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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.

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    $\begingroup$ "These quantities are just powers of c and G so they should have an interpretation within completely classical general relativity" - While this is certainly possible (and even likely based on the magnitudes), there is no sufficient logic in this statement to guarantee that it is true. For example, the origin of G is unknown and may be related to quantum phenonena. $\endgroup$
    – safesphere
    Dec 3, 2017 at 3:15
  • $\begingroup$ I wonder if the interpretation of the black hole horizon as a membrane satisfying Ohm’s law with a surface resistivity of ρ = 4π ≈ 377 Ω is related. Not sure, since apparently it's non relativistic Ohm’s law. Interesting question, though. $\endgroup$
    – Rexcirus
    Dec 3, 2017 at 8:41
  • $\begingroup$ +1: for planck charge & voltage - I hadn't come across them before. $\endgroup$ Dec 3, 2017 at 13:34
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    $\begingroup$ @Rexcirus That's not the surface resistivity of a BH, but rather the surface resistance. And it might be more relativistic than you suspect, because if you stop setting $c = 1$ then it's actually $R = 4 \pi / c = 377\, \Omega$ :-). As suggested by the lack of a $G$, that value is actually just the flat-spacetime impedance of free space and is not particular to black holes. $\endgroup$
    – tparker
    Dec 3, 2017 at 19:12

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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.

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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 $\hbar$.

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  • $\begingroup$ A minor point and a major point in response. The minor point is that while you are right that the elementary charge certainly sets an important charge scale, I completely disagree that it's the natural choice of "Planck charge". The whole point of Planck units is that they characterize the structure of spacetime and the most fundamental laws of physics themselves, not any contingent Standard Model parameters like coupling constants between fields. There is physical content in the statement that $e/\hbar c = \sqrt{\alpha}$ is much smaller than 1; it tells us that QED is a weakly coupled $\endgroup$
    – tparker
    Jul 31, 2021 at 16:17
  • $\begingroup$ theory. By contrast, the value of $\sqrt{\hbar c}$ has no fundamental physical interest, and it's clearly the natural unit to normalize to 1. (It would be possible, but less convenient or conceptually clear, to do QED in units where $\alpha = 1$. Moreover, the Planck charge sets the natural charge scale in QFT for all gauge couplings, including the strong and weak interactions, not just EM. It seems very arbitrary to designate EM as setting "the" Planck scale, especially since the value of $\alpha$ isn't even a fundamental parameter in the SM, but it can be expressed in terms of the more $\endgroup$
    – tparker
    Jul 31, 2021 at 16:21
  • $\begingroup$ fundamental electroweak coupling constants. $\endgroup$
    – tparker
    Jul 31, 2021 at 16:21
  • $\begingroup$ @tparker Well, two things. (a) As far as we know, the electron charge is a fundamental charge scale in nature. But, ok, if you want to define a "Planckian" base charge scale that is not the charge of a particle. that's fine. (b) You could run through my argument with $e$ replaced by $e_{\rm Pl}$. The key step will be whether you say $e_{\rm Pl}$ is a fixed quantity, or if it scales with $\hbar$. If you say $e_{\rm Pl}$ scales with $\hbar$, then it doesn't really have a classical meaning, so I would say any quantities based on $e_{\rm Pl}$ like a voltage also don't have a classical meaning. $\endgroup$
    – Andrew
    Jul 31, 2021 at 16:23
  • $\begingroup$ Also this is a side point but I don't understand how you can "choose units" where the dimensionless quantity $\alpha$ is equal to $1$. $\endgroup$
    – Andrew
    Jul 31, 2021 at 16:23
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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.

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    $\begingroup$ That disclaimer aside, with $c=\hbar=G=4\pi\varepsilon_0=1$ the Planck voltage is $\frac{c^2}{\sqrt{4\pi\varepsilon_0G}}\approx1.0429\times10^{27}\operatorname{V}$, while the Planck current is $c^3\sqrt{\frac{4\pi\varepsilon_0}{G}}\approx3.4789\times{10}^{25}A$. The question is really about whether these, modulo powers of $4\pi$ etc., are significant. $\endgroup$
    – J.G.
    Jul 30, 2021 at 16:58
  • $\begingroup$ Yet, c=ℏ=G=1 is nothing more than a convenient way to compare universal constants. There is absolutely no physical reason for establishing this equality. While apparently some do not like my answer that one needs to be careful in placing too much meaning into Planck values, it none-the-less is the correct statement. Planck values can be useful, but one must be very careful in placing a physical meaning on them. $\endgroup$
    – JRL
    Aug 3, 2021 at 22:05
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    $\begingroup$ You're missing the point. Yes, our definitions of Planck voltages & currents can be adjusted by some small multiplicative constants. But the question is whether anything in that set of options happens to be of physical significance. Given how often the answer is yes (e.g. the Chandrasekhar limit for $L=0$ white dwarfs is $N^2m_\text{Pl}^3$), it's a question worth answering with either "no" or "yes, with this specific definition" (whichever is factually true), not "multiple meanings exist, question invalid". $\endgroup$
    – J.G.
    Aug 4, 2021 at 6:40
  • $\begingroup$ @J.G. Point well taken. And, yes Planck values can be useful, a point to which I fully agree. Perhaps I need to be more specific and state that one must be careful not to assign any fundamental physical meaning to Planck values. $\endgroup$
    – JRL
    Aug 6, 2021 at 19:07
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    $\begingroup$ That's a better point, of which the OP is cognizant. $\endgroup$
    – J.G.
    Aug 6, 2021 at 19:14

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