Understanding why Einstein found Maxwell's electrodynamics not relativistic

[GregAshmore]

I'm trying to understand exactly why Einstein considered Maxwell's electrodynamics to be non-relativistic. As I read Maxwell's paper, it seems to me that it is concerned only with relative motions. I'm thinking that the problem must be with the stationary ether proposed by Lorentz, for then motions must be considered relative to the coordinate system of the ether, not relative to the other poles and conductors in the system under consideration.

Here is Einstein from his 1905 paper:

It is known that Maxwell’s electrodynamics—as usually understood at the
present time—when applied to moving bodies, leads to asymmetries which do
not appear to be inherent in the phenomena. Take, for example, the reciprocal
electrodynamic action of a magnet and a conductor. The observable phenomenon
here depends only on the relative motion of the conductor and the
magnet, whereas the customary view draws a sharp distinction between the two
cases in which either the one or the other of these bodies is in motion. For if the
magnet is in motion and the conductor at rest, there arises in the neighbourhood
of the magnet an electric field with a certain definite energy, producing
a current at the places where parts of the conductor are situated. But if the
magnet is stationary and the conductor in motion, no electric field arises in the
neighbourhood of the magnet. In the conductor, however, we find an electromotive
force, to which in itself there is no corresponding energy, but which gives
rise—assuming equality of relative motion in the two cases discussed—to electric
currents of the same path and intensity as those produced by the electric
forces in the former case.

Einstein says that when the magnet is stationary and the conductor moving, no electric field arises in the neighborhood of the magnet.

As I read Maxwell's equation for electromotive force (equation D in the paper), it seems to me that "electromotive force" and "electric field" are synonymous. For in the equation, electromotive force at any point is a vector quantity, and it gives rise to current; that is the definition of an electric field.

Here is why I do not understand Einstein's claim that no magnetic field arises in the vicinity of the magnet: In the equation for electromotive force, the first term is the product of the strength of the magnetic field and the velocity of the conductor relative to the field. Thus, when the conductor moves and the magnet is stationary, electromotive force increases. Electromotive force is a synonym for electric field. Therefore, the movement of the conductor in the magnetic field gives rise to an electric field.

Einstein also says that there is no energy which corresponds to the electromotive force, in itself. I'm not sure what he means. As to intrinsic energy in the electromagnetic field, Maxwell says that it is proportional to the magnetic intensity, independent of electromotive force; he makes no mention of whether the magnet is moving or not.

Comments? Am I on the right track in thinking it is the stationary ether which causes the problem?

 

[PeterDonis]

I think the key is in that phrase in the Einstein quote: "as usually understood at the present time". The "usual" understanding he was referring to used Newtonian mechanics for mechanical phenomena, so there were going to be issues whenever you had a scenario with both mechanical motions and electromagnetic fields involved. So he wasn't just talking about Maxwell's electrodynamics by itself; he was talking about how Maxwell's electrodynamics is combined with mechanics. Some of the things he says about the combined theory "as usually understood at the present time" don't really make sense; but that's because the usual understanding at that time, trying to combine Maxwell's electrodynamics with Newtonian mechanics, didn't make sense.

 

[Bill_K]

Electromotive force is a synonym for electric field.

No, they are different. The Lorentz force on a moving charge is F = q(E + v x B). qE is the electric force and qv x B is the electromotive force. The electric force arises from an E field and electromotive force arises from motion in a B field. Einstein's point is that when you look at the same situation from different reference frames, one turns into the other.

 

[GregAshmore]

No, they are different. The Lorentz force on a moving charge is F = q(E + v x B). qE is the electric force and qv x B is the electromotive force. The electric force arises from an E field and electromotive force arises from motion in a B field. Einstein's point is that when you look at the same situation from different reference frames, one turns into the other.

The two may be equivalent.

There are three terms in Maxwell's equation for electromotive force:
[velocity of conductor] times [magnetic intensity at that instant]
rate of change of electromagnetic momentum (accounts for movement of poles, or change of intensity of stationary poles)
rate of change of electric potential

All three terms are vector operations. He presents the one equation as a set of three differential equations. I haven't done that kind of math in close to 30 years, so I don't pretend to understand the equations completely. I think I get the gist.

I'm guessing that in the Lorentz equation, v x B covers the first two terms of Maxwell, and E covers the third.

The first term of Maxwell can be eliminated if one chooses to put the origin of the coordinate system on the moving conductor. Then the electromotive force will be due to the change in magnetic intensity, from the second term. This is a trivial transformation--no need for the Lorentz transformation, and no need to make a fuss over it, in my opinion. Seems to me that Einstein had a deeper objection to the asymmetry than that.

 

[Dale]

I'm trying to understand exactly why Einstein considered Maxwell's electrodynamics to be non-relativistic.

At the time "relativistic" meant Galilean relativity. I.e. if something were relativistic then it was invariant under the Galilean transform. Maxwell's electrodynamics are not invariant under the Galilean transform, so they were non-relativistic by that criterion.

Einstein showed that the Lorentz transform also satisfied the principle of relativity and that Maxwell's equations were invariant under the Lorentz transform.

