Time derivative of vector potential - indices help

[Dixanadu]

Hey guys,

So I'm reading something about vector potentials and I've come across this one line which is really annyoing me. Here's how it goes

$\frac{d}{dt}\mathbf{A}=\frac{\partial \mathbf{A}}{\partial t}+\frac{\partial \mathbf{r}}{\partial t}\cdot \frac{\partial }{\partial \mathbf{r}}\mathbf{A}{}$

So everything is fine so far. Now he goes on to say the same thing in index form - so you can find any given component of the derivative. So he writes

$\frac{d}{dt}A_{i}=\frac{\partial A_{i}}{\partial t}+\frac{\partial r_{j}}{\partial t}\cdot\frac{\partial}{\partial r_{j}}A_{i}$

And he says "where a sum over repeated indices is implied". And this is where I get lost. I don't understand the summation over j. Say for instance $\mathbf{r}$ had 3 components, $r_{1},r_{2},r_{3}$. Does this mean that

$\frac{d}{dt}A_{i}=\frac{\partial A_{i}}{\partial t}+(\frac{\partial r_{1}}{\partial t}\cdot\frac{\partial}{\partial r_{1}}+\frac{\partial r_{2}}{\partial t}\cdot\frac{\partial}{\partial r_{2}}+\frac{\partial r_{3}}{\partial t}\cdot\frac{\partial}{\partial r_{3}})A_{i}$?

Thanks a lot guys!

 

[jedishrfu]

That's Einstein notation repeated indices are summed over ie there is an implied Ʃ in the equations second term where the j is used. the j ranges from 1 to 3 for a 3D vector.

http://en.wikipedia.org/wiki/Einstein_notation

 

[Dixanadu]

so what i wrote..was it right or wrong?

 

[jedishrfu]

so what i wrote..was it right or wrong?

Looks right, what do you think?

 

[Dixanadu]

That's what I don't get...how does this work? how can one component of A involve the derivatives of all the components of $\mathbf{r}$?

 

[Chestermiller]

That's what I don't get...how does this work? how can one component of A involve the derivatives of all the components of $\mathbf{r}$?

The problem is not with the second and third equations, it is with the first equation. The symbology is unconventional. In this equation, dr/dt is supposed to be the velocity vector. The expression $\frac{\partial \vec{A}}{\partial \vec{r}}$ is supposed to be $\nabla \vec{A}=\vec{i_{\alpha}}\vec{i_{\beta}}\frac{\partial A_{\alpha}}{\partial r_{\beta}}$

 

[Dixanadu]

Umm. Whats $\vec{i}_{\alpha}\vec{i}_{\beta}\frac{\partial A_{\alpha}}{\partial r_{\beta}}$? what are these $i$'s?

 

[Chestermiller]

Umm. Whats $\vec{i}_{\alpha}\vec{i}_{\beta}\frac{\partial A_{\alpha}}{\partial r_{\beta}}$? what are these $i$'s?

Unit vectors. Del of a vector is a second order tensor, and can be written as a summation involving dyadic (outer) products of unit vectors (unit vectors placed in juxtaposition, with no operation implied between them).

 

[Dixanadu]

riiiiiiight...could you somehow simplify that explanation to an undergraduate's level? T_T

 

[Chestermiller]

