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Gravity and Strings

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3.1 Scalar SRFTs of gravity 49<br />

via the Noether theorem for global transformations: using the infinitesimal form of the<br />

Poincaré transformations,<br />

we obtain the conservation law<br />

δx µ = σ µ νx ν + σ µ , σ µν =−σ νµ , (3.12)<br />

dJ(σ )<br />

dξ = 0, J(σ ) = PµδX µ . (3.13)<br />

The conserved quantity associated with translations is the linear momentum J(σ µ ) ∼ P µ<br />

<strong>and</strong> the conserved quantity associated with Lorentz transformations is the angular momentum<br />

J(σ µν ) ∼ M µν .<br />

Observe that the invariance of the action is due to the fact that it depends only on the<br />

derivatives of the coordinates. In particular, the Minkowski metric does not depend on<br />

the coordinates. A better way to express this fact is to say that the Minkowski metric has<br />

d(d + 1)/2 independent isometries that generate the d-dimensional Poincaré group. This<br />

association between spacetime isometries <strong>and</strong> conserved quantities will still hold in more<br />

complicated spacetimes.<br />

This action is also invariant under non-singular reparametrizations of the worldline ξ ′ (ξ).<br />

These are local (gauge) transformations that infinitesimally can be written δξ = ɛ(ξ).Taking<br />

into account that the X µ s are scalars with respect to these transformations, we find<br />

δξ = ɛ(ξ), δdξ =˙ɛdξ, ˜δX µ = 0, δ ˙X µ =−˙ɛ ˙X µ , (3.14)<br />

<strong>and</strong> it is a simple exercise to check that ˜δS = 0 identically. If we now consider the variation<br />

of the action under just<br />

δX µ =−ɛ ˙X µ , (3.15)<br />

we find that it is invariant only up to a total derivative<br />

<br />

δS = dξ d<br />

<br />

Mcɛ<br />

dξ<br />

ηµν<br />

˙X µ ˙X ν<br />

<br />

. (3.16)<br />

On varying the action with respect to general variations of the coordinates first <strong>and</strong> integrating<br />

by parts, we obtain<br />

<br />

δS = dξ ɛ δS<br />

δX ν ˙X ν + d<br />

<br />

Mcɛ ηµν ˙X µ ˙X ν . (3.17)<br />

dξ<br />

By equating the two results <strong>and</strong> taking into account that the equation is valid for arbitrary<br />

functions ɛ(ξ),weobtain the gauge identity<br />

δS<br />

δX ν ˙X ν = 0, ⇒ ˙Pν ˙X ν = 0, (3.18)<br />

which is satisfied off-shell (trivially on-shell). Since ˙X ν is proportional to the momentum,<br />

this identity is proportional to<br />

d(P µ Pµ)<br />

= 0. (3.19)<br />

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