Magnetic reconnection and physical processes

[Ziggurat]

I really don't understand your objection to my statement. The current density and field strength are related. The movement of the electrons generates the field. The termination of the electron flow causes the field to collapse. It's all related to the movement of charged particles.

Yes: magnetic energy is related to the movement of charge. But kinetic energy is the movement of mass. And mass and charge are not equivalent. Moreover, the relationship of mass to kinetic energy and charge to magnetic energy aren't even the same. The kinetic energy scales as mv². Because m and v are raised to different powers, it makes a difference whether you have a small mass flowing at high velocity or a large mass flowing at low velocity. The mass current (the product mv) isn't enough to determine kinetic energy. But the magnetic field only depends upon the total current (the product qv), so the energy in the field does only depend on electric current, and it scales as B²~q²v². It's a fundamentally different scaling relationship. That's why it matters (in a direct, measurable, physical sense) that we correctly distinguish between kinetic energy and energy in the magnetic field. They do not act the same way.

 

[sol invictus]

We've agreed (repeatedly) that magnetic fields are generated by currents. Not very surprising; it's all B = del cross J as usual. Tell me J and I'll tell you B, and vice versa, modulo a constant of integration.

It's not a constant of integration when going from J and rho to B and E - it's an arbitrary linear combination of solutions to the homogeneous equation. And those solutions are kind of important, since they include all forms of electromagnetic radiation!

 

[sol invictus]

Yes: magnetic energy is related to the movement of charge.

Not necessarily. Permanent magnets are an exception, as are all forms of EM radiation, constant fields, etc.

The problem here is that MM is so utterly off base that it's difficult to even pick out a few correct things and comment on them.

 

[Ziggurat]

Not necessarily.

Well, I guess that depends on whether or not you want to include electron spin under movement of charge. There are reasons not to, but for current purposes I think we can lump that in anyways. Still doesn't fix Michael's error regarding magnetic energy, though, since alignment of spin won't change kinetic energy at all but will change energy stored in the field.

 

[Michael Mozina]

Yes: magnetic energy is related to the movement of charge. But kinetic energy is the movement of mass. And mass and charge are not equivalent. Moreover, the relationship of mass to kinetic energy and charge to magnetic energy aren't even the same. The kinetic energy scales as mv². Because m and v are raised to different powers, it makes a difference whether you have a small mass flowing at high velocity or a large mass flowing at low velocity. The mass current (the product mv) isn't enough to determine kinetic energy. But the magnetic field only depends upon the total current (the product qv), so the energy in the field does only depend on electric current, and it scales as B²~q²v². It's a fundamentally different scaling relationship. That's why it matters (in a direct, measurable, physical sense) that we correctly distinguish between kinetic energy and energy in the magnetic field. They do not act the same way.

In that case the "stored magnetic energy" you're talking about is kinetic energy in the one coil that is transferred to the secondary coil. It's still a "kinetic energy" transfer event. The same is true in all 'magnetic reconnection" events in the solar atmosphere. These are also know as "electrical discharge" events. The "current flow" creates the magnetic field.

Is this the specific statement you took exception to?

 

[Ziggurat]

Is this the specific statement you took exception to?

Yes.

 

[ben m]

the energy in the field does only depend on electric current, and it scales as B²~q²v². It's a fundamentally different scaling relationship. That's why it matters (in a direct, measurable, physical sense) that we correctly distinguish between kinetic energy and energy in the magnetic field. They do not act the same way.

It's not just that B^2 ~ q^2 v^2, it's that B^2 depends on, and only on, the current distribution---not just the current magnitude but its spatial distribution. Take 1 amp of current in 1 meter of wire---imagine the charge-carrier kinetic energy to be whatever you like. This current generates ~ zero B^2 energy if that wire is half-meter of twisted pair with the end shorted out, a small B^2 energy if the wire is straight, and a large B^2 energy if the wire is solenoidal.

 

[Michael Mozina]

Yes.

