# Do stars remain electrically neutral? [duplicate]

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How electrically neutral do stars remain through their lifetime? As an example, I could imagine processes such as coronal mass ejections leaving the Sun in a slightly charged state. Are there such processes that will leave a star with overall charge?

## marked as duplicate by Emilio Pisanty, AccidentalFourierTransform, ACuriousMind♦, Carl Witthoft, user10851 May 2 '16 at 20:02

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• Do you mean any charge at all, or only significant amounts of charge? If you're counting electrons and protons I think you'd be hard pressed to find any macro object that is perfectly neutral. – Asher May 2 '16 at 13:27
• @Asher I'm interested in a characterization of the neutrality of the sun. I don't expect it to be neutral to the electron, I want to know how far it deviates from equilibrium. – anon01 May 2 '16 at 13:35

# Background

Stars are composed of plasmas, which are an ionized gas that exhibit a collective behavior much like a fluid.

There are two important aspects of plasmas to keep in mind. The first is that they act like very highly conductive metals in that the electrons can move very freely in order to cancel out any charge imbalance. The consequence is that they are said to be quasi-neutral over distances larger than the Debye length. By quasi-neutral, I mean: $$n_{e} = \sum_{s} \ Z_{s} \ n_{s} \tag{1}$$ where $n_{e}$ is the total electron number density, $n_{s}$ is the number density of ion species $s$, and $Z_{s}$ is the charge state of ion species $s$ (e.g., +1 for protons). Generally the Debye length is considered to be as microscopic as one would ever care about in most situations.

The second thing is that unless driven, plasmas tend to satisfy a zero current condition given by: $$\sum_{s} \ Z_{s} \ n_{s} \ \mathbf{v}_{s} = 0 \tag{2}$$ where $s$ in this case includes electrons and $\mathbf{v}_{s}$ is the bulk flow velocity of species $s$ (i.e., the first velocity moment). This is specifically referring to the flow of plasma out of the sun (i.e., the solar wind), which is derived from the continuity equation for electric charge. There are, of course, large localized currents all through the sun from its interior to its upper atmosphere.

How electrically neutral do stars remain through their lifetime?

Very. If they "charged up" they would produce tremendous Coulomb potentials that would prevent particles of certain charge signs from leaving the surface. Meaning, much like in a conductor, electric fields will do work to get rid of themselves. This is why plasmas are found to be quasi-neutral over distances larger than the Debye length.

The second condition given by Equation 2 results in relative drifts between different particle species in order to maintain a zero net current flow out of the sun (except during active periods like in coronal mass ejections or solar flares). Yet even so, these relative drifts do not act to increase the net charge of the sun.

As an example, I could imagine processes such as coronal mass ejections leaving the Sun in a slightly charged state.

No, generally they do not alter the macroscopic charge state of the sun. They consist of plasmas, which as I said before, are quasi-neutral.

Are there such processes that will leave a star with overall charge?

None of which that I am aware. As I said before, if you suddenly "charged up" a star the resultant electric fields would do work on the system until those electric fields no longer existed.

# Update

While I maintain that Equation 1 generally holds, for various reasons it does appear that there is a small net charge to the sun, as mentioned in this answer:
https://physics.stackexchange.com/a/73773/59023.

# Update 2

After further discussion with colleagues and my finally gaining access to the paper in question [i.e., Neslusan, 2001], I have a few comments.

The electric field to which that article refers, now called the Pannekoek and Rosseland (P-R) electric field, is only valid for a stellar corona in hydrostatic equilibrium. This field is not consistent with observations because it only produces a sort of breeze, not a supersonic wind like the one that has been observed since the 1960s. This field would also not remain stable. Meaning, it would eventually accelerate protons away from the star and eliminate any net charging.

The more correct approach, which is now generally accepted, is called an exospheric model [e.g., Zouganelis et al., 2005]. This model, unlike the P-R model, can include the multiple components of the solar wind electron velocity distributions (i.e., core, halo, and strahl). It still includes the mass of the star through a gravitational term but more importantly, it is actually consistent with the observations.

# References

• Neslusan, L. "On the global electrostatic charge of stars," Astron. & Astrophys. 372, pp. 913-915, doi:10.1051/0004-6361:20010533, 2001.
• Zouganelis, I. et al., "Acceleration of Weakly Collisional Solar-Type Winds," Astrophys. J. 626, pp. L117-L120, doi:10.1086/431904, 2005.
• I find the electromagnetic arguments much more compelling than electrostatic ^^ these are good considerations. But unless I'm mistaken, the sun does have significant currents, no? – anon01 May 2 '16 at 14:14
• @anon0909 - Oh yes, sorry if I misled you in my answer. I was meaning the solar wind flow. I will clarify this. – honeste_vivere May 2 '16 at 14:35
• @honeste_vivere fyi I've just asked Are there measurements of, or experimental limits to the residual charge of the Sun? – uhoh Oct 28 '17 at 20:21

Overall, a star stays more or less neutral. This is true for all stellar objects beside black holes. I am using a simple calculation that can be found in a footnote of https://arxiv.org/abs/1001.3294 on p. 11 chap. 2.

Suppose the star has an overall charge of Z times the elementary charge, $Ze$, and we consider the Coulomb repulsion of a test particle, say a proton, with mass $m$ and charge $e$ .The Coulomb force, seeking to expel the test particle, has to be smaller than the gravitational force, seeking to keep the test particle within the star. This gives the condition $\frac{(Ze)e}{R^2}\le \frac{GMm}{R^2}$ where M is the mass of the star R its radius. For the moment we assume that $M=Am$, which means that we neglect all effects of the binding energy, which would lower the mass and actually make the case for our test particle even worse. Therefore we find $\frac{(Ze)e}{R^2}\le \frac{GAm^2}{R^2}$ which leads to $Z<G\frac{m^2}{e^2}A$. If you plug in some numbers you find that $Z<10^{-37}A$ i.e., the average charge per nucleon has to be extremely small in order to ensure the stability of the star. This argument only relies on the huge difference of the strength of the two opposing forces.($A$ is the number of nucleons)

• This is an awfully simple model for a star. Firstly, its an electromagnetic system, not electrostatic. And this says nothing of the convective, thermal or nuclear considerations – anon01 May 2 '16 at 18:09
• But on the other hand if it built up a charge potential anything like 1 per nucleons the electrostatic force would dominate. – Joshua May 2 '16 at 18:25
• I don't see why one should take into account how the charge builds up. It can't stay there in a relevant amount as shown. – Noldig May 3 '16 at 13:03
• @Noldig fyi I've just asked Are there measurements of, or experimental limits to the residual charge of the Sun? – uhoh Oct 28 '17 at 20:21