Charged particle in a magnetic field (1 Viewer)

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Emanresu

When two current carrying wires are place beside each other they
are are either attracted or repelled by the magnetic fileds generated
and these fields are just electrostatic forces caused by the relativistic
effect.

Why does a stationary charged particle not feel this force from a
single current carrying wire ? If I have understood the two wire case
correctly then the charged particle should see a net negative charge
in the current carrying wire due to lorentz contraction of the current.

E.

Meir Achuz

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Emanresu said:
When two current carrying wires are place beside each other they
are are either attracted or repelled by the magnetic fileds generated
and these fields are just electrostatic forces caused by the relativistic
effect.

Why does a stationary charged particle not feel this force from a
single current carrying wire ? If I have understood the two wire case
correctly then the charged particle should see a net negative charge
in the current carrying wire due to lorentz contraction of the current.

E.
I think the first example is one of Feynman's clever arguments that works when it works, but is not really general.

George Jones

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When applied to this situation, doesn't this argument give that the stationary charge sees equal Lorentz contractions for positive and negative charges in the wire, since these charges move with equal speeds (opposite directions) with respect to the stationary charge?

pervect

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Emanresu said:
When two current carrying wires are place beside each other they
are are either attracted or repelled by the magnetic fileds generated
and these fields are just electrostatic forces caused by the relativistic
effect.

Why does a stationary charged particle not feel this force from a
single current carrying wire ? If I have understood the two wire case
correctly then the charged particle should see a net negative charge
in the current carrying wire due to lorentz contraction of the current.

E.
You probably have missed some of the subtle points if you are reaching this conclusion. Note that relativity is completely compatible with Maxwell's equations, specifically Gauss's law.

One could in fact argue that Maxwell's equations are the inspiration for relativity ,predicting as they do that light has a constant speed, but for our purposes it is sufficient to note that Gauss's law must be valid relativistically as well as non-relativistically.

If one has an electrically neutral wire, Gauss's law shows that the radial electric field of that wire must be zero, regardless of whether or not it is carrying a current.

While Lorentz contractions do explain the magnetic field, it is important to realize that the correct boundary conditions for an uncharged wire is "no electric field". One can express the same conclusion in the language of potential by saying that the electric potential of an uncharged current-carrying wire is 4-volts.

A similar treatment based on potential can explain magnetic fields in detail, but one must use the electromagnetic 4-potential, which is a 4-vector. See for example

http://en.wikipedia.org/wiki/Electromagnetic_four-potential

A treatment in terms of potential is convenient, though not necessary. However, it is always necessary that an uncharged object satisfy Gauss's law - the intergal of the normal force over an enclosing boundary is always zero when the boundary contains no charge.

marcusl

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The densities of positive charges (positive ions of the copper atoms in the metal) and negative charges (free electrons) in the wire are equal in the lab frame [there can be no net charge], so a stationary test charge sees no net electric field and feels no force. If the test charge moves towards or away from the wire (perpendicular), there is no Lorentz contraction and the situation is the same as a stationary charge. If the charge moves parallel to the wire, however, its velocity relative to the ions differs from that relative to the electrons since ions are stationary in lab frame while the valence electrons drift, so the Lorentz contractions differ and so do the apparent charge densities. The wire appears electrically charged and the test charge is attracted or repelled depending on sign. In the lab frame we call the force a magnetic field.

Note that the key features of the Lorentz force are explained: The force is felt by a test charge proportional only to the portion of velocity parallel to the current, and the force is normal to the motion.

Look at Mel Schwartz's beautiful derivation in his undergrad book "Principles of Electrodynamics" for full details.

pmb_phy

Emanresu said:
When two current carrying wires are place beside each other they
are are either attracted or repelled by the magnetic fileds generated
and these fields are just electrostatic forces caused by the relativistic
effect.

Why does a stationary charged particle not feel this force from a
single current carrying wire ? If I have understood the two wire case
correctly then the charged particle should see a net negative charge
in the current carrying wire due to lorentz contraction of the current.

E.
The physics is given in my website at
http://www.geocities.com/physics_world/em/rotating_magnet.htm

It also gives the math to understand it.

Pete

Emanresu

Okay. Say we take two cases, one where the particle is at rest in
the lab's frame of reference and the other where the particle is at
rest in the currents (moving electrons) frame of reference.

In the second case the particle sees a Lorentz contraction for
the positive ions and thus is acted upon by a net positive charge.

In the first case I would have thought that the particle would have
seen a Lorentz contraction for the electrons and thus would have
been acted upon by a net negative charge but this is not the case
and I don't understand why not.

E.

George Jones

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Emanresu said:
Okay. Say we take two cases, one where the particle is at rest in the lab's frame of reference
In this case, the lab electron sees a Lorentz contraction for the moving electrons, just as you say. However, there are positively charge holes flowing down the wire in the other direction, which the lab electron also sees Lorentz contracted. The net result is that the lab electron sees the charge density of the wire to be zero.

and the other where the particle is at rest in the currents (moving electrons) frame of reference.

