# B Against length contraction explanation of magnetism

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1. Nov 8, 2017

### jartsa

When a test charge stands next to a wire, there are coulomb forces between the test charge and the protons of the wire.

1: When the test charge starts to move, the Coulomb forces stays the same, because distances from test charge to protons stay the same. There is no magnetism in this case.

2: When the test charge stays still and the wire starts to move, the Coulomb forces increase, because distances from test charge to protons decrease. But we do not call this magnetism. We call it decrease of distance and increase of Coulomb force.

If we ask the test charge, it says there is no difference between the case 1 and the case 2. In both cases the wire looks the same to the test charge.

3: When two test charges placed side by side start to move there is no change of distance, but there is a change of total force between the charges. This case we call magnetism. And we explain the 'magnetism' by the change of the electric fields. Right?

4: And the same change of fields can explain the total force between two wires with opposite currents.

Last edited: Nov 8, 2017
2. Nov 8, 2017

### Staff: Mentor

Yes, but in this case we also have a nonzero magnetic field due to the nonzero current from the moving wire, so the increased Coulomb force is counterbalanced by a nonzero magnetic force in the opposite direction. So the net force on the charge is the same.

Yes, but if you only take the Coulomb force into account in case 2, you can't explain why this is true. You need the magnetic force as well.

Yes, because of the current due to the moving charges. Not because of any change in the charges themselves.

Wrong. We explain it by the presence of a nonzero current. Go look at Maxwell's Equations.

3. Nov 8, 2017

### Staff: Mentor

Are you talking about a wire with a net charge but no current in the lab frame?

What direction? And by “start to move” are you asking about a sudden acceleration to a constant speed or a gradual acceleration or do you just mean that it is moving inertially? Please describe your scenarios a little better. I am not following you here.

Last edited: Nov 8, 2017
4. Nov 8, 2017

### Staff: Mentor

Veritasium explored this very topic in an excellent video on YouTube.

5. Nov 9, 2017

### jartsa

The test charge does not care about magnetic fields, because it's not in any way magnetic, because it's not moving. In the lab frame.

As I said, the test charge does not feel the magnetic field of the moving protons. In the lab frame.

(There was the other case where the test charge was moving in the lab frame and the protons where still in the lab frame, same reasoning applies in that case; no magnetic force felt by test charge in the lab frame, because the test charge is the only charge that is moving in the lab frame)

That is one explanation.

Last edited: Nov 9, 2017
6. Nov 9, 2017

### Staff: Mentor

I am still confused about your scenarios. In addition to what I wrote before, when you say “increase of force” are you talking about an increase over time or a comparison to the previous scenario or a comparison to the reference scenario.

I think that @PeterDonis is answering a different question than you think you are asking.

7. Nov 9, 2017

### pervect

Staff Emeritus
I didn't read any further, because I'm pretty sure this is wrong. The transverse field increases. See for instance wiki, https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity

The field component of interest is $E_{\bot}$, you can see from the wiki (or a different reference) that if there is no magnetic field, for a moving charge, $E'_{\bot} = \gamma E_{\bot}$.

8. Nov 9, 2017

### jartsa

Here's an edited version of post #1:

When a test charge stands next to a wire, there are coulomb forces between the test charge and the protons of the wire.

1: When the test charge is slowly accelerated, the Coulomb forces stays the same, because distances from test charge to protons stay the same, in the lab frame. I mean, there are always some constant number of protons inside some radius around the test charge, in lab frame. There is no magnetism in this case, in lab frame.

2: When the test charge stays still and the wire is slowly accelerated, the Coulomb forces are increasing during the acceleration, in lab frame, because distances from test charge to protons are decreasing during the acceleration, in lab frame. We in the lab frame are observing the length of the wire getting shorter. But we in the lab frame do not call this magnetism. We call it decrease of distance and increase of Coulomb force.

If we ask the test charge, it says there is no difference between the case 1 and the case 2. In both cases the wire looks the same to the test charge.

3: When two test charges placed side by side are slowly accelerated, there is no change of distance between the particles, in any frame, but total force between the charges is decreasing during the acceleration, in the lab frame. This case we in the lab frame call magnetism. And we explain the 'magnetism' by the change of the electric fields. Right?

