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# Chapter 7. Symmetries

This chapter is not required in order to understand the later material.

## 7.1 Killing vectors

a / The two-dimensional space has a symmetry which can be visualized by imagining it as a surface of revolution embedded in three-space. Without reference to any extrinsic features such as coordinates or embedding, an observer on this surface can detect the symmetry, because there exists a vector field $$\boldsymbol{\xi}du$$ such that translation by $$\boldsymbol{\xi}du$$ doesn't change the distance between nearby points.

b / Wilhelm Killing (1847-1923).

c / Vectors at a point P on a sphere can be visualized as occupying a Euclidean plane that is particular to P.

d / Example 3: A cylinder has three local symmetries, but only two that can be extended globally to make Killing vectors.

The Schwarzschild metric is an example of a highly symmetric spacetime. It has continuous symmetries in space (under rotation) and in time (under translation in time). In addition, it has discrete symmetries under spatial reflection and time reversal. In section 6.2.6, we saw that the two continuous symmetries led to the existence of conserved quantities for the trajectories of test particles, and that these could be interpreted as mass-energy and angular momentum.

Generalizing, we want to consider the idea that a metric may be invariant when every point in spacetime is systematically shifted by some infinitesimal amount. For example, the Schwarzschild metric is invariant under $$t\rightarrow t+dt$$. In coordinates $$(x^0,x^1,x^2,x^3)=(t,r,\theta,\phi)$$, we have a vector field $$(dt,0,0,0)$$ that defines the time-translation symmetry, and it is conventional to split this into two factors, a finite vector field $$\boldsymbol{\xi}$$ and an infinitesimal scalar, so that the displacement vector is

$\begin{equation*} \boldsymbol{\xi}dt = (1,0,0,0)dt . \end{equation*}$

Such a field is called a Killing vector field, or simply a Killing vector, after Wilhelm Killing. When all the points in a space are displaced as specified by the Killing vector, they flow without expansion or compression. The path of a particular point, such as the dashed line in figure a, under this flow is called its orbit. Although the term “Killing vector” is singular, it refers to the entire field of vectors, each of which differs in general from the others. For example, the $$\boldsymbol{\xi}$$ shown in figure a has a greater magnitude than a $$\boldsymbol{\xi}$$ near the neck of the surface.

The infinitesimal notation is designed to describe a continuous symmetry, not a discrete one. For example, the Schwarzschild spacetime also has a discrete time-reversal symmetry $$t \rightarrow -t$$. This can't be described by a Killing vector, because the displacement in time is not infinitesimal.

##### Example 1: The Euclidean plane
The Euclidean plane has two Killing vectors corresponding to translation in two linearly independent directions, plus a third Killing vector for rotation about some arbitrarily chosen origin O. In Cartesian coordinates, one way of writing a complete set of these is is
\begin{align*} \boldsymbol{\xi}_1 &= (1,0) \\ \boldsymbol{\xi}_2 &= (0,1) \\ \boldsymbol{\xi}_3 &= (-y,x) . \end{align*}
A theorem from classical geometry1 states that any transformation in the Euclidean plane that preserves distances and handedness can be expressed either as a translation or as a rotation about some point. The transformations that do not preserve handedness, such as reflections, are discrete, not continuous. This theorem tells us that there are no more Killing vectors to be found beyond these three, since any translation can be accomplished using $$\boldsymbol{\xi}_1$$ and $$\boldsymbol{\xi}_2$$, while a rotation about a point P can be done by translating P to O, rotating, and then translating O back to P.

In the example of the Schwarzschild spacetime, the components of the metric happened to be independent of $$t$$ when expressed in our coordinates. This is a sufficient condition for the existence of a Killing vector, but not a necessary one. For example, it is possible to write the metric of the Euclidean plane in various forms such as $$ds^2=dx^2+dy^2$$ and $$ds^2=dr^2+r^2d\phi^2$$. The first form is independent of $$x$$ and $$y$$, which demonstrates that $$x\rightarrow x+dx$$ and $$y\rightarrow y+dy$$ are Killing vectors, while the second form gives us $$\phi\rightarrow \phi+d\phi$$. Although we may be able to find a particular coordinate system in which the existence of a Killing vector is manifest, its existence is an intrinsic property that holds regardless of whether we even employ coordinates. In general, we define a Killing vector not in terms of a particular system of coordinates but in purely geometrical terms: a space has a Killing vector $$\boldsymbol{\xi}$$ if translation by an infinitesimal amount $$\boldsymbol{\xi}du$$ doesn't change the distance between nearby points. Statements such as “the spacetime has a timelike Killing vector” are therefore intrinsic, since both the timelike property and the property of being a Killing vector are coordinate-independent.

Killing vectors, like all vectors, have to live in some kind of vector space. On a manifold, this vector space is particular to a given point, figure c. A different vector space exists at every point, so that vectors at different points, occupying different spaces, can be compared only by parallel transport. Furthermore, we really have two such spaces at a given point, a space of contravariant vectors and a space of covariant ones. These are referred to as the tangent and cotangent spaces.The infinitesimal displacements we've been discussing belong to the contravariant (upper-index) space, but by lowering and index we can just as well discuss them as covariant vectors. The customary way of notating Killing vectors makes use of the fact, mentioned in passing on p. 199, that the partial derivative operators $${\partial_0,\partial_1,\partial_2,\partial_3}$$ form the basis for a vector space. In this notation, the Killing vector of the Schwarzschild metric we've been discussing can be notated simply as

$\begin{equation*} \boldsymbol{\xi} = \partial_t . \end{equation*}$

The partial derivative notation, like the infinitesimal notation, implicitly refers to continuous symmetries rather than discrete ones. If a discrete symmetry carries a point $$\text{P}_1$$ to some distant point $$\text{P}_2$$, then $$\text{P}_1$$ and $$\text{P}_2$$ have two different tangent planes, so there is not a uniquely defined notion of whether vectors $$\boldsymbol{\xi}_1$$ and $$\boldsymbol{\xi}_2$$ at these two points are equal --- or even approximately equal. There can therefore be no well-defined way to construe a statement such as, “$$\text{P}_1$$ and $$\text{P}_2$$ are separated by a displacement $$\boldsymbol{\xi}$$.” In the case of a continuous symmetry, on the other hand, the two tangent planes come closer and closer to coinciding as the distance $$s$$ between two points on an orbit approaches zero, and in this limit we recover an approximate notion of being able to compare vectors in the two tangent planes. They can be compared by parallel transport, and although parallel transport is path-dependent, the difference bewteen paths is proportional to the area they enclose, which varies as $$s^2$$, and therefore becomes negligible in the limit $$s\rightarrow0$$.

Self-check: Find another Killing vector of the Schwarzschild metric, and express it in the tangent-vector notation.

It can be shown that an equivalent condition for a field to be a Killing vector is $$\nabla_a \boldsymbol{\xi}_b+\nabla_b \boldsymbol{\xi}_a = 0$$. This relation, called the Killing equation, is written without reference to any coordinate system, in keeping with the coordinate-independence of the notion.

When a spacetime has more than one Killing vector, any linear combination of them is also a Killing vector. This means that although the existence of certain types of Killing vectors may be intrinsic, the exact choice of those vectors is not.

##### Example 2: Euclidean translations

The Euclidean plane has two translational Killing vectors $$(1,0)$$ and $$(0,1)$$, i.e., $$\partial_x$$ and $$\partial_y$$. These same vectors could be expressed as $$(1,1)$$ and $$(1,-1)$$ in coordinate system that was rescaled and rotated by 45 degrees.

##### Example 3: A cylinder
The local properties of a cylinder, such as intrinsic flatness, are the same as the local properties of a Euclidean plane. Since the definition of a Killing vector is local and intrinsic, a cylinder has the same three Killing vectors as a plane, if we consider only a patch on the cylinder that is small enough so that it doesn't wrap all the way around. However, only two of these --- the translations --- can be extended to form a smooth vector field on the entire surface of the cylinder. These might be more naturally notated in $$(\phi,z)$$ coordinates rather than $$(x,y)$$, giving $$\partial_z$$ and $$\partial_\phi$$.
##### Example 4: A sphere
A sphere is like a plane or a cylinder in that it is a two-dimensional space in which no point has any properties that are intrinsically different than any other. We might expect, then, that it would have two Killing vectors. Actually it has three, $$\boldsymbol{\xi}_x$$, $$\boldsymbol{\xi}_y$$, and $$\boldsymbol{\xi}_z$$, corresponding to infinitesimal rotations about the $$x$$, $$y$$, and $$z$$ axes. To show that these are all independent Killing vectors, we need to demonstrate that we can't, for example, have $$\boldsymbol{\xi}_x=c_1\boldsymbol{\xi}_y+c_2\boldsymbol{\xi}_z$$ for some constants $$c_1$$ and $$c_2$$. To see this, consider the actions of $$\boldsymbol{\xi}_y$$ and $$\boldsymbol{\xi}_z$$ on the point $$P$$ where the $$x$$ axis intersects the sphere. (References to the axes and their intersection with the sphere are extrinsic, but this is only for convenience of description and visualization.) Both $$\boldsymbol{\xi}_y$$ and $$\boldsymbol{\xi}_z$$ move P around a little, and these motions are in orthogonal directions, wherease $$\boldsymbol{\xi}_x$$ leaves P fixed. This proves that we can't have $$\boldsymbol{\xi}_x=c_1\boldsymbol{\xi}_y+c_2\boldsymbol{\xi}_z$$. All three Killing vectors are linearly independent.

This example shows that linear independence of Killing vectors can't be visualized simply by thinking about the vectors in the tangent plane at one point. If that were the case, then we could have at most two linearly independent Killing vectors in this two-dimensional space. When we say “Killing vector” we're really referring to the Killing vector field, which is defined everywhere on the space.

##### Example 5: Proving nonexistence of Killing vectors

$$\triangleright$$ Find all Killing vectors of these two metrics:

\begin{align*} ds^2 &= e^{-x}dx^2+e^xdy^2 \\ ds^2 &= dx^2+x^2dy^2 . \end{align*}

$$\triangleright$$ Since both metrics are manifestly independent of $$y$$, it follows that $$\partial_y$$ is a Killing vector for both of them. Neither one has any other manifest symmetry, so we can reasonably conjecture that this is the only Killing vector either one of them has. However, one can have symmetries that are not manifest, so it is also possible that there are more.

One way to attack this would be to use the Killing equation to find a system of differential equations, and then determine how many linearly independent solutions there were.

But there is a simpler approach. The dependence of these metrics on $$x$$ suggests that the spaces may have intrinsic properties that depend on $$x$$; if so, then this demonstrates a lower symmetry than that of the Euclidean plane, which has three Killing vectors. One intrinsic property we can check is the scalar curvature $$R$$. The following Maxima code calculates $$R$$ for the first metric.

load(ctensor);
dim:2;
ct_coords:[x,y];
lg:matrix([exp(-x),0],[0,exp(x)]);
cmetric();
R:scurvature(); /* scalar curvature */


The result is $$R=-e^x$$, which demonstrates that points that differ in $$x$$ have different intrinsic properties. Since the flow of a Killing field $$\xi$$ can never connect points that have different properties, we conclude that $$\xi_x=0$$. If only $$\xi_y$$ can be nonzero, the Killing equation $$\nabla_a \xi_b+\nabla_b \xi_a = 0$$ simplifies to $$\nabla_x\xi_y=\nabla_y\xi_y=0$$. These equations constrain both $$\partial_x\xi_y$$ and $$\partial_y\xi_y$$, which means that given a value of $$\xi_y$$ at some point in the plane, its value everywhere else is determined. Therefore the only possible Killing vectors are scalar multiples of the Killing vector already found. Since we don't consider Killing vectors to be distinct unless they are linearly independent, the first metric only has one Killing vector.

A similar calculation for the second metric shows that $$R=0$$, and an explicit calculation of its Riemann tensor shows that in fact the space is flat. It is simply the Euclidean plane written in funny coordinates. This metric has the same three Killing vectors as the Euclidean plane.

It would have been tempting to leap to the wrong conclusion about the second metric by the following reasoning. The signature of a metric is an intrinsic property. The metric has signature $$++$$ everywhere in the plane except on the $$y$$ axis, where it has signature $$+0$$. This shows that the $$y$$ axis has different intrinsic properties than the rest of the plane, and therefore the metric must have a lower symmetry than the Euclidean plane. It can have at most two Killing vectors, not three. This contradicts our earlier conclusion. The resolution of this paradox is that this metric has a removable degeneracy of the same type as the one described in section 6.4. As discussed in that section, the signature is invariant only under nonsingular transformations, but the transformation that converts these coordinates to Cartesian ones is singular.

### 7.1.1 Inappropriate mixing of notational systems

Confusingly, it is customary to express vectors and dual vectors by summing over basis vectors like this:

\begin{align*} \mathbf{v} &= v^\mu \partial_\mu \\ \boldsymbol{\omega} &= \omega_\mu dx^\mu . \end{align*}

This is an abuse of notation, driven by the desire to have up-down pairs of indices to sum according to the usual rules of the Einstein notation convention. But by that convention, a quantity like $$\mathbf{v}$$ or $$\boldsymbol{\omega}$$ with no indices is a scalar, and that's not the case here. The products on the right are not tensor products, i.e., the indices aren't being contracted.

