8 Fields

In Section 7 we recapped the Euler-Lagrange equations to see how they simplify calculations for geodesics. To go further, we need just a bit more formal development to see how the equations work for fields, as opposed to single particles.

We first replace the single coordinate $q(t)$ by a set of spacetime-dependent fields ${\mathrm{\Phi }}^i(x^{\mu })$, and the action becomes a functional of these fields. Such fields could be, for example, electric or magnetic fields, or gravitational fields, or scalar fields (like the Higgs field) and so on… in any case, $i$ labels individual fields. A functional is a function of an infinite number of variables – in this case, the values of a field at each point in spacetime.

The Lagrangian can now be expressed as an integral over the space of a Lagrangian density, $ℒ$, which is a function of the fields ${\mathrm{\Phi }}^i$ and their spacetime derivatives ${\partial }_{\mu }{\mathrm{\Phi }}^i$:

$$L=\int d^{3}xℒ({\mathrm{\Phi }}^i,{\partial }_{\mu }{\mathrm{\Phi }}^i).$$

(217)

Then the action becomes,

$$S=\int 𝑑tL=\int d^{4}xℒ({\mathrm{\Phi }}^i,{\partial }_{\mu }{\mathrm{\Phi }}^i).$$

(218)

We should note in passing that, if we’re to have a hope of building a coordinate-independent theory, we need to make sure that $S$ is independent of coordinate system. We will adopt this as a requirement.

The Euler-Lagrange equations, just like when we considered them for individual particles, come from requiring that $S$ be invariant under small variations – but this time around, variations of the field:

$${\mathrm{\Phi }}^i\to {\mathrm{\Phi }}^i+\delta {\mathrm{\Phi }}^i,{\partial }_{\mu }{\mathrm{\Phi }}^i\to {\partial }_{\mu }{\mathrm{\Phi }}^i+\delta ({\partial }_{\mu }{\mathrm{\Phi }}^i)={\partial }_{\mu }{\mathrm{\Phi }}^i+{\partial }_{\mu }(\delta {\mathrm{\Phi }}^i).$$

(219)

The expression for variation in ${\partial }_{\mu }{\mathrm{\Phi }}^i$ is the derivative of the variation of ${\mathrm{\Phi }}^i$, just as we saw in Exercise 7. Since $\delta {\mathrm{\Phi }}^i$ is assumed to be small, we can Taylor-expand the Lagrangian under this variation,

	$ℒ({\mathrm{\Phi }}^i,{\partial }_{\mu }{\mathrm{\Phi }}^i)$	$\to$	$ℒ({\mathrm{\Phi }}^i+\delta {\mathrm{\Phi }}^i,{\partial }_{\mu }{\mathrm{\Phi }}^i+{\partial }_{\mu }\delta {\mathrm{\Phi }}^i),$		(220)
		$=$	$ℒ({\mathrm{\Phi }}^i,{\partial }_{\mu }{\mathrm{\Phi }}^i)+{\displaystyle \frac{\partial ℒ}{\partial {\mathrm{\Phi }}^i}}\delta {\mathrm{\Phi }}^i+{\displaystyle \frac{\partial ℒ}{\partial ({\partial }_{\mu }{\mathrm{\Phi }}^i)}}{\partial }_{\mu }(\delta {\mathrm{\Phi }}^i).$

Correspondingly, the action goes to $S\to S+\delta S$, with

$$\delta S=\int d^{4}x\left[\frac{\partial ℒ}{\partial {\mathrm{\Phi }}^i}\delta {\mathrm{\Phi }}^i+\frac{\partial ℒ}{\partial ({\partial }_{\mu }{\mathrm{\Phi }}^i)}{\partial }_{\mu }(\delta {\mathrm{\Phi }}^i)\right].$$

(221)

