# Why does the universe have no net charge?

1. Nov 18, 2015

### Buzz Bloom

Questions re: Matter-Antimatter Annihilation​
prompts this question since the discussion there raised the point that a zero net charge seems very unlikely. This new thread's title question might be rephrased as follows:
If the universe did have a non-zero charge, how large would the average charge density have to be before it would be noticed by astronomical and/or physical instruments?​
If the answer to this question is that a very tiny charge density would be detectable, then the mystery of how the universe came to have a (almost?) zero net charge remains.

2. Nov 18, 2015

### Chalnoth

Because charge conservation is an exact conservation law.

Conservation laws stem from symmetries. Charge conservation stems from phase angle symmetry: changing the phase angles of all particles by the same amount has zero impact on the physics. So this says that phase angle symmetry is a fundamental, exact symmetry.

3. Nov 19, 2015

### Buzz Bloom

Hi @Chalnoth:

Thanks for you post.

It took me quite a while to digest this, but i think I understand what you mean. It will take me a bit longer to figure out how to express the idea and my understanding of the consequences clearly, so I will make a followup post as soon as I can.

Regards,
Buzz

4. Nov 20, 2015

### Buzz Bloom

Hi @Chalnoth:

The following is a description of what I understand from your post #2. Please let me know if I have still have something wrong.

Let t1 be the time that proton anti-proton annihilation starts.
Let t2 be the time that proton anti-proton annihilation ends.
Let t3 be the time that electron positron annihilation starts.
Let t4 be the time that electron positron annihilation ends.

During this entire process, the net charge of the universe remained zero.

Before t1:
A very frequent interaction, say an N interaction, with inputs having a net zero charge, created a proton and anti-proton pair. A short time later each such pair was destroyed by mutual annihilation, but at any given time some of these pairs of particles were not yet destroyed. There were also some other (strong force?) interactions (say P interactions) that created a proton without a corresponding anti-proton, and these interactions also created one or more other particles, mesons, with a smaller mass than a proton's, and the net charge of these mesons created by a P interaction was -1. After a while the charged mesons decayed to uncharged mesons and their charges were then carried by electrons and positrons, with one net extra electron. Thus the net effect of the creation of a proton (without an anti-proton) together with mesons, at least one of which having a -1 charge, was a proton in the universe without a corresponding anti-proton, and also an corresponding electron in the universe without a corresponding positron. The P interactions were much less frequent than the N interactions.

For almost every P interaction, an A interaction, the opposite of a P interaction, also occurred. The A interaction created one anti-proton and mesons which later decayed, yielding one extra positron. Due to an asymmetry between the A and P interactions there were slightly more occurrences of P interactions than A interactions.

Other interactions, say EP interactions, created a much large number density of electron positron pairs, most of which later annihilated each other, but at any given time, some pairs had not yet been destroyed.
During this period, there were also some other, relatively rare interactions, say E and E', that created electrons without a corresponding positron, and vice versa. These interactions will be described later.

The net result of the N, P, A, and EP interactions was a large number density of protons, anti-protons, and a much large number density of electrons and positrons, with a relatively tiny number of extra protons and electrons unmatched by a corresponding anti-protons and positrons. Just before t2, the ratio of the number of protons that had a corresponding anti-proton to the number of protons that didn't is estimated to be about 10,000,000,000. (From http://scienceline.ucsb.edu/getkey.php?key=114 .)

There were also meson particles and antiparticles of less mass than protons which were more numerous because of their lesser mass. The mesons were less numerous than electrons because they has less mass.

Between t1 and t2:

All the matched protons and anti-protons annihilated each other leaving just a much smaller number density of surviving protons. Also surviving were about 2000 times as many electron and positron pairs, and in addition a very small number of extra unpaired electrons, one for each surviving proton. For each electron not matched by a proton, there were about 20,000,000,000,000 electrons that were.

As the number of protons and anti-protons grew smaller, the relative number of mesons grew larger.

Between t2 and t3:

There were no more N, P, and A interactions, but EP, E and E' interactions continued. Some of the E and E' interactions had asymmetries similar to the P and A asymmetries. For example: (also from http://scienceline.ucsb.edu/getkey.php?key=114 ):
Experiments, for instance, show that a certain type of decay of long-lived kaons produce 301 positrons for every 299 electrons.​
(A kaon is a kind of meson.) However the E and E' interactions also maintained charge conservation, so for each extra positron/electron created there was a negatively charged output particle, which later decayed to produce a balancing electron/positron.

