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Planck constant is Lorentz invariant?

  1. Aug 16, 2011 #1
    It is widely recognized in physics textbooks that Planck constant is a "universal constant". But I nerver see a proof. As we know, in the special theory of relativity, c is a universal constant, namely a Lorentz invariant, which is Einstein's hypothesis. But How do we know the Plack constant h is also a Lorentz invariant?

    Suppose E=hv in the lab frame, while E'=h'v' in the moving frame.
    Usually, one assume that photon's momentum and energy forms a momentum-energy 4-vector, generalized from a real particle, like an electron which has non-zero rest mass and of which the velocity is less than c. However the derivation of electron's 4-vector is not valid for a photon. (k, w/c) is Lorentz covariant from the invariance of phase, but we don't know if h is Lorentz invariant. [Of course, if h is Lorentz invariant, (h_bar*k, h_bar*w/c) is a Lorentz covariant momentum-energy 4-vector.]

    Therefore, the Lorentz invariance of Planck constant is only an artificial assumption. Am I right?
     
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  3. Aug 16, 2011 #2

    vanhees71

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    Planck's constant is assumed to be a Lorentz scalar, and quantum theory can be built in an explicitly Poincare-covariant way with this assumption. The resulting theory (which is relativistic quantum field theory) is one of the most successful scientific results ever, and thus we can be pretty sure that our assumption of [itex]\hbar[/itex] being a scalar universal constant is good. That's the nature of any model building in the natural sciences: You make assumptions and look where they lead you in terms of observable predictions. Then you do an experiment to check, whether these predictions are correct and within which limits of physical circustances they are valid etc.

    Of course, the momentum-four vector of a photon is Lorentz covariant. Otherwise it would not be a four vector to begin with! How do you come to the conclusion, it's not?
     
  4. Aug 16, 2011 #3

    PeterDonis

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    Can you elaborate on this? I'm not sure what you mean by "the derivation of electron's 4-vector".
     
  5. Aug 16, 2011 #4
    The electron's momentum-energy 4-vector is set up based on the 4-velocity V_4=(d/dt0)(x,ct) where t0 is the proper time and it is a Lorentz invariant. Since V_4 is Lorentz covariant and the electron's rest mass m0 is a Lorentz scalar from the principle of relativity, we conclude m0*V_4 = (p=m0*gamma*v, E/c=m0*gamma*c**2/c) is Lorentz covariant, and defined as momentum-energy 4-vector. But the derivation of V_4=(d/dt0)(x,ct) is not valid for a photon, because the photon's gamma-->infinity, a sigularity; in addition, the photon's rest mass is defined to be zero. Thus mathematically speaking, the derivation of electron's 4-vector is not applicable to a photon.
     
  6. Aug 16, 2011 #5
    It is found that, Einstein’s Doppler formula is not applicable when a moving point light source is close enough to the observer; for example, it may break down or cannot specify a determinate value when the point source and the observer overlap. From this, we judge that (k,w/c) for a moving point light source is NOT Lorentz covariant, and if [itex]\hbar[/itex]*(k,w/c) IS still Lorentz covariant, then [itex]\hbar[/itex] must NOT be a Lorentz invariant.

    I think to experimentally demonstarte the invariance of Planck constant is never easier than to demonstrate the invariance of one-way light speed.
     
  7. Aug 16, 2011 #6

    PeterDonis

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    Ah, I see. I'm not sure that what you have given is an actual "derivation" of the electron's momentum-energy 4-vector; you are assuming that 4-velocity is primary, which I'm not sure is a valid assumption. For one thing, it involves that "position vector" (x,ct), which, while it is Lorentz "covariant", as you note, is not, IMO, a good candidate for a "fundamental" quantity.

