As I understand them, Cosmic rays mainly consist of high energy charged particles. I began to wonder if they would eventually net charge the Earth and then assumed that they must come in roughly equal amounts of charge. If they don't I suppose the charged Earth would deflect like charge particles and attract the opposite until it is neutral once more. Any thoughts on the matter would be interesting to hear.

# [Physics] Are Cosmic Rays net neutral in charge

chargecosmic-rays

#### Related Solutions

This is an area where there is a lot of confusion, and a lot of opinion cannot be supported by sound mathematical theory.

### Not-analytic continuation

The theory of black holes requires continuation beyond regions which are accessible. These continuations are often called "analytic continuations", but this is ridiculous, because the fields do not have to be analytic in order to do it. Hyperbolic equations don't have general analytic solutions, and generally covariant equations can't have solutions which are analytic in non-analytic coordinate systems, which are in no way bad for GR.

The term I would use is "Minkowski horizon equivalence principle continuation", because it is continuation past a horizon by assuming that the horizon looks like Minkowski space locally, that is, by assuming the equivalence principle includes horizons too. When you reach a horizon which is locally like a Rindler horizon, like special relativity in accelerating coordinates, so that the curvature does not blow up, then you can continue a little ways past the horizon by extending the solution to be locally flat. In this way, you can continue into the interior of a black hole knowing the outside solution.

A Rindler horizon is vertical sheet in time for the accelerated observer. There is another kind of horizon which is locally a horizontal sheet. The Rindler horizon is formed by the spatial light cone going to the right in a 2d space-time diagram where time is vertical. If you tip this cone by 90 degrees clockwise so that it goes into the past, this is called a Cauchy horizon. A Cauchy horizon is a limit to predictability, because it is where new light rays come from infinity. This Cauchy horizon in question is the limit of the future development of a past pointing half-hyperbola which is just the 90 degree rotated trajectory of an accelearated observer.

Cauchy horizons are also locally flat, so one may continue past them by assuming the equivalence principle, just as for Rindler horizons. The issue is that when one does this, there is always new information coming in from someplace that you haven't seen before, so it is difficult to predict what happens, at least in classical GR.

But the full continuation is not in principle any more difficult than the continuation past the event horizon. The equivalence principle should hold at all horizons. But the results are strange, and require interpretation

### Charged black hole solutions

I will restrict myself to 4 dimensions here, and to classical General Relativity. In string theory, there is a dilaton and the charges are d-form fields, so the "charged" black holes are extended objects in higher dimensions, sometimes with very different interior structure.

The charged black hole is nice because it is still spherically symmetric, and yet looks like the generic case of an arbitrary rotating black hole. The reason charge and rotation look alike is Kaluza Klein theory--- a charged black hole can be thought of as moving in a small circular fifth dimension. This is similar to rotation.

But the solution in GR is just as simple as the Schwartschild solution: for general mass M and charge Q:

$ ds^2 = - f(r)dt^2 + {dr^2\over f(r)} + r^2 dS_2^2$

Where

$f(r)= 1-{2m\over R} + {Q^2\over R^2}$

The important thing is that there are now two radii where f is zero, and these are the two horizons. The outer horizon is continuously deformable to the Schwartschild horizon, and is an Event horizon. The inner horizon is a Cauchy horizon. As M=Q, the black hole is extremal, because the two zeros of f(r) have the same value of r, so you get a double root. Although the two horizons are collided in the r coordinates, they are not at the same point! The radius r breaks down, because the distance between the horizons approaches a finite limit at extremality.

For $M<Q$ there is no solution to GR without a naked singularity. Such solutions are absurd, they cannot be reached by continuous deformation of the uncharged black hole, and are forbidden by the Censorship conjecture.

The black hole singularity is at r=0, past both horizons, and trajectories in the interior of the Cauchy horizons are repelled by it. So it is impossible for a massive object to hit the singularity. The Cauchy horizon itself is an outgrowth of the singularity, it is the first light rays from the singularitly

The structure: outer event horizon- inner Cauchy horizon - timelike singularity is generic. It is the situation for rotating/charged/rotating-charged black holes. But it is considered unstable. I will discuss why in the next section, and why I think these arguments are bogus.

### Bad arguments that the Cauchy Horizon is the singularity

There are, in the literature, arguments that the Cauchy horizon becomes the singularity in realistic solutions. These arguments are hand-wavy, and based on the intuition that everything should be mushed somewhere. In my opinion they are completely wrong.

The arguments are based on the following properties:

- Cauchy horizons are unstable: you can see that it is possible to deform the solution to get whatever you like there, because you can start emitting arbitrary gravitational waves from the singularity, and these will go out and influence up to the Cauchy horizon.
- The Cauchy horizon has a property that it can see the entire future development of the black hole exterior, in principle. When you approach a Cauchy horizon, you see the entire future development of your past, which includes the entire future development of the outer region.

The second point is very important, because it means that photons which come to you from outside are infinitely blue-shifted at the Cauchy horizon, leading people to expect a wall of hard-radiation to appear at this point under generic perturbations, which will make a singularity.

I don't buy these arguments, because the black hole is not eternal. It is not clear to me that a decaying black hole has the same Cauchy horizon structure as a regular black hole. It is also not clear to me why the singularity can't send out mollifying fields to make the Cauchy horizon *non-singular* rather than singular.

But most damningly, we have models of near extremal black holes in d-branes, slightly off from extremality, and these are shiny and thermodynamically nearly time-reversible. There are arguments by Gubser that d-branes will absorb other d-branes irreversibly by going off-diagonal in the non-commutative description, that is, by producing strings between the branes, but I don't buy these arguments at all.

I think that one should take the story in rotating/charged black holes at face value.

