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velocityxphase velocity=cc is discussed (derived?). Thanks

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- Thread starter bernhard.rothenstein
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- #1

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velocityxphase velocity=cc is discussed (derived?). Thanks

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invariance to what?

- #3

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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

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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

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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

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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

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.

Please comment.

- #7

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Sorry for the question. So, in wave mechanics, it's wrong to write:

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.

p = mv*gamma and E = mc^2*gamma for a particle?

- #8

Meir Achuz

Science Advisor

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Whoops, I made a silly mistake. Thank you light--> for questioning the result.

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.

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

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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

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I would instead write: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!

wavelength L =h/mv*gamma;

energy E = mc^2*gamma.

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