 Solution

ID: past-exam-of-the-symmetry-and-relativity-course-of-the-university-of-oxford/2015/1/c/i/solution

Interpret

3

GeV as the electron's total energy; the alternative kinetic-energy reading changes the estimate only slightly at this scale. The Lorentz factor and orbital acceleration are

γ = \frac{3000}{0.510999} = 5.871 \times 1 0^{3}, a = \frac{v ^{2}}{R}, a_{0} = γ^{2} \frac{v ^{2}}{R} .

Using the supplied Larmor formula,

μ_{0} = 1/ (ϵ_{0} c^{2})

and the revolution time

2 π R / v

P = \frac{e ^{2} γ ^{4} v ^{4}}{6 π ϵ _{0} c ^{3} R ^{2}}, U_{turn} = \frac{e ^{2}}{3 ϵ _{0} R} γ^{4} β^{3} .

For

R = 561

m this gives

P = 1.7405 \times 1 0^{- 10}

W per electron and

T_{turn} = 1.17577 \times 1 0^{- 5}

s. Their product is

U_{turn} = 2.0464 \times 1 0^{- 15} J = 1.2773 \times 1 0^{4} eV ≃ 12.8 keV .

This is the energy loss of one electron on a full circular orbit, not its energy loss per metre or the total loss of the whole bunch. The fractional loss is only about

4.26 \times 1 0^{- 6}

of the

3

GeV energy each revolution. Maintaining a steady stored energy requires replacing that energy through the accelerator's rf system.

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