- #1

etotheipi

The first two parts to the question were as follows. If a vector field ##\xi## preserves the Maxwell field then ##\mathcal{L}_{\xi} F = 0## so by Cartan's magic formula ##i_{\xi} \mathrm{d}F + \mathrm{d}(i_{\xi} F) = 0##. But since ##\mathrm{d}F = 0## then ##\mathrm{d}(i_{\xi} F) = 0 \implies i_{\xi} F = \mathrm{d}\Phi## for some ##\Phi##, since ##\mathrm{d} \circ \mathrm{d} = 0##.

Then we need to show that ##\xi \cdot u - \frac{q}{m} \Phi## is conserved along the worldline of a charged particle. We check\begin{align*}

u^a \nabla_a (\xi_b u^b - \frac{q}{m} \Phi) &= u^a \xi_b \nabla_a u^b + u^a u^b \nabla_a \xi_b - \frac{q}{m} u^a \nabla_a \Phi

\end{align*}but by Killing's equation ##\nabla_{a} \xi_b + \nabla_b \xi_a = 0## we have that ##u^a u^b \nabla_a \xi_b = u^b u^a \nabla_b \xi_a = - u^a u^b \nabla_a \xi_b \implies u^a u^b \nabla_a \xi_b = 0##. Furthermore the equation of motion in the field is ##u^a \nabla_a u^b = \frac{q}{m} {F^b}_c u^c## and therefore\begin{align*}u^a \nabla_a (\xi_b u^b - \frac{q}{m} \Phi) = \frac{q}{m} \left( \xi_b {F^b}_c u^c - u^a \nabla_a \Phi \right) = \frac{q}{m} u^a (\xi_b {F^b}_a - \nabla_a \Phi)

\end{align*}but ##\xi_b {F^b}_a## is nothing but ##\xi_b {F^b}_a = F(\xi, \mathbf{e}_a) = (i_{\xi} F)(\mathbf{e}_a)## hence as per the first part we have ##\xi_b {F^b}_a = \nabla_a \Phi## which proves the result. I'm stuck on the next part:

Because the RN solution is spherically symmetric it has an ##\mathrm{SO}(3)## isometry group and hence admits a Killing field ##m = \partial / \partial \phi## which gives rise to a conserved quantity\begin{align*}

h := m \cdot u = g_{\mu \nu} m^{\mu} u^{\nu} = g_{\phi \phi} u^{\phi} = r^2 \frac{\mathrm{d} \phi}{\mathrm{d}\tau}

\end{align*}and from parts a) and b) we can define another conserved quantity ##-\mathcal{E}## by\begin{align*}

-\mathcal{E} := g_{\mu \nu} \xi^{\mu} u^{\nu} - \frac{q}{m} \Phi &= g_{tt} u^t - \frac{q}{m} \Phi \\

&= \frac{-\Delta}{r^2} \frac{\mathrm{d}t}{\mathrm{d}\tau} - \frac{q}{m} \Phi

\end{align*}or otherwise written ##\mathcal{E} = \frac{\Delta}{r^2} \frac{\mathrm{d} t}{\mathrm{d} \tau} + \frac{q}{m} \Phi##. Firstly, why can we then write ##\Phi## in the form ##\Phi = Q/r##? I'm also not sure how to get the equation of motion, nor what to write for d); I suppose that ##h = 0## corresponds to zero angular momentum and thus a radial path, but I'm not sure about the other parts.

I wondered if someone could help explain these last few points, then I'll have a go at e). Thanks!

Then we need to show that ##\xi \cdot u - \frac{q}{m} \Phi## is conserved along the worldline of a charged particle. We check\begin{align*}

u^a \nabla_a (\xi_b u^b - \frac{q}{m} \Phi) &= u^a \xi_b \nabla_a u^b + u^a u^b \nabla_a \xi_b - \frac{q}{m} u^a \nabla_a \Phi

\end{align*}but by Killing's equation ##\nabla_{a} \xi_b + \nabla_b \xi_a = 0## we have that ##u^a u^b \nabla_a \xi_b = u^b u^a \nabla_b \xi_a = - u^a u^b \nabla_a \xi_b \implies u^a u^b \nabla_a \xi_b = 0##. Furthermore the equation of motion in the field is ##u^a \nabla_a u^b = \frac{q}{m} {F^b}_c u^c## and therefore\begin{align*}u^a \nabla_a (\xi_b u^b - \frac{q}{m} \Phi) = \frac{q}{m} \left( \xi_b {F^b}_c u^c - u^a \nabla_a \Phi \right) = \frac{q}{m} u^a (\xi_b {F^b}_a - \nabla_a \Phi)

\end{align*}but ##\xi_b {F^b}_a## is nothing but ##\xi_b {F^b}_a = F(\xi, \mathbf{e}_a) = (i_{\xi} F)(\mathbf{e}_a)## hence as per the first part we have ##\xi_b {F^b}_a = \nabla_a \Phi## which proves the result. I'm stuck on the next part:

Because the RN solution is spherically symmetric it has an ##\mathrm{SO}(3)## isometry group and hence admits a Killing field ##m = \partial / \partial \phi## which gives rise to a conserved quantity\begin{align*}

h := m \cdot u = g_{\mu \nu} m^{\mu} u^{\nu} = g_{\phi \phi} u^{\phi} = r^2 \frac{\mathrm{d} \phi}{\mathrm{d}\tau}

\end{align*}and from parts a) and b) we can define another conserved quantity ##-\mathcal{E}## by\begin{align*}

-\mathcal{E} := g_{\mu \nu} \xi^{\mu} u^{\nu} - \frac{q}{m} \Phi &= g_{tt} u^t - \frac{q}{m} \Phi \\

&= \frac{-\Delta}{r^2} \frac{\mathrm{d}t}{\mathrm{d}\tau} - \frac{q}{m} \Phi

\end{align*}or otherwise written ##\mathcal{E} = \frac{\Delta}{r^2} \frac{\mathrm{d} t}{\mathrm{d} \tau} + \frac{q}{m} \Phi##. Firstly, why can we then write ##\Phi## in the form ##\Phi = Q/r##? I'm also not sure how to get the equation of motion, nor what to write for d); I suppose that ##h = 0## corresponds to zero angular momentum and thus a radial path, but I'm not sure about the other parts.

I wondered if someone could help explain these last few points, then I'll have a go at e). Thanks!

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