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_
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_