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DeBroglie and Time Dilation Paradox

  1. May 25, 2009 #1
    I need someone to resolve this paradox for me.

    Before I start, here are the basic ideas:
    Okay, so every moving particle has a DeBroglie wavelength: E=hc/[tex]\lambda[/tex].
    This also means that each particle has a frequency: E=hf
    So it also has a period: E=h/[tex]\tau[/tex]
    So any given mass has a DeBroglie period of [tex]\tau[/tex]=h/mc^2

    Consider the train problem with the light clock. The guy on the train sees the light clock ticking normally. Now an outside observer sees the train moving and thus will see the light clock ticking slower. This is due to time dilation: t' = t[tex]\gamma[/tex].

    Okay, so here is the paradox: Now imagine you have an electron on the train instead of the light clock and you are monitoring it's DeBroglie period. Now since it is a time period, it should appear to increase in duration to the observer outside of the train. This cannot happen however, because if the time period increase, the frequency would decrease meaning the energy would decrease. A moving particle must have more energy than it does at rest.

    What is the deal here?
  2. jcsd
  3. May 25, 2009 #2
    If we talk use the example of photons (visible reigion) instead of electrons (since both behave as waves) we would observe a shift towards the red end of the spectrum- the doppler shift. So this leads to another question; is the energy of red-shifted light lower than it would be (if observed at rest relative to the source) because it appears to have a lower frequency?
  4. May 25, 2009 #3

    Jonathan Scott

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    Gold Member

    As usual with relativity, you need to consider the whole picture, not just one part of it.

    The phase of the de Broglie wave with energy E and momentum p (ignoring unit conversion constants of hbar) is given by the integral of (E dt - p.dx), which for constant velocity can be rewritten as

    phase = E (1-v2/c2) t.

    If the energy in the rest frame is E0 = mc2 then the increased energy E of the moving object is given by

    E = E0/sqrt(1-v2/c2).

    Substituting for E in the previous expression, this gives:

    phase = E0 sqrt(1-v2/c2) t

    From this, you can see that when the space-like phase change due to momentum is taken into account, the rate of change of phase with time does indeed decrease with the time dilation factor, as previously mentioned, but the energy increases by the same factor.
  5. May 25, 2009 #4


    Staff: Mentor

    Here is a good page to start learning the four-vector de Broglie relation which unifies the usual three-vector de Broglie relation and Planck's relation.

    http://physics.nmt.edu/~raymond/classes/ph13xbook/node73.html [Broken]

    Let's take the example of an electron. Given the four-momentum p = (E/c,px,py,pz) and the wave-four-vector k = (f/c,kx,ky,kz) where E is the total energy of the electron, px is the x component of the electron's momentum, f is the (temporal) frequency and kx is the wavenumber (spatial frequency). The four-vector formulaiton of the DeBroglie relation is:
    p = hk

    So, for an electron at rest
    p = (2.7E-22,0,0,0) N s
    k = p/h = (4.1E11,0,0,0) m-1
    f = 1.2E20 Hz

    Boosting to the frame where the electron is moving at 0.6 c in the x' direction.
    p = (3.4E-22,2.0E-22,0,0) N s
    k = p/h = (5.2E11,3.1E11,0,0) m-1
    f = 1.5E20 Hz

    So we see that the full formulation of the DeBroglie relation correctly predicts the higher frequency in the frame where it has higher energy.
    Last edited by a moderator: May 4, 2017
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