Continuum mechanics: Finding the material derivative of plasma subjected to a decaying electric field

Question

I have been given the fact that a fluid is subjected to a decaying electric field of (scalar) magnitude: $$ e(\boldsymbol{x},t)=r^{-1}e^{-At},$$$$r^2=x_1^2+x_2^2+x_3^2$$ where $A$ is a positive constant, and the Eulerian velocity of the fluid is: $$v(\boldsymbol{x},t)=(x_1x_3,tx_2^2,tx_2x_3)$$ and I have to work out the material rate of change of $e$ at time $t=1$ of the particle at the point $\boldsymbol{x}=(2,-2,1)$. I haven't really done much mechanics in maths so the 'decaying electric field' is kind of new to me. I know that the material derivative is given by $$\nabla{e}{\cdot}v+e_t$$ So my questions is basically... how do I work out the gradient of $e$ when I have been given it in this form? Do I need to put in the given $x$ co-ordinates first, to obtain $r^2$ and then work out $\nabla{e}$ from there? Any help would be much appreciated!

Answer 1

First off MHD is a topic close to my heart!!

The material derivative is not as you have given (or a typo) and instead is given by $$ \partial_t e + \mathbf{v}\cdot\nabla e $$ To compute the $\mathbf{v}\cdot \nabla$ we go as follows $$ \left(x_{1}x_{3},tx_{2}^{2},tx_{2}x_{3}\right)\cdot\left(\frac{\partial}{\partial x_{1}},\frac{\partial}{\partial x_{2}},\frac{\partial}{\partial x_{3}}\right) $$ This equates to $$ x_{1}x_{3}\frac{\partial}{\partial x_{1}} + tx_{2}^{2}\frac{\partial}{\partial x_{2}} + tx_{2}x_{3}\frac{\partial}{\partial x_{3}} $$

for which you can sub back into the original "material derivative". Then to complete your answer, you can compute the derivatives of r with respect to $x_{i}$ as normal. Alternatively you could of converted all the derivatives into cylindrical or spherical geometry, but I think the way i outlined above is more straight forward.

To explicitly show how to take the derivative, I will compute the first term. $$ x_{1}x_{3}\frac{\partial e}{\partial x_{1}} = x_{1}x_{3}\frac{\partial }{\partial x_{1}}\left(-\frac{1}{r}\mathrm{e}^{-At}\right) = \mathrm{e}^{-At}\frac{-x_{1}x_{3}}{r^{2}}\frac{\partial r}{\partial x_{1}} $$ You can finish off the differentiation. So now since i decided to stay in these cartesian co-ordinates, it makes compute the derivatives more cumbersome, but not impossible. You will have to do the same for the derivatives with respect to $x_{2}$ and $x_3$.