Length contraction in a current carrying wire?

[maartenrvd]

Reading http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Length_Contraction I reminded something that always bothered me about this explanation: I would think that a current carrying wire would only appear electrically neutral when the observer is moving along the wire with half the speed of the electrons, and not when it is stationary relative to the protons in that wire.

Is this correct or am i missing something?

 

[Dale]

Hi maartenrvd, welcome to PF!

It is a measured fact that the wire is electrically neutral in the lab frame. As the link says "A current-carrying wire is observed to be electrically neutral in its rest frame" (emphasis added). The charge densities in other frames have to conform to that fact.

 

[maartenrvd]

Then how is that possible?
I would assume that a stationary observer would see length contraction for the free electrons and not for the stationary positive charge in the wire. To me this should logically result in a net observed electric field.
However, since the velocities are very small this would only have a marginal effect.

I'm still puzzled about this.

 

[Dale]

Again, the fact that the wire is neutral and has no E field in the lab frame is observed. It is a fact. It is under experimental control. Consider it like an initial condition or a "given" in the problem. It has nothing to do with relativity.

What relativity explains is: Given the fact that a wire has a current in the lab frame and given the fact that a wire is neutrally charged in the lab frame, then what does it look like in other reference frames?

 

[maartenrvd]

I appreciate the effort but your answer doesn't help me: I did not ask the question to just accept facts but to understand them.

My logic says otherwise so if anybody can show me the error I would appreciate that.

 

[vin300]

I would think that a current carrying wire would only appear electrically neutral when the observer is moving along the wire with half the speed of the electrons,

Why do you say that?

 

[vin300]

With relativistic considerations or not, the number of electrons on the conductor does not increase. They are only replaced and without a net charge there is no E field

 

[Dale]

My logic says otherwise so if anybody can show me the error I would appreciate that.

Your logic is wrong because you think that you should be able to derive a given or a boundary condition.

Look, a scientist is sitting in a lab. They can drive a current through a wire (by connecting one end to the positive terminal and the other end to the negative terminal of a battery) and/or they can charge the wire (by connecting both ends to the positive terminal of a battery). It is something that they can choose arbitrarily at their whim (within the limits of their apparatus).

The question is: If the scientist chooses to have a current-carrying uncharged wire, then what does that wire look like in other frames? You cannot derive the fact that it is uncharged, that is simply part of the specification of the problem.

 

[Hans de Vries]

I appreciate the effort but your answer doesn't help me: I did not ask the question to just accept facts but to understand them.

My logic says otherwise so if anybody can show me the error I would appreciate that.

You can find the math here:

section 2: Neutral wire condition from the relativistic charge field.
http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf

The electron density is equal to the ion density for a neutral current carrying wire.

Regards, Hans.

 

[Dale]

Hi Hans,

I was looking at the derivation. For equation 1 why did you set the magnetic force equal to the electric force? They are not equal in general.

 

[Hans de Vries]

Hi Hans,

I was looking at the derivation. For equation 1 why did you set the magnetic force equal to the electric force? They are not equal in general.

It's the derivation that the magnetostatic force is a relativistic effect of electrostatics,
that's why they have to be the same. It assumes that the wire is neutral in the rest frame.Somebody did place a link to this paper at Wikipedia a long time ago
http://en.wikipedia.org/wiki/Relativistic_electromagnetism
and it seems that the derivation is published recently here
http://dx.doi.org/10.1119/1.3098206 by somebody who
came independently to the same conclusion.Regards, Hans

It was first discussed here: https://www.physicsforums.com/showthread.php?t=133587

 

[DrGreg]

Then how is that possible?
I would assume that a stationary observer would see length contraction for the free electrons and not for the stationary positive charge in the wire. To me this should logically result in a net observed electric field.
However, since the velocities are very small this would only have a marginal effect.

I'm still puzzled about this.

I think you probably have a misunderstanding of what Lorentz contraction is. I think you have the idea that whenever you have two objects A and B that are initially stationary in some frame, and you then accelerate both objects, then the distance between A and B automatically shortens relative to the original frame. Well, the shortening isn't automatic. It all depends on what the relationship between A and B is. If there is a reason why the distance between A and B in the new frame must be equal to the original distance in the original frame, then, yes, their distance in the original frame will decrease.

