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Deriving del cross A in Electrodynamics

  1. Jun 16, 2013 #1
    Hello,

    I am trying to derive the equation for the B-field due to a moving charge. ~ Griffiths Chapter 10, equation 10.66.

    I have been trying to “do” the del cross A and simplify . Things get messy and I am uncertain on some of my vector operations.

    In searching the internet I find statements like, from del cross A and a little algebra you get …

    I have tried 3 different times and can get some of it and a good bit of mess.

    Below is an attempt.

    The final result is (what I want to end up with):

    [tex] \nabla \times \vec{A} = - \frac{1}{c}\frac{q}{4\pi e_0}\frac{1}{(\vec{u}*\vec{r})^3}\vec{r} \times [(c^2 – v^2)\vec{v} + (\vec{r}\bullet \vec{a})\vec{v} + (\vec{r}\bullet \vec{u})\vec{a}][/tex]


    Some background equations:

    [tex] \vec{A} = \frac{\vec{v}}{c^2} V[/tex]

    [tex]V = \frac{1}{4\pi e_0}\frac{qc}{(rc - \vec{r}\bullet \vec{v})} [/tex]

    Using a product rule for a cross product:

    (1) [tex] \nabla \times \vec{A} = \frac{1}{c^2}\nabla \times (V\vec{v}) - \vec{v} \times (\nabla V)[/tex] (1)



    [tex] \nabla \times \vec{v} = ( -\vec{a} \times \nabla t_r)[/tex]

    [tex] \nabla t_r = -\frac{\vec{r}}{rc - \vec{r} \bullet \vec{v}} [/tex]


    [tex] \nabla V = \frac{1}{4\pi e_0}\frac{qc}{(rc - \vec{r}*\vec{v})^3} [(rc - \vec{r}\bullet \vec{v})\vec{v} - (c^2-v^2 + \vec{r}\bullet \vec{a})\vec{r}][/tex]

    Plug these into (1)


    [tex] \nabla \times \vec{A} = \frac{1}{c^2}[\frac{1}{4\pi e_0}\frac{qc}{(rc - \vec{r}\bullet \vec{v})}](-\vec{a} \times \nabla t_r))-\vec{v}\times [\frac{1}{4\pi e_0}\frac{qc}{(rc - \vec{r}\bullet \vec{v})^3}[((rc - \vec{r}\bullet \vec{v})\vec{v}) - ((c^2-v^2 + \vec{r}\bullet \vec{a})\vec{r})]] [/tex]


    Factor out a [tex]\frac{qc}{4\pi e_0} [/tex]


    [tex] \nabla \times \vec{A} = \frac{1}{c^2}\frac{qc}{4\pi e_0} [\frac{1}{(rc - \vec{r}\bullet \vec{v})}](-\vec{a} \times \nabla t_r))-\vec{v}\times [\frac{1}{(rc - \vec{r}\bullet \vec{v})^3}((rc - \vec{r}\bullet \vec{v})\vec{v}) - ((c^2-v^2 + \vec{r}\bullet \vec{a})\vec{r})] [/tex]

    Can I factor out a [tex]\frac{1}{(rc - \vec{r}\bullet \vec{v}}) [/tex] ?

    And end up with:

    [tex] \nabla \times \vec{A} = \frac{1}{c^2}\frac{qc}{4\pi e_0}\frac{1}{(rc - \vec{r}\bullet \vec{v})} (-\vec{a} \times \nabla t_r))-\vec{v}\times [\frac{1}{(rc - \vec{r}\bullet \vec{v})^2}[((rc - \vec{r}\bullet \vec{v})\vec{v}) - ((c^2-v^2 + \vec{r}\bullet \vec{a})\vec{r})]] [/tex]


    Using the equation for [tex] \nabla t_r [/tex]

    [tex] \nabla \times \vec{A} = \frac{1}{c^2}\frac{qc}{4\pi e_0}\frac{1}{(rc - \vec{r}\bullet \vec{v})} [(-\vec{a} \times -\frac{\vec{r}}{(rc - \vec{r} \bullet \vec{v})})-\vec{v}\times [\frac{1}{(rc - \vec{r}\bullet \vec{v})^2}[((rc - \vec{r}\bullet \vec{v})\vec{v}) - ((c^2-v^2 + \vec{r}\bullet \vec{a})\vec{r})]]] [/tex]

    Can I factor out another [tex]\frac{1}{(rc - \vec{r}\bullet \vec{v}}) [/tex] ?

    Yielding:

    [tex] \nabla \times \vec{A} = [\frac{1}{c^2}\frac{qc}{4\pi e_0}\frac{1}{(rc - \vec{r}\bullet \vec{v})^2}] [(-\vec{a} \times -\vec{r}) -\vec{v}\times [\frac{1}{(rc - \vec{r}\bullet \vec{v})}[((rc - \vec{r}\bullet \vec{v})\vec{v}) - ((c^2-v^2 + \vec{r}\bullet \vec{a})\vec{r})]]] [/tex]

    I see that I have a
    [tex] \vec{v} \times \vec{v}[/tex] in the middle which results in zero.

    [tex] \nabla \times \vec{A} = [\frac{1}{c^2}\frac{qc}{4\pi e_0}\frac{1}{(rc - \vec{r}\bullet \vec{v})^2}] [(-\vec{a} \times -\vec{r}) + \vec{v}\times [ \frac{((c^2-v^2 + \vec{r}\bullet \vec{a})\vec{r})}{(rc - \vec{r}\bullet \vec{v})}]] [/tex]

    Introducing [tex] \vec{u} = c\hat{r} – \vec{v} [/tex]

    [tex] \vec{r} \bullet \vec{u} = (rc - \vec{r} \bullet \vec{v}) [/tex]


    [tex] [/tex]
    Assuming I am correct at this point, here is where for me things start getting messy.
    [tex] \nabla \times \vec{A} = [\frac{1}{c^2}\frac{qc}{4\pi e_0}\frac{1}{(\vec{r} \bullet \vec{u})^2}] [(-\vec{a} \times -\vec{r}) + \vec{v}\times [ \frac{((c^2-v^2 + \vec{r}\bullet \vec{a})\vec{r})}{(\vec{r} \bullet \vec{u})}]] [/tex]

    Frustration is showing up and I could use some help.

    Bottom line is I am trying to derive and see how to get the del cross A result at the top. Searching the internet I find that equation and how it is used for the B field but not how it was derived.

    Thanks
    Sparky_
     
  2. jcsd
  3. Jun 17, 2013 #2

    Simon Bridge

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    It works better of you change the size of your brackets as you go out, and use \cdot instead of \bullet for a dot-product. vis:
    [tex] \nabla \times \vec{A} = - \frac{1}{c}\frac{q}{4\pi e_0}\frac{1}{(\vec{u}*\vec{r})^3}\vec{r} \times \left [ (c^2 – v^2)\vec{v} + (\vec{r}\cdot \vec{a})\vec{v} + (\vec{r}\cdot \vec{u})\vec{a}\right ][/tex]
    ... I don't know what the asterisk is supposed to represent.

    Some background equations:

    [tex] \vec{A} = \frac{\vec{v}}{c^2} V[/tex]

    [tex]V = \frac{1}{4\pi e_0}\frac{qc}{(rc - \vec{r}\cdot \vec{v})} [/tex]

    ... but you seem to be missing a ##\vec{u}##.

    The product rule for curl is:
    $$\nabla \times (\vec{u}\cdot\vec{v}) = (\nabla \cdot \vec{u}) \times \vec{v} + \vec{u}\cdot (\nabla \times \vec{v})$$
    ... so, on the whole, I'm not sure I follow what you are doing here.

    In general, when a text says "... and then, after some algebra, you get..." the "some algebra" tends to be pretty ugly.
     
  4. Jun 17, 2013 #3
    I went through the exercise (on pages 435-437, Griffiths) of taking the gradient of V and the time derivative of A and arriving at E. The text then states that after del cross A you arrive at:

    [tex] \nabla \times \vec{A} = - \frac{1}{c}\frac{q}{4\pi e_0}\frac{1}{(\vec{u}*\vec{r})^3}\vec{r} \times \left[(c^2 – v^2)\vec{v} + (\vec{r}\bullet \vec{a})\vec{v} + (\vec{r}\bullet \vec{u})\vec{a}\right][/tex]

    I want to turn the crank and arrive at the above equation.

    In my scratch work I did not introduce or bring in u at the start-

    [tex] \vec{u} = c\hat{r} – \vec{v}[/tex]

    u sort of cleans up the [tex] (rc - \vec{r}\cdot \vec{v})[/tex]

    I brought in u after a step or two into my work.


    also – I missed the asterisk – that also is a dot product.

    Thanks
    -Sparky_
     
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