😇 Exercise 8A Factor out the $\delta {\mathrm{\Phi }}^i$ term from the integrand, by integrating the second term by parts. Hence show that $$\delta S=\int d^{4}x\left[\frac{\partial ℒ}{\partial {\mathrm{\Phi }}^i}-{\partial }_{\mu }\left(\frac{\partial ℒ}{\partial ({\partial }_{\mu }{\mathrm{\Phi }}^i)}\right)\right]\delta {\mathrm{\Phi }}^i.$$ (222) Hint: You will obtain one term which is a total derivative – the integral of something of the form ${\partial }_{\mu }{V}^{\mu }$ – that can be converted to a surface term by the four-dimensional version of Stokes’ Theorem. As when we derived Euler-Lagrange equations for particles, we choose to consider variations that vanish at the boundary, along with their derivatives.

As before, we now argue that since this is to be true for any possible change $\delta {\mathrm{\Phi }}^i$, the term multiplying $\delta {\mathrm{\Phi }}^i$ must be zero everywhere:

$$\boxed{\frac{\partial ℒ}{\partial {\mathrm{\Phi }}^i}-{\partial }_{\mu }\left(\frac{\partial ℒ}{\partial ({\partial }_{\mu }{\mathrm{\Phi }}^i)}\right)=0}.$$

(223)

Before moving on, let’s introduce a slightly more compact notation for this kind of argument. Namely, let $\delta S/\delta {\mathrm{\Phi }}^i$ be the functional derivative of a functional $S$ with respect to a function ${\mathrm{\Phi }}^i$, defined to satisfy

$$\delta S=\int d^{4}x\frac{\delta S}{\delta {\mathrm{\Phi }}^i}\delta {\mathrm{\Phi }}^i$$

(224)

The actual function denoted by $\delta S/\delta {\mathrm{\Phi }}^i$ has to be extracted by the exact same series of manipulations that we’ve explored above, so there’s nothing actually new here. It’s just that, for example, saying “$\delta S/\delta {\mathrm{\Phi }}^i=0$” will be a helpful short-hand for saying “$\delta S=0$ for all possible small changes to ${\mathrm{\Phi }}^i$”.

In addition to all the benefits of the Lagrangian formulation for particle mechanics that we discussed above, in field theory we also get a unique definition for the energy-momentum tensor. This is essential to have in hand as we move from familiar content for the universe (matter, radiation) on to more abstract content required for study of the early Universe and inflation.

A scalar field is simply an association of a single number with every point in spacetime. To turn this into a physical theory, we need to specify an equation of motion for the field – or, equivalently, a Lagrangian. We’ll go right ahead and start by writing down a Lagrangian:

$$S_{\phi }=\int {\mathrm{d}}^{4}x\left(\frac{1}{2}{\eta }^{\mu \nu }({\partial }_{\mu }\phi )({\partial }_{\nu }\phi )-V(\phi )\right)$$

(225)

The equations of motion for this field are

$$\mathrm{\square }\phi +\frac{\mathrm{d}V}{\mathrm{d}\phi }=0\text{,}$$

(226)

where the box operator (also known as the d’Alembert operator) is defined by $\mathrm{\square }\phi \equiv {\eta }^{\mu \nu }{\partial }_{\mu }{\partial }_{\nu }\phi =\ddot{\phi }-{\nabla }^2\phi$.

😇 Exercise 8B Derive equation (226) from equation (225) using expression (223).

What on earth does it mean? First, set $V=0$. Then $\phi$ obeys a wave equation. Recalling that $c=1$ in this course, the waves will propagate at the speed of light. You can see this explicitly by substituting the trial wave solution $\phi \propto \exp (ik(x-t))$.

So, localized features in the field travel along and start to look rather like a quantum particle (more in the exercise below).

