Does a DC supplied superconductive coil give off radiation?

[binis]

A DC supplied superconductive electric coil must give off EM radiation according to Maxwell's law because rotation is acceleration. Does it?

 

[vanhees71]

A DC doesn't radiate, no matter whether the current flows in a superconducting or normal conducting material. I don't know what you mean when you say "rotation is acceleration". Any static vector field is a pure solenoidal magnetic field, because of Gauss's Law, $\vec{\nabla} \cdot \vec{B}=0$, but static fields don't describe electromagnetic radiation but static fields ;-)).

 

[binis]

I don't know what you mean when you say "rotation is acceleration".

Because the electrons are moving in circles (a spiral path) and by this are accelerated.

 

[Ibix]

An individual electron moving in a circle would radiate, but a continuous stream of them does not. I suspect that if you work out the field from an elementary charge moving in a circle you will see radiation, but if you then integrate the fields from charges around the ring to get the field from a current loop then the total contributions to the radiation field cancel out.

Or you could just observe that $\partial E/\partial t=0$ from symmetry and hence there can be no radiation.

 

[hutchphd]

I am perplexed by this thread:

A DC doesn't radiate, no matter whether the current flows in a superconducting or normal conducting material. I don't know what you mean when you say "rotation is acceleration". Any static vector field is a pure solenoidal magnetic field, because of Gauss's Law, ∇→⋅B→=0, but static fields don't describe electromagnetic radiation but static fields ;-)).

By this definition a cyclotron should not radiate? Is this because

An individual electron moving in a circle would radiate, but a continuous stream of them does not.

Is there a specific treatment of this you can reference? I don't recall seeing this addressed and wonder how it actually falls out...it is not clear to me at first blush although your continuum supposition is likely correct..
Edit; Or is it fundamentally a quantum mechanical issue?
/

 

[Ibix]

By this definition a cyclotron should not radiate?

Well, cyclotron electrons move in an expanding spiral, not a circle. Also they can't reasonably be called DC. Their speed is increasing, and they must be pulsed, since electrons moving from the currently positive D to the currently negative one will be decelerated while ones moving the other way (at the opposite side of the chamber) are accelerated.

Is there a specific treatment of this you can reference?

It's just a current loop with a constant current, as far as I can see. That's a textbook solenoid, no?

 

[dRic2]

@Ibix @hutchphd I googled it and found this link where the author addresses this issue. https://physics.princeton.edu/~mcdonald/examples/steadycurrent.pdf
I just skimmed through it. It seems to consider only the non-relativistic limit.

 

[hutchphd]

Well, cyclotron electrons move in an expanding spiral, not a circle. Also they can't reasonably be called DC. Their speed is increasing, and they must be pulsed, since electrons moving from the currently positive D to the currently negative one will be decelerated while ones moving the other way (at the opposite side of the chamber) are accelerated.

The longitudinal acceleration is not necessary for the "cyclotron" radiation. In fact the (non-relativistic) Larmor Equation pasted from Wikipedia

gave rise to the question "why don't electrons in atoms radiate" when they circulate in atoms. So that ain't it.

@Ibix @hutchphd I googled it and found this link where the author addresses this issue. https://physics.princeton.edu/~mcdonald/examples/steadycurrent.pdf
I just skimmed through it. It seems to consider only the non-relativistic limit.

The answer in this paper is in fact a quantum argument. They use stationary phase to "do" the integration: I quote from the conclusion
" If the charge carriers in a wire were localized to distances much smaller than their sepa-ration, radiation of “steady” currents could occur. However, in the quantum view of metallic conduction, such localization does not occur."

Thanks to all. Makes sense to me.

/

 

[vanhees71]

In other words: A DC current in a wire is in a classical picture described as the motion of a fluid of electrons with the time-independent velocity field $\vec{v}(\vec{x})$. Also the corresponding density is not changing with time, i.e., also $\vec{j}=\rho \vec{v}$ is time-independent, and thus you have a magnetostatic situation, and there are no em. waves.

A single electron moving in a circle provides however a time-dependent current density. In a classical picture you have
$$\rho(t,\vec{x})=-e \delta^{(3)}[\vec{x}-\vec{y}(t)], \quad \vec{j}(t)=-e\dot{\vec{y}}(t) \delta^{(3)}[\vec{x}-\vec{y}(t)],$$
which is time-dependent.

If $\ddot{\vec{y}} \neq 0$ there is "cyclotron radiation". For $\dot{\vec{y}}=\text{const}$ there's no radiation, but a Lorentz-boosted Coulomb field, which of course doesn't describe em. waves. Both cases can be calculated using the Lienard-Wiechert potentials. See Sect. 4.5ff in

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

 

[Vanadium 50]

Since this is A-level...

