# B Electromagnets are a relativistic phenom?

1. Oct 7, 2015

### DaveC426913

This is a new one one me:
http://www.iflscience.com/physics/4-examples-relativity-everyday-life

2. Oct 7, 2015

### SlowThinker

This is a popular "explanation" of magnetism, but it's hard to apply to, e.g., individual electrons affecting each other, or electrons moving towards/away from a wire.
It may be correct, but don't take it too seriously.

3. Oct 7, 2015

### DaveC426913

I don't see how it can be.

It suggests that "even if the electrons are flowing" there is a negligible amount of length contraction, yet I can wave the electromagnet fast enough (like, 2mph) to create length contraction?

4. Oct 7, 2015

### SlowThinker

The important thing is that if you just wave a piece of copper, both electrons and nuclei are contracted in the same way, so you have no effect.
Only if a current is flowing through a wire, the speeds, and thus contractions, of electrons and nuclei are different, and you get a magnetic field.

5. Oct 7, 2015

### DaveC426913

Electrons flowing in a wire don't generate length contraction, but moving the wire by hand does. That seems to be what the paragraph is saying.

6. Oct 7, 2015

### Staff: Mentor

Edward Purcell used this model in his classic E&M textbook (I believe he came up with it in the first place), and it's mentioned in the FAQ linked from our sticky thread on experimental verification of SR - perhaps as close to a practical demonstration of length contraction that we can find, as macroscopic objects moving at relativistic speeds are (thankfully) scarce. All that's necessary to make it work is that the charge density be frame-dependent.

Google for "Purcell magnetism relativity" and you'll find many hits. This one is good: http://physics.weber.edu/schroeder/mrr/MRRtalk.html

7. Oct 7, 2015

### DaveC426913

All right. Fair enough.

I think the problem I'm having is with the murky explanation in the article. It seems to suggest that an electrified wire will not have an attractive force, but waving the wire with my hand will.

8. Oct 7, 2015

### Staff: Mentor

That's a problem with the article, not the concept.

9. Oct 7, 2015

### DaveC426913

Yes. That's what I said.

10. Oct 7, 2015

### bcrowell

Staff Emeritus
There is length contraction in the lab frame, and in this frame the wire is electrically neutral with the length contraction taken into account. In the moving frame, there is a different amount of length contraction for the electrons, and there is also length contraction for the protons. The net effect of these two changes in length contraction is to produce an attraction or repulsion.

The original quote is a somewhat garbled and incorrect explanation, but the general idea is correct. Magnetism is a purely relativistic phenomenon. For a (hopefully) correct version of the explanation, see http://www.lightandmatter.com/lm/ , section 23.2.

11. Oct 9, 2015

### vanhees71

It's put in a bit an unclear way. It's all about Lorentz-transformation properties of the electromagnetic field and the Lorentz force.

The situation discussed is the following: In the inertial reference frame A you have a wire with a DC current density $\vec{j}$ producing a magnetic field. Neglecting the very small relativistic effects (or the Hall effect for the conduction electrons) in this reference frame the charge density is $\rho=0$. Then you have no electric field $\vec{E}=0$ but a magnetic field $\vec{B}=B \vec{e}_{\varphi}$, where I use cylinder coordinates $(\rho,\varphi,z)$ with the wire along the $z$ axis.

A charged particle moving with a velocity $\vec{v}=v \vec{e}_z$ along the wire feels a purely magnetic Lorentz force,
$$m c \mathrm{d}_{\tau} \vec{u}=q \vec{u} \times \vec{B}.$$
Here $\tau$ is the proper time of the charged particle and $\vec{u}=\gamma \vec{v}$ are the spatial components of the four-velocity
$$u^{\mu} = \frac{1}{c} \mathrm{d}_{\tau} x^{\mu}.$$
The forth component of the equation of motion follows from $u_{\mu} u^{\mu}=1=\text{const}$, which implies
$$u_{\mu} \mathrm{d}_{\tau} u^{\mu}=0 \; \Rightarrow \; u_0 \mathrm{d}_{\tau} u^0=\vec{u} \cdot \mathrm{d}_{\tau} \vec{u}=0$$
in our case, i.e., the 4D invariant Minkowski force on the particle at the initial time is
$$K^{\mu}=\begin{pmatrix} 0 \\ q \vec{u} \times \vec{B} \end{pmatrix}=\begin{pmatrix}0 \\ -\frac{q B v \gamma}{c} \vec{e}_{\rho} \end{pmatrix}.$$
Here we've used the usual notation $\gamma=1/\sqrt{1-v^2/c^2}=u^0$.

Now consider the situation in another inertial frame B, where the particle is initially at rest. This is described by a Lorentz boost in $z$ direction with boost velocity $v$.

