Special Relativity Paradox

[M S La Moreaux]

Consider a current carrying circuit with a straight side.  Because of length contraction, in the straight side the moving electrons will appear closer together than they would be if there were no current.  This gives a greater charge density, so the straight side should have a negative charge.  But this is not observed.  How come?

[Markus Hanke]

3 hours ago, M S La Moreaux said:

This gives a greater charge density, so the straight side should have a negative charge.

I don’t really understand what you are getting at. First of all, it isn’t just one side that gets length-contracted, but the entire circuit in the direction of motion. Secondly, both the circuit itself and the classical electron radius in the direction of motion are shortened by the same factor, thus the total amount of charge (ie number of electrons in the wire) in any section of wire remains conserved. Particle number is a conserved quantity under Lorentz transformations.

I don’t see an issue here?

Quote

Special Relativity Paradox

SR is just the geometry of Minkowski spacetime; specifically, in the special case of relationships between inertial frames, it concerns hyperbolic rotations of the coordinate system. Let A be some coordinate-dependent description of an inertial frame, and L be a Lorentz transformation parametrised by rotational angle (=rapidity, a measure of relative velocity). The action of L on A is thus simply

\[A’=L(\omega)A\]

Furthermore, we know that, from definition, the matrix L is a square matrix and must preserve the metric:

\[L^{-1}gL=g\]

which implies that \(|det(L)|=1\). From linear algebra we know that all square matrices with determinant other than zero are invertible, and thus:

\[L^{-1}(\omega)L(\omega)A=A\]

So it is mathematically and logically impossible to construct physical paradoxes using just the axioms of SR, no matter what kind of physical scenario you try to concoct. Any purported SR “paradox” is by the above automatically an erroneous conclusion based on some misapplication of SR. 
 

Edited February 27 by Markus Hanke

[M S La Moreaux]

13 hours ago, Markus Hanke said:

I don’t really understand what you are getting at. First of all, it isn’t just one side that gets length-contracted, but the entire circuit in the direction of motion. Secondly, both the circuit itself and the classical electron radius in the direction of motion are shortened by the same factor, thus the total amount of charge (ie number of electrons in the wire) in any section of wire remains conserved. Particle number is a conserved quantity under Lorentz transformations.

I don’t see an issue here?

SR is just the geometry of Minkowski spacetime; specifically, in the special case of relationships between inertial frames, it concerns hyperbolic rotations of the coordinate system. Let A be some coordinate-dependent description of an inertial frame, and L be a Lorentz transformation parametrised by rotational angle (=rapidity, a measure of relative velocity). The action of L on A is thus simply

 

A′=L(ω)A

 

Furthermore, we know that, from definition, the matrix L is a square matrix and must preserve the metric:

 

L−1gL=g

 

which implies that |det(L)|=1 . From linear algebra we know that all square matrices with determinant other than zero are invertible, and thus:

 

L−1(ω)L(ω)A=A

 

So it is mathematically and logically impossible to construct physical paradoxes using just the axioms of SR, no matter what kind of physical scenario you try to concoct. Any purported SR “paradox” is by the above automatically an erroneous conclusion based on some misapplication of SR. 
 

 

[studiot]

Just now, M S La Moreaux said:

 

I'm sorry, what did you want to say ?

11 hours ago, M S La Moreaux said:

Consider a current carrying circuit with a straight side.  Because of length contraction, in the straight side the moving electrons will appear closer together than they would be if there were no current.  This gives a greater charge density, so the straight side should have a negative charge.  But this is not observed.  How come?

An explanatory diagram would help greatly her.

Observed ?

Observed by whom, in what frame ?

greater charge density ?

greater than what ?

length contraction again observed by whom in what frame ?

[swansont]

19 hours ago, M S La Moreaux said:

Consider a current carrying circuit with a straight side.  Because of length contraction, in the straight side the moving electrons will appear closer together than they would be if there were no current.  This gives a greater charge density, so the straight side should have a negative charge.  But this is not observed.  How come?

What happens to two parallel wires with current flow?

[M S La Moreaux]

This really should not be this difficult.  I assume a straight side to avoid acceleration so special relativity will apply.  The circuit is stationary.  The only things moving are the free electrons.  Their charge density appears greater while the charge density of the protons remains unchanged, thus resulting in a net negative charge for the segment of the circuit under discussion.  This should be observed by someone in the frame of reference of the circuit, but is not.

[Markus Hanke]

52 minutes ago, M S La Moreaux said:

Their charge density appears greater

No. Charge as well as particle number are conserved quantities under Lorentz transformations, as I have pointed out above. The ratio of electrons to protons doesn’t change just because something is in relative motion. How could it? Inertial motion doesn’t magically create or annihilate particles in a wire.

Consider, as a very primitive and purely classical analogy, a length of transparent plexiglass tube filled with - say - 100 ping pong balls, initially at rest. The balls are all perfectly spherical, and their number is constant and doesn’t change. Let’s say it’s a random mix of red and yellow balls.

Now you put the tube into motion at relativistic speeds, relative to an observer. What happens? For that observer, the tube becomes length-contracted in the direction of motion, and so does the radius of each ball. The balls no longer appear spherical to him, but instead look like flattened disks; however, both their total number and the ration between red to yellow balls remains the same.

It’s similar with the particles in the wire, with the caveat that those aren’t classical objects, so for a precise description you need to use relativistic quantum mechanics, which means fields of bispinors. But it all works out. As I said above, you cannot construct physically paradoxes with the axioms of SR.

[exchemist]

2 hours ago, Markus Hanke said:

No. Charge as well as particle number are conserved quantities under Lorentz transformations, as I have pointed out above. The ratio of electrons to protons doesn’t change just because something is in relative motion. How could it? Inertial motion doesn’t magically create or annihilate particles in a wire.

Consider, as a very primitive and purely classical analogy, a length of transparent plexiglass tube filled with - say - 100 ping pong balls, initially at rest. The balls are all perfectly spherical, and their number is constant and doesn’t change. Let’s say it’s a random mix of red and yellow balls.

Now you put the tube into motion at relativistic speeds, relative to an observer. What happens? For that observer, the tube becomes length-contracted in the direction of motion, and so does the radius of each ball. The balls no longer appear spherical to him, but instead look like flattened disks; however, both their total number and the ration between red to yellow balls remains the same.

It’s similar with the particles in the wire, with the caveat that those aren’t classical objects, so for a precise description you need to use relativistic quantum mechanics, which means fields of bispinors. But it all works out. As I said above, you cannot construct physically paradoxes with the axioms of SR.

This isn't my field at all, but I wonder if the conceptual difficulty is that, while the electrons are in motion relative to the observer, the wire in which they travel (i.e. the array of atoms of which it is made) is not. So it seems to me the external dimensions of the wire are unchanged to the observer, even though the contents, i.e. the moving electrons, are foreshortened in the direction of travel as you have explained. 

I'm not sure myself how I should picture this. In terms of your analogy, the balls inside the plexiglas tube are in motion, but the tube itself is not. 

[KJW]

11 hours ago, swansont said:

What happens to two parallel wires with current flow?

There is a force acting on the wires due to the magnetic field. Magnetic field??? Where did THAT come from???

 

 

[Genady]

5 hours ago, M S La Moreaux said:

to avoid acceleration so special relativity will apply

Special relativity applies with acceleration as well. Albeit the application is more involved.

[Markus Hanke]

3 hours ago, exchemist said:

This isn't my field at all, but I wonder if the conceptual difficulty is that, while the electrons are in motion relative to the observer, the wire in which they travel (i.e. the array of atoms of which it is made) is not.

I think you might be right, I misunderstood, see below.

7 hours ago, M S La Moreaux said:

The circuit is stationary. 

Ok, I just saw this while re-reading the thread. Apologies, I initially misunderstood your scenario. 

7 hours ago, M S La Moreaux said:

The only things moving are the free electrons.

So the observer is at rest in the circuit hardware frame.

7 hours ago, M S La Moreaux said:

Their charge density appears greater while the charge density of the protons remains unchanged

The drift velocity of electrons in a DC circuit is typically nowhere near relativistic, so you wouldn’t see any relativistic effects.

If that velocity somehow was made relativistic, the resolution is still the same - in the circuit frame, the classical radius of the electrons is length-contracted in the direction of motion; in the electron rest frame, it is the wire length and the classical radius of the protons that is length-contracted by the same factor. Crucially, in both cases, you furthermore have to consider relativity of simultaneity when figuring out how many particles are actually there in total, since the events “reached beginning” and “reached end” of wire are no longer simultaneous in both frames - the adjustment happens by the same factor, thus the ratio between the number of electrons and the number of protons is the same in both frames.

This is really just a variation of the classical ladder paradox, and is resolved similarly.

As an aside - an experimental example of where you’d see a type of length-contraction of a roughly spherical object into something resembling more of a flattened disk, is the collision of gold ions at the Relativistic Heavy Iron Collider.

Edited February 28 by Markus Hanke
Ladder paradox mentioned

[studiot]

1 hour ago, M S La Moreaux said:

This really should not be this difficult.  I assume a straight side to avoid acceleration so special relativity will apply.  The circuit is stationary.  The only things moving are the free electrons.  Their charge density appears greater while the charge density of the protons remains unchanged, thus resulting in a net negative charge for the segment of the circuit under discussion.  This should be observed by someone in the frame of reference of the circuit, but is not.

Thank you for responding to my query about frames, even if the reply was short.

When I first read this I wondered if you are confusing the relativistic origin of magnetism .....

But We still have no diagram.

[exchemist]

42 minutes ago, Markus Hanke said:

I think you might be right, I misunderstood, see below.

Ok, I just saw this while re-reading the thread. Apologies, I initially misunderstood your scenario. 

So the observer is at rest in the circuit hardware frame.

The drift velocity of electrons in a DC circuit is typically nowhere near relativistic, so you wouldn’t see any relativistic effects.

If that velocity somehow was made relativistic, the resolution is still the same - in the circuit frame, the classical radius of the electrons is length-contracted in the direction of motion; in the electron rest frame, it is the wire length and the classical radius of the protons that is length-contracted by the same factor. Crucially, in both cases, you furthermore have to consider relativity of simultaneity when figuring out how many particles are actually there in total, since the events “reached beginning” and “reached end” of wire are no longer simultaneous in both frames - the adjustment happens by the same factor, thus the ratio between the number of electrons and the number of protons is the same in both frames.

This is really just a variation of the classical ladder paradox, and is resolved similarly.

As an aside - an experimental example of where you’d see a type of length-contraction of a roughly spherical object into something resembling more of a flattened disk, is the collision of gold ions at the Relativistic Heavy Iron Collider.

Aha. So the resolution of the seeming paradox is  actually quite subtle. To be honest, the ladder paradox almost does my head in. I'm not surprised @M S La Moreaux hadn't thought of it that way. 

[swansont]

6 hours ago, KJW said:

There is a force acting on the wires due to the magnetic field. Magnetic field??? Where did THAT come from???

You can look at it solely from electrostatics and relativity, as Markus has detailed.

[Markus Hanke]

17 hours ago, exchemist said:

Aha. So the resolution of the seeming paradox is  actually quite subtle. To be honest, the ladder paradox almost does my head in. I'm not surprised @M S La Moreaux hadn't thought of it that way. 

You can think of the number of electrons in the wire to correspond to the number of steps on the ladder. This number does not change as you switch between frames (so no excess of either electrons or protons), and despite the circuit being length-contracted, the “electron-ladder” still fits in the wire, due to relativity of simultaneity.

[M S La Moreaux]

There is no cutoff velocity of relativistic effect.  Although the drift velocity of the electrons is minuscule, the relativistic increase in their density multiplied by the huge number of free electrons is palpable, on a par with magnetism.

I do not see that the ladder paradox applies.  In the ladder paradox, the paradox involves the discrepancy between the views of two different reference frames.  The relativistic length contraction of the ladder, and thus the increased density, as it were, of the rungs in the barn's frame is not denied.

[Markus Hanke]

1 hour ago, M S La Moreaux said:

There is no cutoff velocity of relativistic effect.

That’s technically true. However, I’d like to invite you to yourself calculate the gamma factor for a typical electron drift velocity, which is on the order of 10^(-4)m/s, and draw your own conclusions as to the order of magnitude of any length contraction effects this would induce. Not that it matters, see below.

1 hour ago, M S La Moreaux said:

the relativistic increase in their density

There is no such increase, see below.

1 hour ago, M S La Moreaux said:

The relativistic length contraction of the ladder, and thus the increased density, as it were, of the rungs in the barn's frame is not denied.

It is also not denied that the total number of rungs on the ladder, and thus the ratio between rungs and (eg) number of boards used to build the barn walls, does not change.

1 hour ago, M S La Moreaux said:

In the ladder paradox, the paradox involves the discrepancy between the views of two different reference frames.

If you are merely comparing “current on” vs “current off” in the same stationary circuit frame, then you have nothing to resolve, because

On 2/27/2025 at 4:22 AM, M S La Moreaux said:

in the straight side the moving electrons will appear closer together than they would be if there were no current.

is wrong. What is contracted is each electron’s radius in the direction of motion, but distances in the stationary (from the POV of the observer) wire don’t change - it’s only the electrons that are moving in the frame of the observer, not the wire, so a unit of distance in the wire is the same whether current is on or off. So the total number of electrons seen to be moving through that wire does not change either, just the mix of E and B fields resulting from the presence of these charges changes.

Applying an electric field to a distribution of charges in a wire, all else remaining the same, does not change the number of charges, it just sets them in motion, relativistic or not. This is a trivial scenario, there’s nothing to be resolved here. It’s much more interesting to compare circuit frame to electron frame (where distances really do become contracted), and that’s resolved the same way as the ladder paradox. Either way, there’s no difference in outcomes, and thus no paradoxes. Like I showed in my first response, there can never be any physical paradoxes arising from SR; any apparent paradoxes always indicate some error in applying the axioms. You can’t get around this.

Edited Tuesday at 06:02 AM by Markus Hanke
Further clarifications

[swansont]

On 2/28/2025 at 12:55 AM, Markus Hanke said:

Consider, as a very primitive and purely classical analogy, a length of transparent plexiglass tube filled with - say - 100 ping pong balls, initially at rest. The balls are all perfectly spherical, and their number is constant and doesn’t change. Let’s say it’s a random mix of red and yellow balls.

Now you put the tube into motion at relativistic speeds, relative to an observer. What happens? For that observer, the tube becomes length-contracted in the direction of motion, and so does the radius of each ball. The balls no longer appear spherical to him, but instead look like flattened disks; however, both their total number and the ration between red to yellow balls remains the same.

What happens to the number of balls per unit length (the ball density)?

[Markus Hanke]

11 hours ago, swansont said:

What happens to the number of balls per unit length (the ball density)?

Note first that that response of mine you quoted was based on a misunderstanding on my part of what the OP was actually suggesting, so it missed the point of the scenario.

But as to your question, I don’t think anything would happen to the ball density in that particular case, since the radius of each ball (in the direction of motion) is length-contracted by the same factor as the tube itself, so the ball density remains unchanged, unless I’m overlooking something. This is provided that the balls are at rest relative to the tube, or else things become more complicated. In general though, a spatial density is not always automatically an invariant quantity - which is why in relativistic scenarios it is wise to use current density J instead, which is a 4-vector.

[swansont]

12 hours ago, Markus Hanke said:

Note first that that response of mine you quoted was based on a misunderstanding on my part of what the OP was actually suggesting, so it missed the point of the scenario.

But as to your question, I don’t think anything would happen to the ball density in that particular case, since the radius of each ball (in the direction of motion) is length-contracted by the same factor as the tube itself, so the ball density remains unchanged, unless I’m overlooking something. This is provided that the balls are at rest relative to the tube, or else things become more complicated. In general though, a spatial density is not always automatically an invariant quantity - which is why in relativistic scenarios it is wise to use current density J instead, which is a 4-vector.

You’re still misunderstanding the scenario.

