Can a magnetic field exist without a moving electric field? [Archive]

Zeuzzz

5th July 2009, 11:14 PM

Full answer: Too long to be bothered. Read the rest of the above page.

This might help, though we're (unavoidably) getting into electric currents instead of fields. http://en.wikipedia.org/wiki/Magnet#Two_models_for_magnets:_magnetic_poles_and_ atomic_currents

Magnetic pole model: Although for many purposes it is convenient to think of a magnet as having distinct north and south magnetic poles, the concept of poles should not be taken literally: it is merely a way of referring to the two different ends of a magnet. The magnet does not have distinct "north" or "south" particles on opposing sides. (No magnetic monopole has yet been observed.) If a bar magnet is broken in half, in an attempt to separate the north and south poles, the result will be two bar magnets, each of which has both a north and south pole.

The magnetic pole approach is used by professional magneticians to design permanent magnets. In this approach, the pole surfaces of a permanent magnet are imagined to be covered with 'magnetic charge', little 'north pole' particles on the north pole and 'south poles' on the south pole, that are the source of the magnetic field lines. If the magnetic pole distribution is known, then outside the magnet the pole model gives the magnetic field exactly. In the interior of the magnet this model fails to give the correct field (see Units and Calculations, below). This pole model is also called the "Gilbert model" of a magnetic dipole.[2] Griffiths suggests (p. 258): "My advice is to use the Gilbert model, if you like, to get an intuitive "feel" for a problem, but never rely on it for quantitative results."

Ampère model: Another model is the "Ampère model", where all magnetization is due to the effect of microscopic, or atomic, circular "bound currents", also called "Ampèrian currents" throughout the material. For a uniformly magnetized cylindrical bar magnet, the net effect of the microscopic bound currents is to make the magnet behave as if there is a macroscopic sheet of electric current flowing around the surface, with local flow direction normal to the cylinder axis. (Since scraping off the outer layer of a magnet will not destroy its magnetic field, it can be seen that this is just a model, and the tiny currents are actually distributed throughout the material). The right-hand rule tells which direction the current flows. The Ampere model gives the exact magnetic field both inside and outside the magnet. It is usually difficult to calculate the Amperian currents on the surface of a magnet, whereas it is often easier to find the effective poles for the same magnet.

I think that most of the confusion in this area comes from the fact that although the electic field and magnetic field are pretty much analogous, people have proposed different extensions to both E field and B fields, which often have differences. This is also where a lot of the confusion over magnetgic reconnection comes in... but thats another matter completely.

Perpetual Student

5th July 2009, 11:21 PM

A magnetic field exists around a magnet. No electricity required. A permanent magnet, just like every other material in the normal world save plasma, requires electron movement in the orbits as predicted probabilistically by quantum mechanics. The key thing about a permanent magnet is that the atoms charge dipoles (which exist because of the existence of electrons themselves) are aligned so that they reinforce each other. Any other magnetic field requires current flowing in a conductor.

So, this magnetic field is caused by an intrinsic property of the electrons, not their motion?

nathan

6th July 2009, 12:32 AM

So, this magnetic field is caused by an intrinsic property of the electrons, not their motion?

Yes. the electrons are not forming an electric currrent.

Floyt

6th July 2009, 01:00 AM

Hmmm... does absolute zero inhibit magnetic fields?

Actually, I'm probably wrong in a previous conjecture - does absolute zero equate to no movement of electrons?

MattusMaximus

6th July 2009, 01:03 AM

Yes. the electrons are not forming an electric currrent.

Um, electrons in motion is the very definition of electric current. At least it is in classical E&M.

ETA: This assumes, of course, that there is a net direction of motion of the electrons. Gah, it's late...

MattusMaximus

6th July 2009, 01:07 AM

Hmmm... does absolute zero inhibit magnetic fields?

Actually, I'm probably wrong in a previous conjecture - does absolute zero equate to no movement of electrons?

No. Read up on zero-point energy (http://en.wikipedia.org/wiki/Zero_point_energy), especially in regards to electrons in an atom following the behavior of a quantum harmonic oscillator. (http://en.wikipedia.org/wiki/Quantum_harmonic_oscillator) Even if you were to cool something to absolute zero, the ZPE would still be there due to the quantum "wiggling" of the electrons.

Evilgiraffe

6th July 2009, 01:09 AM

So, this magnetic field is caused by an intrinsic property of the electrons, not their motion?

Electrons are magnetic because of their spin (http://en.wikipedia.org/wiki/Spin_(physics)). They can be thought of as little spinning tops, (This is not entirely accurate and I'm sure a physicist will be along in a moment to tell me off for saying so) and classical treatment of a spinning charged particle results in a magnetic field.

In quantum mechanics, electrons are point particles and, as such, have no axis to rotate around. Therefore they can't be like little tops. However, they can still possess angular momentum and this causes them to have a magnetic moment. The angular momentum of an electron, like many of its other properties, is quantised and was first demonstrated by the Stern-Gerlach experiment.

Linky (http://en.wikipedia.org/wiki/Stern_gerlach) (Warning, mind-bending physics)

As for permanent magnets, these are created because the crystal structure of the material allows electrons to communicate their spin directions to one another and line up, thereby reinforcing each others magnetic field. Individually electrons have a pretty puny magnetic field, but lots of them pointing in the same direction (say 1023 or so) leads to stuff like this (http://video.google.co.uk/videoplay?docid=1109174457096582405&ei=HKJRSqm7O83q-AbDlNmWDg&q=super+magnets&hl=en).

shadron

6th July 2009, 01:17 AM

So, this magnetic field is caused by an intrinsic property of the electrons, not their motion?

Yes (I believe that one should never contradict a lady avatar). The electrons are, like all electrons, moving about the nuclei they are more-or-less bound to according to quantum mechanics as they do in all atoms, but the atoms certainly are not moving; they'd have to be in some other state than solid to do that. There is no net current (the loose outer ellectrons may be skipping from nuclei to nuclei, but without an EMF applied the movement is random and has a net current of zero, and that movement does not therefore have a net magnetic effect.

There are nine ferromagnetic elements that can be attracted to a magnet and be magnetized; they form Group VII of the periodic table. They all have a single electron in their outermost electron shell.

tusenfem

6th July 2009, 01:21 AM

Short answer: No, the electric field and magnetic are inextricably linked. Where there is one, there will always be the other component.

Although electric and magnetic fields are inter-related, the answer is YES, Zeuzzz is clearly wrong here. A simple example:

take a conducting wire and put a voltage drop over it
a current will start to flow in the wire
the current will create a magnetic field
hence we have a stationary (i.e. not moving) electric field and a magnetic field

tusenfem

6th July 2009, 01:25 AM

Um, electrons in motion is the very definition of electric current. At least it is in classical E&M.

ETA: This assumes, of course, that there is a net direction of motion of the electrons. Gah, it's late...

And you forget that the Bohr model of the atom is not correct. The electrons do not orbit the nucleus, but have a probability cloud around it. Therefore, the "orbit" of the electron around the nucleus does not lead to a "current."

nathan

6th July 2009, 02:29 AM

Um, electrons in motion is the very definition of electric current. At least it is in classical E&M.

ETA: This assumes, of course, that there is a net direction of motion of the electrons. Gah, it's late...

The question was about a permanent magnet. Not an electro-magnet. Yes a moving electron is an electric current. You are too sleepy :)

nathan

6th July 2009, 02:38 AM

Talking about permanent magnets, I and my boiler maintainer recently had an 'interesting' time fixing my boiler. The cause of the failure was eventually traced to a failed permanent magnet. There's a water flow measuring device -- a diaphram with a venturi and a plunger. the plunger contains a permanent magnet and actuates a reed switch. When water flows correctly, the plunger overcomes a spring and actuates the switch. Because of the venturi, small amounts of water flow around the plunger.

Replacing the plunger fixed the problem. We took the whole assembly out and could play with it on the counter. It was very definite. Old plunger == not working, new plunger == working. All the plunger is is a lump of plastic with a magnet embedded in it.

The question is how did the permanent magnet become demagnetized? We also found quite a bit of iron crud in the system (which we cleaned out and added suitable degunking fluids to the system). My supposition is that this crud could move past the magnet, and my guess is this caused the crud to become magnetized and reduced the magnetization of the permanent magnet. Over the 8 or so years of the boiler's life, this reduced sufficiently for the device to no longer function.

Does that sound plausible?

Of course this was no where listed in the manufacturer's fault finding flow chart.

Aitch

6th July 2009, 02:41 AM

Heat demagnetizes permanent magnets; could that have been the cause.

So does banging them with a hammer, IIRC. Could be wrong on that one, 'cause it's also a way of magnetizing iron bars. Depends on the bar/magnets alignment with a magnetic field (the Earth's?). :confused:

nathan

6th July 2009, 02:48 AM

Heat demagnetizes permanent magnets; could that have been the cause.

True. Doesn't that involve reaching a transition temperature?

But because this device is inside a boiler, I'd've thought the manufacturer might have considered it could get hot. The water would only go past the plunger when it was in the process of changing state from one end of the tube it was in to the other. This water would be hot when turning off, and cold when turning on.

stevea

6th July 2009, 02:58 AM

I believe this has been discussed on other threads. I never did see a definitive answer, however. If there was one I missed it -- could someone summarize any conclusions, please.

Here is a definitive answer. A magnetic field in nothing but a relativistic view of an electrical (electrostatic) field. I was taught this as a 2nd year physics undergrad. I am shocked at all the silly classical (pre-1920s) physics rubbish still filling people heads (especially wrt thermodynamics).

Einstein's most famous 1905 paper was titled, "On the Electrodynamics of Moving Bodies", for a reason. If you consider the charges moving in a wire for example, they each have an isotropic field in their own frame of reference, but as they move there is a Lorentzian contraction of the field. Lorentz was hot on Einstein's heels with a better mathematical representation.

I know the know-nothings will object that electrons in wires aren't travelling at a significant velocity. If you do the math you'll (somewhat against intuition) see the velocity and charge density terms do not seperately matter, but only the charge flux combination.

Any serious undergrad text will fill in the blanks.

stevea

6th July 2009, 03:12 AM

WRT demagnetization - if you get a magnetic or paramegnetic material above it's curie temperature it loses all magnetization, but typically that's an extremely high temperature (~1400F for iron for example).

But note that the magnetization of tiny individual domains becomes random if the local energy is sufficient, so banging it or heating it has a negative impact.

nathan

6th July 2009, 03:20 AM

Here is a definitive answer. A magnetic field in nothing but a relativistic view of an electrical (electrostatic) field. I was taught this as a 2nd year physics undergrad. I am shocked at all the silly classical (pre-1920s) physics rubbish still filling people heads (especially wrt thermodynamics).

Einstein's most famous 1905 paper was titled, "On the Electrodynamics of Moving Bodies", for a reason. If you consider the charges moving in a wire for example, they each have an isotropic field in their own frame of reference, but as they move there is a Lorentzian contraction of the field. Lorentz was hot on Einstein's heels with a better mathematical representation.

I know the know-nothings will object that electrons in wires aren't travelling at a significant velocity. If you do the math you'll (somewhat against intuition) see the velocity and charge density terms do not seperately matter, but only the charge flux combination.

In a permanent magnet, where is the current then?

jimgerrish

6th July 2009, 03:21 AM

None of these theories takes into account the consequenses of accelerated space/time expansion.

sol invictus

6th July 2009, 05:34 AM

I believe this has been discussed on other threads. I never did see a definitive answer, however.

Yes, it can. For example, a current creates a magnetic field but no electric field. And you don't need any electric field to maintain a current under some circumstances (for example in a superconductor, or in an ionized gas).

Here is a definitive answer. A magnetic field in nothing but a relativistic view of an electrical (electrostatic) field. I was taught this as a 2nd year physics undergrad. I am shocked at all the silly classical (pre-1920s) physics rubbish still filling people heads (especially wrt thermodynamics).

You're confused. What relativity tells us is that IF there is an inertial frame in which one has a magnetic field but zero electric field, then in any other frame there will also be an electric field. Both are elements of a second rank tensor, so they transform into each other when one changes frames. But that doesn't change the fact that one can certainly have a magnetic field with zero electric field.

None of these theories takes into account the consequenses of accelerated space/time expansion.