 

[Ibix]

As an undergrad, I was given an example of what DaleSpam is talking about. I have to admit that I've never actually done the maths, but I am told that if one solves for a B-field, then transforms it to a Galilean moving frame (x'=x-vt, t'=t), it no longer satisfies ∇B=0. In other words, Maxwell+Galileo say that a common bar magnet should gain magnetic monopole characteristics just by being moved. That is not exactly consistent with observation.

 

[Naty1]

Sounds to me like you are the right track:

Since the Maxwell equations imply a fixed speed of light, they were long believed to be only valid for an observer at rest with respect to a postulated "ether". Einstein, in the special theory of relativity, postulated instead that the Maxwell equations are valid for arbitrary observers, and showed that this implies that separate notions of space and time have no observer-independent physical reality. Since the mid-20th century, it has been understood, however, that Maxwell's equations are not exact laws of the universe but an approximation to the more accurate and fundamental theory of quantum electrodynamics.

http://en.wikipedia.org/wiki/Maxwell's_equations

but understanding historical context is even worse than trying to understand current perspectives.

such as: "At the time "relativistic" meant Galilean relativity."
Oh good grief!

 

[bcrowell]

Einstein also says that there is no energy which corresponds to the electromotive force, in itself. I'm not sure what he means.

When he refers to energy, he's talking about the energy density that exists in a field, which is proportional to E² or B². In the frame where there's an induced E field, he's emphasizing that this field is a real thing by pointing out that it has energy. In the frame where there's no E field, he's emphasizing that its nonexistence is a concrete, verifiable fact, because there's no energy.

I think the simplest way of understanding why Maxwell's equations aren't compatible with Galilean relativity is to look at what happens when you take an electromagnetic plane wave and transform into the frame in which the wave is at rest. In that frame, it clearly violates Maxwell's equations.

 

[GregAshmore]

When he refers to energy, he's talking about the energy density that exists in a field, which is proportional to E² or B².

For Maxwell (1864 paper), the total energy stored in the electromagnetic field is the sum of two components. One component is due to the magnetization of the field; the other is due to the displacement current. I'm 99.9% sure that B corresponds exactly to Maxwell's magnetic intensity, which he labels I. I'm reasonably sure that the energy due to the displacement current (that is, electric charge pumped into the field by electromotive force) corresponds to E.

As I read your statement, the energy is determined by either E or B; not the combination of the two. I'm not saying you are wrong; I trying to understand how what Maxwell says corresponds to what Einstein says.

Whether it is "E and B" or "E or B", in the case under discussion there is energy in the field due to the magnet.

In the frame where there's an induced E field, he's emphasizing that this field is a real thing by pointing out that it has energy. In the frame where there's no E field, he's emphasizing that its nonexistence is a concrete, verifiable fact, because there's no energy.

But there is energy in the field, due to the magnetization of the field by the magnetic pole.

That energy produces the electromotive force. Here I would go back to my original post and argue that Maxwell's equation says that the movement of the conductor in the magnetic field produces an electric field. For the electromotive force that the equation predicts has exactly--no more, no less--the characteristics of an electric field: a force vector at any point in space.

In fact, the same equation, the one for electromotive force, is the one that predicts the rise of an electric field due to the movement of a magnet relative to a stationary conductor. The electric field due to the movement of the conductor in a magnetic field is in the first term of the equation; the electric field due to the movement of a magnet is in the second term of the equation. (The second term is the rate of change of electromagnetic momentum, which is magnetic energy stored in the field. Maxwell says that a moving magnetic pole, or moving current, or a change in intensity of pole or current, changes the electromagnetic momentum at a point in space.)

I think the simplest way of understanding why Maxwell's equations aren't compatible with Galilean relativity is to look at what happens when you take an electromagnetic plane wave and transform into the frame in which the wave is at rest. In that frame, it clearly violates Maxwell's equations.

I'd have to think about that a while. It is not immediately obvious to me that there is any physical significance to that transformation.

As far as the case under discussion is concerned, it seems to me that Maxwell's equation for electromotive force is invariant with respect to a Galilean transformation. It must be: as noted above, this equation is used to calculate the electromotive force in both cases, when the magnet is stationary and when the conductor is stationary. The answer is the same in both cases.

 

[Naty1]

Greg: I am also having trouble how to interpret frames here...Crowell's comment. there is some interesting easy to follow background here:

http://en.wikipedia.org/wiki/Relativistic_electromagnetismand I think Crowell's statement is consistent with this:

Purcell argued that the question of an electric field in one inertial frame of reference, and how it looks from a different reference frame moving with respect to the first, is crucial to understand fields created by moving sources. In the special case, the sources that create the field are at rest with respect to one of the reference frames. Given the electric field in the frame where the sources are at rest, Purcell asked: what is the electric field in some other frame?
He stated that the fundamental assumption is that, knowing the electric field at some point (in space and time) in the rest frame of the sources, and knowing the relative velocity of the two frames provided all the information needed to calculate the electric field at the same point in the other frame. In other words, the electric field in the other frame does not depend on the particular distribution of the source charges, only on the local value of the electric field in the first frame at that point. He assumed that the electric field is a complete representation of the influence of the far-away charges.