But in this case there is no reason for that. In fact, we already know that the charge remains neutral, in the wire's rest frame, which means that the average separation of moving electrons must the the same as the average separation of (almost) stationary electrons. Lorentz contraction therefore tells us that in the frame in which the "moving" electrons are at rest, the electrons must be further apart.

 

[Per Oni]

wikipedia:

In the simple model of events in a wire stretched out horizontally, a current can be represented by the evenly spaced positive charges, moving to the right, whilst an equal number of negative charges remain at rest. If the wire is electrostatically neutral, the distance between adjacent positive charges must be the same as the distance between adjacent negative charges.

Hans de Vries, end of part 2:

We arrived at this result by assuming that the number of electrons per unit length stays the same, regardless of if they do move or do not move.

Both statements defeat the object of explaining magnetism as a result of relativity. According to relativity there should occur a length contraction in a moving train in the direction of movement. Charge per unit length should increase for moving ions according to relativity.

Which takes me to a related subject. I read here, I think it was in this forum, that in particle accelerators just because of length contraction it is possible to push in more ions in the direction of travel the higher the ionic speed.
However I it could be I misread or the post was wrong.

 

[Dale]

It's the derivation that the magnetostatic force is a relativistic effect of electrostatics,
that's why they have to be the same. It assumes that the wire is neutral in the rest frame.

But they do not have to be the same in general. In fact, as you say, the wire is neutral in the rest frame meaning that the electric force is 0 (and the magnetic force is not 0). Your equation certainly doesn't hold for the general case, I believe it only holds for one specific reference frame.

 

[Hans de Vries]

Both statements defeat the object of explaining magnetism as a result of relativity. According to relativity there should occur a length contraction in a moving train in the direction of movement. Charge per unit length should increase for moving ions according to relativity.

First of all. The effect at low speed is almost all due to non-simultaneity
rather than to due to Lorentz contraction, The magnetic force is linear in
the velocity while Lorentz contraction is approximately quadratically
dependent on the velocity.

Secondly, The wire is neutral in the restframe. Which means that the
electron density is equal to the ion density. This taking into account
that the fields of moving electrons are transformed. The proof of this
is given in section 2. Remember that the electrons drift all at different
velocities, which are also constantly changing.

More over moving relativistic fields can be found here:
http://physics-quest.org/Book_Chapter_EM_LorentzContr.pdf

Regards, Hans

 

[Hans de Vries]

I believe it only holds for one specific reference frame.

That's of course correct. The electric force is in the rest frame of the test charge.

Regards, Hans

 

[Per Oni]

First of all. The effect at low speed is almost all due to non-simultaneity
rather than to due to Lorentz contraction

,
Which effect are you pointing at?

The magnetic force is linear in
the velocity while Lorentz contraction is approximately quadratically
dependent on the velocity.

True but what is the reason you mention this?

The proof of this
is given in section 2.

In section 2 you assume that the electrons have the same density as the ions at all speeds and then you go to great lengths to prove that therefore the electrical fields around the wire are the same.

Remember that the electrons drift all at different
velocities, which are also constantly changing.

We are talking about macroscopic effects so this statement serves no one.

More over moving relativistic fields can be found here:
http://physics-quest.org/Book_Chapter_EM_LorentzContr.pdf

26 pages? I've put forward a simple question.

Regards Willum.

 

[maartenrvd]

I would think that a current carrying wire would only appear electrically neutral when the observer is moving along the wire with half the speed of the electrons,

vin300: Why do you say that?

Because only than the length contraction for the negative charge is the same as the length contraction for the positive charge.

@Hans

How do you come to the assumption that the electrons are evenly spaced (as seen from a moving observer). Would that also be the case for neutral particles? (and please do not assume that the speed of the electrons can be neglected just because it is very low).

Hans: That's of course correct. The electric force is in the rest frame of the test charge.

I agree: that is the reference frame I am talking about. That is also why the force on the test-charge has a pure electrostatic and not a magnetic origin.

