# Time derivative of vector potential - indices help!

1. Dec 3, 2013

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 dont 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!

2. Dec 3, 2013

### Staff: Mentor

3. Dec 3, 2013

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

4. Dec 3, 2013

### Staff: Mentor

Looks right, what do you think?

5. Dec 3, 2013

That's what I dont get...how does this work? how can one component of A involve the derivatives of all the components of $\mathbf{r}$?

6. Dec 3, 2013

### Staff: Mentor

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 $\nabla \vec{A}=\vec{i_{\alpha}}\vec{i_{\beta}}\frac{\partial A_{\alpha}}{\partial r_{\beta}}$

Last edited: Dec 3, 2013
7. Dec 4, 2013

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

8. Dec 4, 2013

### Staff: Mentor

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).

9. Dec 4, 2013

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

10. Dec 4, 2013

### Staff: Mentor

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

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.

11. Dec 4, 2013

### stevendaryl

Staff Emeritus
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 $A^z$.

12. Dec 5, 2013