w or kbook a cosmol ogy - libretexts · 2021. 1. 14. · s ol ut i ons p rogra m , a nd me rl ot ....

A COSMOLOGY WORKBOOK

Lloyd KnoxUniversity of California, Davis

University of California, Davis

A Cosmology Workbook

Lloyd Knox

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TABLE OF CONTENTSIn this quarter-long course, we will learn how to think about the expanding universe using concepts from Einstein's theory of generalrelativity. We will use Newtonian gravity to derive the dynamical equations that relate the expansion rate to the matter content of theuniverse. Connecting the expansion dynamics to observables such as luminosity distances and redshifts, we will see how astronomersuse observations to probe these dynamics, and thereby the contents of the cosmos.

OVERVIEWWe provide an overview of our subject, broken into two parts. The first is focused on the discovery of the expansion of the universe in1929, and the theoretical context for this discovery, which is given by Einstein's general theory of relativity (GR). The second is onthe implications of this expansion for the early history of the universe, and relics from that period observable today: the cosmicmicrowave background and the lightest chemical elements. The consistency of such observations with

1: EUCLIDEAN GEOMETRYThis chapter is entirely focused on the Euclidean geometry that is familiar to you, but reviewed in a language that may be unfamiliar.The new language will help us journey into the foreign territory of Riemannian geometry. Our exploration of that territory will thenhelp you to drop your pre-conceived notions about space and to begin to understand the broader possibilities -- possibilities that arenot only mathematically beautiful, but that appear to be realized in nature.

2: CURVATUREWe introduce the notion of "curvature'' in an attempt to loosen up your understanding of the nature of space, to have you betterprepared to think about the expansion of space.

3: GALILEAN RELATIVITYWe now extend our discussion of spatial geometry to spacetime geometry. We begin with Galilean relativity, which we will thengeneralize in the next section to Einstein (or Lorentz) relativity.

4: EINSTEIN RELATIVITYWhile the principle of relativity holds, its specific implementation as Galilean relativity does not. As you know, because you havestudied special relativity, this is indeed the correct solution to the puzzle of the Maxwell Equations lack of invariance under a Galileantransformation.

5: THE SIMPLEST EXPANDING SPACETIMEIn this chapter we begin our exploration of physics in an expanding spacetime. We start with a spacetime with just one spatialdimension that is not expanding: a 1+1-dimensional Minkowski spacetime. We then generalize it slightly to describe a spacetime withone spatial dimension that is expanding. With additional assumptions we then calculate the age of this spacetime as well as the "pasthorizon."

6: REDSHIFTSWe begin to work out observational consequences of living in an expanding spatially homogeneous and isotropic universe.

7: DISTANCES AS DETERMINED BY STANDARD CANDLESMeasuring the flux (energy/unit time/unit area) give us a way to figure out the distance to the object assuming we know its luminosityand that it is emitting isotropically. This is the so-called standard candle method of distance determination. Here we work out thetheoretical relationship between flux, luminosity, curvature constant k, coordinate distance between observer and source, and theredshift of the source.

8: THE DISTANCE-REDSHIFT RELATIONFor a given scale factor history, a(t) , one can work out a relationship between luminosity distance and redshift. This will be useful tous because it shows how we can infer a(t) from measurements of luminosity distance and redshift, over a range of redshifts.

9: DYNAMICS OF THE EXPANSIONIn the following set of chapters we will derive the dynamical equations that relate the matter content in a homogeneous and isotropicuniverse to the evolution of the scale factor over time.

10: A NEWTONIAN HOMOGENEOUS EXPANDING UNIVERSEConsider a homogeneous expanding universe in a Newtonian manner, where space is fixed, and the universe is filled with a fluid thatis moving in such a manner as to keep the density spatially uniform. If homogeneity is to be preserved over time, then the motionmust be such that the separation between any pair of fluid elements must scale up with time in the same way.

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11: THE FRIEDMANN EQUATIONSticking with our Newtonian expanding universe, we will now derive the Friedmann equation that relates how the scale factorchanges in time to the mass/energy density. We will proceed by using the Newtonian concept of energy conservation. (You may besurprised to hear me call this a Newtonian concept, but the fact is that energy conservation does not fully survive the transition fromNewton to Einstein).

12: PARTICLE KINEMATICS IN AN EXPANDING UNIVERSE - NEWTONIAN ANALYSISWe want to understand how an observer, at rest in their local rest frame, will observe the evolution of peculiar velocities of freeparticles. In a Newtonian analysis, the local rest frame will be an inertial frame (one in which Newton's laws of motion apply) only ifacceleration is a constant.

13: THE EVOLUTION OF MASS-ENERGY DENSITY AND A FIRST GLANCE AT THE CONTENTS OF THE COSMOSWe've seen that the rate of change of the scale factor depends on the mass density rho. In order to determine how the scale factorevolves with time, we thus need to know how the density evolves as the scale factor changes.

14: ENERGY AND MOMENTUM CONSERVATIONThe lack of energy conservation in an expanding universe is quite surprising to people with any training in physics and thereforemerits some discussion, which we present here in this chapter. The student could skip this chapter and proceed to 15 without seriousharm. If, subsequently, the lack of energy conservation becomes too troubling, know that this chapter is here for you.

15: PRESSURE AND ENERGY DENSITY EVOLUTIONThere is a sense in which energy is conserved in general relativity. We say it is locally conserved, which effectively means that in asufficiently small region of spacetime, the change in energy is equal to the flux of energy across the boundary of the region, includingthat via any work being done on the region.

16: DISTANCE AND MAGNITUDEWe have the invariant distance equation for a homogeneous and isotropic universe. We now introduce no fewer than five kinds ofspatial distances: Coordinate distance, Physical distance, Comoving distance, Luminosity distance, and Comoving angular diameterdistance.

17: PARALLAX, CEPHEID VARIABLES, SUPERNOVAE, AND DISTANCE MEASUREMENTKey to observing the consequences of this expansion is the ability to measure distances to things that are very far away. Here wecover the basics of how that is done. We have to do it in steps, getting distances to nearby objects and then using those objects tocalibrate other objects that can be used to get to even further distances. We refer to this sequence of distance determinations as thedistance ladder.

18: COSMOLOGICAL DATA ANALYSISThere are two distinct aspects of data analysis: model comparison and parameter estimation. In model comparison we try to determinewhich model is better than another. In parameter estimation, we have one assumed model and we are estimating the parameters of thatmodel. We focus here on parameter estimation.

19: THE EARLY UNIVERSETo understand the "primordial soup" and its relics, we now turn our attention from a relativistic understanding of the curvature andexpansion of space, to statistical mechanics. We begin with equilibrium statistical mechanics, before moving on to a discussion ofdepartures from equilibrium. We will come to understand the production in the big bang of Helium, photons, other "hot" relics such asneutrinos, and "cold" relics such as the dark matter. We will also discuss the observations that test

20: EQUILIBRIUM STATISTICAL MECHANICSOut of the early Universe we get the light elements, a lot of photons and, as it turns out, a bunch of neutrinos and other relics of ourhot past as well. To understand the production of these particles we now turn to the subject of Equilibrium Statistical Mechanics.