DrGreg: Lorentz contraction therefore tells us that in the frame in which the "moving" electrons are at rest, the electrons must be further apart.

When I read this it exactly confirms my statement about the observed electric field around a current carrying wire: the electrons show contraction (at least the electrons closes to the test-charge) and the protons don't.
I don't see the difference of a test-charge being stationary relative to the 'moving' electrons or being stationary relative to the protons in the wire. Except of course that the polarity changes. As seen from the test-charge there is no magnetic force. So there has to be an e-field to create the force on the test-charge.

About experimental confirmation: This observed E-field would be quite small. By approximation derived from simple induction formula e=vxB * l, about 1 micro volt per cm for a local magnetic field strength of .1 tesla and an electron drift of 1 mm/s. You would have a hard time measuring this and distinguish this from induction effects.

 

[Dale]

About experimental confirmation: This observed E-field would be quite small. By approximation derived from simple induction formula e=vxB * l, about 1 micro volt per cm for a local magnetic field strength of .1 tesla and an electron drift of 1 mm/s. You would have a hard time measuring this and distinguish this from induction effects.

Your calculations are way off. You didn't specify the configuration, but let's say we are measuring the E and B field 1 cm from a long straight wire.

The B-field is given by http://hyperphysics.phy-astr.gsu.edu/HBASE/magnetic/magcur.html#c2. For a B field of .1 T at a distance of 1 cm requires a current of 5 kA.

5 kA/(1 mm/s) gives a linear charge density of 5E6 C/m. Now, for the calculation of the E-field we only care about the excess charge. The Lorentz factor for v = 1 mm/s is γ-1 = 1.1E-16. That would give an excess charge of 5.6E-10 C/m (that is if there really were excess charge in the lab frame due to Lorentz contraction from the electron's rest frame).

The E-field is given by http://hyperphysics.phy-astr.gsu.edu/Hbase/electric/elecyl.html. For a charge density of 5.6E-10 C/m and a distance of 1 cm we get an E-field of not just 1 uV/cm as you stated but about 1E7 uV/cm, or 1000 V/m. This would be easily detectable (if it were to happen).

 

[maartenrvd]

@Dalespam
This would indeed be the case when the whole wire would show the calculated length contraction, but that only counts for the part that is orthogonal to the observer. Then a small part of the wire shows length contraction adding to the observed field and the rest shows length "extension".

It would take me considerable time do the math but the overall observed E-field would be much less that what you mentioned and would indeed have to be the same as what I said it should be.
The reason that I used the that simple formula is just because of what I earlier mentioned: what is the difference of being stationary to the negative charge as being stationary to the positive charge in a wire (except that the polarity will change). For example: moving along the mentioned wire with 1 mm/sec and thus being stationary to the moving electrons, would produce a E-field of 1uV/cm, the other way around when being stationary to the protons would have to produce an E-field of the same magnitude but in opposite direction.
I still haven't heard conclusive reasons to assume otherwise. So again: what is wrong with this reasoning?

 

[Per Oni]

Hi all.
Sometime ago I started a similar post so I’m quite interested in this one.
My understanding of the problem is this:

It is a fact that a stationary (dc) current carrying wire has no electric field around it (ignoring the one which causes the current to run in the first place).
The electric field we are interested in here is zero.

For me this is a paradox because according to relativity a length contraction takes place in a continuous line of moving charges. Then using Gauss’s law there should be an increase in the E-field. However, as stated E=0.

To solve this problem people have suggested that, when a current flows the line of charges spread out length ways and get contracted again because of relativity so that we end up with exactly the same charge density before and after.

I’ve got various reasons to doubt that explanation, for one: there’s no reason why these electrons should feel a force which spreads them out.

Now look at chapter 12.3.1 of Griffiths, adapted from Purcell. There’s a straight forward solution to this problem. First of all they accept that length contraction takes place in the normal relativistic way, both for stationary and moving wires. No spread out theory’s required.

OK, I know they cheated, they are using a +ve and a –ve current to balance things out. However I've got a feeling that they didn't like the spread out theory either. So here we are, for me the paradox goes on…….

 

[Dale]

This would indeed be the case when the whole wire would show the calculated length contraction, but that only counts for the part that is orthogonal to the observer.

