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Vector Calculus with Maxwell's Equations

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Dick

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The things you are writing down are products of two vectors, since u is already a vector. The result should just be a vector. How does ##\frac{dr}{dt} \cdot (\nabla u)## sound? The ##(\nabla u)## part will be a product of two vectors or as the wikipedia article said, a 'tensor derivative', but the dot product with dr/dt will cut that down to a single vector.
 

pasmith

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Question: The velocity of an element of fluid at position r(t) at time t is described by a vector field ##\underline{u}(\underline{r}(t),t)##. Use the chain rule to show that the total derivative of this velocity field with respect to time is$$\frac{du}{dt} = \frac{\partial u}{\partial t} + (u \cdot \nabla)u$$
That doesn't look right at all. How can u appear twice in the result? This looks something like http://en.wikipedia.org/wiki/Material_derivative I think the result should be something more like $$\frac{du}{dt} = \frac{\partial u}{\partial t} + (\frac{dr}{dt} \cdot \nabla)u$$.
Yes, and then one substitutes [itex]\dot r(t) = u(r(t),t)[/itex] to obtain the result. The material derivative is what you get when you differentiate with respect to time under the assumption that position is a function of time whose derivative is the local fluid velocity.

I guess I don't see why dr/dt would be u.
Because it's stated to be so in the question: "The velocity of an element of fluid at position r(t) at time t is described by a vector field u(r(t),t)".

You can't really have u appearing twice in that last term. If you multiply u by 2 then the first two terms get multiplied by u, but the last one gets multiplied by 4. All the terms need to scale the same way.
There is no basis, either in physics or mathematics, for such a requirement here. So long as the expression is dimensionally consistent (as it is) there is no problem in it being non-linear.
 

Dick

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Perhaps I got sucked into this without completely understanding the context. I meant to answer just the original posted question. Apologies for my confusion. Please take over, I've really been hoping somebody else would step in.
 

pasmith

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Hi Dick,
I have another question relating to this material:
Consider some vector field ##\underline{u}(\underline{r}(t),t)##. Find the total derivative ##\frac{du}{dt}##. What I have is $$\frac{du}{dt} = \frac{\partial u}{\partial r} \frac{dr}{dt} + \frac{\partial u}{\partial t}\frac{dt}{dt}$$

This is a 'show that' question so what I need to show is that ##\frac{\partial u}{\partial r} \frac{dr}{dt} = (u \cdot \nabla) u##
There is an abuse of notation in which one can write [itex]\frac{\partial}{\partial \mathbf{x}}[/itex] to mean [itex]\nabla[/itex], but this is best avoided.

The chain rule for a real-valued function [itex]f(t, x_1(t), \dots, x_n(t))[/itex] is
[tex]
\frac{\mathrm{d}f}{\mathrm{d}t} = \frac{\partial f}{\partial t} + \sum_{k=1}^n
\frac{\partial f}{\partial x_k} \frac{\mathrm{d}x_k}{\mathrm{d}t}
[/tex]
The sum on the right can be interpreted as [itex]\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} \cdot
\nabla f[/itex] if the [itex]x_i[/itex] are treated as cartesian coordinates.

For a vector-valued function, it's best to look component-by-component in cartesian coordinates, and if you do you will discover that
[tex]
\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} = \frac{\partial \mathbf{u}}{\partial t} + \sum_{k=1}^n
\frac{\partial \mathbf{u}}{\partial x_k} \frac{\mathrm{d}x_k}{\mathrm{d}t}[/tex]
where the sum can be interpreted as [itex](\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} \cdot \nabla)\mathbf{u}[/itex] in cartesian coordinates.
 

CAF123

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I have the comment on my work that the gradient of a vector is 'ill defined'.
So when I wrote ##\nabla(\underline{a})## and gave the expression $$\frac{\partial a_1}{\partial x_1}e_1 + \frac{\partial a_2}{\partial x_2}e_2 + \frac{\partial a_3}{\partial x_3}e_3$$that was wrong
 

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