Re: de broglie wave velocity


richardskaa@xxxxxxxxxxxxx a écrit :

Introduction

According to the theory of Special Relativity;

m = m0/(1 - v2/c2)1/2

m = mass of particle

m0 = rest mass of particle

v = velocity of particle

c = velocity of light

p = mv

p = momentum of particle

I see a little disconnect here

According to SR, the momentum of a moving particle is

p = mv/(1-v^2/c^2)1/2

E = mc2

E = total energy of particle

These equations have two implications:

A) Since a photon has a rest mass of zero, photons must travel at the
speed of light, otherwise the momentum and the total energy of a photon
would both be zero which can't be so! Having a velocity of the speed
of light means that a photon has an undefined mass (m), since the
formulae for mass becomes 0/0, and so their momentum and their total
energy both have non-zero values.

B) A particle with a non-zero rest mass must travel slower than the
speed of light, otherwise their: mass, momentum, and total energy would
be infinite!


According to the De Broglie theory, all particles (whether they have a
zero rest mass, i.e. are photons, or a non-zero rest mass) have dual
wave properties. The following equations come into play here;

E = hf

E = total energy

h = planck's constant

f =frequency of the wave

So mc2 = hf

So f = mc2/h

L = h/p

L = wavelength

so L = h/mv

Wave velocity = fL

So wave velocity = mc2/h. h/mv

= mc2/ mv

= c2/v

Well, if you consistantly substitute with relativistic p,
this is not what you get.

Mixing non relativistic and relativistic equations although tempting,
seems hazardous to me.

André Michaud

.

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