Question: Interpretation of Current and Magnetic Field Using Relativity

  • #1
StoneHengeAlive
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The current in a wire is said to produce a magnetic field and a force on a charge. It is said that the correct way to interpret the effect is because of relativity. So does that mean that that the most basic non-relativistic equations that offer basic calculations of the forces and fields are implicitly a first order approximate accounting of relativity effect? That is, the very basic magnetic field and force calculations are in reality merely a backdoor first level relativity correction?
The current in a wire is said to produce a magnetic field and a force on a charge. It is said that the correct way to interpret the effect is because of relativity. So does that mean that that the most basic non-relativistic equations that offer basic calculations of the forces and fields are implicitly a first order approximate accounting of relativity effect? That is, the very basic magnetic field and force calculations are in reality merely a backdoor first level relativity correction?
 
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  • #2
Maxwell equations are not an approximation. They are Lorentz invariant. The theory of relativity provides a correct interpretation.
 
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  • #3
No. It means that any calculation done with any of the maths of any part of electromagnetic theory is inherently a relativistic calculation. You may not need to use relativity explicitly in the calculation, but if you want to change frames then you will have to use the Lorentz transforms explicitly.

So the point you are asking about is that a charge has an electrostatic field in its rest frame and both an electric field and a magnetic field in any other. That cannot be understood without relativity. But you can use force equations in one frame without understanding relativity.
 
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  • #4
StoneHengeAlive said:
The current in a wire is said to produce a magnetic field and a force on a charge.
More precisely: The magnetic field causes a force on a moving charge. Therefore, in the rest frame of this charge, the force on in cannot be caused by a magnetic field. In this frame, it must be an electric field.

 
  • #5
Honestly, I wold not go down this path. A better path, IMHO is to say there is an electromagnetic field, and that behaves as it must under relativity. By convention and history, we call the velocity-dependent part of this field "electric" and the velocity-dependent part "magnetic".

Deciding that the electric fiedl is "the real one" and the magnetic field is just something you get by a Lorentz transformation does not really work. It just makes a mess.
 
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  • #6
StoneHengeAlive said:
The current in a wire is said to produce a magnetic field and a force on a charge. It is said that the correct way to interpret the effect is because of relativity. So does that mean that that the most basic non-relativistic equations that offer basic calculations of the forces and fields are implicitly a first order approximate accounting of relativity effect? That is, the very basic magnetic field and force calculations are in reality merely a backdoor first level relativity correction?

As another poster has already mentioned, Maxwell's equations are not an approximation. The difficulty with Maxwell's equations is that they are not compatible with non-relativistic mechanics. This can be seen in sevral ways, one of the simplest is the fact that Maxwell's equation predict that the speed of light is a constant. This is possible for all observers only in special relativity, it's not compatible with classical Gallilean mechanics and the Gallielean transform, which implies linear velocity addition.

There's another approach to discuss at an elementary level how magnetism arises from special relativity due to Edward Purcell.

Purcell talks about the origins of magnetism as a consequence of length contraction. It might be of some help, but it also might be confusing. If you're not familiar with the ideas, there's Purcell's textbook on E&M, of which some earlier versions are public domain, and has an informal web summary at https://physics.weber.edu/schroeder/mrr/MRRtalk.html.

Basically, the issue that makes Purcell's treatment tricky is the relativity of simultaneity, and Purcell doesn't really discuss this. The way I would approach his treatment, which is a bit different than the original, is that if you have a current loop, which we approximate as being long. And we assume that we have a moving frame, where the direction of motion of the frame is in the "long" direction of the loop.

The relativity of simultaneity in this moving frame makes one side of the loop have a net plus charge i, while the other side of the loop have a net minus charge. Conservation of charge makes the net charge zero of course, in all frames, but the relativity of simultaneity makes the wires that have no net charge in the lab frame split into a segment which has a positive charge, and a segment which has a negative charge.

The distribution of charge in the moving frame thus gives rise to electric fields in the moving frame, which exert an electric force on a charge co-moving with the frame. In the stationary lab frame, there is no electric force, of course - but there must still be a force, which we interpret as the magnetic force.

A more advanced approach may take more work, but is probably less confusing in the end. This approach stresses Lorentz invariance as the foundation of special relativity, and notes that the coulomb force by itself is not Lorentz invariant. Hence, there must be something else. This "something else" is the magnetic force.
 
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  • #7
Vanadium 50 said:
By convention and history, we call the velocity-dependent part of this field "electric"
I think you mean the velocity-independent part is "electric".
 
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  • #8
The is also the intermediate version which I will call magnetostatics. While useful, it is the cause of much confusion and is neither really exact nor incorrect within its domain.
 
  • #9
PeterDonis said:
I think you mean the velocity-independent part is "electric".
You are correct.
 
  • #10
Vanadium 50 said:
say there is an electromagnetic field, and that behaves as it must under relativity.
I agree.

Vanadium 50 said:
Deciding that the electric fiedl is "the real one" and the magnetic field is just something you get by a Lorentz transformation does not really work. It just makes a mess.
I don't know, to what you refer. Veritasium doesn't say this in his video. He doesn't use the word "real". He shows correctly at an scenario, to which also the OP refers, that the magnetic force is a relativistic effect.

If you calculate the limit for ##c \to \infty## of the LT, then you get the GT.

GT of x to x':
https://www.wolframalpha.com/input?i=Limit[(x+-+v+t)/Sqrt[1+-+v^2/c^2],+c+->+Infinity]

GT of t to t':
https://www.wolframalpha.com/input?i=Limit[(t+-+(v+x)/c^2)/Sqrt[1+-+v^2/c^2],+c+->+Infinity]

If you calculate the same limit ##c \to \infty## of the Lorentz force, then you get rid of it's magnetic part.
In cgs units:

##\lim_{c \rightarrow \infty} (q(\mathbf {E} +\frac{\mathbf {v}}{c} \times \mathbf {B})) = q\mathbf {E}##.

The magnetic force exists only, if the invariant/maximum speed ##c## is finite. So, it is a relativistic effect.
 
  • #12
Sagittarius A-Star said:
I don't know, to what you refer.
The parade of undergraduates outside my door who were first exposed to this in Purcell. :smile:
 
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  • #14
I am not talking about mistakes. It's more conceptual.

One cannot have an electric field in a relativistic theory without also having a magnetic field. That's an important idea, and I give Purcell a lot of credit for introducing it at the 1st year level. (Not that he needs my approval)

The mistake that students walk away with is that they think there is always a frame where an arbitrary electromagnetic field is purely electric. Purcell does not say this, but the students look at the pictures and that's what they come away with.

The more sophisticated students say, I start with Ex, Et and Ez and include vx, yv, vz and get Bx, By, Bz. Therefore for (Ex, Ev, Ez, Bx, By, Bz) I can find a (vx, vy, vz) so that I get (Ex, Ey, Ez, 0, 0, 0). And that's not how it works.
 
  • #15
Vanadium 50 said:
The mistake that students walk away with is that they think there is always a frame where an arbitrary electromagnetic field is purely electric.
Misner, Thorne, & Wheeler, of all places, is where I first learned the canonical form for a generic electromagnetic field, in terms of two wedge products (Exercise 4.1). But even a simple radiation field should be sufficient to refute the idea that any EM field must be purely electric in some frame.
 
  • #16
PeterDonis said:
But even a simple radiation field should be sufficient to refute the idea that any EM field must be purely electric in some frame.
##E^2-B^2## being an invariant is enough, isn't it?
 
  • #17
Ibix said:
##E^2-B^2## being an invariant is enough, isn't it?
Not by itself, no, since that does not rule out the possibility of finding a frame in which ##B = 0## (and in this frame ##E##'s magnitude would be the square root of the invariant).
 
  • #18
vanhees71 said:
And Purcell's treatment of the DC current makes the same mistake as almost all textbooks about this subject, see
https://itp.uni-frankfurt.de/~hees/pf-faq/relativistic-dc.pdf
and the references cited therein.
In the following file by Kirk T. McDonald this is calculated in the rest frame of the wire simply with the Lorentz force law. It is demanded, that the radial component of the Lorentz force on the moving electrons must vanish (equations 2 to 5):
http://kirkmcd.princeton.edu/examples/wire.pdf
 
  • #19
PeterDonis said:
But even a simple radiation field should be sufficient to refute the idea that any EM field must be purely electric in some frame.
Remember, these are freshmen (or sometimes sophomores). They can't just whip out an example from their back pockets. Same with looking at Lorentz invariants and reasoning from that.