This muddle is the result of trying to make the Einstein notation do too many things at once and of trying to preserve a clumsy and outdated system of notation and terminology originated by Sylvester in 1853. In pure abstract index notation, there are not six flavors of objects as in the two equations above but only two: vectors like $$v^a$$ and dual vectors like $$\omega_a$$. The Sylvester notation is the prevalent one among mathematicians today, because their predecessors committed themselves to it a century before the development of alternatives like abstract index notation and birdtracks. The Sylvester system is inconsistent with the way physicists today think of vectors and dual vectors as being defined by their transformation properties, because Sylvester considers $$\mathbf{v}$$ and $$\boldsymbol{\omega}$$ to be invariant.

Mixing the two systems leads to the kinds of notational clashes described above. As a particularly absurd example, a physicist who is asked to suggest a notation for a vector will typically pick up a pen and write $$v^\mu$$. We are then led to say that a vector is written in a concrete basis as a linear combination of dual vectors $$\partial_\mu$$!

### 7.1.2 Conservation laws

Whenever a spacetime has a Killing vector, geodesics have a constant value of $$v^b \xi_b$$, where $$v^b$$ is the velocity four-vector. For example, because the Schwarzschild metric has a Killing vector $$\boldsymbol{\xi} = \partial_t$$, test particles have a conserved value of $$v^t$$, and therefore we also have conservation of $$p^t$$, interpreted as the mass-energy.
##### Example 6: Energy-momentum in flat 1+1 spacetime

A flat 1+1-dimensional spacetime has Killing vectors $$\partial_x$$ and $$\partial_t$$. Corresponding to these are the conserved momentum and mass-energy, $$p$$ and $$E$$. If we do a Lorentz boost, these two Killing vectors get mixed together by a linear transformation, corresponding to a transformation of $$p$$ and $$E$$ into a new frame.

In addition, one can define a globally conserved quantity found by integrating the flux density $$P^a=T^{ab}\xi_b$$ over the boundary of any compact orientable region.2 In case of a flat spacetime, there are enough Killing vectors to give conservation of energy-momentum and angular momentum.

## 7.2 Spherical symmetry

b / Penrose diagram for flat spacetime.

A little more work is required if we want to link the existence of Killing vectors to the existence of a specific symmetry such as spherical symmetry. When we talk about spherical symmetry in the context of Newtonian gravity or Maxwell's equations, we may say, “The fields only depend on $$r$$,” implicitly assuming that there is an $$r$$ coordinate that has a definite meaning for a given choice of origin. But coordinates in relativity are not guaranteed to have any particular physical interpretation such as distance from a particular origin. The origin may not even exist as part of the spacetime, as in the Schwarzschild metric, which has a singularity at the center. Another possibility is that the origin may not be unique, as on a Euclidean two-sphere like the earth's surface, where a circle centered on the north pole is also a circle centered on the south pole; this can also occur in certain cosmological spacetimes that describe a universe that wraps around on itself spatially.

We therefore define spherical symmetry as follows. A spacetime $$S$$ is spherically symmetric if we can write it as a union $$S=\cup s_{r,t}$$ of nonintersecting subsets $$s_{r,t}$$, where each $$s$$ has the structure of a two-sphere, and the real numbers $$r$$ and $$t$$ have no preassigned physical interpretation, but $$s_{r,t}$$ is required to vary smoothly as a function of them. By “has the structure of a two-sphere,” we mean that no intrinsic measurement on $$s$$ will produce any result different from the result we would have obtained on some two-sphere. A two-sphere has only two intrinsic properties: (1) it is spacelike, i.e., locally its geometry is approximately that of the Euclidean plane; (2) it has a constant positive curvature. If we like, we can require that the parameter $$r$$ be the corresponding radius of curvature, in which case $$t$$ is some timelike coordinate.

To link this definition to Killing vectors, we note that condition 2 is equivalent to the following alternative condition: ($$2'$$) The set $$s$$ should have three Killing vectors (which by condition 1 are both spacelike), and it should be possible to choose these Killing vectors such that algebraically they act the same as the ones constructed explicitly in example 4 on p. 246. As an example of such an algebraic property, figure a shows that rotations are noncommutative.

a / Performing the rotations in one order gives one result, 3, while reversing the order gives a different result, 5.

##### Example 7: A cylinder is not a sphere

$$\triangleright$$ Show that a cylinder does not have the structure of a two-
[3]sphere.

$$\triangleright$$ The cylinder passes condition 1. It fails condition 2 because its Gaussian curvature is zero. Alternatively, it fails condition $$2'$$ because it has only two independent Killing vectors (example 3).

##### Example 8: A plane is not a sphere

$$\triangleright$$ Show that the Euclidean plane does not have the structure of a two-
[3]sphere.

$$\triangleright$$ Condition 2 is violated because the Gaussian curvature is zero. Or if we wish, the plane violates $$2'$$ because $$\partial_x$$ and $$\partial_y$$ commute, but none of the Killing vectors of a 2-sphere commute.

## 7.3 Penrose diagrams

Many of the most difficult and interesting topics in general relativity can be studied within the context of two types of spherically symmetric spacetimes: (1) a nonrotating black hole, such as the Schwarzschild spacetime; and (2) a cosmological spacetime that is homogeneous and isotropic. (In the latter case, there is a higher level of symmetry, since there is no central point.) When a spacetime is spherically symmetric, it can be represented using a Penrose diagram, also known as a Penrose-Carter diagram or causal diagram.

### 7.3.1 Flat spacetime

As a warmup, figure b shows a Penrose diagram for flat (Minkowski) spacetime. The diagram looks $$1+1$$-dimensional, but the convention is that spherical symmetry is assumed, so two more dimensions are hidden, and we're really portraying $$3+1$$ dimensions. A typical point on the interior of the diamond region represents a 2-sphere. On this type of diagram, light cones look just like they would on a normal spacetime diagram of Minkowski space, but distance scales are highly distorted. The diamond represents the entire spacetime, with the distortion fitting this entire infinite region into that finite area on the page. Despite the distortion, the diagram shows lightlike surfaces as 45-degree diagonals. Spacelike and timelike geodesics, however, are distorted, as shown by the curves in the diagram.

The distortion becomes greater as we move away from the center of the diagram, and becomes infinite near the edges. Because of this infinite distortion, the points $$i^-$$ and $$i^+$$ actually represent 3-spheres. All timelike curves start at $$i^-$$ and end at $$i^+$$, which are idealized points at infinity, like the vanishing points in perspective drawings. We can think of $$i^+$$ as the “Elephants' graveyard,” where massive particles go when they die. Similarly, lightlike curves end on $$\mathscr{I}^+$$ (which includes its mirror image on the left), referred to as null infinity.3 The point at $$i^0$$ is an infinitely distant endpoint for spacelike curves. Because of the spherical symmetry, the left and right halves of the diagram are redundant.

It is possible to make up explicit formulae that translate back and forth between Minkowski coordinates and points on the diamond, but in general this is not necessary. In fact, the utility of the diagrams is that they let us think about causal relationships in coordinate-independent ways. A light cone on the diagram looks exactly like a normal light cone.

Since this particular spacetime is homogeneous, it makes no difference what spatial location on the diagram we pick as our axis of symmetry. For example, we could arbitrarily pick the left-hand corner, the central timelike geodesic (drawn straight) or one of the other timelike geodesics (represented as if it were curved).

a / Penrose diagram for Schwarzschild spacetime: a black hole that didn't form by gravitational collapse.

### 7.3.2 Schwarzschild spacetime

Figure a is a Penrose diagram for the Schwarzschild spacetime, i.e., a Minkowski spacetime in which a black hole exists, but in which the black hole did not form by gravitational collapse. This spacetime isn't homogeneous; it has a specific location that is its center of spherical symmetry, and this is the vertical line on the left marked $$r=0$$. The triangle is the spacetime inside the event horizon; we could have copied it across the $$r=0$$ line if we had so desired, but the copies would have been redundant.

The Penrose diagram makes it easy to reason about causal relationships. For example, we can see that if a particle reaches a point inside the event horizon, its entire causal future lies inside the horizon, and all of its possible future world-lines intersect the singularity. The horizon is a lightlike surface, which makes sense, because it's defined as the boundary of the set of points from which a light ray could reach $$\mathscr{I}^+$$.

b / Penrose diagram for a black hole formed by gravitational collapse.

c / According to the distant observer, does the infalling matter ever reach the singularity?

### 7.3.3 Astrophysical black hole

Figure b is a Penrose diagram for a black hole that has formed by gravitational collapse. Using this type of diagram, we can address one of the most vexing FAQs about black holes. If a distant observer watches the collapsing cloud of matter from which the black hole forms, her optical observations will show that the light from the matter becomes more and more gravitationally redshifted, and if she wishes, she can interpret this as an example of gravitational time dilation. As she waits longer and longer, the light signals from the infalling matter take longer and longer to arrive. The redshift approaches infinity as the matter approaches the horizon, so the light waves ultimately become too low in energy to be detectable by any given instrument. Furthermore, her patience (or her lifetime) will run out, because the time on her clock approaches infinity as she waits to get signals from matter that is approaching the horizon. This is all exactly as it should be, since the horizon is by definition the boundary of her observable universe. (A light ray emitted from the horizon will end up at $$i^+$$, which is an end-point of timelike world-lines reached only by observers who have experienced an infinite amount of proper time.)

People who are bothered by these issues often acknowledge the external unobservability of matter passing through the horizon, and then want to pass from this to questions like, “Does that mean the black hole never really forms?” This presupposes that our distant observer has a uniquely defined notion of simultaneity that applies to a region of space stretching from her own position to the interior of the black hole, so that she can say what's going on inside the black hole “now.” But the notion of simultaneity in general relativity is even more limited than its counterpart in special relativity. Not only is simultaneity in general relativity observer-dependent, as in special relativity, but it is also local rather than global.

In figure c, E is an event on the world-line of an observer. The spacelike surface $$\text{S}_1$$ is one possible “now” for this observer. According to this surface, no particle has ever fallen in and reached the horizon; every such particle has a world-line that intersects $$\text{S}_1$$, and therefore it's still on its way in.

$$\text{S}_2$$ is another possible “now” for the same observer at the same time. According to this definition of “now,” all the particles have passed the event horizon, but none have hit the singularity yet. Finally, $$\text{S}_3$$ is a “now” according to which all the particles have hit the singularity.

If this was special relativity, then we could decide which surface was the correct notion of simultaneity for the observer, based on the observer's state of motion. But in general relativity, this only works locally (which is why I made all three surfaces coincide near E). There is no well-defined way of deciding which is the correct way of globally extending this notion of simultaneity.

Although it may seem strange that we can't say whether the singularity has “already” formed according to a distant observer, this is really just an inevitable result of the fact that the singularity is spacelike. The same thing happens in the case of a Schwarzschild spacetime, which we think of as a description of an eternal black hole, i.e., one that has always existed and always will. On the similar Penrose diagram for an eternal black hole, we can still draw a spacelike surface like $$\text{S}_1$$ or $$\text{S}_2$$, representing a definition of “now” such that the singularity doesn't exist yet.

## 7.4 Static and stationary spacetimes

### 7.4.1 Stationary spacetimes

When we set out to describe a generic spacetime, the Alice in Wonderland quality of the experience is partly because coordinate invariance allows our time and distance scales to be arbitrarily rescaled, but also partly because the landscape can change from one moment to the next. The situation is drastically simplified when the spacetime has a timelike Killing vector. Such a spacetime is said to be stationary. Two examples are flat spacetime and the spacetime surrounding the rotating earth (in which there is a frame-dragging effect). Non-examples include the solar system, cosmological models, gravitational waves, and a cloud of matter undergoing gravitational collapse.

Can Alice determine, by traveling around her spacetime and carrying out observations, whether it is stationary? If it's not, then she might be able to prove it. For example, suppose she visits a certain region and finds that the Kretchmann invariant $$R^{abcd}R_{abcd}$$ varies with time in her frame of reference. Maybe this is because an asteroid is coming her way, in which case she could readjust her velocity vector to match that of the asteroid. Even if she can't see the asteroid, she can still try to find a velocity that makes her local geometry stop changing in this particular way. If the spacetime is truly stationary, then she can always “tune in” to the right velocity vector in this way by searching systematically. If this procedure ever fails, then she has proved that her spacetime is not stationary.

Self-check: Why is the timelike nature of the Killing vector important in this story?

Proving that a spacetime is stationary is harder. This is partly just because spacetime is infinite, so it will take an infinite amount of time to check everywhere. We aren't inclined to worry too much about this limitation on our geometrical knowledge, which is of a type that has been familiar since thousands of years ago, when it upset the ancient Greeks that the parallel postulate could only be checked by following lines out to an infinite distance. But there is a new type of limitation as well. The Schwarzschild spacetime is not stationary according to our definition. In the coordinates used in section 6.2, $$\partial_t$$ is a Killing vector, but is only timelike for $$r>2m$$; for $$r\lt2m$$ it is spacelike. Although the solution describes a black hole that is going to sit around forever without changing, no observer can ever verify that fact, because once she strays inside the horizon she must follow a timelike world-line, which will end her program of observation within some finite time.

