Does constant DC current in curved wire produce EM wave?

[user079622]

Accelerating electric charge produce EM wave

Does constant DC current in curved wire produce EM wave, because el.charge has radial acceleration in curved wire?

 

[hutchphd]

Cyclotron radiation

 

[user079622]

Cyclotron radiation

So in case of curved wire, centripetal acceleration can cause EM waves but very weak?

 

[renormalize]

Does constant DC current in curved wire produce EM wave, because el.charge has radial acceleration in curved wire?

No, there is no radiation. For $N$ uniformly-spaced charges flowing in a DC circuit, the radiation from each of the individual accelerating charges destructively interferes with that from the others and the total emitted radiation vanishes as $N\rightarrow\infty$. See Why Doesn’t a Steady Current Loop Radiate? and A Steady Loop Current Does Not Radiate.

 

[berkeman]

Does constant DC current in curved wire produce EM wave, because el.charge has radial acceleration in curved wire?

It may help to think of the example where the wire curves all the way around on itself to form a coil. After the DC current is switched on and stabilizes, do you see any change in the field that would propagate?

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html

 

[user079622]

No, there is no radiation. For $N$ uniformly-spaced charges flowing in a DC circuit, the radiation from each of the individual accelerating charges destructively interferes with that from the others and the total emitted radiation vanishes as $N\rightarrow\infty$. See Why Doesn’t a Steady Current Loop Radiate? and A Steady Loop Current Does Not Radiate.

What if wire is not closed loop, just half of circle?

It may help to think of the example where the wire curves all the way around on itself to form a coil. After the DC current is switched on and stabilizes, do you see any change in the field that would propagate?

View attachment 345673
http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/curloo.html

What is time to reach current from 0km/h to c?
zero seconds, than acceleration is ∞?

 

[Vanadium 50]

Still zero. See @berkeman 's message.

No changing electromagnetic field, no electromagnetic radiation.

 

[renormalize]

What if wire is not closed loop, just half of circle?

It's still ultimately a loop, since for current to flow a circuit has to close back on itself somewhere. In the real world, you can't place your source (battery) at infinity.

 

[user079622]

@Vanadium 50
@renormalize

When you shake back and forth Alternating Current Gaussmeter ,he significantly increase reading, because device is moved in Earth static magnetic field.

Does his acceleration or velocity cause increase in reading and why he even increase reading if magnetic field from Earth is static and constant in magnitude?

I notice, if device is in car, it doesnt mater if I drive 70km/h or 150km/h, so it seems he respond only for acceleration?

It's still ultimately a loop, since for current to flow a circuit has to close back on itself somewhere. In the real world, you can't place your source (battery) at infinity.

But if DC current change in magnitude(fluctating), then EM are produced?

 

[Vanadium 50]

Alternating Current

constant DC current

Pick one.

 

[user079622]

Pick one.

First question is for DC, second question is for AC guassmeter in static magnetic field.

 

[tech99]

Still zero. See @berkeman ;s message.

No changing electromagnetic field, no electromagnetic radiation.

This does not seem to explain cyclotron radiation.
Referring to Electromagnetic Vibrations, Waves and Radiation, by Bekefi and Barrett, page 288, it says that centripetal acceleration results in radiation.
However for a current in a wire, the speed of the electrons is very small, so the centripetal acceleration and therefore the radiation must also be small. In addition, having once found the frequency, which will be low, we need the radius to be comparable to the wavelength, which will be large, in order to avoid cancellation of radiation from opposite sides of the circle. A larger circle results in a lower frequency, so it seems that whatever we do, the radiation will be very small. In the case of electrons in a vacuum the speed can be much higher and so significant cyclotron radiation can be obtained.

 

[user079622]

This does not seem to explain cyclotron radiation.
Referring to Electromagnetic Vibrations, Waves and Radiation, by Bekefi and Barrett, page 288, it says that centripetal acceleration results in radiation.
However for a current in a wire, the speed of the electrons is very small, so the centripetal acceleration and therefore the radiation must also be small. In addition, having once found the frequency, which will be low, we need the radius to be comparable to the wavelength, which will be large, in order to avoid cancellation of radiation from opposite sides of the circle. A larger circle results in a lower frequency, so it seems that whatever we do, the radiation will be very small. In the case of electrons in a vacuum the speed can be much higher and so significant cyclotron radiation can be obtained.

I find this:
"Radial Acceleration and Radiation: When a charged particle moves in a curved path, it experiences centripetal acceleration. This centripetal acceleration can indeed lead to the emission of electromagnetic radiation. This is the principle behind cyclotron radiation, where the emitted radiation is due to the acceleration of charged particles in a circular or spiral trajectory."

 

[renormalize]

This does not seem to explain cyclotron radiation.

