I A bit confused with this question about the Penrose process

[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:

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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!

 

[etotheipi]

I figured out how to derive the equation of motion, but now I'm stuck on part e). Since $u^{\mu} = dx^{\mu} / d\tau$ parameterised by proper time $\tau$ is a unit vector we have $g_{\mu \nu} u^{\mu} u^{\nu} = -1$ and use that $\theta = \pi/2 = \mathrm{constant}$, i.e.\begin{align*}

g_{tt} u^t u^t + g_{\phi \phi} u^{\phi} u^{\phi} + g_{rr} u^r u^r &= -1 \\

\frac{-\Delta}{r^2} \left( \frac{\mathrm{d} t}{\mathrm{d} \tau} \right)^2 + r^2 \left( \frac{\mathrm{d} \phi}{\mathrm{d} \tau} \right)^2 + \frac{r^2}{\Delta} \left( \frac{\mathrm{d} r}{\mathrm{d} \tau} \right)^2 &= -1
\end{align*}then using $\mathrm{d} \phi / \mathrm{d} \tau = h/r^2$ and $\mathrm{d}t / \mathrm{d} \tau = \frac{r^2}{\Delta}(\mathcal{E} - \frac{qQ}{mr})$ you get \begin{align*}

1 + r^2 \left(\frac{h^2}{r^4}\right) + \frac{r^2}{\Delta} \left( \frac{\mathrm{d} r}{\mathrm{d} \tau} \right)^2 &= \frac{\Delta}{r^2} \cdot \left( \frac{r^2}{\Delta} \left[ \mathcal{E} - \frac{qQ}{mr} \right] \right)^2 \\

\frac{\Delta}{r^2} \left( \frac{h^2}{r^2} + 1 \right) + \left( \frac{\mathrm{d} r}{\mathrm{d} \tau} \right)^2 &= \left( \mathcal{E} - \frac{qQ}{mr} \right)^2

\end{align*}as is required. Not sure about part e) still. Let's consider particle $P_1$; it starts from infinity, presumably from rest i.e. $(\mathrm{d} r / \mathrm{d} \tau) \big{|}_{r = \infty} = 0$. If we further assume that it has zero angular momentum, and note that $\lim_{r \rightarrow \infty} \Delta/ r^2 = 1$, then we will obtain $\mathcal{E}_1 = 1$ and consequently $E_1 = m_1\mathcal{E}_1 > 0$.

As for the second particle, we are told that immediately after disintegration it has $\mathrm{d} r / \mathrm{d} \tau \sim 0$ so I suppose we can write:\begin{align*}
\frac{\Delta \big{|}_R}{R^2} \left(\frac{h_2^2}{R^2} + 1 \right) = \left( \mathcal{E}_2 - \frac{q_2Q}{m_2 R} \right)^2

\end{align*}where $R$ is the radius of the horizon. Does that look right - I'm not so sure why that implies $E_2 < 0$, because I'm not sure what to write for $h_2$...?

 

[etotheipi]

There's also a follow-up to calculate the largest possible energy extracted per unit initial mass of black hole, $\eta = E/M$. How can we go about doing that? I suppose we first need to work out $E_2 = m_2 \mathcal{E}_2$ but it's not clear how to do that because\begin{align*}

\left| \mathcal{E}_2 - \frac{q_2 Q}{m_2 R} \right| = \sqrt{\frac{\Delta \big{|}_R}{R^2} \left( \frac{h_2^2}{R^2} + 1\right)} \implies \mathcal{E}_2 = \frac{q_2 Q}{m_2 R} \pm \sqrt{\frac{\Delta \big{|}_R}{R^2} \left( \frac{h_2^2}{R^2} + 1\right)}

\end{align*}and it's still not clear to me why we can conclude $\mathcal{E}_2 < 0$... we're told that $q_2 < 0$ but we also need to evaluate the quantity inside the square root to confirm the RHS is negative in both the $\pm$ cases. I'm not sure what to put for $h_2$, mainly.

 

[PeterDonis]

The first two parts to the question

Question from where?

Also, what is $\Delta$?

 

[etotheipi]

Question from where?

Also, what is $\Delta$?

It is question 3 of this example sheet. Also here $\Delta :=r^2 - 2Mr + e^2$ where $e = \sqrt{Q^2 + P^2}$. The metric is$$\mathrm{d}s^2 = \frac{-\Delta}{r^2} \mathrm{d}t^2 + \frac{r^2}{\Delta} \mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2$$

 

[PeterDonis]

It is question 3 of this example sheet. Also here $\Delta :=r^2 - 2Mr + e^2$ where $e = \sqrt{Q^2 + P^2}$. The metric is$$\mathrm{d}s^2 = \frac{-\Delta}{r^2} \mathrm{d}t^2 + \frac{r^2}{\Delta} \mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2$$

Ok, got it.

