RevModPhys.84.671

# m2 h 0 thus the equation of motion 210 implies

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Unformatted text preview: motion (2.10) implies the three equations ðh À m2 Þh ¼ 0; @ h ¼ 0; h ¼ 0: (2.11) Conversely, it is straightforward to see that these three equations imply the equation of motion (2.10), so Eqs. (2.10) and (2.11) are equivalent. The form (2.11) makes it easy to count the degrees of freedom as well. For D ¼ 4, the ﬁrst of Eq. (2.11) is an evolution equation for the ten components of the symmetric tensor h , and the last two are constraint equations on the initial conditions and velocities of h . The last determines the trace completely, killing 1 real space degree of freedom. The second gives four initial value constraints, and the vanishing of its time derivative, Kurt Hinterbichler: Theoretical aspects of massive gravity i.e., demanding that it be preserved in time, implies four more initial value constraints, thus killing 4 real space degrees of freedom. In total, we are left with the 5 real space degrees of freedom of a four-dimensional spin 2 particle, in agreement with the Hamiltonian analysis of Sec. II.A. The ﬁrst equation in Eq. (2.11) is the standard KleinGordon equation, with the general solution Z dd p h ðxÞ ¼ qﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ½h ðpÞeipÁx þ hÃ ðpÞeÀipÁx : ð2Þd 2!p (2.12) pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ Here p are the spatial momenta, !p ¼ p2 þ m2 , and the D momenta p are on shell so that p ¼ ð!p ; pÞ. Next we expand the Fourier coefﬁcients h ðpÞ over some basis of symmetric tensors, indexed by , &quot; h ðpÞ ¼ ap;  ðp; Þ: (2.13) We ﬁx the momentum dependence of the basis elements &quot; &quot;  ðp; Þ by choosing some basis  ðk; Þ at the standard momentum k ¼ ðm; 0; 0; 0; . . .Þ and then acting with some standard boost3 LðpÞ, which takes k into p, p ¼ LðpÞ  k . This standard boost chooses for us the basis at p, relative to that at k. Thus we have &quot; &quot;  ðp; Þ ¼ LðpÞ LðpÞ  ðk; Þ: (2.15) Imposing the conditions @ h ¼ 0 and h ¼ 0 on Eq. (2.12) then reduces to imposing &quot; k  ðk; Þ ¼ 0; &quot;   ðk; Þ ¼ 0: (2.16) &quot; &quot; The ﬁrst says that  ðk; Þ is purely spatial, i.e., 0 ðk; Þ ¼ &quot;i 0. The second says that it is traceless, so that i ðk; Þ ¼ 0 also. Thus the basis need only be a basis of symmetric traceless spatial tensors,  ¼ 1; . . . ; dðd þ 1Þ=2 À 1. We demand that the basis be orthonormal, &quot;  &quot;  ðk; ÞÃ ðk; 0 Þ ¼ 0 : (2.17) This basis forms the symmetric traceless representation of the rotation group SOðdÞ, which is the little group for a massive particle in D dimensions. If R  is a spatial rotation, we have 0 &quot; &quot; R 0 R 0  ðk; 0 Þ ¼ R   ðk; 0 Þ; (2.18) 677 0 where R  is the symmetric traceless tensor representation of R 0 . We are free to use any other basis  ðk; Þ, related to &quot; the  ðk; Þ by 0 &quot; ...
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## This document was uploaded on 09/28/2013.

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