The Equilibrium Point of Shrinking Charged Shell: A Classical Analysis

In summary, the spherical shell gains mass by taking in energy from the electrostatic field and gains a negative gravitational mass from the gravitational energy. The mass is located outside of the shell, but if the shell were to shrink to the equilibrium point its gravitational and electrostatic forces would cancel out and the shell would be stable.
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I've been looking into this relatively simple problem out of curiosity and I have to say I've thoroughly confused myself, hopefully someone can provide some insight:

Imagine you have a massless (for simplicity) spherical shell with some overall charge on it that starts at a radius of "infinity". What happens to the intertial and gravitational masses as you slowly shrink the shell?

I eventually want to be able to answer this within the general relativistic framework, but to start I'm just focusing on the simpler classical aspects (i.e. Minkowski space-time, Newtonian gravity, spherical instead of cylindrical symmetry) to get a feel for it. The way I see it:

1) At first it has 0 mass (all kinds) and some constant charge Q.
2) Assume it is contracted at a constant velocity to avoid radiation
3) As it shrinks, it builds mass from the electrostatic energy being put into it
4) As the mass grows, a negative contribution to the mass from the gravitational energy is also put into it
5) At some point (either a finite radius or at 0 radius) you should reach an equilibrium where the gravitational and electrostatic forces cancel out and you're left with a stable spherical shell of charge Q and some mass M.

Now my confusion is about this equilibrium point, and whether or not its stable. Clearly it is unstable against perturbations in the +r direction. The -r perturbations are far less trivial though, and it seems to me that either:
a) It is a saddle point equilibrium and the shell once again begins to gain mass and collapse to a point
b) The mass starts to decrease and the shell is pushed back to equilibrium by the electrostatic pressure
c) The most sensical to me: The equilibrium point is at r=0, M is some function of Q, and there don't exist any perturbations in the -r direction

The second point of confusion is on the negative energy contribution from gravitational effects. If you were to set Q=0, and start with some non-zero mass it seems that if you compressed the shell enough it would end up with a negative mass? Clearly I'm doing something wrong, but I don't see what..

The third thing I'm confused about is where the mass is located in an originally massless spherical shell. Clearly the energy density is entirely outside of the shell, so it would seem to me that what I'm viewing as a shell with electromagnetic contributions to its own inertial mass is actually more like a massless ball with some mass density surrounding it (from the electrostatic field). If this were the case it would seem to me that an observer on the surface of the shell would feel no gravitational attraction towards it, as all of the mass is outside of the shell and spherically symmetric. So while it makes sense to say that the inertial mass of the ball increases as you compress it, is it incorrect to say the same about it's gravitational mass? Because if that's the case it is clearly in conflict with the equivalence principle (granted, I know this is a classical discussion atm but I would expect the equivalence principle to hold up at least approximately!)
 
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  • #2
2) Assume it is contracted at a constant velocity to avoid radiation
With a perfect spherical symmetry, it will never radiate.
3) As it shrinks, it builds mass from the electrostatic energy being put into it
As seen from far away, yes.
4) As the mass grows, a negative contribution to the mass from the gravitational energy is also put into it
What does that mean?
5) At some point (either a finite radius or at 0 radius) you should reach an equilibrium where the gravitational and electrostatic forces cancel out and you're left with a stable spherical shell of charge Q and some mass M.
I don't see anything which would cancel.

If this were the case it would seem to me that an observer on the surface of the shell would feel no gravitational attraction towards it, as all of the mass is outside of the shell and spherically symmetric. So while it makes sense to say that the inertial mass of the ball increases as you compress it, is it incorrect to say the same about it's gravitational mass?
That depends on the way you define "the object" and correspondingly "the mass of the object". On the surface, you would not feel a gravitational attraction.
 
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Hmm ok thanks for clarifying a lot of my mistakes. But now I just want to focus on the gravitational effects from a finite sized charged shell. Clearly from far away it has gravitational mass equal to its inertial mass (which is from the electrostatic energy). However why would there be no gravitation on the surface?

I can understand that the electric field is "attached" to the shell, so that the shell picks up that inertial mass. And since the electric field is an extended structure at the surface it would not cause any gravitation. HOWEVER pressure generates gravity too and there is definitely electrostatic pressure ON the shell!
 
