# Electric and magnetic fields

1. Nov 28, 2006

### questionmark

hi,

i'm puzzled.

if an observer stands next to a length of wire with a dc current in it, he will detect a circular magnetic field around the wire due to the movement of the electrons, but no electric field due to the cancellation of the electrons' and protons' electric fields.

what happens if the observer now hitches a ride on one of the moving electrons in the wire (cet par), what does he measure now and why?

2. Nov 28, 2006

### Meir Achuz

In practice, it would be a very bumpy ride, because the electrons actually bounce between molecules in the wire, resulting in a "drift velocity" that is much slower than their velocity between collisions.
If she moves outside the wire at this drift velocity, a Lorentz transformation would give the fields she sees in her rest system.
B would change to \gamma B, and there would be an E field given by
E=\gamma(vXB).

3. Nov 28, 2006

### masudr

Hang on: she would see positive charges (the ions) moving in the opposite direction at the drift velocity and so still see a magnetic field, but of the opposite sign.

Meir Achuz's analysis only holds for beams of electrons, where moving with them truly means seeing a stationary E-field.

4. Nov 28, 2006

### pervect

Staff Emeritus
Meir's analysis is completely general - the electromagnetic field will transform according to the standard Faraday tensor transformation laws: see

http://scienceworld.wolfram.com/physics/ElectromagneticFieldTensor.html

I believe the analysis is correct, too, though it's possible we've both made an error.

The transformation laws on the webpage above are given for a boost in the z direction.

If we take a point at some distance x=L, y=0 from the wire, the electromagnetic field in the rest frame of the wire will be

Ex = Ey = Ez = 0 = Bx = Bz = 0, By=B

(depending on the direction of current flow in the wire for the sign of B)

The transformation laws say that the only field components in the boosted frame are

By' = gamma*B, Ex' = -gamma*v*B

There have been a lot of threads on how to explain this, one of the better explanations it that the Lorentz transformation squashes the field of each moving electron so that it is not round, but ellipsoidal. Various other related effects happen as well, the distance between charges gets Lorentz contracted.

If you look, you'll see some past threads on this issue - one of them is

and there have been many others as well.

5. Nov 28, 2006

### masudr

I'm not denying that. All I'm saying is that if you transform to an inertial frame so that the electrons/charge carriers are at rest, there will be a net motion of positive charges in the opposite direction.

If the electrons in a wire did not come from atoms, then indeed a frame moving with the electrons will not perceive any magnetic fields. But most wires are electrically neutral (i.e. every negative charge carrier has a corresponding positive charge), so if we transform to an inertial frame where the electrons are at rest, we will see that there are ions moving in the opposite direction, so we will still get a magnetic field.

I know how that the EM tensor is indeed a tensor, and transforms under a change of inertial co-ordinates with 2 copies of the Lorentz transformation. This is not what is at debate. Instead I am talking about something far more basic -- essentially that a wire is usually neutral, and the consequences of that.

6. Nov 28, 2006

### marcusl

I'm not following your point. Did someone say there wouldn't be a magnetic field if a test charge moved parallel to and at the same speed as the charge carriers?

7. Nov 28, 2006

### cesiumfrog

..but isn't the drift velocity always incredibly small?

8. Nov 28, 2006

### marcusl

Sure. Isn't that why the magnetic force between two current carrying wires is incredibly small compared to that between two electrically charged wires? In fact it's down by order (v_drift / c)^2.

9. Nov 30, 2006

### swimmingtoday

Your method is corect, but you should have tried to explain the discrepancy between your result and Meir's. Two seemingly correct ways of doing the problem gives different results, yet you are not troubled by this. (Se my next post)

10. Nov 30, 2006

### swimmingtoday

11. Nov 30, 2006

### marcusl

No, that's not quite right. The magnetic field must be in the same direction regardless of whether the test charge is moving more slowly or more quickly than the electron drift velocity. It's not the apparent direction of ions that matters; it's the relative charge densities of electrons and ions after Lorentz contractions are applied to each.

I think you got this backwards. There's no E in the lab frame, if there were then free electrons in the circuit would be attracted until the excess charge were neutralized. This agrees with observation, too (your hairs don't stand up when you stand next to house wiring, nor next to a battery cable carrying 200A as you crank you car, nor next to a superconducting magnet carrying thousands of amps). The neutrality in the lab frame is precisely why you see a field once you move.

12. Nov 30, 2006

### pervect

Staff Emeritus
I'm not quite sure I'm following your point, unfortunately.

Note that the total net charge on the wire is a physical quantity that's Lorentz invariant. The total charge on the entire wire must be specified for the problem to be analyzed and for the electric and magnetic fields to be calculated in the rest frame. It is usually assumed that the total charge on the wire is zero in the rest frame of the wire, but it's certainly possible to have a charged loop of wire. However, the initial poster explicitly said that the electric field in his sample problem was zero.

By Gauss's law, no electric field implies no net charge on the wire, by drawing a Gauss surface around the entire loop of wire.

So we know from the problem statement that the wire is on the average, neutral. We can think of this neutral electrical wire as having +N positive charges and -N negative charges, for a total charge of zero.

These charges re-destribute under a Lorentz boost due to the relativity of simultaneity. This allows charges to redistribute like this for a sample square loop of wire (the boost in this diagram is in the left-right direction, one can assume the current flow is a clockwise loop).

Code (Text):

++++++++
x       x
x       x
x       x
----------

There can and will be electric fields after the boost, but this is and must be compatible with the total charge of the wire being constant. There will still be N positive charges and N negative charges in the loop above, but their distribution will be different after a boost.

13. Nov 30, 2006

### swimmingtoday

<<I think you got this backwards. There's no E in the lab frame, if there were then free electrons in the circuit would be attracted until the excess charge were neutralized.>>

The E field is perpendicular to the wire, so it would not attract charges from the battery.

:<<This agrees with observation, too (your hairs don't stand up when you stand next to house wiring,>>

That is probably due to the quantitative weakness of the effect.

If I walk past a magnet there will be an effective eelectric field (for example viewed in the inertial frame where I am at rest), but my hairs have never stood up when walking past a magnet--the effect is too quantitatively small.

<<The neutrality in the lab frame >>

Consider a wire with no current. The density of the positive charges is equal to the density of the negative ones. So the densities of the negative charges must be different from the density of the positive charges when the negative charges start moving. For the densities to be the same in the wire with current, you must either claim that they are not the same in the wire with no current, or you must reject the Lorentz contraction--neither of which I think you would want to do.

14. Nov 30, 2006

### Staff: Mentor

Consider a loop of wire that initially has no current. Current carrying wires must ultimately be part of a complete circuit, at least in steady-state current situations. Induce a current in the loop by using a changing magnetic field. If the number density of the moving electrons increases because of Lorentz contraction, we need to add more electrons in order to maintain that increased density all around the wire. Where do those electrons come from?

This seems to be related to the sometimes-posed question that starts with a series of stationary objects laid out along a line, not connected to each other. Accelerate the objects up to a relativistic speed. Does the distance between them shrink because of Lorentz contraction, or does it stay constant? The answer turns out to depend on the details of how the objects are accelerated, because of the relativity of simultaneity.

15. Dec 1, 2006

### swimmingtoday

<<Consider a loop of wire that initially has no current. Current carrying wires must ultimately be part of a complete circuit, at least in steady-state current situations. Induce a current in the loop by using a changing magnetic field. If the number density of the moving electrons increases because of Lorentz contraction, we need to add more electrons in order to maintain that increased density all around the wire. Where do those electrons come from?>>

Consider a loop spinning around an axis perpendicular to the plane of the loop going thru the center. The ratio of the circumference to the radius is no longer 2 pi, because of Lorentz contractions of the pieces of the moving circumference. Now back to the scenario you propoesed. Treat the wire as two loops, a loop of positive charge and a loop of negative charge. When there is no current the loops have equal circumferences. When there is current, the negative loop is spinning and has a decreased circumference. So the two loops occupy the same space, having the same radius, yet one has a smaller circumference. Freaky, eh?

So the negative loop has a greater magnitude of charge density because it has the same magnitude of charge but a shorter length

Now we argue that the charge density is the thing determining the electric field. This is a bit uncomfortable for me, and for regions far from the loop, it actually might not even be true but it is true "close" to the loop, which is the region the matters in the original problem of making the Lorentz transformation to a frame where the electrons are at rest. So we see that, at least in the relevant region (close to the wire), the E field from the positive charges is not large enough to cancel out the field from the negative charges.

If you are uncimfortable with this, consider that if what I said does not work out we have a REAL unresolvable problem, the original paradox I pointed out several posts ago--that making a Lorentz transformation of the charges in a wire so as to have the electrons at rest gives a different magnetic field than one would get by performing that Lorentz transafornation directly on the electric and magnetic fields.

16. Dec 1, 2006

### ZapperZ

Staff Emeritus
swimmingtoday: Could you please use the QUOTE function that is available in each post (see lower right-hand corner of each post)? This will make your replies a lot easier and clearer to read, especially when you make multiple replies to a number of people.

Zz.

17. Dec 1, 2006

### swimmingtoday

I'm quite ignorant about computer stuff, so I'll tell you what I do, and you can tell me specifically how to do it better.

I hit the quote button whenever I respond to a post, so the person I respond to will get an email notifying him. Then when I want to post an actual quote, I usually do not want to quote the *whole* post I am referring to, but rather specific excerpts. So I just delete out everything but the specific excerpts.

If I do not delete out certain things, like the word "QUOTE" will that make it work the way you want? Do I need to write out "QUOTE" both before and after something I want to quote? As I said, I know almost nothing about computer stuff.

18. Dec 1, 2006

### ZapperZ

Staff Emeritus
You seem to have done it corectly here. However, I want to correct something. The person you are replying to will NOT get an e-mail notification that you have replied, unless he/she chosed to have e-mail notification (which I believe is not the default setting).

If you simply want to make specific quotes of a part of the post, then make sure you bookend the phrase with a [...quote] and [.../quote] (no periods). If you want to make it clear it was a quote from someone, then start with [..quote=ZapperZ] and end it with [../quote] as usual. This will put that member's nickname in the quotation.

Is that all clear now?

Zz.

19. Dec 1, 2006

### marcusl

Untrue for the case you consider of a localized region where the charges don't cancel.

20. Dec 1, 2006

### pervect

Staff Emeritus
The electric field inside a perfect conductor is zero pretty much by defintion. We'd have to jump through some hoops to handle the case of a real wire with an ohmic drop - I'm not really in the mood to do that, though, hopefully you aren't either.

Which result is that? COuld you quote the post in question?

There is no net electric field in the rest frame of an uncharged wire when the wire is a perfect conductor. The fact that there is no radial electric field is obvious from Gauss's law. See for instance:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecyl.html#c2

where the Gauss surface is a cylinder enclosing the entire wire loop. When the net enclosed charge is zero, the radial electric field around the wire is zero. This relies on symmetry - for a symmetrical circualr loop, the radial electric field must be independent of the angle theta.

So if we have an uncharged wire with no current flowing through it, then we complete the circuit, the wire will still be uncharged, and there will be no net radial electric field after we complete the circuit.

you'll see this has been discussed (and that wasn't the first time, either).

Another simple way to express this fact is that the voltage in a perfect conductor is constant. For an uncharged wire loop, the voltage is everywhere zero. When you run a current through the wire, the voltage is still zero. It doesn't change.

The electric field is the gradient of the voltage - but this is easy to calculate, since the voltage is everywhere zero. This implies that the electric field is zero, too.

In short, solve the problem classically first, using the methods of Maxwell's equations and/or electrical engineering, then make sure that your relativistic analysis gives the same result.

