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!


1 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 $x_3$.

  • $\begingroup$ haha you're completely right, sorry about that, haven't got this whole latex thing down... okay that makes a lot of sense, thanks! $\endgroup$
    – Lucy
    Feb 11, 2014 at 19:17
  • $\begingroup$ I think it is counterintuitive to compute $\mathbf{v}\cdot\nabla$ first, wouldn't you get $\nabla e$ first and then compute the dot product? $\endgroup$
    – kleineg
    Feb 11, 2014 at 19:28
  • $\begingroup$ @kleineg The material derivative is stated as $$\frac{D}{Dt} = \partial_t +\mathbf{v}\cdot \nabla $$ and then apply this operator to the quantity you want to look at. In physics, it is used to look for conserved quantities over that particular flow i.e. when it is set to zero. But having looked at your profile, you know this :). $\endgroup$
    – Chinny84
    Feb 11, 2014 at 19:32
  • $\begingroup$ so how would I apply this operator to e when it is in this form? I'm confused by the $r^2$ $\endgroup$
    – Lucy
    Feb 11, 2014 at 19:36
  • $\begingroup$ I have modified the answer to include how you would compute the derivatives. $\endgroup$
    – Chinny84
    Feb 11, 2014 at 19:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .