Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

I Do these relations have any physical significance

  1. Jan 5, 2017 #1

    ftr

    User Avatar

    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. jcsd
  3. Jan 5, 2017 #2

    PeterDonis

    User Avatar
    2016 Award

    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.
     
  4. Jan 5, 2017 #3

    ftr

    User Avatar

    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.
     
  5. Jan 5, 2017 #4

    PeterDonis

    User Avatar
    2016 Award

    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.
     
  6. Jan 5, 2017 #5

    ftr

    User Avatar

    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.
     
  7. Jan 5, 2017 #6
    Indeed, r is missing a factor of 2 pi. Then all is well.
     
  8. Jan 5, 2017 #7

    ftr

    User Avatar

    See post #5, I just wrote the formulas quickly but if you use h_bar instead of h then all is well.
     
  9. Jan 5, 2017 #8

    PeterDonis

    User Avatar
    2016 Award

    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.
     
  10. Jan 5, 2017 #9

    ftr

    User Avatar

    That was my question.:biggrin:

    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.:smile:

    http://home.claranet.nl/users/benschop/electron.pdf
     
  11. Jan 5, 2017 #10

    PeterDonis

    User Avatar
    2016 Award

    Staff: Mentor

    Then you shouldn't have said it was obvious.

    I'll take a look.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted