# I At what point does classical EM become relativistic EM?

1. Jul 19, 2016

### Electric to be

By classical E&M I mean the general basics. I understand that even classical EM has clear relativistic undertones, but is anything taught classically actually wrong under the context of relativistic EM? For example, does Faraday's law continue to hold etc.

I know this is kind of a broad question, but I'm just concerned if whether things I learned in my intro to EM course could actually be considered wrong under this broader scope.

2. Jul 19, 2016

### Staff: Mentor

It depends on what your intro to EM course taught you were the "general basics" of EM. If you learned Maxwell's Equations, those are exactly correct even in a relativistic context. But some "laws" of EM that are taught are approximations that are only valid in certain situations (for example, Coulomb's Law). Those approximations might break down in a relativistic context (as Coulomb's Law, for example, does).

3. Jul 19, 2016

### Electric to be

4. Jul 19, 2016

### Staff: Mentor

It's one of Maxwell's Equations (at least it is when put in differential form), so it is still valid in relativity.

5. Jul 19, 2016

### Orodruin

Staff Emeritus
On a historical note, the fact that Maxwell's equations were not Galilei invariant was one of the big issues that led to the development of SR in the first place. It was known before Einstein that they were invariant under Lorentz transformations (hence not "Einstein transformations").

6. Jul 19, 2016

### Electric to be

Well the changing magnetic flux is Maxwell's equation (the transformer part), but what about the part due to motion? This is due to a magnetic force acting on the charges, but doesn't the entire concept of force change under SR anyways?

7. Jul 19, 2016

### haushofer

No; force is still a time derivative of momentum. The conceptual change is that one turns to 4-vectors.

By the way, you can obtain the non-rel.limit of the maxwell-eqns. by sending c to infinity. See e.g. papers of Levy-Leblond. The resulting equations are covariant with respect to the Galilei group.

8. Jul 19, 2016

### Electric to be

Sorry but what does it mean for the Maxwell's equations to be invariant? I understand invariance in terms of event interval invariance (intervals among different frames are the same) but how does it apply to the equations?

9. Jul 19, 2016

### Staff: Mentor

When you switch from one inertial reference frame to another, the electric and magnetic fields transform under a Lorentz transformation, analogously to the way position and time transform. The fields are different in the two frames, but nevertheless in each frame they obey Maxwell's equations.

https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

10. Jul 20, 2016

### haushofer

"Covariant". It means the EOM (eqns. of motion) have the same form for the corresponding group of observers. This means that the EOM in the new frame of reference are a linear combination of the EOM in the old frame. We call this a tensor equation.

E.g., Newton's equations are usually not covariant with respect to accelerations (time-dependent rotations and/or translations): you obtain extra terms in your EOM which we call 'inertial forces'. So Newton's F=ma is not a tensor equations under accelerations. You can check this by yourself: e.g.,take

$$F = m \ddot{x}^i(t) \ \ \ (1)$$

and go to a rotating frame of reference,

$$x^i \rightarrow x^{'i} = R^i{}_k (t) x^{k}$$

where R is an element of SO(3) with time-dependent angle. Now write (1) in terms of the primed coordinates. You should obtain two extra terms, which we call centrifugal and Coriolis terms. You can also check which transformations do leave (1) covariant.

11. Jul 22, 2016

### vanhees71

I guess you mean electrodynamics where the motion of (charged) matter is treated non-relativistically. Classical electrodynamics is a relativistic field theory since Maxwell has written down his equations. This was only noticed around 1900 (with the final conclusion drawn by Einstein in 1905). Many apparent paradoxes are solved by this breakthrough in the history of science.

What I don't understand is the relcutance of modern textbook writers to present classical electrodynamics as a relativistic theory from the very beginning and instead copying the old-fashioned didactics from 19th-century textbook writers. My favorite exception is Landau, Lifshitz vol. II.

There are in fact two non-relativistic limits of the full Maxwell theory (the electric and magnetic limits):

https://en.wikipedia.org/wiki/Galilean_electromagnetism

12. Jul 22, 2016

### Electric to be

So could the electric and B fields of a moving charge, with whatever velocity, acceleration, and past behavior, be derived solely as a solution to Maxwell's equations without any form of explicit relativity?

13. Jul 22, 2016

### vanhees71

That doesn't make sense. The exact solutions of Maxwell's equations are relativistic. In your case you just write down the retarded potentials/fields; the former are known as "Lienard-Wiechert potentials", the latter are given by equations named "Jefimenko equations" (although for sure Jefimenko is not the first who wrote down them):

https://en.wikipedia.org/wiki/Jefimenko's_equations

14. Jul 22, 2016

### Orodruin

Staff Emeritus
Maxwell's equations are an explicit form of relativity.

15. Jul 22, 2016

### robphy

Let me just chime in here to suggest caution
that term "Maxwell's Equations" or "E&M" needs to be clarified to avoid confusion.

I'm not splitting hairs...
There are subtleties that I am actively trying to unravel and resolve for myself.
(It is a research problem [not just for me]... and it is in the literature...
and it's really about the structure of physical theories... and how structures may be related.
This could be useful in guiding us to generalizations and extensions [e.g. see Jammer & Stachel below] .)

