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In synchrotrons electrons lose energy while emitting light so accelerating fields are used to boost up their speed back to >99% of light speed.

Why RF (time varying) fields are used for accelerating electrons? Why not constant field?

In this reference it is said that the energy gain from a varying field is: $$\Delta W = q V_0 T \cos\phi = \Delta W_{\text{DC}} T \cos\phi$$

where $\Delta W_{\text{DC}}$ is the the energy gain from a static DC field and $T$ is the transit time factor, $T = \frac{\beta\lambda}{\pi L} \sin\frac{\pi L}{\beta\lambda}$.

But isn't $T\cos\phi$ always smaller than 1?

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  • $\begingroup$ While I love my original Van de Graaff-built electrostatic accelerator, you might contemplate just how you would use a static field to continue to accelerate electrons (or ions) that are going round and round and round again. One quickly comes to the realization that you need a time varying field, timed so that it is in phase with a packet of electrons to accelerate them (which is why the particles are bunched - anything not in the bunch doesn't get accelerated, and may be decelerated). $\endgroup$
    – Jon Custer
    Commented Oct 27, 2016 at 13:16
  • $\begingroup$ @JonCuster isn't acceleration take place within small RF-cavities, several times along the packet trajectory? $\endgroup$
    – Sparkler
    Commented Oct 27, 2016 at 13:26
  • $\begingroup$ Well, of course it does. Now, try to place a static electric field across a pair of plates in a 'cavity' - you let an electron go at the negative plate, it accelerates to the ground plate and off it goes. It comes around the beam line towards that negative plate (assume no losses) - it gradually coasts to a stop at the negative plate, then accelerates again towards ground. If you have energy losses during a round-trip, it doesn't even make it back to the fixed voltage negative plate. $\endgroup$
    – Jon Custer
    Commented Oct 27, 2016 at 13:32
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    $\begingroup$ Because it is climbing back up the potential from ground, back up the potential that accelerated it in the first place. Consider a marble on a round track with a hill on it. Start the marble at the top of the hill and let it go. It happily accelerates down the hill and around and then... has to climb up the hill again. So, you need a time-varying 'hill' - the track is flat until the electron is in the right place, then you push the hill up, making the marble accelerate more, then lower it so it is flat again until the marble rolls by... $\endgroup$
    – Jon Custer
    Commented Oct 27, 2016 at 13:44
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    $\begingroup$ Possible duplicate of Accelerating electrons via microwaves $\endgroup$
    – John Rennie
    Commented Oct 27, 2016 at 14:18

1 Answer 1

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As the electrons circle around, the field oscillations are timed to keep accelerating them forward. With a constant electric field, they would accelerated around half their orbit, then decelerated over the other half, leading to no net gain in energy. With the oscillating field, the acceleration always pushes the electrons forward, so they gain energy over the whole revolution.

With a nonrelativistic cyclotron, the frequency at which the field should oscillate is constant, because the orbital period turns out to be independent of the energy. However, when relativistic corrections are added, the oscillations need to be slowed down as the electrons gain energy. This gives rise to the name: "synchrotron" is short for "synchro-cyclotron," so called because the oscillations have to the synchronized with the orbiting particle's periods.

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  • $\begingroup$ To my understanding, acceleration takes place in RF-cavities, within which the electrons move approximately straight so I'm not sure what you mean by "With a constant electric field, they would accelerated around half their orbit, then decelerated over the other half". $\endgroup$
    – Sparkler
    Commented Oct 27, 2016 at 13:24
  • $\begingroup$ As a general matter, you cannot use any configuration of static electric fields to raise the energy of a charge moving around a circular orbit. The reason is simply that when the charge comes around to where it started, which means that it is at the same potential; hence its kinetic energy cannot have changed. With static cavities, the charge would be slowed down by the fringing fields outside the cavities before it could be accelerated back to its original speed inside the cavity. So ideally (but not in real accelerators), the fields shouldn't cycle on until the charges are already inside $\endgroup$
    – Buzz ♦
    Commented Oct 27, 2016 at 14:41

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