Timeline for A question about canonical momentum and arbitrariness for potential in magnetism
Current License: CC BY-SA 3.0
12 events
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| Feb 7, 2014 at 18:34 | comment | added | qfzklm | @ANDREW, Thank you very much for your explanation, it is exactly that the canonical momentum is not conserved in the case. But I still have one point that I am not understand. I have written down the hamiltonion before, and I think it is conserved and it will lead to the canonical momentum conserved. So, here, where am I wrong? | |
| Feb 7, 2014 at 15:44 | history | edited | ANDREW | CC BY-SA 3.0 |
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| Feb 7, 2014 at 15:31 | comment | added | Ali | @ANDREW, please edit your answer so it will reflect your reasoning. As now, it looks like a comment and it can be deleted(I am not voting on deletion, but it makes sense to do so). | |
| Feb 7, 2014 at 15:22 | comment | added | ANDREW | @Hunter: OK. Let us examine case of homogeneous magnetic field. Then A=xB. We can add constant vector A0 (parallel X axes) without magnetic strength changing. Resulting potential A1=A+A0. Thus canonical momentum equals (mv+qA1,0) at initial position. After quote of circulation the partical velosity is parallel Y axes and canonical momentum equals (qA1,mv). | |
| Feb 7, 2014 at 15:21 | comment | added | ANDREW | @qfzklm: OK. Let us examine case of homogeneous magnetic field. Then A=xB. We can add constant vector A0 (parallel X axes) without magnetic strength changing. Resulting potential A1=A+A0. Thus canonical momentum equals (mv+qA1,0) at initial position. After quote of circulation the partical velosity is parallel Y axes and canonical momentum equals (qA1,mv). | |
| Feb 7, 2014 at 15:00 | comment | added | John Rennie | @ANDREW: could you edit your answer to expand on the stuff in your comment. As it stands I note a couple of people have already flagged it as not an answer. | |
| Feb 7, 2014 at 14:52 | comment | added | ANDREW | OK. Let us examine case of homogeneous magnetic field. Then A=xB. We can add constant vector A0 (parallel X axes) without magnetic strength changing. Resulting potential A1=A+A0. Thus canonical momentum equals (mv+qA1,0) at initial position. After quote of circulation the partical velosity is parallel Y axes and canonical momentum equals (qA1,mv). | |
| Feb 7, 2014 at 14:35 | comment | added | qfzklm | I think the canonical momentum is conserved because the Hamiltonion of the particle $H=P^2/2m$ is conserved where the $P$ is the canonical momentum. Sorry, I don't have the book. Could you tell me more details about it? | |
| Feb 7, 2014 at 14:25 | comment | added | Hunter | Could you explain this in more depth? | |
| S Feb 7, 2014 at 14:23 | review | Low quality answers | |||
| Feb 7, 2014 at 15:32 | |||||
| S Feb 7, 2014 at 14:23 | review | First posts | |||
| Feb 7, 2014 at 14:25 | |||||
| Feb 7, 2014 at 14:05 | history | answered | ANDREW | CC BY-SA 3.0 |