That's because cyclotron radiation is not a product of the "uniformly-spaced charges" that occur in a DC circuit. From the first reference in post #4:
"Note, however, that this nearly complete destructive interference depends on the electrons being uniformly distributed around the ring. Suppose instead that they were distributed with random azimuths $\phi_n$. ...for random azimuths the power radiated by $N$ electrons (at any order) is just $N$ times that radiated by one electron.
If the charge carriers in a wire were localized to distances much smaller than their separation, radiation of “steady” currents could occur. However, in the quantum view of metallic
conduction, such localization does not occur.
The random-phase approximation is relevant for electrons in a so-called storage ring, for which the radiated power is a major loss of energy – or source of desirable photon beams
of synchrotron radiation, depending on one’s point of view."

 

[renormalize]

I find this:
"Radial Acceleration and Radiation: When a charged particle moves in a curved path, it experiences centripetal acceleration. This centripetal acceleration can indeed lead to the emission of electromagnetic radiation. This is the principle behind cyclotron radiation, where the emitted radiation is due to the acceleration of charged particles in a circular or spiral trajectory."

Yes, but that's for one or more accelerated particles that are randomly spaced relative to each other. If the charges move in lock-step with uniform separation, as in a DC circuit, the net emitted radiation vanishes.

 

[sophiecentaur]

If the charges move in lock-step with uniform separation,

Just to check. Cyclotron radiation can occur 'only' when the spacing of the charges in non-uniform?

This could well be right but that would imply that the radiated power would depend on the (non)-uniformity of the charge density. I guess that containing electrons etc into a uniform ring would be difficult and the unavoidable non-uniformity could be detected by a variation in the radiation level? Could this be used (or even is it) as an error signal to control a circular beam of electrons?

 

[Vanadium 50]

The question says "constant DC current". Individual electrons in an accelerator is not "constant DC current". Alternating current is not "constant DC current".

 

[sophiecentaur]

Individual electrons in an accelerator is not "constant DC current".

but the output beam of an actual cyclotron is uniform in density - isn't it? The particles are steadily increasing in velocity but the acceleration is relatively low - much less than the cyclotron frequency due to the circular path. This would be equivalent to a DC current round a loop.

 

[Vanadium 50]

but

The question is about DC.
Well, what about a particle beam?
That's not DC - its bunched and also composed of discrete, countable electrons.
But its like DC.
And the radiation is "like" zero. But its not zero.
Exactly! Its not zero!
But its not DC.
But its like DC.

 

[sophiecentaur]

But its like DC.

Spot the difference - apart from the numbers involved. In any case what's wrong with stretching an analogy?
The DC supply to a crt electron gun suddenly becomes 'notDC' when the electrons leave the gun with a white picture (Ignore the blanking as it's not essential)??

 

[Vanadium 50]

If I have one electron, I get a certain amount of radiation. If I two, I have less (the dipole term drops out and I have only quadrupole and higher). If I have 4, I have even less (the quadrupole term drops out and I have only octopole and higher) In the DC limit, it is zero.

You can argue "DC is not really DC" because of shot noise and the like, but I would say that piles quibble on top of quibble. A DC circuit, as described, does not radiate.

 

[sophiecentaur]

A DC circuit, as described, does not radiate.

I wouldn't argue with that bit of theory. I was just extending the idea to a 'real' cyclotron and wondering how much CR could be detected from the non-continuous and asymmetrical electron beams in a simple cyclotron. The are not used for electrons because of relativistic effects. (Betatron is used, apparently)

 

[SredniVashtar]

Can the op question be explained by considering that the drift velocity of electrons in a wire is 'not real', but it is (classically) just an average of the actual velocity the electrons have while accelerating between collisions with the lattice?

 

[cianfa72]

When a charged particle moves in a curved path, it experiences centripetal acceleration. This centripetal acceleration can indeed lead to the emission of electromagnetic radiation.

The point being made here is that electrons flowing in a curved wire have centripetal acceleration w.r.t. an inertial frame, hence they should radiate energy as EM field.

Btw in the case of Cyclotron, in which an electron radiate energy, where does that energy (transferred into EM field energy) come from ?

 

[Vanadium 50]

The point being made here is that electrons flowing in a curved wire have centripetal acceleration w.r.t. an inertial frame, hence they should radiate energy as EM field.

No they don't.

Why did you feel compelled to jump into a quiet thread with an incorrect answer?

 

[cianfa72]

Why did you feel compelled to jump into a quiet thread with an incorrect answer?

It is incorrect from the point of view of the DC flow of electrons inside the curved path.

Mine was not an answer: I was just reconsidering the point made by the OP, in the case of the Cyclotron.

 

[Vanadium 50]

As said upthread, a cyclotron is not a DC circuit. Surely we don't have to repeat the entire thread.

 

[Baluncore]

When a stable DC current is established in a conductor, a magnetic field is generated that propagates away from the conductor, and out into space. The magnetic field then remains stable in the vicinity of the conductor.

The electric and magnetic fields remain stable, as does the Poynting vector that represents a continuous flow of energy. As fast as that energy radiates away, the space behind it is filled with more of the same.