### 7.4.2 Isolated systems

#### Asymptotic flatness

This unfortunate feature of our definition of stationarity --- its empirical unverifiability --- is something that in general we just have to live with. But there is an alternative in the special case of an isolated system, such as our galaxy or a black hole. It may be a good approximation to ignore distant matter, modeling such a system with a spacetime that is almost flat everywhere except in the region nearby. Such a spacetime is called asymptotically flat. Formulating the definition of this term rigorously and in a coordinate-invariant way involves a large amount of technical machinery, since we are not guaranteed to be presented in advance with a special, physically significant set of coordinates that would lead directly to a quantitative way of defining words like “nearby.” The reader who wants a rigorous definition is referred to Hawking and Ellis.

#### Asymptotically stationary spacetimes

In the case of an asymptotically flat spacetime, we say that it is also asymoptotically stationary if it has a Killing vector that becomes timelike far away. Some authors (e.g., Ludvigsen) define “stationary” to mean what I'm calling “asymoptotically stationary,” others (Hawking and Ellis) define it the same way I do, and still others (Carroll) are not self-consistent. The Schwarzschild spacetime is asymptotically stationary, but not stationary.

### 7.4.3 A stationary field with no other symmetries

Consider the most general stationary case, in which the only Killing vector is the timelike one. The only ambiguity in the choice of this vector is a rescaling; its direction is fixed. At any given point in space, we therefore have a notion of being at rest, which is to have a velocity vector parallel to the Killing vector. An observer at rest detects no time-dependence in quantities such as tidal forces.

Points in space thus have a permanent identity. The gravitational field, which the equivalence principle tells us is normally an elusive, frame-dependent concept, now becomes more concrete: it is the proper acceleration required in order to stay in one place. We can therefore use phrases like “a stationary field,” without the usual caveats about the coordinate-dependent meaning of “field.”

Space can be sprinkled with identical clocks, all at rest. Furthermore, we can compare the rates of these clocks, and even compensate for mismatched rates, by the following procedure. Since the spacetime is stationary, experiments are reproducible. If we send a photon or a material particle from a point A in space to a point B, then identical particles emitted at later times will follow identical trajectories. The time lag between the arrival of two such particls tells an observer at B the amount of time at B that corresponds to a certain interval at A. If we wish, we can adjust all the clocks so that their rates are matched. An example of such rate-matching is the GPS satellite system, in which the satellites' clocks are tuned to 10.22999999543 MHz, matching the ground-based clocks at 10.23 MHz. (Strictly speaking, this example is out of place in this subsection, since the earth's field has an additional azimuthal symmetry.)

It is tempting to conclude that this type of spacetime comes equipped with a naturally preferred time coordinate that is unique up to a global affine transformation $$t\rightarrow at+b$$. But to construct such a time coordinate, we would have to match not just the rates of the clocks, but also their phases. The best method relativity allows for doing this is Einstein synchronization (p. 359), which involves trading a photon back and forth between clocks A and B and adjusting the clocks so that they agree that each clock gets the photon at the mid-point in time between its arrivals at the other clock. The trouble is that for a general stationary spacetime, this procedure is not transitive: synchronization of A with B, and of B with C, does not guarantee agreement between A with C. This is because the time it takes a photon to travel clockwise around triangle ABCA may be different from the time it takes for the counterclockwise itinerary ACBA. In other words, we may have a Sagnac effect, which is generally interpreted as a sign of rotation. Such an effect will occur, for example, in the field of the rotating earth, and it cannot be eliminated by choosing a frame that rotates along with the earth, because the surrounding space experiences a frame-dragging effect, which falls off gradually with distance.

Although a stationary spacetime does not have a uniquely preferred time, it does prefer some time coordinates over others. In a stationary spacetime, it is always possible to find a “nice” $$t$$ such that the metric can be expressed without any $$t$$-dependence in its components.

### 7.4.4 A stationary field with additional symmetries

Most of the results given above for a stationary field with no other symmetries also hold in the special case where additional symmetries are present. The main difference is that we can make linear combinations of a particular timelike Killing vector with the other Killing vectors, so the timelike Killing vector is not unique. This means that there is no preferred notion of being at rest. For example, in a flat spacetime we cannot define an observer to be at rest if she observes no change in the local observables over time, because that is true for any inertial observer. Since there is no preferred rest frame, we can't define the gravitational field in terms of that frame, and there is no longer any preferred definition of the gravitational field.

### 7.4.5 Static spacetimes

In addition to synchronizing all clocks to the same frequency, we might also like to be able to match all their phases using Einstein synchronization, which requires transitivity. Transitivity is frame-dependent. For example, flat spacetime allows transitivity if we use the usual coordinates. However, if we change into a rotating frame of reference, transitivity fails (see p. 108). If coordinates exist in which a particular spacetime has transitivity, then that spacetime is said to be static. In these coordinates, the metric is diagonalized, and since there are no space-time cross-terms like $$dxdt$$ in the metric, such a spacetime is invariant under time reversal. Roughly speaking, a static spacetime is one in which there is no rotation.

### 7.4.6 Birkhoff's theorem

Birkhoff's theorem, proved below, states that in the case of spherical symmetry, the vacuum field equations have a solution, the Schwarzschild spacetime, which is unique up to a choice of coordinates and the value of $$m$$. Let's enumerate the assumptions that went into our derivation of the Schwarzschild metric on p. 211. These were: (1) the vacuum field equations, (2) spherical symmetry, (3) asymptotic staticity, (4) a certain choice of coordinates, and (5) $$\Lambda=0$$. Birkhoff's theorem says that the assumption of staticity was not necessary. That is, even if the mass distribution contracts and expands over time, the exterior solution is still the Schwarzschild solution. Birkhoff's theorem holds because gravitational waves are transverse, not longitudinal (see p. 349), so the mass distribution's radial throbbing cannot generate a gravitational wave. Birkhoff's theorem can be viewed as the simplest of the no-hair theorems describing black holes. The most general no-hair theorem states that a black hole is completely characterized by its mass, charge, and angular momentum. Other than these three numbers, nobody on the outside can recover any information that was possessed by the matter and energy that were sucked into the black hole.

It has been proposed4 that the no-hair theorem for nonzero angular momentum and zero charge could be tested empirically by observations of Sagittarius A*. If the observations are consistent with the no-hair theorem, it would be taken as supporting the validity of general relativity and the interpretation of this object as a supermassive black hole. If not, then there are various possibilities, including a failure of general relativity to be the correct theory of strong gravitational fields, or a failure of one of the theorem's other assumptions, such as the nonexistence of closed timelike curves in the surrounding universe.

