Suppose a particle of mass m in special relativity. Suppose the de Broglie wave-length is (non-relativistic) case: $$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}$$ In the case of RELATIVISTIC particle, the momentum is $p = \gamma m v$. Therefore a way to recast the de Broglie wavelength is: $$\lambda_{r}=\dfrac{h\sqrt{1-v^2/c^2}}{mv}$$ Suppose now that we focus on the kinetic energy. For a free particle, we get in the nonrelativistic case, $\mathrm{KE}=T=p^2/2m$, and thus $p=\sqrt{2Tm}$, and so $$\lambda=\dfrac{h}{\sqrt{2Tm}}$$ 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 $T=E-mc^2$, and from $E^2=(pc)^2+(mc^2)^2$, by simple substitution of $E=T+mc^2$, you get $(pc)^2=T^2+2Tmc^2$, $$p=\sqrt{\dfrac{T^2}{c^2}+2Tm}=\sqrt{2Tm\left(1+\dfrac{T}{2mc^2}\right)}$$ or $$\lambda=\dfrac{h}{\sqrt{\dfrac{T^2}{c^2}+2Tm}}=\dfrac{h}{\sqrt{2Tm \left(1+\dfrac{T}{2mc^2}\right)}}$$ Well, I have seen a couple of places where the relativistic de Broglie wavelength for a kinetic colliding partice is assumed to be $$\lambda=\dfrac{hc}{T}=\dfrac{hc}{\sqrt{(pc)^2+(mc^2)^2}-mc^2}$$ Is this last relativistic consistent in certain limit (it seems is the ultra-relativistic case) to the previous one?

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    $\begingroup$ What do you mean by "better" or "worse"? $\endgroup$
    – user174832
    Commented Mar 21, 2018 at 16:28
  • $\begingroup$ Consistency...I can not guess why the author/s defined the relativistic de Broglie waveglength for colliding particles...with the P=T/c instead of $(Pc)^2=(T^2+2Tmc^2)$... $\endgroup$
    – riemannium
    Commented Mar 21, 2018 at 16:44
  • $\begingroup$ Vocabulary term of the day "ultra-relativistic" regime. Occasionally also seen as "highly relativistic" and other variants. But if you are reading at an advanced level the authors probably expect you to be able to think like a physicist and that means paying attention to scale as it applies to the problem in front of you. $\endgroup$ Commented Mar 21, 2018 at 17:39
  • $\begingroup$ OH, I see...If $T^2>>2Tmc^2$, I get the second expression. Oh, I was so dumb...However, how can it be to do the approximation and then reinsert the masses? $\endgroup$
    – riemannium
    Commented Mar 21, 2018 at 18:02

1 Answer 1


Thinking about this answer: $$KE(rel)=(\gamma-1)mc^2$$ Thus, if T=KE(rel): $$\gamma=1+T/mc^2$$ If $T>>2mc^2>mc^2$: $$\gamma\approx T/mc^2$$ By the other hand, from the de Broglie fundamental relationship: $$\lambda=\dfrac{h}{p}=\dfrac{hc}{pc}=\dfrac{hc}{m\gamma v c}$$ and then, with $\gamma\approx T/mc^2$, and $v\approx c$ we obtain $$\lambda=\dfrac{h mc^3}{m T v c}=\dfrac{h c}{T}$$ From this last equation we get the above one from a simple use of the dispersion relationship. However, as the kinetic energy is much bigger than the rest mass, my doubt concerning the presence of mass remained...Until I did some numbers...


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