Understanding why Einstein found Maxwell's electrodynamics not relativistic

1. Dec 24, 2012

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:
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?

2. Dec 24, 2012

Staff: Mentor

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.

3. Dec 24, 2012

Bill_K

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.

4. Dec 24, 2012

GregAshmore

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.

5. Dec 24, 2012

Staff: Mentor

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.

6. Dec 24, 2012

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.

7. Dec 24, 2012

Naty1

Sounds to me like you are the right track:

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!!

8. Dec 24, 2012

bcrowell

Staff Emeritus
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.

9. Dec 25, 2012

GregAshmore

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.

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'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.

10. Dec 25, 2012

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_electromagnetism

and I think Crowell's statement is consistent with this:

[also some neat historical perspective:]

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 E2 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 [Broken]

[I gave up following it after the first few pages.]

Last edited by a moderator: May 6, 2017
11. Dec 25, 2012

bcrowell

Staff Emeritus
Yes, of course.

No, that's not what I meant.

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.

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

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

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

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

Last edited: Dec 25, 2012
12. Dec 25, 2012

Staff: Mentor

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.

13. Dec 25, 2012

bcrowell

Staff Emeritus
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.

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.

Last edited: Dec 25, 2012
14. Dec 25, 2012

GregAshmore

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 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:
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.

15. Dec 25, 2012

Staff: Mentor

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.

Last edited: Dec 25, 2012
16. Dec 25, 2012

bcrowell

Staff Emeritus
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.)

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 agree.

17. Dec 25, 2012

bcrowell

Staff Emeritus
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).

18. Dec 26, 2012

GregAshmore

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.

19. Dec 26, 2012

Staff: Mentor

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

20. Dec 27, 2012

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