CHAPTER 20 FOOTNOTES21: EQUILIBRIUM PARTICLE ABUNDANCESAt sufficiently high temperatures and densities, reactions that create and destroy particles can become sufficiently rapid that anequilibrium abundance is achieved. In this chapter we assume that such reaction rates are sufficiently high and work out the resultingabundances as a function of the key controlling parameter. We will thus see how equilibrium abundance changes as the universeexpands and cools.

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22: HOT AND COLD RELICS OF THE BIG BANGAs the temperature and density drops, the reactions necessary to maintain chemical equilibrium can become too slow to continue todo so. This departure from equilibrium can occur while the particles are relativistic, in which case we have "hot relics," or when theparticles are non-relativistic in which case we say we have "cold relics." The cosmic microwave and neutrino backgrounds are hotrelics. The dark matter may be a cold relic.

23: OVERVIEW OF THERMAL HISTORY24: BIG BANG NUCLEOSYNTHESIS - PREDICTIONSBig Bang Nucleosynthesis is the process by which light elements formed during the Big Bang. The agreement between predictedabundances and inferences from observations of primordial (pre-stellar) abundances is a major pillar of the theory of the hot big bangand reason we can speak with some confidence about events in the primordial plasma in the first few minutes of the expansion.Elements created at these very early times include Deuterium, Helium-3, Lithium-7, and, most abundantly, Helium-4.

25: THE COSMIC MICROWAVE BACKGROUNDCosmic Microwave Background Radiation, or CMB radiation, is a prediction of Big Bang theory. This theory asserts that the earlyuniverse was occupied by a hot, dense plasma of photons, electrons and baryons that was opaque to electromagnetic radiation. Thephotons were constantly scattering off particles, particularly electrons, in the plasma. As the universe underwent adiabatic expansion,the plasma cooled until it became favorable for electrons to combine with protons and form hydrogen atoms.

26: THE FIRST FEW HUNDRED THOUSAND YEARS: OBSERVATIONS OF THE PRIMORDIAL PLASMAHere we will introduce you to a physical system that is a beautiful gift of nature: the plasma that existed from the first fractions of asecond of the Big Bang until it transitioned to a neutral gas 380,000 years later. Gently disturbed away from equilibrium bymysterious, very early-universe processes, the plasma is an unusually simple, natural dynamical system. Due to its simplicity, we cancalculate the evolution of these disturbances with high accuracy and use these calculations to predict

27: THE FIRST FEW HUNDRED THOUSAND YEARS: DYNAMICS OF THE PRIMORDIAL PLASMABACK MATTER

INDEXGLOSSARY

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OverviewCosmologists speak with a high degree of confidence about conditions that existed billions of years ago when the universe wasquite different from how we find it today: times hotter than today, over times denser, and much much smoother, withvariations in density from one place to another only as large as one part in 100,000. We claim to know the composition of theuniverse at this early time, dominated almost entirely by thermal distributions of photons and subatomic particles calledneutrinos. We know in detail many aspects of the evolutionary process that connects this early universe to the current one. Ourmodels of this evolution have been highly predictive and enormously successful.

In this chapter we provide an overview of our subject, broken into two parts. The first is focused on the discovery of theexpansion of the universe in 1929, and the theoretical context for this discovery, which is given by Einstein's general theory ofrelativity (GR). The second is on the implications of this expansion for the early history of the universe, and relics from thatperiod observable today: the cosmic microwave background and the lightest chemical elements. The consistency of suchobservations with theoretical predictions is why we speak confidently about such early times, approximately 14 billion years inour past.

Overview Part I: The Expansion of Space and the Contents of the UniverseSpace is not what you think it is. It can curve and it can expand over time. We can observe the consequences of this expansionover time, and from these observations infer some knowledge of the Universe's contents.

Newton-Maxwell Incompatibility and Einstein's TheoryIt would be difficult to overstate the impact that Einstein’s 1915 General Theory of Relativity has had on the field ofcosmology. So we begin our review with some discussion of the General Theory, and its origin in conflicts between Newtonianphysics and Maxwell’s theory of electric and magnetic fields. We end our discussion of Einstein’s theory with its predictionthat the universe must be either expanding or contracting.

Newton’s laws of motion and gravitation have amazing explanatory powers. Relatively simple laws describe, almost perfectly,the motions of the planets and the moon, as well as the motions of bodies here on Earth -- at least at speeds much lower thanthe speed of light. The discovery of the planet Neptune provides us with an example of their predictive power. The discoverybegan with calculations by Urbain Le Verrier. Using Newton’s theory, he was able to explain the observed motion of Uranusonly if he posited the existence of a planet with a particular orbit, an extra planet beyond those known at the time. Without thegravitational pull of this not-yet-seen planet, Newton's theory could not account for the motion of Uranus. Following up on hisprediction, made public in 1846, Johannes Gottfried Galle looked for a new planet where Le Verrier said it was to be foundand, indeed, there it was, what we now call Neptune. The time from prediction to confirmation was less than a month.

Less than 20 years after the discovery of Neptune, a triumph of the Newtonian theory, came a great inductive synthesis:Maxwell’s theory of electric and magnetic fields. The experiments that led to this synthesis, and the synthesis itself, haveenabled the development of a great range of technologies we now take for granted such as electric motors, radio, television,cellular communication networks and microwave ovens. More important for our subject, they also led to radical changes to ourconception of space and time.

These radical changes arose from conflicts between the synthesis of Maxwell with that of Newton. For example, in theNewtonian theory velocities add: if A is sees B move at speed v to the west, and B sees C moving at speed v to the westrelative to B, then A sees C moving at speed v+v = 2v to the west. But one solution to the Maxwell equations is a propagatingdisturbance in the electromagnetic fields that travels with a fixed speed of about 300,000 km/sec. Without modification, theMaxwell equations predict that both A and B would see electromagnetic wave C moving away from them at a speed of300,000 km/sec, violating the velocity addition rule that one can derive from Newtonian concepts of space and time.

Einstein’s solution to these inconsistencies includes an abandonment of Newtonian concepts of space and time. Thisabandonment, and discovery of the replacement principles consistent with the Maxwell theory, happened over a considerableamount of time. A solution valid in the absence of gravitation came out first in 1905, with Einstein’s paper titled (in translationfrom German) “On the Electrodynamics of Moving Bodies.” His effort to reconcile gravitational theory with Maxwell’s theorydid not fully come together until November of 1915, with a series of lectures in Berlin where he presented his General Theoryof Relativity.