This page is talking about that change of electric fields, in the part entitled Transformation of electrostatic fields:

4: And the same change of fields can explain the total force between two wires with opposite currents.

9. Nov 9, 2017

### Staff: Mentor

If the wire is uncharged and current-free then the force is zero. If there are electrons present then you cannot just ignore them as they also produce a force and the test charge can only measure the force from the net charge.

10. Nov 9, 2017

### jartsa

The transverse field of a test charge, whose speed is increasing in the lab frame, is increasing in the lab frame?

Yes, but the force exerted on the test charge in the lab frame depends on the electric field of the protons, which stay still in the lab frame in my case number 1.

11. Nov 9, 2017

Staff Emeritus
Jartsa, I don't understand where you are coming from. Are you arguing:

A. Relativity is wrong?
B. Relativity is right, but this calculation is wrong?
C. Relativity and the calculation are right, but it's pedagogical value is low?
D. Something else?

12. Nov 9, 2017

### jartsa

D. "Hey I noticed some problems with the length contraction explanations"

I'm not sure what calculation we are talking about now.

But does this not hold, if a test charge with charge q is in motion:

F=qE

The E there is the field measured in the lab frame, the velocity of the field is zero in the lab frame. And F means force in the lab frame.

Last edited: Nov 9, 2017
13. Nov 9, 2017

### Ibix

I don't think anybody is clear on your scenarios, which is a problem. Do I take it that the two scenarios are meant to be initially identical, with a charge at rest near an un-charged current-carrying wire? Then either (1) the charge starts to move, or (2) the wire starts to move? In which direction?
If q is in motion then this is only true if $\vec B=0$.

14. Nov 9, 2017

### Staff: Mentor

But your description says there should be: in case 1 the Coulomb force stays the same, in case 2 it increases. So whatever alternative model you think you are describing, it gives the wrong answer, so it can't be right.

15. Nov 9, 2017

### Staff: Mentor

I think that your examples are not correct. I don’t think that the problem is with the explanation.

Besides the incomplete description, one really big issue is that accelerating the charge vs accelerating the wire is physically different. You cannot change from one scenario to the other simply by a Lorentz boost. They are not symmetric. Furthermore, the charge’s perspective is inertial in one case and non inertial in another meaning that the standard Maxwell’s equations will not work in both.

Assuming that the wire is charged with no current and assuming that the test charge is accelerated parallel to the wire and assuming that the wire is infinite and straight (all of which would have been nice for you to clarify) then I agree.

I assume that you mean that the wire accelerates parallel to its length. Having charges accelerating at infinity may bring in some mathematical problems. I would call it increase in charge density.

The laws of physics (Maxwell’s equations) are different.

You can explain the magnetism any number of ways. The one most in keeping with the Purcell explanation would be The Lorentz transform. Those can be used to figure out how forces transform.

Last edited: Nov 9, 2017
16. Nov 9, 2017

### jartsa

So we have explanations of magnetism by length contraction. Usually there are protons that are just standing still, in the lab frame. So there is no length contraction effects there, in the lab frame. Because there is no motion.

And then there are electrons that usually are moving, in the lab frame, but the moving does not affect the density of electrons, in the lab frame. Why the density is not affected, that I will not discuss now.

Then usually we go the frame of moving electrons, where we can observe some length contraction effects. Quite usual is that proton density is increased, if "proton density is increased" is unclear, see this video at time 2:10

As I said usually the protons are standing still in the lab frame, so the protons do not have any magnetic effect on anything, in the lab frame. So now as we have finally managed to find a length contraction effect, it does not correspond to a magnetic effect in the lab frame.

Last edited: Nov 9, 2017
17. Nov 9, 2017

### Staff: Mentor

That is why you cannot neglect the electrons if they are present, as I mentioned above in post 9. Please be explicit, are you considering a charged wire or a neutral wire, as I asked you to clarify back in post 3.