  • #4
However why would there be no gravitation on the surface?
For the same reason a hollow shell with finite mass has no (net) gravitational forces inside. Attraction towards all sides cancels.
Pressure is an energy density, you need this over some volume to get a non-zero energy.
 
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michael879 said:
1) At first it has 0 mass (all kinds) and some constant charge Q.

I'm not sure this is possible. If there is charge present, there is electrostatic energy associated with the charge, which means nonzero mass should be present.
 
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In the limit of an infinite diameter, the total energy of the system vanishes. Field strength directly at the surface drops with 1/r^2, so energy density drops with 1/r^4, while the volume scale just increases with r^3.
 
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PeterDonis said:
I'm not sure this is possible. If there is charge present, there is electrostatic energy associated with the charge, which means nonzero mass should be present.

Yep. If you calculate the Komar mass for the isolated static charged sphere you will find that the regular mass parameter and the charge show up independently in a linear combination so even if the regular mass parameter is vanishingly small, the charge contribution to the Komar mass will be there.
 
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WannabeNewton said:
Yep. If you calculate the Komar mass for the isolated static charged sphere you will find that the regular mass parameter and the charge show up independently in a linear combination so even if the regular mass parameter is vanishingly small, the charge contribution to the Komar mass will be there.
We all agree on that, (1) in the first post just means the electromagnetic contribution goes to zero if the radius goes to infinity.
 
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mfb said:
Field strength directly at the surface drops with 1/r^2, so energy density drops with 1/r^4, while the volume scale just increases with r^3.

Ah, got it.
 
  • #10
WannabeNewton said:
Yep. If you calculate the Komar mass for the isolated static charged sphere you will find that the regular mass parameter and the charge show up independently in a linear combination so even if the regular mass parameter is vanishingly small, the charge contribution to the Komar mass will be there.

Yes, but only if the radius of the shell is finite
 
  • #11
mfb said:
For the same reason a hollow shell with finite mass has no (net) gravitational forces inside. Attraction towards all sides cancels.
Pressure is an energy density, you need this over some volume to get a non-zero energy.

Yes INSIDE there are no gravitational or electrostatic forces. However, ON the surface there is an outward pressure that increases as the radius decreases. Since pressure contributes to gravitational force I would assume that at the surface of the sphere there would be SOME gravitational force (even if its not the one you would expect for a sphere of mass M where M is the total energy in the electrostatic field)
 
  • #12
The volume of the sphere is zero, and pressure is finite. A finite energy density and zero volume lead to zero energy in that volume.
 

1. What is the equilibrium point of a shrinking charged shell?

The equilibrium point of a shrinking charged shell is the point at which the electrostatic force, due to the charge on the shell, is equal and opposite to the surface tension force, due to the shrinking of the shell. At this point, the shell will stop shrinking and remain at a constant size.

2. How is the equilibrium point of a shrinking charged shell calculated?

The equilibrium point of a shrinking charged shell is calculated using classical analysis, which involves applying the principles of electrostatics and surface tension to the system. The equations used to calculate the equilibrium point take into account the charge of the shell, the radius of the shell, and the surface tension of the material.

3. What factors affect the equilibrium point of a shrinking charged shell?

There are several factors that can affect the equilibrium point of a shrinking charged shell. These include the charge on the shell, the radius of the shell, the surface tension of the material, and the dielectric constant of the surrounding medium. Additionally, any external forces, such as gravity or an applied electric field, can also impact the equilibrium point.

4. Can the equilibrium point of a shrinking charged shell be manipulated?

Yes, the equilibrium point of a shrinking charged shell can be manipulated by changing the factors that affect it. For example, by increasing the charge on the shell or decreasing the surface tension of the material, the equilibrium point can be shifted to a different location. Additionally, by applying an external force, the equilibrium point can be moved to a new position.

5. Why is the equilibrium point of a shrinking charged shell important to study?

Understanding the equilibrium point of a shrinking charged shell is important because it has practical applications in fields such as materials science and engineering. It can also provide insights into the behavior of other systems, such as biological membranes, that experience both electrostatic and surface tension forces. By studying the equilibrium point, scientists can gain a better understanding of the forces at play and how they can be manipulated for various purposes.

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