Lorentz Force (magnetic force term)

[DJ_Juggernaut]

Hi all. In the Expert Notes forum a post (Electricity is Magnetism) addresses the magnetic force term in the Lorentz Force equation as follows:

But what happens if you jump to the frame where the particle is not moving? With a velocity of zero, the magnetic force is zero.

As I understand it, the v in the Lorentz force equation is the relative velocity between the magnet and a moving charge. It does not matter whether the magnet or the charged particle moves. All that matters is the relative velocity. Therefore, the v is never zero in either frame in the context of a moving charge (q) in a magnetic field (B). Therefore, the magnetic force term [qv * B] is not zero in either frame.

If not, my question is: why is the v in Lorentz Force equation not relative? That is, in the frame where the particle is not moving, why should you ignore the velocity of the moving magnet (v), in the context of the magnetic force term [qv * B].

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Electricity is Magnetism:
http://sciencechatforum.com/viewtopic.php?f=84&t=8621

Moving conductor in a magnetic field:
https://en.wikipedia.org/wiki/Moving_magnet_and_conductor_problem#Conductor_frame

A charge q in the conductor will be at rest in the conductor frame. Therefore, the magnetic force term of the Lorentz force has no effect,

[DJ_Juggernaut]

If you have a charged particle moving in the presence of a current, you will get a magnetic force.

Yes, the moving charged particle experiences a force given by F = qv * B, where v is the relative velocity between the particle (q) and the wire/magnet (B).

When you switch the frame: The particle is at rest. It is the current carrying wire (or magnet) that is moving at v, and therefore the stationary particle experiences a magnetic force via F = qv * B even though it is not moving. My question is, how can the magnetic force term be zero when you switch to the particle's frame?

Isn't the v in F = qv * B, the relative velocity; meaning the v is never zero between the particle and the magnet/wire?

On the other hand, if you are moving along at the same speed as the current, there is no current and, consequently, you will have no magnetic force.

As I understand it, it does not matter what direction a charged particle (q) moves relative to a current (carrying wire or magnet). The magnetic force will not be zero, as long as there exists a relative velocity (v) between the wire/magnet (B) and an external charged particle (q).

[TheVat]

Lincoln, over here.

Oops, sorry, DJ. Thought permissions included that forum, but I see now they don't. I should have put the tag over here.

[Lincoln]

The thing to remember is that it is possible to have a particle moving at the same velocity as the current. In that frame, there is no current.

That's the key point.

When you're talking about ordinary physics, you don't talk about the velocity of the current. You do define your velocity with respect to the wire. And current could be gobs of charge moving slowly or small amounts moving fast.

You have to distinguish between the wire, the motion of charge in the wire, and the relative motion of the charged particle with respect to both. But there are frames in which a particle in motion with respect to the wire is moving at a velocity such that when you jump to that particle's frame, there is no current.

[DJ_Juggernaut]

The thing to remember is that it is possible to have a particle moving at the same velocity as the current. In that frame, there is no current.

Lincoln, I think you are missing my question. My question hinges on whether v is relative in the following equation (particularly the magnetic force term):

F = qE + qv * B

Here's how I see it and I like to keep the scenario simple:

A current carrying wire is essentially a magnet. It has a field (B). An external charged particle (q) moves relative to this field (B). And by extension it is also moving relative to the wire. The velocity of the current is irrelevant here as it does not figure in the magnetic force term. All that matters is the magnetic field (B) and its relative velocity with respect to the charged particle (q). Hence you get:

F = qv * B in the magnet frame where the magnet is at rest and the particle is moving. And
F = Bv * q in the particle frame where the charged particle is at rest and the field lines B or the wire is moving.

Here v is the relative velocity between the field lines (B) and the charged particle (q). An indirect measure of the relative velocity v is indeed the relative velocity between the wire and the charged particle. Is any of this incorrect? If so, why?

You have to distinguish between the wire, the motion of charge in the wire, and the relative motion of the charged particle with respect to both.

Hmm. Why? Why is the relative velocity between a charged particle q and a magnetic field B not sufficient here? Where does the velocity of the current figure in the magnetic force term? Aren't q, v and B sufficient? The v in the Lorentz force equation is not the relative velocity between the current and a charged particle. It is the relative velocity between the field lines (B) and the charged particle (q). If you disagree please explain why.

