About divergence, gradient and thermodynamics

[fluidistic]

At some point, in Physics (more precisely in thermodynamics), I must take the divergence of a quantity like $\mu \vec F$. Where $\mu$ is a scalar function of possibly many different variables such as temperature (which is also a scalar), position, and even magnetic field (a vector field).

My question is, how to evaluate that divergence? I am tempted to set it equal to $\nabla \cdot (\mu \vec F)=\nabla \mu \cdot \vec F + \mu \nabla \cdot \vec F$. I know it doesn't matter here, but it turns out that thanks to some physical fact, the divergence of $\vec F$ vanishes, so we can focus solely on the first term if we want.

And that is where my doubt lies. Precisely, the gradient of $\mu$. Is it like a total derivative? So that if $\mu$ depends on temperature, magnetic field and position, then I should evaluate $\nabla \mu$ as $\left ( \frac{\partial \mu}{\partial T}\right)_{\vec B,x}\frac{\partial T}{\partial x} + \left ( \frac{\partial \mu}{\partial x}\right)_{\vec B,T}\frac{\partial x}{\partial x} + \left ( \frac{\partial \mu}{\partial B_x}\right)_{B_y, B_z,x,T}\frac{\partial B_x}{\partial x}+...$? I am a bit confused on the number of terms and whether what I wrote is correct.

 

[Chestermiller]

To answer the first part of the question, just write it out in Cartesian component form.

For the second part, you correctly determined the partial with respect to x at constant y and z.

 

[fluidistic]

To answer the first part of the question, just write it out in Cartesian component form.

Done, I confirm what I wrote, namely $\nabla \cdot (\mu \vec F)=\nabla \mu \cdot \vec F + \mu \nabla \cdot \vec F$

However, more precisely I get (for the first term on the right side): $\left( \frac{\partial \mu}{\partial x} \right) F_x + \left( \frac{\partial \mu}{\partial y} \right) F_y + \left( \frac{\partial \mu}{\partial z} \right) F_z$. As if only spatial derivatives mattered.

So, even if $\mu$ depends on temperature, I do not see how to reach what I wrote, say the term $\left ( \frac{\partial \mu}{\partial T}\right)_{\vec B,x}\frac{\partial T}{\partial x}$. How could I reach this?

 

[pasmith]

Done, I confirm what I wrote, namely $\nabla \cdot (\mu \vec F)=\nabla \mu \cdot \vec F + \mu \nabla \cdot \vec F$

However, more precisely I get (for the first term on the right side): $\left( \frac{\partial \mu}{\partial x} \right) F_x + \left( \frac{\partial \mu}{\partial y} \right) F_y + \left( \frac{\partial \mu}{\partial z} \right) F_z$. As if only spatial derivatives mattered.

So, even if $\mu$ depends on temperature, I do not see how to reach what I wrote, say the term $\left ( \frac{\partial \mu}{\partial T}\right)_{\vec B,x}\frac{\partial T}{\partial x}$. How could I reach this?

Apply the chain rule: if $\mu(u_1(x,y,z), \dots, u_n(x,y,z))$ then
$$\frac{\partial \mu}{\partial x} = \sum_{i=1}^n \frac{\partial \mu}{\partial u_i} \frac{\partial u_i}{\partial x}$$ and thus $$\nabla \mu = \sum_{i=1}^n \frac{\partial \mu}{\partial u_i} \nabla u_i.$$

 

[fluidistic]

Thank you, I mathematically get it.