What is the effect of temperature on electrostatic-gravitational balance?

We have two identical massive metal spheres at the same temperature at rest in free space. Both have an identical charge and the Coulomb force [plus the black-body radiation pressure if the temperature is non-zero] exactly counteracts the gravitational force between them, resulting in no net forces on either object. They are electrostatically levitating at rest in space. It is my understanding that the charge distribution in each sphere exists only at the surface, and should be concentrated on the side facing away from the other sphere.

Now heat one of the spheres uniformly with an external energy source. What happens in the instant following?

Could it be: The increased mass-energy of the hot sphere increases the gravitational force, and the cold sphere starts to fall inward.

Or else: The increased temperature modifies the charge distribution on the hot sphere by giving it more variance and bringing it closer on average to the cold sphere, increasing the Coulomb force. The cold sphere starts to fall away. [Carl's answer says there is no effect on the charge distribution like I describe, but DarrenW points out that there will be an increased black-body radiation pressure between the spheres, similarly causing an outward force.]

Which direction is correct, and what is the best explanation? Does it depend on the amount of temperature change or the initial conditions?

-
By including a realistic charge distribution on the spheres You open a pandora box of difficulties (see DarenW's answer) To get an answer of the core question omit that! Either You assume point masses/charge or assume a distance to diameter ratio which makes the charge density homogenous on the spheres. –  Georg Mar 13 '11 at 5:41
Georg, I am not sure what would be the meaning of temperature if we assume point mass-charges. Does the weaker assumption d>>r imply an attractive force or is does it still depend on the temperature increase? –  Dan Brumleve Mar 13 '11 at 21:54
Dan, then use the alternative I named: distance/radius big enough to assume homogenous charge on the speres. –  Georg Mar 14 '11 at 9:40

First thing to say is that this system is unstable. Though the simple gravitational and electrostatic forces may be balanced, there is an additional effect for the electrical interaction that won't exist for gravity: the charge of each sphere polarizes the other. The charge distributions on each sphere adjust themselves for minimal energy, and that leads to a feeble but non 1/r^2 attractive force. If anything at all brings the spheres closer together even the tiniest bit, with the plain vanilla Coulomb repulsive force effectively nulled by the attractive gravity (at any distance as both are 1/r^2), this polarizing effect will lead to attraction. We could try to avoid that by tweaking the charges up for a wee little more repulsion, just enough for balance. But the system is unstable, like a pencil balanced on its point, since we're balancing forces that follow different power-of-r laws.

Having a nonzero temperature implies random fluctuations. Even at zero temperature, QM says there will be fluctuations. The net effect of fluctuations over all location on one sphere will lead to random fluctuations in the electric field sensed by the other sphere, and not long you will have to wait for a random extra bit of attraction to start pulling the spheres together. Will random attractions be cancelled by an equal number of random repulsions? Short answer: no. This is like the Van der Waals force between molecules.

Making one sphere hotter will add to its time component of energy-momentum, and increase the size and vigor of those random fluctuations. That would increase the rate at which the random fluctuations and induced polarization do their work. That, and increased gravity, both increase attraction.

OTOH, a hotter sphere would emit more thermal radiation - which would push the other sphere away. How much hotter? The Stefan-Boltzman law says we can get a lot of bang for the buck, so to speak. Without a particular physical system to discuss, I imagine that a big increase in temperature would lead to overall repulsion, the thermal radiation winning out over the other effects. But a small temperature increase? That might take calculation with particular numbers.

The question asks about metal spheres. So What if the spheres were made of an insulator like ceramic or rubber?

(I probably stupidly left off some additional physical effects. Have at it, commenters!)

-
The initial system is stable by hypothesis: charge is increased on each sphere until it exactly counteracts gravity, then we begin heating one of them. The question is will they start pulling or pushing? I am not sure what happens with a dielectric object but I will think about that too about maybe that will help. –  Dan Brumleve Mar 13 '11 at 3:39
Edited question to include force of radiation pressure. –  Dan Brumleve Mar 13 '11 at 20:37
Of course, the system cannot be completely stable because of black-body radiation lost to the vacuum. But it seems like if the time interval is short enough then this can be neglected. –  Dan Brumleve Mar 13 '11 at 20:49
@Dan: you can't make this system "stable by hypothesis." Being stable doesn't just mean that the forces are balanced; it means that if you "bump" (perturb) part of the system, the force will change in such a way as to push that part of the system back into place. In this case, Daren is saying that a small perturbation will change the forces in such a way as to push the system further away from equilibrium; that's an instability. (I haven't done the math myself to check whether this system would be stable or not) –  David Z Mar 14 '11 at 1:14

Heat has no effect on E&M until you get hot enough that the spheres are modified in shape, or hot enough that the charges are able to leave the sphere. So before these sorts of things happen, yes, the increased gravitational attraction will increase the attraction.

In any conceivable practical implementation, these two effects will occur long, long before the gravitational force is changed sufficient to take them significantly out of balance.

Practical example: 1000kg spheres, 0.5m in radius, nearly touching. The gravitational attraction is $G(1000)^2/1m^2 = 6.67428 \times 10^{-5}$ Newtons, say at 300K.

To compute the energy change of 1 degree Kelvin, we need to know the substance. Let's use iron so that an increase of 1 degree Kelvin (at 25 °C) gives $25.10$ Joules per mole. At 55.845 grams per mole, 1000kg has 17900 moles so each degree Kelvin increases the energy by about 45,000 Joules. Using $E=mc^2$ so $E/c^2=m$ this gives a mass increase of $5.0\times 10^{-12}$ kg, or about $5\times 10^{-15}$ change in mass. Assuming that both masses are heated up, the change in force is twice this, or $1.0\times 10^{-14}$ per degree Kelvin.

Current measurements of gravity using spheres gives the gravitational constant to an accuracy of 4 or 5 digits, for example, $6.693(27)(21)\times 10^{-11}$ in a recent measurement. In order to get an error in the smallest (6th) digit of 6.67428 you need to have a temperature change of about $10^8$ degrees Kelvin. At that temperature, you will have to include fusion (and plasma) effects.

-
en.wikipedia.org/wiki/Capacitor#Capacitance_instability says that capacitance depends on temperature. Wouldn't a change in capacitance cause a re-arrangement of the charges? Would that change the net Coulomb force? –  Dan Brumleve Mar 13 '11 at 23:08
There is no change in capacitance unless (1) the dielectric changes, or (2) the geometry changes. In capacitors both these can happen. The geometry changes when things expand (typically) due to temperature increase. To see dielectric as a function of temperature see: doitpoms.ac.uk/tlplib/dielectrics/temperature.php for example. None of this applies to the given problem. Uh, the iron balls might get larger with heat but the gravitational attraction is to their centers. –  Carl Brannen Mar 14 '11 at 1:02