The current density together with the charge density makes up a four-vector field. In frame A it has the components
$$(j^{\mu})=\begin{pmatrix} 0 \\ 0 \\ 0 \\ j \end{pmatrix},$$
and in frame B, according to the Lorentz transformation the components
$$(\tilde{j}^{\mu})=\begin{pmatrix} -\beta \gamma j \\ 0 \\ 0\\ \gamma j \end{pmatrix}.$$
Since in this frame the charged particle is at rest at the moment it moves parallel to the wire in A it only fields the electric force. Since the wire is charged now, in this frame is also an electric field, which is also given by the Lorentz transformation (the electromagnetic field is represented as an antisymmetric four-tensor field in relatitivity which determines its transformation properties). In our case we have
$$\vec{\tilde{E}}=\frac{\gamma}{c} \vec{v} \times \vec{B}=\frac{\gamma v B}{c} \vec{e}_z \times \vec{e}_{\varphi}=-\frac{\gamma v B}{c} \vec{e}_{\varrho}.$$
The Minkowski force on the particle at that moment as measured in frame B thus is
$$\vec{\tilde{K}}=q \vec{\tilde{E}} = -\frac{q\gamma v B}{c} \vec{e}_{\varrho}.$$
The temporal component follows again from
$$\tilde{u}_0 \mathrm{d}_{\tilde{t}} \tilde{u}^0=\mathrm{d}_{\tilde{t}} \tilde{u}^0 = \vec{u} \cdot \mathrm{d}_{\tilde{t}} \vec{u}=0.$$
This you get also directly from the Lorentz boost of $K^{\mu}$.

Note that for the usual non-covariant Force $\vec{F}=\vec{K}/u^0$ there's a $\gamma$ factor
$$\tilde{\vec{F}}=\vec{\tilde{K}}{\tilde{u}^0}=\vec{\tilde{K}}=\gamma \vec{F}.$$

The relation of all this to Lorentz contraction is a bit indirect. You can derive everything from the Lorentz transformation properties of the charge-current-density four vector. In frame A the positively charged ions making up the wire are at rest and the conduction electrons are moving. In our approximation, neglecting the Hall effect, in this frame the total charge density (positive ions + conduction electrons) is set to 0. Now in frame B the positively charged ions are moving and thus their charge density gets a $\gamma$ factor compared to frame A because of the length contraction of the volume elements used to define the charge density. The conduction electrons are moving in frame A and thus their charge density does not only get a Lorentz factor in frame B but there's also a component from the spatial components of the corresponding current density, and thus the charge density of the conduction electrons does not compensate precisely the charge density of the positive ions when measured in frame B. The net result is the above given negative charge density of the wire as a whole, which implies the existence of the radial electric field, which is the cause of the purely electric Lorentz force in frame B.

A very nice discussion of the relativistic formulation of electromagnetism (which simplifies a lot compared to the usual non-relativistic heuristics used in conventional textbooks) can be found in

M. Schwartz, Principles of Electrodynamics, Dover (1987)

or

L. D. Landau, Course of Theoretical Physics, vol. 2, Butterworth-Heinemann (1996)

12. Oct 9, 2015

### A.T.

13. Oct 9, 2015

### Demystifier

14. Oct 9, 2015

### vanhees71

Yep, that's something I never understood: Why is the oldfasioned historical approach to electromagnetism so much favored compared to the relativistic approach? Usually you go through a long pretty dull set of special cases, starting from electrostatics, magnetostatics (sometimes then even the quasi-stationary approximation), and finally the full Maxwell equations and waves. This does not bring the entire beauty of electromagnetism into sight until the very last part of the lecture. Usually there's not even time to discuss the fully relativistic formulation (including relativity for the mechanical part, i.e., the charges and currents making up the sources of the field, let alone the very exciting topic of radiation reaction). The relativistic approach as in Schwartz's book is much more lucid. However, the standard textbook (at least in Germany) for the theory lecture on E&M is usually still Jackson, which is of course very good, but follows this traditional approach. Then there are newer books like Griffiths, who emphasizes the apparent paradoxes like "hidden momentum", instead of doing the full relativistic treatment right away. Then it's very clear that there is no hidden momentum but just momentum, and the apparent "problems" or "paradoxes" have been already well understood by von Laue in 1911 (if not even earlier in 1904-1906 by Poincare)!

15. Oct 9, 2015

### Demystifier

That's probably because most students who learn classical electrodynamics will eventually be something like condensed-matter experimentalists rather than something like particle-physics theorists. For them (the former ones) it probably makes a lot of sense to think of electricity and magnetism as two different phenomena.

16. Oct 9, 2015

### Demystifier

17. Oct 9, 2015

### harrylin

18. Oct 9, 2015

### vanhees71

That's a great book (also bit plagued with typos), but I'd not recommend it as a primary source for the introductory E&M-theory lecture.

You may be right that the reason for doing the "non-relativistic electrodynamics" (i.e., treating the matter part non-relativistically) is that for many physicists that's what they primarily need in applications. However, I think the relativistic consistent picture should be taught to all physicists, not only those specializing in some research dealing with relativistic issues.

19. Oct 9, 2015

### vanhees71

It's clear that you cannot always transform away the magnetic field. This is clear without any specific example from the fact that the two scalar invariants you can construct from the electromagnetic field are $\vec{E} \cdot \vec{B}$ and $\vec{E}^2-\vec{B}^2$, which are the contraction of the Faraday tensor with its Hodge dual and with itself, respectively.

If you have $\vec{E}^2-\vec{B}^2<0$ you never can transform to a system where there is no magnetic field, because there you'd have $\vec{E}'^2-\vec{B}'^2=\vec{E}'^2 \geq 0$.

That's why you need to start from the full electromagnetic field when treating at as a relativistic field theory. The arguments, leading from electrostatics via the demans of Lorentz invariance to the necessity of the additional magnetic field components is not a "derivation" in the strict sense but a heuristic argument. As such it is pretty convincing (see, e.g., the above cited book by Schwartz).

20. Oct 9, 2015

### bcrowell

Staff Emeritus
In the Purcell pedagogy, we're talking about transforming from a frame in which a particular *force* is purely magnetic to another frame in which the *force* is purely electric.