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  1. We all know that the relativistic mass of a moving object in Special relativity increases for an observer who is measuring it for a moving object.
  2. We also know the the concept of particle-wave duality.
  3. We also know that the observed frequency of a wave changes according to where it is moving (away or near, transverse etc...)

Is this concept of relativistic mass increase, related to the concept of Doppler effect of matter waves?

Can other implications of Doppler effect for waves be seen for matter waves and were there any experiments done for them?

Historically, was this one of the reason for developing the concept of matter waves? (We know other reasons that are Compton effect, Interference etc....)

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  • $\begingroup$ The matter wave in some interpretations is considered as the probability amplitude wave. Hence, it wont have physical implications, but there are other interpretations which consider them physical. An experimental test of this can settle the dispute? $\endgroup$
    – Chetan Waghela
    Nov 29, 2015 at 8:38
  • $\begingroup$ AFAIK, the wave interpretation of a single moving object has no inherent frequency, which makes the Doppler effect on the wave moot. $\endgroup$
    – Daniel Griscom
    Nov 29, 2015 at 8:57
  • $\begingroup$ We do mention equations like E = hcross omega. Where omega relates to the frequency. $\endgroup$
    – Chetan Waghela
    Nov 29, 2015 at 9:20
  • $\begingroup$ "3) We also know that the observed frequency of a wave changes according to where it is moving" 1) No, it's not so at all. 2) We have a rule here: One question at a time. I'm annoyed because I feel there's some interesting question lurking there. $\endgroup$
    – stuffu
    Nov 30, 2015 at 1:06
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    $\begingroup$ There are no "matter wave" in quantum mechanics. The probability of finding a particle at an(x,y,z,t) spacetime point has a sinusoidal variation. The particle is always observed whole. $\endgroup$
    – anna v
    Dec 1, 2015 at 5:53

2 Answers 2

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Below I will use some simple formulas, that's why I must make a distinction between longitudinal and transverse relativistic mass.

The transverse relativistic mass of an object has very much to do with the energy of the object: We just multiply the energy by a constant to get the transverse relativistic mass.

The energy of an object has very much to do with the Compton frequency of the object: We just multiply the Compton frequency by a constant to get the energy.

If we set the two aforementioned constants to 1, we get:

  1. frequency = transverse relativistic mass

As Relativistic mass quite often means transverse relativistic mass, this is quite often true:

  1. frequency = relativistic mass

There's no simple way to express frequency as longitudinal relativistic mass

The formula for transverse relativistic mass is: $$m'=\gamma m$$

The formula for longitudinal relativistic mass is: $$m'=\gamma^{3} m$$

How do we calculate Doppler shift of a thrown baseball? Simply:

change of frequency = change of energy

The energy of a baseball doubles if it's thrown at speed 0.86 c. So frequency of a baseball doubles if it's thrown at speed 0.86 c.

At throwing speed 40 m/s the Doppler shift of a baseball is very small, it's smaller than Doppler shift of a radar beam reflecting from the baseball. Therefore it is reasonable to put quote marks around "Doppler shift" when it's a matter wave that is Doppler shifting.

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  • $\begingroup$ Is your answer affirmative to the the suggestion I had? $\endgroup$
    – Chetan Waghela
    Dec 1, 2015 at 6:06
  • $\begingroup$ I don't know. You decide :) But remember that matter waves don't Doppler shift, they just change frequency. (An object with rest frequency f has a relativistic frequency gamma * f) $\endgroup$
    – stuffu
    Dec 1, 2015 at 6:27
  • $\begingroup$ Hi user7027 I was reading a wikipedia artcile which speaks about this tranverse and longitudinal mass and I found criticism to the concept of relativistic mass itself ! The section refers to arxiv.org/abs/physics/0504110v2 $\endgroup$
    – Chetan Waghela
    Dec 1, 2015 at 6:39
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Is this concept of relativistic mass increase, related to the concept of Doppler effect of matter waves?

No. Doppler's effect also happens for non-relativistic waves, including "non-relativistic matter waves", by which I meant Schroedinger's waves. The effect is in fact trivial. When you change the reference frame, the momentum of the particle changes. By de Broglie's formula, $$\lambda = \frac{h}{p},$$ one finds that the wave length also changes. This is manifestly the Doppler's effect.

Can other implications of Doppler effect for waves be seen for matter waves and were there any experiments done for them?

Yes. This is again classical. When a particle scatters from a moving wall (or potential), its energy changes due to the collision. This is also the Doppler effect. In fact, Compton's effect can be considered as some sort of Doppler's effect of photon. I am not aware of any direct experiment with electron waves.

Historically, was this one of the reason for developing the concept of matter waves? (We know other reasons that are Compton effect, Interference etc....)

To what I know, this was not the case.

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  • $\begingroup$ The total mass-energy of a moving body increases so you would expect de Broglie wavelength to decrease, whereas light wavelength increases due to relativistic doppler shift effects. $\endgroup$
    – Peter R
    Jan 7, 2016 at 19:31
  • $\begingroup$ @PeterR Depending on the wall moving towards or away from the particle, you can have the particle's energy (or more correctly, the particle's momentum) increases or decreases, respectively. If you calculate such change in energy for photons, you find exactly the Doppler's formula. $\endgroup$
    – cnguyen
    Jan 8, 2016 at 7:41
  • $\begingroup$ I think the standard terminology is "classical" for "non-quantum" and "non-relativistic" for stuff that's, well, not relativistic. You would definitely not call a wavefunction "classical". $\endgroup$
    – Daniel
    Feb 7, 2016 at 15:17
  • $\begingroup$ @Daniel yeah, that sounds better. I'll fix (thanks). $\endgroup$
    – cnguyen
    Feb 8, 2016 at 13:31

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