 

[PeterDonis]

Let's consider particle $P_1$; it starts from infinity, presumably from rest

I think that is the intent of the question, yes.

$E_1 = \mathcal{E}_1 / m_1$

You have this backwards; it's $E_1 = \mathcal{E}_1 m_1$ (i.e., $\mathcal{E}$ is the "energy per unit mass").

As for the second particle, we are told that immediately after disintegration it has $\mathrm{d} r / \mathrm{d} \tau \sim 0$ so I suppose we can write:\begin{align*}
\frac{\Delta \big{|}_R}{R^2} \left(\frac{h_2^2}{R^2} + 1 \right) = \left( \frac{E_2}{R} - \frac{q_2Q}{m_2 R} \right)^2

\end{align*}where $R$ is the radius of the horizon.

Not quite. First, the thing that is squared on the RHS should be $\mathcal{E}_2 - q_2 Q / m_2 R$. Second, that expression can be simplified further by looking at the formula for $\mathcal{E}$, and moving a term from the RHS to the LHS. That should allow you to write the equation quoted above with an expression involving $dt / d\tau$ on the RHS, with a factor that will cancel with a factor on the LHS.

I'm not sure what to write for $h_2$...?

I think you should be able to assume that the entire process is radial only, i.e., that $h_1 = h_2 = h_3 = 0$.

 

[PeterDonis]

it's still not clear to me why we can conclude $\mathcal{E}_2 < 0$

What is the value of $\Delta$ on the horizon? Given that value, and the assumption that $dr / d\tau = 0$ for $P_2$, what does the radial equation of motion tell you about $\mathcal{E}_2 - q_2 Q / m_2 R$? And if $q_2 < 0$, what does that tell you about $\mathcal{E}_2$?

 

[PeterDonis]

What is the value of $\Delta$ on the horizon? Given that value, and the assumption that $dr / d\tau = 0$ for $P_2$, what does the radial equation of motion tell you about $\mathcal{E}_2 - q_2 Q / m_2 R$? And if $q_2 < 0$, what does that tell you about $\mathcal{E}_2$?

In fact, you don't even need to look at the radial equation of motion; you can just look at the formula for $\mathcal{E}$ and see what it says for the value of $\Delta$ on the horizon and $q < 0$.

 

[etotheipi]

the thing that is squared on the RHS should be $\mathcal{E}_2 - q_2 Q / m_2 R$

You have this backwards; it's $E_1 = \mathcal{E}_1 m_1$ (i.e., $\mathcal{E}$ is the "energy per unit mass").

Thanks, you're right, these are quite silly mistakes! I've fixed up those equations

In fact, you don't even need to look at the radial equation of motion; you can just look at the formula for $\mathcal{E}$ and see what it says for the value of $\Delta$ on the horizon and $q < 0$.

I see! I'd forgotten that $\Delta$ vanishes on the horizon so from the definition of $\mathcal{E}_2$ we can see that it's negative. I also worked out that $dt/d\tau = r / \sqrt{\Delta}$ but as you point out, we don't actually need this!

 

[etotheipi]

@PeterDonis I wondered if you looked at question 5? It seems like a natural extension of this one, but I'm not sure how to approach it.

When $P_2$ crosses $\mathcal{H}^+$ the black hole gains mass $\delta M = \mathcal{E}_2$ and angular momentum $\delta J = h_2 \, (=0)$. I think we may also write that $-u_2 \cdot \xi \geq 0$ where $u_2$ is the 4-velocity of $P_2$, because both $u_2$ and $\xi$ are future-directed causal vectors, and from the definition $-\mathcal{E}_2 = u_2 \cdot \xi - \frac{q_2}{m_2} \Phi$ we may write down the inequality $\mathcal{E}_2 \geq \frac{q_2}{m_2} \Phi$! The maximum decrease of black hole energy then appears to be of magnitude $(|q_2|/m_2 )\Phi$.

 

[PeterDonis]

@PeterDonis I wondered if you looked at question 5? It seems like a natural extension of this one

Sort of; it's in Kerr spacetime, not Reissner-Nordstrom, so the details of the math are somewhat different. Also, as I read it, it's asking about the maximum possible energy that can be extracted globally, using as many individual operations as required; whereas question 3 is about the energy extracted in just one operation (one object falling into the hole).