Students are often confused, and this is all new to them so they don't immediately see the example that will unconfuse them.
 
  • #20
Sagittarius A-Star said:
In the following file by Kirk T. McDonald this is calculated in the rest frame of the wire simply with the Lorentz force law. It is demanded, that the radial component of the Lorentz force on the moving electrons must vanish (equations 2 to 5):
http://kirkmcd.princeton.edu/examples/wire.pdf
I've just glanced at it. It looks as if he does it right. The point is to use the correct relativistic version of Ohm's Law, which of course follows from the full Lorentz force on the electrons (which is a relativistic force law too, of course).
 
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  • #21
Vanadium 50 said:
Remember, these are freshmen (or sometimes sophomores). They can't just whip out an example from their back pockets. Same with looking at Lorentz invariants and reasoning from that.

Students are often confused, and this is all new to them so they don't immediately see the example that will unconfuse them.
Yes, fair point.
 
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  • #22
Well, current didactics research tells that you remember the very things about a subject that you learn first. That's true also from subjective experience: It took me quite a while to unlearn the Bohr-Sommerfeld model of the H-atom with its trajectories subject to quantum conditions, because in highschool we were quite well busy with it, and that although our very good teacher at the time always told us that it's not the true thing, and she went as far to teach the basics about the (time-independent) Schrödinger equation (of course not the full solutions of the hydrogen problem, but at least it was covered qualitatively).

I can imagine that it's very hard to learn relativity from Purcell, because this book buries the physics under a carpet of didactical handwaving, instead of teaching the adequate math in parallel (4D tensor calculus), and the treatment of the straight DC-current carrying wire is simply wrong (which is, however, also the case in the Feynman lectures). The above quoted paper by McDonald has it right: the wire is neutral in the rest frame of the electrons and not in the rest frame of the wire (positive ion lattice making up the solid wire).
 
  • #23
vanhees71 said:
teaching the adequate math in parallel (4D tensor calculus),
And what do you drop so they can learn this?

The majority of people taking frosh E&M are not physics majors. Is it better to teach them a little about the relativistic basis of E&M or nothing? That's the practical choice.

If a chemical engineer knows "Electric fields plus relativity implies magnetic fields" I would count that as a win.
 
  • #24
Vanadium 50 said:
And what do you drop so they can learn this?

The majority of people taking frosh E&M are not physics majors. Is it better to teach them a little about the relativistic basis of E&M or nothing? That's the practical choice.

If a chemical engineer knows "Electric fields plus relativity implies magnetic fields" I would count that as a win.
I talk about physics majors and the theory course lecture. Usually at our university the intro-lecture in electrodynamics is taught in the traditional way, and the relativistic formulation only at the end, but then using the 4D tensor calculus, which has already been introduced in the mechanics lecture, which also covers relativity. I think that's better than this pretty confusing vol. 2 of the Berkeley physics course.
 
  • #25
vanhees71 said:
I've just glanced at it. It looks as if he does it right. The point is to use the correct relativistic version of Ohm's Law, which of course follows from the full Lorentz force on the electrons (which is a relativistic force law too, of course).

In the book "Electricity and Magnetism, 3rd edition" of Purcell and Morin, the chapter 5.9 about the scenario with the infinite wire contains a disclaimer for the sentence "That is, any given length of "wire" contains at a given instant the same number of electrons and protons 9":
9 It doesn't have to, but that equality can always be established, if we choose, by adjusting the number of electrons per unit length. In our idealized setup, we assume this has been done.

Maybe a solution for a real setup would be, if Purcell would use a neutral, closed wire near a permanent magnet, cools it down and induces current in it by removing the permanent magnet, as he describes it in Figure I.1 (page 819) of his Appendix I "Superconductivity".
 
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  • #26
It's simply the Hall effect, which shows that in the rest frame of the wire the wire must be negatively charged.

Of course, a superconductor can not be described with Ohm's Law. You need the London equations as effective constitutive equations. Microscopically they follow from the BCS theory leading to "Higgsing" of the electromagnetic gauge symmetry. A pretty simple treatment can be found in vol. 3 of the Feynman Lectures.
 
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