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Biermann mechanism says that due to different masses of electrons and protons, the motion of former leads to the latter. This effect produces non-zero circular currents which induce vertical magnetic fields.
I wanted to prove if density and pressure gradient are misaligned(not parallel) then magnetic field arises from zero even if the flow is irrotational. I approached by writing the Euler's equation in magnetic field:

$$\frac{\partial \vec{v}}{\partial t}+(\vec{v} \cdot\nabla)\vec{v}=-\frac{\nabla P}{\rho}+\frac{(\nabla \times \vec{B})\times \vec{B}}{\rho \mu_0}.$$

Now I am not sure how to proceed.

Indrayudh Das
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1 Answers1

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It is the combination of Ohm's law, perfect conductivity and Faraday's law that leads you to the ideal MHD equation, $$\frac{\partial\mathbf{B}}{\partial t}=\nabla\times\left(\mathbf{v}\times\mathbf{B}\right)\tag{1}.$$ It is because of this equation that if you start with a zero magnetic field then you will always have a zero magnetic field, regardless of what you want to say about the momentum, density, pressure or their gradients.

The Biermann mechanism solves this problem by adding an additional component to Ohm's law based on the collective motions of the electron population's evolution. This turns Eq (1) for the ideal case into a non-ideal MHD evolution, $$\frac{\partial\mathbf{B}}{\partial t}=\nabla\times\left(\mathbf{v}\times\mathbf{B}+\mathbf{E}_\text{ind}\right),$$ where $\mathbf{E}_\text{ind}$ is the induced electric field from the motions. For instance, Martínez-Gómez et al (A&A 650 (2021) A123; DOI: https://doi.org/10.1051/0004-6361/202039113) says $\mathbf{E}_\text{ind}\sim n_H\nabla P_e$ (with $n_H$ spatially-dependent; cf Eq 20 there).

Kyle Kanos
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