HW9_soln

HW9_soln - PHY 5667 Problem Set no. 9 solution Problem 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This note was uploaded on 12/14/2011 for the course PHY 5667 taught by Professor Okui during the Fall '10 term at FSU.

Page1 / 6

HW9_soln - PHY 5667 Problem Set no. 9 solution Problem 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online