Does bunching reduce synchrotron radiation? 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?
 A: 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.
A: 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.
A: 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).
