I Do these relations have any physical significance

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1. Jan 5, 2017

ftr

Photon energy

E_p= hv=hc/lamda

taking lamda= h/mc which is the electron Compton wavelength and substituting in above

E_p=mc^2

L(angular momentum)=r X P=(Lamda/2)*(E_p/c)=h/2

are these results coincident or have any physical meaning, they relate a photon wavelength equal to an electron Compton wavelength.

Last edited: Jan 5, 2017
2. Jan 5, 2017

Staff: Mentor

The energy result just means that a photon with wavelength equal to an electron's Compton wavelength will have an energy equal to the electron rest energy.

I'm not sure what you are calculating with the angular momentum result; the spin of a photon (or an electron) is intrinsic angular momentum, but the formula you give is for orbital angular momentum.

3. Jan 5, 2017

ftr

Ok thanks. the energy is obvious but it does have a strange aura about it. The intrinsic spin also coincide with the angular momentum, that is a strange double coincidence.

4. Jan 5, 2017

Staff: Mentor

Why? It's just a consequence of the definition of the Compton wavelength, which was defined as a convenience for analyzing experiments where photons scatter off electrons (Compton was one of the first to do them).

No, it doesn't. First, your value of $r$ doesn't make sense, because $r$ is the distance of the particle from some arbitrarily chosen point, not a wavelength or half wavelength. Second, your result is $h / 2$, but the spin of an electron is $\hbar / 2$, where $\hbar = h / 2 \pi$. And the spin of a photon is just $\hbar$. Your result doesn't match either of those.

5. Jan 5, 2017

ftr

I could have written h-->h_bar

obviously we must take the distance from the center of wavelength (i.e. divide by 2)

The formulas imply it calculates the spin of the electron "as if" the electron is a photon with the said wavelength.

6. Jan 5, 2017

Jilang

Indeed, r is missing a factor of 2 pi. Then all is well.

7. Jan 5, 2017

ftr

See post #5, I just wrote the formulas quickly but if you use h_bar instead of h then all is well.

8. Jan 5, 2017

Staff: Mentor

Which would change the equations, since you would have to insert factors of $2 \pi$ in some places but not in others. Try it.

It's not obvious to me. What physical meaning do you think this has?

It implies no such thing. If anything, it implies that you are thinking of the electron as a little spinning sphere with radius equal to half of its Compton wavelength, and calculating the orbital angular momentum of a point on the sphere's surface. And getting an answer that is not the same as the electron's intrinsic angular momentum.

Not in the angular momentum formula. By ftr's explicit assumption, he is using half the wavelength as the radius of the sphere, not its circumference.

9. Jan 5, 2017

ftr

That was my question.

Anyway Googling a bit I found a paper that tries to take advantage of that. I know no such models worked, but still it is interesting. Besides, I noticed it myself, it helps oil my old brain.

http://home.claranet.nl/users/benschop/electron.pdf

10. Jan 5, 2017

Staff: Mentor

Then you shouldn't have said it was obvious.

I'll take a look.