Question

When do we use relativistic de broglie wavelength formula

Answer 1

Suppose a particle of mass m in special relativity. Suppose the de Broglie wave-length is (non-relativistic) case: λ=hp=hmv In the case of RELATIVISTIC ...

Answer 2

Suppose a particle of mass m in special relativity. Suppose the de Broglie wave-length is (non-relativistic) case:

λ=hp=hmv

In the case of RELATIVISTIC particle, the momentum is p=mγv. Therefore a way to recast the de Broglie wavelength is:

λr=h1−v2/c2−−−−−−−−√mv

Suppose now that we focus on the kinetic energy. For a free particle, we get in the nonrelativistic case, K.E.=T=p2/2m, and thus p=2Tm−−−−√, and so

λ=h2Tm−−−−√

I have a doubt concerting to the relativistic case. The natural election for the de Broglie wave-length in the relativistic case is well known: you take T=E−mc2, and from E2=(pc)2+(mc2)2, by simple substitution of E=T+mc2, you get (pc)2=T2+2Tmc2,

p=T2/c2+2Tm−−−−−−−−−−−√=2Tm(1+T/2mc2)−−−−−−−−−−−−−−−√

or

λ=hT2/c2+2Tm−−−−−−−−−−−√=h2Tm(1+T/2mc2)−−−−−−−−−−−−−−−√

Well, I have seen a couple of places where the relativistic de Broglie wavelength for a kinetic colliding partice is assumed to be

λ=hcT=hc(pc)2+(mc2)2−−−−−−−−−−−−√−mc2

Is this last relativistic consistent in certain limit (it seems is the ultra-relativistic case) to the previous one?