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.

enter image description here

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?


$\nabla B^2 \sim B^2/l,$ 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.

  • $\begingroup$ Yes, I got something like that too. I think its the modulus that's maybe causing some difficulty... $\endgroup$
    – user29126
    Aug 24 '19 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
    Aug 24 '19 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
    Aug 25 '19 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.