Homework 3 s Free graviton Hamiltonian Show that the free graviton action we discussed in class (before making it gauge- and Lorentzinvariant), S 0 = α d 4 x µ h ij µ h ij, () yields the correct free Hamiltonian operator: H 0 =Λ+ d 3 p (π) 3 a σ( p)a σ ( p) p 0 ( p), with p 0 ( p) = p, () σ=± where Λ is an infinite constant, if α =. Follow the standard derivation, which you can find e.g. for a scalar field in any QFT textbook. You will need the commutation relations: aσ ( p ),a σ ( p ) =(π) 3 δ 3 ( p p ) δ σ σ (3) Notice that with this normalization of single-particle states the correct field expansion is h µν (x) = d 3 p e µν (π) 3 p 0 σ ( p)a σ ( p) e ip x +h.c., (4) σ=± which is a factor of (π) 3/ different from what you have in your notes (sorry for the mismatch!). The canonical momentum of h ij following from the action () is Π ij = L = αḣij. (5) ḣij The Hamiltonian therefore is H 0 = d 3 x Π ij ḣ ij L = α Plugging the field operator (4) into H 0 we get H 0 = α d 3 d 3 p x (π) 3 p 0 d 3 p (π) 3 p 0 d 3 x ḣ ij + k h ij k h ij. (6) ( ip 0 e ij σ (p)a σ(p) e ip x +h.c. )( ip 0 e ij σ (p )a σ (p ) e ip x +h.c. ) σ,σ + ( ip k e ij σ (p)a σ(p) e ip x +h.c. )( ip k e ij σ (p )a σ (p ) e ip x +h.c. ). (7)
If we expand the product, the integral in x selects the zero-momentum mode of each term: d 3 xe i q x =(π) δ 3( q ). (8) This makes the aa terms and the a a ones vanish. Indeed these appear in the integral multiplied by e i(p+p ) x and e i(p+p ) x, respectively. The integral in x then selects p = p, which implies p 0 = p 0 = p. This yields a cancellation of the aa and a a terms between the second and the third line of eq. (7). We thus need consider just the aa and a a terms. They will appear with identical coefficients, because they are the hermitian conjugate of each other, and the Hamiltonian is hermitian. We get H 0 = α = α d 3 p (π) 3 p 0 σ,σ (p 0 + p k p k ) e ij σ (p)eij σ (p) a σ (p)a σ (p)+h.c. d 3 p (π) 3 p0 σ,σ e ij σ (p)e ij σ (p) a σ (p)a σ (p)+h.c. (0) where in the first step we integrated in x and p, and in the second we used that p 0 = p. Nowwe make use of the fact that, from their very definition, the polarization tensors for given momentum p obey e ij σ (p)e ij σ (p) =δ σσ, () which gets rid of the σ sum in H 0. Also we rewrite the h.c. part as the other one plus the commutator (3), which gives an additive infinite constant, formally defined as Λ αδ 3 (0) d 3 pp 0. () (9) Overall we get H 0 =Λ+ which for α = / matches eq. (). σ=± d 3 p (π) 3 a σ ( p)a σ( p) p 0, (3) Einstein-Hilbert action The Lorentz-invariant (and gauge-invariant) action for free gravitons we derived in class is: Sg free = d 4 x α h µν α h µν ( µ h µν ) + ν h µ h µν ( µ h). (4) Show that this is precisely what you get up to an overall proportionality factor if you expand the Einstein-Hilbert action d 4 x gr (5) 6πG
for the metric g µν η µν + h µν up to quadratic order in h µν. What happened to the first-order term? As usual expanding the determinant of a matrix is straightforward if you recall det(a + δa) = exp log det(a + δa) =exp Tr(log(A + δa)) = (deta) exp Tr(log( + A δa) (6) (valid for any symmetric, real, invertible A), and the expansions for the log and exp: log( + x) =x x +..., exp x =+x + x +... (7) To expand gr up to second order in h µν, we need R up to second order, but g up to first order only, because its second order piece would multiply R at zeroth order, that is the Ricci scalar for flat space, which vanishes. Therefore, for g we just need (see eq. (6)) g =+ hµ µ + O(h ) (8) with the understanding that all contractions