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1st stress-energy tensor component for an electric field

  1. Jul 19, 2014 #1
    I have recently gone over the derivation of the stress energy momentum tensor elements for the special case of dust. This case just used a swarm of particles. Now that I understand that case, I am now trying to see if I can derive the components for an electric field. I just want you guys to please tell me if you agree with what I came up with thus far or if I'm off.

    1st: I know the T00 component to be the relativistic energy density. The classical energy density of an electric field (which can be used to find the amount of energy stored in a capacitor) is:

    nE = [itex]\frac{1}{2}[/itex][itex]\epsilon[/itex]E2

    where: [itex]\epsilon[/itex] is the relative permeability of the material and E is the magnitude of the electric field.

    Well, I noted that E=(KQ)/r2 where K is Coulomb's constant, Q is charge and r is the distance from the charge that is responsible for the electric field.

    K and Q are invariants, but r is not invariant because r is a length and length can be contracted due to lorentz contraction.

    r= r0/[itex]\gamma[/itex]

    r0 is the length that is seen in the electric field's rest frame of reference.
    [itex]\gamma[/itex] is the typical 1/[itex]\sqrt{1-(v2/c2)}[/itex]

    Having said this, if r = r0/[itex]\gamma[/itex] then

    r2 = (r0)2 / [itex]\gamma[/itex]2

    Now going back to the formula for the magnitude of the electric field, the formula would change to:

    E= (KQ) / ((r0)2 / [itex]\gamma[/itex]2) = (KQ[itex]\gamma[/itex]2)/ (r0)2

    This would mean that E2 would be (K2Q2[itex]\gamma[/itex]4)/ (r0)4

    Finally going back to the energy density expression nE = [itex]\frac{1}{2}[/itex][itex]\epsilon[/itex]E2 , this would change to:

    nE = [itex]\frac{1}{2}[/itex][itex]\epsilon[/itex] * (K2Q2[itex]\gamma[/itex]4)/ (r0)4

    Therefore my derived quantity for the energy density T00 = [itex]\frac{1}{2}[/itex][itex]\epsilon[/itex] * (K2Q2[itex]\gamma[/itex]4)/ (r0)4


    What do you guys think? Am I right, a little off or way off base? If it is either of the latter two, then can you please explain where I went wrong?
     
  2. jcsd
  3. Jul 19, 2014 #2

    Dale

    Staff: Mentor

    That is it, you are done. Remember that Maxwell's equations are already relativistic, so the classical energy density is the same as the relativistic energy density. There is no need to do any of the rest of the work that you did.

    However, note, that is the energy density only in the frame where there is a pure electric field. In other frames there is also a magnetic field, and the energy density in other frames is ##1/2 (\epsilon_0 E^2 + B^2/\mu_0)##. In other frames, the relativistic correction does not come about due to some factor of γ but simply due to the presence of the B field in the other frames.
     
  4. Jul 19, 2014 #3
    Thank you for this information. I have not yet studied Maxwell's equations so I did not know that the energy density formula was already relativistic. I shall study them now.
     
  5. Jul 19, 2014 #4

    bcrowell

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    Staff Emeritus
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    Gold Member

  6. Jul 20, 2014 #5
    Thank you for this source. I have a question about something I read in the book.

    In section 10.6 you said that the energy density U for the electromagnetic field was:

    U= (E2 + B2)/(8[itex]\pi[/itex]K) where K is Coulomb's constant.

    K= 1/ (4[itex]\pi[/itex][itex]\epsilon[/itex]0)

    Therefore:

    8[itex]\pi[/itex]K = (8[itex]\pi[/itex])/(4[itex]\pi[/itex][itex]\epsilon[/itex]0)= 2/[itex]\epsilon[/itex]0

    This leads to the original expression for U to become:

    U = (E2 + B2)/(2/[itex]\epsilon[/itex]0)= [itex]\frac{1}{2}[/itex][itex]\epsilon[/itex]0(E2 + B2)

    This is slightly different from the energy density expression that I learned about just prior to this post:

    U = [itex]\frac{1}{2}[/itex][ε0E2 + (B2/[itex]\mu[/itex]0)]

    Can you please explain where the discrepancy between the formulas comes in and why it is one and not the other?

    Thank you.
     
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