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LT of the magnetic vector potential when the scalar potential=0

  1. Feb 15, 2012 #1
    Special relativity predicts that electric fields transform into magnetic fields via Lorentz transformations and that the vice versa also occurs. It also has been argued, since experiments verifying the quantum mechanical phenomenon of the Aharonov–Bohm effect, that the vector potentials are more fundamental than the field concepts.

    Mathematically, how then does the magnetic field [itex]\mathbf{B} = \nabla \times \mathbf{A},[/itex] transform into an electric field [itex]\mathbf{E_A} = - \frac { \partial \mathbf{A} } { \partial t },[/itex] (or vice versa) in the case where the scalar potential is zero (thus [itex]- \nabla \phi=0[/itex])?
  2. jcsd
  3. Feb 15, 2012 #2


    Staff: Mentor

    This may be a little overstated. One of the important invariants of the field is [itex]P=|B|^2-|E|^2[/itex]. So if P is positive there is no reference frame where B is 0, and if P is negative there is no reference frame where E is 0.

    If you have a frame where there is a purely electric field then in other frames there will be a magnetic field and an even larger electric field. Similarly with purely magnetic fields.

    The vector potential and the scalar potential together form a four-vector called the four-potential. If the scalar potential is 0 in some frame (where the vector potential is non-zero) then in other frames the scalar potential will be non-zero.
    Last edited: Feb 15, 2012
  4. Feb 15, 2012 #3
    So in other words, if P does not equal 0, it is not always possible to transform one field an entirely into another, that there will be some residual field, such as a magnetic field that you can't transform into an electric field, or vice versa?

    What are some theories about these "untransformable" residues of fields?
  5. Feb 15, 2012 #4
    I just noticed something.

    The electric field intensity and magnetic field intensity of a photon should both increase or both decrease as a result of Lorentz transformations.

    It's rather hard to see how [itex]P=|B|^2-|E|^2[/itex] could somehow be interpreted as "B is exchanged for E, or vice versa". The equation seems to imply the exact opposite of that - that both B and E change in tandem with one another, except when their sign is "opposite".

    Alternatively if by "exchange" we mean to say that the the real value of B approaches zero as the real value of E moves away from zero, then the only way I could see how this formula could remain an invariant would be for B and/or E to have an imaginary component. I'm not so sure that an imaginary component can even apply with it.
    Last edited: Feb 15, 2012
  6. Feb 15, 2012 #5


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    Science Advisor

    This is not as mysterious as you're making it sound, kmarinas86. Just as the length of a vector is the same in every rest frame, the electromagnetic field has two invariants that are always the same: E2 - B2 and E·B. Since they're both scalars they're clearly invariant under a spatial rotation. Try applying a Lorentz boost to E and B and you'll see that these quantities stay the same under a boost as well.

    For example for a plane wave we have |E| = |B| and EB. Consequently for a plane wave both of the invariants are zero. This is an example of what we call a null field. And it implies that no matter how you boost a plane wave you always get a plane wave - you can't make just one of E or B go away.
  7. Feb 15, 2012 #6
    I thought E was exchanged for B depending on the Lorentz boosts. Do these formulas suggest that this is not the case or that this is not always the case? How am I to see that in these formulas you gave?
  8. Feb 15, 2012 #7
    By "invariant" do you mean just "Lorentz invariant" (unchanging with respect to Lorentz transforms), or does this invariance mean that P is a constant of a system as time passes? I'm having the impression that it just means the former, while P may change with time. Is this correct? If not, I don't understand how this formula can agree with claims about how B turns into E, and vice versa without B and/or E having values in the complex domain.
    Last edited: Feb 15, 2012
  9. Feb 15, 2012 #8


    Staff: Mentor

    Correct. And if P is 0 then there will always be both fields in proportion to each other.
  10. Feb 15, 2012 #9


    Staff: Mentor


    I wouldn't interpret it that way either, which is why I mentioned that I thought your OP overstated the case.

    It is the magnitude of E and B, so even if they did have an imaginary part it wouldn't be relevant for P.
  11. Feb 15, 2012 #10


    Staff: Mentor

    A quantity which is constant as time passes is called "conserved". A quantity which is constant under some transform is called "invariant". P is invariant under arbitrary diffeomorphisms (coordinate changes) including the Lorentz transform.

    P may indeed change with time.
  12. Feb 15, 2012 #11
    That's exactly what I need to know. Thanks!
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