A bit confused with this question about the Penrose process

• I
• etotheipi
In summary, the conversation discusses the proof that a vector field that preserves the Maxwell field also preserves the conservation of a charged particle's energy and angular momentum. From this, the question arises of how to write the potential in the form of ##\Phi = Q/r##. The equation of motion is derived and it is shown that ##\mathcal{E} = -\Delta / r^2 \cdot d_t / d\tau - q/m \cdot \Phi##. It is then noted that particle ##P_1## has zero angular momentum and a positive energy while particle ##P_2## has a negative energy. Finally, the conversation delves into calculating the largest possible energy extracted per unit initial mass of a black
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!

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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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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.

etotheipi said:
The first two parts to the question
Question from where?

Also, what is ##\Delta##?

etotheipi
PeterDonis said:
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$$\mathrm{d}s^2 = \frac{-\Delta}{r^2} \mathrm{d}t^2 + \frac{r^2}{\Delta} \mathrm{d}r^2 + r^2 \mathrm{d}\Omega^2$$

etotheipi said:
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.

etotheipi said:
Let's consider particle ##P_1##; it starts from infinity, presumably from rest
I think that is the intent of the question, yes.

etotheipi said:
##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").

etotheipi said:
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.

etotheipi said:
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##.

etotheipi
etotheipi said:
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##?

etotheipi
PeterDonis said:
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
PeterDonis said:
the thing that is squared on the RHS should be ##\mathcal{E}_2 - q_2 Q / m_2 R##
PeterDonis said:
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

PeterDonis said:
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!

@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##.

etotheipi said:
@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).

1. What is the Penrose process?

The Penrose process is a theoretical concept in astrophysics that describes a way to extract energy from a rotating black hole.

2. How does the Penrose process work?

The Penrose process involves sending particles into the ergosphere of a rotating black hole. These particles split into two, with one particle falling into the black hole and the other escaping with increased energy. This energy is extracted from the rotation of the black hole.

3. Is the Penrose process possible?

While the Penrose process is theoretically possible, it has not been observed in nature. It requires precise conditions and a source of particles with enough energy to enter the ergosphere of a black hole.

4. What are the implications of the Penrose process?

If the Penrose process were to occur, it could potentially be a source of immense energy. It could also have implications for our understanding of black holes and their role in the universe.

5. How is the Penrose process related to Hawking radiation?

The Penrose process and Hawking radiation are both ways in which energy can be extracted from black holes. However, the Penrose process involves extracting energy from the rotation of a black hole, while Hawking radiation involves the spontaneous emission of particles from the event horizon of a black hole.

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