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Relativistic Invariance

  1. Dec 5, 2003 #1
    If I were to attempt to prove that the dot product of an electric and magnetic field is invariant under the conditions of Einstein's Special Theory of Relativity, how would I do this? Would the proof be very involved and complicated? Or should I just use hypothetical magnetic and electric fields and demonstrate how the dot product is unchanged when dealing with relativistic frames of reference?
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  3. Dec 7, 2003 #2


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    First of all, what makes you think that the dot product is invariant? I don't know for sure that it is not, but I would imagine that it is not. Did you hear or read this somewhere, or is this your own idea?

    I would personally use the Faraday tensor (in fact, after I'm done typing, I will do this and see for myself if the dot product is invariant). It shouldn't be too complicated. I will just consider some arbitrary Faraday tensor, Lorentz transform it in an arbitrary direction at an arbitrary speed, extract the components for the E & H fields to calculate the dot products, and then compare the results. I have a feeling the dot product will be inconstent from one frame to the next.

    Do you have formulas for how the field vectors change from one frame to the other?
  4. Dec 7, 2003 #3
    Re: Re: Relativistic Invariance

    [itex]\vec{E}\cdot\vec{B}[/itex] and [itex]E^2-B^2[/itex] are the only quadratic invariants you can build out of the electric and magnetic fields.
  5. Dec 7, 2003 #4
    Re: Re: Re: Relativistic Invariance

    i didn t know this but it seems believable.... basically, to get a lorentz scalar, we would have to contract [itex]F^{\mu\nu}F^{\rho\sigma}[/itex] with something. i think there are only two choices available [itex]\epsilon_{\mu\nu\rho\sigma}[/itex] and [itex]\eta_{\mu\rho}\eta_{\nu\sigma}[/itex]

    the first choice will give you [itex]E^2-B^2[/itex] and the second [itex]E\cdot B[/itex] (i would imagine. i haven t checked it)

    i had a question come up the other day... i was trying to remember the formula for the energy density in terms of the electromagnetic 2 form. i know the lagrangian is [itex]F\wedge*F=(E^2-B^2)\text{vol}[/itex], and since energy density is nearly the same thing, with a plus instead of a minus, i thought i could make a similar formula.

    of course i was mistaken, energy density is not a lorentz covariant object, so there will be no way to write down the energy density from the field strength in a covariant way.

    but can you write the stress energy tensor? i have seen the Noether formula for the stress energy tensor, but i was hoping for a concise formula from the field strength 2 form. i am suspecting that that doesn t exist either, since the stress tensor is symmetric...
  6. Dec 7, 2003 #5
    Re: Re: Re: Re: Relativistic Invariance

    As you say, you can write the electromagnetic stress-energy tensor in terms of the field strength tensor, but not solely by means of operations on differential forms, because the stress-energy tensor isn't a differential form.
  7. Dec 7, 2003 #6


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    Re: Re: Re: Re: Relativistic Invariance

    As I started writing to work things out, things started to look like the dot product may be invariant. That does surprise me, given the way the magnetic field arises from the Faraday tensor. From my cursory inspection, though, I would rather have expected contraction with [itex]\epsilon_{\mu\nu\rho\sigma}[/itex] to give the dot product. I have:

    [itex]E\cdot B[/itex] = (1/c)(F01F32 + F02F13 + F03F21).

    (Sorry, I don't know how to make that math looking text.)

    Anyway, there are clearly permutations of the indices that suggest nontrivial elements would arise only in the case of a permutation. Are you sure you have the appropriate contractions identified with the corresponding invariants?

    That is certainly above my level. I still don't have a satisfactory understanding of what a Lagrangian is.
  8. Dec 7, 2003 #7
    Re: Re: Re: Re: Re: Relativistic Invariance

    no, i had them backwards. you are correct... the Levi-Civita contraction gives the dot product, and the metric contraction gives the Lagrangian.
  9. Dec 7, 2003 #8
    Re: Re: Re: Re: Re: Re: Relativistic Invariance

    actually, i should have guess more cleverly. the lagrangian should be invariant under the full Lorentz group (i.e. parity transformations, as well as Lorentz rotations), whereas the dot product, since if involves a pseudovector, should transform nontrivially under parity transformations.

    the Levi-Civita tensor reverses sign under parity transformations, so i should have been able to guess which is which...

    silly me.
  10. Dec 7, 2003 #9
    I was told to prove the following in my electromagnetic theory class:
    1) that the dot product of E and B would be relatvistically invariant
    2) that the quantity E^2 - c^2B^2 is relativistically invariant.

    We will assume the two relativistic frames S and S' where x'1=x1, x'2=x2,x'3= j(x3-vt) where j = 1/(square root of 1 - B^2) where B= v/c, and t'=j(t-B/cx3).

    When defining the electric and magnetic fields in relativistic frames, current density J and charge density p are not separable nor distinct since a static charge distribution in one frame of reference will be a current density in a moving frame.

    In the static frame of reference S we will assume a charge density po in a certain volume, were the change in charge dq=po(dV). Thus in the moving frame of reference S', p=poj. This lorentz contraction and thus change of charge density only occurs in the x3 direction since this is the only dimension which is moving in relativistic frame S'. This is all the information we need to calculate E in the frame of reference S.

    Since reference S is static, B=0 because there is no current density. So now we can define the current density J (in frame S' only) as (pu,icp) pu corresponds to the three spatial dimensions, while icp corresponds to the dimension of time. My question is this. How do we calculate E and B in the moving frame of reference S' so I can begin to take the dot products of E and B in both references and see if it is indeed invariant. Sorry if I was confusing or unclear.
  11. Dec 7, 2003 #10


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    I wouldn't do it this way.

    It doesn't matter what direction; density is density.

    Are you allowed to use the Faraday tensor? Just transform that, take the appropriate components, and compare dot products.

    Ei = Fi0, cB1 = F32, cB2 = F13, cB3 = F21.

    E'i = F'i0, cB'1 = F'32, cB'2 = F'13, cB'3 = F'21.

    F'μν = aμαaνβFαβ.

    where aμα are the components of the lorentz transformation matrix.
    Last edited: Dec 7, 2003
  12. Dec 8, 2003 #11
    if you are allowed to use relativistic tensors, then it can be even easier.

    it is a fact that any tensor expression with all indices contracted must be a (restricted) lorentz scalar. thus, just take the tensor expression above, check that it is equal to E*B, and you re done. you don t even have to check how it transforms, since it is guaranteed to be a scalar.
  13. Dec 8, 2003 #12


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    This makes sense to me -- if you contract with the metric, you get the Langrangian; if you contract with the Levi-Civita tensor, you get the dot product.

    I do have to ask, though -- what exactly is a "Lorentz scalar?" What properties does it have beyond those of the usual scalar?

    - Warren
  14. Dec 8, 2003 #13
    the (restricted) lorentz scalar is invariant under (restricted) lorentz transformations. and, i could compare this with "the usual scalar", as soon as you tell me which usual scalar you mean.
  15. Dec 8, 2003 #14


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    Hmmm... I guess I always thought that ANY scalar was unchanged by Lorentz transforms.

    And I suppose I mean scalar, as per MathWorld: "A one-component quantity which is invariant under rotations of the coordinate system."

    - Warren
  16. Dec 8, 2003 #15
    yeah, that s about right. so Lorentz scalars are scalars that are invariant under rotations in Minkowski space (the set of all such rotations is called the Lorentz group. hence the name)
  17. Dec 8, 2003 #16
    note that according to this definition, you might think that energy was a scalar.

    and it is, under euclidean rotations. but it is not a Lorentz scalar.

    once you choose some coordinate system, i can talk about scalars that are invariant under rotations of that coordinate system.
  18. Dec 8, 2003 #17


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    Not necessary, I understand the distinction now. Energy is not a scalar when the group of coordinate transformations in question is the Lorentz group. The best you can say about it that it is the zeroth component of the 4-momentum. Not all "one-component quantities" are scalars -- only those that are invariant under the relevant coordinate transformations.

    Thanks as always,

    - Warren
  19. Dec 8, 2003 #18
    I guess my main problem is calculating the 16 components of the electromagnetic field tensor. Once I calculate this, I can just multiply it by the matrix |1 0 0 0 | and |1 0 0 0 |.
    |0 1 0 0 | |0 1 0 0 |
    |0 0 j iB| |0 0 j-iB|
    |0 0-iB j| |0 0 iB j|

    This is in accordance to the equation F'= lambda(F)lambda'. Once again I apologize for lack of notation. So in other words, once I can calculate electromagnetic field tensor F, I can calculate the relativistically transformed field tensor F'. And then I should be able to take the dot product of E and B in both relativistic frames,correct?
  20. Dec 8, 2003 #19
    sometimes a photon with a time-like polarization is called a scalar photon, not because its a Lorentz scalar, but because its a euclidean scalar.

    or, when you do Kaluza-Klein reductions, things that are tensors under SO(4,1) reduce to a tensor, a scalar and a vector under SO(3,1). so one mans scalar is another mans zeroth component. it just depends on which coordinates you feel like working with.
  21. Dec 8, 2003 #20


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    - Warren
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