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10

To extend jk88's answer, I'd like to say that in addition to the "prompt" radiation there is the possibility of activating material near the beamline. Prompt effects are what jk88 is talking about and they go away when the beam shuts off, except that any neutrons generated hang about for a short time. But the prompt effects can activate material near the ...


10

Yes, off the top of my head I can think of two major sources of harmful radiation: Synchrotron radiation: When charged particles are accelerated in a ring they emit EM synchrotron radiation. Depending on the frequency, this radiation can be dangerous. Beam halo effects and internal beam interactions: Beam halos, are particles in the accelerating bunch ...


7

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 ...


5

Not to worry, it's fairly easy: right now you have a differential equation which can be written $$\frac{\mathrm{d}E}{\mathrm{d}t} = -CE^2$$ for some constant $C$. You need to solve that differential equation for $E(t)$. (If you're wondering how to do that, you can find more information at the math site.) Then you can determine the electron's energy at the ...


4

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 $f_0$ ...


4

Cyclotron radiation is the radiation emitted by a non-relativistic charge when it is accelerated by magnetic field. Synchrotron is similar for a relativistic charge with relativistic beaming and characteristic frequency approximately $\gamma^2$ times the cyclotron frequency. Bremsstrahlung is the radiation emitted when a charge is accelerated as it ...


3

I am posting this since it is too long for comments and it starts on the way to gamma ray FEL. It seems that they have managed at SLAC up to X-ray wavelengths.The facility is called LCLS, (Linac Coheren Light Source). The LCLS uses the final third of SLAC's two-mile linear accelerator to drive electrons to high energy and through an array of "undulator" ...


1

X-ray tubes are generally designed such that the electron beam knocks out inner shell electrons in the target (e.g. copper K shell), resulting in well-defined lines in the output spectrum. There is also a bremsstrahlung spectrum but depending on what you want to do, it may be less useful due to its broadband nature. Wiggler/undulator radiation is purely due ...


1

The X-ray radiation is Bremsstrahlung radiation created when the electrons are accelerated in the wiggler. In an X-ray tube the electrons are accelerating (well, decelerating) from their initial energy to zero when they hit the target. So the energy of the Bremsstrahlung radiation is comparable to the initial electron energy. In the wiggler the acceleration ...


1

Synchrotron radiation is a high energy phenomenon. It is a form of bremsstrahlung, this being the electromagnetic radiation produced by the deceleration of a charged particle when deflected by another charged particle, typically an electron by an atomic nucleus. Synchrotron radiation is produced in the deceleration of a charged particle in a magnetic ...


1

Brilliance is as stated in literature: number of photons per second per mm$^2$ per rad$^2$ per $0.001BW$, where $BW = \Delta\omega/\omega$ is the like the "binning" size over which the equation was integrated over. So the number of photons that will arrive at your sample depends on what frequency range you are measuring over. If you are measuring the ...


1

$$\frac{dE}{dt} = -CE^2$$ $$d\left (\frac{1}{E}\right ) = Cdt$$ $$\frac{1}{E_2}-\frac{1}{E_1}= C(t_2-t_1)$$ $$E_2=E_1/2$$ $$\frac{1}{E_1} = C\Delta t$$ $$\Delta t = \frac{1}{CE_1}$$



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