Acceleration doesn't cause radiation. Changing multipoles cause radiation, and a point charge accelerating is but one example.

DC means no radiation.

 

[hutchphd]

Acceleration doesn't cause radiation. Changing multipoles cause radiation, and a point charge accelerating is but one example.

But by this definition there is no DC ? Do N electrons rotating unbunched in a uniform B field emit radiation ? Is this DC? (What is the appropriate continuum limit)

 

[tech99]

According to Electromagnetic Vibrations, Waves and Radiation, by Bekefi and Barrett, p 289, a charge traveling in a circle is subject to acceleration towards the centre and emits cyclotron radiation with a frequency equal its rotational frequency.

 

[hutchphd]

According to Electromagnetic Vibrations, Waves and Radiation, by Bekefi and Barrett, p 289, a charge traveling in a circle is subject to acceleration towards the centre and emits cyclotron radiation with a frequency equal its rotational frequency.

There is no question about that. My particular issue is a "continuous" ring of current and the inconvenient notion that electron-carried charge is not continuous. I believe there may be a problem with any treatment of the free electron absent QM.

 

[vanhees71]

There's no self-consistent description of a classical charged point particle interacting with the em. field. The best approximate description of this notorious radiation-reaction problem is the socalled Landau-Lifschitz approximation of the Lorentz-Abraham-Dirac equation, which can also be motivated by quantum mechanical arguments (leading to a non-Markovian quantum-Langevin equation).

 

[hutchphd]

New to me. Is there a good compendium treatment ? Otherwise I will hunt it down (maybe!)

 

[binis]

I googled it and found this link where the author addresses this issue. https://physics.princeton.edu/~mcdonald/examples/steadycurrent.pdf

This study considers a uniform circular motion in a ring of radius α. Provided that these math apply to a perfect circle circuit, then does not apply to an ellipsoid electric coil or an irregular circuit.

 

[Vanadium 50]

But by this definition there is no DC ?

Frictionless plane, stretchless rope...

 

[Vanadium 50]

...spherical cow.

Since DC started infinitely far in the past and goes on forever, yes, there is no DC.

 

[hutchphd]

This study considers a uniform circular motion in a ring of radius α. Provided that these math apply to a perfect circle circuit, then does not apply to an ellipsoid electric coil or an irregular circuit.

Yes I believe it does (I didn't really dissect it) but it fundamentally uses the quantum nature of the electron in matter to reach that conclusion. (It is the same reason electrons in an atom don't"spiral in": because of QM) I apologize for slightly hijacking the post by asking if a classical electron would similarly behave as a rejoinder to some questionable answers.

/

 

[vanhees71]

New to me. Is there a good compendium treatment ? Otherwise I will hunt it down (maybe!)

The best derivation of the Lorentz-Abraham-Dirac equation can be found in the textbook

K. Lechner, Classical Electrodynamics, Springer International
Publishing AG, Cham (2018),
https://doi.org/10.1007/978-3-319-91809-9

The ingenious trick is to use a particular Poincare covariant regularization of the Lienard-Wiechert potentials. I combined this in my SRT FAQ writeup with the covariant treatment of the radiation power, but it's not yet complete. Particularly non-trivial examples are missing, but these can be found in Landau&Lifschitz vol. 1.

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

The treatment of a non-relativistic quantum particle in the quantized electromagnetic field, particularly the Planck equilibrium radiation as a heatbath, in the sense of an open quantum system with the derivation of a corresponding non-Markovian Langevin equation can be found in

https://doi.org/10.1016/0375-9601(91)90054-C

 

[alan123hk]

A DC supplied superconductive electric coil must give off EM radiation according to Maxwell's law because rotation is acceleration. Does it?

The simple understanding is that charge acceleration is not a sufficient condition for radiation. To generate radiation, the accelerating charge must cause a change in current, and the change in current must cause a change in the electromagnetic field in the surrounding space, and then these electromagnetic waves are propagated without being canceled or blocked.

It can be imagined that there are two closed loops of arbitrary shape closely superimposed, and then the same AC current is applied to these two loops, but their phases are opposite, so the electromagnetic fields generated by them cancel each other out, and the radiation becomes extremely small. For another example, twisted-pair wires and coaxial cables can also effectively reduce the radiation of high-frequency currents, even if the charge has extremely high acceleration in its conductors. In similar situations, the radiation resistance will not appear in the conductor.

It can be inferred from this that, because the electromagnetic field around the DC loop does not change, in this special case, the changes in the electromagnetic field caused by the acceleration of the charge must completely cancel each other out.

However, it is a bit surprising and unimaginable to say that a DC circuit of any shape will completely automatically cancel the radiated electromagnetic field generated by the acceleration of the charge in the loop. .