No, it "counts" for the part which is parallel to the velocity. Since I specified a straight wire that implies that it "counts" for the whole wire.

I don't even know what you mean by "orthogonal to the observer". Usually an observer is considered to be a point, and there is no such thing as orthogonal to a point.

It would take me considerable time do the math but the overall observed E-field would be much less that what you mentioned and would indeed have to be the same as what I said it should be.

Then I suggest you take the time to do so, your present calculation is way off as I already demonstrated. Also, it appears that your simple formula is dimensionally inconsistent (the units on the left don't match the units on the right).

 

[Dale]

It is a fact that a stationary (dc) current carrying wire has no electric field around it (ignoring the one which causes the current to run in the first place).
The electric field we are interested in here is zero.

I don't understand your confusion here. You know the fact, that is it, end of story.

Do you understand what a boundary condition is and why boundary conditions are given instead of derived?

Here is an example that may help. Say you are doing a projectile problem. You can use Newton's laws of motion and a uniform force of gravity to derive a general form showing that projectiles travel in parabolic paths. But that, by itself, is insufficient to describe the path of any specific projectile. For that you need to provide two boundary conditions, typically an initial velocity and position. Say now that you are given that the projectile is initially launched horizontally at 100 mph from a height of 6 ft above the ground. You cannot go back and use Newton's laws of motion and a uniform force of gravity to explain why the projectile was launched horizontally. There is no explanation for it, it is simply a given boundary condition. It could have been some other value; the guy launching the projectile simply made a choice that he wanted to launch it horizontally. You cannot use Newton's laws to explain it, but given that condition you can say what will happen at other times.

Similarly here. Relativity describes the relationship in different inertial reference frames, but you have to start out with a complete description in one frame first. That complete description is the boundary condition. There is no relativistic reason for it, and you cannot use relativity to explain it. In this case, the current and the charge on the wire in the lab frame. The guy running the lab simply made a choice that he wanted an uncharged current-carrying wire; it could have been some other value but that was his choice. You cannot use relativity to explain it, but given that condition you can say what will happen in other reference frames.

 

[Per Oni]

Here is an example that may help. Say you are doing a projectile problem. You can use Newton's laws of motion and a uniform force of gravity to derive a general form showing that projectiles travel in parabolic paths. But that, by itself, is insufficient to describe the path of any specific projectile. For that you need to provide two boundary conditions, typically an initial velocity and position. Say now that you are given that the projectile is initially launched horizontally at 100 mph from a height of 6 ft above the ground. You cannot go back and use Newton's laws of motion and a uniform force of gravity to explain why the projectile was launched horizontally. There is no explanation for it, it is simply a given boundary condition. It could have been some other value; the guy launching the projectile simply made a choice that he wanted to launch it horizontally. You cannot use Newton's laws to explain it, but given that condition you can say what will happen at other times.

Similarly here. Relativity describes the relationship in different inertial reference frames, but you have to start out with a complete description in one frame first. That complete description is the boundary condition.

No problem so far.

There is no relativistic reason for it, and you cannot use relativity to explain it. In this case, the current and the charge on the wire in the lab frame. The guy running the lab simply made a choice that he wanted an uncharged current-carrying wire; it could have been some other value but that was his choice.

That was’nt his choice. He could have charged the wire if he wanted to by using the capacitance of this wire wrt another object or even wrt Earth but we are not interested in any such charges. Ignoring those he cannot produce a charged wire simply by sending a current through. Nature therefore determines the neutrality of that wire and not some guy. Nature determines this boundary condition.

You cannot use relativity to explain it,

That’s your private opinion?

I'm just one of those guys who wants to know why nature behaves as it does.

 

[Dale]

That was’nt his choice. He could have charged the wire if he wanted to by using the capacitance of this wire wrt another object or even wrt Earth but we are not interested in any such charges. Ignoring those he cannot produce a charged wire simply by sending a current through. Nature therefore determines the neutrality of that wire and not some guy. Nature determines this boundary condition.