Proof of Birkhoff's theorem: Spherical symmetry guarantees that we can introduce coordinates $$r$$ and $$t$$ such that the surfaces of constant $$r$$ and $$t$$ have the structure of a sphere with radius $$r$$. On one such surface we can introduce colatitude and longitude coordinates $$\theta$$ and $$\phi$$. The $$(\theta,\phi)$$ coordinates can be extended in a natural way to other values of $$r$$ by choosing the radial lines to lie in the direction of the covariant derivative vector5 $$\nabla_a r$$, and this ensures that the metric will not have any nonvanishing terms in $$drd\theta$$ or $$drd\phi$$, which could only arise if our choice had broken the symmetry between positive and negative values of $$d\theta$$ and $$d\phi$$. Just as we were free to choose any way of threading lines of constant $$(\theta,\phi,t)$$ between spheres of different radii, we can also choose how to thread lines of constant $$(\theta,\phi,r)$$ between different times, and this can be done so as to keep the metric free of any time-space cross-terms such as $$d\thetadt$$. The metric can therefore be written in the form6

$\begin{equation*} ds^2 = h(t,r)dt^2 - k(t,r)dr^2 - r^2(d\theta^2+\sin^2\theta d\phi^2) . \end{equation*}$

This has to be a solution of the vacuum field equations, $$R_{ab}=0$$, and in particular a quick calculation with Maxima shows that $$R_{rt}=-\partial_t k/k^2r$$, so $$k$$ must be independent of time. With this restriction, we find $$R_{rr}=-\partial_rh/hkr-1/r^2-1/kr^2=0$$, and since $$k$$ is time-independent, $$\partial_rh/h$$ is also time-independent. This means that for a particular time $$t_\text{o}$$, the function $$f(r)=h(t_\text{o},r)$$ has some universal shape set by a differential equation, with the only possible ambiguity being an over-all scaling that depends on $$t_\text{o}$$. But since $$h$$ is the time-time component of the metric, this scaling corresponds physically to a situation in which every clock, all over the universe, speeds up and slows down in unison. General relativity is coordinate-independent, so this has no observable effects, and we can absorb it into a redefinition of $$t$$ that will cause $$h$$ to be time-independent. Thus the metric can be expressed in the time-independent diagonal form

$\begin{equation*} ds^2 = h(r)dt^2 - k(r)dr^2 - r^2(d\theta^2+\sin^2\theta d\phi^2) . \end{equation*}$

We have already solved the field equations for a metric of this form and found as a solution the Schwarzschild spacetime.7 Since the metric's components are all independent of $$t$$, $$\partial_t$$ is a Killing vector, and it is timelike for large $$r$$, so the Schwarzschild spacetime is asymptotically static.

The no-hair theorems say that relativity only has a small repertoire of types of black-hole singularities, defined as singularities inside regions of space that are causally disconnected from the universe, in the sense that future light-cones of points in the region do not extend to infinity.8 That is, a black hole is defined as a singularity hidden behind an event horizon, and since the definition of an event horizon is dependent on the observer, we specify an observer infinitely far away. The theorems cannot classify naked singularities, i.e., those not hidden behind horizons, because the role of naked singularities in relativity is the subject of the cosmic censorship hypothesis, which is an open problem. The theorems do not rule out the Big Bang singularity, because we cannot define the notion of an observer infinitely far from the Big Bang. We can also see that Birkhoff's theorem does not prohibit the Big Bang, because cosmological models are not vacuum solutions with $$\Lambda=0$$. Black string solutions are not ruled out by Birkhoff's theorem because they would lack spherical symmetry, so we need the arguments given on p. 234 to show that they don't exist.

### 7.4.7 The gravitational potential

When Pound and Rebka made the first observation of gravitational redshifts, these shifts were interpreted as evidence of gravitational time dilation, i.e., a mismatch in the rates of clocks. We are accustomed to connecting these two ideas by using the expression $$e^{-\Delta\Phi}$$ for the ratio of the rates of two clocks (example 12, p. 59), where $$\Phi$$ is a function of the spatial coordinates, and this is in fact the most general possible definition of a gravitational potential $$\Phi$$ in relativity. Since a stationary field allows us to compare rates of clocks, it seems that we should be able to define a gravitational potential for any stationary field. There is a problem, however, because when we talk about a potential, we normally have in mind something that has encoded within it all there is to know about the field. We would therefore expect to be able to find the metric from the potential. But the example of the rotating earth shows that this need not be the case for a general stationary field. In that example, there are effects like frame-dragging that clearly cannot be deduced from $$\Phi$$; for by symmetry, $$\Phi$$ is independent of azimuthal angle, and therefore it cannot distinguish between the direction of rotation and the contrary direction. In a static spacetime, these rotational effects don't exist; a static vacuum spacetime can be described completely in terms of a single scalar potential plus information about the spatial curvature.

There are two main reasons why relativity does not offer a gravitational potential with the same general utility as its Newtonian counterpart.

The Einstein field equations are nonlinear. Therefore one cannot, in general, find the field created by a given set of sources by adding up the potentials. At best this is a possible weak-field approximation. In particular, although Birkhoff's theorem is in some ways analogous to the Newtonian shell theorem, it cannot be used to find the metric of an arbitrary spherically symmetric mass distribution by breaking it up into spherical shells.

It is also not meaningful to talk about any kind of gravitational potential for spacetimes that aren't static or stationary. For example, consider a cosmological model describing our expanding universe. Such models are usually constructed according to the Copernican principle that no position in the universe occupies a privileged place. In other words, they are homogeneous in the sense that they have Killing vectors describing arbitrary translations and rotations. Because of this high degree of symmetry, a gravitational potential for such a model would have to be independent of position, and then it clearly could not encode any information about the spatial part of the metric. Even if we were willing to make the potential a function of time, $$\Phi(t)$$, the results would still be nonsense. The gravitational potential is defined in terms of rate-matching of clocks, so a potential that was purely a function of time would describe a situation in which all clocks, everywhere in the universe, were changing their rates in a uniform way. But this is clearly just equivalent to a redefinition of the time coordinate, which has no observable consequences because general relativity is coordinate-invariant. A corollary is that in a cosmological spacetime, it is not possible to give a natural prescription for deciding whether a particular redshift is gravitational (measured by $$\Phi$$) or kinematic, or some combination of the two (see also p. 312).

## 7.5 The uniform gravitational field revisited

This section gives a somewhat exotic example. It is not necessary to read it in order to understand the later material.

In problem 6 on page 200, we made a wish list of desired properties for a uniform gravitational field, and found that they could not all be satisfied at once. That is, there is no global solution to the Einstein field equations that uniquely and satisfactorily embodies all of our Newtonian ideas about a uniform field. We now revisit this question in the light of our new knowledge.