The number of balls (N) is invariant but L decreases, so I don’t see how N/L can do anything but increase. Same with charge density. 

The answer to the OP’s question is that we do notice the situation, as I hinted at with my first response. Two current-carrying wires will attract or repel. We call it magnetism, but that’s just the name we give it, since we know magnetism is just electrodynamics that includes relativity.

https://physicsworld.com/a/the-invisibility-of-length contraction/#:~:text=Length contraction has never been,in the other wire's frame.

“Length contraction has never been directly measured. But its effects show up in the magnetic force that acts between parallel, current-carrying wires. Bizarrely, this force, which is purely magnetostatic, appears in one wire due to length contraction as experienced by the charge carriers in the other wire’s frame. (It’s complicated)”

(references are given, for anyone interested)

[studiot]

Hi @Markus Hanke have you read the attachments in my posts a couple back.?

That might help.

[M S La Moreaux]

Copper contains 10^22 free electrons per cubic centimeter.

A paradox within special relativity may be impossible, but that is not the situation I am describing.  It is instead a discrepancy between what special relativity predicts and what  is actually observed.

A magnetic field affects a moving charge, not a stationary one.  The effect the I am describing would affect a stationary charge external to the wire.

[studiot]

Just now, M S La Moreaux said:

Copper contains 10^22 free electrons per cubic centimeter.

A paradox within special relativity may be impossible, but that is not the situation I am describing.  It is instead a discrepancy between what special relativity predicts and what  is actually observed.

A magnetic field affects a moving charge, not a stationary one.  The effect the I am describing would affect a stationary charge external to the wire.

 

I am inclined to wonder how you can do everything in your head, whereas lesser mortals like myself need a diagram.

Perhaps because I am a plodder I need to know what electrons are travelling at speed in that copper wire to produce observable relativistic effects.

For instance say 10 V is applied to the ends of a 1mm diametre copper wire 100m long what is the electron velocity at 290^oK  ?

 

Spoiler

answer 0.4 millimetre per second  , hardly relativistic.

 

[swansont]

2 hours ago, M S La Moreaux said:

Copper contains 10^22 free electrons per cubic centimeter.

A paradox within special relativity may be impossible, but that is not the situation I am describing.  It is instead a discrepancy between what special relativity predicts and what  is actually observed.

A magnetic field affects a moving charge, not a stationary one.  The effect the I am describing would affect a stationary charge external to the wire.

In the lab frame, what is length-contracted? Not the wire, because it’s in the lab frame.

The length contraction giving the charge imbalance is the electron frame (the one giving rise to the current)

 

[Markus Hanke]

13 hours ago, swansont said:

The number of balls (N) is invariant but L decreases, so I don’t see how N/L can do anything but increase. Same with charge density. 

I can see your point, and you are of course right.
The kind of density I had in mind though was a different one - I took the balls to be extended objects, and mentally considered the ratio between ball radius and tube length, ie which proportion of the tube volume is occupied by each ball. Since both are length-contracted by the same factor, this ratio does not change.

In my defense, I tend to have a tendency to seek out invariant quantities when looking at relativistic scenarios.

13 hours ago, swansont said:

Two current-carrying wires will attract or repel. We call it magnetism, but that’s just the name we give it, since we know magnetism is just electrodynamics that includes relativity.

But I don’t think that’s what the OP has in mind, unless I’m still misunderstanding him. He is comparing the same circuit from the same frame, only with current off and on, and argues that because the electrons are in motion, the distance between them decreases, and thus there’s a larger net negative charge in that section of the circuit because there are more electrons in that same length of wire. He never mentions the rest frame of the electrons, nor the EM fields.

What I’m saying here is that the distance between the electrons doesn’t change just because you turn on the current, because the observer is still stationary with respect to the circuit; there’s no length contraction of distances in the wire in the observer’s frame. In other words, the total amount of charge is the same, it’s just that this charge is flowing rather than standing still.

 

13 hours ago, studiot said:

Hi @Markus Hanke have you read the attachments in my posts a couple back.?

That might help.

I have, but I don’t think that’s what the OP had in mind, see my comments above.