This is still precisely analogous to the projectile problem. You could say:

"That was’nt his choice. The projectile could have been launched vertically if he wanted to by elevating the barrel of the launcher, but we are not interested in any such elevation. Ignoring those he cannot produce a vertically launched projectile simply by using a different speed of projectile. Nature therefore determines the horizontal initial velocity and not some guy. Nature determines this boundary condition."

In any case, I would disagree in both cases with the statements "That was’nt his choice" and "Nature determines this boundary condition" since in both cases there is something that the researcher could choose to do to obtain a different boundary condition. The fact that you are not interested in that change is also an arbitrary choice, determined by you and not nature.

 

[Dale]

Per Oni and maartenrvd, is it possible that you are actually asking a question about Maxwell's equations and I am mistakenly giving you the relativity answer? Is your question perhaps why in this case
(+V) -/\/\/\/\/- (-V)
there is a current but no net charge and in this case
(+V) -/\/\/\/\/- (+V)
there is a net charge but no current? Is that what you are really asking? Are you really asking how Maxwell's equations lead to that result in the wire?

 

[Per Oni]

This is still precisely analogous to the projectile problem.

I couldn't disagree more.

In any case, I would disagree in both cases with the statements "That was’nt his choice" and "Nature determines this boundary condition" since in both cases there is something that the researcher could choose to do to obtain a different boundary condition.

Such as?

Hey DaleSpam thanks for your efforts but I've seen these kind of discussions here ones too often. Were going towards a yes no debate with no real new input. So you can have the last word.

I've also seen your next post and no that's not the discussion I'm after.
Thanks anyway.

 

[Hans de Vries]

First of all. The effect at low speed is almost all due to non-simultaneity
rather than to due to Lorentz contraction

Which effect are you pointing at?

I'm talking about the magnetic force F = v x B

The magnetic force is linear in the velocity while Lorentz contraction is approximately quadratically
dependent on the velocity.

True but what is the reason you mention this?

Because the magnetic force F = v x B is linear in v and not quadratic.

This is because what I said: The effect of non-Simultaneity is linear while
the effect of Lorentz contraction is quadratic (at low speeds)

The wire becomes electrically charged because of non-simultaneity. More
electrons enter the wire at one side while less exit at the other side. This is
a linear effect at low speeds where \gamma \approx 1

At drift speeds of 1 mm/s the length contraction can be neglected compared
to the effect due to non simultaneity.

In section 2 you assume that the electrons have the same density as the ions at all speeds
and then you go to great lengths to prove that therefore the electrical fields around the
wire are the same.

Yes, it is far from obvious that equal-density means equal-fields. The shape
of the fields of the individual electrons change at different speeds. The total
field does not change.

About the neutrality of the wire:

The (average) voltage of the wire is determined by the experimenter. If he
puts 1000.0 V at one end and 1000.1 V at the other end then the wire is not
neutral but has a voltage of ~1000 V.

The voltage of the wire, and it being neutral in the rest frame is a boundary
condition determined by the experimenter, ignoring the voltage difference
between both end which causes the current. Regards, Hans

 

[Dale]

Such as?

Such as elevating the barrel to obtain a different initial velocity for the projectile or such applying an overall voltage (i.e. "using the capacitance" as you said) to obtain a net charge in the lab for the wire.

I am sorry that you are dissatisfied with the answer, but I think that the analogy is solid and (if you don't really want the Maxwell's discussion) the boundary condition answer is complete wrt electromagnetism in relativity.

 

[Per Oni]

@Hans,
First of all I want you to have a look at posts # 12 of DrGreg

Lorentz contraction therefore tells us that in the frame in which the "moving" electrons are at rest, the electrons must be further apart.

and # 21 of myself.

To solve this problem people have suggested that, when a current flows the line of charges spread out length ways and get contracted again because of relativity so that we end up with exactly the same charge density before and after.

Both of us are pointing to a existing explanation often found on web pages.

Now all I want to know of you at this stage do you agree or disagree with this theory.
If you disagree, and you know of a different theory can you give us some references.
Thanks.

