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What is the dielectric momentum density 4-vector?

  1. Jul 7, 2013 #1
    Assume that there is a dielectric material with a mass density of ρ0 observed in the dielectric-rest frame. And further, it is assumed that observed in the lab frame, v(x,y,z,t) is the velocity distribution, β=v/c is the normalized velocity, and γ=(1-β2)-1/2 is the relativistic factor, with c the vacuum light speed.

    In my opinion, ρ0 is a Lorentz scalar, and γ(v,c) is a Lorentz 4-velocity, and thus ρ0γ(v,c) also is a 4-vector.

    My question is:

    Can ρ0γ(v,c) be defined as the momentum density 4-vector of the dielectric material?

    What I mean is: If ρ0γ(v,c) is defined as the momentum density 4-vector, is it compatible with the principle of relativity?

    Note: I do not claim ρ0 is a constant, otherwise the material would be rigid, not consistent with the relativity.
     
    Last edited: Jul 7, 2013
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  3. Jul 7, 2013 #2

    Jano L.

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    One can define many things. The important issue is, whether the definition adopted is useful. What question are you trying to resolve?
     
  4. Jul 7, 2013 #3

    PeterDonis

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    If it were, then it couldn't be a component of a Lorentz 4-vector (or a tensor, which is what it's actually a component of--see below). A Lorentz scalar doesn't change at all under a Lorentz transformation; the components of 4-vectors (and tensors) do.

    No, because there is no such thing as a momentum density 4-vector. There is a 4-momentum vector, but it only applies to point particles, or objects being treated as point particles, not to continuous substances, which is what you are assuming when you use a density. To describe a continuous substance, you have to use a tensor, the stress-energy tensor; the mass density is the 0-0 component of the stress-energy tensor. See here for some more discussion:

    http://en.wikipedia.org/wiki/Stress–energy_tensor
     
  5. Jul 7, 2013 #4
    Please note, in ρ0γ(v,c), ρ0 is NOT a component of a Lorentz 4-vector.

    Let me show you an example. In free space, the wave 4-vector Kμ is a Lorentz 4-vector, and hbar*Kμ is the 4-momentum of a photon, where hbar is the Planck constant, which is a Lorentz scalar.
     
  6. Jul 7, 2013 #5

    PeterDonis

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    Ah, ok, I see, you meant it to just be a scalar multiplier in front of the 4-velocity vector, which is what γ(v,c) is. However, my second comment is still valid: the 4-velocity vector describes a point particle, but a point particle can't have a density. If you are trying to use a density, you are trying to describe a continuous substance, which requires a stress-energy tensor.
     
  7. Jul 8, 2013 #6

    Jano L.

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    Peter, I am not sure what you mean. The description of matter by a stress-energy tensor does not automatically preclude description by momentum density.

    In the case of a dielectric, i.e. pure water, limited by a closed boundary surface, the stress-energy tensor will presumably contain contribution

    $$
    T_0^{\mu\nu} = \rho_0 u^{\mu}u^{\nu},
    $$

    similar to above expressions, and perhaps also the contribution related to internal stresses in the liquid (in the simplest case, the pressure ##p##).

    Now, integrating the equation of conservation of momentum over whole domain where the liquid is, it may be possible to get an equation for a four-momentum of the liquid defined as

    $$
    P^{\nu} = \int_V T_0^{0\nu}\,d^3\mathbf x.
    $$

    I did not check the details, but in the case we replace liquid by dust (no pressure), it is possible.
     
  8. Jul 8, 2013 #7

    Bill_K

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    Which is the energy-momentum vector, not a density.
     
  9. Jul 8, 2013 #8

    Jano L.

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    Yes, that is right. The integrand may be called density of four-momentum. In the case of dust, it is

    $$
    T_0^{0\nu} = \gamma c \rho_0 u^{\nu},
    $$

    so it is not a four-vector, because ##\gamma## is not Lorentz invariant, but surprisingly it is close to the expression guessed by phy12345.
     
  10. Jul 8, 2013 #9

    PeterDonis

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    You've got this backwards; the observed mass density ##\rho_0## is the 0-0 component of the stress-energy tensor in the fluid's rest frame, i.e., it's what you get when you double contract the SET with the fluid 4-velocity:

    $$
    \rho_0 = T_{\mu \nu} u^{\mu}u^{\nu}
    $$

    Contracting the SET with other appropriately chosen basis vectors gives the other components, such as the pressure.

    Also, as Bill_K said, the integral you wrote down gives a 4-momentum, not a momentum density. (Also, the integral is not an invariant, because the integration measure d^3 x depends on your choice of coordinates.)
     
