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A continuous charge distribution flowing as a constant current in a closed loop doesn't radiate. Is it therefore true that as you increase the number of proton bunches in the LHC, while keeping the total charge constant, the synchrotron radiation decreases?

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Please somebody correct the question. The LHC has circulating protons. Either change "the LHC" to "an accelerator" or the "electron" to "proton". –  anna v Jan 30 '11 at 15:41
    
By the way, a remark to the title: it is actually debunching not bunching that reduces the SR. –  Igor Ivanov Jan 30 '11 at 18:47
    
@Anna, I've made the requested edit. –  John McVirgo Jan 30 '11 at 18:50
    
@Igor i'm interested in whether it's possible to collide charged particles together, while reducing synchrotron radiation. You would still need bunched charged particles to do this, correct? –  John McVirgo Jan 30 '11 at 19:03
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@John — just to make sure: the term "bunching" means "the act of grouping particles in bunches", while "debunching" means "spreading out initially bunched particles into a more homogeneous distribution". You seem to be using "bunching" as an equivalent of "the number of bunches", which is not the correct usage. –  Igor Ivanov Jan 30 '11 at 22:32
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3 Answers

up vote 7 down vote accepted

Synchrotron radiation can be coherent and incoherent. Coherent SR arises when electrons are grouped into short bunches so that the entire bunch emits SR as a whole. Quantum mechanically, in coherent SR the photon emission from different electrons in a bunch sum up at the amplitudes level and constructively interfere. In the incoherent SR they sum up at the level of intensity, and there is no interference.

Incoherent SR does not care how electrons are distributed along the ring, while the coherent SR is obviously boosted up in the presence of strong bunching. So, the more homogeneously you distribute the electrons, the less the effect of coherent SR will be and the less overall SR you'll have.

Now let's look at the incoherent SR. Theoretically, you are right: if we managed to create the absolute homogeneous charge distribution along the ring, we would (classically) have no SR at all because charge distribution does not change in time. The point is that this is not feasible experimentally, at least for the accelerators and the beams we have. That would require putting electrons in a well-defined quantum state of the radial motion and a well-defined angular quantum number m for the azimuthal dependence, and the accelerator technology is very far from that.

However, there is another thing which mimics that situation closely. People have managed recently to put freely propagating electrons in states with well-defined orbital angular momentum (m as high as 75), see this paper in Science for details, and they really see the annular distribution for the electron density. For such a state there exists a reference frame where the electron does not move along the z axis but just rotated as a whole in the transverse plane (with some radial distribution) around the symmetry axis. This rotation is not driven by any force, it's just the peculiar superposition of plane waves that creates this steady pattern. So in this case you can say that the electron indeed circulates but does not emit any SR.

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Ahh, what a relief...:=) Otherwise I wouldn'have slept coming nights. –  Georg Jan 30 '11 at 16:05
    
I'll look further into the points you've made, thanks. –  John McVirgo Jan 30 '11 at 20:41
    
What a wonderful answer! –  Carl Brannen Jan 31 '11 at 0:44
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Dear John, a good question. You may want to read a relevant paper about the closely related question for the late SSC collider:

http://mafurman.lbl.gov/SSC-N-143.pdf
Bunch-Length Dependence of Power Loss for the SSC

The beam has $M$ bunches in the orbit. Each of them carries $Ne$ of electric charge. All of the particles orbit by frequency $f_0$ (revolutions per second, in Hertz). We define the product, the bunch current, to be $I_b=N e f_0$.

In equation 12, you will see the result: $$Power = 1.101 Z_0 M I_b^2 \sigma_{\phi}^{-4/3} $$ Here, $Z_0$ is just the impedance of the vacuum, $4\pi/c = 377 \Omega$; they use some Gaussian units.

More importantly, $\sigma_\phi$ is just $\sigma_z/\rho$, the angular root mean square size of the bunch. You see that it's the only quantity whose increase makes the power decrease. If you spread the bunches around the ring, you're getting closer to your "closed loop current" that doesn't radiate, indeed. In practice, you don't want to spread the bunches completely because you wouldn't know the timing of the collisions. In real applications, $\sigma_\phi$ is much smaller than one, giving you a significant increase to the synchrotron radiation.

The formula is simply proportional to the number of bunches. If they're separated, each of them loses the same energy per revolution. Without a loss of generality, you may imagine that we only consider one bunch, $M=1$.

In that case, for a fixed $f_0$ - which is given by the size of the tunnel and the speed of light, assuming that the particles are near the speed of light - the power radiated by the bunch is actually proportional to $N^2$. If you double the number of charged particles in the bunch, the synchrotron radiation quadruples!

That's because the energy density (and flux) is proportional to the squared electric (and magnetic) fields, and those - derived from the Liénard-Wiechert potentials - are linear in the charges (and currents) that produce the electromagnetic fields.

