4 The Einstein Equation

We first need to figure out what $\mathrm{d}/\mathrm{d}\tau$ really means. With the equivalence principle in mind, the safest thing to do is to start with a definition for a locally Minkowski set of coordinates (i.e. free-falling, inertial, cartesian coordinates), just as we did in Section 2. Let’s call those coordinates $x_M^{\mu }$ and the components of the vector $X_M^{\mu }$; then we can straight-forwardly apply the chain rule:

$$\frac{\mathrm{d}{X}_M^{\mu }}{\mathrm{d}\tau }\equiv \frac{\mathrm{d}{x}_M^{\nu }}{\mathrm{d}\tau }\frac{\partial X_M^{\mu }}{\partial x_M^{\nu }}\mathit{\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}}\text{(locally Minkowski coordinate system).}$$

(62)

From a physical perspective that’s all there is to it – we could work directly with definition (62). But it’s a nuisance to be forever finding transformations into a specific coordinate system. We want instead to write a version of equation (62) that is applicable in any coordinates. Such an equation is called covariant.

To find a covaraint expression, recall that equation (12) tells us how $X^{\mu }$ transforms between coordinate systems. To define $\mathrm{d}{X}^{\mu }/\mathrm{d}\tau$ in coordinate systems other than the free-falling one employed in equation (62), the best option is to insist that it, too, transforms as a vector. This provides the definition for the derivative of a vector in non-inertial coordinate systems in a way that is consistent with the equivalence principle. To put it another way, it ensures that any equation involving vectors and derivatives of vectors will have the same physical meaning regardless of which coordinate system it’s expressed relative to.

But can we do this without forever transforming back and forth to locally free-falling coordinate systems? Thankfully the answer is yes, and the logic is very similar to that from Section 2. Let’s keep $x_M^{\mu }$ as locally inertial coordinates and imagine a new, general coordinate system $x^{\mu }$. Starting from the definition of $\mathrm{d}{X}_M^{\mu }/\mathrm{d}\tau$, we now have:

$$\frac{\mathrm{d}{X}_M^{\mu }}{\mathrm{d}\tau }\equiv \frac{\mathrm{d}{x}_M^{\nu }}{\mathrm{d}\tau }\frac{\partial X_M^{\mu }}{\partial x_M^{\nu }}=\frac{\mathrm{d}{x}^{\nu }}{\mathrm{d}\tau }\frac{\partial x_M^{\mu }}{\partial x^{\alpha }}\frac{\partial X^{\alpha }}{\partial x^{\nu }}+\frac{\mathrm{d}{x}^{\nu }}{\mathrm{d}\tau }\frac{{\partial }^2{x}_M^{\mu }}{\partial x^{\nu }\partial x^{\beta }}{X}^{\beta }$$

(63)

😇 Exercise 4A Verify equation (63), starting from the definition (62) and using the appropriate transformation laws such as (12) and (11). Hint: First argue that $$\frac{\mathrm{d}{x}_M^{\nu }}{\mathrm{d}\tau }\frac{\partial X_M^{\mu }}{\partial x_M^{\nu }}=\frac{\mathrm{d}{x}^{\nu }}{\mathrm{d}\tau }\frac{\partial X_M^{\mu }}{\partial x^{\nu }}\text{.}$$ (64)

Comparing with equation (12), the first term in equation (63) is precisely the transformation that I just argued we want. What of the second term? We can rewrite it in terms of our old friends, the Christoffel symbols (37), as follows:

$$\frac{\mathrm{d}{x}^{\nu }}{\mathrm{d}\tau }\frac{{\partial }^2{x}_M^{\mu }}{\partial x^{\nu }\partial x^{\beta }}{X}^{\beta }=\frac{\mathrm{d}{x}^{\nu }}{\mathrm{d}\tau }\frac{\partial x_M^{\mu }}{\partial x^{\alpha }}{\mathrm{\Gamma }}_{\nu \beta }^{\alpha }{X}^{\beta }.$$

(65)

Putting everything together we now have:

	${\displaystyle \frac{\mathrm{d}{X}^{\alpha }}{\mathrm{d}\tau }}$	$={\displaystyle \frac{\partial x^{\alpha }}{\partial x_M^{\mu }}}{\displaystyle \frac{\mathrm{d}{X}_M^{\mu }}{\mathrm{d}\tau }}$	(stating the equivalence principle)
		$={\displaystyle \frac{\mathrm{d}{x}^{\nu }}{\mathrm{d}\tau }}\left({\displaystyle \frac{\partial X^{\alpha }}{\partial x^{\nu }}}+{\mathrm{\Gamma }}_{\nu \beta }^{\alpha }{X}^{\beta }\right)$	(the final result),		(66)

where the second line follows from substituting (63) and (65) for $\mathrm{d}{x}_M^{\mu }/\mathrm{d}\tau$. Note that in the result (66) all explicit reference to a special Minkowski system of coordinates is gone. It has been fully absorbed into the Christoffel symbols ${\mathrm{\Gamma }}_{\nu \beta }^{\alpha }$ which can be computed directly from the metric in any coordinate system, equation (39).

We normally abbreviate the combination in brackets as the covariant derivative $\nabla$; for any vector $V^{\beta }$ we can write

$${\nabla }_{\alpha }{V}^{\beta }\equiv \frac{\partial V^{\beta }}{\partial x_M^{\alpha }}+{\mathrm{\Gamma }}_{\alpha \gamma }^{\beta }{V}^{\gamma }.$$

(67)

Then we have a neat way to express equation (66):

$$\frac{\mathrm{d}{X}^{\alpha }}{\mathrm{d}\tau }\equiv \frac{\mathrm{d}{x}^{\nu }}{\mathrm{d}\tau }{\nabla }_{\nu }{X}^{\alpha }.$$

(68)

We also sometimes use a semi-colon to indicate a covariant derivative, and a comma to indicate a partial derivative with respect to the coordinates $x$ — in summary:

$$V_{,\beta }^{\alpha }\equiv \frac{\partial V^{\alpha }}{\partial x^{\beta }}\mathit{\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}}\text{and}\mathit{\hspace{1em}\hspace{1em}\hspace{0.5em}\hspace{1em}}{V}_{;\beta }^{\alpha }\equiv {\nabla }_{\beta }{V}^{\alpha }\equiv V_{,\beta }^{\alpha }+{\mathrm{\Gamma }}_{\beta \gamma }^{\alpha }{V}^{\gamma }\text{.}$$

(69)

Our derivation shows, in a very general way, that the equivalence principle allows us to simply write what we would have wanted in flat Minkowski space then replace any partial derivatives $\partial /\partial x_M^{\alpha }$ with the covariant derivative ${\nabla }_{\alpha }$. That’s a very powerful statement: for example, it allows us to re-derive the properties of geodesics almost instantaneously; we can conveniently re-write equation (38) as

$$\frac{\mathrm{d}{x}^{\alpha }}{\mathrm{d}\lambda }{\nabla }_{\alpha }\frac{\mathrm{d}{x}^{\mu }}{\mathrm{d}\lambda }=0.$$

(70)

☞ Exercise 4B Explain why equation (70) must be a valid description of particle motion, using the equivalence principle in the form of replacing partial by covariant derivatives. Then explicitly expand equation (70) using the definition (67) and show it is indeed the same as equation (38).

In a locally free-falling coordinate system, we have ${\nabla }_{\nu }=\partial /\partial x_{\nu }$ (or, using the other notation, $V_{;\nu }^{\mu }=V_{,\nu }^{\mu }$) and everything boils back to our first definition, equation (62).

We can work out a covariant derivative for any type of tensor with any number of indices. For example, given a covector $W_{\beta }$ we have

$${\nabla }_{\alpha }{W}_{\beta }=\frac{\partial W_{\beta }}{\partial x^{\alpha }}-{\mathrm{\Gamma }}_{\alpha \beta }^{\lambda }{W}_{\lambda }\text{,}$$

(71)

which can be obtained in a number of ways – most obviously, by following the logic for vectors $V^{\mu }$ above, but with the appropriate transformation $W_{\mu }=(\partial x_M^{\nu }/\partial x^{\mu })W_{\nu }^M$.