Alternatively, imagine the field starts out completely homogeneous. Then, ${\nabla }^2\phi =0$ (and it remains zero for all time). All we have, at each point, is $\ddot{\phi }+\mathrm{d}V/\mathrm{d}\phi =0$. This equation is also familiar; for example, if we make the potential $V(\phi )\propto {\phi }^2$ we just have a ‘mass on a spring’ which oscillates back and forth forever. The only difference is that $\phi$ is an abstract number, not an actual position. This is a useful analogy: quite often when discussing inflation we will talk about things like “balls rolling down a hill”. The thing to remember is that in reality there is no ball, no rolling, and no hill! There is just an abstract mean field value represented by the ball’s height, an abstract rate of change of that value represented by the motion of the ball, and an abstract potential represented by the hill.

☞ Exercise 8C What if we combine the two scenarios? Let’s say we have $V(\phi )=m^2{\phi }^2/2$ and wish to find a travelling wave solution. Substitute $\phi \propto \exp (i(kx-\omega t))$ and show that $${\omega }^2=k^2+m^2\text{.}$$ (227) Recall from your quantum mechanics courses that $⟨E⟩=\mathrm{\hslash }\omega$, and that $⟨p⟩=\mathrm{\hslash }k$. The relation you have derived then becomes the relativistic equation $E^2=m^2+p^2$ (with $c=\mathrm{\hslash }=1$ as always).

The result from the exercise above is not a coincidence – we’re seeing that relativistic particles can be understood as oscillations in quantum fields. We’ll touch on this later but the detail is beyond the scope of this course (see e.g. PHAS0073 instead). Note that in the limit $k\to 0$ – when the fluctuations in the field are on very long wavelengths – we get ${\omega }^2\approx m^2$, i.e. we automatically revert to the homogeneous, coherent oscillator picture outlined just above the exercise. This sort of coherent behaviour of a scalar field (suitably modified for expanding spacetime) will be our main focus for cosmology.

Before looking at cosmological implications, we need to generalise the field theory framework into curved space. One might imagine that we can use the equivalence principle to promote the fixed metric of Minkowski space ${\eta }^{\mu \nu }$ into a general metric $g^{\mu \nu }$:

$$S_{\phi }^{\mathrm{first}\mathrm{guess}}=\int {\mathrm{d}}^{4}x\left(\frac{1}{2}{g}^{\mu \nu }({\nabla }_{\mu }\phi )({\nabla }_{\nu }\phi )-V(\phi )\right)$$

(228)

Sadly, it’s not true. The problem is that ${\mathrm{d}}^{4}x$ is dependent on our choice of coordinates. For example, suppose we transform coordinates such that $x^1\to x^{1\prime }=2x^1$. This isn’t permitted in special relativity, but in general relativity we have to allow it – remember the whole construction is about keeping physics invariant even in different coordinate systems. If $S_{\phi }$ is to have an actual physical meaning, it has to be invariant under this sort of transformation.

The integrand is correctly invariant under our trial transformation, because the $g^{11\prime }$ component of the metric picks up a factor $4$ that cancels against a $1/2$ attached to each of the $\partial /\partial x^{1\prime }$s. But the integral still changes because $\mathrm{d}{x}^1\to \mathrm{d}{x}^{1\prime }=2\mathrm{d}{x}^1$. This needs to be cancelled off, and it turns out the way to do it in general is to write:

$$\boxed{S_{\phi }=\int {\mathrm{d}}^{4}x\sqrt{-g}\left(\frac{1}{2}{g}^{\mu \nu }({\nabla }_{\mu }\phi )({\nabla }_{\nu }\phi )-V(\phi )\right)}$$

(229)

where $g\equiv det{g}_{\mu \nu }$ is the determinant of $g_{\mu \nu }$. The point is, this determinant exactly cancels the effect of coordinate transformations on the ${\mathrm{d}}^{4}x$ measure, as we’ll now see…

😇 Exercise 8D Check that this works in our specific test case of $x^1\to x^{\prime 1}=2x^1$. Start by verifying that, for the original metric $g^{\mu \nu }={\eta }^{\mu \nu }$, the action reduces back to (225). Then, calculate $det{g}_{\mu \nu }$ and so write the action in the primed coordinates. Finally transform your new expression back into the unprimed coordinate system to once again recover (225).