They don't take chopped liver into account either.

ponderingturtle

6th July 2009, 05:46 AM

Sorry for the ambiguity. Let me clarify.
A static electric field can exist with no accompanying or causative magnetic field. In a similar manner, can a magnetic field exist without an accompanying or causative electric field? As you say, a permanent magnet depends on moving electrons, so it would not be the kind of example I am looking for, assuming there is one.

This depends, do you believe in magnetic monopoles?

ponderingturtle

6th July 2009, 05:54 AM

You're confused. What relativity tells us is that IF there is an inertial frame in which one has a magnetic field but zero electric field, then in any other frame there will also be an electric field. Both are elements of a second rank tensor, so they transform into each other when one changes frames. But that doesn't change the fact that one can certainly have a magnetic field with zero electric field.

But does this negate the point?

ponderingturtle

6th July 2009, 05:56 AM

In a permanent magnet, where is the current then?

Well if the orbits of the electons line up you would have a lot of moving electrons with out any real current. Each atom would be a small magnetic diapole and if you line them up in a crystal they could add together.

sol invictus

6th July 2009, 06:01 AM

But does this negate the point?

If someone asked whether it's possible to have energy but zero momentum, I presume you'd agree the answer is yes (for example, a massive particle at rest). But in any frame other than the rest frame, the particle will be moving and the momentum will be non-zero (and the energy will be greater). That's because energy and momentum fit into a 4-vector, so they mix with each other when you change frames.

Something very similar (but slightly more complicated) happens with electric and magnetic fields.

nathan

6th July 2009, 06:12 AM

Well if the orbits of the electons line up you would have a lot of moving electrons with out any real current. Each atom would be a small magnetic diapole and if you line them up in a crystal they could add together.

What orbits?

edd

6th July 2009, 06:22 AM

This depends, do you believe in magnetic monopoles?

Why do you think a magnetic monopole is needed? Wouldn't a magnetic dipole be enough?

sol invictus

6th July 2009, 06:24 AM

What orbits?

You're being a little picky. Electrons are in orbitals, to be absolutely correct, but those orbitals in general carry angular momentum just like classical orbits. The electron's intrinsic spin adds to that, giving most atoms a magnetic dipole moment. When those dipoles align, the material is magnetized.

Is there a current? Actually there is, at least in a technical sense. Maxwell's equations tell us that a magnetic dipole field must have a source, and that source is generally referred to as a current - even when, as in this case, it's not a flow of charged particles.

MetalPig

6th July 2009, 06:55 AM

I have a somewhat related question. Does an electromagnetic field bend light that passes through it?
I told someone it doesn't, but only because I couldn't think of a reason why it would.

sol invictus

6th July 2009, 07:04 AM

I have a somewhat related question. Does an electromagnetic field bend light that passes through it?
I told someone it doesn't, but only because I couldn't think of a reason why it would.

Your logic should go the other way: if you can't think of a reason it wouldn't, it does. :)

And it does, although the effect is extremely small at optical wavelengths and reasonable field strengths. The largest interaction between light and light (or EM fields, same thing) arises from a quantum process where two photons interact through a virtual electron-positron pair.

The effect is small because the energy required to create an electron-positron pair is much larger than the energy available in optical frequency photons. It's dominated by electron/positron because those are the lightest charged particles. It has never been observed directly, although it probably will be soon with improved laser technology, but its indirect effects on various electrodynamic processes have been seen.

ponderingturtle

6th July 2009, 07:12 AM

If someone asked whether it's possible to have energy but zero momentum, I presume you'd agree the answer is yes (for example, a massive particle at rest). But in any frame other than the rest frame, the particle will be moving and the momentum will be non-zero (and the energy will be greater). That's because energy and momentum fit into a 4-vector, so they mix with each other when you change frames.

Something very similar (but slightly more complicated) happens with electric and magnetic fields.

When I was taking E&M we did learn that you could think of all magnetic effects as reletivistic electric effects. We also learned that this approach was not very useful.

So I am not convinced it is wrong to think of it in that way, but it is not likely to help you solve a problem.

ponderingturtle

6th July 2009, 07:17 AM

You're being a little picky. Electrons are in orbitals, to be absolutely correct, but those orbitals in general carry angular momentum just like classical orbits. The electron's intrinsic spin adds to that, giving most atoms a magnetic dipole moment. When those dipoles align, the material is magnetized.

Is there a current? Actually there is, at least in a technical sense. Maxwell's equations tell us that a magnetic dipole field must have a source, and that source is generally referred to as a current - even when, as in this case, it's not a flow of charged particles.

This all depends on how one defines current.

Of course all most all currents are circular if you think of them as being large enough, otherwise you get a buildup of charge and will quickly reach equilibrium.

I don't see why looking at a current as only being bigger than the atomic scale movements is justified.

MetalPig

6th July 2009, 07:22 AM

Your logic should go the other way: if you can't think of a reason it wouldn't, it does. :)
D'oh!
The effect is small because the energy required to create an electron-positron pair is much larger than the energy available in optical frequency photons.
I don't entirely understand the technical explanation, but I'm pretty sure this means that the idea of cloaking planet Niburu with an EM field that bends light around it is pretty far-fetched. :D

MattusMaximus

6th July 2009, 09:59 AM

And you forget that the Bohr model of the atom is not correct. The electrons do not orbit the nucleus, but have a probability cloud around it. Therefore, the "orbit" of the electron around the nucleus does not lead to a "current."

That's why I said in classical E&M ;)

Perpetual Student

8th July 2009, 07:19 AM

In summary then, a stationary electron has a few attributes, which include electric charge, mass, and because of its spin, a magnetic field. Is that a correct description?

Cuddles

8th July 2009, 07:52 AM

In summary then, a stationary electron has a few attributes, which include electric charge, mass, and because of its spin, a magnetic field. Is that a correct description?

To a certain extent. However, it's worth remembering that there's no such thing as a stationary electron, or any other particle for that matter.

Ziggurat

8th July 2009, 08:13 AM

To a certain extent. However, it's worth remembering that there's no such thing as a stationary electron, or any other particle for that matter.

If it's in an energy eigenstate, I think it's safe to call it stationary.

Evilgiraffe

8th July 2009, 11:59 AM

If it's in an energy eigenstate, I think it's safe to call it stationary.

Umm, no... (http://hyperphysics.phy-astr.gsu.edu/hbase/hyde.html#c3)

Aren't electrons in each of the energy levels of the hydrogen atom in energy eigenstates?

It would be surreal to suggest that electrons in atomic hydrogen are stationary. Yes, we know their energy and can pin down their location more or less, but to say that they are stationary is just weird.

Ziggurat

8th July 2009, 12:17 PM

Aren't electrons in each of the energy levels of the hydrogen atom in energy eigenstates?

Yes.

It would be surreal to suggest that electrons in atomic hydrogen are stationary. Yes, we know their energy and can pin down their location more or less, but to say that they are stationary is just weird.

Depending on what one means by "stationary", one can consider an electron in a hydrogen orbital to be non-stationary, but since the only thing that changes over time in an eigenstate is the phase (and absolute phase never matters), it is indeed stationary in the sense that expectation values have no time dependence. And while this isn't exactly a classical definition of "stationary", I see no problem with using such a definition. In fact, energy eigenstates are often referred to as "stationary states". Weird? Perhaps. But you can't avoid weirdness in quantum mechanics, so I don't see much reason to bother trying.

But even if we try to adopt a more classical approach, how is the ground-state wave function of an electron around a hydrogen atom not stationary? Not only is the expectation value of position static (and hence the expectation value of momentum is zero), but even the expectation value of angular momentum is zero. We have to go to the expectation value of momentum squared to find anything that looks like motion. And while this is non-zero, that too is weird, since the distinction between <p>2 and <p2> is intrinsically quantum mechanical.

Evilgiraffe

8th July 2009, 12:21 PM

Fair enough, I think I understand that. My exposure to QM is rather limited being a chemist and I've never really explored time dependence and phases in QM.

Thanks for the explanation. :D

Dr. Trintignant

8th July 2009, 04:23 PM

Ziggurat

8th July 2009, 04:40 PM

Granted, these individual paths are inaccessible, and we can't really say that they "really" exist, but just the same, I don't think it's completely unreasonable to view the electron cloud as having some kind of fundamental motion to it.

I'm not trying to suggest it is unreasonable to use a definition of "stationary" in which an electron never is, only that it is reasonable (and quite simple) to use a definition in which the electron can be. And as I suggested above, one need not even go all the way to path integrals to come up with a sense in which electrons are never stationary (namely, non-zero <p2> ). But this really becomes a matter of preference. I'm satisfied with considering eigenstates as stationary.

Perpetual Student

8th July 2009, 04:43 PM

What if you look at it from a path integral formulation? The probability cloud may be static, but the paths which sum up to that cloud are not.

Granted, these individual paths are inaccessible, and we can't really say that they "really" exist, but just the same, I don't think it's completely unreasonable to view the electron cloud as having some kind of fundamental motion to it.

- Dr. Trintignant

OK then, if so, should we conclude that it is this fundamental motion that gives rise to a magnetic field or is it electron spin or both -- or something else (like an intrinsic magnetic field) or some combination of these?

Dr. Trintignant

8th July 2009, 05:34 PM

OK then, if so, should we conclude that it is this fundamental motion that gives rise to a magnetic field or is it electron spin or both -- or something else (like an intrinsic magnetic field) or some combination of these?

Both, although I couldn't tell you the relative contributions (I guess it depends on the scenario). Anyway, with the path integral formulation, you have to account for the different paths canceling each other out. The ground state hydrogen electron is totally isotropic--all paths have a symmetric copy, and so you end up with no net angular momentum, and thus no magnetic field. The same is not necessarily true of other orbitals.

- Dr. Trintignant

Dr. Trintignant

8th July 2009, 05:39 PM

I'm not trying to suggest it is unreasonable to use a definition of "stationary" in which an electron never is, only that it is reasonable (and quite simple) to use a definition in which the electron can be. And as I suggested above, one need not even go all the way to path integrals to come up with a sense in which electrons are never stationary (namely, non-zero <p2> ). But this really becomes a matter of preference. I'm satisfied with considering eigenstates as stationary.

Agreed. I just find the path integral approach to be handy in getting an intuitive sense of why a "stationary" orbital can nevertheless possess angular momentum, and thus a magnetic moment. There may not "really" be zillions of electrons whirring around, summing up to a finite electric current, but the math certainly works out the same.

- Dr. Trintignant

Pmb

8th July 2009, 07:19 PM

No. Read up on zero-point energy, especially in regards to electrons in an atom following the behavior of a quantum harmonic oscillator. Even if you were to cool something to absolute zero, the ZPE would still be there due to the quantum "wiggling" of the electrons.
I read that page and I found the statement

According to this expression, an atomic system at absolute zero retains an energy of hv/2.

to be a bit confusing.

The Wikipedia quote refers to a system of harmonic oscillators--specifically, the vibrational modes of diatomic hydrogen molecules. Picture them as billiard balls connected in pairs with springs. You can talk about the temperature of any large system, so you can apply the concept of absolute zero to any large system. It could be a system of atoms, or photons, or something even more exotic. So when one speaks of zero absolute temperature one normally means that the kinetic energy of all the atoms is zero. So what degrees of freedomg that page refers to when it speaks of a zeropoint energy of hv/2 is unclear to me. What do you think it means?

Reality Check

8th July 2009, 09:10 PM

I read that page and I found the statement

to be a bit confusing.

The Wikipedia quote refers to a system of harmonic oscillators--specifically, the vibrational modes of diatomic hydrogen molecules. Picture them as billiard balls connected in pairs with springs. You can talk about the temperature of any large system, so you can apply the concept of absolute zero to any large system. It could be a system of atoms, or photons, or something even more exotic. So when one speaks of zero absolute temperature one normally means that the kinetic energy of all the atoms is zero. So what degrees of freedomg that page refers to when it speaks of a zeropoint energy of hv/2 is unclear to me. What do you think it means?
It is the difference between classical and quantum mechanics.
In classical mechanics, a system of harmonic oscillators has a zero point energy of zero.

In quantum mechanics, a system of harmonic oscillators has a zero point energy of hv/2. The result is that even at absolute zero (http://en.wikipedia.org/wiki/Absolute_zero) there is kinetic energy.

Cuddles

9th July 2009, 08:25 AM

If it's in an energy eigenstate, I think it's safe to call it stationary.

Unless it's at absolute zero, it's not stationary.

sol invictus

9th July 2009, 08:44 AM

Unless it's at absolute zero, it's not stationary.

