2 The Metric

On the largest scales, the universe is assumed to be uniform. This idea is called the cosmological principle. There are two aspects of the cosmological principle:

• •

  the universe is homogeneous. There is no preferred observing position in the universe.

• •

  the universe is isotropic. The universe looks the same in every direction.

☞ Exercise 2A To illustrate the difference between homogeneous and isotropic, draw a sketch of a universe that’s homogeneous but not isotropic; and another sketch of a universe that’s isotropic but not homogeneous.

There is an overwhelming amount of observational evidence that the universe is expanding. This means that early in the history of the universe, the distant galaxies were closer to us than they are today. It is convenient to describe the scaling of the coordinate grid in an expanding universe by the scale factor. In a smooth, expanding universe, the scale factor connects the coordinate distance with the physical distance.

In general, we need to separate the idea of coordinates from physical distances. In normal cartesian coordinates, these two ideas are almost the same. But more generally they are different. Think about polar coordinates, for example; the angle $\theta$ is not a distance. But it can be turned into a distance using a suitable formula.

In fact a formula turning coordinate differences into a physical distance is known as a metric. The metric is an essential tool to make quantitative predictions in an expanding universe.

The most familiar metric is that of Cartesian coordinates (Figure 2); in 2D,

$$\mathrm{d}{s}^2=\mathrm{d}{x}^2+\mathrm{d}{y}^2\mathit{\hspace{1em}}=\sum _i{\left(\mathrm{d}{x}^i\right)}^2\text{,}$$

(1)

where $x^i$ refers to the two components of the Cartesian coordinates, i.e. $x^1=x$ and $x^2=y$. The content should be extremely familiar but the notation looks a little weird if you haven’t seen it before; in particular note that indices are written as superscripts for reasons that will become clear later (subscripts are reserved for a different type of index). The context should normally distinguish between $x^2$ meaning “the square of $x$” and $x^2$ meaning “the second component of the $x$ vector”. Do be aware of this potential for confusion, though. To be clear, ${(\mathrm{d}{x}^i)}^2$ is the square of the change in the $i$th coordinate, so ${(\mathrm{d}{x}^1)}^2={(\mathrm{d}x)}^2$ and ${(\mathrm{d}{x}^2)}^2={(\mathrm{d}y)}^2$ for Cartesian coordinates.

To switch to polar coordinates, we now imagine that $x^1=r={(x^2+y^2)}^{1/2}$ and $x^2=\theta ={\tan }^{-1}(y/x)$. Thinking back to previous courses on polar coordinates, it should be clear that calculating a physical distance $\mathrm{d}s$ now requires a different approach:

$$\mathrm{d}{s}^2=\mathrm{d}{r}^2+r^2\mathrm{d}{\theta }^2\mathit{\hspace{1em}}\ne \sum _i{\left(\mathrm{d}{x}^i\right)}^2\text{.}$$

(2)

In fact, things can get more messy still. Suppose I define a new coordinate system $(u,v)$ where $u=\sqrt{2}x$ and $v=y-x$. The right panel of Figure 2 shows how the displacements $\mathrm{d}u$ and $\mathrm{d}v$ no longer form a right-angled triangle at all. How could we calculate the physical distance $\mathrm{d}s$ if forced to work in these coordinates?

The trick is to relate it back to what we do know, the cartesian coordinate system. We can immediately find that $\mathrm{d}u=\sqrt{2}\mathrm{d}x$ and $\mathrm{d}v=(\mathrm{d}y-\mathrm{d}x)$. Inverting this relationship, we find

$$\mathrm{d}x=\frac{1}{\sqrt{2}}\mathrm{d}u\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}\mathrm{d}y=\frac{1}{\sqrt{2}}\mathrm{d}u+\mathrm{d}v\text{,}$$

(3)

which we can substitute into equation (1) to obtain

$$\mathrm{d}{s}^2=\mathrm{d}{u}^2+\mathrm{d}{v}^2+\sqrt{2}\mathrm{d}u\mathrm{d}v\text{.}$$

(4)

Note the appearance of cross-terms in $\mathrm{d}u\mathrm{d}v$; this is the sign of a non-orthogonal coordinate system because it violates Pythagoras’ theorem. But don’t be fooled: so far there is no hint of “curvature”, which is an intrinsic property of spaces and not dependent on the coordinate system in use.

We can generalise rules (1), (2) and (4) using the concept of the metric tensor $g_{ij}$. In 2D (Fig. 2),

$$\mathrm{d}{s}^2=\sum _{i,j=1,2}{g}_{ij}\mathrm{d}{x}^i\mathrm{d}{x}^j,$$

(5)

where the metric $g_{ij}$ is a $2\times 2$ symmetric matrix. Note that this allows for cross-terms of the form seen in Equation (4). In Cartesian coordinates,

(6)

while in polar coordinates,

(7)

For our example $u$–$v$ coordinate system,

(8)

Note that $g_{ij}$ is defined to be symmetric in $i\leftrightarrow j$.

☞ Exercise 2B Check that substituting equations (6), (7) and (8) into (5) correctly regenerates the expressions (1), (2) and (4) for the distances in these example coordinate systems. Optional: Show that the concept of a metric tensor expresses a completely general way of describing distances in any coordinate system that can be written as a non-singular function of the cartesian coordinates. (In particular, you may like to show that all terms must be quadratic in the coordinate displacements.)

In classical Newtonian mechanics, gravity is a force, and particles travel along curved trajectories because of this force. In general relativity, gravity is instead encoded in the metric; particles actually move along the closest equivalent to a straight line, but this begins to look curved because of curvature built into the space itself. To start formalising and using this concept, we need metrics in 4 dimensions (1 time + 3 space). The physically invariant spacetime distance is then defined by

$$d{s}^2=\sum _{\mu ,\nu =0}^3{g}_{\mu \nu }d{x}^{\mu }d{x}^{\nu },$$

(9)

where $\mu ,\nu ⟶\{0,1,2,3\}$. Typically $0$ is chosen as the time and $1\to 3$ as the space coordinates; however there is no unique choice of coordinates (not even for time) and therefore we refer to coordinate $0$ as timelike and $1\to 3$ as spacelike respectively.

As a final adjustment to our notation, we will adopt the Einstein summation convention which states that repeated indices in a single term are summed over. So, we can rewrite Equation (9) more compactly as:

$$d{s}^2=g_{\mu \nu }d{x}^{\mu }d{x}^{\nu },$$

(10)

The displacement $\mathrm{d}{x}^{\mu }$ is an example of a vector. However we can generalise the concept to a generic vector $A^{\mu }$. Vectors are, by definition, any object that behaves like a small displacement under transformations of the coordinate system (for example: a rotation, or a change from cartesian to polar coordinates). In particular, in a new coordinate system $x^{\prime \mu }$ the coordinate displacements transform via the chain rule:

$$\mathrm{d}{x}^{\prime \mu }=\frac{\partial x^{\prime \mu }}{\partial x^{\nu }}\mathrm{d}{x}^{\nu }\text{,}$$

(11)

via the chain rule. (Don’t forget we are now using the Einstein summation convention so there is an implicit sum over $\nu$ from $0$ to $3$ – but not over $\mu$ even though it appears twice, as the appearances are in different terms). All vectors transform in the same way:

😇 Exercise 2C Any 4-vector $V^{\mu }$ can be thought of as an arrow connecting two very close points in spacetime – specifically, let’s imagine it connects $x^{\mu }$ and $x^{\mu }+ϵV^{\mu }$ for a very small $ϵ\ll 1$. Using a Taylor expansion show that, under the coordinate transformation $x^{\mu }\to x^{\prime \mu }$, the vector $V^{\mu }$ transforms according to the rule $$V^{\prime \mu }=\frac{\partial x^{\prime \mu }}{\partial x^{\nu }}{V}^{\nu },$$ (12) Note that the inverse relation also immediately follows, $$V^{\mu }=\frac{\partial x^{\mu }}{\partial x^{\prime \nu }}{V}^{\prime \nu }.$$ (13) just by a trivial relabelling of which coordinate system we think of as the “primed” one.