 

[Hans de Vries]

@Hans,
First of all I want you to have a look at posts # 12 of DrGreg

Yes, #12 of DrGreg is all correct: Lorentz contraction of a "many particle object"
when accelerated only occurs if there is a reason that such an object has a
certain size and shape in it's rest frame. A line of electrons spaced 10 cm apart
for instance is not such an object. However when such a line of electrons is not
accelerated but simply viewed from another reference frame then you have to
take Lorentz contraction into account.

and # 21 of myself.

Both of us are pointing to a existing explanation often found on web pages.

Now all I want to know of you at this stage do you agree or disagree with this theory.
If you disagree, and you know of a different theory can you give us some references.
Thanks.

I'm not exactly sure what this theory is from the posts but I've already said
several times that Lorentz contraction can be neglected at low speed.

The magnetostatic force is above all the result of non-simultaneity and very simple.

Please read section 1 of my paper here:
http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf
http://www.chip-architect.org/news/2007_02_27_Magnetism_as_a_Relativistic_side_effect.html

There is also a peer reviewed version of section 1 here:
http://dx.doi.org/10.1119/1.3098206
Charles L. Adler, the author of this paper did send me a preprint in September
last year and he explained that he had independently come to the same
conclusion as I had and that he was about to publish it.Regards, Hans

 

[Per Oni]

Yes, #12 of DrGreg is all correct: Lorentz contraction of a "many particle object"
when accelerated only occurs if there is a reason that such an object has a
certain size and shape in it's rest frame. A line of electrons spaced 10 cm apart
for instance is not such an object. However when such a line of electrons is not
accelerated but simply viewed from another reference frame then you have to
take Lorentz contraction into account.

Where did DrGreg say all that?

I'm not exactly sure what this theory is from the posts but I've already said
several times that Lorentz contraction can be neglected at low speed.

Everything we have delt with so far is low speed. Why is DrGreg then correct?
Why do you refer to articles with Lorentz contraction in at the first place?

I've got some more quarrels but got to dash for now.

 

[maartenrvd]

@Hans:
From your paper "The simplest, and the full derivation of Magnetism as a Relativistic side effect of electroStatics" after equation 8: "We see that the field of the moving electrons is the same as the field for the electrons at rest. The result is independent of the speed of the electrons."

Don't get me wrong, I'm just trying to understand, but how does this match with the fact that a magnetic force can be transformed into an electric force in the correct reference frame, i.e. the frame of the observer?

I also question your assumption that length contraction can be ignored. As dalespam calculated:

Now, for the calculation of the E-field we only care about the excess charge. The Lorentz factor for v = 1 mm/s is γ-1 = 1.1E-16.

.
I think he than made an error in the calculation of the excess charge but I still think this can't be ignored.

You also said:

The voltage of the wire, and it being neutral in the rest frame is a boundary condition determined by the experimenter, ignoring the voltage difference between both end which causes the current.

I agree, but as I understand it, this does not mean that it is observed as such (due to relativistic effects from the viewpoint of an observer with a different reference frame).
For example, take an infinite long uncharged straight current carrying wire (exact the same amount of electrons and protons). Than what is the correct reference frame for a charge particle to feel no force? Is it when it is stationary with the electrons or is it when it is stationary with the protons?

 

[Dale]

For example, take an infinite long uncharged straight current carrying wire (exact the same amount of electrons and protons). Than what is the correct reference frame for a charge particle to feel no force? Is it when it is stationary with the electrons or is it when it is stationary with the protons?

The wire is only uncharged in a single reference frame. If the test charge is stationary in that reference frame then it will experience no force. Therefore, in any other frame moving at velocity v wrt the uncharged frame the net force on the test charge must also be 0 if the charge is moving with velocity v.

 

[Per Oni]

For example, take an infinite long uncharged straight current carrying wire (exact the same amount of electrons and protons). Than what is the correct reference frame for a charge particle to feel no force? Is it when it is stationary with the electrons or is it when it is stationary with the protons?

For a normal copper conductor: when it's stationary with the protons ie visible wire.

 

[Hans de Vries]

@Hans:
From your paper "The simplest, and the full derivation of Magnetism as a Relativistic side effect of electroStatics" after equation 8: "We see that the field of the moving electrons is the same as the field for the electrons at rest. The result is independent of the speed of the electrons."