  11. Jul 10, 2013 #10
    In my understanding, you agree the momentum-energy 4-vector for a particle: m0γ(v,c).

    Now I am trying to show m0γ(v,c) and ρ0γ(v,c) are equivalent under some conditions.

    Suppose the dielectric is a solid which is made up of identical particles (molecules or atoms). In a given differential volume dV, all particle have approximately the same velocity. Suppose the proper particle number density is N0, which is a Lorentz invariant. Thus we have ρ0 = N0*m0, leading to

    ρ0γ(v,c) = N0m0γ(v,c).

    Therefore, the dielectric momentum-energy density 4-vector is equivalent to your single particle momentum-energy 4-vector, because they are only different by a Lorentz scalar. Am I right?

    Can you prove the single particle momentum-energy 4-vector is equivelanet to that obtained from your "stress-energy tensor"?
     
  12. Jul 10, 2013 #11
    I wonder why $$
    P^{\nu} = \int_V T_0^{0\nu}\,d^3\mathbf x.
    $$

    is a 4-vector?
     
  13. Jul 10, 2013 #12

    PeterDonis

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    No, it isn't, because the differential volume dV is not Lorentz invariant.
     
  14. Jul 10, 2013 #13
    I do not claim dV is a Lorentz invariant, but I do claim the proper particle number density N0 is a Lorentz invariant.
     
  15. Jul 10, 2013 #14

    PeterDonis

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    Yes, and I am saying that claim can't be correct because number density, which is the number of particles in a differential volume element, must depend on the volume element, which is not Lorentz invariant.
     
  16. Jul 10, 2013 #15

    PeterDonis

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    I should respond to this, too: a single point particle, strictly speaking, does not have a stress-energy tensor. For an isolated object all of whose particles are at rest relative to each other, one can choose an appropriate frame (the one in which all of the particles are at rest) and integrate the stress-energy tensor over the object's volume in that frame to obtain a 4-momentum vector for the object, yes. That 4-momentum vector will transform appropriately under Lorentz transformations, so its length, the rest mass of the object, will be Lorentz invariant. But there will still be no way to define a Lorentz invariant density for the object, because its volume, as I said before, will not be Lorentz invariant.
     
  17. Jul 10, 2013 #16

    WannabeNewton

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  18. Jul 10, 2013 #17
    The proper differential element dV0 (dV used before) and the proper particle number dn0 in dV0 are all invariants: N0 = dn0/dV0. Proper quantities are all invariants.
     
  19. Jul 10, 2013 #18

    PeterDonis

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    How are these "proper" quantities defined? What is their physical meaning? As far as I can see, these "proper" quantities are basically the quantities in a specific frame, the fluid's rest frame. What you are calling the "density" is the fluid's density in its rest frame. So all you have shown is that the numerical value of that particular observable, the fluid's density as measured in its rest frame, must be agreed on by all observers. That's true, but you appear to be making a much stronger claim, that there is some Lorentz scalar "density" that will be measured to have the same value in any frame. The latter claim is *not* true.
     
  20. Jul 10, 2013 #19

    Jano L.

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    You can find proof in the paper

    H. Ohanian: Klein's theorem and proof of E_0 = mc^2

    Am. J. Phys. 80, 1067 (2012); doi: 10.1119/1.4748812
    http://dx.doi.org/10.1119/1.4748812

    You seem to think that it ##\rho_0 \gamma (c, \mathbf v)## is density of four-momentum, but that is not the case. I will use the convention where the temporal component of four-vector has index 0 and is placed first in a row.

    The tensor of momentum-energy of the particle is the distribution

    $$
    T^{\mu\nu} = \rho_0 u^{\mu} u^{\nu}
    $$

    where

    $$
    \rho_0 (\mathbf x) = m \sqrt{1-v^2/c^2} \delta(\mathbf x-\mathbf r),
    $$
    is the Lorentz invariant distribution, ##\mathbf r## is the radius vector of the particle and

    $$
    u^{\mu} = \gamma (c, \mathbf v)
    $$
    is its four-velocity.

    The total four-momentum is

    $$
    P^{0\nu} = \frac{1}{c}\int T^{0\nu}\,d^3\mathbf x.
    $$

    For the above tensor, this gives

    $$
    P^{0\nu} = \int \rho_0 \gamma u^{\nu} d^3\mathbf x,
    $$

    so the density of four-momentum can be defined as ## \gamma \rho_0 u^{\nu}##, which is equal to ##\rho_0 \gamma^2(c,\mathbf v)##. It is not a four-vector - the gamma is not Lorentz invariant.
     
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