Can the magnetic field from a single charge be converted into an electric field?

In summary, the conversation discusses the relationship between electricity and magnetism under special relativity, specifically focusing on the asymmetry between electric and magnetic fields and the role of magnetic monopoles. The discussion also touches on the cancellation of electric fields in current-carrying wires and the idea that all objects have a four-velocity in spacetime.
  • #1
Trifis
167
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).
 
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  • #2
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.
 
  • #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.
 
  • #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.
 
  • #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".
 
  • #6
Muphrid said:
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 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?
Muphrid said:
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".
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).
 
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  • #7
Trifis said:
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?

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.
 
  • #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 separate 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).
 
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  • #9
Trifis said:
So loosely speaking, based on the current formulation of EM under SR (no magnetic monopoles): 1)the are no pure B-Fields
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.
 
  • #10
DaleSpam said:
However, for configurations involving multiple charges it is certainly possible to get this quantity to be negative.
Have you got maybe a simple example handy?
 
  • #11
PAllen said:
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.

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.
 
  • #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].
 
  • #13
PeterDonis said:
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.

See, for example:

http://arxiv.org/abs/math-ph/0203043

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

http://arxiv.org/abs/math-ph/0203043

You have to introduce another potential, another tensor, and another current. See esp. eqn. 14-16 in above.

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".)
 
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  • #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]
 
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  • #17
Trifis said:
Have you got maybe a simple example handy?
A current in an uncharged wire.
 
  • #18
DaleSpam said:
A current in an uncharged wire.
Macroscopically the wire seems to possesses 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...
 
  • #19
Trifis said:
Macroscopically the wire seems to possesses 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 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.

Trifis said:
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...
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.
 
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  • #20
Trifis said:
Have you got maybe a simple example handy?

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.
 
  • #21
DaleSpam said:
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.
Yes that is well understood. What I meant to say is that the ideal wire of magnetostatics is not possible even with a theoretical assumptions when analyzed in microscopic level.

DaleSpam said:
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.
Yes I would have added some positive charges or distributions if you like, but the whole attempt would be pointless since the flow of electrons must be at least constant in order to give the observer a chance to detect a pure B-Field.

pervect said:
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.
Am I right to think that something like that could not be analyzed within the scope of SR, since the reference frame of the charges would have to accelerate (circular orbit) ?
 
  • #22
Trifis said:
Yes that is well understood. What I meant to say is that the ideal wire of magnetostatics is not possible even with a theoretical assumptions when analyzed in microscopic level.
I think you are thinking of a EM field as something monolithic that covers all space as a single object. A field has a different value at each point. The fact that there may be some points where the invariant is negative does not change the fact that there are definitely points where the invariant is positive. At those points there is no reference frame where you will ever see an E field only.

Trifis said:
Am I right to think that something like that could not be analyzed within the scope of SR, since the reference frame of the charges would have to accelerate (circular orbit) ?
SR can handle acceleration just fine:
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html
 
  • #23
DaleSpam said:
I think you are thinking of a EM field as something monolithic that covers all space as a single object. A field has a different value at each point. The fact that there may be some points where the invariant is negative does not change the fact that there are definitely points where the invariant is positive. At those points there is no reference frame where you will ever see an E field only.
In order to detect a pure magnetic field at a specific point the invariant has to be 2B2 not just positive right? How could that be possible?
DaleSpam said:
"In SR it is still possible to use co-ordinate systems corresponding to accelerating or rotating frames of reference just as it is possible to solve ordinary mechanics problems in curvilinear co-ordinate systems. This is done by introducing a metric tensor. The formalism is very similar to that of many general relativity problems but it is still special relativity so long as the space-time is constrained to be flat and Minkowskian."

That is extremely interesting! Have you any books or papers, which elaborate that, to suggest?
 
  • #24
Trifis said:
In order to detect a pure magnetic field at a specific point the invariant has to be 2B2 not just positive right? How could that be possible?
Where are you getting the factor of 2 from? As long as it is positive in any frame there exists some frame where E=0.

