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This question arose in a seminar today about the solar wind...

This is my vagueish understanding of the problem - please correct if you see errors!

The 'classical' picture of atmospheric electricity is that the Earth as a whole is neutral, but that thunderstorms maintain a voltage of around +300kV at the electrosphere with respect to the Earth's surface, with a current of around 1 kA slowly discharging around 500 kC of total charge separation. The solar wind is supposed to neutralise any net charge that might be there between the Earth as a whole and the solar wind.

However, positive and negative charges in the solar wind are differently trapped in the van Allen belts, from which they can then descend to the Earth's atmosphere, which implies that a net charge can be developed due to this differential leakage. This begs the question of whether there are any estimates of the total net charge. I've hunted in the literature but have found little useful material other than Dolezalek's 1988 paper: http://www.springerlink.com/content/u057683112l148x5/

Can anyone offer an explanation, or point me to some more relevant papers?

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2 Answers 2

However, positive and negative charges in the solar wind are differently trapped in the van Allen belts, from which they can then descend to the Earth's atmosphere, which implies that a net charge can be developed due to this differential leakage. This begs the question of whether there are any estimates of the total net charge. I've hunted in the literature but have found little useful material other than Dolezalek's 1988 paper: http://www.springerlink.com/content/u057683112l148x5/

Right I see an important misunderstanding here! Positive and negative charges from the solar wind are NOT differently trapped. When we talk about the van Allen belts we are talking about high energy particles (MeV); low energy particles are still there, though in the majority of the outer belt the density is very low. Its fair to say that the electrons and protons go through different acceleration processes (its also worth noting that there are two inner belts, a positive and a negative that overlap as well as sitting on the region of the magnetosphere called the plasmasphere - a cold dense plasma that extends to about 4 Earth radii)

Some of the same waves that scatter relativistic waves into the loss cone and cause them to precipitate (electromagnetic ion-cyclotron waves) also scatter lower energy protons into the atmosphere. Over very short time scales there will be a charge difference, but on average quasi-neutrality will be maintained.

When precipitation occurs you have to remember that you are dealing with the ionosphere which is itself a plasma and therefore quasi-neutral. Current systems do form due to the collisions of ions with the neutral atmosphere. In the D-region, where MeV electrons deposit when they precipitate (in fact they cause extra ionisation and extend the D-layer downwards) there is a whole mixed bag of high collisions, attachment processes and recombination processes involving all sorts of complex ion-chemistry, the less energetic protons will deposit at higher altitudes so there would be a small charge gains in different regions but due to high conductivity in the E layer (120 km) and possible current systems that would not be the case for long. *

Now the atmosphere stuff is not my bag but you might like to go and look at the work of Dr. Martin Fullerkrug of the university of Bath who does some very cool work on sprites and whatnot.

Hope that helps

*I am not saying that exactly equal numbers of protons and electrons precipitate at a given moment but looking at the system over even relatively small time-scales the net charge would be damn close to zero.

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Thanks! By 'trapped differently' I meant that protons are most dense at L~1.4, electrons at L~4, which is what I learnt from Richard Horne's slides. Since the acceleration processes are different, I still don't get why a priori you would get the same rate of precipitation. I see how initially the ionosphere would be uncharged, but surely with differential precipitation there could be accumulation of net charge. –  Hywel Jan 27 '11 at 15:52
    
I guess the point is that the ionosphere is in electrical 'contact' with the solar wind? i.e. net charge can be conducted away. Therefore no significant charge can be accumulated - the net charge is nearly zero. Obviously it would be hard to measure, but is there a sensible upper bound on the value? ~1 MC, ~1 GC etc? –  Hywel Jan 27 '11 at 15:54
    
Just to be clear, energetic (MeV) protons are most dense at L~1.4 and MeV electrons at L~4. There are lower energy counterparts (eV) there as well, hence quasi-neutrality. Richard did not really explain that but only because it wasn't massively relevant for the main part of his talk. Your second comment gets to the nub of the matter, though it does simplify things quite a bit :-) I do not know if anyone has tried to work it out it would take a lot of assumptions (knowing the approx rate of precipitation for both cold ions and MeV electrons (scattered by the same waves sometimes) for example. –  user1621 Jan 27 '11 at 16:08

My understanding for this is that electrons weigh a lot less than anything else so they have higher velocities (at the same temperature, because of thermodynamics). For matter to escape the earth requires a minimum velocity. So the result is that the earth loses more electrons than positive ions (or negative ions). The result is that the earth is positively charged.

On the other hand, in the earth's atmosphere, the transfer of charge is due to rain. Charged ions and electrons are attracted to raindrops for the same reason that charges are attracted to a conducting surface, that is, the image charges. But electrons are lighter than ions and so electrons get to the raindrops before ions and so raindrops are net negatively charged.

Since raindrops fall to earth, the earth ends up negatively charged while the atmosphere above it is positively charged. This is the source of the "sky voltage":
http://en.wikipedia.org/wiki/Sky_voltage

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