Simplifying the geodesic equation

Guth's review of Lecture 13 (slide 1) adds one word: the observers who give $ds^2$ its meaning are local free-falling observers. The equivalence principle cancels gravity only in a small region, so the special-relativity measurements behind $ds^2$ are made by an observer arbitrarily close to the events. A special case of "$ds^2 < 0$" is someone reading their own wristwatch, which is always at the same place relative to them.

Asked whether time always enters as just $-c^2dt^2$, Guth says no: that relied on isotropy. To describe the real, lumpy universe rather than the ideal one, you would probably need $dt\,dr$ terms (transcript p. 3).

Why care about geodesics? Because according to general relativity, particles with no forces on them except gravity move along geodesics of spacetime. Applying this to the Robertson–Walker metric is an 8.286 homework problem; later in the lecture (p. 13) Guth says it shows how massive particles slow down as they travel through the expanding universe, just as photons lose energy.

Guth adds some history (p. 5). The method of varying a path, the calculus of variations, dates from 1696, when Johann Bernoulli posed the brachistochrone problem: the curve along which a frictionless bead slides between two points in the least time. The answer is a cycloid, the same curve as the closed matter-dominated universe. Newton is said to have solved it overnight after a day at the Royal Mint. [Guth has him coming home at 4 a.m.; the usual version of the story has him home at 4 p.m. and done by 4 a.m.]

The parameter $\lambda$ was arbitrary, and the equation holds for every choice. So pick the most convenient one: let $\lambda$ be the arc length $s$ along the path. Since $ds^2 = A\,d\lambda^2$, this makes $A = 1$. Guth stresses that what really matters is that $A$ is constant: then it is not differentiated, and it cancels between the two sides. The square roots disappear:

Guth boxes this: it is the final geodesic equation, and for solving problems it is almost always the best form.

Most books instead give a formula for $d^2x^i/ds^2$. To get it, expand the left side with the product rule. The metric depends on $s$ only through the position $x^k(s)$, so the chain rule gives $\frac{d}{ds}g_{ij} = \partial_k g_{ij}\,\frac{dx^k}{ds}$, where $\partial_k$ is shorthand for $\partial/\partial x^k$. Moving that term to the right:

(On the board this took several attempts — students caught a misplaced bracket and two wrong indices before it came out right.)

The left side is a matrix $g_{ij}$ times a vector $d^2x^j/ds^2$. To isolate the vector, multiply by the inverse matrix. In general relativity the inverse metric is written with upper indices, $g^{ij}$:

Summing over the neighbouring index $l$ is exactly matrix multiplication, and $\delta^i_j$ (1 if $i = j$, 0 otherwise) is the identity matrix.

Multiplying through by $g^{il}$ and renaming dummy indices gives the textbook form. The coefficients of the two first derivatives get their own symbol:

$\Gamma^i_{jk}$ is the affine connection (Weinberg's name), also called the Christoffel symbol, and — according to Sean Carroll's book, which Guth quotes — sometimes the Riemann or Levi-Civita connection. Lots of names, Guth jokes, presumably because lots of people invented it. Everyone agrees on the definition up to a sign, and books differ on that sign. [The transcript does not make the sign on the board audible; with $\Gamma$ defined as above, the derivation gives the minus sign shown, which is the most common convention.]

For Guth this form is "window dressing" to connect with other books; in practice the arc-length form above is usually easier to solve.

The graph shows what the latitude circle loses. Between two points on the same latitude, the geodesic — a great circle — bulges towards the pole and is shorter.

A particle moves along a timelike path. So instead of $s$ it is natural to use proper time $\tau$, defined by $ds^2 = -c^2d\tau^2$, and — a near-universal convention — Greek indices $\mu, \nu, \lambda, \sigma$ for spacetime (with $0$ for time), keeping Latin $i, j, k$ for space alone. The geodesic equation becomes

[Guth says "the partial of $g_{\lambda\sigma}$ with respect to $x^\nu$"; the free index must be $\mu$, as written here.]

Nothing in the derivation used a positive-definite metric, so it goes through unchanged. The difference is that spacetime geodesics can be maxima or other stationary points as well as minima.

• Choosing the arc length as parameter makes $A = 1$ (constant is what matters), giving $\frac{d}{ds}\left[g_{ij}\frac{dx^j}{ds}\right] = \frac12\partial_i g_{jk}\frac{dx^j}{ds}\frac{dx^k}{ds}$, usually the best form for solving problems.
• The inverse metric $g^{ij}$ satisfies $g^{il}g_{lj} = \delta^i_j$; with it, $\frac{d^2x^i}{ds^2} = -\Gamma^i_{jk}\frac{dx^j}{ds}\frac{dx^k}{ds}$, where $\Gamma^i_{jk} = \frac12 g^{il}(\partial_j g_{lk} + \partial_k g_{lj} - \partial_l g_{jk})$ is the Christoffel symbol (sign conventions vary).
• For particles, use proper time $\tau$ and Greek indices: $\frac{d}{d\tau}\left[g_{\mu\nu}\frac{dx^\nu}{d\tau}\right] = \frac12\partial_\mu g_{\lambda\sigma}\frac{dx^\lambda}{d\tau}\frac{dx^\sigma}{d\tau}$.