Use the Larmor formula (Eq. 11.70) and special relativity to derive the Liénard formula (Eq. 11.73).
P=\frac{\mu_{0} q^{2} a^{2}}{6 \pi c} (11.70)
P=\frac{\mu_{0} q^{2} \gamma^{6}}{6 \pi c}\left(a^{2}-\left|\frac{ v \times a }{c}\right|^{2}\right) (11.73)
We know that (proper) power transforms as the zeroth component of a 4-vector K^{0}=\frac{1}{c} \frac{d W}{d \tau} . The Larmor formula says that for v=0, \frac{d W}{d \tau}=\frac{1}{4 \pi \epsilon_{0}} \frac{2}{3} \frac{q^{2}}{c^{3}} a^{2} (Eq. 11.70). Can we think of a 4-vector whose zeroth component reduces to this when the velocity is zero?
Well, a^{2} \text { smells like }\left(\alpha^{\nu} \alpha_{\nu}\right), but how do we get a 4-vector in here? How about \eta^{\mu}, whose zeroth component is just c, when v = 0? Try, then:
K^{\mu}=\frac{1}{4 \pi \epsilon_{0}} \frac{2}{3} \frac{q^{2}}{c^{5}}\left(\alpha^{\nu} \alpha_{\nu}\right) \eta_{\mu}This has the right transformation properties, but we must check that it does reduce to the Larmor formula when v = 0:
\frac{d W}{d t}=\frac{1}{\gamma} \frac{d W}{d \tau}=\frac{1}{\gamma} c K^{0}=\frac{1}{\gamma} c \frac{\mu_{0} q^{2}}{6 \pi c^{3}}\left(\alpha^{\nu} \alpha_{\nu}\right) \eta^{0}, \text { but } \eta^{0}=c \gamma, \text { so } \frac{d W}{d t}=\frac{\mu_{0} q^{2}}{6 \pi c}\left(\alpha^{\nu} \alpha_{\nu}\right) . [Incidentally, this tells us that the power itself (as opposed to proper power) is a scalar. If this had been obvious from the start, we could simply have looked for a Lorentz scalar that generalizes the Larmor formula.]
In Prob. 12.39(b) we calculated \left(\alpha^{\nu} \alpha_{\nu}\right) in terms of ordinary velocity and acceleration:
\begin{aligned}\alpha^{\nu} \alpha_{\nu} &=\gamma^{4}\left[a^{2}+\frac{( v \cdot a )^{2}}{\left(c^{2}-v^{2}\right)}\right]=\gamma^{6}\left[a^{2} \gamma^{-2}+\frac{1}{c^{2}}( v \cdot a )^{2}\right] \\&=\gamma^{6}\left[a^{2}\left(1-\frac{v^{2}}{c^{2}}\right)+\frac{1}{c^{2}}( v \cdot a )^{2}\right]=\gamma^{6}\left\{a^{2}-\frac{1}{c^{2}}\left[v^{2} a^{2}-( v \cdot a )^{2}\right]\right\}\end{aligned}Now v \cdot a =v a \cos \theta, \text { where } \theta is the angle between v and a, so:
v^{2} a^{2}-( v \cdot a )^{2}=v^{2} a^{2}\left(1-\cos ^{2} \theta\right)=v^{2} a^{2} \sin ^{2} \theta=| v \times a |^{2}.
\alpha^{\nu} \alpha_{\nu}=\gamma^{6}\left(a^{2}-\left|\frac{ v \times a }{c}\right|^{2}\right).
\frac{d W}{d t}=\frac{\mu_{0} q^{2}}{6 \pi c} \gamma^{6}\left(a^{2}-\left|\frac{ v \times a }{c}\right|^{2}\right) , which is Li´enard’s formula (Eq. 11.73).