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One indicator that Einstein was on the right track was his realization, in September 1915, that his theory provided anexplanation for a longstanding problem in solar system dynamics known as the anomalous perihelion precession of Mercury.Given Newtonian theory, and an absence of other planets, Mercury would orbit the Sun in an ellipse shape. However, theinfluence of the other planets is to make Mercury follow almost an ellipse, in a pattern that is well approximated as that of aslowly rotating ellipse. One way of expressing this rotation is to say how rapidly the location of closest approach, calledperihelion, is rotating around the Sun. Mercury’s perihelion precession is quite slow. In fact, it’s less than one degree percentury. More precisely, it’s 575 seconds of arc per century, where a second of arc is a degree divided by 3,600 (just as asecond is an hour divided by 3,600).

Urbain Le Verrier, following his success with Neptune, took up the question of this motion of Mercury: could the perihelionprecession be understood as resulting from the pulls on Mercury from the other planets. He found that he could ascribe about532 seconds of arc to the other planets, but not all of it. There is an additional, unexplained (“anomalous”) precession of 43seconds of arc per century. Le Verrier, of course, knew how to handle situations like this. He proposed that this motion iscaused by a not-yet-discovered planet. This planet was proposed to have an orbit closer to the Sun than Mercury’s andeventually had the name Vulcan.

But “Vulcan, Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, Pluto” is probablynot the list of planets you learned in elementary school! Unlike the success with Neptune, Newtonian gravity was not going tobe vindicated by discovery of another predicted planet. Rather than an unaccounted-for planet, the anomalous precession ofMercury, we now know, as Einstein figured out in September of 1915, is due to a failure of Newton’s theory of gravity. Atslow speeds and for weak gravitational fields the Newtonian theory is an excellent approximation to Einstein’s theory -- so thelargest errors in the Newtonian theory show up for the fastest-moving planet orbiting closest to the Sun.

Amazingly, thinking about a theory that explains experiments with electricity and magnets, and trying to reconcile it withNewton’s laws of gravitation and motion, had led to a solution to this decades-old problem in solar system dynamics. Thetheory has gone on from success to success since that time, most recently with the detection of gravitational waves firstreported in 2016. It is of practical importance in the daily lives of many of us: GPS software written based on Newtoniantheory rather than Einstein’s theory would be completely useless.

The Expansion of SpaceMore important to our subject, Einstein’s theory allowed for more-informed speculation about the history of the universe as awhole. In the years following Einstein’s November 1915 series of lectures, a number of theoreticians calculated solutions ofthe Einstein equations for highly-idealized models of the universe. The Einstein field equations are extremely difficult to solvein generality. The first attempts at solving these equations for the universe as a whole thus involved extreme idealization. Theyused what you might call “the most spherical cow approximation of all time.” They approximated the whole universe ascompletely homogeneous; i.e., absolutely the same everywhere.

We now know that on very large scales, this is a good approximation to our actual universe. To illustrate what we mean byhomogeneity being a good approximation on very large scales, we have the figure below which slows a slice from a large-

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volume simulation of the large-scale structure of our universe. In the image, brighter regions are denser regions. The image hastwo sets of sub-boxes: large ones and small ones. We can see that the universe appears different in the small boxes. Box 4 isunder dense, Box 5 is over dense, and Box 6 is about average. If we look at the larger boxes, the universe appears morehomogeneous. Each box looks about the same. This is the sense by which we mean that on large scales the universe is highlyhomogeneous.

On large-scales the Universe is highly homogeneous. The large boxes (boxes 1, 2, and 3) are about 200 Mpc across (that'sabout 600 million light years). No matter where you put down such a large box, the contents look similar. For the smallerboxes this is not the case. Images are from the Millenium Simulation.

Gif of an expanding grid (below) Image by Alex Eisner

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D vs. z: Low z

In 1929, Edwin Hubble made an important observation by measuring distances to various galaxies and by measuring their"redshifts." Hubble inferred distances to galaxies by using standard candles, which are objects with predictable luminosities.Since farther away objects appear dimmer, one can predict distances by comparing the object’s expected luminosity with howbright it appears. The redshift of a galaxy, which cosmologists label as “ ”, tells us how the wavelength of light has shiftedduring its propagation. Mathematically, is equal to the ratio of the wavelength of observed light to the wavelength ofemitted light: / . At least for low z, one can think of this as telling us how fast the galaxy is movingaway from us according to the Doppler Effect: a higher redshift indicates a larger velocity. If a galaxy were instead movingtowards us, its light would appear blueshifted. Hubble found that not only were nearly all the galaxies redshifted, but there wasa linear relationship between the galaxies’ distances and redshifts. This is represented by Hubble’s law, v= d. This simplelaw had profound consequences for our understanding of the cosmos: it indicated that the universe was expanding.

To understand this, take a look at the image of an expanding grid. Each of the two red points is “stationary”; i.e., they eachhave a specific defined location on the grid and do not move from that location. However, as the grid itself expands, thedistance between the two points grows, and they appear to move away from each other. If you lived anywhere on thisexpanding grid, you would see all other points moving away from you. It's easiest to see this is true for the central location onthe grid. Try placing your mouse over different points on the grid and following them as they expand. You will notice thatpoints farther from the center will move away faster than points close to the center. This is how Hubble’s law and theexpansion of space works. Instead of viewing redshifts as being caused by a Doppler effect from galaxies moving throughspace, we will come to understand the redshift as the result of the ongoing creation of new space.

Hubble’s constant, , tells us how fast the universe is expanding. It was first estimated to be about 500 km/s/Mpc, but aftereliminating errors and obtaining more accurate measurements, we know this initial estimate was off, and not by any smallamount. It's wrong by about a factor of 7! The exact value of the Hubble constant is actually somewhat controversial today, buteveryone agrees it's somewhere between 66 and 75 km/s/Mpc. This means that on large length scales, where theapproximation of a homogeneous universe becomes more accurate, about 70 kilometers of new space are created each secondin every Megaparsec.

Figure A: Hubble’s law shown for low redshifts, where we can think of the redshift as arising due to a Doppler effect and sov=cz. Blue line is actual data from supernovae, green lineis a best-fit line for a Hubble constant of 70 km/s/Mpc. Orange andred lines show how this relationship would differ for other Hubble constants. Image by Adrianna Schroeder.

D vs. z: High zAs we will see, from Einstein’s theory of space and time we can expect the rate of expansion of space to change over time. Thehistory of these changes to the expansion rate leave their impact on the distance-redshift relation if we trace it out tosufficiently large distances and redshifts. As we measure out to larger distances, the relationship is no longer governed by cz =

z

1 +z

(1 +z = λobserved )λemitted

H0

H0

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d. Instead, we will show that the redshift tells us how much the universe has expanded since light left the object we areobserving.

where a is the “scale factor” that parameterizes the expansion of space, is the scale factor today and is the scale factorwhen the light was emitted. We can observe quasars so far away that the universe has expanded by a factor of 7 since the lightleft them that we are receiving now. For such a quasar we have / = 7 so the wavelength of light has been stretched by afactor of 7, and by definition of redshift z we have z = 6.

The distance from us to such a quasar depends on how long it took for the universe to expand by a factor of 7. If the expansionrate was slower over this time, then it would have taken longer, so the quasar must be further away. Measurements of distancevs. redshift are thus sensitive to the history of the expansion rate.

As we will see, how the expansion rate changes over time depends on what the universe is made out of. Therefore, studying Dvs. z out to high distances and redshifts can help us determine the composition of the universe. We can see such measurementsin Fig. B, together with some model curves. The models all have the same expansion rate today, , but differ in the mix ofdifferent kinds of matter/energy in the universe. A model that’s purely non-relativistic matter, the green dashed line, does avery poor job of fitting the data. The data seem to require a contribution to the energy density that we call “vacuum energy” or“the cosmological constant.” The “Lambda CDM” model has a mixture of this cosmological constant and non-relativisticmatter. The cosmological constant causes the expansion rate to accelerate. Thus, compared to the model without acosmological constant, the expansion rate in the past was slower, it thus took a longer time for the universe to expand by afactor of 1+z, and thus objects with a given z are at further distances. This acceleration of the expansion rate, discovered via Dvs. z measurements at the end of the 20th century, was a great surprise to most cosmologists. Why there is a cosmologicalconstant, or whether there is something else causing the acceleration, is one of the great mysteries of modern cosmology.

Figure B: Redshift versus distance up to higher redshifts. The shape of the graph can tell us about the contents of the universe--current data best fits the “Lambda-CDM” model, which we will learn about later. Note: we use z here for our x axis ratherthan the recessional velocity v = cz. Although for low z, we can think of redshift as arising from a Doppler effect, thatinterpretation only makes sense for z


Figure C: Relative abundances of oxygen-to-hydrogen and iron-to-hydrogen. We see a trend in which low-iron stars tend tohave low-oxygen, while iron-rich stars are also oxygen-rich. This is because iron and oxygen were formed together overbillions of years of stellar processing.

Observations of helium abundances gives us a different relationship, which can be seen in Figure D. As with iron and oxygen,an oxygen-rich star is likely to contain more helium, which indicates that both helium and oxygen have been created over timewith stellar processing. The difference is that we see a significant abundance of helium even in very old stars formed by gasclouds containing little to no heavy elements. This points to a primordial abundance of helium that existed even before stellarprocessing. Where did this helium come from?

In 1948, Gamow explored the very early universe as a source of elements heavier than hydrogen. He extrapolated Einstein’stheory of an expanding universe with certain assumptions of what the universe is made of, and concluded that the universe was

infinitely dense at a finite time in the p ast. He theorized that this early universe could be a prodigioussource of heavier elements. If the universe began at infinitely high density and temperatures and underwent rapid expansionand cooling, atomic nuclei would form within a window of just a few minutes. Before this window, high-energy radiation didnot permit any nuclei to survive, and after this window, temperatures were too low for the nuclear collisions to overcome theCoulomb repulsion.

Gamow’s important discovery was that in order to avoid overproduction of helium and other heavy elements, the ratio ofnucleons to photons had to be small. Since the number of photons in black body radiation is proportional to temperature cubed,this means that the “Big Bang” had to be very, very hot. From these parameters, Gamow theorized that we should see abackground of heat and light from this period of high temperature and photon density. This background was discovered later in1964 and is known as the Cosmic Microwave Background.

Figure D: Relative abundances of oxygen-to-hydrogen and helium-to-hydrogen in stars. While there is still a trend in whichstars with low O/H also have lower He/H, even the most oxygen-poor stars contain a significant abundance of helium. This isbecause all these objects formed with a primordial abundance of helium, whereas there is essentially no primordial abundanceof oxygen. (Ned Wright, UCLA)

Map of the CMB

CMB shows universe when it was very young, and very simple. (High degree of uniformity/ homogeneity - 1 part in100,000)Diagram of past light cone- show where/when the CMB came fromPhotons could not travel freely until the universe had expanded and cooled such that protons and neutrons combined tomake atoms. These photons have traveled since this last-scattering and redshifted with the expansion of the universe. Theones that reach us now are 46 billion light years away in the microwave spectrum of light (temp = 2.7K)

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Although its existence had been predicted around 1950, the Cosmic MicrowaveBackground was discovered in 1964. It was uncovered by accident while two radio astronomers, Arno Penzias and RobertWoodrow Wilson, were experimenting with an ultra-sensitive radio antenna. Even after removing all known possible sourcesof noise and interference, they detected a consistent noise of microwave radiation coming from all directions, day and night.They inferred that this noise had to be coming from a source outside our galaxy. The characteristics of this radiation perfectlyfit predictions for the Cosmic Microwave Background, a background of radiation left over from the Big Bang that permeatesall of space.

For the first few hundred thousand years, the universe was opaque because photons were constantly scattered off freeelectrons, unable to travel freely. As the temperature decreased, matter was able to recombine into atoms, and electrons boundto nuclei. Photons stopped scattering and were now able to travel freely: this time period is called the Recombination Era.Those same photons have traveled across the universe ever since this time, and the CMB we observe is the photons that arejust now reaching us. The region of space where they started traveling freely, we call the “surface of last scattering,” areceding spherical shape around the observer which is the boundary on our past light cone between when the universe wasopaque and when the universe became transparent.

The CMB currently has a temperature of about 2.73 K, cooled by the expansion of space by over a factor of 1000 since thetime it was released, and is uniform to 1 part in 100,000. If one had eyes highly sensitive to microwave radiation, the aboveimage shows how the sky over Chicago would look, with the colors encoding tiny variations in brightness. The high degree ofuniformity reflects the high degree of uniformity in the early universe. The small departures from uniformity eventually grewover time to produce the diversity of structures we see in the universe today. Without these departures from uniformity therewould be today no galaxies, stars, planets, or physics students.

As we have seen above, the CMB was predicted based on consideration of production of light elements when the universe wasjust a few minutes old, very hot, and expanding very rapidly. It is thus strong evidence in support of a hot, dense, and rapidlyexpanding phase of the evolution of the universe; i.e., a Big Bang.

Cosmic Pie ChartWe know this now from data like the above and our knowledge of physicsWe obtain a lot of information from light freely falling down on us, and we can interpret this information by using simplenatural laws that we know to be true from laboratory experiments here on EarthThe standard cosmological model has this particular mix of ingredients

Conclusion / Wrapping It Up

The agreement between cosmological predictions and observations give us a high degree of confidence. However, we stilldon’t know what most of the universe is made of, and there are many more mysteries remaining to be solved.Examples: dark matter, cause of cosmic acceleration, how did Big Bang beginOur measurements are constantly improving, and one never knows when a combination of precise measurements andpredictions will reveal something new and interesting about the universe

Assignment: Write a brief summary of the contents of the chapter.