In the second case the particle sees a Lorentz contraction for the positive ions and thus is acted upon by a net positive charge.
Right.

Emanresu

George Jones said:
In this case, the lab electron sees a Lorentz contraction for the moving electrons, just as you say. However, there are positively charge holes flowing down the wire in the other direction, which the lab electron also sees Lorentz contracted. The net result is that the lab electron sees the charge density of the wire to be zero.
Are the positive ions not the holes and we only 'say' they move ?
If they don't actually move can they be Lorentz contracted ?

E.

George Jones

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Emanresu said:
Are the positive ions not the holes and we only 'say' they move ?

If they don't actually move can they be Lorentz contracted ?
If this is the case, I guess, then, that the electron spacing must be Lorentz contracted in such a way that the total charge density is zero in the lab frame, as others have said.

I think I saw these types of arguments first in Purcell, and then in Griffiths.

As marcusl has said, Schwartz is very good at relating electromagnetic effects and relativity.

George Jones

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Interesting - I just had a look at Purcell and Griffiths.

Griffiths's treatment has both the positive and negative charges moving in the lab frame as I said, whille in Purcell the positive ions are stationary in the lab frame, as you said.

marcusl

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George, thanks for your comments. Actually, I looked at my copy of Schwartz at work today and he also assumes equal plus and minus currents flowing in opposite directions (not like I said!). This is indeed interesting! I always understood holes to exist in semiconductors where conduction occurs when an electron from the filled shells near the Fermi surface is boosted into the valence band, and not in metals which have free valence electrons even at zero temperature. Can any solid-state experts comment on this? Do holes carry currents in metals such as copper?

I made another mistatement in my earlier email:
Schwartz observes that there is still a Lorentz force if the test charge is moving perpendicular to the current (doh! what was I thinking when I said otherwise?). The electrostatic analogy fails here, so a relativistic four-vector treatment is required and the electric and magnetic fields are written as elements of a second rank tensor. The mixture of fields we observe in an experiment comes from the transformation applied to the tensor, which in turn depends on the geometry and dynamics of the situation. It is a pretty theory, and I had forgotten all of these details in the thirty years since I took the class. I have learned a lesson here: check the reference before typing!

Emanresu

marcusl said:
Schwartz observes that there is still a Lorentz force if the test charge is moving perpendicular to the current (doh! what was I thinking when I said otherwise?). The electrostatic analogy fails here, so a relativistic four-vector treatment is required and the electric and magnetic fields are written as elements of a second rank tensor.
I thought Lorentz contraction was an explanation not an analogy. Is
it then the case that you need a brain the size of a planet to
understand this - you can see these equations in four-vectors the way
Neo sees the Matrix ?

It's obviously more complicated than I thought if perpendicular movement
results in a force but no movement doesn't.

Come on guys, I know you're out there, I can feels the gravitational
pull of your brains.

E.

marcusl

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Yeah, it's more complicated than I remembered as I sat at my computer at home on Saturday night. Sorry! I don't know about needing a planetary sized brain to get it all, but maybe meteoric...

I'll quote Schwartz, pg. 125:
"So we have made a major discovery. A charged particle moving in the proximity of a current distribution has a force acting on it, even though there is no other charge density present! Unfortunately, though, life is not quite so elementary, for if we try to project the charge q in the y direction (toward the axis of the current distribution), we run into trouble with our simple reasoning." ... "What do we do now? Do we conclude that the charge q will now have no force on it? Of course not! We have gone outside the area of validity of electrostatics because we now see a current distribution which is changing with time." ... "So we must go back and think out our whole problem from scratch."

The new approach, then, is relativistic and fully time dependent. Poisson's equation is replaced with one that is invariant under Lorentz transformation: the vector potential becomes a four-vector, the Laplacian is replaced by the D'Alembertian operator
d^2 / dx^2 + d^2 / dy^2 + d^2 / dz^2 - 1/c^2 * d^2 / dt^2,
and electric and magnetic fields are components in a tensor. The Lorentz force, and all four Maxwell equations, follow quickly. The approach is elegant, but not elementary.

I recommend looking at Schwartz, "Principles of Electrodynamics", which has been reprinted by Dover. George mentioned that it's also in Griffith's E&M book, as well as Purcell--that would be vol. 2 of the Berkeley Physics series. (Since Purcell is a first-year book, he probably covers only the motion parallel to the wire).

selfAdjoint

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Emanresu said:
I thought Lorentz contraction was an explanation not an analogy. Is
it then the case that you need a brain the size of a planet to
understand this - you can see these equations in four-vectors the way
Neo sees the Matrix ?

It's obviously more complicated than I thought if perpendicular movement
results in a force but no movement doesn't.

Come on guys, I know you're out there, I can feels the gravitational
pull of your brains.

E.