In summary, one can think of vectors as little arrows living at a particular point in the spacetime. Mathematicians express this idea by saying that vectors belong to the tangent space which can be defined more formally.

A vector has a length $|𝑨|$ given by ${|𝑨|}^2\equiv g_{\mu \nu }{A}^{\mu }{A}^{\nu }$. One can generalise this notion to a dot product of two vectors $𝑨\cdot 𝑩\equiv g_{\mu \nu }{A}^{\mu }{B}^{\nu }$. These constructions are all independent of the coordinate system in which they are evaluated because, under a coordinate transformation $x^{\mu }\to x^{\prime \mu }$, the metric becomes

$$g_{\mu \nu }^{\prime }=\frac{\partial x^{\alpha }}{\partial x^{\prime \mu }}\frac{\partial x^{\beta }}{\partial x^{\prime \nu }}{g}_{\alpha \beta }\text{,}$$

(14)

which follows from the invariance of $\mathrm{d}{s}^2$ in equation (10).

😇 Exercise 2D Flesh out this argument: show that $g_{\mu \nu }$ must transform this way given the transformation (11). Then show that, as a result, $A^{\mu }{B}^{\nu }{g}_{\mu \nu }$ is also invariant under coordinate transformations for any vectors $𝑨$ and $𝑩$.

There is another way to think about the dot product: we can imagine that to each vector $A^{\mu }$ corresponds a covector $A_{\mu }$ defined by

$$A_{\mu }\equiv g_{\mu \nu }{A}^{\nu }\text{.}$$

(15)

Then the dot product $𝑨\cdot 𝑩$ is expressed simply as $𝑨\cdot 𝑩=A_{\mu }{B}^{\mu }$. Covector indices transform under coordinate transformations as follows:

$$A_{\mu }\to A_{\mu }^{\prime }=\frac{\partial x^{\nu }}{\partial x^{\prime \mu }}{A}_{\nu }\text{,}$$

(16)

which you can verify from the definition (15) and existing transformation laws (12) and (14).

Finally, it is sometimes useful to be able to make a vector again starting from a covector. For this purpose we use the inverse metric $g^{\mu \nu }$:

$$A^{\mu }=g^{\mu \nu }{A}_{\nu }\mathit{\hspace{1em}}\text{where}\mathit{\hspace{1em}}{g}^{\alpha \beta }{g}_{\beta \gamma }={\delta }_{\gamma }^{\alpha }\text{.}$$

(17)

Here ${\delta }_{\gamma }^{\alpha }$ is the Kronecker delta:

$${\delta }_{\gamma }^{\alpha }=\{\begin{array}{c}\hfill 1\hspace{1em}(\alpha =\gamma )\hfill \\ \hfill 0\hspace{1em}(\alpha \ne \gamma )\hfill \end{array},$$

(18)

i.e. the identity matrix in component form. All this does is to define $g^{\mu \nu }$ as the matrix inverse of $g_{\mu \nu }$, so that going from a vector to a covector and then back again gives you the vector you started with.

As we will now see, the special nature of spacetime (as opposed to just ordinary space extended to four dimensions) is that space and time coordinate carry opposite signs in the metric. In this course we will use the following convention for the signature of the metric: $(+,-,-,-)$. So, the physical distance associated with a spacelike coordinate displacement is $\sqrt{-\mathrm{d}{s}^2}$. The time interval associated with a timelike coordinate displacement is $\sqrt{\mathrm{d}{s}^2}$.

Beware: while this convention is commonly used by particle physicists, the convention used in relativity and cosmology is sometimes $(-,+,+,+)$. This is one of a number of conventions in cosmology that are not universally agreed upon and leads to frustrating confusion when comparing results from different authors.

The Minkowski metric ${\eta }_{\mu \nu }$ is the metric of special relativity, and it describes flat space. Its line element is given by

	$\mathrm{d}{s}^2$	$=$	${\eta }_{\mu \nu }\mathrm{d}{x}^{\mu }\mathrm{d}{x}^{\nu },$		(19)
		$=$	$\mathrm{d}{t}^2-\left(\mathrm{d}{x}^2+\mathrm{d}{y}^2+\mathrm{d}{z}^2\right),$		(20)

and so the components of the metric are

$$\boxed{{\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)}.$$

(21)

Note that in this course we set the speed of light $c=1$; otherwise $c^2$ would appear in the first entry (generating $c^2\mathrm{d}{t}^2$ as the first term in (20)). SR applies in inertial frames (where there is no gravity). We’ll shortly see that it also applies locally in GR frames that are free-falling. Because it is only locally equivalent, we can’t say in general that there is a GR frame where

$$\mathrm{\Delta }{s}^2=\mathrm{\Delta }{t}^2-\mathrm{\Delta }{x}^2-\mathrm{\Delta }{y}^2-\mathrm{\Delta }{z}^2$$

(22)

for finite coordinate displacements $\mathrm{\Delta }t$, $\mathrm{\Delta }x$, $\mathrm{\Delta }y$, $\mathrm{\Delta }z$. In fact, if equation (22) holds regardless of the size of the coordinate changes, one may show there is no gravitational field. Conversely, when a gravitational field is present, (22) cannot apply everywhere. The relationship between the individual locally flat coordinate systems in neighbouring patches is what generates curvature and gravity.

Let’s consider two forms of the key principle that Einstein used to develop general relativity.

Weak Equivalence Principle (WEP): At any point in a gravitational field, in a frame moving with the free fall acceleration at that point, the laws of motion of free test particles have their usual Special Relativity (SR) form, i.e. particles move in straight lines with uniform velocity locally.

Strong Equivalence Principle (SEP): At any point in a gravitational field, in a frame moving with the free fall acceleration at that point, all the laws of physics have their usual Special Relativity (SR) form, except gravity, which disappears locally.

Instead of thinking of particles moving under a force, the WEP allows us to think of them in a frame without gravity, but moving with the free fall acceleration at that point. The SEP goes further and claims that all laws of physics (e.g. those governing electromagnetic fields) also behave in this way. Clearly, the WEP follows from the SEP. Despite its name, the WEP is in itself a very powerful idea which has two important consequences. First, it explains the equivalence of gravitational mass and inertial mass. Second, it says that the motion of a test body in a gravitational field only depends on its position and instantaneous velocity in spacetime.

The equivalence principle works only locally (technically, in an infinitessimally small patch) where the gravitational field can be taken to be uniform. If the gravitational field varies across a patch, tidal forces appear which cannot be ignored. This is just like curvature – on a small enough scale, any surface looks flat. But as one takes a broader view, variations in the surface become significant.

True gravity as opposed to just an accelerating frame manifests itself via the separation or coming together of test particles initially on parallel tracks. This is equivalent to what we said above: there is no global Lorentz frame in the presence of a non-uniform gravitational field.

What is the metric of an expanding universe?