The following isn't exhaustive... just illustrative of what clarifications might be needed.
There are certainly the four differential equations...
but in what form? Vector-calculus? Differential-forms? tensorial? Maxwell's original? other?
In terms of E and B (i.e. one field tensor F and its metric-related Hodge dual *F)?
or E,B,D,H (or two field tensors F, G with specific index-positions)?
are the constitutive relations specified or included among the "Equations"? (some argue that, when unraveled, the metric actually lives here)

This issue arises because there is a notion of a metric-independent (sometimes called "pre-metric") electrodynamics:
van Dantzig (1934) - Electromagnetism. independent of metrical geometry. ( http://www.dwc.knaw.nl/DL/publications/PU00016602.pdf )
Hehl (2008) - Maxwell's equations in Minkowski's world: their premetric generalization and the electromagnetic energy-momentum tensor ( http://arxiv.org/abs/0807.4249 )
which is
somehow related to @haushofer's mention of Levy-Leblond and @vanhees71's link to the wikipedia page on https://en.wikipedia.org/wiki/Galilean_electromagnetism (*)
which describes
( see also Jammer & Stachel (1980) - If Maxwell had worked between Ampère and Faraday: An historical fable with a pedagogical moral (Am. J. Phys. 48, 5 (1980); http://dx.doi.org/10.1119/1.12239 )

Fred Hehl is quite active on pre-metric electrodynamics.

So, depending on what is specifically meant by "Maxwell's Equations"... some statements are true and some are not.

Here are some links to old posts of mine [with more links to the literature]

*note that one should not confuse Levy-Leblond's Galilean Electromagnetism
with a dubious journal with a similar name Galilean Electrodynamics.

Thus, "pre-metric electromagnetism" or "pre-metric electrodynamics" (meaning it doesn't make use of a metric [as much as possible]) are better terms.

16. Jul 22, 2016

### vanhees71

The constitutive equations are not part of the fundamental equations of physics, but derivable approximately in (quantum) many-body theory. Fully consistent is only a fully relativistic theory, although for many purposes the non-relativistic limit for the matter part is sufficient.

17. Jul 22, 2016

### robphy

In the context of Hehl's work on pre-metric electrodynamics ( http://arxiv.org/abs/physics/0610221 ) [which I haven't worked out for myself],
part of your statement could be rephrased to say
"The constitutive equations are not part of the fundamental equations of electrodynamics"
... but (if I understand Hehl correctly) these equations are where the spacetime metric would first make contact with pre-metric electrodynamics. [That is to say, it may be possible that, given a constitutive equation, one may be able to extract (recover or deduce) the metric (or aspects of the metric) associated with it.] Afterwards, presumably, use of additional structures derived from the metric (like Hodge) could be used to reformulate all of the differential-form quantities into ordinary vector quantities. If this is actually how it turns out, it would be nice to see a clean sequence from first principles (e.g. an action principle).

[EDIT: It might be worth making a connection to the original question in the thread...
(of course, assuming everything is consistent)
in its most abstract form, pre-metric electrodynamics is independent of the metric... that is to say, the distinction of where it becomes relativistic or non-relativistic doesn't lie in the field equations.. but in the constitutive relations.... that is, how E&B are related to D&H. How this is seen practically, I don't know yet.]

18. Jul 23, 2016

### vanhees71

I don't know much about this approach to electrodynamics without a metric (it's of course always a pseudo-metric of signature (1,3) we are talking about). I guess, this makes indeed only sense for free classical electromagnetic fields, which obeys a larger symmetry group than the Poincare group, because there's no scale in it, i.e., you have symmetry also under space-time dilations. The corresponding Noether theorem tells you that the energy-momentum tensor's covariant trace vanishes.

This has not too much to do with the original question, because that was about "non-relativistic electrodynamics". As I stressed before, this can only mean an approximation, where the matter, consisting of charged particles, are treated non-relativistically, which allows to use certain approximations to get two types of Galilean electrodynamics (electric of magnetic, depending on the physical situation). See the famous paper by Leblond and LeBellac cited in the Wikipedia article quoted above.

19. Aug 1, 2016

### dragoo

20. Aug 1, 2016

### Staff: Mentor

Its more than that.

Coulombs law plus relativity gives Maxwell's equations:

Historically is was done ass about - Maxwell's equations before relativity. We now know that relativity is a theory about space-time symmetries leading to a certain geometry (Minkowski geometry) and it is that geometry that is responsible for EM.

That symmetries underlies a lot of physics is one of the great discoveries of 20th century physics.

Thanks
Bill

21. Aug 2, 2016

### vanhees71

It's important to get the convention straight. Nowadays, "classical physics" means a treatment (approximation) that doesn't include quantum theory. No matter whether a problem is relativistic or non-relativistic that you call "classical physics". Classical electrodynamics is Maxwell's equations, and these are the paradigmatic example for a relativistic classical field theory. There are two kinds of non-relativistic limits of the full relativistic Maxwell theory, depending on the situation, where these approximations are applicable (see the discussion above).

Sometimes, apparently "non-relativistic" situations lead to wrong results and to much confusion when not treated with the full relativistic theory. One example is the homopolar generator (like "Faraday's Disk" or the like):

http://th.physik.uni-frankfurt.de/~hees/pf-faq/homopolar.pdf