[also some neat historical perspective:]

In 1953 Albert Einstein wrote to the Cleveland Physics Society on the occasion of a commemoration of the Michaelson–Morley experiment. In that letter he wrote:[1]
What led me more or less directly to the special theory of relativity was the conviction that the electromotive force acting on a body in motion in a magnetic field was nothing else but an electric field.
This statement by Einstein reveals that he investigated spacetime symmetries to determine the complementarity of electric and magnetic forces.

and a more formal mathematical discussion in a link from the above article:

http://en.wikipedia.org/wiki/Formulation_of_Maxwell's_equations_in_special_relativity
and here:

http://en.wikipedia.org/wiki/Formul...lativity#Electromagnetic_stress-energy_tensor

the electromagnetic stress energy tensor seems to be dependent on BOTH E<sup>2 B2...

a lot of this math is above my pay grade, but I am guessing the frame of T sets E and B??

I knew I had seem something like Crowell's explanation:
from Dalespam:
"See here for a rigorous proof based on established mainstream science demonstrating that the power density delivered by an EM field is given by E.j in all cases" : http://farside.ph.utexas.edu/teachin...es/node89.html

this thread:
https://www.physicsforums.com/showthread.php?t=660589&highlight=magnetic+field+energy

[I gave up following it after the first few pages.]</sup>

 

[bcrowell]

For Maxwell (1864 paper), the total energy stored in the electromagnetic field is the sum of two components.

Yes, of course.

As I read your statement, the energy is determined by either E or B; not the combination of the two.

No, that's not what I meant.

One component is due to the magnetization of the field; the other is due to the displacement current. I'm 99.9% sure that B corresponds exactly to Maxwell's magnetic intensity, which he labels I.

Is there a reason that you want to discuss the archaic terminology and notation? The passage we're discussing is the opening paragraph of Einstein's 1905 paper, which doesn't even use any mathematical symbols.

I'm reasonably sure that the energy due to the displacement current (that is, electric charge pumped into the field by electromotive force) corresponds to E.

No, this is wrong. The energy density is $\frac{E^2}{8\pi k}+\frac{B^2}{8\pi k/c^2}$.

But there is energy in the field, due to the magnetization of the field by the magnetic pole.

I don't know what you mean by this. Fields don't get magnetized, materials get magnetized.

That energy produces the electromotive force. Here I would go back to my original post and argue that Maxwell's equation says that the movement of the conductor in the magnetic field produces an electric field. For the electromotive force that the equation predicts has exactly--no more, no less--the characteristics of an electric field: a force vector at any point in space.

This is wrong. A field isn't defined by the existence of a force.

As far as the case under discussion is concerned, it seems to me that Maxwell's equation for electromotive force is invariant with respect to a Galilean transformation. It must be: as noted above, this equation is used to calculate the electromotive force in both cases, when the magnet is stationary and when the conductor is stationary. The answer is the same in both cases.

You're using "electromotive force" for what would today be called the Lorentz force. If you read Jeffery's translation of the Einstein paper closely, you'll see that the term is used in a way that is consistent with modern English usage (i.e., it has units of energy per unit charge) except in section 6, where it's in quotes, presumably indicating that even then, physicists didn't really use the term that way. The equation for the Lorentz force is not one of the four equations that are today referred to as Maxwell's equations. Here are Maxwell's equations: http://en.wikipedia.org/wiki/Maxwell's_equations#Conventional_formulation_in_SI_units

 

[Dale]

As far as the case under discussion is concerned, it seems to me that Maxwell's equation for electromotive force is invariant with respect to a Galilean transformation.

This is easy enough to check. The Lorentz force is given by $f=q(E+v \times B)$. When we do a Galilean boost by a velocity u then q is unchanged as are E, and B, but v transforms to v+u. So in the boosted frame $f=q(E+(v+u) \times B)$ which is not the same as in the original frame. So the electromotive force is definitely not invariant wrt a Galilean transform.

 

[bcrowell]

This is easy enough to check. The Lorentz force is given by $f=q(E+v \times B)$. When we do a Galilean boost by a velocity u then q is unchanged as are E, and B, but v transforms to v+u. So in the boosted frame $f=q(E+(v+u) \times B)$ which is not the same as in the original frame. So the electromotive force is definitely not invariant wrt a Galilean transform.

I think this is a little bit of an oversimplification. Even before Einstein, I don't think anyone believed that E and B were invariant. That would be impossible, because a charge that was at rest in one frame would be moving in another, and therefore produce a magnetic field that wasn't present in the original frame. Section 6 of Einstein's 1905 paper presents the correct relativistic transformation of E and B. Before 1905, I think they must have had some muddled understanding of how this transformation worked, presumably without the factors of γ (i.e., β in Einstein's notation). They would have probably conceived of this in terms of an approximate property of Maxwell's equations that made a velocity u relative to the ether undetectable to first order in u/c.

As far as the case under discussion is concerned, it seems to me that Maxwell's equation for electromotive force is invariant with respect to a Galilean transformation. It must be: as noted above, this equation is used to calculate the electromotive force in both cases, when the magnet is stationary and when the conductor is stationary. The answer is the same in both cases.

If you want this claim to be a meaningful statement, you're going to have to define how you think E and B transform under a Galilean transformation. Your argument is based on the assumption that "The answer is the same in both cases." There is no reason to believe that this is true. The first paragraph of the Einstein paper is appealing to the fact that experiments had failed to detect the existence of a preferred frame for electromagnetism, and it discusses how the prevailing theory, based on Galilean transformations, was able to approximately explain the relevant forces, to some order in u/c, in two different Galilean frames.

 

[GregAshmore]

This is easy enough to check. The Lorentz force is given by $f=q(E+v \times B)$. When we do a Galilean boost by a velocity u then q is unchanged as are E, and B, but v transforms to v+u. So in the boosted frame $f=q(E+(v+u) \times B)$ which is not the same as in the original frame. So the electromotive force is definitely not invariant wrt a Galilean transform.

Yes. I misused the term "Galilean transformation" earlier. I only meant to shift the origin of the coordinate system, not introduce a frame in which both magnet and conductor are moving relative to the ether. But thinking about it, the origin of the coordinate system has nothing to do with it either.

What I had in mind is this: So long as one or the other of the two bodies under consideration (magnet or conductor) is at rest relative to the ether, Maxwell's equation for electromotive force predicts one and the same outcome regardless of which one moves, and the energy which drives the outcome is the same in either case. I did not think about it in terms of "at least one body at rest in the ether" because Maxwell never stipulates the rest frame of the ether. But, it is implicit in the equations that the ether is at rest in the frame of the coordinate system.

Part of my difficulty in understanding Einstein's claim of asymmetry is that Maxwell never uses the term "electric field" in any of his published papers. (That claim is true if a search of the pdf file is to be trusted.) I have not seen "electric field" in the 1864 paper on electrodynamics. He talks mostly about the electromagnetic field; sometimes he says magnetic field; once (in On Faraday's Lines of Force) he says magneto-electric field.

Maxwell does talk about electromagnetic force. As noted earlier, Maxwell's electromotive force is a vector quantity at any point in space that forces movement of electric charge. My understanding of an electric field is precisely that: a vector quantity at any point in space that forces movement of electric charge. If my definition of electric field is correct, then I believe I am on solid ground when I say that the equation for electromotive force is the equation for the electric field.

In that equation, there are three contributors to the electric field at any point: motion of a conductor, change of intensity of magnetization (includes motion of a magnetic pole), and change of electric potential. So for Maxwell, movement of a conductor in a magnetic field produces an electric field, as does movement of a magnet. Here is the equation (in three parts) for electromotive force, from the paper:

Maxwells' explanation:

The first term in each equation represents the electromotive force arising from the motion of the conductor itself.

The second term in each equation indicates the effect of changes in the position or strength of magnets or currents in the field.

The energy which produces the electric field is in the magnetic field, regardless of whether it is the conductor moving, or the magnet, or both. Maxwell says that the electromagnetic energy is stored in the field; he also says that the field is magnetized. He means this literally, because he believes that the ether is a material substance:

In speaking of the Energy of the field, however, I wish to be understood literally. All energy is the same as mechanical energy, whether it exists in the form of motion or in that of elasticity, or in any other form. The energy in electromagnetic phenomena is mechanical energy. The only question is, Where does it reside ? On the old theories it resides in the electrified bodies, conducting circuits, and magnets, in the form of an unknown quality called potential energy, or the power of producing certain efiects at a distance. On our theory it resides in the electromagnetic field, in the space surrounding the electrified and magnetic bodies, as well as in those bodies themselves, and is in two different forms, which may be described without hypothesis as magnetic polarization and electric polarization, or, according to a very probable hypothesis, as the motion and the strain of one and the same medium.

To summarize: I understand why Einstein might have started out by saying that Maxwell's equations are not valid when all bodies in system are moving relative to the ether. But he did not say that. He said (or seems to say) that an electromotive force arises unbidden by energy when a conductor moves in a stationary magnetic field. I don't believe Maxwell says that; perhaps Lorentz says it.

 

[Dale]

I think this is a little bit of an oversimplification. Even before Einstein, I don't think anyone believed that E and B were invariant.

Hmm, I was just taking the relativistic transformations and letting c→∞ which is the usual relationship between the Lorentz and Galilean transformation and which produces an invariant E and B. So I believe that it is correctly simple. If someone had proposed a different transformation for the fields then I don't know what it would be.

However, honestly my interest in historical matters is pretty minimal so I wouldn't be surprised that there was one that I am just not aware of. So I will let it rest with that. I think that the only necessary historical point is that it was clear to scientists at the time that Maxwell's equations were not (Galilean) relativistic, but the specific details of the attempted patches are not so critical to understand.

 

[bcrowell]

Hmm, I was just taking the relativistic transformations and letting c→∞ which is the usual relationship between the Lorentz and Galilean transformation and which produces an invariant E and B.

I don't think that's the right way to go here, because in that limit, magnetic fields don't exist. (The source terms in Maxwell's equation for curl B both have a factor of 1/c^2 in front.)

So I believe that it is correctly simple. If someone had proposed a different transformation for the fields then I don't know what it would be.

It's possible that they simply didn't think in those terms. It's a nontrivial assumption to imagine that there is some universal law for transforming E and B to E' and B' at a given point in space, regardless of the nature of the sources that created the fields.

I think that the only necessary historical point is that it was clear to scientists at the time that Maxwell's equations were not (Galilean) relativistic[...]

I agree.

 

[bcrowell]

Turns out both DaleSpam and I were wrong in our guesses about the history and physics of how E and B would transform and were thought to have transformed under a Galilean transformation. This is apparently quite subtle, and was not worked out until recently.

Marc De Montigny, Germain Rousseaux, "On the electrodynamics of moving bodies at low velocities," http://arxiv.org/abs/physics/0512200

See pp. 2-5, 22, 23 (pdf page numbers).

 

[GregAshmore]

It's possible that they simply didn't think in those terms. It's a nontrivial assumption to imagine that there is some universal law for transforming E and B to E' and B' at a given point in space, regardless of the nature of the sources that created the fields.

My sense from reading Maxwell, and from the equations he proposes, is that there would be no conception of such a transformation. In evaluating any system of (possibly moving) magnetic poles, currents, and conductors, one would use the equations to solve for the state of the electromagnetic field at any location and time. The state of the electromagnetic field is the state of the ether: its magnetic and electric polarization (energy physically stored in the ether) and the rate of change of that polarization. The electromotive force (electric field) in a conductor is determined by the state of the ether and the motion of the conductor (if any) relative to the ether. At any time, there can only be one state of the ether; the notion of transforming to different frames would have no physical significance, I think.

 

[Dale]

Turns out both DaleSpam and I were wrong in our guesses about the history and physics of how E and B would transform and were thought to have transformed under a Galilean transformation. This is apparently quite subtle, and was not worked out until recently.

Marc De Montigny, Germain Rousseaux, "On the electrodynamics of moving bodies at low velocities," http://arxiv.org/abs/physics/0512200

See pp. 2-5, 22, 23 (pdf page numbers).

Excellent paper. It never occurred to me that there would be two limits.

 

[GregAshmore]

I read the paper. My math is not good enough for anything more than a superficial appreciation of the argument.

I'd like to go back to my understanding of the electric field, and Ben's response.
Me:

For the electromotive force that the equation predicts has exactly--no more, no less--the characteristics of an electric field: a force vector at any point in space.

Ben:

This is wrong. A field isn't defined by the existence of a force.

No, not in general. But I was talking about an electric field, which is a force field, as I understand it.
From Georgia State University:

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge.

From the Physics Classroom:

The direction of the electric field is always directed in the direction that a positive test charge would be pushed or pulled if placed in the space surrounding the source charge. Since electric field is a vector quantity, it can be represented by a vector arrow. For any given location, the arrows point in the direction of the electric field and their length is proportional to the strength of the electric field at that location. A more useful means of visually representing the vector nature of an electric field is through the use of electric field lines of force.

It seems pretty clear that force is at least a component of the electric field. Is there some other component in addition to force?

 

[bcrowell]

No, not in general. But I was talking about an electric field, which is a force field, as I understand it.
From Georgia State University:

Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge.

The force is the force that hypothetically *would* act if a charge were placed there. If there's no charge there, there's no force, but there's still a field.

 

[dextercioby]

[...]No, not in general. But I was talking about an electric field, which is a force field, as I understand it.[...]

The electric field (or its intensity $\displaystyle{\vec{E}\left(\vec{r},t\right)}$ according to the former SI nomenclature) is defined through Maxwell's equations (for simplicity you can choose vacuum) in the presence or in the absence of sources (stationary/moving electric charges). In this view, the Coulomb electrostatic force is a derived concept.

The definitions you quoted are essentially operational (experiment based) but misleading, in the sense that one initially defines the force, then the field which is wrong.

 

[GregAshmore]

Okay--no force if no charge present.

The result of the electromotive force equation in Maxwell's 1864 paper must be the electric field.

In that case, why would Einstein say that there is no electric field when the magnet is stationary and the conductor moves? The first term of the equation says that there is an electric field due to the motion of the conductor in the magnetic field.

 

[Dale]

In that case, why would Einstein say that there is no electric field when the magnet is stationary and the conductor moves?

Because it is correct. There is no electric field when the magnet is stationary.

The first term of the equation says that there is an electric field due to the motion of the conductor in the magnetic field.

Which equation?

 

[GregAshmore]

Because it is correct. There is no electric field when the magnet is stationary.

The electric field is the electromotive force which would be on a unit charge at a given location if there is a charge present.

Experiment shows that there is an electromotive force at a given location when the magnet is stationary and there is a conductor moving.

Therefore, experiment shows that there is an electric field when the magnet is stationary and there is a conductor moving.

Which equation?

See #14 above.

The equation is for electromotive force ay any point. So, by the definition of the electric field, it is the equation for the electric field.

In this equation, the first term is the contribution to the electric field from a charge moving in the presence of a magnetic field. The second term is the contribution to the electric field from a moving magnet. If by this equation a moving magnet produces an electric field, then by the same equation a charge moving in the presence of a magnetic field also produces an electric field.

 

[Dale]

The electric field is the electromotive force which would be on a unit charge at a given location if there is a charge present.

Not necessarily. The electromotive force can also be due to a magnetic field, as in the case under consideration. The mere presence of an elecromotive force does not provide enough information to determine if it is an electric or a magnetic field or both. In the case under consideration it is due purely to the magnetic field, there is no electric field.

See #14 above.

The notation is archaic, and I don't know what all the variables are, so I cannot comment much on that equation. However, only three variables could possibly be the components of the electric field, and there are more than three variables. So clearly, the mere fact of an electromotive force does not imply an electric field by that equation either.

 

[GregAshmore]

Not necessarily. The electromotive force can also be due to a magnetic field, as in the case under consideration. The mere presence of an elecromotive force does not provide enough information to determine if it is an electric or a magnetic field or both. In the case under consideration it is due purely to the magnetic field, there is no electric field.

I don't think I made myself clear. I'll try again.

1. The electric field is the electromotive force at a given location that hypothetically *would* act if a charge were placed there. [The italics are from BCrowell's response a few posts back.] The point is that the electric field is the map of the electromotive force, irrespective of the agent(s) that bring about the force.

2. Experiment shows that there is an electromotive force when a magnet is stationary and a conductor is moving. The electromotive force is acting on the charges in the conductor. Considering a single charge in the conductor, experiment shows that there is an electromotive force at a given location when a charge is present and moving.

3. Therefore, experiment shows that there is an electric field when the magnet is stationary and there is a conductor moving. [There is a charge; there is electromotive force on the charge; by definition there is an electric field.]

The three statements above are presented as proof that there is an electric field present when the magnet is stationary. In short: It has been said that there is no electric field present when the magnet is stationary. I respond, give me a magnet and a conductor and I will demonstrate that there is an electric field.

The notation is archaic, and I don't know what all the variables are, so I cannot comment much on that equation. However, only three variables could possibly be the components of the electric field, and there are more than three variables. So clearly, the mere fact of an electromotive force does not imply an electric field by that equation either.

Addressing the text in bold: The electric field is defined as the map of electromotive force. By definition, then, the mere fact of an electromotive force is the very same thing as an electric field.

Consider the case when the magnet is moving and the conductor is stationary. The assertion is that there is an electric field. On what basis is the assertion made? On the basis of an observed electromotive force. In both cases, electromotive force is observed. Why is the observed electromotive force taken as proof of an electric field in the one case, but not the other?

 

[Dale]

I don't think I made myself clear. I'll try again.

You were clear, just wrong.

1. The electric field is the electromotive force at a given location that hypothetically *would* act if a charge were placed there. [The italics are from BCrowell's response a few posts back.] The point is that the electric field is the map of the electromotive force, irrespective of the agent(s) that bring about the force.

Close, but not quite. The electric field is proportional to the force that would act if a stationary charge were located there.

2. Experiment shows that there is an electromotive force when a magnet is stationary and a conductor is moving. The electromotive force is acting on the charges in the conductor. Considering a single charge in the conductor, experiment shows that there is an electromotive force at a given location when a charge is present and moving.

Yes.

3. Therefore, experiment shows that there is an electric field when the magnet is stationary and there is a conductor moving. [There is a charge; there is electromotive force on the charge; by definition there is an electric field.]

No, there is an electromotive force on moving charges, which is not the definition of an electric field.

The three statements above are presented as proof that there is an electric field present when the magnet is stationary. In short: It has been said that there is no electric field present when the magnet is stationary. I respond, give me a magnet and a conductor and I will demonstrate that there is an electric field.

Not if you move the conductor.

Addressing the text in bold: The electric field is defined as the map of electromotive force. By definition, then, the mere fact of an electromotive force is the very same thing as an electric field.

The electric field is a map of the electromotive force on stationary charges, so the mere fact of an electromotive force on a moving charge is not the very same thing as an electric field.

Consider the case when the magnet is moving and the conductor is stationary. The assertion is that there is an electric field. On what basis is the assertion made? On the basis of an observed electromotive force. In both cases, electromotive force is observed. Why is the observed electromotive force taken as proof of an electric field in the one case, but not the other?

Because the charges are moving in one case but not the other.

 

[Dale]

In modern notation the electromotive force, aka the Lorentz force, is given by:
$F = q (E + v \times B)$

Note that this force consists of a term arising from an E field and a velocity dependent term arising from a B field. If you want to claim that an emf is only due to an E field, then you must get the B field term to drop out, which happens if v=0.

I generally think of this equation as defining the E and B fields.

 

[vanhees71]

I think one has to elaborate this a bit since it's important to make clear what's meant by "force" here. Force is a pretty difficult concept if it comes to relativity. It's better to derive everything from the action principle. As will become clear, the "Lorentz force", as defined in the previous posting is not a manifestly Lorentz covariant quantity and must be handled with some care.

Let's take, as the most simple example, a single classical charged particle moving in a given (external) electromagnetic field. The latter is given by the four-vector potential $A^{\mu}$. The relativistically covariant description of the field itself is given by the Faraday tensor,
$$F_{\mu \nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}.$$
In conventional 1+3-dimensional writing we have
$$\vec{E}=\frac{1}{c} \partial_t \vec{A}-\vec{\nabla} A^0, \quad \vec{B}=\vec{\nabla} \times \vec{A},$$
where $$\vec{A}=(A^1,A^2,A^3).$$

