For translation in a fixed direction n^, take K=n^ and the preceding Fn. The triple-product identity gives Fn=−(q/2)n^⋅(B×x). ThusChoosing the three Cartesian directions gives the three constantsThese are magnetic pseudomomentum components, not the three components of kinetic momentum mx˙. A direct check is dKmag/dt=qx˙×B+qB×x˙=0. In the stated gauge, Kmag=p+(q/2)B×x, consistent with the Noether boundary correction.
Qn=p⋅n^−Fn=n^⋅(mx˙+qB×x).
Kmag=mx˙+qB×x=constant vector.
Ancestors (11)
- 3
- S7
- 2012
- Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford
- Short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford
- Physics course of the University of Oxford
- Undergraduate course of the University of Oxford
- Course of the University of Oxford
- University of Oxford
- List of universities
- Home
Discussion (0)
New discussionThere are no discussions about this article yet.
Articles by others on the same topic (0)
There are currently no matching articles.