Trifis said:
That is extremely interesting! Have you any books or papers, which elaborate that, to suggest?
I like the first chapter here:
http://arxiv.org/abs/gr-qc/9712019/
 
  • #25
DaleSpam said:
Where are you getting the factor of 2 from? As long as it is positive in any frame there exists some frame where E=0.

I think you are referring to the EM Lorentz scalar FμνFμν=2(Β2-[itex]\frac{Ε^2}{c^2}[/itex])
 
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  • #26
Trifis said:
I think you refer to the EM Lorentz scalar FμνFμν=2(Β2-[itex]\frac{Ε^2}{c^2}[/itex])
I was referring to half that quantity, but both quantities are invariant. The point remains that if it is positive in any frame then there exists some frame where E=0. Do you understand that now?
 
  • #27
DaleSpam said:
I was referring to half that quantity, but both quantities are invariant. The point remains that if it is positive in any frame then there exists some frame where E=0. Do you understand that now?
Hmmm so theoretically we can assume that there are somewhere some observers that happen to detect no E-Field. I was just wondering if we could examine analytically such a bizarre configuration.
 
  • #28
Trifis said:
Hmmm so theoretically we can assume that there are somewhere some observers that happen to detect no E-Field. I was just wondering if we could examine analytically such a bizarre configuration.
I don't know what would make you use the word "bizarre", but yes, you can do it analytically for simple current distributions (straight wires and circular loops). For more complicated geometries you need to do it numerically.
 
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  • #29
An obvious point: there are pure, macroscopic B fields, but they are dipole plus higher moments. There is no B analog of a coulomb field without monopoles.
 
  • #30
There is no B analog of a coulomb field without monopoles.

This is...I recognize it's the common view, but let me provide an alternative perspective on this.

We already know that the fundamental object representing the EM field is the Faraday tensor, [itex]F_{\mu \nu}[/itex]. This object can be called a bivector field--a field of oriented planes. The six components represent the six planes in a 3+1D spacetime: tx, ty, tz, yz, zx, xy. The first three, of course, correspond to the electric field; the other three correspond to the magnetic field. The full EM tensor is just a linear combination of these basis planes.

This should also be suggestive: maybe we're wrong to consider the magnetic field a vector field. After all, the yz, zx, and xy planes exist in ordinary 3D space already.

With that in mind, consider the magnetic field outside a straight current carrying wire and, in your mind, imagine instead the planes perpendicular to those field lines. What do you get? You get planes extending radially outward from this wire in a manner similar to a Coulomb field. The difference is that there is a translational symmetry along the direction of the wire.

But, remember the electric field can be interpreted as planes also, just with one of the directions in the plane being the t direction. In this way, a stationary charge looks exactly like the current-carrying wire. It just goes in the t direction instead of a spatial direction. And the electric field (as planes) looks exactly like the magnetic field (as planes).

This is why the straight current-carrying wire is as fundamental to magnetic fields as the stationary point charge is to electric fields. And while in most circles it's probably good enough to say there are no magnetic monopoles, I think this connection between the magnetic field from a wire and the electric field from a point charge is too significant to ignore.
 

1. Can a single charge produce both a magnetic and electric field?

Yes, a single charge can produce both a magnetic and electric field. According to Maxwell's equations, a moving charge creates a magnetic field and a stationary charge creates an electric field.

2. How can the magnetic field from a single charge be converted into an electric field?

The magnetic field from a single charge can be converted into an electric field through the process of electromagnetic induction. This occurs when a changing magnetic field induces an electric field in a nearby conductor.

3. Is it possible to completely convert the magnetic field from a single charge into an electric field?

No, it is not possible to completely convert the magnetic field from a single charge into an electric field. This is because the two fields are fundamentally different and cannot be fully converted into each other.

4. Can the strength of the electric field produced by a single charge be controlled?

Yes, the strength of the electric field produced by a single charge can be controlled by altering the distance between the charge and the point of measurement. The electric field strength decreases as the distance increases.

5. What are some real-world applications of converting magnetic fields into electric fields?

One common application is in electrical generators, where the motion of a conductor through a magnetic field converts the magnetic field into an electric field, producing electricity. Another application is in transformers, where changing magnetic fields are used to induce electric currents in a secondary coil. This is used in power distribution systems to increase or decrease the voltage of electricity.

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