The Lagrangian for the particle's motion in the given em. field reads
$$L=-m c \sqrt{c^2-\vec{v}^2}-\frac{q}{c} A^{\mu} \frac{\mathrm{d} x_{\mu}}{\mathrm{d} t}.$$
This is of course not covariant, but the action is
$$A[x]=\int \mathrm{d} t L.$$
This can be written as a functional of the world line $x^{\mu}(\lambda)$, where $\lambda$ is an arbitrary scalar parameter
$$A[x]=\int \mathrm{d} \lambda \left (-m c \sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}-\frac{q}{c} \dot{x}_{\mu} A^{\mu} \right ).$$
Here the dot means a derivative with respect to the "world parameter" $\lambda$. The equations of motion follow from Hamilton's principle. To derive them we note that
$$\frac{\partial L}{\partial \dot{x}^{\mu}}=-m c \frac{\dot{x}_{\mu}}{\sqrt{\dot{x}_{\mu} \dot{x}^{\mu}}}-\frac{q}{c} A_{\mu}, \quad \frac{\partial L}{\partial x^{\mu}}=-\frac{q}{c} \dot{x}^{\nu} \partial_{\mu} A_{\nu}.$$
To simplify the further calculation we note that we now can use the proper time of the particle $\mathrm{d} \tau = 1/c \sqrt{\mathrm{d} x_{\mu} \mathrm{d} x^{\mu}}$ as the "world parameter". This leads to $\dot{x}_{\mu} \dot{x}^{\mu}=c^2=\text{const}$, and the equations of motion finally read
$$-m \ddot{x}_{\mu} -\frac{q}{c} \dot{x}^{\nu} \partial_{\nu} A_{\mu} = - \frac{q}{c} \dot{x}^{\nu} \partial_{\mu} A_{\nu}$$
or
$$m \ddot{x}_{\mu}=\frac{q}{c} F_{\mu \nu} \dot{x}^{\nu}:=K_{\mu}.$$
This is the equation of motion for a particle in an electromagnetic field in manifestly covariant form. Both sides of the equation are obviously four vectors. The right-hand side is what's usually called the Minkowski-force vector.

As expected, because we have derived the equation of motion from a world-parameter-independent action, it fulfills the necessary constraint
$$\dot{x}^{\mu} \ddot{x}_{\mu}=\frac{1}{2} \frac{\mathrm{d}}{\mathrm{d} \tau} (\dot{x}_{mu} \dot{x}^{\mu})=0.$$
Thus the constraint
$$\dot{x}_{\mu} \dot{x}^{\mu}=c^2=\text{const}$$
is compatabile with the equation of motion.

Thus only three of the four equations of motion are independent. It's thus sufficient to only consider $\vec{x}$ as independent degrees of freedom. Written in terms of the proper time the equation of motion for these components read, rewritten in three-dimensional notation,
$$m \frac{\mathrm{d}^2 \vec{x}}{\mathrm{d} \tau^2}=q \left (\vec{E} \frac{\mathrm{d} t}{\mathrm{d} \tau}+\frac{1}{c} \frac{\mathrm{d} \vec{x}}{\mathrm{d} \tau} \times \vec{B} \right )=\vec{K}.$$
Of course, we can further rewrite this in terms of the coordinate time of an arbitrary inertial observer, using
$$\mathrm{d} \tau=\mathrm{d} t \sqrt{1-\vec{v}^2/c^2}=\mathrm{d} t/\gamma,$$
where we have set
$$\vec{v}=\frac{\mathrm{d} \vec{x}}{\mathrm{d} t} \quad \text{and} \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.$$
This gives after some algebra the equation of motion in the non-covariant form
$$m \frac{\mathrm{d}}{\mathrm{d} t} \left (\gamma \vec{v} \right ) = q \left (\vec{E}+\frac{1}{c} \vec{v} \times \vec{B} \right)=:\vec{F}.$$
From the derivation it's clear that $\vec{F}$ is not the spatial part of a four-vector. That's the case only for
$$\vec{K}=q \left (\vec{E} \frac{\partial t}{\partial \tau} +\frac{1}{c} \gamma \vec{v} \times \vec{B} \right )=q \gamma \left (\vec{E} +\frac{1}{c} \vec{v} \times \vec{B} \right )=q \gamma \vec{F}.$$
It's of course correct that the physical meaning of the electromagnetic field $(\vec{E},\vec{B})$ is defined by its action in terms of the "Lorentz force", but one should keep in mind that the Lorentz force itself is a non-covariant quantity, while the electromagnetic field is, because it's nothing but the components of the Faraday tensor, which is a 2nd-rank antisymmetric tensor in Minkowski space-time.

 

[GregAshmore]

In modern notation the electromotive force, aka the Lorentz force, is given by:
$F = q (E + v × B)$

Note that this force consists of a term arising from an E field and a velocity dependent term arising from a B field. If you want to claim that an emf is only due to an E field, then you must get the B field term to drop out, which happens if v=0.

I generally think of this equation as defining the E and B fields.

Okay. I did not realize that the electric field applies to a stationary charge.

In the Maxwell equation for emf, the second and third terms will be included in E. These are the forces on a stationary charge from a change in the magnetic field and from the potential gradient. The first term is v × B.

Even with the knowledge that the electric field applies to a stationary charge, I am still at a loss to understand Einstein's statement, "In the [moving] conductor, however, we find an electromotive force, to which in itself there is no corresponding energy."

Clearly, the energy for the emf on the charges in the moving conductor comes from the magnetic field through which they are moving: v × B.

If Einstein has a legitimate quarrel on that point, it is not with Maxwell. Maxwell attributes the cause of the emf to the magnetic field, regardless of whether the charge is moving through the field or the field is changing in strength (as when a pole moves):

This, then, is a force acting on a body caused by its motion through the
electromagnetic field, or by changes occurring in that field itself; and the effect
of the force is either to produce a current and heat the body, or to decompose
the body, or, when it can do neither, to put the body in a state of electric
polarization...

