Do the Planck voltage and the Planck current have a natural physical interpretation in classical general relativity?

Question

Most Planck units are a product of powers of all three of $\hbar$, $c$, and $G$, so we will not be able to fully understand their physical significance until we have a full theory of quantum gravity. But some of them are only powers of one or two of those three quantities, and we should be able to understand those units in a simpler physical regime. For example, the Planck charge $\sqrt{\hbar c}$ (under Lorentz-Heaviside conventions) sets the natural scale for the coupling constants of a relativistic quantum field theory in a non-dynamical $(3+1)$-dimensional spacetime (and indeed, at low energies the Standard Model's coupling constants are all within an order of magnitude or so of this charge scale).

Other examples are the Planck force $c^4/G$, the Planck power $c^5/G$, the Planck voltage $c^2/\sqrt{G}$, and the Planck current $c^3/\sqrt{G}$. These quantities are just powers of $c$ and $G$, so they should have an interpretation within completely classical general relativity. Indeed, as the Wikipedia page explains, the Planck force sets the scale of the gravitational force between any two Schwarzchild black holes (regardless of their masses) whose event horizons just touch, and also sets the scale for the effective gravitational forces that spacetime curvature induces on matter.

The physical significance of the Planck voltage and the Planck current in classical GR is less clear to me. In what sense do they set the natural scales for voltage and current in GR?

Two guesses, which seem to be in a similar spirit as the physical interpretation of the other Planck units, is that the effective current circulating around the equator of any Kerr-Newman black hole is on the order of the Planck current, and that the Planck voltage is the largest possible voltage difference before the electrostatic energy self-gravitation causes the system to collapse into a Reisser-Nordstrom black hole. (But those are basically just wild guesses.)

Edit: to clarify, the proposition that "the Planck voltage and current do not depend on $\hbar$" perhaps requires some subtlety of interpretation, but it is not important to my question. I'm just wondering whether there's any sense in which the quantities $c^2/\sqrt{G}$ and $c^3/\sqrt{G}$ set natural scales for voltage and current in the purely classical theory of EM+GR (with no charge quantization). You don't have to call those quantities the "Planck voltage and current" if you don't want to; that was just motivation.

Answer 1

To avoid the issues raised in Andrew's answer, let us work in SI units. Then the Planck voltage is

$$\frac{c^2}{\sqrt{4\pi \varepsilon_0 G}} \approx 1.04 \times 10^{27} \text{ V}$$

and the Planck current is

$$\frac{c^3}{\sqrt{G/4\pi \varepsilon_0}} \approx 3.48 \times 10^{25} \text{ A}$$

Planck Voltage

What is the maximum voltage that one could theoretically achieve in the classical, EM+GR universe? Consider a spherically symmetric charged object with charge $Q$, mass $M$, and radius $R$. (Of course, objects do not have to be spherically symmetric, but the result from the spherically symmetric case should be within a small constant factor of the true limit.) The voltage at its surface is

$$V = \frac{Q}{4\pi\varepsilon_0 R}$$

Let us try to increase the voltage by reducing the radius of the object. If we do this, the object will eventually collapse into a black hole. This occurs at the Schwarzschild radius

$$R_s = \frac{2GM}{c^2}$$

when the voltage is

$$V = \frac{Qc^2}{4\pi\varepsilon_0 \cdot 2GM}$$

We can now increase the voltage further by increasing the charge $Q$. According to the Reissner–Nordström metric, a charged black hole has two event horizons, the (outer) true event horizon and an (inner) Cauchy horizon. The radii of these horizons are given by the roots of a quadratic equation. As the black hole's charge increases, the horizons move closer and closer together until their radii are equal. This happens when

\begin{align} \sqrt{\frac{Q^2G}{4\pi\varepsilon_0 c^4}} = R_Q &\geq \frac12 R_S = \frac{GM}{c^2} \\ \Rightarrow \frac{Q}{M} &\geq \sqrt{4\pi\varepsilon_0 G} \end{align}

Beyond this point, the solutions to the quadratic equation become imaginary, indicating that there is no event horizon and the black hole's singularity has become naked. If we assume that is not physically meaningful, the voltage has reached a limit of

$$V = \frac{c^2\sqrt{4\pi\varepsilon_0 G}}{8\pi\varepsilon_0 G} = \frac{c^2}{2\sqrt{4\pi \varepsilon_0 G}}$$

which is half the Planck voltage.

