Quantum Gravity

80 PARAMETRIZED AND RELATIONAL SYSTEMSIn order that only a surface term remains one has to chooseδN(τ) =˙ɛ(τ) . (3.32)This leads to[ (δS = ɛ(τ) p ∂H )] τ2S∂p − H S . (3.33)τ 1Since the term in brackets gives p 2 − m 2 ≠ 0, one must demandɛ(τ 1 )=0=ɛ(τ 2 ) , (3.34)that is, the boundaries must not be transformed.We note that for a constraint of the form H S = α(x)p, the term in bracketswould vanish and there would be in this case no restriction at the boundaries.Constraints of this form arise in electrodynamics and Yang–Mills theories (Gaussconstraint) provided the sources are treated dynamically too (otherwise, theconstraint would no longer be homogeneous in the momenta, see e.g. ∇E = ρ inelectrodynamics).We shall now show how the gauge can be fixed for the relativistic particle.If a gauge is independent of the lapse function N, it is called ‘canonical gauge’,otherwise it is called ‘non-canonical’. Consider first a canonical gauge,χ(x, p, τ) ≈ 0 . (3.35)An example would be x 0 −τ ≈ 0 (such a gauge was used in the deparametrizationof the non-relativistic particle, see the paragraph before (3.12)). A potentialproblem is that (3.35) holds at all times, including the endpoints, and may thusbe in conflict with ɛ(τ 1 )=ɛ(τ 2 ) = 0—since there is no gauge freedom at theendpoints, χ ≈ 0 could restrict physically relevant degrees of freedom.For reparametrization-invariant systems, a canonical gauge must depend explicitlyon τ. From the condition that (3.35) be invariant under time evolution,0 ≈ dχdτ = ∂χ∂τ + N{χ, H S} ,a τ-independent gauge χ would lead to the unacceptable value N = 0 (‘freezing’of the motion). (In order for the gauge to break the reparametrization invariancegenerated by H S , {χ, H S } must be non-vanishing.) For the relativistic particle,this yields0 ≈ ∂χ∂τ + N ∂χ ∂H S∂x µ = ∂χ ∂χ+2Npµ∂p µ ∂τ ∂x µ .For the example x 0 − τ ≈ 0, one has N =1/2p 0 , in accordance with (3.28).To avoid potential problems with the boundary, one can look for an equationof second order in ɛ (since there are two conditions ɛ(τ 1 )=0=ɛ(τ 2 )). As x andp transform proportional to ɛ, one would have to involve ẍ or ¨p, which would