OK. Let's be specific:

In that case the "stored magnetic energy" you're talking about is kinetic energy in the one coil that is transferred to the secondary coil. It's still a "kinetic energy" transfer event.

So basically you do not believe that this is a kinetic energy transfer event? Would a photon transfer kinetic energy and do photons play a role as the carrier particle of the EM field? Could they (do they) transfer kinetic energy to electrons in the secondary coil?

 

[ben m]

So basically you do not believe that this is a kinetic energy transfer event? Would a photon transfer kinetic energy and do photons play a role as the carrier particle of the EM field? Could they (do they) transfer kinetic energy to electrons in the secondary coil?

No, this is not a kinetic energy transfer event. To make sure we stay on target, for concreteness we're talking about (say) a current-carrying inductor; it requires energy (measure it: current driven against a back-EMF) to initiate current in an inductor; the inductor can then discharge this energy (into a resistor for heat, into another inductor, into a motor---whatever you like.)

No, there is no way to call this kinetic energy. In the initial state, the quantity U = Integral(B^2/mu) is nonzero and is equal to the energy you measured having put into the inductor (or the energy you measure that you can get out). At all times, the quantity S = Integral (E cross B/mu0), called the pointing vector, is zero, so there is no momentum or kinetic energy in the field. At any time you like, you can measure the sum of the kinetic energies of every component of the system; in general you'll find it to be vanishingly small, but even if you engineer it to be large (which you can do; invent your own charge carriers and you can do whatever you like) this hand-inserted "kinetic energy" is unrelated to U and has entirely different behavior.

So we've got a system with no kinetic energy and no momentum, it discharges via an intermediate state with no kinetic energy and no momentum, and you end up with any sort of energy you like depending how you load the circuit.

There is no kinetic energy involved; it is not a kinetic energy issue; there is such a thing as a kinetic energy transfer, but this isn't an example. What makes you think it is?

Nor is it a photon problem; if you were to give a correct QED treatment of the system (i.e using photons to represent the B field itself) you will find that this treatment makes exactly the same predictions as the Maxwell's treatment: it's still U=B^2, S=ExB. (So, if you think you can say, "Hey, QED makes everything photons", and you think that this gives you some photon-related excuse for changing your story: you can't, and it doesn't.)

 

[tusenfem]

First of all, none of this is the least bit relevant in *plasma* where the charged particles are whizzing by at a million miles per hour! In the case of the coil, you have current flow in the coil and EM energy being transferred via induction. There is absolutely no form of "magnetic reconnection" going on in the coil.

Wow, Really!?!?!?!?!

I am happy when my plasma moves at like 1000 km/s at a so called "reconnection" event, maybe 2000, sort of the limit of the Alfven velocity in the Earth's magnetotail. But, I guess your theory only works in the solar corona.

 

[sol invictus]

It's not just that B^2 ~ q^2 v^2, it's that B^2 depends on, and only on, the current distribution---not just the current magnitude but its spatial distribution.

Ben, you need to be a little more precise - that statement as written is incorrect. The current distribution at any given time does not determine B, nor does it determine the energy or momentum in it.

For example - a radio transmitter switches on for an hour, then turns off permanently. 4.3 years later, aliens on Alpha Centauri use it to cook an egg (it was a very tight beam). There are no currents or charges anywhere.

 

[Ziggurat]

So basically you do not believe that this is a kinetic energy transfer event?

I do not. Energy is stored in the static magnetic field. There is simply no sensible way to call that energy "kinetic energy". It is not. Moreover, it's not even in the coil itself, it's in the magnetic field.

Would a photon transfer kinetic energy

Photons are energy. You can calculate their energy just like you can for static fields: it's the volume integral of the field squared. Calling it "kinetic energy" is silly. And you do not need to (and should not) consider photons in a case like this. You can handle it completely and correctly with classical electrodynamics. In fact, even trying to do it with photons will be a nightmare, because the number of photons in a case like this is indeterminate. How many photons make up a 5-gauss magnetic field? As I've pointed out to you before, it's just not a sensible question. So how many photons get absorbed, and what's the energy of each of them? That too is just not a sensible question.