Of course we’ve just shown it works for a simple, specific coordinate transformation but it’s not too much extra work to show that it actually works for any coordinate transformation (see Carroll for help with this if you want to give it a go)²²²² 22 It also turns out that measure theory gives you a more fundamental justification for why ${\mathrm{d}}^{4}x\sqrt{-g}$ has to be the correct way to integrate in a curved space; see https://en.wikipedia.org/wiki/Volume_form..

By Einstein’s equivalence principle, since the action (229) is coordinate-invariant and boils down to (225) in locally Minkowski coordinates, it’s the correct extension of scalar field theory into curved space.

Following the arguments given above for the specific case, it’s correct to generalise and say that actions in curved space look like this

$$S=\int {\mathrm{d}}^{4}x\sqrt{-g}\widehat{ℒ}({\mathrm{\Phi }}^i,{\nabla }_{\mu }{\mathrm{\Phi }}^i),$$

(230)

where $\widehat{ℒ}$ is a scalar (i.e. invariant under coordinate transformations). Here, we don’t consider $g_{\mu \nu }$ to be one of the fields because we have to treat it on a different footing (later). So in the scalar field example, there is just ${\mathrm{\Phi }}^1=\phi$ and no other fields to consider.

What do the Euler-Lagrange equations now look like? Well, there’s nothing special about this new expression, so all the old arguments go through and the correct equations of motion for the field ${\mathrm{\Phi }}^i$ are:

$$\frac{\partial \left(\sqrt{-g}\widehat{ℒ}\right)}{\partial {\mathrm{\Phi }}^i}-{\partial }_{\mu }\left[\frac{\partial \left(\sqrt{-g}\widehat{ℒ}\right)}{\partial ({\nabla }_{\mu }{\mathrm{\Phi }}^i)}\right]=0.$$

(231)

We’ve derived this equation from a coordinate-invariant expression so it’s got to be coordinate-invariant itself. Knowing that, we can just throw caution to the wind and have a guess at what the above statement really means:

$$\boxed{\frac{\partial \widehat{ℒ}}{\partial {\mathrm{\Phi }}^i}-{\nabla }_{\mu }\left(\frac{\partial \widehat{ℒ}}{\partial ({\nabla }_{\mu }{\mathrm{\Phi }}^i)}\right)=0.}$$

(232)

This is a correct guess. The technical details of why it works out are unnecessary for our future purposes, but if you’re interested, the next exercise guides you through it.

Exercise 8E Show that equation (232) is correct. 1. First you take the $\sqrt{-g}$’s out side their immediate derivatives just by noting that $g_{\mu \nu }$ is not one of the ${\mathrm{\Phi }}^i$ field(s) under consideration; 2. Next, you prove that ${\partial }_{\mu }(\sqrt{-g}{A}^{\mu })=\sqrt{-g}{\nabla }_{\mu }{A}^{\mu }$ for any vector field $A^{\mu }$. If you have trouble with this step, start by trying to show the following preliminary result: ${\displaystyle \frac{\partial \sqrt{-g}}{\partial g_{\alpha \beta }}}$ $=$ ${\displaystyle \frac{\sqrt{-g}}{2}}{g}^{\alpha \beta },$ (233) which takes a fair bit of work. It arises most naturally from the variation of the identity $\ln (det𝐌)=\mathrm{Tr}(\ln 𝐌)$ for $𝐌$ a square matrix. It’s equally worth remembering that $\exp (\ln 𝐌)=𝐌$. 3. Then convince yourself that $\partial \widehat{ℒ}/\partial ({\nabla }_{\mu }{\mathrm{\Phi }}^i)$ is indeed a vector field $A^{\mu }$, and so transform the second term appropriately; 4. Finally you cancel the $\sqrt{-g}$ that is now outside all the derivatives in both terms.