You might suspect that it's possible to sneak around this limit by not allowing the object to collapse into a black hole. However, that doesn't work either: even if the rest mass is kept low, if we increase $Q/R$ too much, we still have too much electrostatic potential energy concentrated into a small space. A black hole forms once again.

Planck Current

In relativity, current is a relative quantity: it changes from one reference frame to another. Therefore, instead of a current-carrying wire, let us examine the Planck current by considering a free-fall trajectory approaching the extremal black hole from the last section. As before, the black hole's charge is limited to

$$\frac{Q}{M} = \sqrt{4\pi\varepsilon_0 G}$$

By dimensional analysis, as we approach the black hole's event horizon and increase our speed towards the speed of light, we would expect to observe a current increasing towards approximately

$$Q \left(\frac{c}{R_s}\right) = \frac{Mc\sqrt{4\pi\varepsilon_0 G}}{2GM/c^2} = \frac{c^3\sqrt{4\pi\varepsilon_0}}{2 \sqrt{G}}$$

which is the Planck current (within a factor of 2). However, this result does not imply any clear physical limit since, unlike voltage, current can be spread out across large regions of space.

Answer 2

I don't think this question really has a unique answer, but my take is that the premise of the question is a little confused because of the use of Heaviside units.

I would actually find it more natural to introduce the fine-structure constant \begin{equation} \alpha = \frac{e^2}{4\pi \epsilon_0 \hbar c} \end{equation} Then the "Planck charge" is really just the standard base charge unit \begin{equation} e = \sqrt{4\pi \epsilon_0 \hbar c \alpha} \end{equation} And the Planck voltage (Planck energy per Planck charge) is \begin{equation} V_{\rm Pl} = \frac{E_{\rm pl}}{e} = \sqrt{\frac{\hbar c^5}{4 \pi \epsilon_0 \alpha \hbar c G}} = \sqrt{\frac{c^4}{4\pi \epsilon_0 \alpha G}} \end{equation} Or, in units where $4\pi\epsilon_0=1$, \begin{equation} V_{\rm Pl} = \sqrt{\frac{c^4}{\alpha G}} = \sqrt{\frac{\hbar c^5}{e^2 G }} \end{equation} To me this is more natural, since there is explicitly an electromagnetic parameter appearing in this equation.

In this form, it's maybe more clear that the presence or absence of $\hbar$ in this expression depends on whether you think of $\alpha$ or $e$ as being more fundamental. The straightforward classical logic would say that the electric charge $e$ is a basic quantity. I have two main arguments: (1) the charge is something you can measure classically, and (2) there's no reason to use $\alpha$ in electrodynamics unless you are doing quantum field theory.

The point in (2) that the natural interpretation of a parameter is different in classical and quantum theory is a pretty common situation actually. For example, we normally write the mass term of a scalar field theory Lagrangian as $\frac{1}{2} m^2\phi^2$, but this is only because we set $\hbar=c=1$. Dimensionally, the parameter $m^2$ is really a frequency squared, and if you restore $\hbar$ and $c$ this term is really $\frac{1}{2} \omega^2 \phi^2 = \frac{1}{2} \frac{m^2 c^4}{\hbar^2}\phi^2$. Classically there's no reason at all to think of this term as a mass; the interpretation comes from quantizing the scalar field and identifying the quanta with particles. Similarly, classically the natural way to write the covariant derivative of a charged scalar is $D_\mu \Phi = (\partial_\mu - i e A_\mu) \Phi$, where $e$ is the charge. It's only convenient to introduce $\alpha$ when we start calculating Feynman diagrams or other quantum effects. Indeed, wikipedia gives several interpretations of $\alpha$, which all involve some quantum effect, for example: "The ratio of the velocity of the electron in the first circular orbit of the Bohr model of the atom, which is $\frac{e^2}{4\pi\epsilon_0\hbar}$, to the speed of light in vacuum, $c$."

Anyway, if you accept that $e$ is a more natural choice than $\alpha$, classically, then the Planck voltage does depend on $\hbar$.

Answer 3

One needs to be very careful in assigning any fundamental physical meaning to any of the Planck values. They are all based on arbitrary decisions as to which of the physical constants contain a value of unity. While Planck values can be useful, they do not represent fundamental values of nature in that they can be defined in many different ways.