Relativistic Velocity Addition: Calculating Electron Speed

[lindberg]

TL;DR

Does the relativistic velocity addition apply to De Broglie matter waves?

If we imagine launching an electron wave in a reference frame S with speed v, should someone viewing the electron from frame S1, which is in inertial motion referring to S, use the relativistic velocity addition to calculate the speed of the electron?

 

[Orodruin]

First of all, any thread on de Broglie matter waves need a compulsory disclaimer about it being a concept on the way to quantum mechanics that has been superceded since about a hundred years ago.

With that out of the way, relativistic velocity composition holds for any speeds. However you need to choose what you mean by ”speed” when it comes to a wave that is not lightlike: group or phase velocity?

 

[Vanadium 50]

To add to Orodruin's message, matter waves are not just a step on the way to quantum mechanics, it's a non-relativistic step on the way to non-relativistic quantum mechanics. Mixing this with relativity is unlikely to be sensible.

 

[Sagittarius A-Star]

Summary: Does the relativistic velocity addition apply to De Broglie matter waves?

If we imagine launching an electron wave in a reference frame S with speed v, should someone viewing the electron from frame S1, which is in inertial motion referring to S, use the relativistic velocity addition to calculate the speed of the electron?

The group velocity of the matter wave is equal to the velocity of the electron. As already mentioned by Orodruin, the relativistic velocity composition holds for any speeds. Accordingly, the "relativistic velocity addition" formula can be applied to both, the group velocity and the phase velocity.

In a beautiful application of SR, de Broglie proposed the following relation between the particle's 4-momentum $\mathbf P$ and the wave 4-vector of the associated wave ... :
$$ \mathbf P= h \mathbf L, \ \ \text{that is,} \ \ E(\frac{\mathbf u}{c^2},\frac{1}{c})=h\nu(\frac{\mathbf n}{w},\frac{1}{c}). \ \ \ \ \ \ \ \text{(51)}$$
In fact, if Planck's relation (50) is to be maintained for a material particle and its associated wave, then (51) is inevitable. For then the 4th components of the 4-vectors on either side of (51) are equal; by our earlier "zero-component lemma", the entire 4-vectors must therefore be equal! From (51) it then follows that the wave travels in the direction of the particle ($\mathbf n$ ∝ $\mathbf u$), but with a larger velocity $w$, given by de Broglie's relation
$$uw=c^2, \ \ \ \ \ \ \ \text{(52)}$$
as can be seen by comparing the magnitudes of the leading 3-vectors. (However, the group velocity of the wave, which carries the energy, can be shown to be still $u$.) The wave must necessarily travel at a speed other than the particle unless that speed is $c$, for waves and particles aberrate differently, and a particle comoving with its wave would slide across it sideways in another frame.

Source:
http://www.scholarpedia.org/article/Special_relativity:_mechanics#Particles_and_Waves

Assume an electron moving in the unprimed frame with velocity $u$ in x-direction. You can transform it's velocity to a primed frame, which is moving with $v$ in x-direction, by applying the "relativistic velocity addition" formula:

$u' = u \oplus (-v) = \frac{u-v}{1-uv/c^2}$

The phase velocity in the unprimed frame is $w = \frac{c^2}{u}$. If you apply the "relativistic velocity addition" formula to this phase velocity, then you get:

$w' = w \oplus (-v) = \frac{(c^2/u)-v}{1-(c^2v/uc^2)} = \frac{1-uv/c^2}{(u/c^2)-(v/c^2)} = c^2/u'$, as it should be.

 

[lindberg]

@Sagittarius A-Star thank you