# De Broglie's relation in general relativity

In his 1924 PhD thesis, de Broglie proposed that just as light has both wave-like and particle-like properties, electrons also have wave-like properties. He derived the relation

$$p=\frac{h}{\lambda}.$$ As far as I understand, this relation is derived in the framework of special relativity. Does it also hold in general relativity?

De Broglie's relation $$p=\frac{h}{\lambda}.$$ is a very old and mostly outdated concept. Nowadays nobody uses it for anything. De Broglie's theory was soon superseded by Schrödinger's theory, and people only use the former as a teaching aid in schools. The range of applicability of de Broglie's theory was very limited when it was proposed, and today we only study it because of its historical relevance, as it laid down the foundations of QM in general, and wave mechanics in particular.

Nowadays, de Broglie's relation is regarded as a definition of $\lambda$, $$\lambda\equiv \frac{h}{p}$$ and as such, is just as valid in SR as it is in GR. On its own, it is devoid of physics. In any case, it is not derived in the framework of special relativity, because it is not a derived relation. It's just a definition.

• I think you are overly harsh with one of the greatest ideas in the history of physics: the wave nature of matter. Romer's estimate of the speed of light is way wrong for being of any use today, still it is one of the most important experimental result ever, namely that the speed of light is finite. – hyportnex Feb 4 '17 at 21:51
• @Accidental FourierTransform In SR we have the identity (d=1): $$\frac{E}{p}\frac{\partial E}{\partial p}=c^2,$$ that is, the product of the group velocity and the phase velocity is equal to $c^2$. Therefore, one can write $$p=\gamma m v_{group}=\frac{E}{c^2}v_{group}=\frac{h\nu}{c^2}v_{group}=\frac{h\nu}{v_{phase}}=\frac{h\nu}{\nu\lambda}=\frac{h}{\lambda}.$$ – user56224 Feb 5 '17 at 9:02
• @hyportnex of course it was a great idea: de Broglie got a Nobel out of it! But it's an outdated idea. Today it's useless. You wouldn't use Romer's estimate for an actual calculation today, would you? – AccidentalFourierTransform Feb 5 '17 at 15:05
• @kaffeeauf your first equation is only valid of a massless particle. For a massive particle, $$\frac{E}{p}\frac{\partial E}{\partial p}=c^2\left(1+\frac{m^2c^2}{p^2}\right)$$ – AccidentalFourierTransform Feb 5 '17 at 15:06
• Let $E^2=(mc^2)^2+(pc)^2$ and $m\neq 0$. Then $$\frac{\partial E^2}{\partial p}=2pc^2.$$ On the other hand we have $$\frac{\partial E^2}{\partial p}=2 E \frac{\partial E}{\partial p}.$$ Equating both it follows $$\frac{E}{p}\frac{\partial E}{\partial p}=c^2.$$ – user56224 Feb 5 '17 at 17:56