😇 Exercise 4C Perform this derivation.

Each Christoffel $\mathrm{\Gamma }$ term appears because of the transformation associated with an index. Consequently, as indices get added, the additional Christoffel terms associated with each index simply add up; for example, for a rank-2 mixed tensor $T_{\beta }^{\alpha }$ we have

$${\nabla }_{\gamma }{T}_{\beta }^{\alpha }=\frac{\partial T_{\beta }^{\alpha }}{\partial x^{\gamma }}+{\mathrm{\Gamma }}_{\gamma \lambda }^{\alpha }{T}_{\beta }^{\lambda }-{\mathrm{\Gamma }}_{\gamma \beta }^{\lambda }{T}_{\lambda }^{\alpha }\text{.}$$

(72)

For convenience, we define the covariant derivative of a scalar quantity to be just

$${\nabla }_{\alpha }q\equiv \frac{\partial q}{\partial x^{\alpha }}\text{,}$$

(73)

which is neatly self-consistent: when there are zero indices, there will be zero $\mathrm{\Gamma }$ terms.

A nice property of the covariant derivative is that it satisfies the product rule, just like the partial derivative:

$${\nabla }_{\alpha }\left(S_{\delta ϵ\mathrm{\cdots }}^{\beta \gamma \mathrm{\cdots }}{T}_{\rho \sigma \mathrm{\cdots }}^{\mu \nu \mathrm{\cdots }}\right)=\left({\nabla }_{\alpha }{S}_{\delta ϵ\mathrm{\cdots }}^{\beta \gamma \mathrm{\cdots }}\right)T_{\rho \sigma \mathrm{\cdots }}^{\mu \nu \mathrm{\cdots }}+S_{\delta ϵ\mathrm{\cdots }}^{\beta \gamma \mathrm{\cdots }}\left({\nabla }_{\alpha }{T}_{\rho \sigma \mathrm{\cdots }}^{\mu \nu \mathrm{\cdots }}\right)\text{.}$$

(74)

for tensors $T$ and $S$ which can have any number of indices (including zero). You can straight-forwardly verify this property for yourself if interested; it basically follows because the covariant derivative is the combination of a standard partial derivative (which follows the product rule) and an extra term for each index. We will make use of the product rule several times in derivations. The covariant derivative is also said to be metric-compatible:

$${\nabla }_{\alpha }{g}_{\beta \gamma }=0={\nabla }_{\alpha }{g}^{\beta \gamma }\text{,}$$

(75)

a property that you can again verify for yourself if keen. This is again extremely helpful in practical derivations and we will take it for granted.

Now we need one last mathematical ingredient to characterise gravity, which was our original aim. The missing insight is about the vector $X^{\mu }$. I was a bit slapdash in saying $X^{\mu }$ connects two nearby worldlines – what does that really mean?

Figure 7 shows how to define $X^{\mu }$ more carefully. We imagine setting up a series of test geodesics labelled by a continuous parameter $\sigma$. Then we can define $X^{\mu }\equiv \partial x^{\mu }/\partial \sigma$, where the partial derivative is taken at fixed $\tau$. Where we had derivatives with respect to $\tau$ we must now take them at fixed $\sigma$ (to unambiguously take a derivative down a particular particle’s worldline). With this in place, we can expand ${\partial }^2{X}^{\mu }/\partial {\tau }^2$ in terms of covariant derivatives:

${\displaystyle \frac{{\partial }^2{X}^{\mu }}{\partial {\tau }^2}}\equiv {\displaystyle \frac{{\partial }^3{x}^{\mu }}{\partial {\tau }^2\partial \sigma }}$

$={\displaystyle \frac{\partial x^{\alpha }}{\partial \tau }}{\nabla }_{\alpha }\left({\displaystyle \frac{\partial x^{\beta }}{\partial \sigma }}{\nabla }_{\beta }{\displaystyle \frac{\partial x^{\mu }}{\partial \tau }}\right)$

(76)

Here I have very conciously chosen to expand a $\tau$ derivative, then a $\sigma$ derivative, and then the second $\tau$ derivative using equation (68). Obviously I could have chosen to expand the derivative in a number of other ways, but the result below will explain why this order is particularly helpful. We can continue simplifying using the product rule:

${\displaystyle \frac{{\partial }^2{X}^{\mu }}{\partial {\tau }^2}}$

$={\displaystyle \frac{\partial x^{\alpha }}{\partial \tau }}{\displaystyle \frac{\partial x^{\beta }}{\partial \sigma }}{\nabla }_{\alpha }{\nabla }_{\beta }{\displaystyle \frac{\partial x^{\mu }}{\partial \tau }}+{\displaystyle \frac{\partial x^{\alpha }}{\partial \tau }}\left({\nabla }_{\alpha }{\displaystyle \frac{\partial x^{\beta }}{\partial \sigma }}\right)\left({\nabla }_{\beta }{\displaystyle \frac{\partial x^{\mu }}{\partial \tau }}\right).$

(77)

Next, we identify that

$$\frac{\partial x^{\alpha }}{\partial \tau }\left({\nabla }_{\alpha }\frac{\partial x^{\beta }}{\partial \sigma }\right)\equiv \frac{\partial x^{\beta }}{\partial \tau \partial \sigma }=\frac{\partial x^{\beta }}{\partial \sigma \partial \tau }=\frac{\partial x^{\alpha }}{\partial \sigma }\left({\nabla }_{\alpha }\frac{\partial x^{\beta }}{\partial \tau }\right)$$

(78)

and furthermore that

$$\left({\nabla }_{\alpha }\frac{\partial x^{\beta }}{\partial \tau }\right)\left({\nabla }_{\beta }\frac{\partial x^{\mu }}{\partial \tau }\right)=-\frac{\partial x^{\beta }}{\partial \tau }{\nabla }_{\alpha }{\nabla }_{\beta }\frac{\partial x^{\mu }}{\partial \tau }$$

(79)

which follows from the geodesic property in the form of equation (70). Putting it all together (and relabelling some dummy indices to simplify the expression) we have the result:

$$\frac{{\partial }^2{X}^{\mu }}{\partial {\tau }^2}=\frac{\partial x^{\alpha }}{\partial \tau }\frac{\partial x^{\beta }}{\partial \sigma }({\nabla }_{\alpha }{\nabla }_{\beta }-{\nabla }_{\beta }{\nabla }_{\alpha })\frac{\partial x^{\mu }}{\partial \tau }.$$

(80)

We choose to expand the result in this particular way because equation (80) has a very clear physical interpretation. In flat spacetime, ${\nabla }_{\alpha }$ reduces back down to partial derivative with respect to $x^{\alpha }$. As a result,

$${\nabla }_{\alpha }{\nabla }_{\beta }=\frac{{\partial }^2}{\partial x^{\alpha }\partial x^{\beta }}=\frac{{\partial }^2}{\partial x^{\beta }\partial x^{\alpha }}={\nabla }_{\beta }{\nabla }_{\alpha }\mathit{\hspace{1em}}\text{(flat spacetime)}$$

(81)

and the acceleration as expressed by equation (80) is obviously zero – this is exactly the intuitive result we were wanting!

In non-flat space, we instead expand (80) in terms of Christoffel symbols, to find that

	${\displaystyle \frac{{\partial }^2{X}^{\mu }}{\partial {\tau }^2}}$	$={\displaystyle \frac{\partial x^{\alpha }}{\partial \tau }}{\displaystyle \frac{\partial x^{\beta }}{\partial \sigma }}{\displaystyle \frac{\partial x^{\nu }}{\partial \tau }}{R}_{\nu \alpha \beta }^{\mu }$		(82)
	$\text{where }{R}_{\nu \alpha \beta }^{\mu }$	$\equiv {\displaystyle \frac{\partial {\mathrm{\Gamma }}_{\beta \nu }^{\mu }}{\partial x^{\alpha }}}-{\displaystyle \frac{\partial {\mathrm{\Gamma }}_{\alpha \nu }^{\mu }}{\partial x^{\beta }}}+{\mathrm{\Gamma }}_{\alpha \lambda }^{\mu }{\mathrm{\Gamma }}_{\beta \nu }^{\lambda }-{\mathrm{\Gamma }}_{\beta \lambda }^{\mu }{\mathrm{\Gamma }}_{\alpha \nu }^{\lambda }$		(83)

and $R_{\nu \alpha \beta }^{\mu }$ is known as the Riemann curvature tensor of the spacetime. Note that $R_{\nu \alpha \beta }^{\mu }$ has no dependence on the particular choice of geodesics $x^{\mu }(\tau ,\sigma )$ – it is purely constructed from the Christoffel symbols $\mathrm{\Gamma }$ that are intrinsic to the spacetime. The Riemann tensor was a widely-studied property of curved spaces long before Einstein started investigating it as a description of gravity.

😇 Exercise 4E Check that expanding the covariant derivatives in equation (80) does indeed result in expression (83).

The physical connection we have made is essential: the curvature of spacetime describes the way that nearby particles accelerate towards or away from each other. Therefore, to complete the general relativistic theory of gravity, we want to connect the curvature back to the contents of spacetime.

The derivations above won’t appear in the examination for this course – but applying the results or understanding the physical content might, so let’s summarise it for convenience:

1. 1.

   We realised we need to take the derivative of a vector over time, to start defining what gravity does;

2. 2.

   We defined this derivative in local inertial coordinates, equation (62);

3. 3.

   We invoked, as always, the equivalence principle to work out how to calculate the same derivative in a general coordinate system, equation (66). This can be written in compact notation as equation (68). By construction we can now write a general relativistic equation of motion just by taking the special relativistic equations and replacing partial derivatives $\partial /\partial x^{\alpha }$ with covariant derivatives ${\nabla }_{\alpha }$.

4. 4.

   Finally, using these results, we worked out how to calculate the relative acceleration of two particles, ${\partial }^2{X}^{\mu }/\partial {\tau }^2$. This required a careful definition of $X^{\mu }$, the vector separating the two particles, and some further cunning manipulations to reach equations (82) and (83). Equation (82) shows that relative acceleration of particles is generated by curvature of spacetime, as parametrised by the Riemann tensor $R_{\nu \alpha \beta }^{\mu }$.

Start with “dust”, by which cosmologists mean massive particles with negligible velocities or pressure. This is the case that we have already been considering in Exercise 2. So we need $T^{00}={\rho }_{\mathrm{rest}}$ where ${\rho }_{\mathrm{rest}}$ is the rest-frame density, and all other components must vanish. The covariant way to write this requirement is

$$T_{\mathrm{dust}}^{\mu \nu }={\rho }_{\mathrm{rest}}{U}^{\mu }{U}^{\nu }\text{,}$$

(87)

where $U^{\mu }$ is the velocity 4-vector of the particles discussed previously (it’s $(1,0,0,0)$ in the rest-frame, but different in other frames).

We can now examine what the other components mean; suppose we switch to a frame moving at speed $v$ along the $x$-axis relative to the original. Then $U^{\prime \mu }=(1/\sqrt{1-v^2},v/\sqrt{1-v^2},0,0)$, so $\rho \equiv T^{\prime 00}={\rho }_{\mathrm{rest}}/(1-v^2)$ (we discussed how this is required above) and, more interestingly, $T^{\prime 0i}={\rho }_{\mathrm{rest}}{v}^i/(1-v^2)=\rho v^i$. We can recognize this quantity as a momentum density, i.e. it is the net momentum per unit volume that the fluid posesses.

