Deriving del cross A in Electrodynamics

[Sparky_]

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):

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


Some background equations:

$$\vec{A} = \frac{\vec{v}}{c^2} V$$

$$V = \frac{1}{4\pi e_0}\frac{qc}{(rc - \vec{r}\bullet \vec{v})}$$

Using a product rule for a cross product:

(1) $$\nabla \times \vec{A} = \frac{1}{c^2}\nabla \times (V\vec{v}) - \vec{v} \times (\nabla V)$$ (1)



$$\nabla \times \vec{v} = ( -\vec{a} \times \nabla t_r)$$

$$\nabla t_r = -\frac{\vec{r}}{rc - \vec{r} \bullet \vec{v}}$$


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

Plug these into (1)


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


Factor out a $$\frac{qc}{4\pi e_0}$$


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

Can I factor out a $$\frac{1}{(rc - \vec{r}\bullet \vec{v}})$$ ?

And end up with:

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


Using the equation for $$\nabla t_r$$

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

Can I factor out another $$\frac{1}{(rc - \vec{r}\bullet \vec{v}})$$ ?

Yielding:

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

I see that I have a
$$\vec{v} \times \vec{v}$$ in the middle which results in zero.

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

Introducing $$\vec{u} = c\hat{r} – \vec{v}$$

$$\vec{r} \bullet \vec{u} = (rc - \vec{r} \bullet \vec{v})$$


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

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_

 

[Simon Bridge]

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:
$$\nabla \times \vec{A} = - \frac{1}{c}\frac{q}{4\pi e_0}\frac{1}{(\vec{u}*\vec{r})^3}\vec{r} \times \left [ (c^2 – v^2)\vec{v} + (\vec{r}\cdot \vec{a})\vec{v} + (\vec{r}\cdot \vec{u})\vec{a}\right ]$$
... I don't know what the asterisk is supposed to represent.

Some background equations:

$$\vec{A} = \frac{\vec{v}}{c^2} V$$

$$V = \frac{1}{4\pi e_0}\frac{qc}{(rc - \vec{r}\cdot \vec{v})}$$

... 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.

 

[Sparky_]

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:

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

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-

$$\vec{u} = c\hat{r} – \vec{v}$$

u sort of cleans up the $$(rc - \vec{r}\cdot \vec{v})$$

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_