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

In summary: 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?Yes, that is correct. The invariants of the electromagnetic field, P and E·B, tell us that there will always be a residual field that cannot be transformed away completely. This is due to the fundamental nature of electric and magnetic fields, as described by special relativity and the Aharonov-Bohm effect.
  • #1
kmarinas86
979
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])?
 
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  • #2
kmarinas86 said:
Special relativity predicts that electric fields transform into magnetic fields via Lorentz transformations and that the vice versa also occurs.
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.

kmarinas86 said:
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])?
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.
 
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  • #3
DaleSpam said:
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.

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?
 
  • #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.
 
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  • #5
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.
 
  • #6
Bill_K said:
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.

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?
 
  • #7
DaleSpam said:
This may be a little overstated. One of the important invariants of the field is [itex]P=|B|^2-|E|^2[/itex].

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.
 
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  • #8
kmarinas86 said:
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?
Correct. And if P is 0 then there will always be both fields in proportion to each other.
 
  • #9
kmarinas86 said:
The electric field intensity and magnetic field intensity of a photon should both increase or both decrease as a result of Lorentz transformations.
Yes.

kmarinas86 said:
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".
I wouldn't interpret it that way either, which is why I mentioned that I thought your OP overstated the case.

kmarinas86 said:
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.
It is the magnitude of E and B, so even if they did have an imaginary part it wouldn't be relevant for P.
 
  • #10
kmarinas86 said:
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?
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.

kmarinas86 said:
I'm having the impression that it just means the former, while P may change with time. Is this correct?
P may indeed change with time.
 
  • #11
DaleSpam said:
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.

That's exactly what I need to know. Thanks!
 

1. What is the magnetic vector potential?

The magnetic vector potential is a mathematical concept used in electromagnetism to describe the magnetic field in a given region. It is a vector quantity that depends on the current distribution and is defined as the curl of the magnetic field.

2. What is the scalar potential in this context?

In this context, the scalar potential refers to the electric potential, which is a scalar quantity that describes the electrical potential energy of a charged particle in an electric field. It is defined as the work done per unit charge in moving a particle from one point to another in an electric field.

3. What does it mean when the scalar potential is equal to 0?

When the scalar potential is equal to 0, it means that there is no net electric potential in the given region. This could be due to the absence of any charges or the presence of equal and opposite charges that cancel each other's electric potential.

4. How is the magnetic vector potential related to the scalar potential?

In the absence of any charges, the magnetic vector potential is related to the scalar potential through the equation: A = μ0φ, where A is the magnetic vector potential, μ0 is the permeability of free space, and φ is the scalar potential. This relationship is known as the magnetic vector potential identity.

5. Why is it useful to have a zero scalar potential in the context of the magnetic vector potential?

Having a zero scalar potential simplifies the equations and calculations involved in describing the magnetic field. It also allows for a more intuitive understanding of the magnetic field in terms of the magnetic vector potential, as it is not affected by any electric potential energy. Additionally, in some cases, the scalar potential can be set to zero by choosing a suitable gauge, making the calculations even simpler.

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