Here are the problems I see with your scenarios:

Did not specify the charge in the lab frame
Did not specify the current in the lab frame
Did not specify the geometry of the wire
Did not specify the direction of the acceleration
Used acceleration rather than inertial motion
Used non inertial frames
Did not consider how force transforms

Last edited: Nov 10, 2017
18. Nov 10, 2017

### Ibix

I remain confused about what you think the problem is. The electrons are responsible for the magnetic field in the lab frame, and they are length contracted (or at least their spacing is). In the electron frame the protons are responsible for the magnetic field, and they are length contracted (or at least their spacing is). Also in this frame there is an electric field because the charge densities don't cancel.

19. Nov 10, 2017

### A.T.

20. Nov 10, 2017

### vanhees71

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I think this confusion is again due to the strange attempt to be didactic by Purcell in volume 2 of the Berkley Physics Course. It's very hard to read and very confusing. Of course electrodynamics is in itself a relativistic theory from the very beginning although Maxwell and the Maxwellians didn't know before Einstein completed the picture with his famous article in 1905.

The right way to clearly teach the subject in my opinion is to start with an introduction of Minkowski space and four-vectors and four-tensors as well as four-vector/tensor fields. Then the behavior of components wrt. to inertial reference frames under Lorentz transformations is clearly stated, and you learnt a lot about the somewhat unfamiliar kinematics implied by the relativistic spacetime structure.

Then your troubles are solved very easily: The source of the electromagnetic field is the electric-current four-vector field with components
$$(j^{\mu})=\vv{c \rho}{\vec{j}},$$
where $j^{\mu}$ in fact is an abbreviation for $j^{\mu}(x)$, where $x=(x^{\mu})=(c t,\vec{x})$ is the space-time four-vector.

If you have a Lorentz-transformation matrix $\hat{\Lambda}={\Lambda^{\mu}}_{\nu}$ with $x'=\hat{\Lambda}(x)$ the transformation for the four-current is as for any other four-vector field:
$$j^{\prime \mu}(x')={\Lambda^{\mu}}_{\nu} j^{\nu}(x)={\Lambda^{\mu}}_{\nu} j^{\nu}(\hat{\Lambda}x').$$
The electromagnetic field is described by the antisymmetric Faraday tensor with components $F_{\mu \nu}$. The time-space components can be uniquely mapped to the electric field components, and the space-space components to the magnetic field components as seen from the perspective of an observer at rest with respect to the chosen inertial reference frame. Also its transformation properties are clear from the Minkowski-space tensor formalism
$$F^{\prime \mu \nu}(x')={\Lambda^{\mu}}_{\rho} {\Lambda^{\nu}}_{\sigma} F^{\rho \sigma}(\hat{\Lambda}^{-1} x').$$
The Lorentz-force law for a charged particle moving in a given electromagnetic field can also be written in manifestly covariant notation, and thus also there you can't have any doubts concerning how the components transform under Lorentz transformations.

For this you introduce proper time of the particle by
$$c^2 \mathrm{d} \tau^2=\mathrm{d} s^2=\eta_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu},$$
where $(\eta_{\mu \nu})=\mathrm{diag}(1,-1,-1,-1)$ (west-coast convention as in the majority of HEP physicists' textbooks and papers) are the components of the Minkowski pseudo-scalar product. Then the manifestly covariant equation of motion reads
$$m \frac{\mathrm{d}^2 x^{\mu}}{\mathrm{d} \tau^2}=K^{\mu}=\frac{q}{c} F^{\mu \nu}(x) \frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}.$$
Since $\mathrm{d} \tau$ is a scalar and $x^{\mu}$ is a vector under Lorentz transformations so is the Minkowski-Lorentz force $K^{\mu}$, and thus its components transform as any four-vector components.

So the little effort of 1-2 lecture hours to introduce this tensor calculus, using Einstein's original arguments, postulating the special principle of relativity and the invariance of the speed of light in any inertial reference frame, is paid back by a very lucid discussion of the physics of electromagnetism and any other subject of relativistic physics (e.g., relativistic hydrodynamics and transport theory, relativistic QFT), and it's also only a small step towards General Relativity, which admittedly is the most beautiful theory discovered yet!