I reviewed several sources for the Hankel contour, but your presentation was the best. Also, I feel better because what I thought was not correct in some outside sources was confirmed by looking at your video. Interesting that the residue theorem remains correct with a branch cut singularity as the effect in crossing the branch cut in a clockwise direction in one part of the contour is undone by crossing the branch cut in a counter clockwise direction in a different part of the contour. Excellent presentation!
Thank you for the positive feedback! I’m glad you enjoyed the video. This is some pretty advanced subject matter and sometimes the sources leave out important steps or ideas. I am an amateur trying to master this material myself, so in the process of learning this math and turning it into presentations, I try to make everything in my videos as clear as possible. Thanks again!
I just have one question, and I am afraid I may not be able to see the obvious here… . In the original Hankel contour (the open contour), we are going from +inf back to +inf, circling the branch point in a counterclockwise (CCW) orientation. Going CCW means that the branch point is INCLUDED inside the contour, and the function is not holomorphic at this point. I believe this is the reason why we can integrate along this contour only for z with Re(z)>0. So here’s my - probably naive - question: Why are we not going round the branch cut in the CLOCKwise orientation - that would exclude it. I know I am missing something obvious but I cant see it. Any comment appreciated. Many thanks - Ralph.
Thanks for your question. The orientation you take around the branch cut affects the sign of the integral, but does not determine whether the branch point is included or excluded. For the branch point to be included in the sense I think you mean it, the contour would have to literally go through it, or be closed around it. The reason on the limit contour we have to make z>0 is just that we are integrating a function whose leading term is r^(z-1) over a circular contour, and importantly, taking r to zero. The function diverges if z
@@FireSermon83 Yes thank you very much. I think the point I was missing is that for the branch point to be INcluded, the contour around it would need to be closed, which of course it is not. One follow-up question: I hope you do not mind: Is it then irrelevant what the direction of travel along the open Hankel contour is? I have always wondered why the contour is specified as +inf to +inf with a CCW open circle round the branch point. Could we also do +inf to +inf with a clockwise open circle?
Yes. We could perform the integration in the opposite orientation- starting below the real line at + infinity, traveling toward the origin, circling the origin clockwise, then traveling out towards + infinity again above the real line. That would just change the sign of the integral without changing its absolute value.
Well that was masterful. Thank you very much for this exceptionally clear explanation.
Thanks so much! I’m glad you enjoyed the video!
I reviewed several sources for the Hankel contour, but your presentation was the best. Also, I feel better because what I thought was not correct in some outside sources was confirmed by looking at your video. Interesting that the residue theorem remains correct with a branch cut singularity as the effect in crossing the branch cut in a clockwise direction in one part of the contour is undone by crossing the branch cut in a counter clockwise direction in a different part of the contour. Excellent presentation!
Thank you for the positive feedback! I’m glad you enjoyed the video. This is some pretty advanced subject matter and sometimes the sources leave out important steps or ideas. I am an amateur trying to master this material myself, so in the process of learning this math and turning it into presentations, I try to make everything in my videos as clear as possible. Thanks again!
Good video, I learned a lot
Thank you for another excellent video.
You’re welcome! I’m glad you continue to find these videos worthwhile.
@@FireSermon83 They are more than worthwhile. I really appreciate the depth and care you put into them.
I just have one question, and I am afraid I may not be able to see the obvious here… . In the original Hankel contour (the open contour), we are going from +inf back to +inf, circling the branch point in a counterclockwise (CCW) orientation. Going CCW means that the branch point is INCLUDED inside the contour, and the function is not holomorphic at this point. I believe this is the reason why we can integrate along this contour only for z with Re(z)>0. So here’s my - probably naive - question: Why are we not going round the branch cut in the CLOCKwise orientation - that would exclude it. I know I am missing something obvious but I cant see it. Any comment appreciated. Many thanks - Ralph.
Thanks for your question.
The orientation you take around the branch cut affects the sign of the integral, but does not determine whether the branch point is included or excluded. For the branch point to be included in the sense I think you mean it, the contour would have to literally go through it, or be closed around it.
The reason on the limit contour we have to make z>0 is just that we are integrating a function whose leading term is r^(z-1) over a circular contour, and importantly, taking r to zero. The function diverges if z
@@FireSermon83 Yes thank you very much. I think the point I was missing is that for the branch point to be INcluded, the contour around it would need to be closed, which of course it is not. One follow-up question: I hope you do not mind: Is it then irrelevant what the direction of travel along the open Hankel contour is? I have always wondered why the contour is specified as +inf to +inf with a CCW open circle round the branch point. Could we also do +inf to +inf with a clockwise open circle?
Yes. We could perform the integration in the opposite orientation- starting below the real line at + infinity, traveling toward the origin, circling the origin clockwise, then traveling out towards + infinity again above the real line. That would just change the sign of the integral without changing its absolute value.
@@FireSermon83 Perfect. Many thanks for your help.
Great lecture
Thanks! I’m glad you enjoyed it!