# Will EM wave be emitted from simple circuit?

1. Oct 18, 2009

### kelvin490

Even in simple circuit, there must be some "turning point", for example the electric wire is bent somewhere. Therefore when electron pass through there is centripetal acceleration. I have heard that accelearation of electrons emits EM wave, so will EM wave be emitted from these positions?

2. Oct 18, 2009

### f95toli

Yes, although it has nothing to do with the charges accelerating.

Also, only AC circuits emit radio-waves although DC circuits will of course generate magnetic fields that can be picked up inductively at a distance.

3. Oct 18, 2009

### kelvin490

Thanks.

At some turning point in the circuit, isn't the electron accelerating?Is that acceleration of electron would produce EM wave?

4. Oct 18, 2009

### cesiumfrog

Sounds reasonable. One reason you're unlikely to measure it is that (on average) electrons drift very slowly around a typical circuit, hence the acceleration may be small. (Picture it like a pipe or hose system pre-filled with water, with low output so that it would take a long time for a dye to pass through, but with high pressure applied from the input; the pressure wave propagates almost instantly through the system.) Can you do an order of magnitude estimate of the radiation?

Last edited: Oct 18, 2009
5. Oct 18, 2009

### Born2bwire

Not really. I don't think I have seen a discussion about the emitted radiation of accelerating charges that wasn't dealing with relativistic charges. Yeah, theoretically it should radiate but the radiated power will be rather minute. If we have an AC signal, then bends in the traces of the transmission line for, say, a microstrip line will stimulate radiation. Conceptually, it has to do with the fact that the guided wave hits the discontinuity of the bend, some of the wave is reflected, some of it is guided around the bend, and some escapes the guided mode into a propagation mode. There are basic design rules when it comes to laying down traces, like having say 45 degree bends as being the largest allowable bend, that help avoid these problems. A patch antenna works in a similar way, the patch antenna is like a resonant cavity with open sides. A standing wave is excited between the patch and the ground plane and when the travelling parts of the standing wave hit the open sides, radiation leaks out.

6. Oct 19, 2009

### cesiumfrog

On second thought, since the circuit has a smooth current (rather than a localised packet of orbiting charge) the charge distribution will be constant in time, so it cannot produce an oscillating field (or any such radiation should be cancelled out by radiation from other points along the circuit).

7. Oct 19, 2009

### kelvin490

Thanks all.

I think there is one thing that I don't really sure, that is whether electromagnetic wave will always be produced whenever there is an accelearation of electron.

When velocity changes(include change of moving direction), there is accelearation. I have read from a book about synchrotron radiation that electron emits EM wave when it accelearates. This is where my question came from.

8. Oct 19, 2009

### Born2bwire

Synchrotron radiation assumes relativistic particles. The total power radiated by an accelerated charge is given by the Lienard formula.

$$P = \frac{2}{3}\frac{e^2}{c} \gamma^6 \left[ \left( \dot{\boldsymbol{\beta}} \right)^2 - \left( \boldsymbol{\beta} \times \dot{\boldsymbol{\beta}} \right) ^2 \right]$$

where

$$\boldsymbol{\beta} = \frac{\mathbf{v}}{c}$$
$$\gamma = \left[ 1-\left( \boldsymbol{\beta} \right)^2 \right]^{\left(-1/2\right)}$$

As we approach non-relativistic speeds, \beta goes to 0 and \gamma goes to 1. So in the non-relativistic limit we regain the Larmor formula,

$$P = \frac{2}{3} \frac{e^2}{c^3} \left( \dot{\mathbf{v}} \right)^2$$

Since the speeds are very low, the acceleration around any bend will also be low compared with c. The acceleration for a circular orbit is v^2/r, r will be on the order of millimeters for most PCB traces given their large width. So really, the total power radiated will be very small given the fact that in comparison to c^3, the charge density, velocities and the inverse of the "radius of orbit" for a common bend in a wire or microstrip line are all small. So for a circuilar orbit,

$$P = \frac{2}{3} \frac{e^2v^4}{c^3r^2} = \frac{2}{3} \frac{e^2\beta^4c}{r^2}$$

For synchrotron, assuming a perfect orbit and doing a quick back of the envelope from Lienard, gives

$$P = \frac{2}{3} \frac{e^2c}{r^2}\gamma^4\beta^4$$

I'm not sure if the above equation is completely valid because the analysis of the synchrotron radiation is rather complex but we are only considering the total power radiated and the actual radiated power is spatially and frequency dependent which requires much more detailed analysis. Anyway, we can see though in the last equation that the power radiated is much much larger.

EDIT: Making the extra reduction we can see that the relativistic properties cause an increase in the power of \gamma^4. Hey, that sounds familiar now that I think about it, probably derived this before. Anyway, so we can see that relativistic speeds will cause an increase of \gamma^4-fold, which can be pretty significant when you consider that \gamma goes to 0 as we approach c.

Last edited: Oct 19, 2009
9. Oct 19, 2009

### kelvin490

It's useful.Thanks.

10. Oct 19, 2009

### Born2bwire

Whoops, finally figured out what I did wrong in my equations. Fixed that for you but the conclusion still stands.

