Relativistic explanation of electromagnetism

  • #1

Summary:

Why a test charge at rest in the lab frame does not experience a force from a current

Main Question or Discussion Point

I am intrigued by the special-relativity explanation of magnetic force discussed here (linked from the physicsforums FAQ): http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction

Naively, from this explanation, it seems that a test charge at rest in the lab frame should experience a force from a current-carrying wire, since the electrons' fields are Lorentz-contracted relative to the test charge, but the nuclei fields are not. And, that the test charge should experience no force only if the positive and negative charges in the wire are moving in equal and opposite directions relative to the test charge, i.e., when the test charge is moving along the wire at 1/2 the drift velocity. But that's not what happens. What am I missing?
 

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  • #2
PeroK
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Summary:: Why a test charge at rest in the lab frame does not experience a force from a current

I am intrigued by the special-relativity explanation of magnetic force discussed here (linked from the physicsforums FAQ): http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction

Naively, from this explanation, it seems that a test charge at rest in the lab frame should experience a force from a current-carrying wire, since the electrons' fields are Lorentz-contracted relative to the test charge, but the nuclei fields are not. And, that the test charge should experience no force only if the positive and negative charges in the wire are moving in equal and opposite directions relative to the test charge, i.e., when the test charge is moving along the wire at 1/2 the drift velocity. But that's not what happens. What am I missing?
If we assume the test charge is positive:

In the rest frame of the current there is: a) a net postive charge density, which produces an electric field; b) a positive current, which produces a magnetic field; and, c) a moving charge, moving in the same direction as the current.

The electric field produces an outward force on the particle, and the magnetic field produces an inward force on the particle. If you do the maths, these two forces are equal and opposite and the test particle remains at rest.
 
  • #3
In the rest frame of the current
Can you please specify what that means — does it mean in the rest frame of the drifting electrons (with the wire moving "backwards"), or in the rest frame of the wire (with the electrons moving "forwards")? Or something else? I'm very weak on electricity.
 
  • #4
PeroK
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Can you please specify what that means — does it mean in the rest frame of the drifting electrons (with the wire moving "backwards"), or in the rest frame of the wire (with the electrons moving "forwards")? Or something else? I'm very weak on electricity.
I meant the rest frame of the drifting electrons.

And, of course, bearing in mind that the current is by convention the flow of positive charge, saying the rest frame of the current was wrong. The rest frame of the negative current, perhaps!
 
  • #5
Why is there a net positive charge density, producing an electric field? I assume this is independent of the Lorentz effect that produces the counterbalancing attractive magnetic field, and that the same electric field exists in the rest frame of the wire, or any other frame.
 
  • #6
PeroK
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Why is there a net positive charge density, producing an electric field? I assume this is independent of the Lorentz effect that produces the counterbalancing attractive magnetic field, and that the same electric field exists in the rest frame of the wire, or any other frame.
The net positive charge comes from length contraction. If we assume the positive charges and negative charges are equally spaced in the rest frame of the wire. Then in the rest frame of the moving charges:

1) They are further apart (in their own rest frame) and this is length contracted in the rest frame of the wire.

2) The positive charges are moving and hence the distance between them is length contracted in the frame of the moving negative charges.

You put these two factors together to get the net positive charge density.
 
  • #7
I must be very stupid because I'm still not getting it. I'd like to understand what is producing the net positive charge density in the rest frame of the wire/test charge. (Combining different frames is confusing me.)

In the rest frame of the wire, wouldn't length contraction of the distance between the moving charges in the wire, but not of the stationary charges, result in a net negative charge density?
 
  • #8
PeroK
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I must be very stupid because I'm still not getting it. I'd like to understand what is producing the net positive charge density in the rest frame of the wire/test charge. (Combining different frames is confusing me.)

In the rest frame of the wire, wouldn't length contraction of the distance between the moving charges in the wire, but not of the stationary charges, result in a net negative charge density?
The assumption is that in the rest frame of the wire before the current starts there are equal numbers of positive and negative charges, equally spaced. Let's call the spacing between changes ##L##.

After the negative charges have started moving (perhaps imagine them hopping from one atom to the next) there are still an equal number of positive and negative charges, equally spaced, but now the negative charges are moving. In the rest frame of the wire the positive and negative charges both have equal spacing ##L##.

The immediate conclusion is that in the rest frame of the moving electrons they must have spacing ##\gamma L##. Which is length-contracted to ##L## in the rest frame of the wire.

The next conclusion is the the positive charges, whose spacing is ##L## in their rest frame, have spacing ##L/\gamma## in the rest frame of the moving electrons.

Therefore, the positive charge density is greater in the rest frame of the moving electrons.
 
  • #9
Thank you for your infinite patience. This must be what's tripping me up:
now the negative charges are moving. In the rest frame of the wire the positive and negative charges both have equal spacing L.
If (in the rest frame of the wire) they had equal spacing L before the current started, then why (also in the rest frame of the wire) would they still have equal spacing L when the electrons are moving but the positive charges remain at rest? Why (in the rest frame of the wire) isn't the spacing of the moving electrons now L/γ, while the positive charges remain at rest with spacing L?
 
  • #10
pervect
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Summary:: Why a test charge at rest in the lab frame does not experience a force from a current

I am intrigued by the special-relativity explanation of magnetic force discussed here (linked from the physicsforums FAQ): http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction

Naively, from this explanation, it seems that a test charge at rest in the lab frame should experience a force from a current-carrying wire, since the electrons' fields are Lorentz-contracted relative to the test charge, but the nuclei fields are not. And, that the test charge should experience no force only if the positive and negative charges in the wire are moving in equal and opposite directions relative to the test charge, i.e., when the test charge is moving along the wire at 1/2 the drift velocity. But that's not what happens. What am I missing?
It's easiest to analyze a complete loop of wire.

To do the analysis properly, we impose the requirement that the loop of wire is electrically neutral, that the total number of positive charges an negative charges over the whole wire must be equal.

Clearly, we could analyze a charged loop of wire, and we'd expect such a charged loop of wire to have an electric field. But we don't want to analyze a charge loop of wire, we want to analyze a neutral loop of wire.

It's easiest to imagine a square loop of wire. Symmetry says that the + and - charge densities will be independent of angle in the lab frame. So, the densities in the lab frame for + and =- charges must be equal if we are to have overall electrical neutrality.

In the frame where the current loop is moving, the charge density is NOT independent of angle. If you carry out Purcell's analysis, you'll see that some sections of the loop , the sections aligned with the direction of a motion, have a net charge. But the total charge of all sections is still zero, since the wire is neutral.

This can be regarded as a consequence of the relativity of simultaneity, also.
 
  • #11
It's easiest to analyze a complete loop of wire.
Oh, that's an interesting way of looking at it. So in the frame in which the loop is moving, say, vertically, one vertical side is net-negative and the other vertical side is net-positive. And the horizontal sides remain neutral. That's a great visual.

If there is no current in the wire, I assume that all sections are electrically neutral in all frames.
 
  • #12
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Yes, with no current density in the wire, given that we've assumed it's neutral, there is no force on a nearby charge - there are no fields of any kind.
 
  • #13
PeroK
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If (in the rest frame of the wire) they had equal spacing L before the current started, then why (also in the rest frame of the wire) would they still have equal spacing L when the electrons are moving but the positive charges remain at rest?
That's a different question. And not a bad one. That depends on how the current is sustained. If you imagine that the electrons bunch up in the rest frame of the wire once they are moving, then you have a net negative charge in the rest frame of the wire. The current, in that case, would attract the test positive charge in the rest frame of the wire.

But, that's then a different problem. In that case you would have a charged wire and a current. You could analyse that scenario in the rest frame of the electrons and check that there is still a net attraction of the positive charge.

The assumption for the problem at hand, however, is that we have a neutral wire with a current, hence the electrons retain their spacing in the frame of the wire even once they are moving.
 
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  • #14
A.T.
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If (in the rest frame of the wire) they had equal spacing L before the current started, then why (also in the rest frame of the wire) would they still have equal spacing L when the electrons are moving but the positive charges remain at rest?
Because the electrons are still repulsing each other, so they still spread out as far as possible. Length contraction doesn't make their E-fields attractive, so they have no reason to reduce their spacing.

As others suggested, it's best to consider the entrie loop. Here is a great figure by @DrGreg:

attachment-php-attachmentid-44016-d-1329434012-png.png


More explanation here:
https://www.physicsforums.com/threads/magnetism-seems-absolute-despite-being-relativistic-effect-of-electrostatics.577456/post-3768045

See also these threads:
https://www.physicsforums.com/threads/sr-and-magnetism.938812
https://www.physicsforums.com/threads/how-to-consistently-explain-electromagnetism-with-relativity.932270
 
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  • #15
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An electron
If (in the rest frame of the wire) they had equal spacing L before the current started, then why (also in the rest frame of the wire) would they still have equal spacing L when the electrons are moving but the positive charges remain at rest? Why (in the rest frame of the wire) isn't the spacing of the moving electrons now L/γ, while the positive charges remain at rest with spacing L?
Length contraction does not happen magically at a distance.

Two distant electrons accelerating in an uniform electric field don't care about each other.

OTOH all the particles that form a steel rod do care about each other. Internal stresses in an accelerated rod cause the rod to contract. Lack of internal stresses cause a contracted speeding rod to stay contracted.


As for accelerating observer feeling like charges of the universe are pulling his charges more and more to the transverse direction - well we could explain that by the internal magnetic fields in the observer. Right? The observer becomes weak in the transverse direction.
 
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  • #16
vanhees71
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Before treating that problem, please check out this Insights article, where the correct relativistic treatment of the current conducting straight wire is made. Then you can transform the fields correctly from one frame of reference (moving particle) to the other (particle at rest) properly:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/

On the other hand it's very easy to see without any such complicated calculations, because in any electromagnetic field, ##F_{\mu \nu}## the four-force on a charge ##q## is
$$K^{\mu}=m \mathrm{d}_{\tau} p^{\mu} = \frac{q} F^{\mu \nu} u_{\nu}$$
with
$$u^{\mu}=\frac{1}{c} \mathrm{d}_{\tau} x^{\mu}=\begin{pmatrix} \gamma \\ \gamma \vec{v} \end{pmatrix} \quad \text{with} \quad \gamma=\frac{1}{\sqrt{1-\vec{v}^2/c^2}}.$$
Thus the four-force is in components
$$K^{\mu} = \begin{pmatrix} \gamma q \vec{E} \cdot \vec{v}/c \\ \gamma (\vec{E}+\vec{v} \times \vec{B}/c) \end{pmatrix}.$$
Thus in a reference frame, where ##\vec{v}=0##, there's no contribution to the force of the magnetic field.
 
  • #17
Here is a great figure by @DrGreg:
Indeed that is a great diagram!
 
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  • #18
PeroK
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Indeed that is a great diagram!
Summing up, the source of my confusion seems to be that we must specify that it's the current-carrying wire that's electrically neutral — my assumption was wrong that if it's electrically neutral before the current starts, it'll also be electrically neutral after.
I'm not sure what you mean here. The assumption is that the wire is neutral with or without the current.

We are free to set this problem up any way we want. We could assume that the electrons bunch up once they are moving and then we have a different problem. You might argue about the physical process that would lead to such a configuration, but there is nothing electrodynamically invalid about that. You simply postulate a very long wire with more electrons than protons near the test charge. Perhaps simply that we had a charged wire in the first place.

The important point is that whatever the configuration is in the rest frame of the wire and although the configuration is different in another frame (in terms of E & M fields) it results in the same force on the test charge in all frames.
 
  • #19
Duly retracted above. Thank you.
 
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  • #20
ETA: I've self-answered this below. Perhaps I even did it correctly this time.

As I noted, the diagram by @DrGreg is excellent and clears up my earlier misunderstandings in this thread (and other threads like it). Thank you, @A.T. There's one thing I don't understand, though: Why are only two electrons shown in the lower-right case? Seems like the shape and spacing of the electrons (in the diagram) should be the same as in the upper-left case, since the electrons are at rest in both frames.

However, maintaining such spacing creates a problem between the lower-left and lower-right cases.

What is the explanation for the spreading-out of electrons from the upper-left to lower-right cases, since in both cases the electrons are seen as being at rest? I realize these are physically different situations and you cannot simply transform between them, but I can't get past the idea that the same electrons are seen as being at rest in each case, and yet their spacing is seen as different.

The only difference between the upper-left and lower-right cases seems to be that in the latter, the positive charges are moving to the left. Everything else is the same. Surely this can't be responsible for the electrons spreading out.

chment-php-attachmentid-44016-d-1329434012-png-png.png
 
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  • #21
Okay, so, I've been reading PF posts on this topic for several days now. Everyone seems to have the same confusions as I did. But I think I can now answer my own question above, as well as others:

When the current is on, the electrons do length-contract in the wire frame (bottom-left). Their separation would also length-contract, except that we've specified the boundary condition that the wire is uncharged, and there remain the same number of electrons in the loop as without current, so necessarily the electrons must remain at the same separation and charge density. However, in the co-moving frame (lower right), the electrons and their spacing un-contract — and thus the initial boundary condition results in a lower charge density than exists in the no-current/wire-frame case (upper left). Perhaps one could say that when a current is applied, the electrons acquire a greater proper length between them, which becomes apparent in the co-moving frame. This manifests as a positive charge on that side of the wire.

This also answers why a straight wire uncharged in the lab frame, for example, doesn't spontaneously acquire a charge due to electron length contraction when the current is switched on. With the current, every electron leaving an arbitrary region of the wire is replaced by an electron entering at the other end of the region. They don't contract closer together, the reason for which becomes clear when the wire is a loop.

Finally, this answers the inevitable question of why the test particle doesn't experience zero electrostatic force only when it is moving at half the velocity of the electrons (in which case, in the test particle's frame, the electrons and + nuclei are moving at the same speed in opposite directions). The electrons and the nuclei don't enjoy the same freedom. The electrons can move around; the nuclei can't. This breaks the symmetry. Electrons and their distribution conform to the boundary condition of 0 net charge by acquiring a greater proper distance between them, but the nuclei cannot do this.

Did I get all of that right?
 
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  • #22
vanhees71
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Again the hint that the usual treatment of the current conducting wire is not correct in the literature. The correct treatment can be found in my insights article and the literature cited therein:

https://www.physicsforums.com/insights/relativistic-treatment-of-the-dc-conducting-straight-wire/

Having the correct relativistic solution you can put a test charge outside the wire and calculate the force (most conveniently in the sense of the relativistically covariant Minkowski four-force vector) on this test particle in any frame of reference you like. You just have to transform the fields of the wire as well as the force, and you see that everything is consistent, because everything is relativistically covariant and thus the Lorentz transformation consistently transforms between the components of the electromagnetic field tensor (Faraday tensor) and the Lorentz force (in covariant Minkowski four-force form) as Minowski-space tensor and vector components.
 
  • #23
If I'm understanding this correctly, then the following statement of @PeroK is inaccurate?
The assumption is that the wire is neutral with or without the current.
...And that if a wire is electrically neutral in the wire frame without current, then with current, it will only be neutral in the electrons' rest frame?
And that if the wire is a loop, in this frame the opposite side becomes positively charged when current flows?
 
  • #24
vanhees71
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Yes, this statement is wrong, as my Insights article shows in two ways.
 
  • #25
PeroK
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Yes, this statement is wrong, as my Insights article shows in two ways.
An assumption is not wrong, per se. It's not physically impossible to accelerate two objects so that they do not get closer together in the original rest frame. The diagram posted in post #14 is not fundamentally physically impossible.

That assumption may not be supported by a valid model of the DC in a wire - although it is an assumption used in the much of the literature - which is what your Insight was aimed at.

Your Insight presents a more realistic model of DC current in a wire, but it takes things beyond a "B" level treatment of EM.
 
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