For some reason I am unable to find a source on the internet about this. I think I have an answer, but I want to be doubly sure about this. All I could find (here), is that for an incompressible fluid, the viscous stress term is written in conservative form as: $$ \nabla\cdot\left[\mu\left(\nabla\boldsymbol{u}+(\nabla\boldsymbol{u})^{\top}\right)\right]. $$ Knowing that this viscous stress acts as diffusion, I assumed that I could simply replace $\boldsymbol{u}$ by any vectorial quantity and $\mu$ by any diffusion coefficient. However, when working this out in spherical coordinates for the MHD induction equation, I find that I should get instead: $$ \nabla\cdot\left[\eta\left(\nabla\boldsymbol{B}-(\nabla\boldsymbol{B})^{\top}\right)\right]. $$ I'm confused. I can't seem to find any mistake I made which would explain the minus sign. Is one of them wrong or are they somehow both right?

At the same time, the advection term in the Navier-Stokes equation reads: $$ -\nabla\cdot (\rho\boldsymbol{u}\otimes\boldsymbol{u}), $$ while in the MHD induction equation, we have instead: $$ -\nabla\cdot (\boldsymbol{u}\otimes\boldsymbol{B}-\boldsymbol{B}\otimes\boldsymbol{u}). $$ I'm not sure I understand why they are both different, if they represent the same physics (advection), even though the flow field obviously advects itself.

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    $\begingroup$ A few questions: How do you get anything conservative that includes diffusion? Navier-Stokes does not include magnetic fields, so should you not get different results (this all also depends upon the assumptions you used to construct these equations)? $\endgroup$ Jan 18, 2022 at 20:49
  • $\begingroup$ 1. well, if you can write diffusion under a divergence, then you have it in conservative form. 2. I‘m talking about the MHD induction equation, but I compare the terms to the Navier-Stokes terms. Normally, if a physical phenomena is universal, it should be written the same way regardless of the quantity you are interested with (momentum, magnetic field). Although I suppose that since these quantities are vectorial, it complicates things. $\endgroup$ Jan 19, 2022 at 12:11
  • $\begingroup$ I assumed that I could simply replace $u$ by any vectorial quantity... Why? ...if they represent the same physics... Do they? $\endgroup$
    – Kyle Kanos
    Jan 29, 2022 at 16:38


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