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Time derivative of vector potential - indices help!

  1. Dec 3, 2013 #1
    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

    [itex]\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}{}[/itex]

    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

    [itex]\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}[/itex]

    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 [itex]\mathbf{r}[/itex] had 3 components, [itex]r_{1},r_{2},r_{3}[/itex]. Does this mean that

    [itex]\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}[/itex]?

    Thanks a lot guys!
     
  2. jcsd
  3. Dec 3, 2013 #2

    jedishrfu

    Staff: Mentor

  4. Dec 3, 2013 #3
    so what i wrote..was it right or wrong?
     
  5. Dec 3, 2013 #4

    jedishrfu

    Staff: Mentor

    Looks right, what do you think?
     
  6. Dec 3, 2013 #5
    That's what I dont get...how does this work? how can one component of A involve the derivatives of all the components of [itex]\mathbf{r}[/itex]?
     
  7. Dec 3, 2013 #6
    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 [itex]\frac{\partial \vec{A}}{\partial \vec{r}}[/itex] is supposed to be [itex]\nabla \vec{A}=\vec{i_{\alpha}}\vec{i_{\beta}}\frac{\partial A_{\alpha}}{\partial r_{\beta}}[/itex]
     
    Last edited: Dec 3, 2013
  8. Dec 4, 2013 #7
    Umm. Whats [itex]\vec{i}_{\alpha}\vec{i}_{\beta}\frac{\partial A_{\alpha}}{\partial r_{\beta}}[/itex]? what are these [itex]i[/itex]'s?
     
  9. Dec 4, 2013 #8
    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).
     
  10. Dec 4, 2013 #9
    riiiiiiight....could you somehow simplify that explanation to an undergraduate's level? T_T
     
  11. Dec 4, 2013 #10
    [itex]\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}}[/itex]
    [itex]\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}})[/itex]
    [itex]\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}[/itex]

    The way dyadics work is summarized as follows: [itex]\vec{a} \centerdot \vec{b}\vec{c}=(\vec{a} \centerdot \vec{b})\vec{c}[/itex]
    Incidentally, I made mistake in my earlier post. The correct equation should have been:
    [itex]\nabla \vec{A}=\vec{i_{\alpha}}\vec{i_{\beta}}\frac{\partial A_{\beta}}{\partial r_{\alpha}}[/itex] with the summation convention implied.
     
  12. Dec 4, 2013 #11

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    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 [itex]T(x, t)[/itex] be the temperature at a distance [itex]x[/itex] from the start at time [itex]t[/itex]. 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 [itex]x[/itex] at time [itex]t[/itex] to the point [itex]x + \delta x[/itex] at time [itex]t + \delta t[/itex]. The change in temperature outside your car is:

    [itex]\delta T = T(x+\delta x, t+\delta t) - T(x,t)[/itex]

    which is approximately:

    [itex]\delta T = \dfrac{\partial T}{\partial x} \delta x + \dfrac{\partial T}{\partial t} \delta t[/itex]

    The rate of change of temperature is

    [itex]\dfrac{dT}{dt} = \dfrac{\delta T}{\delta t} = \dfrac{\partial T}{\partial x} \dfrac{\delta x}{\delta t} + \dfrac{\partial T}{\partial t}[/itex]

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

    [itex]\dfrac{dT}{dt} = \dfrac{\partial T}{\partial x} v^x + \dfrac{\partial T}{\partial t}[/itex]

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

    [itex]\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}[/itex]

    This is written more compactly as:

    [itex]\dfrac{dT}{dt} = \dfrac{\partial T}{\partial x^i} v^i + \dfrac{\partial T}{\partial t}[/itex]

    where [itex]\dfrac{\partial T}{\partial x^i} v^i [/itex] means [itex]\dfrac{\partial T}{\partial x} v^x + \dfrac{\partial T}{\partial y} v^y + \dfrac{\partial T}{\partial z} v^z[/itex]

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

    [itex]\dfrac{dA^j}{dt} = \dfrac{\partial A^j}{\partial x^i} v^i + \dfrac{\partial A^j}{\partial t}[/itex]

    In the above equation, [itex]A^j[/itex] means [itex]A^x[/itex] or [itex]A^y[/itex] or [itex]A^z[/itex].
     
  13. Dec 5, 2013 #12
    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
     
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