The Mathematical Details

To see why equation (2) is equivalent to the usual formulation of Einstein's equation, we need a bit of tensor calculus. In particular, we need to understand the Riemann curvature tensor and the geodesic deviation equation. For a detailed explanation of these, the reader must turn to some of the texts in the bibliography. Here we briefly sketch the main ideas.

When spacetime is curved, the result of parallel transport depends on the path taken. To quantify this notion, pick two vectors $u$ and $v$ at a point $p$ in spacetime. In the limit where $\epsilon \to 0$, we can approximately speak of a 'parallelogram' with sides $\epsilon u$ and $\epsilon v$. Take another vector $w$ at $p$ and parallel transport it first along $\epsilon v$ and then along $\epsilon u$ to the opposite corner of this parallelogram. The result is some vector $w_1$. Alternatively, parallel transport $w$ first along $\epsilon u$ and then along $\epsilon v$. The result is a slightly different vector, $w_2$:

The limit

$$\lim_{\epsilon \to 0} {w_2 - w_1 \over \epsilon^2} = R(u,v)w$$ (5)

is well-defined, and it measures the curvature of spacetime at the point $p$. In local coordinates we can write it as

$$R(u,v)w = R^\alpha_{\beta\gamma\delta} u^\beta v^\gamma w^\delta,$$

where as usual we sum over repeated indices. The quantity $R^\alpha_{\beta\gamma\delta}$ is called the 'Riemann curvature tensor'.

We can use this tensor to compute the relative acceleration of nearby particles in free fall if they are initially at rest relative to one another. Consider two freely falling particles at nearby points $p$ and $q$. Let $v$ be the velocity of the particle at $p$, and let $\epsilon u$ be the vector from $p$ to $q$. Since the two particles start out at rest relative to one other, the velocity of the particle at $q$ is obtained by parallel transporting $v$ along $\epsilon u$.

Now let us wait a short while. Both particles trace out geodesics as time passes, and at time $\epsilon$ they will be at new points, say $p'$ and $q'$. The point $p'$ is displaced from $p$ by an amount $\epsilon v$, so we get a little parallelogram, exactly as in the definition of the Riemann curvature:

Next let us compute the new relative velocity of the two particles. To compare vectors we must carry one to another using parallel transport. Let $v_1$ be the vector we get by taking the velocity vector of the particle at $p'$ and parallel transporting it to $q'$ along the top edge of our parallelogram. Let $v_2$ be the velocity of the particle at $q'$. The difference $v_2 - v_1$ is the new relative velocity. Here is a picture of the whole situation:

The vector $v$ is depicted as shorter than $\epsilon v$ for purely artistic reasons.

It follows that over this passage of time, the average relative acceleration of the two particles is $a = (v_2 - v_1)/\epsilon$. By equation (5),

$$\lim_{\epsilon \to 0} {v_2 - v_1 \over \epsilon^2} = R(u,v)v,$$

so

$$\lim_{\epsilon \to 0} {a \over \epsilon} = R(u,v)v .$$

This is called the 'geodesic deviation equation'. From the definition of the Riemann curvature it is easy to see that $R(u,v)w = -R(v,u)w$, so we can also write this equation as

$$\lim_{\epsilon \to 0} {a^\alpha \over \epsilon} = - R^\alpha_{\beta\gamma\delta} v^\beta u^\gamma v^\delta .$$ (6)

Using this equation we can work out the second time derivative of the volume $V(t)$ of a small ball of test particles that start out at rest relative to each other. For this we must let $u$ range over an orthonormal basis of tangent vectors, and sum the 'outwards' component of acceleration for each one of these. By equation (6) this gives

$$\lim_{V \to 0} {\ddot V\over V}\Bigr\vert _{t = 0} = - R^\alpha_{\beta\alpha\delta} v^\beta v^\delta .$$

In terms of the 'Ricci tensor'

$$R_{\beta\delta} = R^\alpha_{\beta\alpha\delta}$$

we may write this as:

$$\lim_{V \to 0} {\ddot V\over V} \Bigr\vert _{t = 0} = - R_{\beta\delta} v^\beta v^\delta .$$

In local inertial coordinates where the ball starts out at rest we have $v = (1,0,0,0)$, so

$$\lim_{V \to 0} {\ddot V\over V}\Bigr\vert _{t = 0} = - R_{tt}.$$ (7)

In short, the Ricci tensor says how our ball of freely falling test particles starts changing in volume. The Ricci tensor only captures some of the information in the Riemann curvature tensor. The rest is captured by something called the 'Weyl tensor', which says how any such ball starts changing in shape. The Weyl tensor describes tidal forces, gravitational waves and the like.

Now, Einstein's equation in its usual form says

$$G_{\alpha \beta} = T_{\alpha \beta} .$$

Here the right side is the stress-energy tensor, while the left side, the 'Einstein tensor', is just an abbreviation for a quantity constructed from the Ricci tensor:

$$G_{\alpha \beta} = R_{\alpha \beta} - {1\over 2}g_{\alpha \beta} R^\gamma_\gamma.$$

Thus Einstein's equation really says

$$R_{\alpha \beta} - {1\over 2}g_{\alpha \beta} R^\gamma_\gamma = T_{\alpha \beta} .$$ (8)

This implies

$$R^\alpha_\alpha - {1\over 2}g^\alpha_\alpha R^\gamma_\gamma = T^\alpha_\alpha,$$

but $g^\alpha_\alpha = 4$, so

$$-R^\alpha_\alpha = T^\alpha_\alpha .$$

Plugging this in equation (8), we get

$$R_{\alpha \beta} = T_{\alpha \beta} - {1\over 2}g_{\alpha \beta} T^\gamma_\gamma .$$ (9)

This is equivalent version of Einstein's equation, but with the roles of $R$ and $T$ switched! The good thing about this version is that it gives a formula for the Ricci tensor, which has a simple geometrical meaning.

Equation (9) will be true if any one component holds in all local inertial coordinate systems. This is a bit like the observation that all of Maxwell's equations are contained in Gauss's law and $\nabla \cdot B = 0$. Of course, this is only true if we know how the fields transform under change of coordinates. Here we assume that the transformation laws are known. Given this, Einstein's equation is equivalent to the fact that

$$R_{tt} = T_{tt} - {1\over 2}g_{tt} T^\gamma_\gamma$$ (10)

in every local inertial coordinate system about every point. In such coordinates we have

$$g = \left( \begin{array}{cccc} -1 & 0 &0 & 0 \\ 0 & 1 &0 & 0 \\ 0 & 0 &1 & 0 \\ 0 & 0 &0 & 1 \\ \end{array}\right)$$ (11)

so $g_{tt} = -1$ and

$$T^\gamma_\gamma = -T_{tt} + T_{xx} + T_{yy} + T_{zz}.$$

Equation (10) thus says that

$$R_{tt} = {1\over 2} (T_{tt} + T_{xx} + T_{yy} + T_{zz}) .$$

By equation (7), this is equivalent to

$$\lim_{V \to 0} {\ddot V\over V}\Bigr\vert _{t = 0} = - {1\over 2} (T_{tt} + T_{xx} + T_{yy} + T_{zz}),$$

as desired.

© 2006 John Baez and Emory Bunn

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