Feynman Path Integral Formulation

7.2 Lattice Weak Field Expansion and Transverse-Traceless Modes 229thus completely decoupling the (body diagonal) fluctuations ε 7 ,ε 11 ,ε 13 ,ε 14 . Thesein turn can now be integrated out, as they appear in the action with no ω (i.e. derivative)term. As a result the number of dynamical degrees of freedom has been reducedfrom 15 to 10, the same number as in the continuum.The remaining dynamics is thus encoded in the 10×10 dimensional matrix L ω =A 10 −18 1 BB† . By a second rotation, here affected by the matrix T , it can finally bebrought into the form˜L ω = T † L ω T = [ 8 − (Σ + ¯Σ) ]( )12β 0− C † C , (7.12)0 I 6with the matrix β given by⎛⎞1 −1 −1 −1β = 1 ⎜−1 1 −1 −1⎟2 ⎝⎠ . (7.13)−1 −1 1 −1−1 −1 −1 1The other matrix C appearing in the second term is given by⎛⎞f 1 0 0 0 ˜f 2˜f 4 0 ˜f 8 0 0⎜ 0 fC = 2 0 0 ˜f 1 0 ˜f 4 0 ˜f 8 0 ⎟⎝⎠ , (7.14)0 0 f 4 0 0 ˜f 1˜f 2 0 0 ˜f 80 0 0 f 8 0 0 0 ˜f 1˜f 2˜f 4with f i ≡ ω i − 1 and ˜f i ≡ 1 − ¯ω i . Furthermore Σ = ∑ i ω i , and for small momentaone finds8 − (Σ + ¯Σ) =8 −4∑i=1(e ik i+ e −ik i) ∼ k 2 + O(k 4 ) , (7.15)which shows that the surviving terms in the lattice action are indeed quadratic in k.The rotation matrix T involved in the last transformation is given by( )( )Ω4 β 0 I4 0T =, (7.16)0 I 6 Ω 6 γ I 6with Ω 4 = diag(ω 1 ,ω 2 ,ω 4 ,ω 8 ) and Ω 6 = diag(ω 1 ω 2 ,ω 1 ω 4 ,ω 2 ω 4 ,ω 1 ω 8 ,ω 2 ω 8 ,ω 4 ω 8 ), and⎛⎞0 0 1 10 1 0 1γ = − 1 1 0 0 12. (7.17)⎜0 1 1 0⎟⎝⎠1 0 1 01 1 0 0At this point one is finally ready for a comparison with the continuum result, namelywith the Lagrangian for pure gravity in the weak field limit as given in Eq. (1.7)