are now done via the flat metric η µν.forr we have R g µν R µν = η µν R () µν + η µν R () µν h µν R () µν + O(h 3 ), (9) where R µν () and R µν () are the first- and second-order terms in the Ricci tensor, respectively, and we used that the inverse metric is g µν = η µν h µν + O(h ). (0) The Ricci tensor is R µν = R α µαν = α Γ α µν ν Γ α µα +Γ α βαγ β µν Γ α βνγ β µα. () The Christoffel symbols start at linear order in h µν (they vanish in flat space), so that schematically we have R µν () Γ () () R µν () Γ () +Γ () Γ (). (3) Given their definition, We have Γ µ αβ = gµν( ) α g νβ + β g να ν g αβ. (4) ( ) Γ ()µ αβ = α h µ β + β h ν α µ h αβ (5) Γ ()µ αβ = hµν( ) α h νβ + β h να ν h αβ (6) 3
After a straightforward computation we get R () µν = α µ h να + α ν h µα h µν µ ν h (7) R µν () = ( ) µh αβ ν h αβ + β h να β h α µ α h β µ (8) ( ) + h αβ µ ν h αβ + α β h µν β ν h α µ β µ h α ν (9) ( α h αβ β h )( ) µ h νβ + ν h µβ β h µν, (30) where as usual we defined h = h µ µ. Putting everything together we finally have S EH g µν = d 4 x gr (3) 6πG d 4 x ( )( + 6πG hµ µ η µν R µν () + ηµν R µν () ) hµν R µν () (3) d 4 x ( µ ν h µν h ) (33) 6πG + 3πG Sfree g h µν, (34) where to get the last line we had to perform some integration by parts in the quadratic terms. The linear piece, eq. (33), is a total derivative, and thus vanishes if we integrate by parts. In fact the linear piece has to vanish, for the background metric η µν solves the vacuum e.o.m., which means that it is astationary point of the action. In other words, under η µν η µν + h µν, at linear order in h µν the action does not change. Eq. (34) shows that the geometric h µν is related to the canonically normalized graviton field H µν by h µν = κh µν, κ 3πG. (35) 3 Einstein s equations Consider our free graviton action, eq. (4).. Perform an infinitesimal variation h µν (x) h µν (x)+δh µν (x) in eq. (4) and compute the contribution of Sg free to the equation of motion for h µν,thatis δs free g δh µν (x). (36) Pay attention to the symmetry of h µν as well as of δh µν. If you are left with a variation of the form d 4 xδh µν A µν (37) 4
before concluding that this is zero for any variation δh µν (x) if and only if A µν vanishes, you have to symmetrize A µν, because the antisymmetric part of A µν gives a vanishing contribution to the above integral regardless.. Show that eq. (40) is precisely up to an overall proportionality factor the expansion of the Einstein tensor G µν at linear order in h µν. That is, at the order we are working at, our action (4) correctly yields the l.h.s. of Einstein s equations.. Under an infinitesimal variation h µν (x) h µν (x)+δh µν (x) the action changes by δsg free = d 4 x α h µν α (δh µν ) 4 µ h µν α (δh αν ) (38) + ν h µ (δh µν )+ µ h µν ν (δh) µ h µ (δh). (39) Upon integrating by parts, rewriting δh explicitly as δh µν η µν, and renaming some of the indices, we get δsg free =+ d 4 xδh µν h µν 4 µ α h αν +η µν α β h αβ + µ ν h η µν h. (40) As we argued, before removing the δh µν to get the e.o.m., we want to have a symmetric tensor multiplying δh µν. All terms in brackets are symmetric but the second. We remedy this by symmetrizing it We finally get 4 δh µν µ α h αν = δh µν ( µ α h αν + ν α h αµ) (4) δs free g δh µν (x) = hµν µ α h αν ν α h αµ + µ ν h + η µν( α β h αβ h ). (4). To get the linear order approximation for G µν, we use our results of Problem : G () µν = ( R µν g µν R ) () (43) = R µν () η µνη αβ R () αβ (44) ( = hµν µ α h αν ν α h αµ + µ ν h + η µν α β h αβ h ), (45) which is exactly eq. (40), with an extra / overallfactor. 5