The 1+1-dimensional metric

$\begin{equation*} ds^2 = e^{2gz}dt^2-dz^2 \end{equation*}$

is the one that uniquely satisfies our expectations based on the equivalence principle (example 12, p. 59), and it is a vacuum solution. We might logically try to generalize this to 3+1 dimensions as follows:

$\begin{equation*} ds^2 = e^{2gz}dt^2-dx^2 - dy^2 - dz^2 . \end{equation*}$

But a funny thing happens now --- simply by slapping on the two new Cartesian axes $$x$$ and $$y$$, it turns out that we have made our vacuum solution into a non-vacuum solution, and not only that, but the resulting stress-energy tensor is unphysical (ch. 8, problem 8, p. 341).

One way to proceed would be to relax our insistence on making the spacetime one that exactly embodies the equivalence principle's requirements for a uniform field.9 This can be done by taking $$g_{tt}=e^{2\Phi}$$, where $$\Phi$$ is not necessarily equal to $$2gz$$. By requiring that the metric be a 3+1 vacuum solution, we arrive at a differential equation whose solution is $$\Phi=\ln(z+k_1)+k_2$$, which recovers the flat-space metric that we found in example 19 on page 140 by applying a change of coordinates to the Lorentz metric.

What if we want to carry out the generalization from 1+1 to 3+1 without violating the equivalence principle? For physical motivation in how to get past this obstacle, consider the following argument made by Born in 1920.10 Take a frame of reference tied to a rotating disk, as in the example from which Einstein originally took much of the motivation for creating a geometrical theory of gravity (subsection 3.5.4, p. 108). Clocks near the edge of the disk run slowly, and by the equivalence principle, an observer on the disk interprets this as a gravitational time dilation. But this is not the only relativistic effect seen by such an observer. Her rulers are also Lorentz contracted as seen by a non-rotating observer, and she interprets this as evidence of a non-Euclidean spatial geometry. There are some physical differences between the rotating disk and our default conception of a uniform field, specifically in the question of whether the metric should be static (i.e., lacking in cross-terms between the space and time variables). But even so, these considerations make it natural to hypothesize that the correct 3+1-dimensional metric should have transverse spatial coefficients that decrease with height.

With this motivation, let's consider a metric of the form

$\begin{equation*} ds^2 = e^{2z}dt^2-e^{-2jz}dx^2 - e^{-2kz}dy^2 - dz^2 , \end{equation*}$

where $$j$$ and $$k$$ are constants, and I've taken $$g=1$$ for convenience.11 The following Maxima code calculates the scalar curvature and the Einstein tensor:

load(ctensor);
ct_coords:[t,x,y,z];
lg:matrix([exp(2*z),0,0,0],
[0,-exp(-2*j*z),0,0],
[0,0,-exp(-2*k*z),0],
[0,0,0,-1]
);
cmetric();
scurvature();
leinstein(true);


The output from line 9 shows that the scalar curvature is constant, which is a necessary condition for any spacetime that we want to think of as representing a uniform field. Inspecting the Einstein tensor output by line 10, we find that in order to get $$G_{xx}$$ and $$G_{yy}$$ to vanish, we need $$j$$ and $$k$$ to be $$(1 \pm \sqrt{3} i)/2$$. By trial and error, we find that assigning the complex-conjugate values to $$j$$ and $$k$$ makes $$G_{tt}$$ and $$G_{zz}$$ vanish as well, so that we have a vacuum solution. This solution is, unfortunately, complex, so it is not of any obvious value as a physically meaningful result. Since the field equations are nonlinear, we can't use the usual trick of forming real-valued superpositions of the complex solutions. We could try simply taking the real part of the metric. This gives $$g_{xx}=e^{-z}\cos\sqrt{3}z$$ and $$g_{yy}=e^{-z}\sin\sqrt{3}z$$, and is unsatisfactory because the metric becomes degenerate (has a zero determinant) at $$z=n \pi/2\sqrt{3}$$, where $$n$$ is an integer.

It turns out, however, that there is a very similar solution, found by Petrov in 1962,12 that is real-valued. The Petrov metric, which describes a spacetime with cylindrical symmetry, is:

$\begin{equation*} ds^2 = -dr^2 - e^{-2r}dz^2+e^r[2\sin\sqrt{3}r d \phi dt-\cos\sqrt{3}r(d \phi^2-dt^2)] \end{equation*}$

Note that it has many features in common with the complex oscillatory solution we found above. There are transverse length contractions that decay and oscillate in exactly the same way. The presence of the $$d \phi dt$$ term tells us that this is a non-static, rotating solution --- exactly like the one that Einstein and Born had in mind in their prototypical example! We typically obtain this type of effect due to frame dragging by some rotating massive body (see p. 149), and the Petrov solution can indeed be interpreted as the spacetime that exists in the vacuum on the exterior of an infinite, rigidly rotating cylinder of “dust” (see p. 132).

The complicated Petrov metric might seem like the furthest possible thing from a uniform gravitational field, but in fact it is about the closest thing general relativity provides to such a field. We first note that the metric has Killing vectors $$\partial_z$$, $$\partial_\phi$$, and $$\partial_r$$, so it has at least three out of the four translation symmetries we expect from a uniform field. By analogy with electromagnetism, we would expect this symmetry to be absent in the radial direction, since by Gauss's law the electric field of a line of charge falls off like $$1/r$$. But surprisingly, the Petrov metric is also uniform radially. It is possible to give the fourth killing vector explicitly (it is $$\partial_r + z\partial_z + (1/2)(\sqrt{3}t-\phi)\partial_\phi - (1/2)(\sqrt{3}\phi+t)\partial_t$$), but it is perhaps more transparent to check that it represents a field of constant strength (problem 4, p. 265).

For insight into this surprising result, recall that in our attempt at constructing the Cartesian version of this metric, we ran into the problem that the metric became degenerate at $$z=n \pi/2\sqrt{3}$$. The presence of the $$d \phi dt$$ term prevents this from happening in Petrov's cylindrical version; two of the metric's diagonal components can vanish at certain values of $$r$$, but the presence of the off-diagonal component prevents the determinant from going to zero. (The determinant is in fact equal to $$-1$$ everywhere.) What is happening physically is that although the labeling of the $$\phi$$ and $$t$$ coordinates suggests a time and an azimuthal angle, these two coordinates are in fact treated completely symmetrically. At values of $$r$$ where the cosine factor equals 1, the metric is diagonal, and has signature $$(t,\phi,r,z)=(+,-,-,-)$$, but when the cosine equals $$-1$$, this becomes $$(-,+,-,-)$$, so that $$\phi$$ is now the timelike coordinate. This perfect symmetry between $$\phi$$ and $$t$$ is an extreme example of frame-dragging, and is produced because of the specially chosen rate of rotation of the dust cylinder, such that the velocity of the dust at the outer surface is exactly $$c$$ (or approaches it).

