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The De Broglie Wave Equation

  1. Jun 20, 2004 #1
    I recently came across a proof for the De Broglie wave equation in a book, which went as follows:
    E of photon = mc2
    = m*c*c
    = (m*c)c
    = (p)c ( ie - momentum*speed of light)
    = (p)(f*lamda)
    Therefore, hf = p(f*lambda)
    Therefore, p = hf/ f*lambda
    = h/lambda.
    Therefore, Lambda (ie- wavelength) = h/ mv.

    However, I'm not sure if I agree with this, so I wanted to ask a few questions. Firstly, it is true that the energy of a photon = hf, but I dont think that it should equal mc2. After all, this Einsteinian equation literally means energy produce = mass destroyed*speed of light. So, what mass has been destroyed to create the energy of the photon? Furthermore, how can we take mc to be the momentum of the photon, if this is mass destroyed, not mass of the photon?
    As slightly different question, what exactly is the wavelength of a particle? If the wavelength of a wave is "the distance between two crests or adjacent points", then what is the wavelength of a particle?
    Thanks is advance. :-D
  2. jcsd
  3. Jun 20, 2004 #2


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    Really tou need the equation:

    [tex]E^2 = m_0^2c^4 + p^2c^2[/tex]

    As a photon has zero mass:

    [tex]E = pc[/tex]

    In your book they've defined m as relativistic mass.
  4. Jun 20, 2004 #3
    I've had a think and could you say that, since if you accelerate and object and it gains XJoules of KE it will gain X/c2 kg of mass, that in creating the photon you have "destroyed" some mass and thus that the photon must have mass? It must do so that when it gives its energy to something else, that thing will gain energy AND mass.
  5. Jun 20, 2004 #4


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    The photon doesn't have mass as mass is usually defined as the rest mass.
  6. Jun 20, 2004 #5
    The verificiation of the De Broglie formula (I am lead to believe) comes from considering a de Broglie wave:
    [tex]$\psi ({\bf{r}},t) = Ae^{i({\bf{k}}.{\bf{r}} - \omega t)} $

    If you assume the relationshup [tex]E = \hbar \omega [/tex] holds for material particles you then write [tex]E = \hbar \omega = \frac{{m_0 c^2 }}{{\sqrt {1 - \beta ^2 } }}[/tex], use this in the group velocity formula and obtain the required result.

    However to show that [tex]E = \hbar \omega [/tex] is a consitent step, you show the invariance of the quantity [tex]kr - \omega t
    [/tex] in two inertial frames S and S'.
    Last edited: Jun 20, 2004
  7. Mar 18, 2009 #6
    sorry, there is mass/energy and it is conserved. A photon has mass/energy and it may be expressed as mass or frequency. E=hn=mCC
  8. Mar 21, 2009 #7
    Just consider that all physics equations : E = mc2 ; E = hf ; Gij = Tij; can be expressed in a single way depending on th scale. This single law is :

    extensity flux = diffusivity * extensity concentration gradient.
  9. Mar 21, 2009 #8


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    this thread is 5years old..
  10. Mar 21, 2009 #9
  11. Mar 21, 2009 #10
    The mass of a photon is: m=hn/CC; planks constant times the frequency divided by the speed of light squared.
    Mass is mass; mass times accelleration is force but for a photon accelleration is defined as CC ie. C squared. Thus F=mCC.
  12. Mar 21, 2009 #11


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    sammy, this is one of the most discussed things in this forum, Please read the thread that was suggested to you.

    Now, WHERE and WHY should the acceleration of a photon be defined in that way? There is no reason and meaning to that, c^2 does not even have the units of acceleration.

    sammy, there equation is [itex]E=m c^2[/itex], where [itex]m = \gamma m_0[/itex] and [itex]\gamma = 1/\sqrt{1-v^2/c^2}[/itex].

    The equation [itex]E=m c^2[/itex] can also be written as [itex]E^2 = (pc)^2 + (m_0c^2)^2[/itex].

    The energy of a photon is [itex]E = \hbar \omega[/itex]

    Now, try to fit the photon-energy equation to make [itex] \hbar \omega = mc^2 [/itex]... and you will find:

    [itex]E = \hbar \omega = mc^2 = \gamma m_0c^2[/itex] now the rest mass of the photon is zero, and \gamma for the photon is infinite, since the photon moves at c. So what happened to the photon mass?

    we have that [itex]m = m_0 \gamma = 0 \times \infty [/itex] .. nice

    BUT this is nice, since m_0 = 0:

    [itex]E^2 = (pc)^2 \rightarrow [/itex]
    [itex]p = E/c[/itex]
    Last edited: Mar 21, 2009
  13. Mar 21, 2009 #12
    I agree with you, cc has not the unit of acceleration. But cc/l with "l" being wave length has the unit of acceleration. it is then easy to make de broglie and einstein equations equivallent.
  14. Mar 21, 2009 #13


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    in what sense is deBroigle and "Einstein equation" equivalent?
  15. Mar 21, 2009 #14
    They are the same mathematical description of the quantic reality. And you can not imagine how Sammy is close to the final solution when he writes F = m cc !!
  16. Mar 21, 2009 #15


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    Sorry dude, but you have to refer to accepted science, published in peer review journals, not your own opinions and idea what "quantic reality" is etc.

    Force is instantaneous change momentum also, so there is in the first place no need for an equation of the kind F = mc^2, since F = dp/dt will work too, using p = E/c for a photon.
  17. Mar 21, 2009 #16
    ok I stope. But there is no fault when you are doing true calculations directly. You do it yourself F = dp/dt... I can do for example E/V = P (Pa) = mc2/V = mc2/S.l et c2/l (m.s-2) = S/m * P
  18. Mar 21, 2009 #17


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    This thread is from 5 years ago. There is no need to dredge up something that old, thus I am locking this.
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