I'm not sure I understand the distinction you're trying to draw.

Take a single harmonic oscillator. In its ground state - absolute zero - all expectation values are independent of time, but the expectation value of (for example) the root--mean-square distance from the origin is non-zero. You can either call that stationary - since there's no time dependence - or not - since the RMS distance is non-zero. But you should pick one or the other and be consistent.

In an excited energy eigenstate, exactly the same things are true. So you could either call it stationary or not, but you should be consistent with your terminology for the ground state.

At finite temperature in a thermal ensemble, the same things are true again.

Ziggurat

9th July 2009, 12:00 PM

Unless it's at absolute zero, it's not stationary.

How do you assign a temperature to a single-particle system? It doesn't really work. You could put a single particle system in thermal contact with a heat bath at some temperature, and allow it to exchange energy, but what you're really doing is turning on interactions so that single-particle energy eigenstates aren't energy eigenstates of the combined system anymore, and so they're no longer truly stationary. But if you keep the single-particle system thermally isolated, then temperature is neither really definable nor relevant, and even non-ground-state energy eigenstates are still stationary. In practice that can be hard to do, but not in principle.

sol invictus

9th July 2009, 04:27 PM

How do you assign a temperature to a single-particle system? It doesn't really work.

Sure it does. Put it into a thermal density matrix. Physically it means you've weakly coupled the particle to a large heat bath. Mathematically it's perfectly well defined, and absolute zero - the zero-temperature limit of the thermal ensemble - is a pure state, the ground state.

You could put a single particle system in thermal contact with a heat bath at some temperature, and allow it to exchange energy, but what you're really doing is turning on interactions so that single-particle energy eigenstates aren't energy eigenstates of the combined system anymore, and so they're no longer truly stationary.

That's not the case. At least in the above treatment, the finite temperature density matrix is just as stationary as an ordinary energy eigenstate (because it is thermal equilibrium, and equilibrium is static).

Cuddles

10th July 2009, 09:07 AM

How do you assign a temperature to a single-particle system?

At which point is any real electron part of a single particle system?

Ziggurat

10th July 2009, 02:13 PM

Sure it does. Put it into a thermal density matrix. Physically it means you've weakly coupled the particle to a large heat bath. Mathematically it's perfectly well defined

It may be perfectly well defined, but it's not the same thing as an energy eigenstate.

and absolute zero - the zero-temperature limit of the thermal ensemble - is a pure state, the ground state.

Not necessarily. The zero-temperature limit of some systems (such as frustrated Ising magnets) can retain macroscopically large entropies.

Ziggurat

10th July 2009, 02:15 PM

At which point is any real electron part of a single particle system?

All the time in toy problems. Yes, you don't get that in real problems. But by the same token, you don't get pure energy eigenstates in real problems either.

sol invictus

10th July 2009, 02:20 PM

It may be perfectly well defined, but it's not the same thing as an energy eigenstate.

Eh? Who said it was?

You claimed: "How do you assign a temperature to a single-particle system? It doesn't really work" and I gave you an example of a well-defined, physical way to do it - one which is stationary in precisely the same sense an energy eigenstate is.

Not necessarily. The zero-temperature limit of some systems (such as frustrated Ising magnets) can retain macroscopically large entropies.

We were discussing single-particle systems.

Ziggurat

10th July 2009, 02:23 PM

Eh? Who said it was?

I guess nobody. I'm getting sidetracked from my original point, though, which is that an eigenstate, be it the ground state or not, can be considered stationary. Temperature is essentially irrelevant in that picture.

We were discussing single-particle systems.

Not if you want to introduce a heat bath.

sol invictus

10th July 2009, 02:26 PM

I'm getting sidetracked from my original point, though, which is that an eigenstate, be it the ground state or not, can be considered stationary. Temperature is essentially irrelevant in that picture.

I don't agree. States which can be interpreted as corresponding to finite temperature - either canonical or microcanoncal - are precisely as stationary as energy eigenstates, in exactly the same way and for the same reasons.

Not if you want to introduce a heat bath.

But you don't need to. The thermal density matrix can be interpreted that way, but it's a perfectly good state on its own.

Ziggurat

10th July 2009, 02:40 PM

I don't agree. States which can be interpreted as corresponding to finite temperature - either canonical or microcanoncal - are precisely as stationary as energy eigenstates, in exactly the same way and for the same reasons.

That may well be. But now I'm not seeing your point. I was talking about eigenstates when the issue of temperature was introduced. You are introducing states which are not eigenstates. Neither the stability nor the temperature of these other states tells us much about the stability or the temperature of an energy eigenstate in general (the zero-temperature non-degenerate ground-state case being a trivial exception).

sol invictus

10th July 2009, 02:45 PM

That may well be. But now I'm not seeing your point. I was talking about eigenstates when the issue of temperature was introduced. You are introducing states which are not eigenstates. Neither the stability nor the temperature of these other states tells us much about the stability or the temperature of an energy eigenstate in general (the zero-temperature non-degenerate ground-state case being a trivial exception).

Cuddles said: "Unless it's at absolute zero, it's not stationary." My point is that finite temperature states are stationary in exactly the same sense energy eigenstates (such as the ground state, or absolute zero limit of finite T) are.

Hence what Cuddles said was incorrect - not because s/he introduced temperature as you objected, but because s/he drew a distinction which does not exist.

ben m

10th July 2009, 03:19 PM

Cuddles said: "Unless it's at absolute zero, it's not stationary." My point is that finite temperature states are stationary in exactly the same sense energy eigenstates (such as the ground state, or absolute zero limit of finite T) are.

Hence what Cuddles said was incorrect - not because s/he introduced temperature as you objected, but because s/he drew a distinction which does not exist.

Hi Sol,

Finite temperature states are stationary in *some* of the same senses as energy eigenstates. For example: in general, energy eigenstates *also* happen to have a time-independent position expectation value. A wave packet with a well-defined and time-dependent position expectation value must have an energy uncertainty and therefore not be an eigenstate. So the quantum definition of "stationary=eigenstate" coincides with the intuitive definition "stationary=not moving".

Your thermal state is not an eigenstate of the Hamiltonian, and moreover (not in general, but often) its position expectation value will vary---I dare say it will undergo a quantum version of Brownian motion. So it doesn't satisfy either of the two most common "stationary" definitions---it's not an eigenstate and its position is not sitting still. The only thing stationary about it is that its energy distribution does not change over time---that's unusual for an interacting system, but perfectly normal for most non-interacting ones that I wouldn't think to call "stationary".

Just my two bits.

sol invictus

10th July 2009, 03:27 PM

Finite temperature states are stationary in *some* of the same senses as energy eigenstates.

Nope - it's stationary in all senses. If anything it's even more stationary than an energy eigenstate, because it doesn't even have an overall phase that rotates.

Your thermal state is not an eigenstate of the Hamiltonian, and moreover (not in general, but often) its position expectation value will vary---I dare say it will undergo a quantum version of Brownian motion.

No, it won't. It can't, because there is no time dependence in thermal equilibrium. Mathematically that follows immediately from the fact that the thermal density matrix is diagonal in the energy basis, and from cyclicity of the trace. Physically it follows from the fact that Brownian motion is random, hence its average is zero, hence by Ehrenfest's theorem the quantum "expectation value" (trace in the density matrix) must be zero too.

Of course the expectation value of x^2 isn't zero, but it isn't for an energy eigenstate either - and it's still time-independent.

So it doesn't satisfy either of the two most common "stationary" definitions---it's not an eigenstate and its position is not sitting still.

The first is not a definition - it's a consequence of the second - and thermal equilibrium satisfies the second.

ben m

10th July 2009, 11:41 PM

Nope - it's stationary in all senses. If anything it's even more stationary than an energy eigenstate, because it doesn't even have an overall phase that rotates.

Hmm. My mental picture of this state may be wrong. I'm thinking of (e.g.) a particle in a box, say, in a mixed state |Phi> = a x exp(-E0/2kT + i E0 t/hbar) |0> + b x exp(-E1/2kT + i E1 t/hbar) |1> + c exp(-E2/2kT + i E2 t/hbar) |2> + ... so that the probability distribution of the energies is thermal, and where a,b,c are random complex numbers of absolute value 1. Is that wrong? 'Cause it sure doesn't look like <x|Phi> = constant for that state. It's not even obvious that <x|Phi> = constant is possible in special cases.

sol invictus

11th July 2009, 07:16 AM

Hmm. My mental picture of this state may be wrong. I'm thinking of (e.g.) a particle in a box, say, in a mixed state |Phi> = a x exp(-E0/2kT + i E0 t/hbar) |0> + b x exp(-E1/2kT + i E1 t/hbar) |1> + c exp(-E2/2kT + i E2 t/hbar) |2> + ... so that the probability distribution of the energies is thermal, and where a,b,c are random complex numbers of absolute value 1. Is that wrong? 'Cause it sure doesn't look like <x|Phi> = constant for that state. It's not even obvious that <x|Phi> = constant is possible in special cases.

That's not a mixed state - it's a pure state. It has zero entropy, and does not describe anything thermal. If you like, you can think of your state as a "typical" pure state picked from a canonical ensemble. It has time dependence because any individual such state should, since it corresponds to a particular choice of motions of the particles (or, in QM, phases).

A canonical thermal mixed state looks like \rho_T = \Sum_i e^{-E_i/kT} |E_i><E_i| - it's a density matrix, diagonal in the energy basis. It has entropy -Tr(\rho ln \rho).

Pmb

11th July 2009, 07:43 AM

It is the difference between classical and quantum mechanics.
In classical mechanics, a system of harmonic oscillators has a zero point energy of zero.

In quantum mechanics, a system of harmonic oscillators has a zero point energy of hv/2. The result is that even at absolute zero there is kinetic energy.

Thanks but that doesn't answer my question, i.e. in what sense does the article mean absolute zero in temperature?

Pmb

11th July 2009, 08:14 AM

Take a single harmonic oscillator. In its ground state - absolute zero -...

The ground of a single harmonic oscillatorstate is the state of lowest energy whereas absolute zero refers to zero energy (e.g. zero kinetic energy and zero potential energy).

sol invictus

11th July 2009, 08:41 AM

The ground of a single harmonic oscillatorstate is the state of lowest energy

Correct - and that lowest energy is greater than zero.

whereas absolute zero refers to zero energy (e.g. zero kinetic energy and zero potential energy).

No, that's not correct - in fact no such configuration is physically possible. Absolute zero means the lowest possible temperature; i.e. the temperature with the property that heat will never flow out of a system at that temperature, no matter what it is put in contact with. Therefore absolute zero is the ground state, both formally - the zero temperature limit of the finite temperature ensemble is the ground state - and physically - the only state that can have the property I just specified is the ground state.

The only possible subtlety is when there are multiple ground states with degenerate energies, but in that case the best definition of absolute zero is a density matrix with equal coefficients for each state. In that case absolute zero does not have zero entropy - it has entropy equal to the log of the ground state degeneracy.

Perpetual Student

11th July 2009, 04:40 PM

:popcorn1

Reality Check

11th July 2009, 04:59 PM

Thanks but that doesn't answer my question, i.e. in what sense does the article mean absolute zero in temperature?
Absoute zero in termerature is the assignment of the number zero to the state of the system where no heat can be extracted from the system.
As sol invictus's post states this state does not have zero energy.

Pmb

12th July 2009, 07:23 AM

Absolute zero means the lowest possible temperature; ..

Thanks for the correction. I've been reviewing thermo lately and had in mind an extraploation point relating to an ideal gas. I forgot that T was defined in terms of entropy. It's been 20 years since I've thought about that part. Damn thermodynamics. Always my worst subject. :)

How did the subject matter on this thread change so much from the OP?

stevea

15th July 2009, 06:22 PM

In a permanent magnet, where is the current then?

The current are the electrons rotating around the aligned iron atoms (trapped in a crystal structure..

Imagine you caused a point charge to rotate around some axis - say you captured an electron in a jar and mounted it to a bicycle wheel and rotated the wheel. This is a current (similar to charges in a coil) and it causes a mag field to the stationary observer.

If you had 10^26 of these rotating bicycle wheels all aligned and all rotating in the same direction, then the mag fields sum to a huge field strength. If OTOH the wheels are rotating along random axes then the mag fields of the various wheels cancel each other and there is no net mag field.

This is exactly what happens in in a magnetic material. Then the atoms have a net alignment and the tiny mag fields become apparent as a macroscopic effect. When the atoms are randomly aligned then the same magnet loses it's net field.

--

Some wag who stopped reading the physics books at chaper one will quip that the electrons aren't point charges, but distributed quantum fields. This is true, but the Q-field has a net spin and create a mag field despite this.

stevea

15th July 2009, 11:13 PM

Yes, it can. For example, a current creates a magnetic field but no electric field.

A current is by definition moving charge (moving e-fields).

And you don't need any electric field to maintain a current under some circumstances (for example in a superconductor, or in an ionized gas).

I agree you don't need an external E-field to maintain a current (moving charge) in a vacuum or superconductor - but no one ever said otherwise. A hearty "So what ?" Totally irrelevant. Your ionized gas example is dead-wrong - maintaining ionisation requires a high E-field potential. In detail the E-field tmes the mean-free-path must be greater than the ionization energy of the gas molecules.

You're confused. What relativity tells us is that IF there is an inertial frame in which one has a magnetic field but zero electric field, then in any other frame there will also be an electric field. Both are elements of a second rank tensor, so they transform into each other when one changes frames. But that doesn't change the fact that one can certainly have a magnetic field with zero electric field.

I'm not confused at all, but you're confusing mathematics with the physics - a common error on these public forums. Yes you can describe a M & E fields with tensors, and it describes the same universe as the Maxwell's eqn and the Lorentzian view, but it's just a mathematical device. It doesn't change the fact that the M-field is ALWAYS an artifact of the E-field as viewed in a moving frame of reference; not the other way. If it was the "other way" then we would have magnetic monopoles. They don't exist.

When a charge is at rest wrt a static E-field the force is completely described by the E-field. When the charge is in motion relative to the E-field then the Lorentzian contraction creates a different force vector than would otherwise be expected by non-relativistic view. To mathematically explain the modified force we can (and have) created a fictitious a B-field (magnetic field) term and insert it into tensors or the Maxwell eqn but this is just a mathematical device. There is no Magnetic field at all !!! It's just a means to describe the relativistic distortion of the electrostatic field. All magnetic forces can be fully explained as "relativistic electrodynamics".

Here is the simplest example I can think of. We have two parallel wires. each is electrically neutral (no net charge, the protons and electrons match). So an electon in wire A sees no electrostatic force ... it's wire and the parallel ones are neutral.

Now we apply a voltage and cause a current to flow the same direction in both wires. The wires are still "net neutral". We all know the wires are attracted together by the fictitious magnetic force which has been well described by all sorts of Magnetic equations forces. But I've already stated that the M-field is a fiction. Don't get me wrong Maxwell's 1873 equations are perfectly accurate for calculations, but the explanation using fictitious fields is baloney - mathematical artifice that explains nothing wrt physics.

So here is the inside scoop. When a current flows in a wire the protons are fixed and the valence electrons flow. in the "parallel wires, same direction of current flow" case the Protons fixed in wire A (call them protons-A) still see their non-moving neighbor protons-B (in wire B) in the same frame of reference, same electrostatic force as before. They also see the moving electrons-B, and in the protons view the once spherically even electrical field of the electons-B is flattened in the direction of motion, (the Lorentz contraction). So protons-A see a stronger attractive force from the moving electrons-B which are just across from it and just the same force from the still fixed protons-B. ProtonsA are attracted toward wire-B by this extra force.

The moving electronsA see the same repulsive electrostatic force from the electrns of wireB, but they view the field of the protonsB as flattened in their(the electrons) direction of motion, so again there is an added attractive force between the electrons in A and nearest protons in B, but just the same force to the electrons of B. The sum is that the electrons in A also experience an added attraction toward wire B.

The argument is symmetric and wire B sees extra attraction to wire A as well.

Mag fields are entirely explained by electrostatic forces calculates withthe relativistic correction. There is absolutely no need to create hypothetical forces like B-fields and certainly not hypothetical particles like magnetic monopoles.

Einstein stood the world on it's ear when he explained this in 1905 and successfully subsumed magnetic into electrostatics. It seems your physics is more than a century out of date.

sol invictus

16th July 2009, 05:52 AM

A current is by definition moving charge (moving e-fields).

Nope. A current is by definition the source on the right hand side of an equation like Ampere's law. Sometimes it comes from a moving charge, sometimes it doesn't. For example the magnetic dipole moment of an elementary point particle cannot be said to come from moving charges. And some of them are exactly neutral - zero electric field.

But even in a classical theory where we can ignore those facts, moving charge does not mean moving E-fields. If you have equal density of positive and negative ions, with the positive ions moving and the negative ions at rest, the E-field on any scale larger than the ion-ion separation is zero. And since we can make that separation as small as we like, there can be zero E-field.

Your ionized gas example is dead-wrong - maintaining ionisation requires a high E-field potential. In detail the E-field tmes the mean-free-path must be greater than the ionization energy of the gas molecules.

Utter nonsense. Any gas will ionize at sufficiently high temperatures or when exponsed to enough ionizing radiation, all with zero E. In fact most of the matter in the universe is ionized precisely because of that.

I'm not confused at all, but you're confusing mathematics with the physics - a common error on these public forums. Yes you can describe a M & E fields with tensors, and it describes the same universe as the Maxwell's eqn and the Lorentzian view, but it's just a mathematical device.

A common "feature" of these forums are ignorant, arrogant physics crackpots posting misinformed garbage. So now Maxwell's equations are wrong - they're mere mathematics, nothing compared to your secret knowledge - and you will reveal the truth about a 150 year old, incredibly well-understood theory to us.

Einstein stood the world on it's ear when he explained this in 1905 and successfully subsumed magnetic into electrostatics. It seems your physics is more than a century out of date.

Yours is quite simply wrong. Einstein didn't change Maxwell's equations in the slightest - in fact they were the example that led him to special relativity. What he showed is that B and E fit into a second-rank Lorentz tensor. One thing that makes clear is that B and E are on a precisely equivalent footing in vacuum - you can exchange one for the other by a Poincare duality. They are analogous to x and y components of, say, velocity. The only asymmetry comes from the lack of magetic monopoles - but as I've explained, that doesn't preclude B fields with no E fields.

Ziggurat

16th July 2009, 09:03 AM

The current are the electrons rotating around the aligned iron atoms (trapped in a crystal structure..

Except in most magnetic materials, these orbital moments are quenched, and only spin moments (which do not consist of displacement of the electron) remain. Furthermore, how do you explain the magnetic moment of a neutron? By the motion of charge?

Singularitarian

16th July 2009, 09:12 AM

I believe this has been discussed on other threads. I never did see a definitive answer, however. If there was one I missed it -- could someone summarize any conclusions, please.

If you mean by a moving electric field as in its composition i.e. the statistical averages that mediate the force, then yes of course the magnetic field still exists. In relativity, electrons experience the magnetic force when moving through a wire. Photons might not experience any internal charge, but does have a relation to gravity through the curvature it posesses in its momentum through spacetime, curving along its path. Magnetic Fields and Electric Field are synonymous fields representing one single fact in the current Unification in physics. Just because a systems Eigentsate may not describe a value for momentum does not mean that the fields of electromagnetism fail, or magnatism in general. In effect, magnetism is a force arising from the angular momentum of a large group of particles acting towards the same directional vector line. Electric forces exist not only as innate properties of particles like electrons or positrons, but also as a field of background energy which arises from this in the form of photon energy. The absence of any state in a quantum object never means the absence of a quantum field acting in the universe.

Perpetual Student

16th July 2009, 10:01 AM

...

Yes you can describe a M & E fields with tensors, and it describes the same universe as the Maxwell's eqn and the Lorentzian view, but it's just a mathematical device. It doesn't change the fact that the M-field is ALWAYS an artifact of the E-field as viewed in a moving frame of reference; not the other way. If it was the "other way" then we would have magnetic monopoles. They don't exist.

When a charge is at rest wrt a static E-field the force is completely described by the E-field. When the charge is in motion relative to the E-field then the Lorentzian contraction creates a different force vector than would otherwise be expected by non-relativistic view. To mathematically explain the modified force we can (and have) created a fictitious a B-field (magnetic field) term and insert it into tensors or the Maxwell eqn but this is just a mathematical device. There is no Magnetic field at all !!! It's just a means to describe the relativistic distortion of the electrostatic field. All magnetic forces can be fully explained as "relativistic electrodynamics".

...

That is an interesting perspective. Does this represent a larger "school of thought" among physicists or is it solely your opinion? Can you provide any sources further describing this perspective?

soylent

16th July 2009, 05:48 PM

Fair enough, I think I understand that.

Here, have an awful analogy: The eigenstates of a bound electron in a hydrogen atom are a little bit like a set of roleplaying dice. A stationary state is one in which you keep choosing the same dice(e.g. the octahedron numbered 1 through 8) without dependence on time.

soylent

16th July 2009, 06:02 PM

Furthermore, how do you explain the magnetic moment of a neutron? By the motion of charge?

A neutron is not an elementary particle. It contains one up quark(+2/3e charge) and two down quarks(-1/3e charge). I don't know why the magnetic moments of the quarks do not cancel as it does for overall charge, but apparently it does not.

Perpetual Student

16th July 2009, 06:24 PM

Originally Posted by stevea

...

Yes you can describe a M & E fields with tensors, and it describes the same universe as the Maxwell's eqn and the Lorentzian view, but it's just a mathematical device. It doesn't change the fact that the M-field is ALWAYS an artifact of the E-field as viewed in a moving frame of reference; not the other way. If it was the "other way" then we would have magnetic monopoles. They don't exist.

When a charge is at rest wrt a static E-field the force is completely described by the E-field. When the charge is in motion relative to the E-field then the Lorentzian contraction creates a different force vector than would otherwise be expected by non-relativistic view. To mathematically explain the modified force we can (and have) created a fictitious a B-field (magnetic field) term and insert it into tensors or the Maxwell eqn but this is just a mathematical device. There is no Magnetic field at all !!! It's just a means to describe the relativistic distortion of the electrostatic field. All magnetic forces can be fully explained as "relativistic electrodynamics".

...
That is an interesting perspective. Does this represent a larger "school of thought" among physicists or is it solely your opinion? Can you provide any sources further describing this perspective?

Does anyone else have an opinion about this?

Ziggurat

16th July 2009, 06:49 PM

Does anyone else have an opinion about this?

I'd call it half right. Let's ignore the issue of the dipole moments intrinsic to elementary particles for the moment. Magnetic fields are indeed what you get when you take an electric field and transform to another reference frame. But there's no reason to conclude from that that magnetic fields are in any way fictitious. They are not.

Magnetic fields are 3 components of the 4x4 antisymmetric electromagnetic field tensor. Just like 4-vectors, one transforms this field when switching reference frames by applying the Lorentz transformation matrix to it. When you do such a transformation, the electromagnetic field is in some sense still the same field, but the components (which distinguish between electric field and magnetic field) get altered.

Now let's consider some simple examples. Let's suppose we have a reference frame in which E is non-zero and M is zero. We switch reference frames, and lo and behold, M becomes nonzero and E changes. But M is not equivalent to E in this scenario: I can keep switching reference frames, and find all sorts of different values for M and E, but I can never transform to a reference frame in which E becomes zero, even though I can go back to the frame where M is zero. One might naively conclude that this means that E is fundamental and M is artificial.

But let's consider the reverse now. What if we start with a field in which M is non-zero and E is zero? Well, if I switch reference frames from here, I can make E nonzero, but I can never make M zero. So there's a sort of symmetry between them.

What does this have to do with the question of whether or not magnetic fields are mere artifacts? Well, let's take the example of a clear-cut artifact: classical fictitious forces like the centripetal force. The centripetal force exists in rotating reference frames, but it's termed "fictitious". Why? Because it only exists in certain reference frames. If you find that you have a centripetal force in your reference frame, you can always find a reference frame in which it disappears. Its existence is then merely an artifact of our choice of reference frames, and not real in the way that forces which exist in all reference frames are real.

Now from our first example, if we can pick a reference frame which makes the magnetic field disappear, doesn't that mean that the magnetic field is also just an artifact in a similar sense? Well, not quite. Why not? Because of the second example: there are cases in which it is not possible to pick a reference frame in which the magnetic field disappears. You can't get rid of it no matter what frame you pick.

