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Gravity and Strings

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76 A perturbative introduction to general relativity<br />

The 1/r term is not present27 in the Newtonian case28 <strong>and</strong>, as we did before, we exp<strong>and</strong> W<br />

around its Newtonian value,<br />

<br />

ω 2 W ∼ dr −<br />

c<br />

l2<br />

<br />

∂<br />

ω 2 <br />

ω<br />

2 x l2<br />

+ 2RS dr + −<br />

r 2 ∂x c c r r 2<br />

<br />

<br />

<br />

<br />

<br />

x=0<br />

∼ WNewtonian + RSω<br />

<br />

1<br />

dr <br />

c r 2 − ρ2 ∼ WNewtonian + RSω<br />

c arccosh<br />

<br />

r<br />

, (3.164)<br />

ρ<br />

where ρ = cl/ω is clearly the minimal value of r in the path of the massless particle. Following<br />

[644], the variation of W when the particle starts from r = R≫ρ, goes through<br />

r = ρ, <strong>and</strong> again reaches r = R is<br />

<strong>and</strong>, according to Eq. (3.156),<br />

W ∼ WNewtonian + 2RSω<br />

c<br />

ϕ ∼ ∂<br />

∂l WNewtonian + 2R 1<br />

√<br />

ρ 1 − ρ/R<br />

<br />

R<br />

arccosh , (3.165)<br />

ρ<br />

R→∞<br />

−→ π + 2R<br />

, (3.166)<br />

ρ<br />

<strong>and</strong> we find that the deviation from the Newtonian value ϕ = π (which means simply<br />

no bending of the light ray) is δϕ = 2R/ρ, ingood agreement with observation. This is an<br />

encouraging result, which indicates that we have found a reasonable relativistic theory of<br />

gravitation worth studying in more detail.<br />

At this point, we remember that we still have to answer the second question posed on<br />

page 67. The answer will prompt us to seek <strong>and</strong> introduce into our theory corrections that<br />

will make the prediction for the precession of the perihelion of Mercury agree completely<br />

with observations.<br />

3.2.4 The consistency problem<br />

The answer to the second question formulated on page 67 is that, in general, the matter<br />

energy–momentum tensor derived from the free-matter Lagrangian is no longer conserved.<br />

As explained in Chapter 2, the divergence of the energy–momentum tensor is proportional<br />

to the equations of motion derived from the same Lagrangian, but the coupling to gravity<br />

changes these equations. This can be seen in the modified massive-particle action of the<br />

above example but the real scalar field which we studied in Chapter 2 will, however, make<br />

a better example.<br />

27 There are also 1/r 2 corrections, but we take only the most important one.<br />

28 The Newtonian case corresponds to a free massive particle (i.e. vanishing gravitational potential energy)<br />

moving at the speed of light with 2mE ′ = (ω/c) 2 .

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