# How does one find the electric conductivy of a planet's atmosphere?

I am working on an undergraduate project in which we study shaped structures in the atmosphere of the gas giants. We are investigating whether the magnetic field is the source of the instability generating the observed shaped structures. We therefore use spectral methods to make a stability study on the magnetohydrodynamics equations which in our case are

$$\frac{\partial{}\boldsymbol{B}}{\partial{}t}+\left(\boldsymbol{v}\cdot\boldsymbol\nabla\right)\boldsymbol{B}=\frac{1}{\mathrm{Re}_\eta}\boldsymbol\nabla^2\boldsymbol{B}+\left(\boldsymbol{B}\cdot\boldsymbol\nabla\right)\boldsymbol{v}$$

$$\frac{\partial{}\boldsymbol{v}}{\partial{}t}+\left(\boldsymbol{v}\cdot\boldsymbol\nabla\right)\boldsymbol{v}=\frac{1}{\mathrm{Re}}\boldsymbol\nabla^2\boldsymbol{v}+\boldsymbol{f-\nabla}p+\mathrm{N}\left(\boldsymbol{j\times{}B}\right)$$

where $$\rm{}Re_\eta$$ is the magnetic Reynolds number and $$\rm{}N$$ is the Stuart number expressed

$$\mathrm{N} = \frac{\sigma B^2L_c}{\rho{}U}$$

where

• $$B$$ is the magnetic field intensity;
• $$L_c$$ is the characteristic length of the studied structure;
• $$\sigma$$ is the flowing fluid electric conductivity;
• $$\rho$$ is the fluid density;
• and $$U$$ is the characteristic velocity scale of the flow.

Our model is now functional and the next step is to run the calculation with real values for physical parameters. We have all the data except the electric conductivity. Hence, I am looking for this data (or a method to calculate an approximation). I thought that a first approximation could be made from the volume fraction of hydrogen and helium and their respective conductivity.

All ideas, avenues of research and scientific publication references are welcome. Thank you for help.

PS: I don't know much (yet) about astrophysics and the planet's atmosphere.

• The conductivity of neutral gas is negligible, you need to estimate the ion fractions in your gas and a current density model. For the solar system gas giants and the dynamics of their visible atmospheres this is going to be completely irrelevant, as those are very cold. In Hot Jupiters that would be a different issue. But if you are determined on working on this topic, I can write up an answer. – AtmosphericPrisonEscape Dec 4 '19 at 14:36
• Why would it irrelevant for the very cold atmospheres? I am interested and I would appreciate your answer, thank you. – user31132 Dec 4 '19 at 14:45
• Because you're very cold, so your ion conductivity will be very small, and the ion conductivity is the one that counts. Hence your Stuart number will be very small. – AtmosphericPrisonEscape Dec 4 '19 at 15:00
• Indeed it makes sense, thanks. – user31132 Dec 4 '19 at 15:09

## 1 Answer

The conductivity of neutral gas is negligible, you need to estimate the free-electron conductivity in your gas. For this, one could first consult a phase diagram, like referenced in this answer. There you'd see that for the atmospheric pressures and temperature for all gas giants, you will stay in the $$H_2$$-gas regime, so no exotic equation of state is necessary to describe the state of the visible atmosphere of the gas giants.

The updated models with and without Juno-data show that adjustments to structure models in the interior of the planet need to be made but the atmosphere, which you are interested in, is still an ideal gas.

To compute the amount of free electrons in the atmosphere, you need to compute the possible ions, and their ionization fraction. This gets really nasty for cold atmospheres, because molecules form. Mole fractions of those have to be computed by a chemical model, then ionized (possibly by a Saha-type ionization model) and translated into a conductivity (possibly by a Lorentz-Drude-type of model).

For your purposes that could get really involved, but I guess what you could do that is much simpler, is look at some published literature, e.g. here, just take and cite their conductivity profile.
However even in this source, the two models they use for the conductivity differ by 9 orders of magnitude in the neutral atmosphere, so you want to read carefully about their assumptions.

• Thank you! I will have at look at all of this. – user31132 Dec 5 '19 at 6:32
• @Loic: Sure, if you run into trouble, just ask. – AtmosphericPrisonEscape Dec 5 '19 at 11:23