Superconductivity at room temperature??

[ben m]

Let me add a few clarifications to the accelerating-charges scenario. Maxwell's Equations, as we've said repeatedly, don't fundamentally care about acceleration. They care about time-varying EM fields. Nor do they care about the causes of accelerations: a charged pith ball, in (say) uniform circular motion, has certain EM fields that depend on q, r(t), and v(t)---nothing else. The fields don't care whether the pith ball is tied to a string, glued to a toy train, or hovering in a cyclotron. They don't care about the ball's mass.

Let me add that I can construct a linearly accelerating ensemble of charges that, like the spinning disk, does not radiate. The conclusion that they do not radiate follows from Maxwell's Equations. If you want to conclude that I'm wrong, or if you want to invoke imaginary engineers to tell me I'm wrong, then you are telling me that Maxwell's Equations are wrong.

Here's the scenario. Consider a cylindrical rope of uniform volumetric charge density d. The rope-material is incompressible, but deformable, so I can change its radius---envision a wire going through an extruder. I have a straight run of this rope stretched through the region from x=0 to x=L. It's moving, but at a non-constant velocity. When it enters the region, at x=0, the rope has radius r = R0 and is moving at a low velocity v = v0. It gets thinner as it goes along---and since it's incompressible that is only possible if its velocity increases. By the time it gets to x=L, the velocity has doubled to v = 2 v0 and the radius has decreased to r = R0/sqrt(2). At any point x in between, the rope's velocity is v = v0(1+ x/L) and its radius is r = R0 / sqrt(1 + x/L).

(We can make any assumption we like about how the rope behaves outside the box---does it loop back from x=L to x=0? Are we looking at the midpoint of an infinitely long rope? I don't care.)

All of the charges in the rope are indeed accelerating---and accelerating in a straight line, just like (say) a particle beam in an accelerator. Let's look at the electric and magnetic fields of these accelerating charges.

The electric field is given by Coulomb's Law. At any point in space, the charge density is constant---in a quasi-cylindrical approximation where L >> R0, there's more charge per unit x on the left (thick rope) and less on the right (thin rope) but these densities do not vary in time.

The magnetic field is given by the Biot-Savart law. At any point in space, the current density is constant. The current densities are higher on the right (faster-moving charge) and lower on the left (slower-moving), but the net current I(x) is constant in x. There's a small radial component of the current, since charges have to migrate inward, but that component is also constant in time.

Constant E field? Check. Constant B field? Check. Solve Maxwell's Equations---what far-field EM fields match these boundary conditions? Looks like it's static in the far field also. That's a nonradiating linear accelerator, Johan. Run it backwards and it's a nonradiating decelerator.

Nonradiating. Let me say it again:

There is no fundamental classical law saying that accelerating charges radiate. None whatsoever. It is common for simple accelerating charges to radiate, but according to Maxwell's Equations this common observation is not a general rule. If you learned otherwise, you learned wrongly.

 

[Skwinty]

The predicted absence of radiation for hyperbolic motion has sometimes been cited as a way of reconciling the Equivalence Principle with the fact that a charged particle held stationary in a gravitational field (and therefore undergoing constant proper acceleration) does not radiate. For example, in Feynman's "Lectures on Gravitation" he says "we have inherited a prejudice that an accelerating charge should radiate", and then he goes on to argue that the usual formula giving the power radiated by an accelerating charge as proportional to the square of the acceleration "has led us astray" because it applies only to cyclic or bounded motions.

This is what I meant by "old school"

Ziggurat,
What does Feynmann mean then by the above statement.

Surely the cyclic and bounded motion conditions apply to conductors with transferring charges.

 

[Skwinty]

Perhaps my confusion arises from this:

If you take a length of copper wire and lay it over a compass and
you then connect a 12V battery with load to the wire, the radiated electro-magnetic field will cause the compass needle to deflect.

This is as a result of the radiated EM field, If there was no radiation, then the common relay would not work.

The photon is the carrier of EM fields.

Where am I going wrong here?

 

[ben m]

Perhaps my confusion arises from this:

If you take a length of copper wire and lay it over a compass and
you then connect a 12V battery with load to the wire, the radiated electro-magnetic field will cause the compass needle to deflect.

This is as a result of the radiated EM field, If there was no radiation, then the common relay would not work.

Ah, I see. No, that is not radiation---those are static EM fields. (It's true that they are "carried by photons", but these are virtual photons and in fact it's very, very challenging to visualize how this works.) They require no power-consumption to maintain, and they don't "propagate" like real photons.

"Real photons" are what you get when the EM fields, viewed from sufficiently far away, start to look sinusoidal in both time and space. When you get such a field, you find that it's carrying electromagnetic field energy and moving it away from the source.

ETA: in your examples---turning on an electromagnet near a compass, turning on an electromagnet in a relay---you're in a regime partway between the static and the radiating ones. Since you turned the field from off to on, your field does change with time, and if an observer far away could watch you throw that relay they'd see some radiofrequency waves emerge. But you couldn't describe the near field work done---the motion of the needle, for example---in terms of these "the solenoid emits radio waves, which hit the compass needle and turn it" or anything like that. No no no.

And I want to reiterate that no one is saying that nothing radiates---that would mean radio doesn't work, etc. Prins is saying everything that accelerates radiates, and the rest of us are saying, no, Maxwell's Equations tell us that some things radiate and other things do not.

 

[Skwinty]

Ah, I see. No, that is not radiation---those are static EM fields. (It's true that they are "carried by photons", but these are virtual photons and in fact it's very, very challenging to visualize how this works.) They require no power-consumption to maintain, and they don't "propagate" like real photons..

I understand the Feymann diagrams just fine.

"Real photons" are what you get when the EM fields, viewed from sufficiently far away, start to look sinusoidal in both time and space. When you get such a field, you find that it's carrying electromagnetic field energy and moving it away from the source.

You can detect a static EM field from some distance too though.
That static EM field energy also moves or moved away from the source.

An alternating current will start to radiate very strongly above 300 kHz and is detectable from great distance.

Below 300kHz, the radiation is very weak and only detected from very small distances.

Too my simple mind, this all sounds like hair splitting.
I suppose that is why I am not a physicist. I only had to learn and memorise formulae and not derive them.

 

[Skwinty]

And I want to reiterate that no one is saying that nothing radiates---that would mean radio doesn't work, etc. Prins is saying everything that accelerates radiates, and the rest of us are saying, no, Maxwell's Equations tell us that some things radiate and other things do not.

Yes, I appreciate that, hence my post about the stationary charge does not radiate even though is is subject to gravitional acceleration.

However, charges in a conductor are not static and Maxwells equations, in which I have ultimate faith, does not consider gravitational acceleration.

Also, bear in mind, my studies in light current electronics was done more than 30 years ago.

 

[Dilb]

Too my simple mind, this all sounds like hair splitting.

No, they are very different cases. A radiating field propagates itself through space, and as a result the source can continuously emit energy. A non-radiating field only emits enough energy to set up the field around itself. Capacitors and inductors work by creating non-radiating fields, for example.

 

[Skwinty]

Capacitors and inductors work by creating non-radiating fields, for example.

That is true for DC currents, not AC currents.

 

[ben m]

I understand the Feymann diagrams just fine.

I know what they look like, and I know how they work, but haven't convinced myself that I understand them ... not yet.

You can detect a static EM field from some distance too though.
That static EM field energy also moves or moved away from the source.

If I place a charge Q somewhere, you will feel an E field E ~ Q/r^2 at any distance r, and you'll feel it forever and ever---as long as the charge sits there. There's energy in that field but the energy is not moving---the source charge is not consuming power to sustain that field; if you place an absorber in the field, it does not absorb any power and does not heat up. If the charge has some funny shape, the effects of this shape drop off as 1/r^3 or faster.

An alternating current will start to radiate very strongly above 300 kHz and is detectable from great distance.

Below 300kHz, the radiation is very weak and only detected from very small distances.

There's nothing special about 300 kHz---except, perhaps, for some particular technology you were dealing with.

Yes, most AC currents will radiate. Put a charge Q somewhere, grab it, and make it wiggle. It takes power to keep it wiggling; if you don't supply power it will stop. The sinusoidal E fields that result will drop off strictly as 1/r^2, even though they depend on the sort of shape-details that (in the static case) dropped off as 1/r^3. An absorber placed far from an AC source will have power available to it, and will heat up.

Yes, they're all just Maxwell's Equations, and if you had a 3D-numerical-integrator, you could tell it "just calculate the fields of this source" and it could do so, perfectly correctly, without stopping to name some things "static" and other things "radiating".

But for us humans the words do matter. When we talk about "radiation", we're talking specifically about the sinusoidal, 1/r^2, nonzero-Poynting-vector, power-carrying part of the fields---not, generically, about any field that extends through space.

 

[Skwinty]

I know what they look like, and I know how they work, but haven't convinced myself that I understand them ... not yet.

OK, understand was probably too strong a term.

If I place a charge Q somewhere, you will feel an E field E ~ Q/r^2 at any distance r, and you'll feel it forever and ever---as long as the charge sits there. There's energy in that field but the energy is not moving---the source charge is not consuming power to sustain that field; if you place an absorber in the field, it does not absorb any power and does not heat up.

Agreed, otherwise radio would work based on DC currents in the aerial. No transfer of information without alternating sinusoidal EM fields

There's nothing special about 300 kHz---except, perhaps, for some particular technology you were dealing with.

The 300kHz comes from open wire frequency division multiplexing system.
24 simultaneous telephony calls per open wire pair.

But for us humans the words do matter. When we talk about "radiation", we're talking specifically about the sinusoidal, 1/r^2, nonzero-Poynting-vector, power-carrying part of the fields---not, generically, about any field that extends through space.