### Traversable Cauchy horizon implies you can leave a black hole

If you do take the story at face value, objects going into a black hole will come out pretty much unchanged, except for possibly a little singeing when traversing the hard-radiation at the Cauchy horizon. These objects go in, turn around past the Cauchy horizon, and go out.

There are many problems with this picture. The continuation of the solution for a particle which does a full circuit goes past an event horizon, as many cauchy horizons as it wants to (by choosing whether to leave going "up in t" or "down in t" (remember that t is a spatial coordinate inside the black hole) and then out the event horizon going the other way, the way only Hawking radiation can be said to come.

This picture is crazy, because the continuation goes on producing more sheets, just like analytic continuation (but it's not analytic continuation). Each traversing path puts you in another universe, with another direction of time. The result is total nonsense.

This is why people are adament that one has to throw away interior charged solutions. These would naively lead to information loss, other universes, all that nonsense. In my opinion, the correct solution is to do some cutting and pasting and identify the outgoing and incoming paths. I tried a little, but I never found the right pasting.

If you know the pasting, you can predict what will come out of a charged/rotating black hole, mostly, up to a little randomization which becomes more random as the black hole becomes neutral/non-rotating. This type of idea means that black holes are not black at all, not just slightly thermally non-black, but shiny-mirror non-black.

This idea has exactly one supporter, which is me. I give a reasonable probability to this possibility because the counterarguments about a singular Cauchy horizon are all weak. Perhaps I should take solace in the observation that that mysterious anonymous author of the Wikipedia article also harbors suspicions, but I somehow don't think his opinion is independent.

### Heuristic Justification for Pasting

The reason I think pasting is necessary is because of the following classical paradox, which I haven't found in the literature. If you throw a mass into a maximally extended charged black hole, and let it go past the horizons and come out, it comes out in another patch. But the black hole mass went up in your patch after absorption, and the mass of the black hole in the other patch went down after emission. But this means that they no longer fit together as continuations of each other.

So when the object comes out of the other sheet black hole, by consistency, it seems that something must come out of your sheet black hole. But the problem is that coming out can happen in either time direction, depending on how the infalling stuff steers itself, whether it decides to come out going towards positive or towards negative t, and so it means that objects can come out time-reflected, and the only way to make sense of that is to do a full identification with CPT gluing, so matter thrown in can come out as reflected motion-reversed antimatter. This is doubly crazy. So I am not sure if it is true, but I think it is. This might explain some astronomical puzzles of anomalous antimatter production because some astrophysical blackholes are measured to rotate at 99% of extremality.

The electric field of earth is really quite complicated. This paper is old, but presents a nice overview of the issues. That $500\,000\, \mathrm{C}$ number is related a charge *separation* between the surface and the bottom of the ionosphere, rather than the net charge on the earth. And in any case, it is only the result of a rough model whose underpinnings aren't entirely accurate. Your small sphere is embedded inside this massive system (the Earth's atmosphere). As @John Rennie points out in the comments, because the Earth is so large, you can ignore its curvature, and a better model for this system is a small sphere between the two plates of an enormous capacitor with a $\sim 300\,000\, \mathrm{V}$ potential difference.

So, as always, when dealing with potentials, you need to ask yourself: potential relative to what? Well, assuming that the ionosphere is actually at the same potential as a point infinitely far away, the natural interpretation of your question is to assume that your sphere is neutral infinitely far away, is insulated, transported to the surface of the earth, and then touched to the surface of the earth, and ask how much charge is transferred at this last step. My calculations suggest that a $1\,\mathrm{m}$ sphere has a capacitance of about $100\, \mathrm{pF}$. For a voltage difference of $\sim 300\,000\, \mathrm{V}$, that would be a few times $10^{-5}\,\mathrm{C}$.

Incidentally, as my answer on this question suggests, humans are modeled as capacitors with about the same capacitance, so you would get the same amount of charge.

Of course, these calculations are all for isolated objects, which makes them not perfectly valid, but should be reasonable approximations.

## Best Answer

Yes, roughly 90% of all cosmic rays (CRs) are protons, another 9% are alpha particles and the remaining 1% are electrons, positrons, and heavy nuclei. Most of these are in the 1-10 GeV range, dropping off at $N(E)\sim E^{-2}$ from thereon out (this Physics.SE post shows the common CR spectra).

If an over-abundance of positively charged CRs developed, they would start deflecting the same positively charged CRs & attracting the negatively charged CRs; similarly the other way around. Thus, our planet & atmosphere remains roughly quasi-neutral (though, as seen in this Physics.SE post, there is a 500 kC charge on the surface).

The charge however, is related more to the internal processes and the solar wind than galactic/extra-galactic CRs accumulation. For this aspect of earth's charge, see this Physics.SE post. At energies less than ~ 10 GeV, the solar wind will actually be able to deflect the CRs away from earth (hence the slight down-tick on the CR spectrum linked in the top paragraph). At higher energies, the solar wind cannot do this, but the particles come more infrequently (ranging between 1 particle per square-meter per second down near 100 GeV to 1 particle per square-kilometer per century around $10^{21}\,\rm eV$) so the fewer particles would not really affect the electric field.

Not at all. As I stated above, the vast majority are single protons and only a small fraction are electrons. As an aside, the electron-positron anisotropy is a somewhat hot topic in CR Astrophysics (cf. this ADS search of

`electron positron anisotropy`

).We believe that the bulk of galactic CRs are produced in galactic supernova remnants (see also this post of mine for CR protons producing neutrinos). Models produced by the leaders of the field (e.g., Ellison et al and Jones & Kang) show that the ratio $N_p/N_{e^-}\sim10^{-4}$ exist at the sources before diffusing out into the galaxy, matching what is found in the observatories (as it should, assuming a steady-state production of CRs).