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I De Broglie wavelength - relativistic correction

  1. Apr 16, 2016 #1
    According to the wiki page:
    the generalised form for the de Broglie wave is simply:
    [tex] \lambda = h/p[/tex]

    I suppose this not correct, because there is no trasform which can change only one side of the equation.
    In this case we have two variables:
    1. a momentum: [tex]p = mv[/tex]
    2. a wavelength: [tex]\lambda[/tex]

    the h is just a number - constant.

    Therefore the correct, transformed version, of the whole equation is:
    [tex]\gamma\lambda = h/{\gamma mv}[/tex]

    thus the final - general the de Broglie relation is:
    [tex]\gamma^2\lambda = h/{mv}[/tex]
    [tex]mv\lambda = h(1-v^2/c^2)[/tex]

    So, what you think about my discovery, proposition? :)

    Is there possible to resolve (experimentally) which form is correct?
  2. jcsd
  3. Apr 16, 2016 #2


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    The usual formula for the DeBroglie wavelength is correct


    The expression for momentum changes from [itex]mv[/itex] to [itex]\gamma mv[/itex], but the relationship between momentum and wavelength remains the same.

    In fact, the DeBroglie formula was derived with special relativity.

    If you have a stationary wavepacket of mass M oscillating at frequency [itex]f[/itex], such that its energy [itex]E[/itex] is [itex]h f[/itex], then in a different reference frame, moving with respect to the wave packet, the Lorentz transformation, (and the relativity of simultaneity) shows that that wavepacket will have a momentum [itex]p[/itex] equal to [itex]h[/itex], divided by the observed wavelength [itex]\lambda[/itex].
    It's one of the more elegant findings in the early days of quantum physics.
  4. Apr 16, 2016 #3
    Indeed. But this transformed version of the equation is mathemathicaly inconsistent.

    Therefore my claym is: the wavelength should be transformed too.

    Disprove my claim.

    BTW. The radiation of accelerated charge is proportional to the gamma^2 also.

    And the radiation is just prominent example of the wave phenomena.
    So, now I have a very strong argument!
    Last edited: Apr 16, 2016
  5. Apr 17, 2016 #4


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    Don't bother with old theories outdated since 1925! Learn quantum mechanics!
  6. Apr 17, 2016 #5


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    How did you come up with the left-hand side of that equation?
  7. Apr 17, 2016 #6


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    I'm not sure what inconsistency you're hinting about, but let me sketch how something like the de Broglie relation is true relativistically.

    Instead of using wavelength, it's more convenient to deal with wave number [itex]k[/itex], which in one dimension is related to wavelength through:

    [itex]k = \frac{2\pi}{\lambda}[/itex]

    In terms of [itex]k[/itex], the de Broglie relation is just:

    [itex]k = frac{p}{\hbar}[/itex]

    where [itex]\hbar = \frac{h}{2\pi}[/itex]

    Under a Lorentz transformation, it isn't simply that [itex]p \Rightarrow \gamma p[/itex]. The general relationship is (in one spatial dimension):

    [itex]p' = \gamma (p - \frac{v}{c^2} E)[/itex]
    [itex]E' = \gamma (E - v p)[/itex]

    [itex]k[/itex] and [itex]\omega[/itex] transform under Lorentz transforms similarly:

    [itex]k' = \gamma (k - \frac{v}{c^2} \omega)[/itex]
    [itex]\omega' = \gamma(\omega - k v)[/itex]

    What's invariant under Lorentz transformations is the phase of the wave. The phase is given by:

    [itex]\phi = k x - \omega t[/itex]

    The pair [itex](k,\omega)[/itex] and [itex](x,t)[/itex] both transform under Lorentz transformations in a way that makes the phase invariant.
  8. Apr 17, 2016 #7
    I simpy transforemed not the momentum only, but the length too, using the same transformation: the Lorentz's transform.

    I remember an obvious fact:
    a length - distance is not invariant under the Lorentz transformation!