Now we can apply the Euler-Lagrange equations of motion in curved space, equation (232), to the action (229) to show that

$$\mathrm{\square }\phi +\frac{dV}{d\phi }=0,$$

(234)

where now

$$\mathrm{\square }\phi ={\nabla }^{\nu }{\nabla }_{\nu }\phi =g^{\mu \nu }{\nabla }_{\mu }{\nabla }_{\nu }\phi .$$

(235)

It’s all just the same as in flat space, but with partial derivatives replaced by covariant derivatives – a direct consequence of the equivalence principle, of course. It really couldn’t have worked out any other way.

We have developed the machinary to work out how fields behave within a curved spacetime – but now we have to recognise the fact that the metric, $g_{\mu \nu }$, is itself a field (i.e. a function of spacetime position). What scalars can we make out of the metric to serve as a Lagrangian? We know that the metric can be set equal to the Minkowski metric ($g_{\mu \nu }={\eta }_{\mu \nu }$) and its derivatives set to zero at any one point (remember Exercise 4). Consequently any non-trivial, coordinate-invariant scalar must involve at least second derivatives of the metric.

We have already encountered the Ricci scalar. It turns out to be the only independent scalar constructed from the metric, which is no higher than second order in its derivatives. The mathematician David Hilbert figured that this was the simplest possible choice for a Lagrangian, and proposed that the following action might be appropriate for general relativity:

$$S_H=\int {\mathrm{d}}^{4}x\sqrt{-g}R\mathit{\hspace{1em}}\text{(Einstein–Hilbert Action)}.$$

(236)

This turns out to be a correct guess (as we will sketch out below). In some ways it gives a very natural explanation for why Einstein’s equations look the way they do – because their action is as simple as it possibly could be.

How can we check that the Einstein-Hilbert action correctly describes GR? Since $R$ is a function of $g$, the approach that we so carefully developed above – working with $\widehat{ℒ}$ and just applying the “covariantised” Euler-Lagrange equations (232) – does not work. We have to work with the “raw” Lagrangian (231) and replace ${\mathrm{\Phi }}^i$ by $g_{\mu \nu }$ (i.e. we treat each component of the metric tensor as a separate field when deriving the equations of motion).

This becomes a bit of a dreary exercise in algebra, so (although I certainly encourage you to go through the derivation once in your life; see Exercise 5) let’s for now just state the result:

	$\delta S_H$	$=-{\displaystyle \int {\mathrm{d}}^{4}x\sqrt{-g}\left[R^{\mu \nu }-\frac{1}{2}R{g}^{\mu \nu }\right]\delta g_{\mu \nu }},$
		$=+{\displaystyle \int {\mathrm{d}}^{4}x\sqrt{-g}\left[R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }\right]\delta g^{\mu \nu }},$		(237)

meaning that at stationary points of $S_H$ with respect to variations in $g_{\mu \nu }$ we have:

$$\frac{1}{\sqrt{-g}}\frac{\delta S_H}{\delta g^{\mu \nu }}=R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }=0,$$

(238)

which has recovered Einstein’s equation in a vacuum ($T_{\mu \nu }=0$).

To generalise to Einstein’s equation coupled to matter, we just need to include the relevant matter terms in the action. Consider

$$S=\frac{-1}{16\pi G}{S}_H+S_M,$$

(239)

where $S_M$ is the matter action, and we have pre-normalized the gravitational action to get the right answer. (Note our normalization includes a minus sign in front of $S_H$, which is not present in all textbooks – it depends on the metric signature.) The calculation for the revised action goes through just by linearity:

$$\frac{1}{\sqrt{-g}}\frac{\delta S}{\delta g^{\mu \nu }}=\frac{-1}{16\pi G}\left[R_{\mu \nu }-\frac{1}{2}R{g}_{\mu \nu }\right]+\frac{1}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu \nu }}=0\text{,}$$

(240)

from which one recovers the Einstein equation provided that the energy-momentum tensor can be written

$$\boxed{T_{\mu \nu }=\frac{2}{\sqrt{-g}}\frac{\delta S_M}{\delta g^{\mu \nu }}}.$$

(241)

(Note again the different sign to some textbooks, again due to our choice of metric signature.)

