Question

Is \[\lambda= \frac{h}{p} \] an approximation?

Using c=1, v as a fraction of c,
Assuming all particles have a frequency, as are themselves waves:
\[E=\sqrt{p^2+m^2}=(hf)= h\frac{v}{\lambda}\]
\[\lambda= h\frac{v}{\sqrt{p^2+m^2}}\]
Given that for an electron p>>m, this would be a god approximation for it and similar particles.

The rationale for this is that I can't see why we can treat m=0 and apply it to cases where it is not. The above seems more sensible.

Answer 1

No. it is not an approx. de Broglie suggested the equation might be true for particles given that it was true for em radiation which where at that time known to act like particles. He also suggested E=hv where E is the relativistic energy. In both cases the implied m in p and E are not zero. If you want something interesting write the relation among velocity, wavelength and frequency.

Answer 2

The relationship of vel. freq and wavelen for particles that is;

Answer 3

Surely \[E=\sqrt{p^2+0^2}=p\] (c=1) assumes that m=0? I understand the rest, just not why this de Broglie set m=0 in the derivation (to get E=p, which only holds for m=0, as it arises from the equation E^2=p^2+m^2).

Answer 4

for particles de Broglie did not set mo equal to zero. E^2=(pc)^2 +mo^2 and additionally E=moc^2 +KE.

He postulated that E=hv and p=h/lambda.

Answer 5

http://physics.stackexchange.com/questions/41865/matter-waves-debroglies-relations

See the first answer here: there is no mathematical derivation.

Answer 6

correct.

de Broglie postulated the wave relation for particles which was shown to be true by Davisson and Germer who scatter 54eV electrons off a Nickle crystal.