# Invariance of uxw (velocityxphase velocity)

• bernhard.rothenstein
E where p is the momentum of the particle and E is the energy of the particle. In wave mechanics, the equation for a particle's position is described by awave packet, whose peak moves with a group velocity, v_g=dw/dk. This equals k/w for a particle of any mass.

#### bernhard.rothenstein

Please inform me if you know places where the invariance of the product
velocityxphase velocity=cc is discussed (derived?). Thanks

invariance to what?

I don't understand the question. How could it possibly be invariant?

By counter example, choose a frame where a particle is moving: velocity x phase velocity is non-zero. Now choose a frame where that same particle isn't moving: velocity is zero so the product is as well.

electron velocity u and asociated wave w uxw=cc

JustinLevy said:
I don't understand the question. How could it possibly be invariant?

By counter example, choose a frame where a particle is moving: velocity x phase velocity is non-zero. Now choose a frame where that same particle isn't moving: velocity is zero so the product is as well.

Thanks. Please have a look at K. Moller, "The Theory of relativity" Clarendon Press Oxford 1972 Chapter 2.9." Consider that u represents the velocity of a tardyon and w represents the phase velocity of the associated wave. I mean by invariace the fact that uw=u'w'. I am interested if you have found mentioned that fact elsewhere as well, in order to enlarge my reference. Excuse please the inexact formulation of my question.

I haven't read Moller in years, but he must be using the classical expression
v=p/E for the "velocity of a tardyon". This gives v=k/w k and w are divided by hbar. The phase velocity of a wave is v_p=w/k, and he gets his result.
BUT, in wave mechanics, the particle's position is described by a
wave packet, whose peak moves with a group velocity, v_g=dw/dk.
This only equals k/w for a massless particle.
Moller seems to be using classical mechanics for v, and wave mechanics
for v_p.

uxw=cc

Meir Achuz said:
I haven't read Moller in years, but he must be using the classical expression
v=p/E for the "velocity of a tardyon". This gives v=k/w k and w are divided by hbar. The phase velocity of a wave is v_p=w/k, and he gets his result.
BUT, in wave mechanics, the particle's position is described by a
wave packet, whose peak moves with a group velocity, v_g=dw/dk.
This only equals k/w for a massless particle.
Moller seems to be using classical mechanics for v, and wave mechanics
for v_p.
Thanks. For you and for others interested in my thread I recall what Moller does. He starts with the phase of a plane wave propagating with phase velocity w in I and w' in I'. Among others he derives the addtion law of phase velocities and the transformation equation of the angles along which the wave propagates when detected from I and I' respectively. He has derived previously the transformation equation for the angles along which a tardyon moves with velocity u (u') and the addtion law for u and u'. The conclusion is:"A comparison of the transformation equations for the addition law of u and w respectively and of the angles along which the particle moves and the wave propagates become equal to each other respectively when we put u=cc/w and u'=cc/w'. In other words the velocity of a particle u and its direction n are transformed in the same manner as the corresponding quantities for a wave with the phase velocity w=cc/u and direction n. In his wave theory of elementary particles de Broglie made use of the circumstances by attributing to a particle with the direction of propagation n and the phase velocity w=cc/u a procedure which thus is relativistically invariant."

Meir Achuz said:
I haven't read Moller in years, but he must be using the classical expression
v=p/E
for the "velocity of a tardyon". This gives v=k/w k and w are divided by hbar. The phase velocity of a wave is v_p=w/k, and he gets his result.
BUT, in wave mechanics, the particle's position is described by a
wave packet, whose peak moves with a group velocity, v_g=dw/dk.
This only equals k/w for a massless particle.
Moller seems to be using classical mechanics for v, and wave mechanics
for v_p.
Sorry for the question. So, in wave mechanics, it's wrong to write:
p = mv*gamma and E = mc^2*gamma for a particle?

Meir Achuz said:
I haven't read Moller in years, but he must be using the classical expression
v=p/E for the "velocity of a tardyon". This gives v=k/w k and w are divided by hbar. The phase velocity of a wave is v_p=w/k, and he gets his result.
BUT, in wave mechanics, the particle's position is described by a
wave packet, whose peak moves with a group velocity, v_g=dw/dk.
This only equals k/w for a massless particle.
Moller seems to be using classical mechanics for v, and wave mechanics
for v_p.
Whoops, I made a silly mistake. Thank you light--> for questioning the result.
The sentence starting with BUT (I only capitalize when I am wrong.)
A particle's position is described by a
wave packet, whose peak moves with a group velocity, v_g=dw/dk.
This equals k/w for a particle of any mass.
Then scrap the last sentence.
The v in the expressions is v_g.

The algebra is:(d/dk)\sqrt{k^2+m^2}=k/\sqrt{k^2+m^2}.
(All hbare=1=c)
Sometimes I click submit too before I think.

uxw=cc

Meir Achuz said:
Whoops, I made a silly mistake. Thank you light--> for questioning the result.
The sentence starting with BUT (I only capitalize when I am wrong.)
A particle's position is described by a
wave packet, whose peak moves with a group velocity, v_g=dw/dk.
This equals k/w for a particle of any mass.
Then scrap the last sentence.
The v in the expressions is v_g.

The algebra is:(d/dk)\sqrt{k^2+m^2}=k/\sqrt{k^2+m^2}.
(All hbare=1=c)
Sometimes I click submit too before I think.

Do you think that the following derivation holds
Using results of quantum mechanics: wavelength L =h/mv, where v is the
particle velocity.
Using relativity: energy E = mc^2 = hf, where f is the frequency
associated with E.

Multiply: hfL = mc^2 h/mv
fL = c^2/v
u = c^2/v, where u is the phase velocity

so uv=c^2
Note that this uses for frequency the total energy,including the rest
energy, not just the kinetic energy. And the phase velocity gets larger
as the particle velocity gets smaller!

bernhard.rothenstein said:
Do you think that the following derivation holds
Using results of quantum mechanics: wavelength L =h/mv, where v is the particle velocity.Using relativity: energy E = mc^2 = hf, where f is the frequency
associated with E.
Multiply: hfL = mc^2 h/mv
fL = c^2/v
u = c^2/v, where u is the phase velocity
so uv=c^2
Note that this uses for frequency the total energy,including the rest
energy, not just the kinetic energy. And the phase velocity gets larger
as the particle velocity gets smaller!