Lloyd Knox 5/6/2021 1.1 https://phys.libretexts.org/@go/page/5099

1: Euclidean GeometryOne of the great privileges of teaching this class is the opportunity I have to blow your minds with a radically differentunderstanding of the nature of space and time. The shift from a Euclidean/Newtonian understanding of space and time, to aRiemannian/Einsteinian one is centrally important to our understanding of cosmology. This chapter is entirely focused on theEuclidean geometry that is familiar to you, but reviewed in a language that may be unfamiliar. The new language will help usjourney into the foreign territory of Riemannian geometry where space is curved. Our exploration of that territory will thenhelp you to drop your pre-conceived notions about space and to begin to understand the broader possibilities -- possibilitiesthat are not only mathematically beautiful, but that appear to be realized in the natural world.

According to Euclidean geometry, it is possible to label all space with coordinates x, y, and z such that the square of thedistance between a point labeled by , , and a point labeled by , , is given by

. If points 1 and 2 are only infinitesimally separated, and we call the square of thedistance between them , then we could write this rule, that gives the square of the distance as

This rule has physical significance. The physical content is that if you place a ruler between these two points, and it is a goodruler, it will show a length of . Since it is difficult to find rulers good at measuring infinitesimal lengths, we canturn this into a macroscopic rule. Imagine a string following a path parameterized by , from to , then the lengthof the string is . That is, every infinitesimal increment corresponds to some length . If we add them all up,that's the length of the string.

Exercise 1.1.1: Find the distance along a path from the origin to (x,y,z) = (1,1,1) where the path is given by

Answer

We have .

The limits of integration are and since for the path as given, is the origin and is so .

There are many ways to label the same set of points in space. For example, we could rotate our coordinate system about the zaxis by angle (with positive taken to be in the counterclockwise direction as viewed looking down toward the origin frompositive z) to form a primed coordinate system with this transformation rule:

Under such a re-labeling, the distance between points 1 and 2 is unchanged.Physically, this has to be the case. All we've done is used a different labelingsystem. That can't affect what a ruler would tell us about the distance between anypair of points. Further, for this particular transformation, the equation that gives usthe distance between infinitesimally separated points has the same form.

Figure 1: A counterclockwise rotation of the coordinate system about the z axis by creates a new coordinate system which we’ve labeled with primes. The axis comes out of the screen and is identical to the axis. As is true for any point in space, point 1 can be described in either coordinate system, by specifying or

with the relationship between the two given by the equations to the right.

x1 y1 z1 x2 y2 z2

+ +( − )x1 x22

( − )y1 y22

( − )z1 z22

dℓ2

d = d +d +dℓ2 x2 y2 z2 (1.1)

dℓ = + dℓ2−−−

√

λ λ = 0 λ = 1

dλ(dℓ/dλ)∫ 10

dλ dℓ

Box 1.1

x(λ) = λ, y(λ) = λ, z(λ) = λ (1.2)

dℓ = dλ = dλ+ +(dx/dλ)2 (dy/dλ)2 (dz/dλ)2− −−−−−−−−−−−−−−−−−−−−−−−−

√ 3–√

λ = 0 λ = 1 λ = 0 λ = 1

(x, y, z) = (1, 1, 1) ℓ = dλ =∫ 10 3–

√ 3–

√

θ θ

= zz′ (1.3)

= −x sinθ+y cosθy′ (1.4)

= x cosθ+y sinθx′ (1.5)

θ z

z′ ( , , )x1 y1 z1( , , )x′1 y

′1 z

′1

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Exercise 1.2.1: Show that the distance rule of Equation applied to the prime coordinates,

gives the same distance; i.e, show that . [Warning: is not a coordinate here. It specifies the relationshipbetween the coordinate systems. So, e.g., .] Because this distance is invariant under rotations ofthe coordinate system, we call it the invariant distance.

Answer

We want to emphasize that the labels themselves, x, y, z or x', y', z' have no physical meaning. All physical meaning associatedwith the coordinates comes from an equation that tells us how to calculate distances along paths. To drive this point home, notethat we could also label space with a value of x, y, z at every point, but do it in such a way that we would have the distancebetween x, y, z and have a square given by

For many readers, this result would look more familiar if we renamed the coordinates , , and so that we getanother expression for the invariant distance,

This is the usual spherical coordinate labeling of a 3-dimensional Euclidean space by distance from origin, r, a latitude-likeangle, , and a longitudinal angle, . The transformation between the two coordinate systems is given by

Exercise 1.3.1: Show that the invariant distance given by the equation , the 2-D version of , and theinvariant distance given by the equation , the 2-D version of , are consistent if the coordinates arerelated via:

Hint: use the chain rule, so that, e.g., . (Note that the coordinate transformation equations hereare obtained from the 3-dimensional case by setting .)

Answer

Box 1.2

1.1

(d = (d +(d +(dℓ′)2 x′)2 y′)2 z′)2

d = dℓℓ′ θd = dx cosθ+dy sinθx′

dx′

dy′

dz′

= cosθdx−sinθdy

= sinθdx+cosθdy= dz

(dx′)2

(dy′)2

(dz′)2

= θd −2 cosθ sinθdxdy+ θdcos2 x2 sin2 y2

= θd +2 cosθ sinθdxdy+ θdsin2 x2 cos2 y2

= dz2

(dℓ′)2

dℓ′= d +d +dx2 y2 z2

= dℓ

x+dx, y+dy, z+dz

d = d + (d + yd )ℓ2 x2 x2 y2 sin2 z2 (1.6)

r = x θ = y ϕ = z

d = d + (d + θd ) ℓ2 r2 r2 θ2 sin2 ϕ2 (1.7)

θ ϕ

z = r cosθ (1.8)

y = r sinθ sinϕ (1.9)

x = r sinθcosϕ (1.10)

Box 1.3

d = d +dℓ2 x2 y2 1.1d = d + dℓ2 r2 r2 ϕ2 1.7

x

y

= r cosϕ

= r sinϕ

dx = dr cosϕ−r sinϕdϕθ = π/2



In preparation for thinking about non-Euclidean spaces, we are going to go through how one could construct a labeling of atwo-dimensional Euclidean space in polar coordinates, , . Our construction starts with what will look like an unusual way ofdefining . We define based on the circumference of the circle rather than the distance from the origin, for reasons that willbecome clear later.

First we choose a center to our coordinate system. Then we label all points with that are equidistant from that center andform a circle with circumference . Thus to label space with the appropriate value of , one takes a string, ties one enddown at the center, and marks out all the points that can be just reached by the other end of the string, when it is pulledstraight. Then one measures the circumference of the resulting circle and labels the points on this circle with a value of givenby . We take strings of varying lengths and repeat again and again to figure out the value of for every point inthe plane.

Next to label space with , we take one point on one of the circles and arbitrarily label that one as . We pull a string tightfrom the origin out to this point and beyond, and label all the points along the string with . We then march outward fromthe origin and when we get to a point labeled with radial value , we make a 90 turn to the left and advance some smalldistance . We then label this point with . Again we pull a string tight from the origin out to this point and beyond,and label all these points along the string with the same value of . We then advance another around the circle and repeat,now labeling the th iteration with . In this manner we label all points in the space with values of . Note that whenwe have done this times we will have advanced all the way around the circle (because we will have covered a distanceof ) and the change in will be

In a Euclidean space, such a construction leads us to the result (unproven here) that the distance between twoinfinitesimally separated points labeled by and has a square given by

Note that in our construction we never made any measurement of the distance from the origin to a circle with origin as centerand with circumference . All we know so far is the circumference of the circle. To calculate the distance from theorigin to this circle we can apply the above rule for a path that extends from the origin to the circle. Let's say a circle withcircumference .

Exercise 1.4.1: Calculate the distance from the origin to the circle with circumference . Do so along apath of constant .

Answer

Using we get

since we are told the path is along constant . We integrate from the origin, , out to to get .

This is the distance from the origin out to a circle with circumference and is thus the radius of thatcircle.

dx

dy

= cosθdr−r sinθdθ= sinθdr+r cosθdθ

dx2

dy2= θd −2r cosθ sinθdrdθ+ si θdcos2 r2 r2 n2 θ2

= θd +2r cosθ sinθdrdθ+ θdsin2 r2 r2 cos2 θ2

dℓ2

dℓ2= d +dx2 y2

= d + dr2 r2 θ2

r ϕ

r r

r

C = 2πr r

r

r = C/(2π) r

ϕ ϕ = 0ϕ = 0

r ∘

Δ ϕ = Δ/rϕ Δ

n ϕ = nΔ/r ϕ2πr/Δ

2πr ϕ (2πr/Δ) ×Δ/r = 2π.

dℓr,ϕ r+dr,ϕ+dϕ

d = d + d .ℓ2 r2 r2 ϕ2 (1.11)

C = 2πr

= 2πC1 r1

Box 1.4

= 2πC1 r1ϕ

d = d + dℓ2 r2 r2 ϕ2

d = dℓ2 r2

ϕ dℓ = dr r = 0 r = r1ℓ = r1

C = 2πr1



You should have got the unsurprising result that the distance from the origin to the circle with circumference is . In the next chapter this will get more interesting as we examine a space for which this is not the case.

We'll see that the distance to a circle with this circumference could be more than or less than .

We constructed our coordinate system so that as goes from 0 to at constant a distance is traversed of .But we can check that our rule for above is consistent with this construction.

Exercise 1.4.2: Show that the parameterized path as goes from 0 to covers a distance of byintegrating along this path.

Answer

In general in this 2-d Euclidean space with coordinates as we have defined them, . For thegiven path this becomes . Integrating from 0 to 2 we get that the circumference is .

Before going on, we could take a little more care. We have shown that a particular path that takes us from the origin out to at constant has distance . But how do we know this is the shortest path? Here we will demonstrate that there is not

a shorter path; the one prescribed is the shortest path possible. To do so, we use a result from the calculus of variations. Thatresult is as follows:

For where , the path from point 1 to 2 that extremizes satisfies these equations

This is a mathematical result with more than one application. In mechanics, the action is given as an integral over theLagrangian so that

with , and because a system passes from point 1 to point 2 along the path that minimizes the action, the path takenwill satisfy

which you know as the Euler-Lagrange equations.

In the case at hand we have length = where

(note the overdot is differentiation with respect to the independent variable which here is again) so the shortest-length pathbetween any two points should satisfy

These equations are kind of hairy, if you work them out in generality. However, we are testing to see if a particular pathsatisfies them, the path from the origin to , and that proceeds at fixed . We could parameterize this path with

= 2πC1 r1 r1r1 r1

θ 2π r = r1 2πr1dℓ

r = , θ = λr1 λ 2π 2πr1dℓ

d = d + dℓ2 r2 r2 θ2

dℓ = dλr1 π ℓ = 2πr1

r = r1 ϕ r1

J = dμf( , ,μ)∫ 21

qi q̇ i ≡ d /dμq̇ i qi J

( ) =d

dμ

∂f

∂q̇ i

∂f

∂qi(1.12)

S = ∫ dtL( , , t)qi q̇ i (1.13)

≡ d /dtq̇ i qi

( ) =d

dt

∂L

∂q̇ i

∂L

∂qi(1.14)

∫ dμdℓ

dμ

f = =dℓ

dμ+ṙ2 r2ϕ̇

2− −−−−−−−

√ (1.15)

μ

( ) =d

dμ

∂f

∂ṙ

∂f

∂r(1.16)

( ) =d

dμ

∂f

∂ϕ̇

∂f

∂ϕ(1.17)

r = r1 ϕ = ϕ1 ϕ



with running from 0 to 1. Note that which really simplifies the evaluation of the aboveequations. We will just do one term out of the first equation as an example, and leave evaluation of the rest of the terms as an

exercise. In particular, we evaluate .

Exercise 1.5.1: Evaluate the three other terms, in the two equations above, and verify that the given path does indeedsatisfy these equations, thereby demonstrating that it is the shortest possible path.

Answer

Summary1. Space can be labeled with coordinates. The same space can be labeled with a variety of coordinate systems; e.g., Cartesian

or Spherical.2. The coordinate labelings themselves have no physical meaning. Physical meaning resides in the distances between points,

which one can calculate from a rule that relates infinitesimal changes in coordinates to infinitesimal distances.3. Paths through a space can be parameterized by a single variable; we saw several examples of this.4. The Euler-Lagrange equations can be used to prove that a particular path is (or is not) one with an extreme value of

distance between a pair of points on the path. Usually the extreme is a minimum rather than a maximum.

Homework Problems

Starting from prove the Pythagorean theorem that the squares of the lengths of two sides of a righttriangle are equal to the square of the hypotenuse. Start off by proving it for a triangle with the right-angle vertex locatedat the origin, so all three vertices are at and . Be careful to use the distance rule todetermine the length of each leg of the triangle, rather than your Euclidean intuition. Let's call the length of the side alongthe -axis and similarly the other lengths and . Parameterize each path and perform the appropriate integral overthe independent variable you used for the parametrization (like we did with in this chapter). Doing so, you should findthat . Having proved the Pythagorean theorem for this specially located and oriented triangle, note that sincetranslations and rotations of the coordinate system leave our invariant distance rule unchanged, you have effectivelyproved it for all right triangles.

Prove that the hypotenuse, the straight line from to you described in 1.1, is the shortest path between thosetwo points.

Show that for a primed system that is rotated relative to the unprimed system so that

the square of the invariant distance is unchanged; i.e., .

