Physics Asked by Global on March 11, 2021
I have done all the calculation in finding the vector potential leading up to the equation 5.68. But then Prof. Griffiths goes back to the original figure and mentions that the coordinate of the point $mathbf{r}$ is $(r, theta,phi)$. And now I have a problem here. Because, we all know, that in spherical coordinate $theta$ is the angle from the $mathbf{z}$ axis and it remains constant even when the point rotates around the $mathbf{z}$ axis. In the original figure $theta$ actually would be $psi$. But he doesn’t use $psi$ in the ultimate equation 5.69 rather goes on with $theta$ which I am unable to get. Or is he just generalizing the result where the angle is $theta$ instead of $psi$?
What am I missing here in my reasoning?
Below are the relevant figures and equations.
EDIT:
I think the first figure (5.45) should have been the following to match up with his explanations towards the end.
The $theta$ is probably a typo due to the standard notation to call that angle $theta$, but you are right that from the picture it should be $psi$.
Answered by M_kaj on March 11, 2021
Griffiths explains it well: he defines $psi$ as the angle between the radius vector of the point of interest $mathbf r$ and the angular velocity vector $omega$ solely for the purpose of the calculation in the special tilted coordinate system. The vector potential value expressed in that system depends on this angle.
The change from $psi$ to $theta$ is made because he wants to express the result in the original frame, where the coordinates are usually denoted $r,theta,phi$; he could use symbol $psi$ instead of $theta$, but that would be confusing for the users of the formula, because the angle from the polar axis is usually denoted $theta$.
Answered by Ján Lalinský on March 11, 2021
Get help from others!
Recent Questions
Recent Answers
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP