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 An. An=A and B=An−1 as well as B=An+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. An+1,An−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 An−1=An+1 and so on.
side note, the graph was generated with parameters as simple as I could make them, a,k, and m1 being unity, and m2=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 (m1+m2)2m1m2 should be (m1+m2m1m2)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 An. An=A and B=An−1 as well as B=An+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. An+1,An−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 An−1=An+1 and so on.
ReplyDeleteside note, the graph was generated with parameters as simple as I could make them, a,k, and m1 being unity, and m2=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 (m1+m2)2m1m2 should be (m1+m2m1m2)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.
Delete