Consider again the action (229) for a scalar field. Now vary this action with respect to $g_{\mu \nu }$, not $\phi$:

$$\delta S_{\phi }=\int {\mathrm{d}}^{4}x\sqrt{-g}\delta g^{\mu \nu }\left[\frac{1}{2}{\nabla }_{\mu }\phi {\nabla }_{\nu }\phi -\frac{g_{\mu \nu }}{2}\left(\frac{g^{\rho \sigma }}{2}{\nabla }_{\rho }\phi {\nabla }_{\sigma }\phi -V(\phi )\right)\right],$$

(242)

we obtain from (241) the energy-momentum tensor for a scalar field,

$$\boxed{T_{\mu \nu }^{(\phi )}={\nabla }_{\mu }\phi {\nabla }_{\nu }\phi -g_{\mu \nu }\left[\frac{1}{2}{g}^{\rho \sigma }{\nabla }_{\rho }\phi {\nabla }_{\sigma }\phi -V(\phi )\right]}.$$

(243)

Here, it is worth pointing out that you will find different sign conventions to (243) in the cosmology literature. This can be traced to the fact that most of the cosmology literature uses the opposite metric signature to ours: $(-,+,+,+)$.

Exercise 8F If you want to check that the variation of the Hilbert action really gives rise to (237), start by noting that $\delta (\sqrt{-g}R)=R\delta \sqrt{-g}+\sqrt{-g}\delta R$, then tackle the two terms separately. For the first term, remind yourself of result (233). For the second term, further split $\delta R=R^{\alpha \beta }\delta g_{\alpha \beta }+\delta R^{\alpha \beta }{g}_{\alpha \beta }$ and use the expression (105) to show that $$\delta R_{\alpha \beta }={\nabla }_{\mu }\delta {\mathrm{\Gamma }}_{\alpha \beta }^{\mu }-{\nabla }_{\beta }\delta {\mathrm{\Gamma }}_{\alpha \mu }^{\mu }\text{.}$$ (244) Hence argue the term $\propto \delta R_{\mu \nu }$ leads to an integral with respect to the covariant divergence of a vector; by Stokes’ Theorem, this is equal to a boundary term that you can discard. Putting everything together, you should obtain the required result.

Exercise 8G We asserted that the energy-momentum tensor must be given by Equation (241); otherwise we don’t recover Einstein’s equations. This is certainly a valid argument, but we earlier remarked that conservation of the energy-momentum tensor (${\nabla }_{\mu }{T}_{\nu }^{\mu }=0$) is the result of a symmetry, and so can be derived from Noether’s theorem. If you’re curious to see how this works out, start in flat Minkowski space (the generalisation to curved spaces doesn’t change the main thrust of the argument). Consider coordinates $x^{\mu }$ with metric $g_{\mu \nu }={\eta }_{\mu \nu }$. Now change the coordinate system by some infinitessimal amount $x^{\mu }\to {\tilde{x}}^{\mu }=x^{\mu }+{ϵ}^{\mu }$. Allow all field content to follow the transformed coordinates, so that for scalar fields $\phi \to \tilde{\phi }$ with $\tilde{\phi }(\widetilde{𝒙})=\phi (𝒙)$. Show that ${\tilde{g}}_{\mu \nu }(\tilde{x})={\eta }_{\mu \nu }-{ϵ}_{\mu ,\nu }-{ϵ}_{\nu ,\mu }$. Hence argue that, if the action is invariant under the coordinate transformation, $T_{,\nu }^{\mu \nu }=0$ where $T^{\mu \nu }=2\delta S/\delta g_{\mu \nu }$. How and why does this derivation differ from the derivation of the canonical energy-momentum tensor in quantum field theory books?

