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Electromagnetic field 4-momentum density and inertial frames

  1. Jul 23, 2015 #1
    Electromagnetic field has a density of energy

    U = ε/2*E2+ μ/2* H2

    And a density of momentum, given by the Poynting vector

    S = E x H

    For an element of volume dV you have a four vector of energy and momentum which is

    [E,P] = dV * [U, S]

    Being E the energy in the element of volume and P the momentum of inertia.

    If you measure this four vector from another inertial frame Fr' which moves with velocity v respect the initial frame, Fr, you can get [E',P'] by making a boost over [E,P]. Because of the contraction of lengths, the dimension in which the boost is made shortens in a factor 1/γ, γ = 1/(1-(v/c)2)1/2, so the density of energy and momentum must be multiplied by γ.

    So: [U', S'] = γ * Boost( [U, S], v )

    Where Boost is the Lorentz Boost function.

    You can transform the fields E and H themselves to the new frame Fr' by using the transformations described here:

    https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

    I would expect that after transforming the fields I would get the same values for [U', S'] that the ones get by the previous way, however, nothing farther to reality than this:

    If I have a E = Eo x and I apply a boost in x direction, according to the transformations of fields I get the same fields:

    E' = E H' = H = 0

    So I will have U' = U and S' = S whereas if I transform the 4-momentum vector, I will get:

    U' = γ2 * U

    S' = - γ2 * U * v

    Which of these results is the good one? Can anybody help me in finding what is wrong with this reasoning?
     
  2. jcsd
  3. Jul 23, 2015 #2

    Dale

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    This is correct. The energy density and momentum density do not transform as components of a four vector, they transform as components of a rank 2 tensor. Specifically, they are components of the stress energy tensor.
     
  4. Jul 23, 2015 #3
    Thank you DaleSpam for your answer, so the correct four vector would be the one obtained by transforming the fields.

    If want to transform the energy and momentum density I have to do it as a energy stress tensor, which transforms on boosts as explained in this link:

    http://physics.stackexchange.com/qu...-momentum-tensor-under-lorentz-transformation

    Best regards,
    Sergio
     
  5. Jul 23, 2015 #4

    Dale

    Staff: Mentor

    Yes.

    I don't think that there is a correct four vector. It is just a tensor. Trying to find the correct four vector to represent energy momentum density is like trying to find the right scalar to represent momentum. It just doesn't fit.
     
  6. Jul 24, 2015 #5

    vanhees71

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    This is a very important point in relativistic field theory. You have a energy-momentum-stress tensor ##T^{\mu \nu}##, which is symmetric. For a closed system energy-momentum conservation holds (I stick to special relativity; in GR it's more subtle). Then you have the continuity equation
    $$\partial_{\mu} T^{\mu \nu}=0.$$
    Now, using the four-dimensional Gauss integral theorem, you can show that
    $$p^{\nu} = \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} T^{0\nu}$$
    is conserved and a four-vector under Lorentz transformations.

    For an open system, where the continuity equation does not hold, the so defined ##p^{\nu}## is neither a four vector nor conserved. For details, see Jackson, Classical Electrodynamics.
     
  7. Aug 3, 2015 #6
    As I understand it, ##T^{0 i}##, in terms of matter, is "the flux of relativistic mass across the surface ##x^i##". This seems to be shorthand for the infinitesimal surface ##dx^j \wedge dx^k = {\epsilon^{jk}}_i dx^i##.

    Shouldn't the integral be over an area; ## p^i = \int_{R^2} T^{0 i} dx^j \wedge dx^k ##?
     
  8. Aug 4, 2015 #7

    vanhees71

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    The energy-momentum four-vector wrt. to an arbitrary inertial reference frame has the components
    $$(p^{\mu})=\begin{pmatrix} E/c,\vec{p} \end{pmatrix},$$
    where it is a matter of convention, where to put the conversion factor ##c## (speed of light in vacuum).

    There is no such thing as a relativistic mass. This is an old-fashioned idea from a time, when the full mathematical structure of SRT has not yet been fully understood. This was corrected very soon after Einstein's ground-breaking paper of 1905 by Minkowski (1908). The invariant mass of an object is given by
    $$m^2 c^2=p \cdot p=\eta_{\mu \nu} p^{\mu} p^{\nu}=\frac{E^2}{c^2}-\vec{p}^2.$$
    Right now, I'm in the process of writing an FAQ/Insights article on SRT. So stay tuned ;-).
     
  9. Aug 4, 2015 #8
    OK, no such thing as ##E/c^2##.

    The reference to relativistic mass is straight out of the Wikipedia Stress-Energy Tensor article. You may feel inspired to change it.
     
  10. Aug 5, 2015 #9

    vanhees71

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    No ##E/c^2## is just energy, written in units of mass but not mass (at least not according to the modern use in contemporary high-energy-particle and -nuclear physics). It is misleading to mix the concepts up, because invariant mass, which is exclusively used to define the mass of a particle or composite object, is a Lorentz scalar (which is why it's called "invariant"), and energy is the time component of the energy-momentum four-vector wrt. an arbitrarily chosen frame of reference.

    What Wikipedia concerns, it's a great place to start reading about a subject, but it's by no means totally trustworthy. You don't even know who's the author(s) of an article!
     
  11. Aug 5, 2015 #10
    Of course, however if you tried, I think you could find a model where E/c^2 servers as inertial mass. How did I get talked into beating this dead horse?
     
  12. Aug 5, 2015 #11

    vanhees71

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    Unfortunately the horse is not dead. I never understood why!
     
  13. Aug 9, 2015 #12

    Jano L.

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    That sounds like concept police. Of course the relativistic mass is a valid and unambiguous concept. It is given by the formula

    $$
    m_{rel} = \frac{m}{\sqrt{1-\frac{v^2}{c^2}}}
    $$
    where ##m## is rest mass and ##v## is speed of the body. It allows momentum to be written as

    $$
    \mathbf p = m_{rel} \mathbf v
    $$

    similarly to classical mechanics.

    It may be an old-fashioned concept, but it is a concept. Don't impose newspeak on everybody. For people to understand what is written in old books on relativity, they need to understand this concept.
     
  14. Aug 9, 2015 #13

    vanhees71

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    No, it gives rise to misunderstandings, and it's not used in the community anymore. If you wish to stick to old-fashioned ideas, it's your choice, but don't expect to be understood easily by other people working in relativistic physics.

    Again: What you quote is the on-shell energy of a classical particle or asymptotic free quantum (divided by ##c^2##) and thus a temporal component of the energy-momentum four-vector (divided by ##c##). The usually used quantity when one speaks about particle masses is the invariant mass (or for particles which have a non-zero invariant mass the rest mass). Usually one refers to this invariant mass simply as mass.
     
  15. Aug 9, 2015 #14

    Jano L.

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    I do not want to stick to any ideas. I refuse to accept a mindset that present-day fashion is the limit of my vocabulary. That's auto-censorship, not science. Understanding does not come easily, especially when talking about non-fashionable things.
     
  16. Aug 10, 2015 #15

    vanhees71

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    Understanding comes easier with clear concepts rather than delving into problems solved for more than 100 years!
     
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