04.06.2013 Views

Gravity and Strings

Gravity and Strings

Gravity and Strings

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

28 Noether’s theorems<br />

where x ′ <strong>and</strong> x st<strong>and</strong> for the coordinates of the same point in the two different coordinate<br />

systems. The transformation of the fields may contain terms associated with the coordinate<br />

transformations <strong>and</strong> also with other “internal” transformations (see footnote 3 in Chapter 1).<br />

We want to find the consequences of the invariance, possibly up to a total derivative that<br />

depends on the variations, of the action Eq. (2.1) under the above infinitesimal changes of<br />

the field <strong>and</strong> the coordinates (which are, then, symmetry transformations). We express this<br />

invariance as follows:<br />

<br />

˜δS = d d x ∂µs µ (˜δ). (2.9)<br />

<br />

Let us now perform directly the variation of the action explicitly, 2<br />

<br />

˜δS =<br />

<br />

˜δd d x L + d d <br />

x ˜δL . (2.10)<br />

We have<br />

˜δd d x = d d x ∂µ ˜δx µ ,<br />

<br />

˜δL = δL + ˜δx µ ∂µL, (2.11)<br />

δL = ∂L ∂L<br />

δϕ +<br />

∂ϕ ∂∂µϕ δ∂µϕ + ∂L<br />

∂∂µ∂νϕ δ∂µ∂νϕ + ···,<br />

where δ st<strong>and</strong>s for the variation of the field at two different points whose coordinates are<br />

the same in the two different coordinate systems considered,<br />

δϕ(x) ≡ ϕ ′ (x) − ϕ(x), (2.12)<br />

<strong>and</strong> we have used the field-operator identity<br />

˜δ = δ + ˜δx µ ∂µ. (2.13)<br />

δ <strong>and</strong> ∂µ commute since δ does not involve any change of coordinates. Thus<br />

<br />

˜δS = d<br />

<br />

d <br />

x ∂µ ˜δx µ L + ˜δx µ ∂µL + ∂L ∂L<br />

δϕ +<br />

∂ϕ ∂∂µϕ ∂µδϕ + ∂L<br />

∂∂µ∂νϕ ∂µ∂νδϕ<br />

<br />

+ ··· .<br />

(2.14)<br />

On integrating by parts as many times as necessary, we obtain<br />

<br />

˜δS = d<br />

<br />

d <br />

x ∂µ L˜δx µ <br />

∂L<br />

+<br />

∂∂µϕ<br />

− ∂ν<br />

<br />

∂L<br />

δϕ +<br />

∂∂µ∂νϕ<br />

∂L<br />

∂∂µ∂νϕ ∂νδϕ<br />

<br />

+ ··· + δS<br />

δϕ δϕ<br />

<br />

.<br />

(2.15)<br />

On reexpressing δϕ in terms of ˜δϕ inside the total derivative, <strong>and</strong> equating the result with<br />

Eq. (2.9), we arrive at <br />

d d <br />

x ∂µj µ<br />

N1 (˜δ) + δS<br />

δϕ δϕ<br />

<br />

= 0, (2.16)<br />

2 We follow here [795].

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!