The above is all pretty technical, so let’s step back and see the practical consequences. So far we’ve used the Euler-Lagrange formalism to define what is meant by the energy-momentum tensor for a scalar field. Because it has been derived from a coherent, self-consistent framework, we can be assured that (for example) $T_{\nu ;\mu }^{\mu }=0$ for our definition, making it properly compatible with Einstein’s theory.

The single most important result for our purposes is Eq. (243) – the correct energy-momentum tensor for a scalar field. Let’s simplify by writing it out in component form in the case that the field is homogeneous but time-varying (i.e. ${\nabla }_i\varphi =0$, but ${\nabla }_0\varphi =\dot{\varphi }\ne 0$):

(245)

Compare this with Eq. (91), and we can read off the density and pressure of a scalar field in a homogeneous universe:

	${\rho }_{\phi }$	$=\underset{\text{“kinetic” energy}}{\underbrace{{\displaystyle \frac{1}{2}}{\dot{\phi }}^2}}+\underset{\text{“potential” energy}}{\underbrace{V(\phi )}}$		(246)
	$P_{\phi }$	$={\displaystyle \frac{1}{2}}{\dot{\phi }}^2-V(\phi )\text{.}$		(247)

The ‘kinetic energy’ term is named by analogy with the corresponding term if we had just a single particle with position $\phi$. But let’s be clear: $\phi$ is not a position, it’s a field value. It’s important to realise that the existence of kinetic energy in the field does not imply the field configuration represents a moving particle. In that light, the terminology is actually quite unfortunate. But it’s the convention, so we’re stuck with it.

The resulting equation of state

$$w=\frac{\frac{1}{2}{\dot{\phi }}^2-V}{\frac{1}{2}{\dot{\phi }}^2+V}$$

(248)

shows that a scalar field can lead to negative pressure () and accelerated expansion () if the potential energy $V$ dominates over the kinetic energy $\frac{1}{2}{\dot{\phi }}^2$. In the next chapter, we will look at this phenomenology in detail.

As stated at the start of the chapter, the mathematical derivations are non-examinable. You may be asked to briefly explain the physics, as we will apply it to inflation in the next chapter. This can be summarised as follows:

• •

  A field is just an assignment of a number, vector, or other quantity to every point in spacetime; often by ‘field’ we also have in mind that there are some equations of motion that predict its evolution, i.e. the future values of the field can be predicted from the past values of the fields.

• •

  The equations governing classical evolution of fields can be derived from the Euler-Lagrange equations applied to suitable actions.

• •

  Deriving things that way, rather than just writing down an evolution equation, helps us analyse the classical behaviour of the field – and is also essential for quantising the field (Chapter 14).

• •

  In cosmology, we are usually most interested in scalar fields $\phi (𝒙)$ (i.e. fields which assign a single number to each point in spacetime), because they may be responsible for powering the accelerated expansion of inflation and/or dark energy.

• •

  Even in flat space, with the simplest possible non-trivial Lagrangian for a scalar field, the field can exhibit a range of behaviours from coherent oscillations to particle-like excitations. Within cosmology, it is normally coherent behaviour of the field, where the field takes one value throughout space, that is of most interest.

• •

  When describing how this field changes over time, we quite often refer to analogies like “a ball rolling down a hill”. But there is no physical ball or hill! The horizontal position of the ball represents the field value $\phi$ (the same throughout space) and the hill is an abstract potential $V(\phi )$ which you can think of as the ball’s height.

• •

  The metric $g_{\mu \nu }(𝒙)$ can also be thought of as a tensor field. The action for Einstein’s relativity is then remarkably simple, Eq. (236).

• •

  Using this action we can deduce the correct energy-momentum tensor for the fields within the Universe;

• •

  This reveals the density $\rho$ and pressure $P$ of a scalar field are given by equations (246) and (247) respectively, in the limit that the field is completely homogeneous.

• •

  If the potential $V(\phi )$ dominates over the kinetic ${\dot{\phi }}^2/2$ term, we find a negative equation of state $P\simeq -\rho$ which we previously concluded (Section 2) will give rise to accelerated expansion.