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I saw your message on Physics Pages that you had moved it to a new hosting company. Then, I noticed that the link (https://physicspages.com/shankar.html) of the solutions to quantum mechanics by Shankar didn’t work. Please check this link and fix it. Thank you very much.

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Try reloading physicspages.com and then try the link to Shankar’s book again. It should work now. If not, you’ll need to clear your browser’s cache, then try again.

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Regarding homework problem P12.7 in Thomas Moore’s A General Relativity Workbook.

You draw a graph of Sin (critical emission angle) for both (+) and (-) and claim in Equation (36) that it is well defined for all r > 0. The quantity in brackets [ ] in the case of (-) is 0 when r = 2GM. But, we need to take [ ] ^ -1. The total expression is 0/0 which is undefined. I agree with you that we should be using the (+) form of the emission formula for r < 3GM. In which case, we can avoid this problem since the + form is well defined at r = 2GM, to w.i.t. 0 for the emission angle as you correctly compute. But, your blue graph should have an undefined point at r = 2GM. It shows an answer of about 0.7, which I'm unable to understand how you calculated, and I do not think is correct.

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Thanks for the comment. I’ll try to have a look. In the meantime I’ve added your comment to the original post.

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In your solution to Shankar’s Quantum Mechanics, problem 12.03.01, equation 9 and the subsequent ones, you’re missing a factor of rho in the integrand. (The rho is also written in the question, I think it’s a typo)

Also, I think the paragraph after equation 11 does not necessarily prove equation 12.

Thanks for your detailed solutions!

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Fixed now. Thanks.

As for equation 12, in the textbook it says to show that “it is enough if obeys” Shankar’s equation 12.3.6 (my equation 12), which is certainly true, since then the value of in equation 11 is zero.

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