Don't get me wrong, I'm just trying to understand, but how does this match with the fact that a magnetic force can be transformed into an electric force in the correct reference frame, i.e. the frame of the observer?

The electron density is only equal to the ion density in one
single reference frame. If the test charge is at rest in this
reference frame then it feels no force. The speed of the
electrons and ions is not important, They can move both
in this reference frame but the only criteria is that the charge
densities are equal.

The proof that the field of the wire is only dependent on the
densities and not the speed, even though the individual charge
fields are velocity dependent, is given in section 2 here.
http://physics-quest.org/Magnetism_from_ElectroStatics_and_SR.pdf

The reason that the densities of the electrons and ions are
different in all other reference frames is non-simultaneity.
For instance: more electrons have entered the wire if Δt
is positive there, and consequently less electrons have left
the wire at the other end because Δt is negative at that
end of the wire.

Than what is the correct reference frame for a charge particle to feel no force? Is it when it is stationary with the electrons or is it when it is stationary with the protons?

See DaleSpam's post.

Regards, Hans

 

[maartenrvd]

The wire is only uncharged in a single reference frame. If the test charge is stationary in that reference frame then it will experience no force.

I agree. Which frame is it in the mentioned example?

Therefore, in any other frame moving at velocity v wrt the uncharged frame the net force on the test charge must also be 0 if the charge is moving with velocity v.

I disagree: A current carrying wire produces a magnetic field. When a charge is moving true that magnetic field it feels a force.
The thing I am discussing here (and I hope you are too) is that the magnetic force is transformed into an electrostatic force when all coordinates and velocities are transformed to the reference frame of the particle (see http://www.phys.ufl.edu/~rfield/PHY2061/images/Lectures_all.pdf" page 78 to 80)

 

[Per Oni]

Everything we have delt with so far is low speed. Why is DrGreg then correct?
Why do you refer to articles with Lorentz contraction in at the first place?

Hans?

 

[Dale]

I agree. Which frame is it in the mentioned example?

In this example the wire is uncharged in the rest frame of the wire (stationary protons). I described it the way I did because in addition to passing a current through the wire you could also charge it (by applying a high overall voltage) in the lab frame. If you did that then some other frame would be the uncharged frame.

I disagree: A current carrying wire produces a magnetic field. When a charge is moving true that magnetic field it feels a force.

Yes, but it also experiences a force due to the electric field. These two forces must cancel each other out (no net force) in every reference frame if they are zero in one frame.

The thing I am discussing here (and I hope you are too) is that the magnetic force is transformed into an electrostatic force when all coordinates and velocities are transformed to the reference frame of the particle

Yes, it should be clear that this must be the case since in that frame the magnetic force will be 0 regardless of the current.

 

[Per Oni]

In this example the wire is uncharged in the rest frame of the wire (stationary protons). I described it the way I did because in addition to passing a current through the wire you could also charge it

Completely and totally wrong.

 

[Dale]

Completely and totally wrong.

That's pretty amusing since the first sentence said the same thing you did in post 35. Also, it is customary to support such assertions with some evidence or logic.

 

[Per Oni]

That's pretty amusing since the first sentence said the same thing you did in post 35.

This sentence doesn’t make much sense to me.

Also, it is customary to support such assertions with some evidence or logic.

This makes much more sense.

After all that has been said in this thread you are still of the opinion that by sending a current through a wire that wire will become charged.
That is totally wrong.
However I cannot give references of a fact that doesn’t exist. The onus is on you to provide us with independent evidence in which is indicated the sign, location and magnitude of that charge.

 

[maartenrvd]

After all that has been said in this thread you are still of the opinion that by sending a current through a wire that wire will become charged.

No, the wire will exert an EM field: depending on the inertial reference frame of the observer this field is pure magnetic (this is supposed to be the wire frame) or pure electric (the moving charge frame) or anything in between. This is an observational effect and does not change anything to the actual state of the wire, so the wire remains uncharged.

Charging the wire due to connecting the wire to + and - poles is not relevant because that is not what we are talking about: we are talking about a hypothetical infinite long straight wire that has exact the same amount of positive as negative charge for every dx (can I be more specific?).

maartenrvd: I disagree: A current carrying wire produces a magnetic field. When a charge is moving true that magnetic field it feels a force.

