Correct on both accounts! The + solutions are red, which are optical modes, and the - solutions are blue, corresponding to acoustic modes. I changed my coefficient convention partway through writing this and apparently forgot to change my notation from \(A_n\). \(A_n = A\) and \(B=A_{n-1}\) as well as \(B=A_{n+1}\) as you said--if I were more careful I would have added a step that signifies a particular coefficient per n value, e.g. \(A_{n+1}, A_{n-1}\), etc. and then made a note that each atom with the same mass will have the same coefficient by symmetry, reducing it to A and B, so that \(A_{n-1} = A_{n+1}\) and so on.
side note, the graph was generated with parameters as simple as I could make them, \(a\),\( k\), and \(m_1\) being unity, and \(m_2= 2\), so the masses would not be equal. The special case where the masses are equal gives the two branches meeting at the edges.
Also, i just noticed a typo in the second to last equation...the term \(\frac{(m_1+m_2)^2}{m_1m_2}\) should be \((\frac{m_1+m_2}{m_1m_2})^2\).
Thanks for the correction Patrick. I am currently working on calculating a density of states for our function. I will post to the blog once I obtain some results.
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ReplyDeleteCorrect on both accounts! The + solutions are red, which are optical modes, and the - solutions are blue, corresponding to acoustic modes. I changed my coefficient convention partway through writing this and apparently forgot to change my notation from \(A_n\). \(A_n = A\) and \(B=A_{n-1}\) as well as \(B=A_{n+1}\) as you said--if I were more careful I would have added a step that signifies a particular coefficient per n value, e.g. \(A_{n+1}, A_{n-1}\), etc. and then made a note that each atom with the same mass will have the same coefficient by symmetry, reducing it to A and B, so that \(A_{n-1} = A_{n+1}\) and so on.
ReplyDeleteside note, the graph was generated with parameters as simple as I could make them, \(a\),\( k\), and \(m_1\) being unity, and \(m_2= 2\), so the masses would not be equal. The special case where the masses are equal gives the two branches meeting at the edges.
ReplyDeleteAlso, i just noticed a typo in the second to last equation...the term \(\frac{(m_1+m_2)^2}{m_1m_2}\) should be \((\frac{m_1+m_2}{m_1m_2})^2\).
Thanks for the correction Patrick. I am currently working on calculating a density of states for our function. I will post to the blog once I obtain some results.
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