## A Guest Paradox

*Here’s a short guest post by my old friend Anton. As usual, please feel free to discuss the paradox through the comments box!*

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I thought of a physics paradox the other day and Peter has kindly granted me a guest post here about it, as follows. Consider a homogeneous isotropic closed universe as described by general relativity. Let it contain a uniform density of a single species of electrically charged particle, so that this universe has a net charge. The charged particle density is sufficiently low, however, that the perturbation from the regular uncharged metric is negligible. Since this universe is homogeneous and isotropic the electric field in it is everywhere zero. BUT if I consider a conceptual 3-dimensional sphere, small enough for space-time curvature to be neglected, then it contains a finite amount of electric charge, and therefore by Gauss’ theorem a nonzero electric field points out of it at every point on its surface. This contradicts the zero-field conclusion based on the metric.

Here are three responses (one my own) and my further responses to these, in brackets:

- In a closed universe it is not clear what is the outside and what is the inside of the sphere, so Gauss’ law is not trustworthy (tell this to a local observer!);
- the electric field lines due to the charges inside this (or any) conceptual sphere wrap round the universe an infinite number of times (this doesn’t negate Gauss’ theorem!);
- the curved rest of the Universe actually adds a field that cancels out the field in your sphere (neither does this negate Gauss’ theorem!)

February 9, 2018 at 10:22 pm

There is no such thing! There is no solution to the Einstein+Maxwell equations that corresponds to a closed Universe with nonzero net charge. Making the charge density very small, so that it feels like you can treat it perturbatively, is a great way to make the paradox seem very sharp, but there aren’t solutions for any nonzero charge density, so you can’t think of it perturbatively.

By the way, you don’t need homogeneity for this statement to be true. There aren’t any closed solutions with nonzero net charge at all, even inhomogeneous ones. The proof is quite pretty. Draw any closed surface. It divides the space into two volumes V and W. You can think of V as the “inside” of the surface and use Gauss’s Law to determine the total charge in V. But you can also think of W as the inside. The two charges you get will be opposites of each other (because the flux out of V is the flux into W). So the total charge is zero.

Gauss’s Law is still true in curved spacetime, so this argument is completely rigorous.

February 9, 2018 at 10:35 pm

There is no solution to the Einstein+Maxwell equations that corresponds to a closed Universe with nonzero net charge.Are you sure? Add a single electron to a closed neutral universe and smear it out throughout that universe and the charge of just one excess electron in 10^80 will wreck the closedness property?

February 9, 2018 at 11:23 pm

Yes. Surprising but true. Gauss’s Law is an actual law, after all. There are no solutions that violate it.

February 10, 2018 at 7:07 am

I’d be impressed if you can prove that from GR applied to a charged universe.

February 10, 2018 at 8:12 am

If you add one atom to a flat matter-dominated universe it becomes a closed universe…

February 11, 2018 at 6:30 pm

Anton Garrett — I claim that what I wrote above is a proof!

Gauss’s Law, in integral form, is an actual theorem in GR+Maxwell. Apply it to a closed surface to get the charge inside. Apply it to the same surface, with inside and ouside reversed, to get charge outside. Add them up and get zero. Therefore, all compact solutions to Einstein+Maxwell have zero charge. QED.

February 11, 2018 at 6:32 pm

Phillip Helbig —

That certainly doesn’t follow from anything I said.

February 11, 2018 at 9:08 pm

As I said, prove it from GR.

February 12, 2018 at 2:24 am

Anton Garrett — What part of what I said fails to be a proof? I genuinely don’t know what more you want.

February 12, 2018 at 2:51 am

By the way, Googling the phrase “close universe must have zero charge” turns up an excerpt from the book Classical Mathematical Physics, by Walter Thirring (2003), which gives essentially the argument I gave. (Actually, the version it gives is the limit as one of the two volumes — and hence the surface over which the flux is taken — goes to zero. It comes to the same thing.)

I don’t know if the link will work, but here’s the URL that results from the Google search.

https://books.google.ca/books?id=NFdStDHPOfcC&pg=PA319&lpg=PA319&dq=closed+universe+must+have+zero+charge&source=bl&ots=fh_1OnX0C7&sig=WQLiS2ST68hh7NP_dy597aPECNs&hl=en&sa=X&ved=0ahUKEwj6m_z6rZ_ZAhUCyWMKHZMqBtQQ6AEIJzAA#v=onepage&q=closed%20universe%20must%20have%20zero%20charge&f=false

February 12, 2018 at 10:08 am

You wrote:

Making the charge density very small, so that it feels like you can treat it perturbatively, is a great way to make the paradox seem very sharp, but there aren’t solutions for any nonzero charge density, so you can’t think of it perturbatively.This is an assertion that calls for a direct proof, so please apply the field equations of GR to a charged universe and show that no solution is closed.

