The book / appendix e
APPENDIX E

Index practice and three extra calculations

E.1 An index-reading checklist#

Before calculating, read the expression aloud. In AμBμA^\mu B_\mu, one index appears upstairs and downstairs, so sum over it and obtain a scalar. In CμνvνC^\mu{}_\nu v^\nu, the ν\nu is summed and the μ\mu remains free: the result is a vector with one upper index.

Check A valid example The mistake it prevents
Free indices match on both sides aμ=bμa^\mu=b^\mu Equating objects of different tensor type.
A dummy appears twice in a term AμBμA^\mu B_\mu An ambiguous threefold repetition.
Rename an entire dummy pair AμBμ=AαBαA^\mu B_\mu=A^\alpha B_\alpha Changing only half a contraction.
An inverse is a matrix inverse gμαgαν=δμνg^{\mu\alpha}g_{\alpha\nu}=\delta^\mu{}_\nu Taking elementwise reciprocals.
Raising uses the metric vμ=gμνvνv^\mu=g^{\mu\nu}v_\nu Changing an index position without applying the metric.
A trace knows the dimension δμμ=4\delta^\mu{}_\mu=4 Forgetting that the repeated pair is summed.

For an antisymmetric AμνA^{\mu\nu} and symmetric SμνS_{\mu\nu}, the contraction vanishes. Rename μν\mu\leftrightarrow\nu throughout: SμνAμν=SνμAνμ=SμνAμνS_{\mu\nu}A^{\mu\nu}=S_{\nu\mu}A^{\nu\mu}=-S_{\mu\nu}A^{\mu\nu}. A number equal to its own negative is zero. This small argument removes many apparently complicated terms.

E.2 A scalar field: when is it dust, and when is it not?#

Problem. In units c==1c=\hbar=1, a homogeneous canonical scalar field has ϵ=ϕ˙2/2+V(ϕ)\epsilon=\dot\phi^2/2+V(\phi) and p=ϕ˙2/2V(ϕ)p=\dot\phi^2/2-V(\phi). Find its equation of state when V=0V=0. Then explain how an oscillating massive scalar can instead act like dust.

Work it out, then reveal the solution

If V=0V=0 and the field has nonzero kinetic energy, p=ϵp=\epsilon, so w=p/ϵ=1w=p/\epsilon=1. This is called stiff matter, not dust. The continuity equation then gives ϵa6\epsilon\propto a^{-6}.

For a quadratic potential V=m2ϕ2/2V=m^2\phi^2/2, and oscillations much faster than cosmic expansion, a cycle average has ϕ˙2/2=V\langle\dot\phi^2/2\rangle=\langle V\rangle. One way to see it is to approximate ϕ=Acos(mt)\phi=A\cos(mt) over a cycle: sine squared and cosine squared have equal averages. Thus p0\langle p\rangle\simeq0 while ϵ>0\langle\epsilon\rangle>0, the dust-like result. It requires the massive potential and the separation of timescales. A field dominated by a slowly varying potential instead has pϵp\simeq-\epsilon.

E.3 Trace reversal in any dimension#

Problem. Let spacetime have dimension nn. Take the trace of RμνRgμν/2+Λgμν=κTμνR_{\mu\nu}-Rg_{\mu\nu}/2+\Lambda g_{\mu\nu}=\kappa T_{\mu\nu}, then solve for RμνR_{\mu\nu} when n2n\ne2.

Reveal the contraction, one step at a time

Contract with gμνg^{\mu\nu}. Since gμνgμν=ng^{\mu\nu}g_{\mu\nu}=n, we get (1n/2)R+nΛ=κT(1-n/2)R+n\Lambda=\kappa T. Therefore R=2(nΛκT)/(n2)R=2(n\Lambda-\kappa T)/(n-2). Substitute it back:

Rμν=κ(TμνTn2gμν)+2Λn2gμν.R_{\mu\nu}=\kappa\left(T_{\mu\nu}-\frac{T}{n-2}g_{\mu\nu}\right) +\frac{2\Lambda}{n-2}g_{\mu\nu}.

In four dimensions this recovers the familiar half-trace term. In two dimensions division by n2n-2 is forbidden. Instead the trace equation gives 2Λ=κT2\Lambda=\kappa T, and the Einstein tensor vanishes identically. Other two-dimensional gravity theories can still have dynamics; this result concerns the Einstein–Hilbert metric theory.

E.4 Pressure and light bending are different questions#

Problem. In an isotropic rest frame with Λ=0\Lambda=0, compare the initial small-ball focusing source for dust with that for radiation at the same energy density ϵ\epsilon. Does this derive the factor of two in deflection of a ray by the Sun?

Reveal the answer and the important distinction

For an initially comoving infinitesimal ball, the rest-frame focusing source is proportional to ϵ+3p\epsilon+3p. Dust has p=0p=0 and isotropic radiation has p=ϵ/3p=\epsilon/3, so their sources are ϵ\epsilon and 2ϵ2\epsilon respectively. This is a statement about different matter stress tensors sourcing Ricci curvature.

Deflection of a ray around the Sun is a different problem: the exterior is approximately vacuum and its metric has both temporal and spatial weak-field contributions. Their combined effect gives twice the deflection obtained from keeping the temporal contribution alone. The two factors of two should not be identified as the same derivation. Nor does the pressure of radiation double the total mass of a sealed photon box; the container’s stresses must be included.

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