21. Dec 5, 2006

### swimmingtoday

Really? Oh, thank you so very much.

Umm, you do not appear to have been paying attention to what I said. I told you that a wire with curent in it creates an electric field. I was not saying that the fileld was in the wire. The field is in the vacuum region, and there is also an electric field on the surface of the wire perpendicular to the surface.

What appears to be obvious to you is simply wrong. (Indeed, ypou ability to discern "obvious" is empirically demonstrated to not be correct from the fact that you felt you needed to teach me Gauss' Law) )This is what is happening. Consider a wire to be the superposition of a a line of positive charges and a line of negative charges. If there is current in the wire then the line of negative charges is moving, and has become Lorentz contracted due to this movement. The contraction leads to it having a greater charge density, and the charge density, not the charge, is the key quantity, so the negative charges create an E field not balanced by the E field of the postive charges.

The E field in tha vacuum region will simple go as the difference in charge densities, multiplied by 1/r. Any E filed in the wire itself will cause charges to move to the surface to destroy the interneal E field without affecting the external field--the charge densities will still be unbalanced.

Wrong. when the circuit is completed, then even though the wire has no net charge, it will have unbalanced current densities. Should I explain Gauss' Law to you?

I'll explain again. This time LISTEN, so I do not have to sit thru a response from you that only indicates that what I said went way past you. Listen CAREFULLY.

I am going to guide you thru two different ways of analysing the same problem, each way seemingly valid. I am going to do the derivations using your belief that here is no elctric field from the wire. And we will see that the two answers are NOT the same-- a paradox will arise if your belief that there is no E filed is true.

Method 1) A wire has current in it such that the electrons are moving to the East with a speed of v when observed in Inertial Frame 1. The magnitude of the B field will be called B1. We make a Lorentz transformation to an inertial frame where the electrons are at rest, and protons are moving with a speed of v to the East. In this frame we call the magnitude of the B field "B2". In the second frame the velocities are reversed (the protons move to the West) but the charges aree also reversed (it is protons, not electrons doing the moving) and so because negative one times negative one equals positive one, it is trivial to see that B2 must turn out to be equal to B1 by direct use of the Maxwell equation relating the curl of B to the J. Right? So we conclude, using this method that the magntic field is the same in both frames.

2) Now do the problem this alternate way: In the first frame there is a magnetic field, and there is an electric field you claim has a magnitude of zero. To calculate the magnetic field in ther second frame we use the Lorentz transformation formula for electric and magnetic fields. The Lorentz transformation gives

B2 = gamma ( B1 - (v/c)E1)

You think E1 is zero, so you must conclude that B2 = gamma (B1) SO WHEN WE DO THE CALCULATION *THIS* WAY, USING YOUR BELIEF THAT THERE WAS NO ELECTRIC FIELD, WE GET A DIFFERENT RESULT FROM THE RESULT WE GOT USING METHOD1 ABOVE. Instead of the magnetic field being the same in both frames, we now get that hey differ by a factor of gamma!

There is yet another way to generate a paradox if your belief was correct.

Suppose that in frame 2 there was no electric field, as you believe. Let's calculate the elctric field in frame 1 using the Lorentz transformation. (I'm always sloppy with minus signs): E1 = gamma ( E2 - (v/c)B2). So we now have to conclude that the elctric field in frame 1, contrary to your belief, does not vanish. In other words, you cannot believe both that a neutral wire with negative charges moving East and also a neutral wire with positive charges moving West have vanishing electric fields, yet you believe both situations involve no electric field produced by the wires.

22. Dec 5, 2006

### pervect

Staff Emeritus
I may come across as being a bit bored with this topic. That's because, unfortuantely, I am.

It would probably help if I had the patience to sit through yet another incorrect explanation of the same wrong answer to the same problem, and carefully point out the errors. And in some sense that's what this board is for, although some might argue that it is better if the people who don't know the correct answers listen to the people who do, rather than visa-versa :-).