Dalespam: Yes, but it also experiences a force due to the electric field. These two forces must cancel each other out (no net force) in every reference frame if they are zero in one frame.

I agree when you mean that "every reference frame" is not the frame of the test-charge but a random other frame (http://en.wikipedia.org/wiki/Lorentz_force should still be correct though).
The reason that I started this thread was that the usual assumption is that the force is zero when the test-charge is in the same frame as the positive charge in the wire, whereas I reasoned that that would only be the case when the positive charge is moving at the same speed as the negative charge (relative to the test charge).
This is easy to see for a wire without an electric current. Now figure out for yourself in which inertial reference frame this is for a current carrying wire (straight, infinite long, uncharged ...).

 

[Per Oni]

No, the wire will exert an EM field: depending on the inertial reference frame of the observer this field is pure magnetic (this is supposed to be the wire frame) or pure electric (the moving charge frame) or anything in between. This is an observational effect and does not change anything to the actual state of the wire, so the wire remains uncharged.

If the wire exert an EM field there must be charges on the wire responsible for that field. This a logic result of Gauss's law.

Hey maartenrvd you wrongly attributed some words to me which are from DaleSpam.
Its an easily done but can you still rectify them?

 

[maartenrvd]

@Per oni: Sorry about that. I corrected it in post 8.

 

[Dale]

That's pretty amusing since the first sentence said the same thing you did in post 35.

This sentence doesn’t make much sense to me.

OK, the first sentence was:

In this example the wire is uncharged in the rest frame of the wire (stationary protons).

Which is exactly what you said in post 35:

For a normal copper conductor: when it's stationary with the protons ie visible wire.

Surely you can understand my amusement. You and I say the same thing, but when I say it you call it "completely and totally wrong".

After all that has been said in this thread you are still of the opinion that by sending a current through a wire that wire will become charged.
That is totally wrong.
However I cannot give references of a fact that doesn’t exist. The onus is on you to provide us with independent evidence in which is indicated the sign, location and magnitude of that charge.

I will try to be as clear as possible. Any wire has some resistance, capacitance, and inductance. If there is an overall voltage applied then it will be charged (C=Q/V). If a voltage difference is applied across it will have a current (dV=IR). So if you have an arrangement like this:
(V+dV/2) -/\/\/\/\/\/- (V-dV/2)
then V determines the charge and dV determines the current.

Again, there is no relativistic reason for the charge or the current. They are under experimental control and must be supplied as boundary conditions (ie nature does not set them). Once you have specified the charge and the current in one reference frame then you can use relativity to determine the charge and the current in any other frame.

 

[Per Oni]

I will try to be as clear as possible. Any wire has some resistance, capacitance, and inductance. If there is an overall voltage applied then it will be charged (C=Q/V). If a voltage difference is applied across it will have a current (dV=IR). So if you have an arrangement like this:
(V+dV/2) -/\/\/\/\/\/- (V-dV/2)
then V determines the charge and dV determines the current.

I’ve got no problem with this part.

The problem is your statement that you can charge a wire by sending a current through that wire.
What is the relation/dependency between that charge and current?

 

[Dale]

The problem is your statement that you can charge a wire by sending a current through that wire.

When did I say that? In a given reference frame the charge and the current are independent. That is why in one reference frame they must both be supplied as boundary conditions.

Once you have independently specified the charge and the current in one reference frame then you can use the Lorentz transforms to determine the charge and the current in any other reference frame.

 

[Per Oni]

When did I say that?

Post #39

in addition to passing a current through the wire you could also charge it

 

[Dale]

I said you could charge the wire in addition to passing a current through it. I did not say that you could charge a wire by passing a current through it.

Again, the charge and the current are independent boundary conditions determined by the setup, not nature (as I showed earlier and which you apparently agree with). I think the rest was simply a miscommunication.

EDIT: I just noticed that my post 39 was completely explicit on this point if it had not been quoted out of context:

in addition to passing a current through the wire you could also charge it (by applying a high overall voltage) in the lab frame.