Perpetual Student

16th July 2009, 07:21 PM

Ziggurat:

Thanks for your helpful response. Am I wrong in thinking you mean centrifugal force above? I believe centripetal force for a rotating system exists in all frames, unlike centrifugal force.
Also, can you give me a physical example of a non zero M field with a zero E field?

Ziggurat

16th July 2009, 10:05 PM

Ziggurat:

Thanks for your helpful response. Am I wrong in thinking you mean centrifugal force above?

No, you're correct. I screwed up the word. The conceptual difference is easy for me, but I sometimes find myself switching the words.

Also, can you give me a physical example of a non zero M field with a zero E field?

Sure: a current-carrying superconducting wire.

Perpetual Student

16th July 2009, 10:25 PM

Sure: a current-carrying superconducting wire.

I'm missing something. Why does a superconducting wire carrying a current have no E field? Would not the current be electrons moving within the wire with an electric field?

Ziggurat

16th July 2009, 11:40 PM

I'm missing something. Why does a superconducting wire carrying a current have no E field? Would not the current be electrons moving within the wire with an electric field?

Nope. No E field is required for steady-state currents in a superconductor, because they have no resistance.

Perpetual Student

17th July 2009, 12:49 PM

Nope. No E field is required for steady-state currents in a superconductor, because they have no resistance.

How can that be? Why would lack of resistance negate the E field of electrons? Sorry, but I am not a physicist and obviously I'm missing something here.

Ziggurat

17th July 2009, 12:59 PM

How can that be? Why would lack of resistance negate the E field of electrons? Sorry, but I am not a physicist and obviously I'm missing something here.

What happens to electrons in an electric field? They accelerate. Which would cause the current to increase. That doesn't usually happen in a wire with resistance, because that resistance serves to slow the electrons down, like friction. Electrons moving through a resistive wire in an electron are, in effect, basically always "falling" at their terminal velocity in that field. But with no resistance, any electric field will produce ever-increasing electron velocities, and ever-increasing currents. OK, so what happens if we turn on an electric field to get a current moving, and then turn it off? In an ordinary wire, electrons accelerate when we apply a field until they reach terminal velocity (determined by Ohm's law, and that equilibration typically happens VERY fast), and when we turn off the field, they slow down because of this same resistance and stop (also typically very quickly). But in the case of a superconductor, there's no resistance to slow them down. So we use an E field to get them moving, but then we can turn it off, and they keep moving on their own. Why do they keep moving? Because there's nothing to stop them from moving. It's zero resistance.

Perpetual Student

17th July 2009, 04:14 PM

What happens to electrons in an electric field? They accelerate. Which would cause the current to increase. That doesn't usually happen in a wire with resistance, because that resistance serves to slow the electrons down, like friction. Electrons moving through a resistive wire in an electron are, in effect, basically always "falling" at their terminal velocity in that field. But with no resistance, any electric field will produce ever-increasing electron velocities, and ever-increasing currents. OK, so what happens if we turn on an electric field to get a current moving, and then turn it off? In an ordinary wire, electrons accelerate when we apply a field until they reach terminal velocity (determined by Ohm's law, and that equilibration typically happens VERY fast), and when we turn off the field, they slow down because of this same resistance and stop (also typically very quickly). But in the case of a superconductor, there's no resistance to slow them down. So we use an E field to get them moving, but then we can turn it off, and they keep moving on their own. Why do they keep moving? Because there's nothing to stop them from moving. It's zero resistance.

Them are still electrons, and they are moving. So, why do these electrons have no E field? An electron moving along in empty space experiences no resistance, but I assume it still has an E field. Not true?

Ziggurat

17th July 2009, 04:17 PM

Them are still electrons, and they are moving. So, why do these electrons have no E field?

They do. But so do the protons. And the electric fields from all those electrons and all those protons cancel.

Perpetual Student

17th July 2009, 04:33 PM

:idea:

THANKS.

stevea

20th July 2009, 11:13 AM

I agree w/ the 1st 3 paragraphs of Ziggurat, (except the word "naive"). After that he goes off the rails. We have collected experimental evidence of static charges since the ancient Greeks noted that amber holds a charge. To date there is ZERO evidence for magnetic monopoles - the sort of magnetic field w/ no related E-field.

Perhaps he has some fundamental misunderstanding of how science work, but it is NOT about creating a mathematical model and then extrapolating it's results and treating these as though they are real. There must be a grounding in physical evidence. Yes there is a sort of symmetry in the tensor equations, but there is ZERO physical evidence for the M field w/o E field ! Once again a lot of amateur lunacy revolves around confusing the equations for the science.

BTW this stuff (B fields are just E fields viewed w/ relativistic correction) is taught in undergrad physics courses I've taken ~3+ decades ago. It's the gist of Einstein's paper. This isn't some personal opinion.

==
As for the superconductor ... there are still E fields in the moving electrons and stationary protons and it's complete nonsense to ignore these.

When we push electrons thru a conventional wire or thru an ionized gas, the electrons collide (w/ the lattice or the gas; ergo resistance). To continue the net electron motion (current) we may apply an external E field to re-accelerate these again after the collision (any force would do). If OTOH we start an electron along a superconductor or in a vacuum, then the electrons continue unimpeded, w/o collisions, without need of a small external field. THIS EXTERNAL E-FIELD IS NOT THE FIELD CREATING THE MAGNETIC TERM IN ANY CASE !!!

The statement Ziggurat and others are trying to make - that a superconducting coil w/ current has a B-field but there is no E-field is DEAD WRONG ! There is no net charge since the electons and protons match. The individual E-fields for every proton and electron exist and are "distorted" by the lorenztian contraction, so the sum of the ensemble e-fields do not exactly cancel. The electrons and protons are not moving together in any magnetic coil, and it is EXACTLY the difference in these motions that create the "B-field effect" of the coil. The magnetic field term of the relativistically viewed ensemble of moving E-fields is entirely due to the e- and P+ charges and their relative motion, and the observers differential view of these two. Yes superconducting magnetic coils have exactly the same dependence on moving E-fields as does any electromagnetic coil.

I'm not trying to make an argument from authority here, but I have a masters degree in physics and you all would be well served by digging though a good undergrad text.

=======

To correct post#91, in any electromagnet, superconducting or not, the motion of the moving electrons and fixed protons relative to the observer creates a difference in the distribution of their observed electrostatic fields. The once spherically uniform electron's E-fields flatten due to Lorenzian contraction. Ziggurat is wrong, the fields don't cancel. If there was no electron motion (no current) then the fields would cancel. The fields are distorted by their motion and so the difference of the proton and electron fields no longer cancel. The difference in these fields (in the frame of the observer) with their different spatial distributions causes an externally viewed "magnetic field", which is in reality just the difference in the distorted E-fields. That is all i is.. Yes superconductors are reliant on the relative motion of these static E-fields. to create the "B-field effect". The superconducting magnet has many unit charge E-fields, and the sum of the fields is not zero in some frames of reference.

Ziggurat

20th July 2009, 11:47 AM

I agree w/ the 1st 3 paragraphs of Ziggurat, (except the word "naive"). After that he goes off the rails. We have collected experimental evidence of static charges since the ancient Greeks noted that amber holds a charge. To date there is ZERO evidence for magnetic monopoles - the sort of magnetic field w/ no related E-field.

I never said anything about magnetic monopoles. Nothing in my post depends upon or even suggests monopoles.

Perhaps he has some fundamental misunderstanding of how science work, but it is NOT about creating a mathematical model and then extrapolating it's results and treating these as though they are real. There must be a grounding in physical evidence. Yes there is a sort of symmetry in the tensor equations, but there is ZERO physical evidence for the M field w/o E field !

I gave an experimental example of exactly that scenario: a superconducting wire with a constant current.

BTW this stuff (B fields are just E fields viewed w/ relativistic correction) is taught in undergrad physics courses I've taken ~3+ decades ago. It's the gist of Einstein's paper. This isn't some personal opinion.

I never said it was anyone's personal opinion. But where does this notion come from? Why, it comes from mathematical models. You can't simultaneously denigrate the use of such models by others and yet rely upon them yourself for the foundation of your argument.

As the the superconductor ... there are still E fields in the moving electrons and stationary protons and it's complete nonsense to ignore these.

No, it's not nonsense. Step outside the wire, and what do you find? You find places in space where the E field is zero, and the M field is non-zero. Whether there's any microscopic E fields near the electrons is irrelevant: we still have a situation in which there are areas of space with no E field but nonzero B field.

THIS EXTERNAL E-FIELD IS NOT THE FIELD CREATING THE MAGNETIC TERM IN ANY CASE !!!

I never said it was. The point was to create a situation in which the complete absence of an E field was obvious. The fact that even in a resistive wire the E field and the M field are decoupled is not news to me, but it requires a level of analysis for beginners that I hoped to avoid having to deal with.

The statement Ziggurat and others are trying to make - that a superconducting coil w/ current has a B-field but there is no E-field is DEAD WRONG !

No, it really isn't.

There is no net static E-field since the electons and protons match. There ARE individual E-fields for every proton and electron.

And if you go outside the wire, such fields become arbitrarily small, and we can safely take them to be zero. Thus, there are areas of space with zero E field and nonzero B field.

The electrons and protons are not moving together in any magnetic coil, and it is EXACTLY the difference in these motions that create the "B-field effect" of the coil.

I never said otherwise.

Yes superconducting magnetic coils have exactly the same dependence on moving E-fields as does any electromagnetic coil.

Nonetheless, we still have a situation in which no reference frame can be adopted in which the magnetic field is zero. As such, the label of "fictitious" is, well, problematic. That was the heart of my argument, and you have not addressed that point in any way, shape, or form.

I'm not trying to make an argument from authority here, but I have a masters degree in physics

You aren't trying to make an argument from authority... but you're waving your authority around. Sure.

and you all would be well served by digging though a good undergrad text.

I must admit, I've only read one undergrad electrodynamics textbook, Griffiths' "Introduction to Electrodynamics". I've also only read one graduate E&M textbook, Jackson's "Classical Electrodynamics". Are these sources insufficient? Is there a page in either that you think I missed?

The superconducting magnet has many unit charge E-fields, and the sum of the fields is not zero in some frames of reference.

I never suggested otherwise. In fact, I explicitly said that this would happen (post 83): "What if we start with a field in which M is non-zero and E is zero? Well, if I switch reference frames from here, I can make E nonzero, but I can never make M zero."

So rather than illustrating that I don't know what I'm talking about, I'm afraid all you've done is show that you have trouble reading.

sol invictus

20th July 2009, 01:25 PM

BTW this stuff (B fields are just E fields viewed w/ relativistic correction) is taught in undergrad physics courses I've taken ~3+ decades ago.

Either those 3+ decades have either dulled your memory - no shame in that - or you never understood the stuff to begin with.

It's the gist of Einstein's paper. This isn't some personal opinion.

You ignored the post in which I explained precisely what Einstein showed in that paper (not to mention pointing out a number of other glaring errors you've made). He didn't show that B fields are just E fields. He showed that B and E fields are on precisely the same footing. You might as well say E fields are just B fields - you'd be precisely as wrong.

To repeat: a decent analogy for B and E are the x and y components of a vector. You can always choose your x-y axes in such a way that one or the other is zero, but the sum of their squares is always the same. Similarly, E^2-B^2 and E dot B are the same in all inertial frames, but the values of E and B themselves change.

I'm not trying to make an argument from authority here, but I have a masters degree in physics and you all would be well served by digging though a good undergrad text.

You're going to lose any argument based on credentials. Badly.

Ziggurat

20th July 2009, 01:42 PM

You're going to lose any argument based on credentials. Badly.

Lately I've been seeing an increase in such accusations of educational inferiority. Is it some weird phase of the moon thing? Oh well, it's harmless and amusing.

stevea

27th July 2009, 07:04 AM

You ignored the post in which I explained precisely what Einstein showed in that paper (not to mention pointing out a number of other glaring errors you've made). He didn't show that B fields are just E fields. He showed that B and E fields are on precisely the same footing. You might as well say E fields are just B fields - you'd be precisely as wrong.

Do you mean post #76 where you clownishly misinterpret what I clearly said, then make a pass at your strawman argument ? Why is it my task to carefully read your comments when you don't give mine an honest reading ?

Nope. A current is by definition the source on the right hand side of an equation like Ampere's law. Sometimes it comes from a moving charge, sometimes it doesn't. For example the magnetic dipole moment of an elementary point particle cannot be said to come from moving charges. And some of them are exactly neutral - zero electric field.