But there are frames in which a particle in motion with respect to the wire is moving at a velocity such that when you jump to that particle's frame, there is no current.

As I see it, the current in a wire depends on the source or the battery that it is connected to. The only way the current disappears (for all or any frame) is when the battery runs out. Also, the velocity of the current does not figure in the magnetic force term: qv * B. Why bring in parameters that are not needed?

[Faradave]

For what it's worth, this minutephysics episode seems to address your question, though (for convenience) it assumes all electrons in the wire have the same velocity.



Also The Science Asylum



with this frequent question answered.

[Lincoln]

Say you have a particle of charge q, moving collinear with a current carrying wire.

In this frame, F = q v B. Easy.

The wire is neutral, so the electric force is zero.

Now, assume that the particle q is moving at the same velocity as the current. Boost to the frame of q. In this frame, q is stationary. But the current in the wire is also stationary. Thus I = 0, v = 0. and therefore F(magnetic) = 0.

The response is what I put in the expert forum. Relativity contracts so that the positive and negative charges in the wire experience different Lorentz contraction and the wire therefore gains a net charge. Accordingly, the particle q experiences an electric force, not a magnetic one.

That's how it goes.

I like the Science Asylum video even better. It's shorter and even more clear.

https://www.youtube.com/watch?v=p4gCTmlm5RQ

[DJ_Juggernaut]

Say you have a particle of charge q, moving collinear with a current carrying wire. In this frame, F = q v B. Easy.

In the wire frame, it's F = B v q. The v here is the relative velocity between the field lines and the charge. Do you disagree with this?

We could address the qE term in another thread.

[DJ_Juggernaut]

For what it's worth, this minutephysics episode seems to address your question, though (for convenience) it assumes all electrons in the wire have the same velocity.

Taking a look. Thanks for the links.

Edit:
I just looked at them. They don't answer my question. They are missing Biot Savart law. They should start with the premise that the Biot Savart law is valid in all frames. And when they switch to the particle frame, the field lines move relative to the stationary particle giving rise to a force: Bvq.

[DJ_Juggernaut]

Now, assume that the particle q is moving at the same velocity as the current. Boost to the frame of q. In this frame, q is stationary. But the current in the wire is also stationary. Thus I = 0, v = 0. and therefore F(magnetic) = 0. The response is what I put in the expert forum.

At first glance it did seem like you were referring to the particle's velocity as zero in its own frame and therefore the magnetic force term reduces to zero. It seems that other explanations of the Lorentz force, for eg, in the Wiki article here do the same.

Thus I = 0, v = 0. and therefore F(magnetic) = 0.

You are missing the Biot Savart Law and it must be valid in all frames. That is, if an ammeter is connected to a wire, the reading must be the same for all frames. If it's zero for the particle frame, it must be zero for the wire frame as well. It can't be zero for one and a non-zero value for another.

I like the Science Asylum video even better. It's shorter and even more clear.

His video did not mention the Biot Savart Law either.

[Lincoln]

Your statement is obviously not true. Current is frame-specific.

If I have an infinite chain of positive charges, with density 1 coulomb/meter and I have them passing me at a rate of 1 meter per second, that's 1 Amp of current.

If I now jump to a frame where I am comoving with those charges, the current is zero.

Just FYI, I don't follow SPCF anymore, so I probably won't see your response. I just popped in as a courtesy for an admin.

I presume the locals can follow up on this topic.

Cheers...

[DJ_Juggernaut]

A cheap ammeter says current is the same in all frames. Cheers to you too.

[Faradave]

A cheap ammeter says current is the same in all frames. Cheers to you too.

"...the ampere, which is the flow of electric charge across a surface at the rate of one coulomb per second. Electric current is measured using a device called an ammeter. … They also create magnetic fields..."

Lincoln's current could be a beam of positrons in a vacuum. Are you allowing the ammeter to move with Lincoln's current when it jumps to the current frame?

Of course, the above videos would have to be redone to address such a current. That would be an interesting project! I'm scheduled to model magnetism from pure current in my own video series this Sept. (it's currently doing quantum spin, which is related). It's not for the faint-hearted, but if you click subscribe, you won't miss it.

[DJ_Juggernaut]

Lincoln's current could be a beam of positrons in a vacuum.