ϕ = , r(μ) = μϕ1 r1 μ = 0ϕ̇

∂f/∂r = (r/f) = 0ϕ̇2

Box 1.5

d/dμ(∂f/∂ ) = d/dμ(−2 /f) = d/dμ(−2 / ) = 0ṙ ṙ ṙ ṙ (1.18)

d/dμ(∂f/∂ ) = d/dμ(−2 /f) = d/dμ(0) = 0ϕ̇ ϕ̇r2 (1.19)

∂f/∂ϕ = 0 (1.20)

Problem 1.1

d = d +dℓ2 x2 y2

(x, y) = (0, 0), ( , 0),x1 (0, )y1

x ℓx ℓy ℓhλ

= +ℓ2h ℓ2x ℓ

2y

Problem 1.2

( , 0)x1 (0, )y1

Problem 1.3

x′

y′

= x cosθ+y sinθ

= −x sinθ+y cosθ

d +d = (d +(dx2 y2 x′)2 y′)2



2: CurvatureWe introduce the notion of "curvature'' in an attempt to loosen up your understanding of the nature of space, to have you betterprepared to think about the expansion of space.

About 2300 years ago Euclid laid down the foundations of geometry with 23 definitions and five postulates. This was all donein an attempt to capture what were obvious fundamental properties of space, so that they could be used as starting points toprove other things about space -- such as that the sum of the angles of a triangle is 180 degrees, and that the ratio of a circle'scircumference to diameter is the same for all circles. The first four postulates could be put rather succinctly. The fifth onethough is a bit unwieldy. In translation from the original Greek, taken from Wikipedia, we have [in two-dimensionalgeometry]:

If a line segment intersects two straight lines forming two interior angles on the sameside that sum to less than two right angles, then the two lines, if extended indefinitely,meet on that side on which the angles sum to less than two right angles.

There were many attempts to prove this fifth postulate, also known as the "parallel postulate" from the other four. Eventually itwas realized that it can't be done.

We have also come to experimentally determine that space is different from what one gets with all five of Euclid's axiomstogether. Our theoretical understanding of the nature of space (consistent with all measurements) allows for the possibility thattwo lines as described in the fifth postulate might never meet, on either side of the line they both intersect. Throwing out thisfifth postulate leads to the possibility that the angles of a triangle sum up to something different from 180 degrees, and that theratio of circumference to diameter of a circle can vary depending on location of the circle center and the size of the circle.

The ratio of circumference to diameter that is different from is a signature of a property called "curvature." Euclideangeometry is a geometry with zero curvature. In this section we will study both two and three-dimensional curved (andtherefore "non-Euclidean") spaces. One way we will gain some intuition about these spaces is by embedding them in aEuclidean space of one extra dimension. We will emphasize here that there need not be any physical reality to the extra-dimensional space. We can describe curved spaces mathematically without any reference to extra dimensions. Space can"curve" without having to "curve into" any external space. Coming to terms with this idea will, perhaps, make you morecomfortable with the idea that space can expand without expanding into anything else.

Let's get started.

We can label space with coordinates, for example, we could label every point in a 2-dimensional space with an value and a value. These coordinates are just labels, with no physical meaning, until we also say something about the distance betweeninfinitesimally separated pairs of points. For example, in a two-dimensional Euclidean space with which you are familiar, thesquare of the distance between and is given by:

The physical interpretation of is as follows: The length of a ruler with an end on each of the two points is .

Expressions for the distance are not always as simple as the one above. Even for the same physical space, with differentcoordinate schemes the equation for will look different. For example, we could choose polar coordinates instead ofCartesian ones and then distances would be given by:

This is the same space, just described with different coordinates.

Note that one can find the distance along any path through the space by calculating along the path.

The space we are describing in this chapter so far (and its higher- and lower-dimensional versions) we call "Euclidean"because these spaces are consistent with geometry as described by Euclid. It turns out though that space is not Euclidean

π

x y

x, y x+dx, y+dy

d = d +ds2 x2 y2 (2.1)

ds2

ds2−−−

√

ds2

d = d + ds2 r2 r2 ϕ2 (2.2)

∫ ds

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(although a Euclidean description is often a very good approximation). This is a bit startling. If you are not startled by it, youdon't understand it yet. But don't worry; you will. And hopefully your mind will be blown.

From Einstein we learned that space can be quite different from Euclidean. To begin to free your mind from its Euclideanconstraints, let's consider a non-Euclidean space that we label with and using the same construction as we outlined in theprevious chapter. The construction is exactly the same, but the strange nature of the space is revealed by this different rule forthe distance, , between and :

At the moment, the introduction of the factor should just look arbitrary. We will eventually derive this distance rulefrom an assumptions of homogeneity and isotropy of the space. For now, let's explore its geometrical implications.

Exercise 2.1.1: How long would a path be that stretches from to at constant ? Call the length andexpress it as an integral that depends on and . Assume that is much less than . Make a Taylor expansion thatapproximates the integrand so that it contains the first order corrections due to . After this approximation, do theintegral. [Hint: many students in my experience have trouble with this Taylor expansion, hence this hint. Use this firstorder Taylor expansion result: where is some small number and apply it to the term .]

Answer

The path is at constant so . Let's call the radius so .

If for .

Therefore the integrand by the Taylor expansion , where .

Now we have

Exercise 2.2.1: Consider the set of points all at with all values of . Is this a circle? What is the circumference ofthis object as a function of its radius, ? Recall that because of how the coordinate system was constructed, you canassume that and are the same point. First find the circumference as a function of and then use your resultfrom the previous problem to express it as a function of and . [Don't get too hung up on solving for as a function of

. If you take advantage of some additional appropriate approximations the algebra is not too bad, but don't spend toomuch time trying to figure it out.]

Answer

Yes it is a circle. The distance from the center to does not depend on the value of , so all these points are at thesame distance from the center."all at ", so now , and then .

changing by takes us once around the circle so the circumference is

r ϕ

ds r,ϕ r+dr,ϕ+dϕ

d = + ds2dr2

1 −kr2r2 ϕ2 (2.3)

1 −kr2

Box 2.1

r = 0 r = r1 ϕ ℓr1 k r1 1/k

−−−√

k ≠ 0

(1 + ϵ = 1 +nϵ)n ϵ 1/(1 −k )r2− −−−−−−−−

√

d = + ds2dr2

1 −kr2r2 ϕ2

ϕ ds = dr/ 1 −kr2− −−−−−√ ℓ ℓ = ds = dr/∫

r10 ∫

r10 1 −k /2r

2− −−−−−−−√

≪ 1/ ⟹ k ≪ 1 ⟹ k ≪ 1r1 k−−

√ r21 r2 r < r1

≃ 1 + k11−kr2√

12

r2 (1 + ϵ ≃ 1 +nϵ)n n = − 12

ℓ ≃ (1 + k )dr = + k = (1 + k )∫r1

0

1

2r2 r1

1

6r31 r1

1

6r21

Box 2.2

r = r1 ϕℓ

r,ϕ r,ϕ+2π r1k ℓ ℓ

r1

r1 ϕ

r = r1 dr = 0 ds = dϕr1

ϕ 2π

∫ ds = dϕ = 2π∫2π

0r1 r1



Exercise 2.2.2: Discuss the result from Exercises 2.1.1 and 2.2.1 and what it means qualitatively for the circumference-radius relationship for circles in spaces with , and . [Note that you can do this even if you did notmanage to get as a function of in the above exercise.]