    An alternative would be to take 4-momentum as primary; after all, that's what we actually measure in experiments (we measure energies and momenta of particles like electrons, as well as photons; we don't measure positions, proper times, or velocities directly). You still have two "categories" of objects, those with nonzero rest mass and those with zero rest mass; but "rest mass" is now simply the invariant length of the energy-momentum 4-vector, which we obtain by the formula:

    [tex]E^{2} - p^{2} = m^{2}[/tex]

    So if the energy is *larger* than the momentum, we say the particle is "massive" (like the electron), and we define the "mass" as what's "left over" when we've subtracted out the momentum from the energy (physically, this corresponds to being at rest with respect to the object, hence the term "rest mass"). Whereas if the energy is *equal* to the momentum, we say the particle is "massless" (like the photon). We don't have to use 4-velocity at all in any of this; 4-velocity becomes a derived quantity, which we obtain for massive objects by dividing the 4-momentum by its length, so it's kind of a "unit vector" in the direction of the 4-momentum. For massless objects, like the photon, we obviously can't do the division, but we can still define their "4-velocity" as being a "unit" null vector in the direction of the photon's momentum, which does make sense mathematically.

    And (to get back to the question in the OP) if 4-momentum is primary, then it's obvious that Planck's constant has to be Lorentz invariant, because the formula you gave for photons, E = hbar w/c, p = hbar k, holds for *both* massive and massless objects; it's just that now that formula serves to define w and k, which are also derived quantities in this model. But since we can actually measure frequencies and wavelengths (which we can do for electrons as well as photons, for example in an electron diffraction experiment), we can check, as vanhees71 points out, to see that the formulas actually hold, and we find that they do, to high accuracy, for *both* electrons *and* photons (and every other quantum particle).
     
  8. Aug 16, 2011 #7

    PeterDonis

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    What formula are you calling "Einstein's Doppler formula"?
     
  9. Aug 16, 2011 #8
    I think you understand the invariance of Planck constant from the viewpoint of experiments.

    1. For a plane wave in free space, (k,w/c) is Lorentz covariant. There are two methods to develop the photon's momentum-energy 4-vector: (1) Directly assume h_bar*(k,w/c) is a Lorentz covariant 4-vector, then we conclude that h_bar must be a Lorentz invariant mathematically. (2) Directly assume that h_bar is a Lorentz invariant, then we conclude that h_bar*(k,w/c) is a Lorentz covariant 4-vector. The two methods are equivalent, and so the invariance of Planck constant is only an artificial assumption. I think you like method-(1), which actually artificially assumes the invariance of Planck constant.

    2. [tex]E^{2} - p^{2} = m^{2}[/tex] ---Mathematically, even m is a Lorentz scalar, (p,E) is NOT necessarily Lorentz covariant. For example, suppose E is the electric field vector and |E| is its absolute value; (E, |E|) --->|E|**2-E**2=0 but (E, |E|) is not a Lorentz covariant 4-vector, because E is transformed in terms of EM-field strength tensor.
     
  10. Aug 16, 2011 #9
    Einstein's Doppler formula is the Doppler formula for a plane wave: w'=w*gamm*(1-n.beta), which can be seen in university physics textbooks. If it is applied to the moving point light source, when the observer and the point source overlap, n.beta is an inderterminate value, because the angle between n and beta can be arbitrary.
     
  11. Aug 17, 2011 #10

    PeterDonis

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    If the point source is moving, the observer and the source will only overlap for an instant. At any other instant, the angle between n and beta is not arbitrary, so obviously by continuity that angle must be the same, at the one instant where the two overlap, as it is at neighboring instants where they don't. So the instant of overlap doesn't invalidate the Lorentz covariance of the 4-vector (k, w/c).
     
  12. Aug 17, 2011 #11

    PeterDonis

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    Yes, exactly. You don't *assume* that a given set of values forms a Lorentz covariant 4-vector, or that a single value is a Lorentz scalar; you *observe* whether the values behave in the appropriate ways when measured in different frames. Planck's constant is measured to be the same in different frames; therefore it is a Lorentz scalar. (For how we measure it, see next comment.)

    No, I like method (3): directly measure the energy and momentum of photons, and show that they transform as a Lorentz covariant 4-vector (p, E/c), *independently* of any measurements of frequency and wavelength. Then we know, physically, that *both* (p, E/c) *and* (k, w/c) are Lorentz covariant 4-vectors, from which it follows that their ratio, Planck's constant, must be a Lorentz scalar. So by measuring energy and momentum, *and* measuring frequency and wavelength, we can measure Planck's constant, and show that it is the same in all frames, as a Lorentz scalar must be.