 

[Dale]

Even with the knowledge that the electric field applies to a stationary charge, I am still at a loss to understand Einstein's statement, "In the [moving] conductor, however, we find an electromotive force, to which in itself there is no corresponding energy."

Clearly, the energy for the emf on the charges in the moving conductor comes from the magnetic field through which they are moving: v × B.

This too is corrrect, although rather cumbersome in my opinion.

The integral of the work done by a force from a static E field about a closed loop is always 0, so the force is said to have a corresponding energy and scalar potential. The integral of the work done by a force from a static B field is not always 0, so the force is not said to have a corresponding energy or scalar potential.

Although the energy does come from the B field, the force and the B field are not the same thing and are in fact not even proportional. So the fact that the B field does have energy "in itself" does not contradict the fact that the force from the B field does not.

 

[GregAshmore]

This too is corrrect, although rather cumbersome in my opinion.

The integral of the work done by a force from a static E field about a closed loop is always 0, so the force is said to have a corresponding energy and scalar potential. The integral of the work done by a force from a static B field is not always 0, so the force is not said to have a corresponding energy or scalar potential.

Einstein says that the force from the B field is the same as the force from the E field, given the same relative velocity and distance. If the forces and paths are the same, how can the integrals of force over path be different? (The appropriate answer to this may be, "Go read this book.")

Although the energy does come from the B field, the force and the B field are not the same thing and are in fact not even proportional. So the fact that the B field does have energy "in itself" does not contradict the fact that the force from the B field does not.

Without contradicting you, I will note that the first term of Maxwell's equation for emf (the term which deals with a moving charge) consists entirely of the addition and subtraction of terms of this form:
(magnetic intensity along a coordinate axis) * (velocity along a coordinate axis)

On inspection, it appears as though the force is proportional to the strength of the magnetic field. (Again, the appropriate answer to this may be, "Go read this book.")

At any rate, given that the two emfs are the same, as Einstein says they are, it seems to me that Einstein's complaint is along the lines of a legal technicality, and of no practical consequence.

I have been wondering why he did not directly raise the issue that Maxwell requires the velocities of the magnet and conductor to be reckoned relative to the ether, rather than relative to each other. Perhaps the issue he raises (the emf on the moving conductor not having energy in itself) resolves to the larger issue of velocities being reckoned relative to the ether, instead of relative to the bodies under test. I note that in the body of the paper he transforms the coordinates (including time) of the moving conductor to the frame in which the conductor is stationary, taking into account only the relative velocity of the conductor and magnet. It would not be possible to do this (and get the correct answer) if the force is transmitted by the ether, and both the magnet and the conductor are moving relative to the ether.

 

[Dale]

Einstein says that the force from the B field is the same as the force from the E field, given the same relative velocity and distance. If the forces and paths are the same, how can the integrals of force over path be different? (The appropriate answer to this may be, "Go read this book.")

Because the force from the B field also depends on the velocity, which can be different, even over the same path.

EDIT: note, I was talking about an arbitrary closed loop I.e. For defining a conserved energy. Not a straight line at constant speed.

Without contradicting you, I will note that the first term of Maxwell's equation for emf (the term which deals with a moving charge) consists entirely of the addition and subtraction of terms of this form:
(magnetic intensity along a coordinate axis) * (velocity along a coordinate axis)

On inspection, it appears as though the force is proportional to the strength of the magnetic field. (Again, the appropriate answer to this may be, "Go read this book.")

Yes. And you are correct, that doesn't contradict me.

At any rate, given that the two emfs are the same, as Einstein says they are, it seems to me that Einstein's complaint is along the lines of a legal technicality, and of no practical consequence.

That is essentially his point. The equations are non relativistic, but the phenomena are not.

Look, the mathematical fact is that Maxwells equations are non (Galilean) relativistic. The historical fact is that scientists of Einstein's day were aware of that. What else is relevant here?

 

[GregAshmore]

Yes. And you are correct, that doesn't contradict me.

But I'm not sure why I'm not contradicting you. You said, "Although the energy does come from the B field, the force and the B field are not the same thing and are in fact not even proportional." I countered, "On inspection, it appears as though the force is proportional to the strength of the magnetic field." I expected you to tell me I am wrong. Instead you agreed with me.

The B field is the map of the strength of the magnetic field in space. (That is, the map at any instant. Variations in the magnetic field over time are accounted for in the second term of the equation.) That being so, the subject of your statement (the B field) is the same as the subject of my counter (the strength of the magnetic field). I don't see how your statement and my counter can both be true.

That is essentially his point. The equations are non relativistic, but the phenomena are not. Look, the mathematical fact is that Maxwells equations are non (Galilean) relativistic. The historical fact is that scientists of Einstein's day were aware of that. What else is relevant here?

I wanted to understand the particular example Einstein gives in the introduction to his 1905 paper. The biggest part of my difficulty was not understanding that "energy in itself" is a technical term, as you have explained. It sounded to me as though Einstein was asserting that Maxwell's theory does not assign a cause to the rise of the emf in the moving conductor.

I also had some difficulty in understanding why the Maxwell equation for emf is not relativistic, in the sense of depending only on the relative velocity of the magnet and conductor. The equation is relativistic in that sense, so long as the "stationary" body is at rest relative to the ether. But of course, that stipulation opens the whole can of relativistic worms, doesn't it?