Energy-momentum formula and deBroglie wavelength

[sha1000]

Hi everyone,

Im a little bit confused about deBroglies procedure on introducing his famous Matterwave formula.

People already knew that the wavelength of the light was equal to Lambda = h/p. The term p comes from the energy-momentum formula; for the light the restmass = 0 so E =pc etc.

As far as I know deBroglie just postulated that this can be applied not only to photons but also to other particles.

So my question is why don't we take into the account that restmass of other particles is non-zero. In this case the energy should be E = sqrt (m₀²c⁴ + p²c²) and not E = pc? But deBroglie applied the photons formula to other particles (even though the restmass of other particles is nonzero) and it actually worked...

What am I missing? To what precision the deBroglie formula was experimentally tested?

Thank you

 

[PeterDonis]

As far as I know deBroglie just postulated that this can be applied not only to photons but also to other particles.

He postulated that the formula ##\lambda = h / p## can be applied to particles other than photons.

He did not postulate that the formula ##E = pc## can be applied to particles other than photons.

why don't we take into the account that restmass of other particles is non-zero.

We do. So did de Broglie. See above.

 

[sha1000]

He postulated that the formula ##\lambda = h / p## can be applied to particles other than photons.He did not postulate that the formula ##E = pc## can be applied to particles other than photons.

We do. So did de Broglie. See above.

Thank you for your reply.But the photons formula lambda = h/p is constructed from E = h×f and E² = m₀²c⁴ + p²c². From here one can write that: lambda = h/sqrt (m₀²c² + p²) [1]I would say that this is a general expression for any particle and that photon is a special case with m₀ = 0; and the expression becomes:lambda = h/p [2]For me it would be more natural to use the general expression [1] for massive particles. I really don't understand why we are using the second one which is just a special case for massless particles.

 

[PeterDonis]

the photons formula lambda = h/p is constructed from E = h×f

It can be, if you start with E = h f as your assumption. But you can equally well start with lambda = h/p as your assumption and derive E = h f from it for a massless particle. The logic works fine either way.

De Broglie started with lambda = h/p as his assumption, not E = h f. If you start with lambda = h/p, and then apply the energy-momentum relation for massive particles instead of massless particles, you don't end up with E = h f, and de Broglie never claimed that you would.

In fact, as far as I know, de Broglie didn't consider the relativistic energy-momentum relation at all; he just assumed lambda = h/p, and proposed that that relation be used to construct a non-relativistic quantum theory for a massive particle. You don't need any relativistic energy-momentum relation to do that.

I really don't understand why we are using the second one which is just a special case for massless particles.

As far as I can tell, you are not talking about what de Broglie actually did at all, but about some straw man theory that you have constructed. The supposed logic you are saying won't work is not logic that de Broglie, or anyone who built on his work, ever used.

 

[sha1000]

Ok. I see now. I was told that he deduced this expression from lambda=h×f. But if he didnt do it and just postulated this expression, then this makes sense to me.

Thank you for this clarification.