Unfortunately, while this is arguably what the board is for, I don't quite feel like it at the moment, and being told to "listen carefully" to this incorrect explanation isn't quite sitting well for me either.

All I can say is that this is a perfectly standard problem that can be worked out by non-relativistic means, and also by experiment.

And I'll add that I've posted two different ways to get the same answer, both of which you can probably find in a standard textbook, and that this answer is diiffrent from the one you get. Therfore, I am pointing out that as obvious as some of the assumptions that you are making may seem to you, some of them are in fact incorrect.

If we have an uncharged loop of wire, together with an uncharged current source (say a battery + resistor), both at zero volts, they do not have any electric field.

Now, when the conductor is an ideal conductor, and we put a current through it, its voltage does not change. It is an equipotential surface. This means that it is still at zero volts. Real conductors are slightly more complicated due to ohmic resistance, but the voltage generated by a fairly large current through a good conductor is going to be very small (though not quite zero). It's simpler to take the case of an ideal conductor.

This is the EE approach, which to put it briefly is that a conductor is an equipotential surface, and that the potential satisfies Laplace's equation in the absence of charge (Poisson's equaton in the presence of charge). We know that the potential is constant in the conductor, and we know that there are no charges outside the conductor, so the potential satisfies Laplace's equation outside the conductor (and is constant inside the conductor).

The trivial solution to Laplaces eqation for an equipotential surface of V=0 on the conductor and V=0 at infintiy is that V=0 everywhere. This implies no electric field (the field is the gradient of the potential).

I'll add a brief comment that the setup of the problem is important. We are assuming that the wire is initially uncharged, and that it remains uncharged. If these conditions are violated, the above solution doesn't apply. These are key points in the defintion of the problem.

The very standard Gauss law argument which I presented earlier is a more basic physics approach to the same problem, and it gets the same answer. It is more basic (and taught even earlier) than the EE approach involving the potential and Laplace's equation which I outlined above.

In any event, if you have an isolated loop of wire, and you run a currrent through it, the wire does not suddenly pick up an electrical charge, which is something that can be confirmed experimentally. Charge is conserved, it does not suddenly appear out of "thin air".

If you are not following the standard Gauss law argument (which I assume you are in fact not following based on your responses) you have a hole in your education that you need to fill. By skipping the basic stuff, you are not realizing when you are getting an incorrect answer by attempting to apply relativity to a situation that you do not previously understand, having "leaped over" the coursework that would enable you to follow the standard arguments, solve the standard problems, and get the correct results (which can also be confirmed experimentally).

Note that Gauss's law (and Maxwell's equations) are fully covariant (i.e. compatible with relativity). They don't need to be "fixed up" in any way to "correct" for relativity. In fact, Maxwell's equations were in a major sense an inspiration for relativity, as they predicted that the speed of light was always constant.

Last edited: Dec 5, 2006
23. Dec 5, 2006

### masudr

Hum. OK, let's discuss this completely in a basic physics kind of way.

In one frame, the electrons are moving with some velocity down a wire. We experience a perpendicular magnetic field.

Now I am in another inertial frame, where the electrons are moving with me. (Note I have invoked no special relativity). I see positive ions moving in the other direction. Doing some basic classical electromagnetism, I have positive charges, but moving in the opposite direction, I still should have the same magnetic field.

This is before we invoke anything about E-fields, and Lorentz transformations.

So, is this correct?

24. Dec 5, 2006

### pervect

Staff Emeritus

The argument is good enough to show that there is always some magnetic field, but it's not good enough to show that the magnetic field is always precisely constant. A more sophisticated analysis shows that the magnetic field goes up by a factor of "gamma" in the moving frame. You can attribute this to length contraction if you are careful enough about how you go about it.

25. Dec 5, 2006

### swimmingtoday

Exactly correct.

And now try to do the problem by using a Lorentz Transformation of the elctromagnetic field. You will see that if the E field were zero, the B field will be different in the two frames (by a factor of gamma), contradicting the calculation you just laid out. So the E field could not actually be zero!