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I have a tricky problem involving an electron and photon.

  1. May 12, 2013 #1
    1. The problem statement, all variables and given/known data
    The question I have is simple enough: at what energy do an electron and a photon have the same wavelength?

    2. Relevant equations
    I know that for a photon, λ=[itex]\frac{hc}{E}[/itex] and for an electron, λ=[itex]\frac{hc}{\sqrt{E^{2}-mc^{2}}}[/itex]

    3. The attempt at a solution
    I can't just equate the two, as that results in the absurd result mc^2=0. I can't think of a method that works. Are there any relativistic corrections that I've missed?
     
  2. jcsd
  3. May 12, 2013 #2

    mfb

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    At the same energy, photon and electron cannot have the same momentum, their corresponding wavelength cannot be equal.
     
  4. May 12, 2013 #3

    rude man

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    The problem is with your energy of the electron, I believe.

    What is the kinetic energy K of the electron as a function of its deBroglie wavelength?
    (Hint: K = K(p,E0), p = momentum, E0 = rest energy, then substitute deBroglie λ for p and equate this K(λ) to the photon energy hc/λ.)

    Remember that when we speak of the energy of a particle with finite rest mass we mean its Kinetic energy, not its total Energy.

    EDIT: mfb may be right, probably is, but try to solve for lambda anyway. You'll find out soon enough!

    EDIT EDIT: The answer is lambda = infinity.
     
    Last edited: May 12, 2013
  5. May 12, 2013 #4
    Check the units under the square root. Do you think they are considtent?
     
  6. May 12, 2013 #5
    Thanks for the help. Will the momentum need a γ correction?

    It's from E2=(pc)2+(mc2)2

    EDIT:
    Is the equation [itex]\lambda=\frac{hc}{\sqrt{2E_{k}mc^{2}}}[/itex] valid?
     
    Last edited: May 12, 2013
  7. May 12, 2013 #6
    It does not follow from that. Which is not surprising, because it is inconsistent and cannot follow from anything meaningful.
     
  8. May 12, 2013 #7

    rude man

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    You should write the rest mass as m0.
     
  9. May 13, 2013 #8

    mfb

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    Good catch, I missed that, even if it does not change the result.

    @rude man: m is used for mass ("rest mass") everywhere in physics, the usage of m0 (as notation) or a relativistic mass are deprecated.
     
  10. May 13, 2013 #9

    rude man

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    No.

    When dealing in relativity, m is relativistic mass. As in E = mc2.
    Rest mass is then distinguished by giving it the symbol m0.. Check any textbook, e.g. Weidner and Sells.

    Obviously, in dealing with Newtonian mechanics, m is rest mass.
     
  11. May 14, 2013 #10
    mfb is right, the relativistic correction is on the momentum, not the mass.
     
  12. May 14, 2013 #11

    rude man

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    That's what I said, and why I insisted (and insist) that the mass here be denoted m0 instead of m, to clarify the fact that it is not the relativistic but the rest mass.

    The correct expression thus becomes, as I said, E2 = (pc)2 + (m0c2)2.
     
  13. May 14, 2013 #12
    There is no such thing as relativistic mass.
     
  14. May 14, 2013 #13

    rude man

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    Gee, I hope you're kidding ... :surprised
     
  15. May 14, 2013 #14

    Dick

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    Of course there is a notion of relativistic mass. It just isn't used much anymore. So more modern books just use m as rest mass and don't talk about 'relativistic mass'. It causes more confusion than it is useful.
     
  16. May 14, 2013 #15

    rude man

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    Thanks for the clue-in, Dick. Amazing!

    So you can't say m = γm0 any more? So how do they e.g. write kinetic energy? Used to be mc2 - m0c2. What is it now?

    Oh my.

    So when the OP wrote E2 = (pc)2 + (mc2)2, that was right because m is now rest mass? And what symbol do you use for (may God forgive me for saying it) relativistic mass?

    And E = mc2 is now "energy = rest mass times c2"?

    I am impressed. I hope Einstein knows about this ...
     
    Last edited: May 14, 2013
  17. May 14, 2013 #16

    Dick

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    Kinetic energy is just ##mc^2 (\gamma-1)##. I, personally, have always taken E=mc^2 to mean equivalence between rest mass and energy. Energy in a boosted frame is ##\gamma mc^2##. I didn't make the choice to avoid the use of 'relativistic mass'. But that seems to be the way the market is choosing. I don't think Einstein would be upset. It's just a change in notation.
     
  18. May 14, 2013 #17

    rude man

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    I am rendered speechless. And I do think Einstein would seriously not be pleased. But OK, we bend with the wind ...
     
  19. May 14, 2013 #18

    Dick

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    I don't see why. You sprinkle a few more gamma's around. Drop some m0's. It's still just classical special relativity. There's no conceptual change, just notation. If Einstein were rendered speechless by this, I'd say he had a pretty thin skin. :)
     
  20. May 15, 2013 #19
    If you think about it, an objects mass cannot increase as it's an extensive property based on its internal structure, which doesn't change with velocity.
     
  21. May 15, 2013 #20

    mfb

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    @rude man: There is no need to get sarcastic. Check any recent (after 1980-1990) publication, you will not find the concept of a relativistic mass there.
     
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