Since $T$ is symmetric, $T^{i0}=T^{0i}$ so the only part left to understand are the $T^{ij}=T^{ji}$ space-space components. In the example above, where we have particles in uniform motion along the $x$ axis, $T^{\prime 11}={\rho }_{\mathrm{rest}}{v}^2/(1-v^2)=\rho v^2$. If you’ve taken a fluid dynamics course you’ll know that $\rho v^2$ is the ram pressure – it’s the pressure felt as particles flow past an object at speed $v$.

Imagine that we now add up the effects of many different particles, all moving in different directions with a speed $v$, so that on average they’re at rest but any given particle is moving⁷⁷ 7 This is just the microscopic description of a gas, although for simplicity let’s assume that the speed is always $v$ – to be more precise, we’d need to use the Maxwell-Boltzmann distribution, which doesn’t change anything fundamentally… but adds a layer of complexity that we can do without at the moment.. Now we’ll get

$$T^{ii}=\frac{{\rho }_{\mathrm{rest}}{v}^2}{3(1-v^2)}=\frac{\rho v^2}{3}\mathit{\hspace{1em}}\text{ (no sum over }i\text{)}$$

(88)

for $i=1,2,3$. In other words, $T^{ii}=p$, the pressure of the overall distribution of particles. But the $T^{ij}$ are still zero for any $i\ne j$…

☞ Exercise 4G Why must $T^{ij}=0$ for $i\ne j$ in the example given above? Consider two sets of particles, each with density $\rho$. Let one set move along the direction at ${45}^{\circ }$ between the $x$ and $y$ axes, and the other moving at the same speed along the direction at $-{45}^{\circ }$. Show that, while the contribution to the diagonal part of $T^{ij}$ adds, the contribution to the off-diagonal part cancels. By pairing all contributions in this way, we can show the off-diagonal contributions are always zero. Are there arguments you can think of that show this vanishing without even having to make any calculations?

So, by thinking about a series of particles moving in different directions, we have motivated the idea that – in the rest frame – we have

• •

  $T^{00}=\rho$ is the density (which equals ${\rho }_{\mathrm{rest}}$ for dust with zero velocity, $v=0$);

• •

  $T^{0i}$ is the momentum density which is zero in the rest frame;

• •

  $T^{ij}$ is the pressure $P$ for $i=j$, and zero for $i\ne j$.

Such gases are known in relativity as perfect fluids and can be completely defined by an energy density $\rho$ and an isotropic rest frame pressure $P$. Note that this $\rho$ should now be thought of as the “rest frame density of the fluid” – although this rest frame is not the same as the rest frame of individual particles, unless they are all at rest ($P=0$). The rest frame of the fluid is the frame in which there is no net motion. To be able to leave the rest frame of the perfect fluid, we need a covariant expression which turns out to be:

$$T^{\mu \nu }=(\rho +P)U^{\mu }{U}^{\nu }-P{g}^{\mu \nu }.$$

(89)

☞ Exercise 4H Check this is the correct expression by writing out the terms in an inertial frame where $g^{\mu \nu }={\eta }^{\mu \nu }$. Having checked it for just one frame, why must expression (89) be valid in any frame?

A perfect fluid is general enough to describe a wide variety of cosmological fluids, given their equation of state,

$$w=\frac{P}{\rho }.$$

(90)

Dust has $P=0,w=0$. Non-relativistic sources like baryonic and dark matter are often well-approximated as dust since $|P|\ll \rho$ ($|w|\ll 1$). Radiation has $P=\rho /3,w=1/3$. Vacuum energy (which we’ll discuss in more detail later) is proportional to the metric: $T^{\mu \nu }=-{\rho }_{\mathrm{vac}}{g}^{\mu \nu }$, $P_{\mathrm{vac}}=-{\rho }_{\mathrm{vac}}$, $w=-1$.

The energy-momentum tensor of a perfect fluid at rest can also be written with one index lowered in the following metric-independent form:

(91)

Because of its relative simplicity and transparent physical interpretation of $\rho$ and $P$ this is often the most useful way to express the energy-momentum tensor.