By t3 there were a relatively small number of protons, and many more electrons and positrons in almost equal numbers, with one extra electron for each proton. There were also some short lived charged mesons with net charges that balanced any other unmatched electrons or positrons. However, as some time between t2 and t3, the temperature was too small to create new mesons, so by t3, there were no more mesons.

Between t3 and t4:

The temperature was now too small to maintain any equilibrium between the creation and destruction of electrons and positrons. Gradually the pairs of electrons and positrons were destroyed leaving only one electron for each proton.

Regards,
Buzz

Last edited: Nov 20, 2015
5. Nov 20, 2015

### Chalnoth

A much simpler way to look at it is that each individual, microscopic interaction always conserved electric charge.

To see how this works, consider the following microscopic interactions:
1. electron (-1) + positron (+1) -> two photons (0).
2. neutron (0) -> proton (+1) + electron (-1) + electron anti-neutrino (0).
3. proton (+1) + proton (+1) -> proton (+1) + proton (+1) + proton (+1) + anti-proton (-1).

There are many, many more possible interactions. But every single one, due to this fundamental symmetry, will have the exact same amount of charge on the left and right side of the interaction. There's no way to add up any number of perfectly-balanced interactions and get an unbalanced end result.

6. Nov 20, 2015

### Buzz Bloom

Hi @Chalnoth:

I appologize for not being clearer about the intent of my post. I was not only trying to explain what I came to undestand from your post, but also the implications from that central concept to what related physics was going on during those perods of time.

Regards,
Buzz

7. Nov 22, 2015

### snorkack

Which means that if Universe has a net charge, there is exactly no way to get rid of it.

Still no reason for the Universe to not have had a nonzero initial charge.

8. Nov 22, 2015

### Chalnoth

Correct, but the initial charge, if any, would have been on the order of a single electron charge due to inflation. There would be no possible way of detecting a charge imbalance that small.

9. Nov 23, 2015

### snorkack

How is inflation supposed to get rid of initial charge density?

10. Nov 23, 2015

### Chalnoth

Inflation dilutes the universe, so much so that there's not expected to be more than one particle within the observable universe by the time inflation ends.

11. Nov 24, 2015

### marcus

12. Nov 24, 2015

### bcrowell

Staff Emeritus
There is actually no way to define the net charge of the universe. If the universe is spatially infinite, then we would expect integrals over all of space to diverge. If the universe is spatially closed, then Gauss's law doesn't allow us to define the total charge; Misner, Thorne, and Wheeler have a nice discussion of this on p. 457.

13. Nov 25, 2015

### nikkkom

It would not be a mystery in some theories of origin of the Universe. For example, if Universe started as a false vacuum state devoid of any particles (and if this vacuum state was unbroken versus U(1)em), then all subsequent evolution has to preserve zero net charge.

14. Nov 25, 2015

### Buzz Bloom

Hi bcrowell:

Are you saying that in an infinite universe the concept of a net charge is meaningless? How about a net charge density, that is, a net charge per unit volume. Would that be meaningful?

Regards,
Buzz

Last edited: Nov 25, 2015
15. Nov 25, 2015

### bcrowell

Staff Emeritus
Yes.

Not for the closed universe, for the reasons described in the MTW reference.

16. Nov 25, 2015

### Buzz Bloom

Hi Ben:

I can't find the MTW reference. Would you please repost it?

Regards,
Buzz

17. Nov 25, 2015

### bcrowell

Staff Emeritus
#12

18. Nov 25, 2015

### Buzz Bloom

Hi Ben:

I have just reserved a library copy of MTW, but I expect it will be weeks before I get to see it. In the mean time, can you post a brief summary of the reason why the concept of net charge density is meaningless in a closed universe? I am assuming that you intend for "closed" to also imply finite, rather than referring to an spacially infinite torus shaped universe.

Regards,
Buzz

19. Dec 23, 2015

### Buzz Bloom

Hi @bcrowell:

I have the MTW book, and I have read pp 457-459. I think I understand what the text says:
"For a closed universe the total mass-energy M and the total angular momentum S are undefined and undefinable."​
The text also says the same thing regarding charge. I believe I also understand the basis of the argument the text gives for this conclusion:
"To weigh something one needs a platform on which to stand to do the weighing."​
What I fail to understand is the logic of this argument. The only thing that makes sense for me is that MTW is using a different definition of "measure" than that which is commonly used for cosmology, as I understand it.

