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This blog is for comments and reports of errors on my site. I will also post notifications of new articles published on the main site using this blog, so if you want  to be kept up to date, check here regularly, or else  subscribe to it for email notifications. I realize this may look like a roundabout way of organizing things, but I don’t trust WordPress with anything substantial any more.

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123 Responses to Comments for

  1. Anonymous says:

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


    • growescience says:

      Try reloading 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.


  2. Justin says:

    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.


    • growescience says:

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


  3. soroush says:

    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!


    • growescience says:

      Fixed now. Thanks.
      As for equation 12, in the textbook it says to show that “it is enough if \psi obeys” Shankar’s equation 12.3.6 (my equation 12), which is certainly true, since then the value of \left.\psi_{1}^{*}\psi_{2}\right|_{0}^{2\pi} in equation 11 is zero.


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