Question

Derive an expression showing how the of an ultra-relativistic electron emitting synchrotron radiation evolves in time. Assume that 0 is the initial value of and that there is a uniform magnetic field of strength B (and so B _ = B sin a). Your expression should involve only the above variables, time t, and physical constants.

Answer 1

Answer on Question 55062, Physics / Astronomy | Astrophysics

Question:

Derive an expression showing how the of an ultra-relativistic electron emitting synchrotron radiation evolves in time. Assume that 0 is the initial value of and that there is a uniform magnetic field of strength B (and so B _ = B sin a). Your expression should involve only the above variables, time t, and physical constants.

Solution:

The synchrotron power of an electron is given by:

$$P = 2 \sigma_{T} \beta^{2} \gamma^{2} c \times \frac{B^{2}}{8 \pi} \sin^{2} a = A \beta^{2} \gamma^{2} = A (\gamma - 1) \approx A \gamma^{2}, \text{ where:}$$ $$A = 2 \sigma_{T} c \times \frac{B^{2}}{8 \pi} \text{ is a constant.}$$

Equating this power to the loss in the energy of the electron:

$$\frac{d}{dt} \gamma m_{e} c^{2} = - P$$ $$m_{e} c^{2} \frac{d}{dt} \gamma = - A \gamma^{2} \Rightarrow \frac{d \gamma}{\gamma^{2}} = - \frac{A}{m_{e} c^{2}} dt$$ $$\int_{\gamma_{0}}^{\gamma} \frac{d \gamma}{\gamma^{2}} = - \frac{A}{m_{e} c^{2}} \int_{0}^{\gamma} dt$$ $$\left[ \frac{1}{\gamma_{0}} = \frac{1}{\lambda} \right] = - \frac{A}{m_{e} c^{2}} t \Rightarrow \gamma = \gamma_{0} (1 + A' \gamma_{0} t)^{-1}$$

where:

$$A' = \frac{A}{m_{e} c^{2}} = \frac{B_{\perp} \sigma_{T}}{4 \pi m_{e} c} = \frac{2 B_{\perp} \varepsilon^{4}}{3 m_{e}^{3} c^{5}}$$

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