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I know that $-\frac{\mathrm{d}E}{\mathrm{d}t} \propto E^2$ for an electron losing energy to synchrotron radiation, but I can't find how to arrange this to give the time it would take for the electron to lose half of its original energy. How would I go about working that out?

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The half-life description only works for exponential decay. As others have shown, this leads to a power-law decay, as 1/t. So you can ask for the asymptotic half-life in log-t variable, but not in t. – Ron Maimon Nov 8 '11 at 6:48
But the OP does not speak of half-life! His problem has meaning per se. – Vladimir Kalitvianski Nov 9 '11 at 15:21

$$\frac{dE}{dt} = -CE^2$$

$$d\left (\frac{1}{E}\right ) = Cdt$$

$$\frac{1}{E_2}-\frac{1}{E_1}= C(t_2-t_1)$$


$$\frac{1}{E_1} = C\Delta t$$ $$\Delta t = \frac{1}{CE_1}$$

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Not to worry, it's fairly easy: right now you have a differential equation which can be written

$$\frac{\mathrm{d}E}{\mathrm{d}t} = -CE^2$$

for some constant $C$. You need to solve that differential equation for $E(t)$. (If you're wondering how to do that, you can find more information at the math site.) Then you can determine the electron's energy at the initial time $t_0$ as $E(t_0)$, and find the time at which its energy becomes half of that:

$$E(t) = \frac{E(t_0)}{2}$$

and solve for $t$.

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