What you need is not a brain the size of the Andromeda Galaxy, but ALGEBRA! You don't have to "see" the four-vectors if you know how they behave algebraically. Think of playing the game "battleship"; you don't need to visualize your opponent's ships if you can figure out their coordinates. Same here.

marcusl

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I could use more practice in algebra too

How about holes in copper?

Emanresu

selfAdjoint said:
What you need is not a brain the size of the Andromeda Galaxy, but ALGEBRA! You don't have to "see" the four-vectors if you know how they behave algebraically. Think of playing the game "battleship"; you don't need to visualize your opponent's ships if you can figure out their coordinates. Same here.
I don't think doing the math is the same as understanding why. I'm
sure a lot of you guys can do the math for quantum physics but you
don't understand why wave particle duality exists, and in that case
you do have to just accept it. The twin non-paradox, however, can
be explained and I understand the explanation (little or no maths
involved).

So is this wave particle duality or the twin thingy ?

E.

pervect

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Science Advisor
I'll add my \$.02 again.

Would you say that you "understand" Gauss law, or that you just "accept" it?

The key points to the problem the way I see it are this:

1) Gauss law works
2) Gauss law tells you that there will not be any net force away from an uncharged wire - you draw a cylinder around the wire, do the surface intergal, and get zero.
3) Gauss law serves as a defintion of charge, and we can "understand" or "accept" that charge is conserved.

So that answers your original question - at least in my opinion. See my comments on "verbal explanations", however.

Now, you might ask, why doesn't Gauss law tell you there will be no net force when the charge is moving and the wire is carying a current? What is different about the situation that we can have a magnetic force on a moving charge, but not on a stationary charge?

Well, what happens there is a bit more complex, basically the relativity of simultaneity makes a current carying wire which is locally neutral in its own frame appear to be locally charged. Globally, charge is still conserved, so parts of the wire loop (it must be a loop to keep charge conservation) appear to have a postive charge, and other parts appear to have a negative charge. In terms of Gauss's law, if you draw a Gauss surface around the entire wire loop, the enclosed charge is zero, but if you draw a Gauss charge along only part of the wire loop IN THE MOVING FRAME, you get a local non-zero charge density.

So there is nothing really all that strange going on, just the usual confusions about simultaneity. In one frame, the charge density appears to be uniform and zero, in another frame, because of differences in sumultaneity, the charge density doesn't appear to be uniform. In all cases, though, charge is conserved.

The key point is that the wire loop is globally uncharged because we made it that way. Physically, we could also charge the wire loop if we wanted to, then we would have a charged wire loop.

The other key point is that symmetry is what makes the globally uncharged wire loop appear to be locally uncharged, even when it has a current flowing through it, in the lab frame.

I think, though, that if you read enough of these verbal explanations, you'll come to appreciate the mathematical explanations more. Mathematical explanations generally are EASIER to communiate, not harder, as you seem to think.

Diagrams would certainly be valuable (I was talking about Gauss law and the total charge enclosed by various surfaces) but they're hard to draw. I know what diagrams I had in mind when I wrote this, but when you read this you may find it hard to figure out just what diagram I had in mind :-(.

Last edited:

pervect

Staff Emeritus
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George Jones said:
Interesting - I just had a look at Purcell and Griffiths.

Griffiths's treatment has both the positive and negative charges moving in the lab frame as I said, whille in Purcell the positive ions are stationary in the lab frame, as you said.
I can't recall the name of the effect offhand, but when you put a current through a loop, the loop experiences a torque - IIRC.

It seems to me that the correct model has the ions stationary in the lab frame, but I might be fooling myself here. I.e. when we perceive a current carrying wire as "not spinning", I think that the positive ions are stationary, and that the wire remains stationary because we've applied a torque to it while we run the current through it.

marcusl

Science Advisor
Gold Member
Emanresu said:
I don't think doing the math is the same as understanding why. I'm
sure a lot of you guys can do the math for quantum physics but you
don't understand why wave particle duality exists, and in that case
you do have to just accept it. The twin non-paradox, however, can
be explained and I understand the explanation (little or no maths
involved).

So is this wave particle duality or the twin thingy ?

E.
I'd say that your two options don't span the space. In addition to a) axioms that are accepted and b) physical explanations that can be understood intuitively with little math, there are c) explanations that are more complex and require either additional background knowledge or mathematical sophistication or both. (Then there's string theory, which doesn't conform to any of this...) In the present case,
a) Coulomb's law and the constancy of the speed of light are axiomatic, and we accept them as the foundations of E&M theory and relativity
b) for the case of a test charge moving parallel to the wire, the identification of M as a manifestation of E under conditions of motion follows intuitively from relativity as we have discussed
c) the identification of M as a relativistic transformation of E under general conditions, leading to Maxwell's equations and the full E&M theory, requires the above plus greater mathematical sophistication. I would say it is reasonable and believable, however, because of the reasonableness and believability of the special case (b).

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