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Unformatted text preview: PHY 5667 Problem Set no. 9 solution Problem 1 (i) As in the case of the spin1 field, where we needed constraints to get rid of the spin0 part, now the question is which are the contraints on a tensor ij with 9 components that get rid of the spin0 part (1 component) and the spin1 part (3 components). For this, notice that any 3 3matrix A ij can be written as A ij = 1 2 parenleftbigg A ij + A ji 2 3 ij A kk parenrightbigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright spin 2 + 1 2 ( A ij A ji ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright spin 1 + 1 3 ij A kk bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright spin 0 (1) The identification of the various parts with the spin2, spin1 and spin0 has yet to be proven. For this, we need to consider how the various parts transform under rotation of the coordinate basis ( A i j = S T i i A ij S jj , where S T = S 1 ) A kk transforms into itself as the trace of a matrix is always basisindependent. Therefore it can be identified as the spin0 part. The part 1 2 ( A ij A ji ) can be contracted using a LeviCivita tensor v k 1 2 kij ( A ij A ji ) = kij A ij . Then: v k v k = k i j A i j = S T k k S T i i S T j j S T i m S T j n kij A mn = S T k k S ii S T i m bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright im S jj S T j n bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright jn kij A mn = S T k k kij A ij = S T k k v k Which means that v k in fact transforms like a vector, and we can identify 1 2 ( A ij A ji ) with the spin1 part. The part 1 2 ( A ij + A ji 2 3 ij A kk ) then must be the missing spin2 part. It has 5 components, as it should. Therefore, the spin2 part of any matrix has to be traceless (to project out spin 0 parts) and symmetric (to project out spin1 parts). You could also think of the symmetry condition as the consequence of fact that a spintwo object can be constructed as the symmetric combination of two spinone objects. The two conditions on the polarization tensors ij of spintwo particles therefore are i ii = 0 ij = ji for all i, j 1 (ii) We find three indepent symmetric matrices with all diagonal element being zero (such it is automatically traceless) 1 = 1 2 1 1 2 = 1 2 1 1 3 = 1 2 1 1 and two independent traceless matrices with no offdiagonal elements (such automatically symmetric)) 4 = 1 2 1 1 5 = 1 6 1 1 2 These matrices fulfill...
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This note was uploaded on 12/14/2011 for the course PHY 5667 taught by Professor Okui during the Fall '10 term at FSU.
 Fall '10
 Okui

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