What about nuclear radiation.

 

[Ziggurat]

You can detect a static EM field from some distance too though.
That static EM field energy also moves or moved away from the source.

When you turned on the current, the field you created propagates outwards. That's radiation. But if you LEAVE it on, if you establish a DC current that's now fixed, radiation does not continue to be emitted from your wire. The radiation that's already propagating continues to propagate, but you aren't making any new radiation. But the electrons speeding around your wire continue to accelerate, as they must in order to go in a circle. So they are accelerating, even in the case of constant current, but not radiating.

Too my simple mind, this all sounds like hair splitting.
I suppose that is why I am not a physicist. I only had to learn and memorise formulae and not derive them.

It's not splitting hairs. It's quite important. Accelerating charges often, but don't always, radiate.

 

[ben m]

What about nuclear radiation.

You think you're joking, but gamma rays and x-rays are EM radiation just like any other. Unlike radio waves, they're easy to detect as single photons and, for strictly technological reasons, impossible to detect as (say) a wave train visible on an oscilloscope. Gamma rays and x-rays obey Maxwell's Equations, carry energy via Poynting vectors, etc., just like visible light or radio waves or whatever.

Alpha, beta, and neutron radiation are different.

 

[Skwinty]

It's not splitting hairs. It's quite important. Accelerating charges often, but don't always, radiate.

Understood, I am just not differentiating between active and passive EM fields.

 

[Skwinty]

You think you're joking, but gamma rays and x-rays are EM radiation just like any other. Unlike radio waves, they're easy to detect as single photons and, for strictly technological reasons, impossible to detect as (say) a wave train visible on an oscilloscope. Gamma rays and x-rays obey Maxwell's Equations, carry energy via Poynting vectors, etc., just like visible light or radio waves or whatever.

Alpha, beta, and neutron radiation are different.

Yes, alpha, beta and neutrons are particles and not waves.

 

[Ziggurat]

Yes, alpha, beta and neutrons are particles and not waves.

You're wrong about that. Which is why people can do diffraction with electrons, neutrons, and yes, even helium atoms.

 

[Skwinty]

You're wrong about that. Which is why people can do diffraction with electrons, neutrons, and yes, even helium atoms.

They are particles which exhibit wave like characteristics.

Wavicles perhaps a better descriptor.
You can isolate a helium atom, electron or neutron but not a gamma ray.

 

[ben m]

Yes, alpha, beta and neutrons are particles and not waves.

They are waves, just not electromagnetic waves; they obey a wave equation which is not Maxwell's Equations.

 

[ben m]

They are particles which exhibit wave like characteristics.

Wavicles perhaps a better descriptor.
You can isolate a helium atom, electron or neutron but not a gamma ray.

There is no physical distinction (though there are technological ones) between our ability to isolate a gamma ray and our ability to isolate a helium atom --- or a radio-frequency photon for that matter. Each of them is simultaneously a particle and a wave.

 

[W.D.Clinger]

It is clear that I am encountering people on this forum which will do anything in their power to distort anything I am saying and to in this way avoid looking at reality.

We know how you feel.

Obviously people will be able to inundate this thread with irrelevant postings as Clinger, Ben m, Ziggurat and previously Sol Invictus did.

You keep saying that, but several of us have explained the relevance, as in post #557.

My opponents claim that the charge-carriers within a superconductor are accelerated by an electric field which is somehow supernaturally induced by the static magnetic field that is produced around the superconducting current.

That's a grotesque distortion of what your opponents are saying. Perhaps you just don't understand what your opponents are saying.

Indeed that they motion around a ring has nothing to do with acceleration.

In Newtonian mechanics, motion around a ring always involves acceleration.

Whatever I might think of William Schockley in other ways, he was not very flattering about the physics abilities of Bardeen: An I am starting to understand why.

For lurkers: John Bardeen^WP is the only person to have received two Nobel Prizes in Physics. In his biography, True Genius: The Life and Science of John Bardeen, Lillian Hoddeson and Vicki Daitch wrote:

That kind of interaction with Bardeen became legendary---so much so that physicists at Bell Labs developed a number to describe it, the "Bardeen number." Bell Labs physicist Ravin Bhatt (a student of Bardeen's student William McMillan) described it as "the ratio of substance to self-advertisement." Bhatt explained that, in general, if a scientist even made number one of that scale "you're doing good. Most people these days are way below that." But "Bardeen was infinity on that scale."

The very next paragraph of that biography alludes to the falling out between Bardeen and Shockley, and the paragraph after that begins the story of the research that won Bardeen his second Nobel Prize (for the BCS theory of superconduction).

 

[Skwinty]

I can understand you holding an electron, helium atom or a neutron in a microscopic tweezer but not a photon.
Once the photon comes to rest, it disappears.

ie no rest mass. (or am I mistaken here?)