the Earth would balance and the Earth would not slow down.
     
    Last edited: Sep 12, 2004
  15. Sep 12, 2004 #14
    You are right because there is no way to explain how waves of any kind catching up with the Earth could increase their frequency

    Rothie M:
    However,there is a way to explain how PARTICLES could increase their energy by catching up with the Earth:
    A large moving mass would compress a medium of particles ahead of it and it would rarefy (decompress) the medium behind it.In the rarefied zone there would be
    fewer particles for particles that are not part of the medium to scatter off, and so non-medium particles would have a faster
    speed in the medium than they would have had if the large mass was not moving.
    In the compressed zone ahead of the moving mass, the particles that are not part of the medium would scatter off more medium particles than they would if the mass was at rest and so would have a smaller speed.
    So a large mass bombarded by non-medium particles would keep moving at constant speed in a medium of particles because the greater number of non-medium particles encountered,per unit time, at the front of the mass, would have less energy per particle than the smaller number of non-medium particles at the back of the mass.
    On the side of the Earth opposite to the direction it is travelling in,
    medium particles will be less dense.An experiment would have to be devised
    to measure this smaller density, and the existence of the medium,
    and non-medium particles travelling through it, would have to be established.
    Thus the question arises: what is the nature of these particles - do they have charge?what is their rest mass,spin etc.
     
    Last edited: Sep 12, 2004
  16. Sep 13, 2004 #15

    Chronos

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    Physics 101. You need to study more.
     
  17. Sep 13, 2004 #16
    Particles diffusing through the Earth from the front side to the back side and then re-emitted into space would balance the force on the Earth.So the Earth does not heat up, the rate of emission due to particles passing from one side of Earth to the other would have to equal the rate of absorption of the particles
     
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