11. Oct 19, 2009

### Staff: Mentor

I could be mistaken about this, but I remember a discussion here a long time ago in which it turned out that although a single charge moving in a curved path does radiate (synchrotron radiation), a continuous distribution of charge does not. If I remember correctly, it's possible to show this by considering a finite number N of charges, each with charge q, and then taking the limit as N goes to infinity and q goes to zero in such a way as to keep the total charge constant.

12. Oct 19, 2009

Staff Emeritus
We're seeming to drift afield here.

It's not terribly useful to think of the accelerating charge as being the fundamental source of radiation. It's much more useful to think of a changing multipole moment - of which a single accelerating charge is an example - as the fundamental source of radiation. The problem with thinking about accelerating charges as the source is that once you get two of them, you have to understand the relative phases of radiation and you get a real mess.

So, a DC circuit does not radiate. Full stop. No multipoles are changing.

AC circuits do radiate. This is a problem with high speed circuit board design, as there are lots of little things on the board that act as antennas.

The idea that quantum fluctuations cause radiation is just wrong. You can make the same argument about a piece of metal. Don't quantum fluctuations cause electrons inside it to move and thus radiate? Shouldn't every object be radiating like crazy? Of course not - we are talking about stationary states in QM.

13. Oct 19, 2009

Staff Emeritus
Bob, that's very, very misleading. Look at the original question: "Will EM wave be emitted from simple circuit?" Bringing up blackbody EM radiation to answer positively to this question is so misleading as to be practically misinformation. Yes, there is blackbody EM radiation. This is true even if the circuit is off.

14. Oct 19, 2009

### hamster143

Charges are discrete. Therefore, the distribution of charges is not perfectly uniform, therefore there will be radiation.

On one hand, fluctuations are small compared with the total multipole moment of all electrons in the system. On the other hand, the total multipole moment is huge (there are on the order of 1000 coulomb worth of free electrons in one gram of copper wire). On the third hand, the actual velocity of electrons inside wires is tiny and power radiated scales as (v/c)^4.

Someone should probably sit down and plug in numbers and come up with a ballpark estimate.

15. Oct 19, 2009

### Bob_for_short

The problem has already been done. If Vanadium 50 lets me explain....

Discreteness of charges in a metal is not essential for our consideration.

The lattice fluctuations (=phonons) create the positive charge fluctuations depending on the local temperature. The current suffers "friction" or lattice resistance. Some current energy is transmitted to the lattice and pumps the fluctuations. These and other charge fluctuations radiate, from surface or from the volume, depending on the sample size. Let us call it a heat radiation for simplicity. Without feeding the current stops. If there is no current, the proper heat radiation power should be equal to the external heat radiation power in an equilibrium state. Electrical current only increases the former one creating a heat flux to the environment.

Of course, it is not a radiation due to "curvature" of the circuit. It is due to charge thermal fluctuations. On average the current is constant and its fluctuations are taken into account via the sample resistance. The supplied power = radiated power (if there is no mechanical and other types of work in the circuit).

Last edited: Oct 19, 2009
16. Oct 19, 2009

### Bob S

Any current in a ring is composed of discrete charges. One nanoamp continuous current of electrons in a ring (electron synchrotron) has shot (Schottky) noise, with the average space between electrons being about 5 cm. There is plenty of continuous rf noise in the GHz range. Electrons also radiate significant synchrotron radiation, but protons radiate only the shot noise.
Bob S

17. Oct 19, 2009

### Bob_for_short

Bob_S, are you speaking of a metallic conductor or of particle beams in vacuum? Nobody discuses here synchrotrons.

Last edited: Oct 19, 2009
18. Oct 19, 2009

### hamster143

Irrelevant ...

19. Oct 19, 2009

### Bob S

Particle beams in vacuum. But note that shot (discrete charged particle) noise is observable in microwave GaAS and other microwave components. There is also kTB (Boltzmann noise) in wideband microwave amplifiers. So real continuous currents do not exist.
Bob S

20. Oct 19, 2009

### Bob_for_short

That's what I say. The simplest example is a heat radiation. There are, of course, many devices converting the current energy into the radiation in a more efficient way (LEDs, lasers, etc.), at least in a spectral sense.

Last edited: Oct 19, 2009
21. Oct 19, 2009

Staff Emeritus
That's a statement that pure DC can't be realized in the real world, not a statement on how DC behaves.

It might be amusing to calculate the expected radiation from a typical circuit - I wonder how many lifetimes of the universe one has to wait before a single photon is radiated.

22. Oct 19, 2009

### Bob_for_short

Look at a car lamp fed on DC and count photons.

23. Oct 19, 2009

Staff Emeritus
Bob, enough. The OP asked a very specific question on circuits and EM radiation. Stop derailing this with discussions on blackbody radiation. You are not doing the OP a service.

24. Oct 19, 2009

### hamster143

Yeah, I did the calculation and the result is ridiculously low, on the order of one photon per week for a macroscopic copper loop.

25. Oct 19, 2009

### Born2bwire

Ha, we could use it as our source in a Young's Double Slit experiment of single photon emission and give the pitch drop experiment a run for it's money as the longest running experiment.