Giggles - this is SOOOO silly. If that's your definition of current you need a remedial course. Current: I = dQ/dt. Left hand side, right hand side are different ? Where did you learn math - clown college ? An elementary point particle is only a 19th century didactic fiction.

The magnetic moment of neutral (not point) particles like a neutron is indication of it's lower structure consisting of charged quark particles and their intrinsic spin. I learned about this in an undergrad n.phys course 3+ decades ago. What's your excuse for not knowing this ?

To repeat: a decent analogy for B and E are the x and y components of a vector. You can always choose your x-y axes in such a way that one or the other is zero, but the sum of their squares is always the same. Similarly, E^2-B^2 and E dot B are the same in all inertial frames, but the values of E and B themselves change.

Yes, quite accurate with one major caveat that you continue to ignore. The static electric field is a well grounded in physical evidence. A static magnetic field except produced by moving charges has zero evidence in the physical world. Serious people have looked for magnetic monopoles - no go. So the two mathematical aspect of this expression, the E & B fields do not have equal weight in the physical world.

Let me give you a better analogy. Physicist, electrical engineers and signal processing types all use imaginary and real numbers to represent the time domain relationship between classes of functions. We can write that an inductor has an impedence of iwL (hmm where is the omega key?). This is mathematically accurate and it does have a meaning, but it is NOT that there exists some imaginary form of current or voltage that has a direct physical reality. It's a mathematical device that accurately describes the time domain changes. We can just as easily describe the variables of a pendulum using imaginary terms ... A + i*B ... but the imaginary part is just a device. The reality is we are seeing a potential/kinetic energy oscillator moving in 4 space.

Similarly the B field in all these examples is merely a consequence of moving charges, the E-fields viewed in another frame of reference. If has no separate reality, despite it's appearance in equations.

But even in a classical theory where we can ignore those facts, moving charge does not mean moving E-fields. If you have equal density of positive and negative ions, with the positive ions moving and the negative ions at rest, the E-field on any scale larger than the ion-ion separation is zero. And since we can make that separation as small as we like, there can be zero E-field.

This is ridiculous. A charge is by definition an e-field so moving charges are moving e-fields BY DEFINITION. The charge density of the moving vs stationary charges is NOT uniform in all frames of reference which is why your statements are wrong.

Here is a nice little undergrad level description.
http://physics.weber.edu/schroeder/mrr/MRRtalk.html
http://physics.weber.edu/schroeder/mrr/MRRnotes.pdf

If you can read the 1st 3 sections without your ridiculous definitions and the erroneous mental baggage you've picked up you'll realize that a relativistic view of the charges *IS* the B-field. BTW the Purcell, Berkeley series book he mentions as too complex was my 2nd year undergrad text. Apparently the fabulously credential Sol doesn't yet understand what has been taught to undergrads for decades.

Same goes for Ziggurat, statements like this ....
And if you go outside the wire, such fields become arbitrarily small, and we can safely take them to be zero. Thus, there are areas of space with zero E field and nonzero B field.
Is just ignorant of the facts. In your superconducting wire case the B-field is nothing except a manifestation of the contracted E-field differences of the protons & electrons in the wire as observed in the frame of reference of some moving external charge. To claim the E-field is zero, but there is a B-field is just artifice -- the B-field *IS* the summation of all the E-fields in some frame of reference. That these field cancel (sum to zero) when there is no current does NOT mean they cancel when the charges are in in relative motion, in some frames of reference.

You're going to lose any argument based on credentials. Badly.

Not to you I won't, but I will at this point ignore this braying idiotic thread. It's going nowhere thanks to Sol's purposeful distortions and the general failure of anyone here to pick up a book of view a good web source.

good bye.

sol invictus

27th July 2009, 08:05 AM

Giggles - this is SOOOO silly. If that's your definition of current you need a remedial course. Current: I = dQ/dt.

You do understand that current is a vector, right? And that dQ/dt is a scalar? What you wrote there is the divergence of the current for a charged medium, not the current.

Now, having made an elementary mistake (equating a scalar with a vector), will you calm down and listen?

An elementary point particle is only a 19th century didactic fiction.

No, it's part of the 21st century standard model of elementary particle physics. You see, as far as anyone knows particles like the electron are point particles. That doesn't mean they can be localized at a point - quantum mechanics prevents that. But it does mean that they cannot be smashed apart into anything, and that when you probe closer and closer to them (using higher energy scattering experiments, for example) they appear smaller and smaller, with no apparent substructure. That's exactly how a point particle would behave given quantum mechanics, so physicists say "the electron is a point particle".

The magnetic moment of neutral (not point) particles like a neutron is indication of it's lower structure consisting of charged quark particles and their intrinsic spin.

The neutron is not a point particle. The neutrino is (again, as far as anyone knows). Neutrinos are exactly neutral. And guess what - they've got a magnetic dipole moment.

A static magnetic field except produced by moving charges has zero evidence in the physical world.

Where are the moving charges in a refrigerator magnet? Producing the magnetic dipole moment of a precisely neutral particle with no sub-structure?

Similarly the B field in all these examples is merely a consequence of moving charges, the E-fields viewed in another frame of reference. If has no separate reality, despite it's appearance in equations.

If that were true, it would be impossible to have a B field with no E field. But it is possible, so it's not true.

Not very complicated, is it?

This is ridiculous. A charge is by definition an e-field so moving charges are moving e-fields BY DEFINITION.

Nonsense. A charge is certainly not an E-field by anyone's definition. A charge is a source for an E-field. But if there's an equal an opposite charge right next to it, the E-field will be almost precisely zero. So if there's a long chain of charges at rest, and an equally dense chain of opposite charges in motion on top of them, you get a B field with zero E field.

Yes, in any other frame the charge densities would differ due to Lorentz contraction and both E and B would be non-zero - but there is a frame in which E=0, and there is no frame in which B=0.

Ziggurat

27th July 2009, 11:16 AM

Same goes for Ziggurat, statements like this ....
And if you go outside the wire, such fields become arbitrarily small, and we can safely take them to be zero. Thus, there are areas of space with zero E field and nonzero B field.
Is just ignorant of the facts. In your superconducting wire case the B-field is nothing except a manifestation of the contracted E-field differences of the protons & electrons in the wire as observed in the frame of reference of some moving external charge.

Funny, but nothing you've said indicates that there is not an area of space with zero B-field and nonzero E-field.

To claim the E-field is zero, but there is a B-field is just artifice -- the B-field *IS* the summation of all the E-fields in some frame of reference. That these field cancel (sum to zero) when there is no current does NOT mean they cancel when the charges are in in relative motion, in some frames of reference.

And here you're simply wrong. There is a current in all reference frames in this scenario. Hence, there is a B field in all reference frames. There is NO E-field in one reference frame. You may indeed claim that in this reference frame, the E-fields simply cancel, but they do so in a frame in which there is current and hence relative motion of charges. Even if you insist upon considering the B-field as a reference-frame artifact, the fact remains that in this scenario, the B field is present in all reference frames, but the E field is not.

Before you presume to lecture me about my cluelessness, it would be to your advantage to actually get the scenario right in your own head.

Not to you I won't

Yes, actually, you would.

There are other posters on this board who wouldn't lose a credentials fight to Sol, but only with a draw. And when such posters get in debates, you will find that they never bring up their credentials. They don't need to. By trying to play that card, you have shown how weak your hand really is.

Michael Mozina

27th July 2009, 01:13 PM

Perhaps he has some fundamental misunderstanding of how science work, but it is NOT about creating a mathematical model and then extrapolating it's results and treating these as though they are real. There must be a grounding in physical evidence.

Steve, you've hit the nail on the head. They do *NOT* understand what these formulas even relate to in terms of actual physics. Look at how tusenfem responded to my question about what MHD theory actually relates to in terms of actual physics in the Electric Universe thread. Note that nobody corrected him, and nobody else has answered my questions related to actual physics.

They all seem to believe that the math relates to some mythical non-physical math formulas, so mythical monopoles sound nice and that is why they believe in "magnetic reconnection" rather than say "particle reconnection", or more accurately "circuit reconnection" theory. None of them understand what these formulas relate to in terms of real physical particles, and real carrier particles of the EM field. To them is all a mysterious math formula without any physical frame of reference.

Michael Mozina

27th July 2009, 01:16 PM

You do understand that current is a vector, right?

Current is a vector related to what *physical* thing(s)?

sol invictus

27th July 2009, 02:51 PM

They all seem to believe that the math relates to some mythical non-physical math formulas, so mythical monopoles sound nice and that is why they believe in "magnetic reconnection" rather than say "particle reconnection", or more accurately "circuit reconnection" theory.

Utterly wrong - as usual. Magnetic reconnection occurs without monopoles. In fact I gave an explicit example (which was originally Zig's) in another thread:

$\vec B(x,y, t) = a(t) y \hat x + b(t) x \hat y$.

If a and b are constants, nothing changes. But give them any time dependence at all (except a/b is constant) and the field lines reconnect at x=y=0. Draw a picture if you're not capable of doing the math (yes, MM, that means you).

Go ahead, Michael, compute the source for that field and see if it's magnetic monopoles. What's that? You don't know how, because you can't handle the trivial math required?

Current is a vector related to what *physical* thing(s)?

That's what this whole thread is about, Michael. Have you read it?

Dr. Trintignant

27th July 2009, 03:49 PM

The neutrino is (again, as far as anyone knows). Neutrinos are exactly neutral. And guess what - they've got a magnetic dipole moment.

Is this proven? The wiki article on neutrinos says only that the mass of neutrinos strongly suggests a (very small) magnetic moment. Also, even if true, wouldn't the magnetic moment only be a result of quantum corrections from charged virtual particles (talking out of my ass a bit here)?

- Dr. Trintignant

Tubbythin

27th July 2009, 03:49 PM

An elementary point particle is only a 19th century didactic fiction.
!?! In which text book did you read that?

sol invictus

27th July 2009, 05:44 PM

Is this proven? The wiki article on neutrinos says only that the mass of neutrinos strongly suggests a (very small) magnetic moment.

It hasn't been measured - but it shouldn't have been, because the predicted value is too small to be observable given the current experimental sensitivity.

The point is that it IS predicted - the standard model of particle physics (extended to include neutrino masses) tells us it's not zero, even though neutrinos are exactly neutral elementary particles.

Also, even if true, wouldn't the magnetic moment only be a result of quantum corrections from charged virtual particles (talking out of my ass a bit here)?

Yes, that's one way to think about it. Nevertheless, if you measure the average electric field (including the contribution from those virtual particles), it will be zero (because the neutrino charge is zero). So neutrinos in their rest frame are an example of something with zero E and non-zero B.

Michael Mozina

27th July 2009, 05:52 PM

Utterly wrong - as usual. Magnetic reconnection occurs without monopoles. In fact I gave an explicit example (which was originally Zig's) in another thread:

We aren't done with that conversation.

Go ahead, Michael, compute the source for that field and see if it's magnetic monopoles. What's that? You don't know how, because you can't handle the trivial math required?

Er, no because monopoles are another one of those figments of your overactive imagination and have nothing to do with actual "physics". It's a huge waste of effort to do math on invisible friends. That never stops you guys from trying it of course. :)

That's what this whole thread is about, Michael. Have you read it?

Ya. FYI, I wasn't even necessarily working against your position in this particular argument with that specific question. But then you didn't really care about that, you just reacted badly anyway. Is that because the answer works against you in the other thread by any chance? :)

Dr. Trintignant

27th July 2009, 06:05 PM

It hasn't been measured - but it shouldn't have been, because the predicted value is too small to be observable given the current experimental sensitivity.

Thanks; that's about what I thought--~10^-19 of an electron's B-field is a rather small amount, especially for a particle that's so hard to detect in the first place.

Yes, that's one way to think about it. Nevertheless, if you measure the average electric field (including the contribution from those virtual particles), it will be zero (because the neutrino charge is zero). So neutrinos in their rest frame are an example of something with zero E and non-zero B.

Makes sense. I'm just coming from the perspective that charge needs to be involved in some way at the deepest levels, and that QM means that even elementary particles are always in a sense "composite". I certainly don't deny the claim that the E- and B-fields are on totally equivalent ground, and just components of the same fundamental force.

Though it would be nice to spot a monopole...

- Dr. Trintignant

sol invictus

27th July 2009, 06:08 PM

Er, no because monopoles are another one of those figments of your overactive imagination and have nothing to do with actual "physics". It's a huge waste of effort to do math on invisible friends.