Energy continues to propagate away from the wire, until the current in the circuit is cut, when the electric, magnetic and Poynting vectors, all fall to zero. The energy has then ceased to radiate from the DC circuit.

 

[cianfa72]

The electric and magnetic fields remain stable, as does the Poynting vector that represents a continuous flow of energy. As fast as that energy radiates away, the space behind it is filled with more of the same.

We had a thread on this topic some time ago, the take-home message was that on a plane orthogonal to the wire, neglecting the loss in the wires, the Poynting vector flux is null (the entering EM energy flow is equal to the egress flow).

However since $E$ and $B$ do not change in time there is not any EM energy radiates away.

 

[jasonRF]

I agree with others that say in the steady state, the DC current loop will not radiate.

When I am unsure about my physical understanding or intuition, I fall back on the math. The vector potential $\mathbf{A}(\mathbf{r},t)$is related to the current density $\mathbf{J}(\mathbf{r},t)$ by (in MKS units)
\mathbf{A}(\mathbf{r},t) = \frac{\mu_0}{4\pi}\int \frac{\mathbf{J}\left(\mathbf{r}^\prime,t-\frac{|\mathbf{r-r}^\prime|}{c}\right)}{|\mathbf{r-r}^\prime|} \, d^3 \mathbf{r}^\prime

Note that if $\mathbf{J}$ has no time dependence (like for a continuous DC current loop), then neither does $\mathbf{A}$ so the fields will be DC. That is, it will not radiate.

jason

 

[cianfa72]

Note that if $\mathbf{J}$ has no time dependence (like for a continuous DC current loop), then neither does $\mathbf{A}$ so the fields will be DC. That is, it will not radiate.

Why if $\mathbf A$ has not time dependence then the DC current loop does not radiate ?

 

[renormalize]

Why if $\mathbf A$ has not time dependence then the DC current loop does not radiate ?

Because the magnetic field $\vec{B}\left(\vec{x},t\right)=\vec{\nabla}\times\vec{A}\left(\vec{x},t\right)$. If the vector potential has no time dependence then neither does the magnetic field. But radiation fields must oscillate with time, so: no time-dependence $\Rightarrow$ no radiation.

 

[cianfa72]

But radiation fields must oscillate with time, so: no time-dependence $\Rightarrow$ no radiation.

Ok, therefore there is a flow of EM energy in the space surrounding the wires (from the source to the load) as described by the Poynting vector field, however the DC circuit does not radiate.

 

[Baluncore]

A trapezoidal current pulse, with a rise-time, a flat top, and a fall-time, can be described by a Fourier transform. All the harmonics that make up the pulse are present throughout the pulse.

Energy flows continuously to the load during the flat period we call DC. How does that energy reach the load if it is not through a non-zero Poynting vector?

For what duration does the flat top need to be maintained stable, before the harmonics cease to be part of the continuous radiation of energy to the load?

 

[cianfa72]

Energy flows continuously to the load during the flat period we call DC. How does that energy reach the load if it is not through a non-zero Poynting vector?

I'm not sure whether Poynting vector field and Poynting theorem actually makes sense only in case $E$ and $B$ are solutions of the EM wave equation.

Does Poynting vector field actually represents a "real" flux of EM energy only in the above case (for example flowing through a given surface) ?

 

[renormalize]

I'm not sure whether Poynting vector field and Poynting theorem actually makes sense only in case $E$ and $B$ are solutions of the EM wave equation.

As pointed out by Feynman, even static EM fields give rise to Poynting vector fields that represent flows of EM energy that circulate continuously (but without radiating). See the Wikipedia entry Static Fields and its reference to the Feynman lectures.

 

[cianfa72]

See the Wikipedia entry Static Fields and its reference to the Feynman lectures.

From Wikipedia

While the circulating energy flow may seem unphysical, its existence is necessary to maintain conservation of angular momentum. The momentum of an electromagnetic wave in free space is equal to its power divided by c, the speed of light. Therefore, the circular flow of electromagnetic energy implies an angular momentum

Here it talks of momentum of electromagnetic wave in free space (bold is mine). Does it mean that $E$ and $B$ are actually solution of EM wave equation ?

 

[renormalize]

From Wikipedia

Here it talks of momentum of electromagnetic wave in free space (bold is mine). Does it mean that $E$ and $B$ are actually solution of EM wave equation ?

Yes, but since the fields are static, the EM "wave" satisfies a "wave-equation" with vanishing time derivatives, i.e., Laplace's equation: $\nabla^{2}\vec{E}=\nabla^{2}\vec{B}=0$.

 

[cianfa72]

Yes, but since the fields are static, the EM "wave" satisfies a "wave-equation" with vanishing time derivatives, i.e., Laplace's equation: $\nabla^{2}\vec{E}=\nabla^{2}\vec{B}=0$.

There is therefore a flow of EM energy, however it is "confined" in the space surrounding the DC circuit: there is not a flow of EM energy that radiates/propagates out from there.