How do the components of $T_{\nu }^{\mu }$ evolve with time? Consider the case where there is no gravity. Then, from thermodynamical arguments we know that the pressure and energy evolve as:

	$\mathrm{ }\text{Continuity equation:}$	${\displaystyle \frac{\partial \rho }{\partial t}}+\nabla \cdot (\rho 𝒖)$	$=0,$		(92)
	Euler equation:	$\rho {\displaystyle \frac{\partial 𝒖}{\partial t}}+\left(\rho 𝒖\cdot \nabla \right)𝒖+\nabla P$	$=0.$		(93)

where $𝒖$ is the Newtonian velocity vector of the fluid. In special relativity, we turn this into a $4$–component conservation equation for the energy-momentum tensor:

$$\frac{\partial T_{\nu }^{\mu }}{\partial x^{\mu }}=0.$$

(94)

In general relativity, the conservation criterion must be modified because the expression above is not covariant. The discussion of the equivalence principle in Section 1 implies that we just need to “promote” the partial derivative to a covariant derivative to get the general relativistic conservation law:

$$\boxed{{\nabla }_{\mu }{T}_{\nu }^{\mu }=0}.$$

(95)

For our perfect fluid stress-energy tensor there are four separate equations to be considered here. Take first the $\nu =0$ component for the metric (23), generating the continuity equation for an expanding universe:

$$\frac{\partial \rho }{\partial t}+3\frac{\dot{a}}{a}\left(\rho +P\right)=0.$$

(96)

📝 Exercise 4K Recover this expression starting from ${\nabla }_{\mu }{T}_0^{\mu }=0$. The covariant derivative for rank-2 tensors is given by equation (72) and the appropriate Christoffel symbols by equation (5).

Rearranging this equation,

$$a^{-3}\frac{\partial }{\partial t}\left(\rho a^3\right)=-3\left(\frac{\dot{a}}{a}\right)P.$$

(97)

Immediately this gives us some information about how the density of the universe changes as it expands. Pressureless matter has $P=0$ by definition, so

$$\frac{\partial }{\partial t}\left({\rho }_{\mathrm{m}}{a}^3\right)=0\Rightarrow {\rho }_{\mathrm{m}}\propto a^{-3}\mathit{\hspace{1em}}\text{(for matter).}$$

(98)

This is expected if you consider that mass remains constant while number density scales as inverse volume.

Equation (88) shows that radiation ($v=1$) has a pressure $P=\frac{\rho }{3}$. In this case we obtain instead

$$\frac{\partial {\rho }_{\mathrm{r}}}{\partial t}+4\left(\frac{\dot{a}}{a}\right){\rho }_{\mathrm{r}}=a^{-4}\frac{\partial }{\partial t}\left({\rho }_{\mathrm{r}}{a}^4\right)=0.$$

(99)

From equation (99) radiation density changes with scalefactor according to

$${\rho }_{\mathrm{r}}\propto a^{-4}\mathit{\hspace{1em}}\text{(for radiation).}$$

(100)

By considering the energy-momentum tensor and the equivalence principle we are obtaining highly non-trivial results already – and we haven’t even written down the Einstein equation yet!

In some ways, the Einstein equation looks a bit peculiar as a description of gravity. For example:

• •

  $T_{\mu \nu }$ specifies the Ricci tensor $R_{\mu \nu }$; in turn $R_{\mu \nu }$ only partially captures the information in the Riemann tensor $R_{\nu \alpha \beta }^{\mu }$. Where does the rest of the information in $R_{\nu \alpha \beta }^{\mu }$ come from?

• •

  In Newtonian physics, ${\nabla }^2\mathrm{\Phi }=4\pi G\rho$ generates action at a distance because ${\nabla }^2$ is a differential operator. So in general relativity, how does the local value of $\rho$ affect geodesic deviation of particles at a distance – when there are no differential operators in the Einstein equation?

Think about the above, as we’ll discuss it in class.