A bit confused with this question about the Penrose process

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  • #1
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:
1619969797831.png


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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  • #2
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##...?
 
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  • #3
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.
 
  • #5
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$$\mathrm{d}s^2 = \frac{-\Delta}{r^2} \mathrm{d}t^2 + \frac{r^2}{\Delta} \mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2$$
 
  • #6
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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.
 
  • #7
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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##.
 
  • #8
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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##?
 
  • #9
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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##.
 
  • #10
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 :wink:

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!
 
  • #11
@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##.
 
  • #12
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@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).
 

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