# Homework Help: Particle in linear accelerator

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1. Dec 15, 2013

### _Matt87_

1. The problem statement, all variables and given/known data
An electron ($e=1.6x10^(-19)$ Coulomb) of total energy 10.22 MeV ($\gamma$ about 20) is undergoing linear acceleration due to application of a uniform electric field of strenght $E_0=1000 MV/m$
a)how large is the applied force, in Newton?
b) Calculate the instantaneous radiated power in watts

2. Relevant equations

i think that for b) it would be : $P=\frac{2}{3}*\frac{e^2}{m^2*c^3}*(\frac{dE}{dt})^2$
but not so sure after all

3. The attempt at a solution

a) F=$E_0$*e ?

2. Dec 20, 2013

### RobyVonRintein

For (a), I agree, I think it's that simple.

For (b), the Larmor formula does not apply because this is so relativistic (gamma = 20!). You will have to use the relativistic formula; see section 8.2 here:

Note that getting acceleration from force (Newton's second Law) must be corrected for relativity, you can't just use F = ma.

3. Dec 21, 2013

### _Matt87_

so for linear acceleration:

$$P(t')=\frac{2}{3}\frac{e^2\dot{\beta^2}}{3c}\gamma^6$$

right? it's from 8.3 in the pdf you sent.
Although at the very end of that chapter there is still the same equation I wrote earlier. And it says 'Note that the radiation power due to linear
acceleration is independent of the particle energy or the relativity factor $\gamma$'

so what now then?

and what's for circulair motion ?:
$$P(t')=\frac{2}{3}\frac{e^2|\dot{v}|^2}{c^3}\gamma^4$$ ?

4. Dec 21, 2013

### RobyVonRintein

I'm looking at Jackson v3 eqns. 14.26-14.28 (hopefully you have a copy), they derive the formula as in the PDF as being dependent on the acceleration, then write acceleration as dp/dt, then write dp/dt as dE/dx, which they interpret as being entirely from the external field. In other words, the acceleration of the particle is entirely caused by the external fields.

So does it depend on the particle energy? No, only on the particle's acceleration.

Does it depend on gamma? No, it depends on dE/dx. BUT, if you try to write dE/dx in terms of dbeta/dt, then the factor of gamma^6 reappears. This is irrelevant in your problem (as in real experiments), as you have dE/dx given.

So I agree with you that the Larmor formula will give you the right answer. But, as presented in Griffiths for example, it only applies to non-relativistic motion -- if you want to use it for relativitistic particles, you have to explain why it is valid.

As for circular motion, this is correct in terms of the particle acceleration -- if you write it in terms of dE/dx = dp/dt (Jackson 14.47), it will depend only on external parameters as before -- except for the factor of gamma^2, which cannot be gotten rid of. This is why we can't do e-e+ collisions in the LHC for example ("synchrotron radiation"); we need an ILC.