    What do the electric field vector and its absolute value have to do with the energy-momentum 4-vector, (p, E/c)? The latter is the 4-vector that appears in the formula you gave; E in that formula is energy, not electric field, by definition. And as I noted in my previous post, we don't claim that (p, E/c) is a Lorentz covariant 4-vector based on m being a Lorentz scalar; that's backwards. We verify directly, by experiments measuring E and p in different frames, that (p, E/c) transforms as a Lorentz covariant 4-vector. Then it follows that the frame-invariant length of that 4-vector, which is m, must be a Lorentz scalar.
     
  13. Aug 17, 2011 #12

    Dale

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    I agree with this approach. The 4-momentum is a valid 4-vector and obeys all of the usual vector rules such as addition or scalar multiplication, regardless of whether it is a null vector or a timelike vector. The 4-velocity, on the other hand, does not follow vector rules like addition or scalar multiplication.

    A vector cannot have a norm of 1 and 0, so I have a hard time seeing how a unit null vector makes sense mathematically.
     
  14. Aug 17, 2011 #13
    The longitudinal Doppler effect from the Einstein's Doppler formula has a frequency-shift jump between the left-approaching overlap-point and the right-approaching overlap-point. I take the overlap-point as an example to show that Einstein's formula is not the exact formula for a moving point light source, although it is a good approximation when the observer is far away from the point source.
     
  15. Aug 17, 2011 #14
    You suggest a very good, original method to show the invariance of Planck constant. But I never see such experiments reported.
     
  16. Aug 18, 2011 #15

    PeterDonis

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    You're right, that was a poor choice of words. I was trying to express the idea that a photon's 4-momentum does have a "direction", which is simply the direction of its 3-vector spatial component. I'm sure there's a proper mathematical term for what I'm getting at, but I'm not sure what it is.
     
  17. Aug 18, 2011 #16

    PeterDonis

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    I assume that what you really mean is "left-approaching" and "right-receding" (or the reverse), as a description of the relative motion before and after the overlap. The frequency shift does change in this case from a blue shift (approaching) to a red shift (receding). However, this has nothing to do with any discontinuity in the *angle* between n and beta; it has to do with a change in the sign of beta (from positive, approaching, to negative, receding). I agree the change in sign happens, but it's more a matter of definition than physics; the "beta" in the Doppler formula is defined *differently* than the "beta" in the formulas for energy and momentum as they are used in the photon's energy-momentum 4-vector. There is no discontinuity in the actual motion of either object, or in the energy-momentum 4-vector of the photon.

    Well, Compton scattering experiments have clearly shown that photons can exchange momentum with electrons since the 1920's:

    http://en.wikipedia.org/wiki/Compton_scattering

    The photoelectric effect has been known to demonstrate that photons have energy since Einstein's paper on it was published in 1905:

    http://en.wikipedia.org/wiki/Photoelectric_effect

    (Compton scattering also shows that photons have energy, since they exchange energy as well as momentum with electrons.)

    I'm sure there are other more recent experiments as well, but these show that measuring photon energy and momentum is nothing new, and is certainly not "original" with me.

    I take it that measuring photon frequency and wavelength experimentally is not an issue either.
     
  18. Aug 18, 2011 #17
    angle-change and sign-change are the same thing; angle change = pi ---> cos(pi)=-1
     
  19. Aug 18, 2011 #18
    The "beta" in the Doppler formula is the velocity of the moving point light source. In the photon's energy-momentum 4-vector (k,w/c), there is no "beta"; k is the wave vector. The observation angle has a discontinuity, jumping from zero to pi.
     
  20. Aug 18, 2011 #19
    Compton-effect experiment is a strong support to the Einstein's light-quantum hypothesis: A light wave has wave-particle duality. Taking the light wave consisting of particles, using energy- and momentum-conservation laws to have successfully explained the changes of scattered light-wavelengths and electron's momentums. But this experiment has nothing to do with the invariance of Planck constant, because, at least, all physical quantities are observed and measured in the same lab frame, with nothing to support the Lorentz invariance of Planck constant.
     
  21. Aug 18, 2011 #20

    Dale

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    OK, I get that. I also don't know of a correct term for that, so there may not be a succinct way to say it.

    I guess you could divide a null 4-momentum by the time component. That would give a null four-momentum whose spacelike components are a unit 3-vector. I don't know what you would call it though, and it would have all of the bad-behavior properties of the 4-velocity plus it wouldn't transform right even on its own. But it would have the desired spatial direction.
     
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