Is there any electric current in a continuous medium of moving charges?

Question

Consider a continuous charged fluid (unlike in real life where charge is carried in discrete chunks like electrons) in a bottle. Suppose that the fluid is stirred in a circular manner and then left to move on its own. Would there be an electric current given that for any given portion of fluid that moves over another part of the fluid moves to fill the gap and so there is no net change in charge at any given point in space?

I think this comes down to voltage and energy.

Voltage is a potential. A ratio of how much energy would be imparted to some charge placed in a field.

$$V = \frac{\mathrm{d} U}{\mathrm{d} q} = \frac{\mathrm{d} U}{\mathrm{d} t} / \frac{\mathrm{d} q}{\mathrm{d} t} = P / I$$

Current is dual to voltage.

$$I = \frac{\mathrm{d} q}{\mathrm{d} t} = \frac{\mathrm{d} U}{\mathrm{d} t} / \frac{\mathrm{d} U}{\mathrm{d} q} = P / V$$

Suppose one introduced an electron into the fluid explained above. Then there'd be a voltage because there'd be some amount of energy imparted to some amount of charge. Because there is a voltage there is some amount of current if there is any power loss. Because there is no loss of energy and no power there is no current. But this seems wrong to me.

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Since asking this question initially I have come up with two actual physical cases for this problem. They are both related to the systems where movement does not change the appearance of the system, so should not have any effect.

1. Electron has an intrinsic spin and hence magnetic field and current. Unless it is made of preons, how can it possess a rotation if no physical change happens that the rotation is causing?

2. How can a black hole rotate if the composition of a black hole is unobservable? The inside of a black hole might as well be a continuous muck with no internal structure to outside observers.

Answer 1

Current is the amount charge that goes through a particular cross section area per unit of time. It does not matter whether the charge density is uniform or not for this current to be non-zero; the only thing that prevails is whether or not the charges are flowing.

Observable features that a current is passing comprise for instance Joule heating of the wire containing the charges (even with a fluid model), Hall voltage when put in a magnetic field and generation of a magnetic field that can itself induce a current in a distant circuit.

Answer 2

Would there be an electric current given that for any given portion of fluid that moves over another part of the fluid moves to fill the gap left behind and so there is no net change in charge at any given point in space?

What you are describing is loosely related to a plasma approximation called magnetohydrodynamics or MHD.

Plasmas are highly conductive due to the high mobility of electrons. They exhibit collective behavior and are said to be quasi-neutral over distances larger than the Debye length. By quasi-neutral, I mean: $$ n_{e} = \sum_{s} \ Z_{s} \ n_{s} \tag{1} $$ where $n_{e}$ is the total electron number density, $n_{s}$ is the number density of ion species $s$, and $Z_{s}$ is the charge state of ion species $s$ (e.g., +1 for protons). I wrote a more detailed discussion at https://physics.stackexchange.com/a/253491/59023.

The total macroscopic current density in a plasma can be written as: $$ \mathbf{j} = \sum_{s} \ Z_{s} \ n_{s} \ e \ \mathbf{v}_{s} \tag{2} $$ where $s$ in this case includes electrons, $e$ is the elementary charge, and $\mathbf{v}_{s}$ is the bulk flow velocity of species $s$ (i.e., the first velocity moment).

Current is dual to voltage.

This is only true if the simplest approximation to Ohm's law holds true. The more general form, called generalized Ohm's law, is given by: $$ \partial_{t} \mathbf{j} = - \frac{ q_{e} }{ m_{e} } \nabla \cdot \mathbb{P}_{e} + \frac{ n_{e} \ q_{e}^{2} }{ m_{e} } \left( \mathbf{E} + \mathbf{v} \times \mathbf{B} \right) + \frac{ q_{e} }{ m_{e} } \left( \mathbf{j} \times \mathbf{B} \right) - \frac{ n_{e} \ q_{e}^{2} }{ m_{e} } \boldsymbol{\eta} \cdot \mathbf{j} \tag{3} $$ where $n_{e}$, $q_{e}$, and $m_{e}$ are the electron number density, charge, and mass, $\mathbb{P}_{e}$ is the electron pressure tensor, $\boldsymbol{\eta}$ is the resistivity tensor, $\mathbf{E}$ is the electric field, and $\mathbf{B}$ is the magnetic field.

In the absence of pressure gradients, magnetic fields, and time-varying currents, Equation 3 reduces to: $$ \mathbf{E} = \boldsymbol{\eta} \cdot \mathbf{j} \tag{4} $$ which is the more familiar form of Ohm's law. If one integrates over a line element, this would reduce to the more standard form of $V = I \ R$.

Because there is a voltage there is some amount of current if there is any power loss. Because there is no loss of energy and no power there is no current. But this seems wrong to me.

I think you are stumbling upon what is known as Poynting's theorem. I wrote a detailed discussion about this topic at https://physics.stackexchange.com/a/235549/59023.

Note that there can exist a current without any power loss (e.g., superconductors), so power loss is not required.

Unless an electron is made of preons how can it possess a rotation if no physical change happens that the rotation is causing?

I am not a quantum whisperer but I do recall enough to know that what we call particles are excitations of fields and that particle spin is not the same as a sphere spinning, even though this is often taught. So I will let the more quantum adept community members respond to this part of the question.

The first case is that of a single electron that has an intrinsic spin and hence magnetic field and current.

A single electron at rest has an electric field and a dipole moment due to its spin, yes. However, I do not think this is what you seem to be implying it means. Again, I will let the quantum whisperers elaborate on the physical meaning of the dipole moment but it does not derive from a current.