$\nabla \vec{A}=\vec{i_{1}}\vec{i_{1}}\frac{\partial A_{1}}{\partial r_{1}}+\vec{i_{1}}\vec{i_{2}}\frac{\partial A_{2}}{\partial r_{1}}+\vec{i_{1}}\vec{i_{3}}\frac{\partial A_{3}}{\partial r_{1}}+\vec{i_{2}}\vec{i_{1}}\frac{\partial A_{1}}{\partial r_{2}}+\vec{i_{2}}\vec{i_{2}}\frac{\partial A_{2}}{\partial r_{2}}+\vec{i_{2}}\vec{i_{3}}\frac{\partial A_{3}}{\partial r_{2}}+\vec{i_{3}}\vec{i_{1}}\frac{\partial A_{1}}{\partial r_{3}}+\vec{i_{3}}\vec{i_{2}}\frac{\partial A_{2}}{\partial r_{3}}+\vec{i_{3}}\vec{i_{3}}\frac{\partial A_{3}}{\partial r_{3}}$
$\vec{v}\centerdot \nabla \vec{A}=(v_1\vec{i_1}+v_2\vec{i_2}+v_3\vec{i_3})\centerdot (\vec{i_{1}}\vec{i_{1}}\frac{\partial A_{1}}{\partial r_{1}}+\vec{i_{1}}\vec{i_{2}}\frac{\partial A_{2}}{\partial r_{1}}+\vec{i_{1}}\vec{i_{3}}\frac{\partial A_{3}}{\partial r_{1}}+\vec{i_{2}}\vec{i_{1}}\frac{\partial A_{1}}{\partial r_{2}}+\vec{i_{2}}\vec{i_{2}}\frac{\partial A_{2}}{\partial r_{2}}+\vec{i_{2}}\vec{i_{3}}\frac{\partial A_{3}}{\partial r_{2}}+\vec{i_{3}}\vec{i_{1}}\frac{\partial A_{1}}{\partial r_{3}}+\vec{i_{3}}\vec{i_{2}}\frac{\partial A_{2}}{\partial r_{3}}+\vec{i_{3}}\vec{i_{3}}\frac{\partial A_{3}}{\partial r_{3}})$
$\vec{v}\centerdot \nabla \vec{A}=(v_1\frac{\partial A_1}{\partial r_1}+v_2\frac{\partial A_1}{\partial r_2}+v_3\frac{\partial A_1}{\partial r_3})\vec{i_1}+(v_1\frac{\partial A_2}{\partial r_1}+v_2\frac{\partial A_2}{\partial r_2}+v_3\frac{\partial A_2}{\partial r_3})\vec{i_2}+(v_1\frac{\partial A_3}{\partial r_1}+v_2\frac{\partial A_3}{\partial r_2}+v_3\frac{\partial A_3}{\partial r_3})\vec{i_3}$

The way dyadics work is summarized as follows: $\vec{a} \centerdot \vec{b}\vec{c}=(\vec{a} \centerdot \vec{b})\vec{c}$
Incidentally, I made mistake in my earlier post. The correct equation should have been:
$\nabla \vec{A}=\vec{i_{\alpha}}\vec{i_{\beta}}\frac{\partial A_{\beta}}{\partial r_{\alpha}}$ with the summation convention implied.

 

[stevendaryl]

I'm not sure exactly which part you are having conceptual problems with, but let me make the problem a little simpler to start with, just to motivate the concept of a directional derivative.

Suppose that you are driving along a highway that runs from, say, Texas to Canada. Let $T(x, t)$ be the temperature at a distance $x$ from the start at time $t$. How does the temperature outside your car change as you drive along?

Well, there are two reasons that the temperature might change:

1. You could be traveling into a region with colder temperatures.
2. The weather could be changing.

So suppose you drive from the point $x$ at time $t$ to the point $x + \delta x$ at time $t + \delta t$. The change in temperature outside your car is:

$\delta T = T(x+\delta x, t+\delta t) - T(x,t)$

which is approximately:

$\delta T = \dfrac{\partial T}{\partial x} \delta x + \dfrac{\partial T}{\partial t} \delta t$

The rate of change of temperature is

$\dfrac{dT}{dt} = \dfrac{\delta T}{\delta t} = \dfrac{\partial T}{\partial x} \dfrac{\delta x}{\delta t} + \dfrac{\partial T}{\partial t}$

The quantity $\dfrac{\delta x}{\delta t}$ is just the velocity component $v^x$. So we have:

$\dfrac{dT}{dt} = \dfrac{\partial T}{\partial x} v^x + \dfrac{\partial T}{\partial t}$

We were only considering one-dimensional motion. If you are traveling in 3 directions, then the generalization is:

$\dfrac{dT}{dt} = \dfrac{\partial T}{\partial x} v^x + \dfrac{\partial T}{\partial y} v^y + \dfrac{\partial T}{\partial z} v^z + \dfrac{\partial T}{\partial t}$

This is written more compactly as:

$\dfrac{dT}{dt} = \dfrac{\partial T}{\partial x^i} v^i + \dfrac{\partial T}{\partial t}$

where $\dfrac{\partial T}{\partial x^i} v^i$ means $\dfrac{\partial T}{\partial x} v^x + \dfrac{\partial T}{\partial y} v^y + \dfrac{\partial T}{\partial z} v^z$

If instead of a scalar number $T$, you are considering changes to a vector quantity $A^j$, then each component will change in the same way:

$\dfrac{dA^j}{dt} = \dfrac{\partial A^j}{\partial x^i} v^i + \dfrac{\partial A^j}{\partial t}$

In the above equation, $A^j$ means $A^x$ or $A^y$ or $A^z$.

 

[Dixanadu]

wow yea i get it now...i see what you mean, cos each component of the vector depends on all three components of velocity, that's why there's a summation...yea that's what was confusing me. Your explanation rocks man, thanks a lot ! :D