If the distance today is $x_0$, the physical distance between two points at some earlier time $t$ was $a(t)x_0$ (see Figure above). In a flat (as opposed to open or closed) universe, the spatial part of the metric must look like a normal Euclidean metric, except that the distance must be multiplied by the scale factor $a(t)$. Thus, the metric of a flat, expanding universe is the Friedmann-Robertson-Walker metric:

$$\boxed{g_{\mu \nu }=\mathrm{diag}(1,-a^2(t),-a^2(t),-a^2(t))}.$$

(23)

or, equivalently,

$$\boxed{\mathrm{d}{s}^2=\mathrm{d}{t}^2-a^2(t)(\mathrm{d}{x}^2+\mathrm{d}{y}^2+\mathrm{d}{z}^2)}$$

(24)

For once, it’s worth memorising these. You probably will do so naturally by the end of the course if you work through the exercises.

When inhomogeneities are introduced, the metric will become more complicated; the perturbed part of the metric will be determined by the irregularities in the matter and the radiation.

There is some important notation to understand:

• •

  Indices appear in superscript for vectors, e.g. $A^{\mu }$. The time component is $A^0$, while the space components are $A^1$, $A^2$ and $A^3$. Obviously this can look a bit like you’re taking powers of some scalar quantity $A$, but normally the context should make clear what is going on.

• •

  Indices appear in subscript for covectors, e.g. $B_{\mu }$. The time component is $B_0$, while the space components are $B_1$, $B_2$ and $B_3$.

• •

  The components of vectors change when you change coordinate system; so do covectors, but the transformation law is different. For examination purposes you would not be expected to remember the exact transformation laws; but if you need to refer to them, they are given by (12) and (16) respectively.

• •

  When you encounter a superscript and a subscript index in the same term, it means that there is implicitly a sum, i.e. $A^{\mu }{B}_{\mu }\equiv {\sum }_{\mu =0}^3{A}^{\mu }{B}_{\mu }$.

• •

  This in turn means that $A^{\mu }{B}_{\mu }$ is a scalar (i.e. a single number that does not depend on coordinate system).

• •

  Greek indices like $\mu$, $\nu$ etc are spacetime indices which run from $0$ to $3$; regular indices like $a$, $b$, $i$, $j$ etc are spatial indices, running from $1$ to $3$.

• •

  “Tensors” are generalisations of vectors and covectors; they can have any number of independent indices. A tensor with two indices can be thought of as a matrix that changes in a specific way if you change coordinate system. Again, you wouldn’t be expected to remember the transformation law in an exam for this course, but it is a straightforward generalisation of the rule for covectors and vectors.

With this in hand, you need to remember that the spacetime metric $g_{\mu \nu }$ is a tensor that:

• •

  provides the connection between the values of the coordinates and the physically meaningful measure of the spacetime interval $\mathrm{d}{s}^2$, following the rule $\mathrm{d}{s}^2=g_{\mu \nu }\mathrm{d}{x}^{\mu }\mathrm{d}{x}^{\nu }$;

• •

  has values depending on the coordinate system in such a way that $g_{\mu \nu }{A}^{\mu }{B}^{\nu }$ is independent of coordinate system for any vectors $A$ and $B$;

• •

  is defined to be symmetric in $\mu \leftrightarrow \nu$;

• •

  in principle, has $4$ diagonal and $6$ off-diagonal components which must be specified to describe a spacetime in a particular coordinate system;

• •

  can convert a vector $A^{\mu }$ into a covector $A_{\mu }$ via the rule $A_{\mu }=g_{\mu \nu }{A}^{\nu }$;

• •

  has an inverse, $g^{\mu \nu }$ which obeys $g^{\alpha \beta }{g}_{\beta \gamma }={\delta }_{\gamma }^{\alpha }$ and consequently converts covectors back to vectors, $A^{\mu }=g^{\mu \nu }{A}_{\nu }$;

• •

  has a sign difference for spacelike as opposed to timelike directions – in this course we choose the minus sign to be associated with spacelike directions;

• •

  contains the scalefactor (which varies with time) in a homogeneous, expanding universe – see equation (23) and (24);

As the course progresses we will need to use more complicated metrics, but for now we can move onto understanding how the metric of an expanding universe affects physical, measurable quantities.