Invariance of uxw (velocityxphase velocity)

  • #1

Main Question or Discussion Point

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

Answers and Replies

  • #2
529
1
invariance to what?
 
  • #3
894
0
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.
 
  • #4
electron velocity u and asociated wave w uxw=cc

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.
 
  • #5
Meir Achuz
Science Advisor
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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.
 
  • #6
uxw=cc

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."
Please comment.
 
  • #7
1,901
45
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?
 
  • #8
Meir Achuz
Science Advisor
Homework Helper
Gold Member
2,169
63
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.)
should have read:
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.
 
  • #9
uxw=cc

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.)
should have read:
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!
 
  • #10
1,901
45
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!
I would instead write:
wavelength L =h/mv*gamma;
energy E = mc^2*gamma.
 

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