# Tag Info

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

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 ...

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

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" ...

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 $f_0$ (...

3

Conceptually the idea is based on a simple principle: conservation of linear momentum. Every time a charged particle has to be accelerated, a photon has to be involved. If you want to linearly accelerate a charged particle, you have to shoot photons at its back. If you want to stop a charged particle (decelerate), it has to emit photons in the forward ...

3

Light sources usually store an electron beam. The reason is very simple: getting electrons is extremely easy! You can just heat a piece of metal (thermionic guns), or you can shoot a laser on a photo-cathode. If you want a positron beam you need to produce and capture it converting a primary beam. The yield of this process is not very big and you end up with ...

3

From reading various sources, it seems that the main synchrotron at APS is capable of carrying either an electron or positron beam (for example, the introduction of this paper) I gather that the positrons are created by colliding an electron beam from a LINAC with a target, but the target can also be withdrawn to allow the electrons to go directly into the ...

3

Check synchrotron radiation in the wiki article. When high-energy particles are in rapid motion, including electrons forced to travel in a curved path by a magnetic field, synchrotron radiation is produced. bold mine. Charges when accelerated radiate electromagnetic radiation, and the curved paths in the synchrotron give a continuous angular ...

2

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 ...

2

1) The mass of the electron is 0.5 MeV, so, for most pratical purposes, one can ignore the mass when the energy is at 1.2 GeV, the difference, most of the time, would be about 0.05%, which is usually considered small. If there is mass, not all energy is kinectic energy, you are right about that. 2) Depends on the accelerator design, but it's possible to use ...

2

You seem to think of electrons and photons as point particles from relativistic mechanics that have rectilinear trajectory until they collide. This is only one of many different ways to give meaning to these two words. Within this view, if an electron collides with a photon, and if conservation of energy and momentum is assumed, the resulting state after the ...

2

I'm researching synchrotrons for a class project, but I can't seem to find a decent answer to one of my questions. It appears that most synchrotrons use electrons as opposed to some other charged particle, while the Large Hadron Collider uses protons instead.. The first thing that you should know is that there are two completely different uses for ...

2

The quantity that determines what a particle beam may be used for is called gamma ($\gamma$). It is defined as $$\gamma = \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^2}}.$$ As $v$ gets closer to $c$, $\gamma$ gets larger without bound and equals infinity when $v = c$. Since particles in a synchrotron are moving at very close to the speed of light ($0.... 1 I have the feeling that the (partially stripped) ion should be considered as a single particle. The residual electrons should influence the emitted radiation just by altering the charge state of the ion, therefore the$C_\gamma$parameter which enters in the calculation of the total emitted power and power spectrum. The external bending field should be ... 1 The principal advantage of using electrons is that the electron is a fundamental particle so electron-electron (or electron-positron) collisions are are well defined process that is relatively eay to describe mathematically, and very accurate measurements can be made. By contrast the proton is a composite particle. We normally describe the proton as being ... 1 Real photons are emitted during accelerations, in synchrotron radiation and bremsstrahlung and it is a classical electrodynamics prediction . The Millikan experiment : Millikan was the first to determine with great accuracy that the maximum kinetic energy of the ejected electrons obey the equation Einstein had proposed in 1905: namely, 1/2mv2=hf−P, ... 1 X-ray tubes are generally designed such that the electron beam knocks out inner shell electrons in the target (e.g. copper K shell), resulting in well-defined lines in the output spectrum. There is also a bremsstrahlung spectrum but depending on what you want to do, it may be less useful due to its broadband nature. Wiggler/undulator radiation is purely due ... 1 Brilliance is as stated in literature: number of photons per second per mm$^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 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 field,... 1 I usually remember that red light is about 700nm and blue light about 400nm. Ultraviolet is shorter than blue so it's less than 400nm and I suppose extends down to the soft X-ray region. Anyhow, wavelength,$\nu$, and frequency,$f$, are related by the simple equation: $$\nu f = c$$ where$c$is the speed of light. So red light is about$4 \times 10^{14}...

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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