electrodynamics / Companion to J.D. Jackson's Classical Electrodynamics 3rd ed. - R. Magyar

in units of M eV /cm. θc is related to n and β from the results in part a.

I have lot’s of cool Maple plots which I plan on including but for now, I’ll just give you the final numbers.

For an incident electron with T = 1 MeV, the number of Cherenkov photons is about 187. The critical angle is 0.78 rad.

For an incident proton with T = 500 MeV, the number of Cherenkov photons is about 79. The critical angle is 0.50 rad.

For an incident proton with T = 5 TeV, the number of Cherenkov photons is about 208. The critical angle is 0.83 rad.

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Problem 14.5

A non-relativistic particle of charge Zq, mass m, and kinetic energy T makes a head on collision with fixed central force field of infinite range. The interaction is repulsive and described by a potential V (r), which becomes greater than E at close distances.

a. Find the total energy radiated.

The total energy for the particle is constant.

At rmin, the velocity will vanish and E = V (rmin).

From Jackson equation 14.21, we have the power radiated per solid angle for

The total work is the power integrated over the entire trip:

The factor of two comes because the particle radiates as it accelerates to and from the potential. We can solve equation 10 for v.

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The region for the first integral is excluded because the particle will never go there, thus, the first integral vanishes. We are left with

quod erat demonstrandum.

b. For the Coulomb potential, Vc(r) = zZrq2 , find the total energy

radiated.

First, dVdrc = −zZr2q2 = −Vrc . Also, we can solve for dr.

Vc r2dV dVc = − r dr → dr = −zZq2

Plug Vc(r) and dr into equation 11:

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So the integral equals −1615a2 √a, and finally, we have

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