The second case is that of a black hole. How can a black hole rotate if the composition of a black hole is unobservable?

This partly results from an assumption that the conservation of angular momentum holds true. Meaning, the star from which the black hole formed was a sphere of spinning plasma before it collapsed. Things do not stop spinning/rotating because the star collapsed.

The inside of a black hole might as well be a continuous muck with no internal structure to outside observers.

This is dubious teritory and I do not see how it relates to your original question. In some ways, we need not know what goes on inside the event horizon to make predictions about what happens outside of it. Sure, I imagine it would help to know what really happens inside but we have a rather broad and successful field of general relativity without knowing exactly what happens inside black holes.

Answer 3

Let's first look at the system up close. I will show you why Ohm's law has nothing to do with the picture you describe, how it emerges and when is applicable.

Setup

For simplicity let's have a charged fluid flowing at some constant velocity $v$. Your question is formulated in terms of rotation, but we always can look close enough such that it is not different from straight flow. This does not change the conclusion.

Each particle has some change $q$ and mass $m$. Density of the particles $n$ is constant as well. The dynamical equation of the fluid is very simple, because there are no forces:

$$ m n \dot{\vec{v}} = 0 $$

During a period of time $\Delta t$, the flow moves charge on a distance $\Delta L = v \Delta t$. For concreteness let's take some cylinder of length $\Delta L$ and base area $S$. Then, during $\Delta t$ a total charge $\Delta Q = q n \Delta L S$ was moved. Basically this is a definition of a current:

$$ I = \frac{\Delta Q}{\Delta t} = q n v S $$

But nothing changed!

Yes, the system looks exactly the same in time and space and all exited charge has been replaced. Distribution of charges looks the same, so electric field of the fluid is as if it did not move at all. But current induces magnetic field:

$$ \vec{\nabla}\times \vec{B} = \mu_0 \vec{J} $$

where $\vec{J} = q n \vec{v}$ is the current density — the current per small area $S \to 0$.

This magnetic field is detectable and is an experimental proof of the current.

But Ohm's law?

Ohm's is formulated for conductors where the charges do not move by themselves, but are propelled by electric force $q \vec{E}$. The equation of motion changes correspondingly:

$$ m n \dot{\vec{v}} = n q \vec{E} $$

In these conditions, charges would accelerate endlessly which does not happen in a reality. This is due to the collisions between moving charges and other particles that counteract electric force (e.g., electrons collide with atoms in a conductor). We can model this in the following way:

$$ m n \dot{\vec{v}} = n q \vec{E} - \mu \vec{v} $$

where $\mu$ is a drag coefficient due to collisions.

As we want a steady flow, derivative should be $0$ and

$$\mu \vec{v} = n q \vec{E} $$

$$ \vec{E} = \frac{\mu}{n q} \vec{v} = \frac{\mu}{n^2 q^2} \vec{J} = \rho \vec{J} $$

This is indeed our variant of the Ohm's law that we can express in usual form by multiplying equation by some area $S$ and length $L$ of the conductor (i.e, the distance between points where we measure the voltage drop) and dropping the vector notation:

$$ E L S = V S = L \rho I $$

$$ V = \frac{L}{S} \rho I = I R $$

Conclusion

As you see, charged fluid by itself is very different from the current in the conductor and you cannot mix them. Ohm's law is not universal for moving charges, but we found it's analogue for the evenly moving charged fluid under external electric field.

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Bonus round

how can it possess a rotation if no physical change happens that the rotation is causing?

Electron's spin is different from the classical notion of magnetic moment. This is a quantum-mechanical quality of the particles and attempts to interpret electrons as rotating spheres with some charge density meet problems like superluminal surface speed. So you shouldn't understand is so literally.

On the other hand, we saw that even when movement does not cause physical change, there are observable effects like magnetic field.

The second case is that of a black hole. How can a black hole rotate if the composition of a black hole is unobservable? The inside of a black hole might as well be a continuous muck with no internal structure to outside observers.

Essentially same as the previous one, but much more complicated due to even poorer intuition when it comes to black holes. We do perceive black holes as a continuous muck without structure, but theoretical work and experiments have found an effect that distinguishes rotating black holes from stationary: the so-called "frame dragging".

Answer 4

Your intuition is not right that merely specifying the charge distribution as a function of time is enough to determine the current. Here's a thought experiment to show that: suppose you had an infinite line of perfectly uniform (let's say positive) charge. It would produce an electric field pointing away from the line and no magnetic field. Now imagine making a relativistic boost parallel to the line. Then the charge density of the line would increase due to Lorentz contraction, but it would remain completely uniform. So if the charge density were enough to completely determine the current, in the new frame the fields would look qualitatively similar: the electric field would become stronger, but it would still point away from the wire and there would still be no magnetic field. But as any textbook can show, this is not what happens - instead, the boost perpendicular to the electric field partially transforms it into a magnetic field. That's because in your new frame, there is electric current, even you can't tell just from the charge density.

As for your two specific cases, an electron is not actually a uniform sphere of charge physically rotating around one axis. It's actually much more complicated and nonintuitive then that. Again, you can consult any quantum mechanics textbook for more information. And even though we can't look inside a black hole and see whether its contents are rotating, the fact that it has nonzero angular momentum does have physical consequences outside the black hole. The simplest is called "frame dragging" - roughly, it's easier to travel around the black hole in the same direction as it's spinning than in the opposite direction. Perhaps saying that the black hole is "rotating" is a bit of a misnomer, since it's certainly nothing like a rotating rigid body in classical mechanics. It might be better to just say "a black hole with angular momentum," which it gets from the angular momentum of the things that fall in.