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I was reading through the following extract outlining how the Maxwell stress contributes to the $\nabla \cdot \mathbf{P}$ term of the Navier-Stokes equation for fluids in a star. Here $\mathbf{P}$ is a general symmetric stress tensor.

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I am struggling to see how the estimate $\vert \nabla\cdot \mathbf{P}\vert/\rho \approx 3\times 10^{-3}$ ms$^{-2}$ was derived using the information given in the passage. Can someone please provide a quick derivation of this result?

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$\nabla B^2 \sim B^2/l,$ where $l$ is a length scale on which $B^2$ varies.

In SI units, $B^2 \sim 4\times 10^{-8}$ T$^2$ and $\mu_0 = 4\pi \times 10^{-7}$, so an order of magnitude for $\nabla \cdot P_B$ is $4\times 10^{-8}/(8\pi \times 10^{-7}\times 10^{6}) \sim 10^{-8}$ Pa/m.

The density of the solar photosphere (at optical depth unity in the visible spectrum) is around $\rho \sim 10^{-4}$ kg/m$^3$.

Thus $\nabla \cdot P_B/\rho \sim 10^{-4}$ m/s$^2$.

So no, I can't quite reproduce your number.

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  • $\begingroup$ Yes, I got something like that too. I think its the modulus that's maybe causing some difficulty... $\endgroup$
    – user29126
    Commented Aug 24, 2019 at 22:22
  • $\begingroup$ I should also point out that the density specified earlier in the book is $\rho = 1$ g cm$^{-3}$. $\endgroup$
    – user29126
    Commented Aug 24, 2019 at 23:58
  • $\begingroup$ @user29126 I doubt that this is the density of the photosphere, that is more like the average density of the entire Sun. $\endgroup$
    – ProfRob
    Commented Aug 25, 2019 at 14:04

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