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Do particle accelerators emit a lot of radiation? If they do, can someone give me a rough estimate of how much? I'm also curious as to what kind of radiation they give off.

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    $\begingroup$ I was at LANL, and a sign said "30,000 rad/hour when beam on". Pretty sure it was meant to be read when beam off. $\endgroup$ – JEB Mar 19 at 16:30
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    $\begingroup$ @JEB - depends on how fast you read, I guess! Of course, to keep under 5rad you would need to read it (and get out) in under a second, so... $\endgroup$ – Jon Custer Mar 19 at 16:56
  • $\begingroup$ physics humor detected. and ignored ;-) $\endgroup$ – niels nielsen Mar 19 at 17:17
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    $\begingroup$ Remeber boys and girls: keep you access key in you pocket and you live to play another day. At even modestly powerful facilities a great deal of effort goes into making sure that human beings aren't in the tunnels or the experimental halls while the beam is on. The "access key" referred to above a component of the physical interlock system that keeps the beam off. You carry it on your person if you have to enter the controlled areas while the machine is in a idle state (that is, not fully shut down). While you have it the beam can't come back on. Very important safety tip. $\endgroup$ – dmckee Mar 20 at 0:41
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The idea of a storage ring is to store the beam, so giving off radiation means either losing beam, or losing energy. The later is accomplished via synchrotron radiation (accelerating charged particles in a circle), and is easily computed from the current, the energy, and the bending radius.

In the usual LINAC (linear accelerator), the beam is radiation, period. It travels in a vacuum and hits a target, producing secondaries (some of which are the experiment at hand, most of which are radiation).

My recollection of SLAC was that we used 40 $\mu$A at 4 GeV, which means a power of:

$$ P = IV = (40\,\mu A)(4\,GV) = 160\,kW$$

Over the course of an experiment we wound up exposing a temporary building several hundred yards behind End Station A to near the 100 mrem limit of annual radiation.

LAMPF (now LANSCE) was 1 mA of protons at 1 GeV, or 1 MW. There was a target that was only accessible by robots: if you saw it line-of-sight, you died.

The beams generally activate much of the laboratory material, including air ($^{15}$O) and esp. copper in vacuum connections, but the aluminum beam pipe not so much.

For comparison, the 20,000 Ci (!) $^{60}$Co source JPL has for space-radiation testing is "only" making 300 W thermal.

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  • $\begingroup$ Wb converting it to electricity? Will it be efficient? Cus synchotron radiation is electromagnetic radiation right? Any idea how effective it'll be? $\endgroup$ – Physics_gurl Mar 20 at 4:37
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The Large Hadron Collider at CERN emits about six kilowatts of synchrotron radiation, according to the calculations here.

The radiated photons are mostly in the x-ray range. This radiation is due to the protons that travel around the accelerator’s circular ring. Any charged particle that accelerates produces electromagnetic radiation.

You can read all about the interesting physics of synchrotron radiation here.

When the protons collide with each other, they produce a zoo of other kinds of high-energy particles. The only particles that escape the accelerator are neutrinos, which pass harmlessly through anything.

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The radiation power $P$ emitted by an accelerator can be easily computed by Larmor's formula which says in its most general form:

$$ P = -\frac{2}{3}\frac{e^2c}{(mc^2)^2}\left[\frac{dp^\mu dp_\mu}{d\tau d\tau}\right]$$

where $p^\mu=(E,\vec{p})$ is the 4-momentum vector, $\tau$ the proper time of the radiating particle, $e$ the electric charge and $c$ the speed of light. The mass $m$ of the radiating particle is of utmost importance in the formula, because it is the determing factor if the radiation loss is high or low.
For a circular storage ring Larmor's formula can be simplified to (for a linear accelerator the radiation power is orders of magnitudes smaller, as can be seen from this post Does an electron accelerated in a linear accelerator lose any energy?)

$$ P =- \frac{e^2 c}{6\pi \epsilon_0} \left(\frac{E}{mc^2}\right)^4\frac{1}{R^2}$$

where $E$ is the energy of the particle, and $R$ is the bending radius of the dipoles (only if the storage ring were perfectly circular (i.e. without any straight section) $R$ can be identified with the radius of the storage ring) and $\epsilon_0$ is the dielectricity constant of the vacuum. For electron rings the energy loss per turn (and per electron) is between several MeV for "small" machines ($E\sim 1-10$GeV) up to several 100 MeV for "large" machines ($E\sim 10-100$GeV). Elder proton machines, however, produce hardly any radiation power, whereas the projected FCC at CERN can already be considered as perfect synchrotron light source. The radiation (also called synchrotron radiation) is generally considered as a drawback for colliders as it acts as a principal limit of their maximal energy, whereas atom-, condensed matter physicists and chemists consider the radiation as fantastic tool for the structural analysis of matter on atomic scale (for instance for protein cristallography and other applications whose number is almost infinite). One cannot easily transform synchrotron radiation in electricity (actually it could partially be at least transformed to heat which could then be used for electricity production, however, I wonder about the efficiciency of such a procedure.

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