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EM M04 2

< PlasmaWiki | Prelims(Link to this page as [[PlasmaWiki/Prelims/EM M04 2]])

We have the Larmor formula for the total power radiated by an accelerating charge:

$P_{rad}=\frac{2e^{2}a^{2}}{3c^{3}}$

The force doing the accelerating is:

$F_{coulomb}=ma=-\frac{e^{2}}{r^{2}}\Rightarrow a=-\frac{e^{2}}{mr^{2}}$

So that:

$P_{rad}=\frac{2e^{2}}{3c^{3}}\left(-\frac{e^{2}}{mr^{2}}\right)^{2}=\frac{2e^{6}}{3c^{3}m^{2}r^{4}}$

The total energy is:

$E=\frac{1}{2}mv^{2}+\int P_{rad}dt-\frac{e^{2}}{r}$

We find the velocity as a function of radius, since it has a nearly circular orbit, by:

$F_{centrip}=F_{coulomb}\Rightarrow\frac{mv^{2}}{r}=\frac{e^{2}}{r^{2}}\Rightarrow v=\sqrt{\frac{e^{2}}{mr}}$

So we can write the total energy:

$E_{0}=-\frac{e^{2}}{2r}+\int\frac{2e^{6}}{3c^{3}m^{2}r^{4}}dt$

Differentiating with respect to time:

$0=\frac{e^{2}}{2r^{2}}\frac{dr}{dt}+\frac{2e^{6}}{3c^{3}m^{2}r^{4}}$

Rearranging:

$-\frac{3c^{3}m^{2}r^{2}}{4e^{4}}\frac{dr}{dt}=1$

Integrating with respect to time:

$const-\frac{c^{3}m^{2}r^{3}}{4e^{4}}=t$

Since at $t = 0$ , $r = r i$ :

$t=\frac{c^{3}m^{2}}{4e^{4}}\left(r_{i}^{3}-r^{3}\right)\approx.1\,\textrm{ns}$

This page was recovered in October 2009 from the Plasmagicians page on Prelim_M04_EM2 dated 00:15, 6 January 2006.