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E-Field vs B-Field in SR

  1. Jul 13, 2012 #1
    Consider the following "classical" example: [itex]^{}_{}[/itex]

    A system Σ' moves away from a system Σ with relative velocity v in the direction of z'z axis (their origins concur at time t=t'=0). A charge q is located in the origin of Σ' and moves along with it, while an observer sits at (0,yo,0) in Σ. The E-Field as observed in Σ' is according to Coulomb's law: E'=[itex]\frac{q}{4πεοr'3}[/itex](0,yο,-vt) while the B-Field is of course: B'=0.
    Now we make use of Lorentz transformation and obtain for Σ the following relations:
    E=[itex]\frac{q}{4πε0}[/itex](0,[itex]\frac{γyο}{(y[itex]^{2}_{ο}[/itex]+γ2v2t2)3/2}[/itex], [itex]\frac{-γvt}{(y[itex]^{2}_{ο}[/itex]+γ2v2t2)3/2}[/itex]) and B=-[itex]\frac{qv}{4πε0c2}[/itex]([itex]\frac{γyο}{(y[itex]^{2}_{ο}[/itex]+γ2v2t2)3/2}[/itex],0,0) or B=[itex]\frac{μ0}{4π}[/itex][itex]\frac{qγ}{r'3}[/itex](vxr') which for γ[itex]\rightarrow[/itex]1 takes actually the form of Biot-Savart law.

    We observe that a pure E-Field converts into a mixture of E and B fields. Now arises my question: Is the opposite possible for a single charge? If the answer is no that means that electricity has a more fundamental place in relativity than magnetism and respectively Coulomb's law is more general than Biot-Savart, which by the way is not applicable to single charges. Is it therefore possible that the two fields are not symmetrical and no pure B-Field could be ever detected?
    Moreover I would like to know how the leftover E-Field of the above example is compensated for a vast amount of particles (classical example: the straight current-carrying wire).
    Last edited: Jul 13, 2012
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  3. Jul 13, 2012 #2


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    The asymmetry comes not from any difference in the E and B fields, but from the non-existence of magnetic monopoles. If you add these to the classical theory, E and B field, and all laws about them, become perfectly symmetric.
  4. Jul 13, 2012 #3
    There is asymmetry here, but only because it's unrealistic to talk about an entirely spatially-directed current. Theoretically speaking, a particle with purely spatial current would create a magnetic field and not an electric field, but that would require spacelike four-velocity.

    In truth, I think it's something of a misnomer to say there are electric monopoles. While it's handy to say so in 3d, you see in 3+1d that monopoles are just time-directed currents, no different really from other currents, and the fields (when you consider the Faraday tensor and not E and B by themselves) are very symmetrical in this respect.
  5. Jul 13, 2012 #4
    So loosely speaking, based on the current formulation of EM under SR (no magnetic monopoles): 1)the are no pure B-Fields and thus magnetism is a product of electricity and not the other way round, 2)there isn't any "microscopic" version of Biot-Savart and because of that, Coulomb's law is more fundamental.

    What happens then when too many single electrons sum up? How does the B-Field become dominant in real life applications?

    @Murphrid does that mean that even in the unrealistic case of many parallel moving charges with constant velocity there wouldn't be any pure B-Field? That still doesn't explain why we do not detect any significant amount of E-Field around a straight current-carrying wire.
  6. Jul 13, 2012 #5
    I think there's no E-field from a current-carrying wire because the positive charges in the wire create a current purely in the time direction while the negative charges create a current with components in both time and spatial directions. For non-relativistic velocities, the positive and negative currents in the time direction nearly cancel out, and for this reason, current-carrying wires are modeled as carrying purely spatial current.

    I wouldn't say this makes electricity more fundamental, either. EM is a single phenomenon modeled by a single object--the Faraday tensor [itex]F_{\mu \nu}[/itex]. And why do you say Biot-Savart law doesn't apply to single charges? There's a version of it that can do just that, and a single moving charge is a valid "current".
  7. Jul 13, 2012 #6
    I think I do not get your terminology here... What do you imply here with the "direction" terms? Are you just saying that the E-Field cancels out due to the stable (as observed in Σ) positive ions of the wire?

