{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW9_soln

# HW9_soln - PHY 5667 Problem Set no 9 solution Problem 1(i...

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

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 ε 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 δ ij A kk parenrightbigg bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright spin 2 + 1 2 ( A ij A ji ) bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright spin 1 + 1 3 δ 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 ε kij ( A ij A ji ) = ε kij A ij . Then: v k v k = ε 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 ε kij A mn = S T k k S ii S T i m bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright δ im S jj S T j n bracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright δ jn ε kij A mn = S T k k ε 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 δ 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 ε ij of spin-two particles therefore are i ε ii = 0 ε ij = ε ji for all i, j 1

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

View Full Document
(ii) We find three indepent symmetric matrices with all diagonal element being zero (such it is automatically traceless) ε 1 = 1 2 0 1 0 1 0 0 0 0 0 ε 2 = 1 2 0 0 1 0 0 0 1 0 0 ε 3 = 1 2 0 0 0 0 0 1 0 1 0 and two independent traceless matrices with no off-diagonal elements (such automatically symmetric)) ε 4 = 1 2 1 0 0 0 1 0 0 0 0 ε 5 = 1 6 1 0 0 0 1 0 0 0 2 These matrices fulfill ij ε ij s ε ij s = Tr( ε s ε s ) = δ ss .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 6

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

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

View Full Document
Ask a homework question - tutors are online