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Unformatted text preview: PHY 5667 Problem Set no. 9 solution Problem 1 (i) As in the case of the spin-1 field, where we needed constraints to get rid of the spin-0 part, now the question is which are the contraints on a tensor ij with 9 components that get rid of the spin-0 part (1 component) and the spin-1 part (3 components). For this, notice that any 3 3-matrix 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 spin-2, spin-1 and spin-0 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 basis-independent. Therefore it can be identified as the spin-0 part. The part 1 2 ( A ij A ji ) can be contracted using a Levi-Civita 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 spin-1 part. The part 1 2 ( A ij + A ji 2 3 ij A kk ) then must be the missing spin-2 part. It has 5 components, as it should. Therefore, the spin-2 part of any matrix has to be traceless (to project out spin- 0 parts) and symmetric (to project out spin-1 parts). You could also think of the symmetry condition as the consequence of fact that a spin-two object can be constructed as the symmetric combination of two spin-one objects. The two conditions on the polarization tensors ij of spin-two 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 off-diagonal 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