    Have a look at my example again... If that were true, than we could begin by saying that the observed fields in Σ are a zero E-Field and a pure B-Field calculated by this version of Biot-Savart which should hold for single charges... Then the trasformations in Σ' would yield different results than our initial assumptions, which makes me think that there is no stand alone Biot-Savart law and it's just a consequence of the Coulomb's law (by deduction that leads to my previous assessment about the relationship between electricity and magnetism).
    Last edited: Jul 13, 2012
  8. Jul 13, 2012 #7
    There are positive ions moving only in the time direction. There are negative ions moving in a combination of time and spatial directions. For low velocities, they almost completely cancel each other out. This is the same as saying the wire is electrically neutral.

    Every object has a four-velocity in spacetime. Objects we consider to be "not moving" in our frame of reference are still moving through spacetime--they just have a velocity whose direction is the direction of time. This is why even "stationary" charges are really currents, and EM theory is just the theory of currents creating fields. Doesn't matter if that current has a direction in our usual 3D space or if it only goes through time.

    You're right to notice the connection between Coulomb's law and Biot-Savart, but I point out that this doesn't make one more supreme than the other. They're both simplifications of a single law for different types of currents: Coulomb's applies when the current is only in the time direction. Biot-Savart applies when the current is only in a spatial direction. Because Maxwell's equations are linear, you can just chop up the current into time-directed and spatially-directed bits and apply Coulomb and Biot-Savart to the individual pieces and you're done.
  9. Jul 13, 2012 #8
    Ok now I fully comprehend your explanation about the wire and the fact that you speak in terms of tensors to describe current, charge density and fields.

    What we conceive in 3D as seperate B and E fields are not mutually interchangeable. That is due to the very fact that we cannot relinquish the movement in space direction in our minds and therefore the B-Field is never pure. That explains why the E-Field seems to be more fundamental since it is our starting point in order to describe the phenomenon at my example.

    PS: Btw the argument about the absence of magnetic monopoles reflects directly to the Maxwell Equations and is not relevant to my inquiry after all... What changes in this case is that if there were magnetic monopoles we could get a pure B-Field from a pure E-Field and vice versa. In other words the EM tensor would be also different and there would be no entanglement of the two fields after a Lorentz transformation (if and only if each observer detects a pure field).
    Last edited: Jul 13, 2012
  10. Jul 13, 2012 #9


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    One of the invariants of the EM field tensor is [itex]B^2-E^2[/itex]. This quantity is the same in all reference frames, so if it is positive in any reference frame then there exists some reference frame where there is no E field and the EM field is purely a B field.

    For a single charge this quantity is negative, so there is no frame where there is a pure B field. However, for configurations involving multiple charges it is certainly possible to get this quantity to be negative.
  11. Jul 13, 2012 #10
    Have you got maybe a simple example handy?
  12. Jul 13, 2012 #11


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    Can you do this in the covariant formulation of Maxwell's equations? The source term there is the charge-current 4-vector; I'm not sure how you would add magnetic monopoles to that.

    Or, to look at it another way, the two covariant Maxwell equations correspond to the four ordinary 3-vector equations as follows:

    [tex]F_{ab,c} + F_{bc,a} + F_{ca,b} = 0[/tex]

    corresponds to [itex]\nabla \cdot \vec B = 0[/itex] if a, b, c are all spatial indexes, and to [itex]\nabla \times \vec E + d \vec B /dt = 0[/itex] if one of the indexes is a "time" index. So the absence of magnetic monopoles is a consequence of this equation, which is a geometric identity (it has to hold for any tensor F that is the exterior derivative of a 1-form; F = dA, where A is the 4-potential).

    [tex]{F^{ab}}_{,b} = 4 \pi J^{a}[/tex]

    corresponds to [itex]\nabla \cdot \vec E = 4 \pi \rho[/itex] if index a = 0, and to [itex]\nabla \times \vec B - d \vec E/dt = 4 \pi \vec J[/itex] if index a = 1, 2, 3. So the only sources here are electric charges and currents; there are no "magnetic currents" any more than there are "magnetic charges". On this view, magnetism is a sort of "relativistic correction" to electricity (I believe Feynman describes it this way in some of his lectures on physics), not a separate "thing" in itself that can act as a source.
  13. Jul 13, 2012 #12
    In other words, there is only one type of current. If the antisymmetric (exterior) derivative of F had a nonzero source, that three-index tensor would be a second source for the EM field and would affect not only magnetic fields but electric fields. Relativity would forbid just altering the [itex]\nabla \cdot B[/itex] part without also affecting [itex]\nabla \times E + dB/dt[/itex].
  14. Jul 13, 2012 #13


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    See, for example:


    You have to introduce another potential, another tensor, and another current. See esp. eqn. 14-16 in above.
  15. Jul 13, 2012 #14
    Well, it doesn't require a second tensor; seems like the author is keeping them separate arbitrarily.
  16. Jul 13, 2012 #15


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    Thanks, PAllen, I wasn't aware of this. One thing I notice is that the two 4-potentials are not completely symmetric; in equation (3), defining the field tensor F^ik in terms of the two potentials, one appears with the Kronecker delta and the other appears with the completely antisymmetric symbol. And in the definition of the dual tensor the two are reversed. But the Maxwell equations do end up completely symmetric between the "electric" and "magnetic" sources (because both tensors come into play). Interesting!

    Edit: It's also interesting that in the Lagrangian, equation 13, only the field tensor F appears (not G); the only change is to add a second interaction term. (That is, there is only one "kinetic" term, even though there are two "field tensors".)
    Last edited: Jul 13, 2012
  17. Jul 13, 2012 #16
    @PeterDonis Why does the charge and by extension its relative motion has to be an attribute of electricity?

    Anyway the magnetic monopole correction reads as follows:
    Maxwell Equations:
    1) E= 4πρe
    2) Β= 4πρm
    3) ×E = - [itex]\frac{1}{c}[/itex]∂tΒ - [itex]\frac{4π}{c}[/itex]jm
    4) ×B = [itex]\frac{1}{c}[/itex]∂tΒ + [itex]\frac{4π}{c}[/itex]je
    Electromagnetic Tensor: Fμν = ∂μΑ + ∂νΑ + [itex]\frac{1}{ε0}[/itex]ελνστσΑ[itex]^{τ}_{m}[/itex]
    Last edited: Jul 13, 2012
  18. Jul 13, 2012 #17


    Staff: Mentor

    A current in an uncharged wire.
  19. Jul 13, 2012 #18
    Macroscopically the wire seems to possess a pure B-Field. However I really doubt it that it is possible to cancel out all the leftover E-Fields of the current (electrons) with positive ions.

    I will try to examine the problem again adding some more electrons to my initial example, while avoiding any complicated interactions, but I fear that it could hardly yield any satisfying results...
  20. Jul 13, 2012 #19


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    I suspect that you probably have it backwards. In all likelyhood if there is a microscopic region where the invariant is positive it would be near the nucleus where the electrons couldn't cancel out the field of the protons. Remember, the invariant is the same in all reference frames, but it generally varies from place to place and from time to time.

    It won't work unless you include some positive charge distribution also.

    Also, don't forget that there is no such thing as a classical point charge, classical EM is runs into some irritating problems if you use point charges rather than charge distributions. Otherwise you have to do a full QED analysis, in which case you don't have a classical point charge either.
    Last edited: Jul 13, 2012
  21. Jul 14, 2012 #20


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    A superconducting ring with no net charge but carrying a current (or the non-superconducting equivalent with normal wire and a battery) is a good simple example. In the lab frame B is nonzero, and E is zero.
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