So once again, the power that is radiated is not proportional to the "density" of protons in the bunch but to its square! In this sense, bunching makes the synchrotron radiation worse, not better.

However, you shouldn't think that it is a catastrophe. In the designed conditions for the LHC, one proton only loses something like 6.7 keV of energy per revolution which is a billionth of those 7 TeV they ultimately want to get (in 2011, they decided to continue at 3.5 TeV). Why is it so small for the hadron colliders?

Well, for the lepton colliders, you lose a lot because the synchrotron radiation is proportional to $\gamma^6$ and the Lorentz factor $\gamma$ has to be 2,000 times higher for electrons than for protons to achieve the same energy; see the derivation. Take the sixth power of that to see the impact of the light particles.

For the hadron colliders, the main limitation is of course the magnetic field you need to keep the protons on their circular orbit. That's why you need to have all the superconducting magnets. For colliders with light particles that need a huge $\gamma$, the synchrotron radiation is very important. That's also why linear accelerators are often preferred for the leptons. Well, you won't get rid of the full synchrotron radiation because you still need to accelerate the leptons to have some fun - so there will still be a component of the acceleration in the direction of the velocity even though the straight tunnel may liberate you from the "centripetal" acceleration transverse to the velocity.

To return to the closed loop, yes, I do think that you would turn the synchrotron radiation from the circular motion off completely if you distributed the bunches uniformly - even for leptons. It would be just like a wire with a current. However, there would still be a synchrotron radiation from the acceleration in the forward direction that you need to accelerate the particles to high speeds in the first place.

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""yes, I do think that you would turn the synchrotron radiation from the circular motion off completely if you distributed the bunches uniformly "" Hello Lubos, wasn't one of the reasons against Sommerfeld/Bohrs atom models that the electron would radiate its energy and "fall" into the core? –  Georg Jan 30 '11 at 12:25
    
Mhhmm, that multitude of electrons circulating in a synchrotron is something different. It seems that they cancel each others radiatiion. I am irritated :=( –  Georg Jan 30 '11 at 13:53
    
Dear Georg, the whole point of the Bohr-Sommerfeld atom was that it required the "number of de Broglie waves" around the orbit to be integer, so the -13.6 eV state was the lowest possible orbit. The model was never consistent with other properties of the electron, of course, but if one assumed that the electrons can only orbit along closed path with the quantized number of waves, then the electron couldn't fall to the nucleus. It was the whole point of the atom that they wanted to fix this "collapse" problem of the classical atom - one that had no quantization. –  Luboš Motl Jan 30 '11 at 18:05
    
Lubos, I thought of the times when a "planetary-like" atom was proposed, but Broglies "waves" were not known yet. –  Georg Jan 30 '11 at 20:28
    
Great informative answer as usual, Lubos. I was asking for the case of keeping the total charge in the ring constant while increasing the bunching. Therefore, from equation (12), N and therefore Ib is inversely propotional to M giving the power loss inversely proportional to the square of M. So yes, for a constant total charge in the ring, the powerloss does decrease dramatically with increased bunching, at least in theory. –  John McVirgo Jan 30 '11 at 20:32
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I am afraid that the radiation cannot cancel everywhere so it is better to say the radiation does not occur in case of a constant current. This is so because, according to Maxwell equations, it is not acceleration of a single charge that creates the radiation but the current time-dependence at a given point. In other words, different sources radiate differently and it is not reduced to the sum of radiations. The total filed is determined differently: superposition of fields is not a sum of radiations! The same is valid in the opposite case of short bunches where the radiative losses are proportional to the charge squared ( = source-dependent phenomenon).

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Dear downvoters, leave short explanation or a disagreement statement, please! –  Vladimir Kalitvianski Jan 30 '11 at 18:13
    
Have a look at the nonradiation condition: en.wikipedia.org/wiki/Nonradiation_condition. Problem 14.24 in Jackson asks to show there is no radiaton from a closed loop of current. –  John McVirgo Jan 30 '11 at 20:48
    
I did not get your suggestion. Do you mean that radiation of multiple charges can cancel everywhere? Do you mean that in the whole space on can get a purely destructive interference? –  Vladimir Kalitvianski Jan 30 '11 at 21:12
    
Yes, there are some some accelerating charge distributions where the total radiation at all points sums to zero. This goes back to 1910, if you look at the Wikipedia link: "In 1910 Paul Ehrenfest published a short paper on "Irregular electrical movements without magnetic and radiation fields" demonstrating that Maxwell’s equations allow for the existence of accelerating charge distributions which emit no radiation." –  John McVirgo Jan 30 '11 at 22:39
    
I will look through it if it is available but why then it is not present on the textbooks? –  Vladimir Kalitvianski Jan 30 '11 at 23:13
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