There is no mention of measuring the volume of a finite universe. Do you think this is also "undefined and undefinable"? From
I understand that the current best value for Ωk is 0.0008 +0040 -0.0039 (95% limits). Is it possible that sometime on the future the value for Ωk might be then measured to have a positive value of say +0.001 or +0.002 with an smaller error range? From such a value the volume V of the universe (for a = 1) can be calculated.

The current best value for H0 is 67.74 +/-0.46 (68% limits). From this and the value for G, ρc can be calculated.
The current best value for Ωm is 0.3089 +/-0.0062 (68% limits). From this and ρc, ρmcan be calculated.
From this, M = V × ρm can be calculated.

Other than that this method of "measuring" M does not involve a "platform", do you have an alternative explanation for why MTW would conclude that M calculated this way is not a "measurement"?

Regards,
Buzz

20. Dec 23, 2015

### bcrowell

Staff Emeritus
@Buzz Bloom - This thread is labeled as being at the "I" level, which is defined as an undergraduate level. I'm not sure we're going to succeed in resolving your questions with answers at that level. It sounds like you're deeply interested in this question, and if so, your best course might be to start learning special and general relativity at a more technical level. Re the passage from MTW, you've cherry-picked some of the least mathematical statements from it, but it does go into quite a bit more mathematical detail. Those nonmathematical statements are not intended to stand alone or to be decisive by themselves. They are intended to deobfuscate the more technical material and provide a general philosophical framework for it.

There is no problem, in principle, with measuring the spatial volume of a closed universe.

The total charge of a closed universe is trivially zero, for the reasons given by MTW at the top of p. 458.

To address your question about mass, I'm going to resort to a more technical presentation. Maybe someone more talented than I am at exposition can figure out how to do a careful presentation of these ideas without so much math.

In special relativity (flat spacetime), the definition of mass is $m^2=p^\mu p_\mu$, where $p$ is the energy-momentum vector, there is an implied sum over the index $\mu$ (Einstein summation convention), and the units are such that $c=1$. The first thing to note about this definition is that it's clearly nonlinear in character, and therefore mass isn't additive. You can't get the total mass of the system by adding up the masses of all its parts. Furthermore, it doesn't make sense, in curved spacetime, to talk about the total mass contained in a large region.The trouble here is that there is no coordinate system that allows you to express the vector $p$ as a set of components, over such a large region. Another way of putting this is that if you wanted to add up the momentum of the whole region, you would not be able to do so in any unambiguous way, because vectors that exist at different points in a curved space can only be compared through parallel transport, and parallel transport is path-dependent. So we have two independent reasons why we can't define the total mass

Since mass won't work, what might make more sense to attempt would be to determine the total mass-energy of a large region R. In special relativity, we could do this with the timelike component of the following integral:

$p_\text{total}^\mu = \int_R T^\mu_\nu v^\nu dV$

Here $T$ is the stress-energy tensor, $v$ is the velocity vector of some observer, and $dV$ is the element of spatial volume. In general relativity, we have a problem because the observer's velocity vector isn't comparable at different points in space (except by parallel transport, which is path-dependent). In cosmology, however, there is an obvious thing to do, which is to define the whole thing in terms of the Hubble flow. That is, we could let $v$ vary over the region of integration, with $v$ at each point being the velocity of an observer moving with the Hubble flow. The problem is that the thing inside the integral sign is still a vector, and we can't unambiguously define the integral of a vector over a large region of spacetime. This is, in more explicit mathematical terms, what MTW expresses on p. 458, in the paragraphs beginning with "Not having..." and "Imagine a..."

Your suggestion of measuring the mass density and multiplying by the volume will give a number for a closed universe. The problem is that, for the reasons described above, there is no reason to believe that it's a meaningful number.

21. Dec 23, 2015

### Buzz Bloom

Hi Ben:

I much appreciate your informative response to my questions, and your patience.

Regarding the level "I". I apologize for not using "A", but I did not realize that the question involved graduate level physics.

I just read again the 3 pages in MTW, and I failed to find anything resembling the lucid answer for mass which you gave.
If two simple sentences like the above had been in the MTW text, I would have gotten the message at once.

Regards,
Buzz