Could they (do they) transfer kinetic energy to electrons in the secondary coil?

No. The static magnetic field stores potential energy. This energy is not kinetic energy.

 

[ben m]

Ben, you need to be a little more precise - that statement as written is incorrect. The current distribution at any given time does not determine B, nor does it determine the energy or momentum in it.

For example - a radio transmitter switches on for an hour, then turns off permanently. 4.3 years later, aliens on Alpha Centauri use it to cook an egg (it was a very tight beam). There are no currents or charges anywhere.

You're right of course. Since MM's trouble extends all the way down to magnetostatics, I've implicitly been starting there; I should be more explicit.

 

[sol invictus]

You're right of course. Since MM's trouble extends all the way down to magnetostatics, I've implicitly been starting there; I should be more explicit.

I think it's useful to bear in mind that many people other than MM will probably read these posts - in fact if that weren't the case there would be zero point in posting them.

Even for magnetostatics one can store arbitrary amounts of energy in arbitrarily large regions with precisely zero current. Same goes for momentum.

 

[Tubbythin]

I think it's useful to bear in mind that many people other than MM will probably read these posts - in fact if that weren't the case there would be zero point in posting them.

I dunno, I think its pretty amusing watching him [url='http://www.internationalskeptics.com/forums/showpost.php?p=4449752&postcount=382]contradict himself in consecutive sentences[/url].

 

[Michael Mozina]

I do not.

I tried to post a lengthy reply only to get a database connection error and lose it. Rather than go through the whole thing again, I'll just ask a simple question. What *physical exchange* is creating acceleration of electrons in the secondary coil?

Energy is stored in the static magnetic field. There is simply no sensible way to call that energy "kinetic energy". It is not. Moreover, it's not even in the coil itself, it's in the magnetic field.

*What* is "in" the magnetic field? What's the carrier particle of the EM field?

Photons are energy.

Bingo. They also transfer *kinetic* energy and they are also the carrier particle of the EM field!

 

[Ziggurat]

What *physical exchange* is creating acceleration of electrons in the secondary coil?

Induction.

*What* is "in" the magnetic field?

Nothing. The magnetic field itself stores the energy. There are no photons in a static field - that's why you can't answer the question of how many photons it takes to make a 5 gauss field.

What's the carrier particle of the EM field?

Irrelevant: the energy is not stored as photons. It's stored in a static field. And it is not kinetic energy. As has already been pointed out to you, you don't need to ever introduce photons to handle this problem: you can deal with it perfectly well with classical electrodynamics, no photons involved.

Yes, photons can transfer kinetic energy from one electron to another. But that's not applicable here, because the energy is not in the electrons to begin with, but in the field itself. The reasons for this distinction have been pointed out to you multiple times now, and you still show no signs of understanding what has been said to you.

 

[sol invictus]

*What* is "in" the magnetic field? What's the carrier particle of the EM field?

Bingo. They also transfer *kinetic* energy and they are also the carrier particle of the EM field!

OK, so you claim that the energy density in EM fields can be thought of as the kinetic energy in photons. Great - let's try a simple case: constant B field over a large region. Show us how to get B^2/2 out as the energy density by totaling up the energy in photons.

By the way, something like this is possible - but you're not bright enough to figure out how to do it... and if you somehow did, you'd have learned quite a lot (which would be a relief for all of us).

 

[Dancing David]

So, just for us lurkers, the field induces a current in the wire to the wall outlet without the exchange of a photon being needed?

So the spark is a manifestation of the magnetic field?


Stupid question, what makes the spark?

 

[Ziggurat]

So, just for us lurkers, the field induces a current in the wire to the wall outlet without the exchange of a photon being needed?

No photons needed.

So the spark is a manifestation of the magnetic field?

The energy for the spark comes from the magnetic field. The decrease in the magnetic field that occurs after you unplug it induces an electric field, and that electric field pushes the spark across the gap, but the energy to do so was stored in the magnetic field.