A beam of charges in the context of what? A magnet? Or a magnetic field? That would make sense. In the context of a vacuum that beam is meaningless. Recall that this thread is about Lorentz force. There must be three things:

1) A magnetic field B,
2) a charged particle q and
3) a relative velocity v between B and q that is greater than zero.

Then you get a force F = q v B from the magnetic field frame.
And you get a force F = B v q from the particle frame. Pure and simple.

[DJ_Juggernaut]

Of course, the above videos would have to be redone to address such a current.

Those videos are way off. They must start with the Biot Savart law. The laws of physics are the same in all frames. They must keep those field lines in both frames and then the Lorentz force is: qvB for all frames.

[Faradave]

Recall that this thread is about Lorentz force. There must be three things:

Agreed. If velocity is zero there's only electric force.

A beam of charges in the context of what? ... In the context of a vacuum that beam is meaningless.

For simplicity. Imagine all the positrons in a row with velocity v through a perpendicular unit surface establishing constant one ampere of current. According to the right hand rule, there should be a circular magnetic field produced.

"The Biot–Savart law is used for computing the resultant magnetic field B at position r in 3D-space generated by a steady current I (for example due to a wire). A steady (or stationary) current is a continual flow of charges which does not change with time and the charge neither accumulates nor depletes at any point."

[DJ_Juggernaut]

Recall that this thread is about Lorentz force. There must be three things:

Agreed. If velocity is zero there's only electric force.

Yes, but this qE (electric force) is too small and can be ignored, I think. And is not even considered in an electric motor for eg. The Lorentz force is just F = BIL, where B is the magnetic field, I the current, and L the length of the wire.

And I am not sure what E stands for in qE. Wikipedia defines E here. It looks like Coulomb's law. My guess is, the electrons in a magnet exert a force on q. It must be too small to consider it in a electric motor for eg. When the power is off, the electric motor remains stationary. Heh.

[Faradave]

...this qE (electric force) is too small and can be ignored, I think.

With wire perhaps, but not with the pure current (e.g. a positron beam).

E is the Electric field, the potential for force experienced per unit of test charge (q) at a location in the field.
qE is a force as you note.

[DJ_Juggernaut]

...this qE (electric force) is too small and can be ignored, I think.

E is the Electric field, the potential for force experienced per unit of test charge (q) at a location in the field. qE is a force as you note.

I think the term qE assumes that a charged particle q is in the vicinity of an electric field, E, which is essentially another charged particle and therefore Coulomb's law applies. But if the charged particle q is not in the vicinity of another particle, then the term qE is zero as well. That means, q experiences no force.

But you could argue that any charged particle q is always in the presence of some electric field (no matter how faint) given that the electric field of any electron or proton extends to infinity. Or you could argue that any charged particle experiences gravity and therefore you could add that term too. But Lorentz force per se deals with B, v and q. The term qE (or gravity or any other various faint fields) can be and should be ignored.

Lorentz force is: F = qvB
Total Lorentz force is: F = qE + qvB.

Well, it's not quite total yet, cus there are other forces that you can add to the list and compute a resultant force on q.

[Faradave]

Re: Condensation

qE is zero as well. That means, q experiences no force.

qE is zero for a test charge (q) near a neutral particle or no particle (i.e. in an essentially zero E field).

The term qE (or gravity or any other various faint fields) can be and should be ignored.

Assuming a neutral metal wire with no current. The electrons balance the protons in number and density. But when a current is flowing (electrons flow relative to protons) then the density of the charges seen to be moving increases due to relativistic contraction of the distance between them. Whichever charges you view to be moving (electrons in one direction or protons in the other) will have an increased density relative to the non-moving charges. As per the videos, a nearby test charge responds to the imbalance in charge density accordingly. This brings qE back into the equation.

[DJ_Juggernaut]

Assuming a neutral metal wire with no current. The electrons balance the protons in number and density.

In the "current" model of the atom, the electrons from the perimeter of any wire and will interact with a charged particle near it. Take a single atom of copper for eg. Any charged particle q near one of its 29 electrons will experience a force, qE. But we ignore it because it's faint. The idea of a neutral metal wire is somewhat misleading.

But when a current is flowing (electrons flow relative to protons) then the density of the charges seen to be moving increases due to relativistic contraction of the distance between them.