Translation: I don't understand Maxwell's equations and can't do any math, so I can't check to see whether the B field sol posted is sourced by magnetic monopoles or not.

ben m

27th July 2009, 06:14 PM

Yes, that's one way to think about it. Nevertheless, if you measure the average electric field (including the contribution from those virtual particles), it will be zero (because the neutrino charge is zero). So neutrinos in their rest frame are an example of something with zero E and non-zero B.

I haven't thought about this in a while, but IIRC there is room in the Standard Model for a massive neutrino to have an electric dipole moment.

sol invictus

27th July 2009, 06:18 PM

Makes sense. I'm just coming from the perspective that charge needs to be involved in some way at the deepest levels, and that QM means that even elementary particles are always in a sense "composite".

Charge is involved, yes. But actually even if there were no particles with electric charge (virtual or otherwise), photons (and therefore EM fields) would still couple to matter via gravity. I think that means massive particles with spin would still have magnetic dipole moments, and possibly electric moments too... which actually makes me realize I made a mistake earlier - I think the neutrino probably does have an electric dipole moment as well as a magnetic one. But it's also true that one could write down models in which it's exactly zero.

I certainly don't deny the claim that the E- and B-fields are on totally equivalent ground, and just components of the same fundamental force.

One way to see the equivalence is to look at EM waves. There both E and B are non-zero, but they behave in exactly the same way - exchanging E for B (or -B) is just like changing the polarization, or waiting for 1/4 period of oscillation.

sol invictus

27th July 2009, 06:19 PM

I haven't thought about this in a while, but IIRC there is room in the Standard Model for a massive neutrino to have an electric dipole moment.

Yes, you're right - I just realized that (see my post above).

Michael Mozina

27th July 2009, 06:27 PM

Translation: I don't understand Maxwell's equations and can't do any math, so I can't check to see whether the B field sol posted is sourced by magnetic monopoles or not.

Translation: Elves do not exist. There is no point in doing math with elves. Elegant elf math will never make elves exist, even if I am clever and figure out a way to stuff them into Maxwell's equations.

Dr. Trintignant

27th July 2009, 06:47 PM

Charge is involved, yes. But actually even if there were no particles with electric charge (virtual or otherwise), photons (and therefore EM fields) would still couple to matter via gravity.

Interesting. This coupling--is this the cause of the gravitational bending of light, or is it some quantum effect that's insignificant except at (say) the surface of a black hole?

- Dr. Trintignant

sol invictus

27th July 2009, 06:54 PM

Translation: Elves do not exist. There is no point in doing math with elves. Elegant elf math will never make elves exist, even if I am clever and figure out a way to stuff them into Maxwell's equations.

What total gibberish. You sound insane.

The B field I gave you is perfectly reasonable. It doesn't involve magnetic monopoles - at most standard currents of the type that could occur in plasma, a metal, a conductive fluid, etc. You'd know that if you knew any physics.

sol invictus

27th July 2009, 07:15 PM

Interesting. This coupling--is this the cause of the gravitational bending of light, or is it some quantum effect that's insignificant except at (say) the surface of a black hole?

Well, both. Gravity couples to everything, including photons, which is why light bends near stars - but the coupling is very weak compared to other forces. So the effect I'm talking about is going to by really really tiny, because it's both quantum and proportional to the gravitational coupling (which is really small).

Molinaro

27th July 2009, 08:13 PM

Er, no because monopoles are another one of those figments of your overactive imagination and have nothing to do with actual "physics". It's a huge waste of effort to do math on invisible friends. That never stops you guys from trying it of course. :)

You are the one who said reconnection was because of monopoles. Not Sol.

He then gave you an equation with which you can prove that monopoles are not needed for reconnection.

And yet you go on and on saying he's the one claiming monopoles?!

Get a freaking clue and learn how to follow a thread.

CaveDave

27th July 2009, 10:48 PM

Translation: Elves do not exist. There is no point in doing math with elves. Elegant elf math will never make elves exist, even if I am clever and figure out a way to stuff them into Maxwell's equations.

Wasn't Maxwell possessed by demons (http://en.wikipedia.org/wiki/Maxwell's_demon) instead of elves?:D

Dave:boxedin:

Michael Mozina

28th July 2009, 12:55 AM

You are the one who said reconnection was because of monopoles. Not Sol.

Er, no, I'm the one that doesn't believe that they exist.

I'm also the one that said I didn't even necessarily disagree with him in *this* thread. You evidently missed that part.

Tubbythin

28th July 2009, 01:24 AM

Er, no, I'm the one that doesn't believe that they exist.
You were the one who brought up monopoles which have nothing whatsoever to do with anything. You were the one that then failed to answer Sol's question and instead resorted to
Elves do not exist. There is no point in doing math with elves. Elegant elf math will never make elves exist, even if I am clever and figure out a way to stuff them into Maxwell's equations.

I'm also the one that said I didn't even necessarily disagree with him in *this* thread. You evidently missed that part.
Perhaps he was confused about your nonsensical rambling about elves. Or the fact that you haven't made any point whatsoever???

Molinaro

28th July 2009, 06:33 AM

stevea

28th July 2009, 07:17 AM

Oh, I said goodbye, but your parting lunacy is so rich I had to comment.

You do understand that current is a vector, right? And that dQ/dt is a scalar? What you wrote there is the divergence of the current for a charged medium, not the current.

SO WRONG. You really need a HighSchool understanding of electrodynamics before we can have a discussion.

Current is flow, a FLUX, a scalar. That is exactly why we can take the current density vector and in integrate it's dot product (note this is scalar) over a closed surface to calculate current. Your definitions are crackpot.

For the mathematically less inclined ... if we consider a volume of space, and from this volume's surface, one proton shoots out to the left, and another out to the right per unit time - we certainly have a 2q/T current, but any vector sum suggested by "sol_current" would be zero.

Where are the moving charges in a refrigerator magnet? Producing the magnetic dipole moment of a precisely neutral particle with no sub-structure?

What is your malfunction that you ask the same questions and refuse to read the answer ? Go back and read post #74. Do the obvious calculation and post it here. I'll grade your answer. Here is a good reference on ferromagnetism, "Introduction to Solid State Physics" by Kittel.

If that were true, it would be impossible to have a B field with no E field. But it is possible, so it's not true.

Not very complicated, is it?

Very nice - logical fallacies added to your mangled physics. You ASSUME you can create a B-field w/o E-field (which is the very question of this thread title), then claim, "Viola!, not complicated!". This logical error is called 'begging the question' or 'petitio principii'.

It is NOT real-world to have B field w/o E fields, since no M-monopoles have ever been detected. If you have a such a field then Maxwell's eqn, specifically the Maxwell-Faraday eqn would be violated. This is HYPOTHETICALLY possible IF there are magnetic monopoles and IF there is a net monopole currrent flow. It's fantasy talk and not science until you demonstrate monopoles.

[[QUOET]Nonsense. A charge is certainly not an E-field by anyone's definition.[/QUOTE]

A conventional unit charge such as that of an proton means an electric field exactly defined as q/r^2 r'. I apologize for not typesetting the eqn, but it's just amount of charge over R-squared and a vector of r (r-caret) direction. So YES, charge is defined precisely as an E-field (a vector field). There is no other meaning to charge !!!

A charge is a source for an E-field.

No ! There is absolutely no meaning to additional meaning "charge" other than the quantized q/r^2 radial field. It's not a "source", charge IS the field.

But if there's an equal an opposite charge right next to it, the E-field will be almost precisely zero. So if there's a long chain of charges at rest, and an equally dense chain of opposite charges in motion on top of them, you get a B field with zero E field.

OBVIOUSLY. you didn't read or couldn't understand the link pdf I provided, an undergrad level text. Even tho' the charge count matches and even tho' the charges are co-located, the moving electron charge in the superconductor is foreshortened by the Lorentz contraction and the nearly stationary proton field is not. The current places the different charges in different frames of reference. The ONLY force on an external moving charge is described completely by these distorted E fields as seen by the charge.

We can call this "distorted" portion of the observed E-field a "B field" and then all of the classical electrodynamic equations developed before relativity work perfectly, but it's an error to think these represent a real separate force.

--
BTW - the very nicely typset eqn in your post #102 violates 2 of Maxwell's equations. The arrangement of the hypothetical magnetic monopoles to make this possible requires their distribution to become infinite with X and Y axes. Not only does that eqn begin in fantasy-land, it violates reasonable rules that might exist within fantasyland. You should at least try to produce a realistic eqn for your hypothetical monopole current or magnetic-charge distribution.

As for magnetic dipole in neutrinos ... this is a hypothetical and not measured feature of neutrinos. *IF* it's correct this breaks the "standard model" and will certainly have many deeper implications, especially regarding the status of neutrinos as elementary particles. To use this unmeasured, undetected quantity in an foundation to an argument about basic E-M is silly. You have lost the argument on basics so you reach outside experimental results in an attempt to confuse. Lay off the pixie dust ! Take a trip back to the real world where no violation of Maxwell's eqn has been observed and where Maxwell's eqn can be simplified by viewing E fields relativistically thus replacing the B field terms.

sol invictus

28th July 2009, 07:40 AM

\
SO WRONG. You really need a HighSchool understanding of electrodynamics before we can have a discussion.

Current is flow, a FLUX, a scalar. That is exactly why we can take the current density vector and in integrate it's dot product (note this is scalar) over a closed surface to calculate current. Your definitions are crackpot.

No, my definitions are actually correct, unlike yours. You didn't say its dot product with what, but if you meant with the normal vector to the surface, that computes the total charge per unit time flowing through the surface - whatever surface you happened to pick. But that's not the quantity on the right hand side of Maxwell's equations, nor is it referred to as a current in any but very special situations (like when there's a single wire leaving the source of charge, so the two happen to coincide). In fact it's an impossible "definition" to use outside of those very special cases, because it depends entirely on the imaginary surface you pick (as well as on the real current).

For the mathematically less inclined ... if we consider a volume of space, and from this volume's surface, one proton shoots out to the left, and another out to the right per unit time - we certainly have a 2q/T current, but any vector sum suggested by "sol_current" would be zero.

Go ahead and quote any place I suggested performing a "vector sum" to compute current. What's that? You can't? It's a bad idea to lie about what you or others said on internet fora, because it's all right there in black and white...

As for your example, take a small modification - consider a steady stream of electrons which enter the surface on one side and exit out the other. According to you, the current is zero (because the dot product is negative on one side and equal but positive on the other). But all I've described is the simplest possible situation - a steady current flowing down a wire, piercing your closed surface... oops!

Now, let me help you. You've confused two different things - Gauss' law, for which one generally wants a closed surface, and total current (e.g. flowing down a wire), for which one would usually use an OPEN surface - for example, a cross-section of the wire.

What is your malfunction that you ask the same questions and refuse to read the answer ?

The answer was wrong - that's the malfunction. Particularly if we use your "definition" of current.

It is NOT real-world to have B field w/o E fields, since no M-monopoles have ever been detected. If you have a such a field then Maxwell's eqn, specifically the Maxwell-Faraday eqn would be violated. This is HYPOTHETICALLY possible IF there are magnetic monopoles and IF there is a net monopole currrent flow. It's fantasy talk and not science until you demonstrate monopoles.

Excellent - a concrete statement.

So - you claim that a B field with zero E field violates Maxwell's equations unless there are magnetic monopoles. That's a specific, clear, precise claim - and it's utterly, totally wrong.

The simplest example, one you'll find in every text on E&M ever written, is the field of a long straight wire carrying constant current. Guess what - zero E field, non-zero B field, and not a monopole in sight. Of course there are an infinity of other examples, but that will do.

A conventional unit charge such as that of an proton means an electric field exactly defined as q/r^2 r'. I apologize for not typesetting the eqn, but it's just amount of charge over R-squared and a vector of r (r-caret) direction. So YES, charge is defined precisely as an E-field (a vector field). There is no other meaning to charge !!!

No ! There is absolutely no meaning to additional meaning "charge" other than the quantized q/r^2 radial field.

That's too stupid to even comment on. Charge and electric field don't even have the same units. Not to mention that E fields can exist in the absence of any charges at all... ever heard of light? Radio waves?

OBVIOUSLY. you didn't read or couldn't understand the link pdf I provided, an undergrad level text. Even tho' the charge count matches and even tho' the charges are co-located, the moving electron charge in the superconductor is foreshortened by the Lorentz contraction and the nearly stationary proton field is not.