February 12, 2018 at 10:19 am

By the way, Ted, mass is itself a charge in the general sense of the latter word…

February 12, 2018 at 5:25 pm

Yes, and the Visa corporation uses the word to refer to a financial transaction, and the army uses it to refer to a military action. Words have different meanings in different contexts. The context here is quite clearly electric charge.

February 12, 2018 at 5:48 pm

Please don’t play coy. Right or wrong, I’m making a serious point.

February 10, 2018 at 12:28 pm

Isn’t there also an electric field pointing in to it (from a ‘mirror’ sphere on the other side of the tangent plane to that sphere at that point)?

February 13, 2018 at 10:45 am

Oh well… I was going to say that then you should blow up those spheres until they coincide at the boundaries of a pair of 3-balls. But if you can’t have a non-zero charge density in the first place I guess it doesn’t matter.

February 10, 2018 at 8:13 pm

Paradox by definition means that two contradicted statements are CORRECT in terms of all KNOWN knowledge.

If one (or both) of the two statements is not exactly correct (in your case), there is at best a pseudo-paradox; the problem is to remove the ERROR.

For any genuine paradox (such as Kurt Grelling paradox, Russell paradox), it can always be removed in two ways.

One, downward symmetry breaking: removing the paradox by an additional constraining definition.

Two, upward finding a higher symmetry: unification.

The above (with many examples, especially in physics) was discussed in detail in Chapter 3 of {The Divine Constitution, see https://books.google.com/books?id=8MMzPwAACAAJ&dq=inauthor:%22Gong+Jeh-Tween%22&hl=en&sa=X&ei=9oDyT9z8E-PO2wWznf2fAg }.

February 11, 2018 at 5:49 pm

For genuine paradox, it always points to a higher symmetry.

For pseudo-paradox, one of its two parts must be wrong.

In physics, there is one and only one genuine paradox, the {creation (first principle) paradox; C-paradox}.

If X is creation, then Pre-X creates X; the C-paradox.

C-paradox demands a higher symmetry {Pre-X, X} symmetry, the C-symmetry.

Let, Pre-X = nothing, then X must also be ‘nothing’ in ESSENCE.

Is C-symmetry physics?

Can C-symmetry be written in physics equation?

The answers for both are Yes, but I will not discuss them here.

On the contrary, there are many pseudo-paradoxes in physics.

The incomparable-ness between quantum-ness and GR (General Relativity) is a pseudo-paradox. That is, one of them, at least, is wrong or incomplete.

But, which one?

This can be easily found out by finding out which one is the paradox-trouble maker.

Quantum-ness is very much non-intuitive, such as superposition/measuring problem. But, they are issues of interpretation, not paradox.

GR is obviously wrong in calculating the gravity-transmitting between Sun-Earth, and it forms a pseudo-paradox.

Thus, GR is a paradox-trouble maker, must be wrong or incomplete.

Therefore, any attempt of quantizing GR to remove the incomparable-ness is absolutely useless as GR is simply the ERROR-ONE in this pseudo-paradox.

For Newtonian gravity, it is about the attractive force between two masses.

For GR, it is about the LOCAL space-time sheet distortion by a (one) mass.

In this GR/Quantum-ness pseudo-paradox, GR (very good effective theory) is the one which must be ABANDONED.

The true gravity IS that the entire (This) universe is moved from NOW to NEXT with a quantum action (Planck constant, ℏ), and this DERIVES the quantum-ness, see http://prebabel.blogspot.com/2013/11/why-does-dark-energy-make-universe.html .

February 11, 2018 at 8:12 pm

I hope you are not in hospital for anything serious, and that you will be well soon.

February 11, 2018 at 9:14 pm

To add to the fun, the second of the responses I mentioned was: “the electric field lines due to the charges inside this (or any) conceptual sphere wrap round the universe an infinite number of times”. The implied suggestion is that the resultant electric field inside the sphere, when you sum the infinite series arising from each time an electric field line goes round the universe and passes through the sphere again, is zero. But consider what happens when you move a single electron that is inside the conceptual sphere. The change in the electric field lines propagates along them at the speed of light, so there is a change to the value of the resultant field every billion years or so as the change reaches the field line at its Nth passage through the sphere. In the meanwhile Gauss’ theorem still holds…

February 12, 2018 at 9:53 am

beautiful. get well soon.

February 12, 2018 at 10:17 am

Yes, get well soon – and do say, at your leisure, how you consider that application of Harrison’s work resolves this paradox.

February 16, 2018 at 12:48 pm

I don’t think that this has been resolved here as of the time of this present comment!