This is not correct. Say, there are 10 protons and 10 electrons lined up; they are stationary relative to each other. At time t1, the electrons start to move to the right. After a short time t2, the rate of flow of charge is 10. The distance between the protons maybe contracted. It doesn't matter. The rate of flow is 10. Then switch the frame. The electrons are stationary and their distance is contracted, the protons move to the left. After a short time t2, the rate of flow of charge is 10. In each frame, there are 10 moving charges (between two points on a surface). This can't and won't change.

There is no imbalance.

[DJ_Juggernaut]

A cheap ammeter says current is the same in all frames. Cheers to you too.

Lincoln's current could be a beam of positrons in a vacuum. Are you allowing the ammeter to move with Lincoln's current when it jumps to the current frame?

No, obviously. It won't measure anything if it did. By definition, an ammeter is stationary relative to a current. The only way an ammeter reads zero current in a live wire is if Lincoln disconnects the live wire or he disconnects the ammeter.

[Faradave]

The only way an ammeter reads zero current in a live wire is if Lincoln disconnects the live wire or he disconnects the ammeter.

Or if there is zero current! Then its just a nice, accurate reading.* An ammeter moving with a positron beam identifies a frame in which that current is zero, thus the classical "magnetic" force is transformed away there.

Biot-Savart, Lorentz and other considerations accommodate reference frames in which electric and magnetic forces are found. But as far as the test particle is concerned, there's no difference. A force is a force, of course of course (F=ma, regardless). That's our key to a ToE (theory of everything, uniting all four interactions G, EM, Strong, Weak).

Since electric force can't be transformed away (it is invariant to frame), it is considered somewhat more fundamental than magnetism, which in most classical cases can be transformed away. In the end it's still just "force".

*Yes, you can sidestep this by defining a "live" wire as one with a current. But a complete circuit with a dead battery and an LED must be allowed as a " circuit" with legitimately zero current.

[DJ_Juggernaut]

Or if there is zero current!

That's not what Lincoln is saying. He says the current is zero for the particle. But not zero for the circuit. Rig the circuit with an LED bulb. The bulb is on for one and off for another. This is just wrong. LED is an ammeter here.

[Faradave]

I believe Lincoln is using the more general definition of "current":
"...the flow of electric charge across a surface". (Net positive charge flow in the reference direction is positive current.)

I agree that in a physical conductor (such as a wire), a non-zero current means electrons in one direction or protons in the other (through a cross-section of the conductor). To the degree one such current is transformed away, always invokes the other.

However in principle,* even for the situation where relative current can't be transformed away, I believe their magnetism can. By choosing a "center-of-charge-velocity" frame in which electrons have v/2 in one direction while protons have v/2 in the opposite direction (i.e. -v/2), their magnetic fields will essentially cancel out.

*meaning: no one actually wants to rig an ammeter to do this along a wire. Oppositely charged ions might be convincd to flow past each other toward respective electrodes in a fluid conductor with the above result.

[DJ_Juggernaut]

A Summary of Lincoln's errors with regard to Lorentz Force:

F = qE + qv B

1) The v in the Lorentz force equation is zero in at least one frame.
Correction: The v in Lorentz force equation is not zero because it's a relative velocity between field lines B and a charge q. In the particle frame, the field lines B move at v relative to the particle. And in the wire or magnet frame, the particle moves at v relative to the field lines B.

2) The current is zero when you move at the same speed as electrons, for eg.
Correction: The flow of current is not zero in any frame. If you move at the same speed as electrons, the protons appear to move in the opposite direction relative to you. Therefore the current is not zero when you move at the same speed as electrons.

That leaves us with qE, which is essentially Coulomb's law. This is too faint a force to consider it here. In the other words or in the words of Lincoln, "case closed".

[DJ_Juggernaut]

Scott Huges – MIT Lecture 10 (pdf)
http://web.mit.edu/sahughes/www/8.022/lec10.pdf

Suppose we now examine this situation from the point of view of the charge (the “charge frame”). From the charge’s point of view, it is sitting perfectly still. If it is sitting still, there can be no magnetic force!