Classically, we would expect that a test particle released close enough to the cylinder would be pulled in by the gravitational attraction and destroyed on impact, while a particle released farther away would fly off due to the centrifugal force, escaping and eventually approaching a constant velocity. Neither of these would be anything like the experience of a test particle released in a uniform field. But consider a particle released at rest in the rotating frame at a radius $$r_1$$ for which $$\cos\sqrt{3}r_1=1$$, so that $$t$$ is the timelike coordinate. The particle accelerates (let's say outward), but at some point it arrives at an $$r_2$$ where the cosine equals zero, and the $$\phi-t$$ part of the metric is purely of the form $$d\phidt$$. At this location, we can define local coordinates $$u=\phi-t$$ and $$v=\phi+t$$, so that the metric depends only on $$du^2-dv^2$$. One of the coordinates, say $$u$$, is now the timelike one. Since our particle is material, its world-line must be timelike, so it is swept along in the $$-\phi$$ direction. Gibbons and Gielen show that the particle will now come back inward, and continue forever by oscillating back and forth between two radii at which the cosine vanishes.

### 7.5.1 Closed timelike curves

This oscillation still doesn't sound like the motion of a particle in a uniform field, but another strange thing happens, as we can see by taking another look at the values of $$r$$ at which the cosine vanishes. At such a value of $$r$$, construct a curve of the form $$(t=\text{constant},r=\text{constant},\phi,z=\text{constant})$$. This is a closed curve, and its proper length is zero, i.e., it is lightlike. This violates causality. A photon could travel around this path and arrive at its starting point at the same time when it was emitted. Something similarly weird hapens to the test particle described above: whereas it seems to fall sometimes up and sometimes down, in fact it is always falling down --- but sometimes it achieves this by falling up while moving backward in time!

Although the Petov metric violates causality, Gibbons and Gielen have shown that it satisfies the chronology protection conjecture: “In the context of causality violation we have shown that one cannot create CTCs [closed timelike curves] by spinning up a cylinder beyond its critical angular velocity by shooting in particles on timelike or null curves.”

We have an exact vacuum solution to the Einstein field equations that violates causality. This raises troublesome questions about the logical self-consistency of general relativity. A very readable and entertaining overview of these issues is given in the final chapter of Kip Thorne's Black Holes and Time Warps: Einstein's Outrageous Legacy. In a toy model constructed by Thorne's students, involving a billiard ball and a wormhole, it turned out that there always seemed to be self-consistent solutions to the ball's equations of motion, but they were not unique, and they often involved disquieting possibilities in which the ball went back in time and collided with its earlier self. Among other things, this seems to lead to a violation of conservation of mass-energy, since no mass was put into the system to create extra copies of the ball. This would then be an example of the fact that, as discussed in section 4.5.1, general relativity does not admit global conservation laws. However, there is also an argument that the mouths of the wormhole change in mass in such a way as to preserve conservation of energy.13

## Homework Problems

1. Example 3 on page 245 gave the Killing vectors $$\partial_z$$ and $$\partial_\phi$$ of a cylinder. If we express these instead as two linearly independent Killing vectors that are linear combinations of these two, what is the geometrical interpretation?

2. Section 7.4 told the story of Alice trying to find evidence that her spacetime is not stationary, and also listed the following examples of spacetimes that were not stationary: (a) the solar system, (b) cosmological models, (c) gravitational waves propagating at the speed of light, and (d) a cloud of matter undergoing gravitational collapse. For each of these, show that it is possible for Alice to accomplish her mission.

3. If a spacetime has a certain symmetry, then we expect that symmetry to be detectable in the behavior of curvature scalars such as the scalar curvature $$R=R^a_a$$ and the Kretchmann invariant $$k=R^{abcd}R_{abcd}$$.
(a) Show that the metric

$\begin{equation*} ds^2 = e^{2gz}dt^2-dx^2 - dy^2 - dz^2 \end{equation*}$

from page 260 has constant values of $$R=1/2$$ and $$k=1/4$$. Note that Maxima's ctensor package has built-in functions for these; you have to call the lriemann and uriemann before calling them.
(b) Similarly, show that the Petrov metric

$\begin{equation*} ds^2 = -dr^2 - e^{-2r}dz^2+e^r[2\sin\sqrt{3}r d \phi dt-\cos\sqrt{3}r(d \phi^2-dt^2)] \end{equation*}$

(p. 262) has $$R=0$$ and $$k=0$$.

Surprisingly, one can have a spacetime on which every possible curvature invariant vanishes identically, and yet which is not flat. See Coley, Hervik, and Pelavas, “Spacetimes characterized by their scalar curvature invariants,” arxiv.org/abs/0901.0791v2.

4. Section 7.5 on page 260 presented the Petrov metric. The purpose of this problem is to verify that the gravitational field it represents does not fall off with distance. For simplicity, let's restrict our attention to a particle released at an $$r$$ such that $$\cos\sqrt{3}r=1$$, so that $$t$$ is the timelike coordinate. Let the particle be released at rest in the sense that initially it has $$\dot{z}=\dot{r}=\dot{\phi}=0$$, where dots represent differentiation with respect to the particle's proper time. Show that the magnitude of the proper acceleration is independent of $$r$$. (solution in the pdf version of the book)

5. The idea that a frame is “rotating” in general relativity can be formalized by saying that the frame is stationary but not static. Suppose someone says that any rotation must have a center. Give a counterexample. (solution in the pdf version of the book)

(c) 1998-2013 Benjamin Crowell, licensed under the Creative Commons Attribution-ShareAlike license. Photo credits are given at the end of the Adobe Acrobat version.

##### Footnotes
[1] Coxeter, Introduction to Geometry, ch. 3
[2] Hawking and Ellis, The Large Scale Structure of Space-Time, p. 62, give a succinct treatment that describes the flux densities and proves that Gauss's theorem, which ordinarily fails in curved spacetime for a non-scalar flux, holds in the case where the appropriate Killing vectors exist. For an explicit description of how one can integrate to find a scalar mass-energy, see Winitzki, Topics in General Relativity, section 3.1.5, available for free online.
[5] It may seem backwards to start talking about the covariant derivative of a particular coordinate before a complete coordinate system has even been introduced. But (excluding the trivial case of a flat spacetime), $$r$$ is not just an arbitrary coordinate, it is something that an observer at a certain point in spacetime can determine by mapping out a surface of geometrically identical points, and then determining that surface's radius of curvature. Another worry is that it is possible for $$\nabla_a r$$ to misbehave on certain surfaces, such as the event horizon of the Schwarzschild spacetime, but we can simply require that radial lines remain continuous as they pass through these surfaces.
[6] On the same surfaces referred to in the preceding footnote, the functions $$h$$ and $$k$$ may to go to 0 or $$\infty$$. These turn out to be nothing more serious than coordinate singularities.
[11] A metric of this general form is referred to as a Kasner metric. One usually sees it written with a logarithmic change of variables, so that $$z$$ appears in the base rather than in the exponent.
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