1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Deriving del cross A in Electrodynamics

  1. Jun 16, 2013 #1

    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] ?


    [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.

  2. jcsd
  3. Jun 17, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    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.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook