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Another Jackson problem

  • Thread starter dingo_d
  • Start date
  • #1
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Homework Statement



Every time I pick up Jackson I feel dumb :(

So in 5.5 chapter about Circular Current Loop, he says:

[tex]|\vec{r}-\vec{r}'|=[r^2+r'^2-2rr'(\cos\theta\cos\theta'+\sin\theta\sin\theta'\cos\phi')]^{1/2}[/tex]

So it's cosine rule, but how did he got to that sine/cosine thing? It should be the angle between r and r'. I tried drawing but to no avail :\

help...
 

Answers and Replies

  • #2
213
8
in 3d space the vector r has coordinates

[tex] \bold r = (x,y,z) = (rsin(\theta)cos(\phi),rsin(\theta)sin(\phi),rcos(\theta))[/tex]

and vector r' has coordinates

[tex] \bold r' = (x',y',z') = (r'sin(\theta ')cos(\phi '),r'sin(\theta ')sin(\phi '),r'cos(\theta '))[/tex]

so basically you use that and the fact that

[tex]\left|\bold r - \bold r' \right| = \sqrt{ (x-x')^2 +(y-y')^2 +(z-z')^2}[/tex]

and that should be equivalent to what you have
 
  • #3
211
0
I thought of sth like that right now in the morning, only I thought of cylindrical coordinate system.

Altho that would get complicated in the z part.

I'll try this. Thank you :)
 

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