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Lorentz transformation of energy and E = h f

  1. Jul 25, 2013 #1
    Can one explain the relativistic energy transformation formula:

    [itex]E = \gamma\ E',[/itex]

    where the primed frame has a velocity [itex]v[/itex] relative to the unprimed frame, in terms of relativistic time dilation and the quantum relation [itex]E=h\ f[/itex]?

    I imagine a pair of observers, A and B, initially at rest, each with an identical quantum system with oscillation period [itex]T[/itex].

    Now A stays at rest whereas B is boosted to velocity [itex]v[/itex].

    Just as in the "twin paradox" the two observers are no longer identical: B has experienced a boost whereas A has not. Both observers should agree on the fact that B has more energy than A.

    From A's perspective B has extra kinetic energy by virtue of his velocity [itex]v[/itex]. Relativistically A should use the energy transformation formula above.

    But we should also be able to argue that B has more energy from B's perspective as well.

    From B's perspective he is stationary and A has velocity [itex]-v[/itex]. Therefore, due to relativistic time dilation, B sees A's oscillation period [itex]T[/itex] increased to [itex]\gamma\ T[/itex].

    Thus B finds that his quantum oscillator will perform a factor of [itex]\gamma\ T/T=\gamma[/itex] more oscillations in the same period as A's quantum system.

    Thus B sees that the frequency of his quantum system has increased by a factor of [itex]\gamma[/itex] over the frequency of A's system.

    As we have the quantum relation, [itex]E=h\ f[/itex], this implies that B observes that the energy of his quantum system is a factor of [itex]\gamma[/itex] larger than the energy of A's system.

    Thus observer B too, using his frame of reference, can confirm that his system has more energy than observer A's system.

    Is this reasoning correct?
     
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  3. Jul 25, 2013 #2

    PeterDonis

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    The two formulas are about two different interpretations of the word "energy", and I'm not sure they can be combined in the way you're trying to. See below.

    Which means that both will have energy equal to their rest mass ##m## (in units where ##c = 1##), in their mutual rest frame. So if the quantum oscillation formula is applicable, we must have ##m = h f = h / T##.

    No, they won't; B will say that A has more energy. The asymmetry between the observers in the twin paradox does not arise just from their being in relative motion; there's more to it than that. I suggest reading the Usenet Physics FAQ pages on the twin paradox:

    http://www.desy.de/user/projects/Physics/Relativity/SR/TwinParadox/twin_paradox.html

    Which means that, in A's frame, B has energy ##E = \gamma m = \gamma h / T = h / (T / \gamma) = h / T'##. In other words, ##T'## must be *smaller* than ##T## for B's energy to increase due to his motion, if the quantum oscillation formula is applicable here.

    But the same reasoning can be applied to B from A's perspective: from A's perspective, B is time dilated and his oscillations slow down. So by relativistic time dilation, B's oscillation period should increase, relative to A, due to his motion. But that isn't consistent with the quantum formula above; as we just saw, that formula requires B's oscillation period to *decrease*, relative to A, due to his motion, in order for his energy relative to A to increase. So something must be wrong with using the quantum formula this way.
     
  4. Jul 26, 2013 #3
    After thinking about it again it seems to me that from either observer's point of view, A or B, in order to convert the energy of a particular atomic system in a moving frame to his energy scale he must multiply it by a factor [itex]\gamma[/itex]. This is because in his frame such atomic systems are oscillating with [itex]\gamma[/itex] times the frequency of the atom in the moving frame.
     
  5. Jul 26, 2013 #4

    Dale

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    In relativity energy transforms as the timelike component of the four-momentum ##\mathbf{P}=(E/c,\mathbf{p})##. Frequency transforms as the timelike component of the wave four-vector ##\mathbf{K}=(\omega/c,\mathbf{k})##. The relativistic deBroglie relationship holds in all frames: ##\mathbf{P}=\hbar\mathbf{k}##
     
  6. Jul 26, 2013 #5
    In my last post I made the statement that transformation of energy between inertial frames is just a matter of multiplying by a factor [itex]\gamma[/itex]. This statement is true for rest energies. In the unprimed frame, in which the particle is moving with velocity [itex]v[/itex], one must subtract the energy of the particle due to its motion before it can be transformed to the primed frame.

    Here are the details. The Lorentz transformation equations for energy and x-momentum between an unprimed frame and a primed frame moving with velocity [itex]v[/itex] relative to it are:
    [tex]
    E' = \gamma (E - v p) \\
    E = \gamma(E' + v p')
    [/tex]
    As the object is at rest in the primed frame we have [itex]p' = 0[/itex] so the second equation becomes:
    [tex]
    E = \gamma E'
    [/tex]
    So to convert rest energy measured in the primed frame to the unprimed frame we multiply by a factor [itex]\gamma[/itex].

    It is my contention that due to the symmetry of the equations we should also convert rest energy measured in the unprimed frame to the primed frame by simply multiplying by [itex]\gamma[/itex].

    What is the rest energy in the unprimed frame?

    Let us assume that the rest energy in the primed frame is given by:
    [tex]
    E' = m c^2
    [/tex]
    where [itex]m[/itex] is the rest mass of the particle.

    Then by my hypothesis the rest energy of the particle in the unprimed frame should be:
    [tex]
    \frac{mc^2}{\gamma} = \gamma m c^2(1-\frac{v^2}{c^2}) = \gamma m c^2 - \gamma m v^2
    [/tex]

    Thus the term [itex]\gamma m c^2[/itex] is the total energy of the particle in the unprimed frame and the term [itex]\gamma m v^2[/itex] is the kinetic energy of the particle in the unprimed frame.

    I admit this is not the usual definition of relativistic kinetic energy which is usually defined as:

    [tex]
    KE = \gamma m c^2 - m c^2.
    [/tex]
     
    Last edited: Jul 26, 2013
  7. Jul 26, 2013 #6

    jtbell

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    [Oops, I had to go off and do something else for a while while I was writing this, and some posts slipped in ahead of me.]

    That's not the correct equation for transforming energy. You have to transform energy and momentum together using a Lorentz transformation, just like you have to transform time and position together:

    $$p^\prime c = \gamma (pc - \beta E) \\
    E^\prime = \gamma (E - \beta pc)$$

    For a photon, pc = E and p'c = E', so the second equation becomes

    $$E^\prime = \gamma (1 - \beta) E$$

    Substituting E = hf and E' = hf' and doing some algebra gives you the relativistic Doppler shift formula.
     
  8. Jul 26, 2013 #7
    Ok - I was taking [itex]p\prime = 0[/itex].
     
  9. Jul 26, 2013 #8

    jtbell

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    Graphing the two formulas versus v (actually I did it versus β=v/c) shows that they don't give the same results.
     
  10. Jul 26, 2013 #9

    PeterDonis

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    But the energy in the unprimed frame is not "rest" energy, because the object is not at rest in the unprimed frame.

    Rest energy in the unprimed frame of what object? An object can't be at rest in two frames that are in relative motion. I don't understand what you're trying to do here.
     
  11. Jul 26, 2013 #10
    By rest energy in the unprimed frame I mean total energy minus the energy due to motion.
     
  12. Jul 26, 2013 #11

    Dale

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    The rest energy is invariant. It is equal to the invariant mass times c^2 in all frames.
     
  13. Jul 27, 2013 #12

    PeterDonis

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    But there's no subtraction in your formulas anywhere, so I don't understand how this corresponds to the math you were trying to do. If you're trying to work backwards, so to speak, to find a formula for the kinetic energy, then you would take the total energy, ##\gamma m c^2##, and subtract off the rest energy, ##m c^2##, to get the kinetic energy, ##( \gamma - 1 ) m c^2##; i.e., you would use what you call the "usual definition". I don't understand why that would pose a problem.
     
  14. Jul 27, 2013 #13

    PeterDonis

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    But the equations aren't symmetrical; ##p' = 0## but ##p## is not zero, so the first equation is not the same as the second.
     
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