 

[alan123hk]

However, it is a bit surprising and unimaginable to say that a DC circuit of any shape will completely automatically cancel the radiated electromagnetic field generated by the acceleration of the charge in the loop.

So another convincing argument may be that the charge in the DC circuit will not accelerate or decelerate in the direction of the wire, otherwise it will not be DC. Therefore, the direction of charge acceleration must be perpendicular to the direction of charge movement in the wire, that is, the electromagnetic force accelerating the charge is perpendicular to the direction of charge movement. Therefore, the charge does not get energy from the electromagnetic force that accelerates it, so it does not generate radiation.

Maybe my statement above is wrong, please correct me if so.

 

[binis]

So another convincing argument may be that the charge in the DC circuit does not accelerate in the direction of the wire,

You are trying to say that the magnitube of the velocity remains stable. So Maxwell's law needs a revision.

 

[alan123hk]

I found another interpretation is that the acccelaterating charge in the DC circuit actually produces radiation, but because the electrons are moving at the drift velocity, which is only around a metre per second, so the amount of radiation released is immeasurably small.

This situation can also be understood from the perspective of radiation resistance. Since the radiation resistance of a wire is inversely proportional to the wavelength, and since the wavelength of DC is infinite, the radiation resistance in all practical DC circuits is effectively zero.

Now I think both effects exist, one is that the electromagnetic waves generated by the circuit cancel each other out, and the other is that the radiation is too small to be measured.

My statements in this post seem to contradict what I've said before, but I don't mind if I find new information that convinces me.

https://en.wikipedia.org/wiki/Radiation_resistance

 

[vanhees71]

The drift velocity is more around ~1mm/s. But that's not the point. As discussed before, stationary sources don't produce radiation fields. If you can describe a situation by a DC, nothing radiates.

The fluctuations due to thermal motion of the atoms/molecules/electrons within the matter of course cause thermal ("Planck") radiation.

 

[alan123hk]

Maybe there's something I haven't figured out yet, but anyway I'm assuming that the radiation from the acceleration of the charge in a small section of wire in the DC circuit isn't canceled out, and did a very rough estimate of the radiation it produces.

 

[vanhees71]

I don't understand the problem. Just look at the electromagnetic field from a stationary current (that's what DC means), and you'll see that there's no radiation part in this field.

Also note that just integrating a partial "segment of current" ala Biot-Savart is not a solution to the magnetostatic Maxwell equations. You have to integrate over the complete current distribution.

 

[Vanadium 50]

Acceleration doesn't cause radiation. Changing multipoles cause radiation, and a point charge accelerating is but one example.

Why does nobody ever believe this?

 

[vanhees71]

I believe it. Why shouldn't I? The multipole moments of time-independent charge-current distributions, however are also time-independent. So there's no contradiction.

 

[alan123hk]

Since I don't understand why in a DC circuit with arbitrarily complex spatial structure, how can the radiation of many single accelerated electrons cancel each other out. So I searched the Internet and found another physics discussion site, someone said that the radiation generated by the acceleration caused by the curved path is too small to measure, because the drift speed of the electrons is too low, and several replies agreed with him. I thought it made sense, So I made that argument here as well.

But after reading the valuable replies from the experts here, I reconsidered and felt I should revise my thinking a little more , this claim that the radiation is too low to detect is unnecessary and probably not true.

Now I think of another simple and intuitive way of explaining it. I don't have to imagine how the electromagnetic fields cancel each other out from the whole complex structure of space circuits. I only need to consider the local current element. Although there is charge acceleration in the local current element, since the charge in the DC circuit flows continuously and stably, when the charges accelerates away from a place, it is instantly replenished by other charges without interruption, so this acceleration will not cause change in the distribution of charge anywhere in the circuit, nor does it result in a change in current flow in any direction throughout the circuit. So the final conclusion is that there is no electromagnetic radiation.

 

[binis]

The drift velocity is more around ~1mm/s.

I read that drift velocity in superconductors is about 20 meters per second.

 

[Vanadium 50]

No.

For the third time, the A-level answer is radiation is caused by time-varying multipoles and these are zero for DC.

 

[artis]

@hutchphd I think the cyclotron analogy is not usable here because in a coiled wire with steady current the actual electrons move slowly , very slowly so from the standpoint of a single electron the curled wire seems almost straight at the speed it is going. Much like we don't see and don't feel the curve we are actually making while traveling along a long road even though the Earth is curved and we are traveling along a curved path.

In cyclotrons on the other hand you have a relativistic electron beam being deflected by a perpendicular B field. But I have another example that i myself would like to ask, for steady DC current doesn't it also matter in what type of "conductor" the current is flowing? It seems to me a toroidal plasma does emit cyclotron radiation with a steady current, I suspect this has something to do with the confining fields which are present for a plasma and not present for a solid copper wire with DC.