Walter Lewin – MIT Lecture 11 (Youtube)
https://www.youtube.com/watch?v=X4dXXnUMHbQ&t=1537s

Magnetic fields can never do work on a moving charge. And the reason is the force is always perpendicular to the velocity v. And if the force is always perpendicular to the motion, you can change the direction of the motion, but you can't change the kinetic energy.

Don Lincoln – SCF Expert Notes (Forum)
http://sciencechatforum.com/viewtopic.php?f=84&t=8621

But what happens if you jump to the frame where the particle is not moving? With a velocity of zero, the magnetic force is zero.

if you are moving along at the same speed as the current, there is no current and, consequently, you will have no magnetic force.

Now, assume that the particle q is moving at the same velocity as the current. In this frame, q is stationary. But the current in the wire is also stationary. Thus I = 0, v = 0. and therefore F(magnetic) = 0.

1. On Lincoln's statements
The v in Lorentz Force F = qvB is the relative velocity between the field lines B and a charged particle q. It does not matter whether the charged particle moves or the field lines move. All that matters is the relative velocity v, which is not zero in either frame. If the charged particle is at rest, the field lines B move at v relative to the particle. If the field lines B are at rest, the charged particle q moves at v relative to the field lines B.

The current in a wire is not zero if you move at the same speed as electrons. If you move at the same speed as electrons, the protons in the wire appear to move in the opposite direction. Therefore, the current is never zero when you move at the same speed as electrons. Therefore, I <> 0; we already know v <> 0; and therefore F(magnetic) <> 0.

2. On Walter Lewin's statements
Magnetic field B does indeed do work on a moving charge. It's the essence of Lorentz force. A moving charge in a magnetic field experiences a force. And indeed, its kinetic energy changes upon interaction as does its direction.

3. On Scott Huges' statements
In the particle frame, where the charged particle is sitting still, the field lines B move at v relative to the particle. Therefore, a force is experienced by the particle, given by F = qv * B.

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PS. Even the Wikipedia entry on the magnetic force term here, asserts that v = 0, in the charged particle frame and therefore, Lorentz force (magnetic) is zero.

A charge q in the conductor will be at rest in the conductor frame. Therefore, the magnetic force term of the Lorentz force has no effect

[TheVat]

SPCF doesn't allow duplication of posts by starting redundant new topics. You may continue your discussion here, where it started.

[bangstrom]

I don’t quite follow the video explaining magnetic attraction as an SR effect.
https://www.youtube.com/watch?v=p4gCTmlm5RQ

In SR, the contraction of either the flow of electrons or of the nuclei of the atoms in the wire relative to an outside observer should be observed in mainly in the forward direction. This would result in a magnetic vector shifted towards the direction of flow. The particles moving towards the observer should appear to be closer together while those moving away should appear to be farther apart. Since the observed polarity of the charge depends on the Lorentz contraction of the particle stream, the charge on the wire would appear to change as it passes by the observer.

Another problem is that, because of the time delay between the wire and the observer, a stationary observer should find his position at right angles to whichever particle stream is in motion as shifted forward from where it would be if all participants were inertially at rest. This would also shift the magnetic vector to the forward direction.

These problems are easier to visualize if you consider the example of two closed superconducting loops placed in close proximity but parallel to each other. Lorentz contraction does not increase the number of either electrons or atoms so there is no bunching of one set of particles relative to the other and either loop by itself can be taken as our test particle observer. The two loop example is one where Lorentz contraction has no effect.

If the current through both loops is in the same direction, the loops will attract but, if the current is in opposite directions, the loops will repel. If the magnetic vector between the two loops is in any direction other than perfectly at right angles, the loops will begin to spin in opposite directions and it is my understanding that this never happens.

Ampere had his own pre-relativistic explanation of magnetism where the spin motions of the electrons and the current flow are the cause for magnetic attraction or repulsion. I have never tried to apply special relativity to Ampere’s explanation but it might work and it would be more economical an explanation than found in the video because it eliminates the presence of the wire from consideration.

[Faradave]

...contraction... mainly in the forward direction.

Length contraction in SR is often described as being "in the direction of motion". However, that does not mean only in the forward direction.

A rocket passing a "stationary" observer will appear length contracted in the direction of motion. Conversely, from that rocket's rest frame, the entire universe is speeding by in the opposite direction. The rocket will find the entire universe length contracted. The rocket will find both the path ahead contracted and the path behind.