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De Broglie+photons

  1. Sep 15, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Assume that electromagnetic waves are a special case of de Broglie's waves. Show that the photons must travel at a speed c and that their rest mass is zero.


    2. Relevant equations
    [itex]E=\sqrt {p^2 c^2+m_0 ^2c^4}[/itex].
    [itex]\lambda _ B =\frac{h}{p}[/itex].

    3. The attempt at a solution
    So I've been playing with 2 stuffs and fell over weird non sense.
    I assumed that the de Broglie's wavelength was worth the wavelength of a photon (it doesn't make any sense I guess since it lead me into non sense).
    p=E/c.
    But for a photon, [itex]E= h \nu[/itex]. This gives me [itex]p=\frac{h \nu}{c}=\frac{h^2}{\lambda}\neq \frac{h}{\lambda _B}[/itex] as the relation of the definition of wavelength.

    Another try I made:
    [itex]p=E/c=\frac{\sqrt {p^2 c^2+m_0 ^2c^4}}{c}[/itex].
    So that [itex]\lambda _B =\frac{hc}{\sqrt {p^2 c^2+m_0 ^2c^4}} \Rightarrow \nu=\sqrt {p^2 c^2+m_0 ^2c^4}[/itex] which is also worth the energy of a photon and makes absolutely no sense...
    So I think I can't assume that [itex]\lambda = \lambda _B[/itex]. Hmm now I don't know any other way to tackle the problem. Any help is appreciated.
     
  2. jcsd
  3. Sep 16, 2011 #2

    vela

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    Actually, you do want to assume [itex]\lambda = \lambda_B[/itex]. I think you're just making algebraic errors, leading to confusion.

    I'm not sure how you're supposed to show the two conclusions, however, without assuming at least one of them is already true.
     
    Last edited: Sep 16, 2011
  4. Sep 16, 2011 #3

    BruceW

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    I think this is wrong:
    [tex]\frac{h \nu}{c} = \frac{h^2}{\lambda}[/tex]
    How did you make this equality? I looks like you tried to use [itex]\nu \lambda = h c[/itex], but this definitely isn't right.

    And on the other try, you wrote:
    [tex]\lambda_B = \frac{hc}{\sqrt{p^2c^2+m_0^2c^4}} \Rightarrow \nu = \sqrt{p^2c^2+m_0^2c^4} [/tex]
    The first bit is right, but to then get the bit on the right of the arrow, it looks like you used [itex]\lambda_B \nu = hc[/itex] again, which isn't right.
     
  5. Sep 16, 2011 #4

    fluidistic

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    Oh right guys, a "nasty" algebraic error, considering I had solved lots of previous problems without doing the error.
    [itex]p=\frac{\sqrt {p^2 c^2+ m_0 ^2 c^4} }{c}\Rightarrow p^2 =p^2+m_0 ^2 c^2 \Rightarrow m_0=0[/itex].
    I take [itex]p=\gamma m_0 v=\frac{m_0v}{\sqrt{1-\frac{v^2}{c^2}}}[/itex]. The only way for this expression to be different from 0 is to have a denominator that "blows up", so v=c... But I see no reason for p to be different from 0, from a mathematical point of view (physically I do know that the momentum of a photon isn't 0, of course).
     
  6. Sep 16, 2011 #5

    vela

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    Assume E=pc, [itex]E=h\nu[/itex], and [itex]p=h/\lambda[/itex] and calculate [itex]\lambda\nu[/itex].

    The problem I have with assuming E=pc is that you're essentially assuming v/c=pc/E=1, which is one of the things you're supposed to prove.
     
  7. Sep 16, 2011 #6
    This might also be useful: v = sqrt(1 - (Erest/Etotal)^2).
     
    Last edited: Sep 16, 2011
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