# Beam particle density

1. Jul 25, 2011

### gbz

Linear collider beams usually carry about 10^25 particles while synchrotron beams considerably less particles, usually 10^11 to 10^14. Why is this? Are there practical (or theoretical) limits to the density of particles in a synchrotron beam? Or is it simply because sychrotron beams can achieve similar luminosities as linear colliders with much fewer particles?

2. Jul 26, 2011

### Drakkith

Staff Emeritus
All I can think of is that Synchrotons have to keep the beam stable and focused against the curvature of the accelerator whereas a linear accelerator does not. That plus the radiation released from these particles as they are acceleratred around may require that less particles be used in order to get them to that velocity.

3. Jul 30, 2011

### Bill_K

gbz, are you sure about the number 1025 you quoted for linear colliders? I can't find any confirmation of that. All colliders accelerate their particles in bunches. I found numbers for the LHC, the old SLC linear collider, and the proposed ILC, and in all three cases the number of particles per bunch was comparable, a few times 1010.

Anyway, a much more important number for colliders is the luminosity (particles per second per cross-sectional area) and this depends greatly on the collision geometry, i.e. how tightly you can focus the beams at the collision point.

4. Jul 30, 2011

Staff Emeritus
gbz is posting nonsense, I'm afraid. 1025 electrons would weigh 10 milligrams and carry 1.6 megacoulombs of charge. With 3000 bunches, you would have a pound of electrons, with nearly 5 gigacoulombs of charge. The current would be 5 gigaamps, and it would take 1021 watts to power this machine. That's a million times more power than humanity produces.

5. Aug 2, 2011

### gbz

I received the 10^25 figure from a physicist, but admittedly I dont have a source for it. I should have double-checked it before posting, my mistake.

But I arrived at a similar number from luminosity figures for linear colliders too. SLC for instance has a luminosity of about 0.002x10^33 /cm^2 sec. Assuming N2 is 1 in the luminosity equation, given the beam is hitting a stationary target, we can derive the particle density in the beam by dividing by c (~10^8). So luminosity (10^30) divided by c (10^8) would give us 10^22 per cm^2 meter --> that is 10^22 particles per cm^2 cross section and 1 meter length of beam. Is the N2 = 1 assumption incorrect? How would you derive the particle density from luminosity for a stationary target beam?

@v50: Wasn't my intention to post 'nonsense', maybe I made some miscalculation. But I'm not sure I understand your math. How do you go from 5 GAmps to 10^21 Watts?

6. Aug 2, 2011

Staff Emeritus
Volts x Amps = Watts

The error in your calculation is that you are assuming continuous beam. Linear colliders are pulsed.

7. Aug 2, 2011

### Bill_K

Of course the beam is not a centimeter across! Beam width is typically a few microns, which lowers your estimate by a factor of 108, and brings it pretty much in line with synchrotrons.

8. Aug 2, 2011

### gbz

@v50: I doubt we can use P = VI here. This is not a circuit current, there is no resistance (except for the infinitesimal resistance of free space), so P=VI can't be applied here as far as I can tell. Anyhow, what would V be here? The voltage from one end of beam to the other is zero, the charge transfer is driven by mechanical momentum, not potential difference.

@Bill K: hmm.. that makes sense. Probably the missing factor..