    Maybe I show some further investigation.
    The standard formula for a circular motion is:
    [tex]pv = k/r[/tex]
    and the common de Broglie condition:
    [tex]pr = n h/2\pi = n \hbar[/tex]

    so, from the first equation: r = k/pv, thus the second eq.:
    [tex]pk/pv = k/v = n\hbar \to\ v = k/n\hbar = c\alpha/n \approx c/137n[/tex]

    And there is just nothing new: the relativistic corrections disappears completely!
    But that is impossible, because we know already the fast - relativistic
    particles behave in other way... due to the relativistic effects,
    therefore these effects can't and don't cancel so silently!
    Last edited: Apr 17, 2016
  9. Apr 17, 2016 #8
    That is nothing, because the phase is invariant under the Galilean transformation too!
  10. Apr 17, 2016 #9


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    The wave equation is not Galilei invariant. Check it!
  11. Apr 17, 2016 #10
    I told a phase is invariant, and under any transformation, therefore it's irrelevant in the context of the relativistic effects.
    But a distance is not invariant under Lorentz, thus the wavelength, in the de Broglie relation,
    which is a longitudinal distance, should be transformed too; not the momentum alone.
  12. Apr 17, 2016 #11


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    It's relevant to the extent that it shows you are wrong: Both [itex]p[/itex] and [itex]k = \frac{2\pi}{\lambda}[/itex] transform under Lorentz transformations in exactly the same way, so it continues to be true that [itex]p = \hbar k = \frac{2 \pi \hbar}{\lambda}[/itex]
  13. Apr 17, 2016 #12


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    So what? The point is that [itex]p = \hbar k = \frac{2 \pi \hbar}{\lambda}[/itex]. [itex]E = \hbar \omega = \frac{2 \pi \hbar}{T}[/itex]. Those equations are always true. Both [itex]p[/itex] and [itex]k[/itex] transform under a Lorentz transformation in exactly the same way:

    [itex]p' = \gamma (p - \frac{E}{c^2} v)[/itex]
    [itex]E' = \gamma (E - pv)[/itex]

    [itex]k' = \gamma (k - \frac{\omega}{c^2})[/itex]
    [itex]\omega' = \gamma (\omega - k v)[/itex]

    So [itex]p = \hbar k[/itex] and [itex]E = \hbar \omega[/itex] are preserved under Lorentz transformations; if they are true in one frame, they are true in every frame.
  14. Apr 17, 2016 #13


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    No, you didn't. The Lorentz transforms convert coordinates from one frame to another, and can be used to derive the correct formulas for transforming wavelengths and momenta from one frame to another. However, you haven't used these formulas. On the right-hand side you're assuming that ##p'=\gamma{p}## but that's not the correct transformation for momentum; and on the left-hand side you've just multiplied the wavelength by ##\gamma## and that's not the correct transformation for wavelengths.

    Take another look at @stevendaryl's analysis, and also pay close attention to the relativity of simultaneity, as @jfizzix suggests.
  15. Apr 18, 2016 #14
    You just promote my idea!
    Namely: momentum depends explicitly on the wavelength,
    therefore the wavelength must be transformed!

    If we leave the wavelength intact, then the momentum can't change too!
  16. Apr 18, 2016 #15


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    Yes, but wavelength also changes under Lorentz transformations.
  17. Apr 18, 2016 #16
    I don't care if the transformation is correct or not - it's just a standard form/transform of the momentum in the relativity.
    Last edited: Apr 18, 2016
  18. Apr 18, 2016 #17
    Indeed. Because that fact I transformed the wavelength, in the de Broglie relation, adequately. :)
  19. Apr 18, 2016 #18


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    That's silly. If you use an incorrect formula of course you'll get incorrect results, and that's what happened here. The wikipedia article does not support the transform you're using; you have to transform the velocity as well multiplying by ##\gamma##: ##p'=\gamma{m_0}v'## not ##p'=\gamma{m}_0v=\gamma{m}_0{p}##.

    We can give you the correct analysis, and you are free not to like it, but the forum rules don't allow continued argument after a mistaken premise has been pointed out. This thread is closed.
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