If that's the case in one frame, go to another where they're moving with equal speeds. Oops!

BTW - the very nicely typset eqn in your post #102 violates 2 of Maxwell's equations. The arrangement of the hypothetical magnetic monopoles to make this possible requires their distribution to become infinite with X and Y axes. Not only does that eqn begin in fantasy-land, it violates reasonable rules that might exist within fantasyland. You should at least try to produce a realistic eqn for your hypothetical monopole current or magnetic-charge distribution.

Ah, another clear, unambiguous, and totally false statement. Utter nonsense. There is nothing at all singular about the X and Y axes.

Tell us, stevea, since you understand E&M so well - what's the current required to produce that field? It's a trivial computation - literally one line. As for why you think it requires monopoles... answering that would probably require a psychologist, because it's certainly not for any mathematical or physics reason. All you have to do is take the divergence of that field - which one can do in one's head, in literally 2 seconds - and see that it's zero. Therefore, no magnetic monopoles: case closed.

sol invictus

28th July 2009, 08:04 AM

By the way, just in case anyone is confused, think about this simple case: a long, straight wire at rest, with some electrons drifting very slowly down it. For simplicity, we can assume the electrons have the same number density as the positive ions in the metal wire in this frame. (If not, in any realistic case one could transform to a frame where that's true.)

Now, in this frame there is a current but zero charge, because at every point the charge density in the electrons cancels the charge density in the positive ions. Of course that's just an approximation - if you went inside the wire, you'd find non-zero charge densities at scales smaller than the inter-molecular separation. But in any case, outside the wire the electric field will be incredibly small - zero if we ignore molecular scale effects - while the magnetic field can be very large.

If the wire has a non-zero resistance, there will be a component of the electric field parallel to the wire (since there is a voltage drop along any section of it). But that has little to do with the charges in the wire, and it's zero for a perfect or super conductor.

Perpetual Student

28th July 2009, 09:08 AM

By the way, just in case anyone is confused, think about this simple case: a long, straight wire at rest, with some electrons drifting very slowly down it. For simplicity, we can assume the electrons have the same number density as the positive ions in the metal wire in this frame. (If not, in any realistic case one could transform to a frame where that's true.)

Now, in this frame there is a current but zero charge, because at every point the charge density in the electrons cancels the charge density in the positive ions. Of course that's just an approximation - if you went inside the wire, you'd find non-zero charge densities at scales smaller than the inter-molecular separation. But in any case, outside the wire the electric field will be incredibly small - zero if we ignore molecular scale effects - while the magnetic field can be very large.

I have been following this discussion closely. Everytime I think I've got it -- something comes up to turn the thing around.
At this point, it seems that magnetic fields do not exist without some underlying electric field at some level (microscopic or macroscopic), whereas electric fields might exist without a magnetic field. The electron itself exists with an inherent E field. Without some sort of motion (linear, rotation, spin) there does not appear to be a B field. Is that not true? Now, I can see that E and B can be seen as equally fundamental since it seems that electrons cannot exist without motion. Is that the bottom line here?

ben m

28th July 2009, 09:16 AM

Current is flow, a FLUX, a scalar.

The magnitude of the current is a scalar; current is a vector. There's a little arrow over the J in Maxwell's Equations.

That is exactly why we can take the current density vector and in integrate it's dot product (note this is scalar) over a closed surface to calculate current.

That (assuming you indeed mean "dot product with the surface normal" is how we calculate the divergence of the current. Divergence is indeed a scalar, but it is different than current. Note, for example, that by your definition a wire carries zero current. (Draw a closed box, the wire enters one side (dot product positive) and exits the other (dot product negative, divergence = 0)

No ! There is absolutely no meaning to additional meaning "charge" other than the quantized q/r^2 radial field. It's not a "source", charge IS the field.

The charge "is" the divergence of the field. The field is the field.

*IF* it's correct this breaks the "standard model" and will certainly have many deeper implications, especially regarding the status of neutrinos as elementary particles.

Incorrect; the magnetic moment is a normal Standard Model prediction for the neutrino. It's small, but it's expected to be there; it would be very, very surprising to find it to be zero.

sol invictus

28th July 2009, 09:30 AM

At this point, it seems that magnetic fields do not exist without some underlying electric field at some level (microscopic or macroscopic),

Only in the sense that there is always an electric field at some level, even in a perfect vacuum (in that case the average E is zero, but the average E^2 is not, and any single measurement of E would give a non-zero value). But precisely the same goes for magnetic fields. And as I mentioned, the same would be true in a world with no electric (or magnetic) charges.

whereas electric fields might exist without a magnetic field.

Not if you're going to include microscopic effects.

The electron itself exists with an inherent E field. Without some sort of motion (linear, rotation, spin) there does not appear to be a B field.

But there is, even for an electron at rest - electrons have an intrinsic magnetic dipole moment. It's there as a result of the electron's spin, but that spin isn't really motion in any classical sense.

Perpetual Student

28th July 2009, 09:42 AM

But there is, even for an electron at rest - electrons have an intrinsic magnetic dipole moment. It's there as a result of the electron's spin, but that spin isn't really motion in any classical sense.

So that would appear to answer the original question. Since the electron's intrinsic angular momentum is associated with (produces?) a magnetic field, the electron's magnetic field is as fundamental to the electron as its electric field. Is that a reasonable way to look at it?

sol invictus

28th July 2009, 10:03 AM

So that would appear to answer the original question. Since the electron's intrinsic angular momentum is associated with (produces?) a magnetic field, the electron's magnetic field is as fundamental to the electron as its electric field. Is that a reasonable way to look at it?

Sure.

In the absence of charged matter, E and B are precisely symmetrical. For example in an EM wave, such as light, they behave in precisely the same way, satisfy identical equations, etc.

Matter breaks this symmetry (for some unknown reason), because there don't appear to be any magnetic monopoles in nature* - and even if they do exist, there certainly isn't a magnetic dual for every particle (i.e. a magnetic version of the electron, quarks, etc.).

*There's actually some very interesting physics connected with that - Dirac noticed that the existence of even a single magnetic monopole would require all electric charges to be quantized. Since all electric charges are quantized as far as we can tell, that's suggestive...

Ziggurat

28th July 2009, 10:05 AM

Current is flow, a FLUX, a scalar.

I'm looking at "Introduction to Electrodynamics", 3rd edition, by David J. Griffiths. On page 208, it says
"Current is actually a vector:
$\bf{I} = \lambda \bf{v}$; (5.14)
since the path of the flow is dictated by the shape of the wire, most people don't bother to display the vectorial nature of I explicitly, but when it comes to surface and volume currents we cannot afford to be so casual"

So clearly you are wrong. Current is a vector.

Very nice - logical fallacies added to your mangled physics. You ASSUME you can create a B-field w/o E-field

No assumptions involved. I already gave a scenario in which exactly that situation occurs.

It is NOT real-world to have B field w/o E fields, since no M-monopoles have ever been detected. If you have a such a field then Maxwell's eqn, specifically the Maxwell-Faraday eqn would be violated.

The divergence of B must be zero. The curl can be nonzero, even if E is zero. Can you honestly not figure out how to get a nonzero B even with a zero E? Even after I told you how to do so? Can you honestly not solve Maxwell's equations for the simple case I gave you?

No, I guess you can't.

This is HYPOTHETICALLY possible IF there are magnetic monopoles and IF there is a net monopole currrent flow.

Uh, no. If monopoles exist, you don't need them to flow to get B without E. In fact, flowing magnetic monopoles would produce an E. But thanks for trying.

A conventional unit charge such as that of an proton means an electric field exactly defined as q/r^2 r'.

Only if the source is stationary. If it's not, then that's the wrong field. And if it's accelerating, then you can't get that field by any change of reference frames either. So you're wrong, once again.

I apologize for not typesetting the eqn

I don't think anyone cares about that. But you can also use superscript and subscript tags outside of latex.

So YES, charge is defined precisely as an E-field (a vector field). There is no other meaning to charge !!!

You really don't understand the difference between a field and the source of that field? Kind of sad, really.

We can call this "distorted" portion of the observed E-field a "B field" and then all of the classical electrodynamic equations developed before relativity work perfectly, but it's an error to think these represent a real separate force.

Nobody is claiming that they are truly separate forces. What's being claimed is that the B field is a very real component of the electromagnetic field.

BTW - the very nicely typset eqn in your post #102 violates 2 of Maxwell's equations.

No it doesn't. Try it yourself.

The arrangement of the hypothetical magnetic monopoles to make this possible requires their distribution to become infinite with X and Y axes.

What on earth are you talking about? No monopoles exist for this field.

Not only does that eqn begin in fantasy-land, it violates reasonable rules that might exist within fantasyland.

There is one reasonable objection to this field (an objection which is easy to solve, BTW), but you have not raised it.

*IF* it's correct this breaks the "standard model" and will certainly have many deeper implications, especially regarding the status of neutrinos as elementary particles.

Uh, no. It's a feature of the standard model.

Take a trip back to the real world where no violation of Maxwell's eqn has been observed and where Maxwell's eqn can be simplified by viewing E fields relativistically thus replacing the B field terms.

Uh, no. Relativistic electrodynamics combines B and E into a single 4x4 antisymmetric tensor, but it most definitely does NOT replace B with E. You need 6 independent components for relativistic electrodynamics.

sol invictus

28th July 2009, 10:08 AM

There is absolutely no meaning to additional meaning "charge" other than the quantized q/r^2 radial field. It's not a "source", charge IS the field.

You really don't understand the difference between a field and the source of that field? Kind of sad, really.

It's like not understanding the difference between a faucet and water.

Perpetual Student

28th July 2009, 10:50 AM

In the absence of charged matter, E and B are precisely symmetrical. For example in an EM wave, such as light, they behave in precisely the same way, satisfy identical equations, etc.

Matter breaks this symmetry (for some unknown reason), because there don't appear to be any magnetic monopoles in nature* - and even if they do exist, there certainly isn't a magnetic dual for every particle (i.e. a magnetic version of the electron, quarks, etc.).

So it seems that this break in symmetry is the basis of the confusion and debate here. The EM wave is a clear manifestation of the complimentary nature of E and B.

sol invictus

28th July 2009, 10:59 AM

So it seems that this break in symmetry is the basis of the confusion and debate here. The EM wave is a clear manifestation of the complimentary nature of E and B.

Actually, looking back at the OP I realize that part of the debate might be due to the term "moving". There isn't really such a thing as a "moving field" - a field is a field, not an object. It's spread everywhere in space. So I ignored the "moving", and just addressed the question "can a magnetic field exist without an electric field" (to which the answer is "yes, if you ignore tiny fields due to quantum fluctuations, molecular level effects, etc.").

But I suppose one could reasonably interpret the question to mean: "Is it possible to have a magnetic field in some inertial frame, with zero electric field in all frames moving with respect to that one?" - to which the answer is "no". There can be at most one frame in which the electric field is zero and the magnetic field is non-zero.

ben m

28th July 2009, 11:07 AM

So it seems that this break in symmetry is the basis of the confusion and debate here. The EM wave is a clear manifestation of the complimentary nature of E and B.

The other source of confusion is:

Stevea's line of thinking seems to have emerged from the idea "the thing that looks like a neutral, current-carrying wire in one frame looks, from another frame, like a nonneutral wire". This is perfectly true; this calculation is done in Chapter 10 of Griffiths. Consider a current-carrying wire and a particle moving parallel to it.

1) In the frame where the wire is neutral, you expect a force on a moving particle due to the Lorentz force law, F = qv x B.

2) In the frame where the particle is at rest, the Lorentz force law predicts zero force because v = 0. In this frame, the wire is no longer neutral, and you predict a force due to the electric field and F = qE.

3) In any other frame, you expect forces due to both fields.

Stevea seems to think that (2) has replaced the magnetic field with an electric field. This is incorrect in this example---the particle-rest-frame has both a magnetic field and an electric field; the magnetic field no longer exerts a force because v = 0, not because B=0, in F = qv x B. It hasn't replaced anything, it's just showing that all of the invariant observables are the same in all frames, and that B and E are not separately invariant.

So: that example is real, stevea is not making it up, and its calculations are correct: "although different observers see different components of the field, they agree on forces". Stevea may be misremembering that the calculation showed that the B-field disappeared in this case and was "just" an E-field all along. This is incorrect.