Answer

It follows then from the previous results that:

: These circles have a bigger circumference than the Euclidean result.: This is the Euclidean result.: These circles have a smaller circumference than the Euclidean result.

Exercise 2.3.1: If you were a two-dimensional creature, and could travel around this space with a measuring tape,describe in a few sentences at least one way for measuring the value of .

Answer1. Pin down one end of the measuring tape and mark the set of points that are all a distance from it.2. Lay the measuring tape along this circle to measure its length. Call that .3. Check that is close to (so approximations we have used so far in above exercises are good).

4. Use to solve for .

or . . .

Use to get (using from (2)) and use with from (1) to get .

EmbeddingSometimes it is possible to visualize a non-Euclidean space, such as the one with invariant distance rule given by Equation

, by embedding it in a higher-dimensional Euclidean space. Such an embedding, into a space with one extra dimension, isshown in the figure for . When using such an embedding diagram it is important to keep in mind that there are extradimensions in the diagram whose sole purpose is visualization -- they have no physical significance. In the figure here, thespace we are describing is the two-dimensional sphere; the radial dimension is fictional, included here only for purposes ofvisualization.

k < 0 k > 0 k = 0ℓ r

k < 0k = 0k > 0

Box 2.3

k

ℓC

C 2πℓ

C = 2πℓ[1 − k ]16

ℓ2 k

C = 2πr1 r1 C ℓ = (1 + k )r116

r21 ℓ k

2.3k > 0



Note that every point on the sphere can be labeled by the latitude-like coordinate and the longitudinal coordinate (not shown). Also, at every point one can convert and to and . In a homework problem you will derive Equation starting from

the assumption that the three-dimensional space used for the embedding is Euclidean.

Exercise 2.4.1: Observe the circle at constant coordinate value in the embedding diagram. The distance (traveled on apath restricted to the 2-dimensional space of the sphere), from the origin (top of the sphere) to any part of the circle, in the2-dimensional space of the sphere, is . Is the circumference of the circle greater than, equal to, or less than ?Compare to the relevant result in the boxes above.

Answer

so . This is consistent with the case in Exercise 2.2.2.

Not all spaces can be embedded by placing them in just one extra dimension. For example, the space requires two extradimensions for embedding. Part of the space is sometimes shown embedded in just one extra spatial dimension, withthe two-dimensional surface having a shape similar to a saddle. One can start to see the problem here because if the diagramwere extended, the saddle would curve into itself. Such self intersection can only be avoided by introduction of yet anotherfictional extra dimension.

Three-dimensional Homogeneous and Isotropic SpacesLet us now take things up one dimension into 3-D.

Previously we asserted that one could label a 3-dimensional Euclidean space with coordinates , , and such that pointsseparated by , , and would be separated by a distance (as one would measure with a ruler) with square given by

A space that can be labeled in this way is homogeneous (invariant under translations) and isotropic (invariant under rotations).The easiest way to see this is to remember that there's a coordinate transformation to Cartesian coordinates for which

Now the homogeneity is more evident, since transforming to would clearly leave the distance rule unchanged.We've also already seen that rotations leave the distance rule unchanged. So, the space is homogeneous and isotropic. If wechoose to label it with spherical coordinates about a particular origin, our labeling obscures the homogeneity and isotropy, butthe space itself is still homogeneous and isotropic.

It turns out that whether one can label space in this way or not is a matter to be settled by experiment. It's not necessarily true.Even if we restrict ourselves to completely homogeneous and isotropic geometries, we can mathematically describe spacesthat cannot be labeled in this way.

What is generally true is that all three-dimensional homogeneous and isotropic spaces can be labeled with coordinates , ,and such that

for a constant that can be positive, negative or zero. Euclidean space is a special case with .

Exercise 2.5.1: You know that in a Euclidean space the relationship between radius and area of a sphere is with specifying the radius. Note that the angular ( and ) parts of the invariant distance equation are unchanged byhaving . Therefore we still have even if . I also claim that the relationship between sphere area andradius does depend on . How can these statements both be true?

θ

ϕ θ ϕ r ϕ 2.3

Box 2.4

r

ℓ 2πℓ

r < ℓ C = 2πr < 2πℓ k > 0

k < 0k < 0

r θ ϕ

dr dθ dϕ

d = d + (d + θd )s2 r2 r2 θ2 sin2 ϕ2 (2.4)

d = d +d +ds2 x2 y2 z2 (2.5)

x = x+Lx′

r θ

ϕ

d = + (d + θd )s2dr2

1 −kr2r2 θ2 sin2 ϕ2 (2.6)

k k = 0

BOX 2.5

A = 4πr2

r dϕ dθ

k ≠ 0 A = 4πr2 k ≠ 0k



Answer

The short answer is that is not the radius if .The statements are both true because while we have even when , whether or not is the radius doesdepend on , since the radius is obtained by integrating not but .

We can construct the homogeneous and isotropic three-dimensional space and derive its invariant distance rule, at least for thecase of , by embedding it in a 4-dimensional Euclidean space. In a 4-dimensional Euclidean space we can have acoordinate system consisting of three dimensions , that are all orthogonal to each other, and a fourth we will call thatis orthogonal to each of the and directions. Impossible as this is to visualize, we can describe it mathematically. Thedistance between and is given by

In this 4-dimensional space, we construct a three-dimensional subspace that is the set of points all the same distance, , from acommon center. Let's center it on the origin so our subspace satisfies this constraint:

This subspace is homogeneous (all points are the same) and isotropic (all directions are the same). You can see that this is trueby imagining it's two-dimensional analog, a sphere, which is the set of all points satisfying .

It will be helpful at this point to swap out the Cartesian for the spherical coordinate system so we have

and our constraint equation can be written as

From this new version of the constraint equation, we can see that if changes by some amount then we will necessarily haveto have a change in in order to continue to satisfy the constraint. The exact relationship between differential changes youcan easily work out to be (because changing by ends up changing by and likewise for and

and since is fixed ). Using this relationship to eliminate from our invariant distance expression, and usingthe constraint equation to eliminate in favor of and we get

We see that our subspace has an invariant distance expression of the form we were intending to derive, and it is exactly the oneintroduced above if we make the identification .

SummaryLet us now summarize the key points in our study of spatial geometry. First, the most general, high-level points:

1. We can label the continuum of points in a space with coordinates.2. These coordinates have no physical meaning on their own.3. Physical meaning comes from the combination of the coordinates and a rule for converting infinitesimal coordinate

differences into an infinitesimal distance.

Now let's summarize at a greater level of detail:

1. We studied two-dimensional and three-dimensional homogeneous and isotropic spaces. Homogeneity means the spaces arethe same everywhere; e.g., no matter where one is in the space, one would find the same relationship betweencircumference and radius of circles. Isotropic means that there are no special directions in the space.

2. We labeled such spaces with polar (for 2D) and spherical (for 3D) coordinates. We used an explicit construction for the 2Dcase. We defined the radial coordinate as a ``circumferential coordinate'' that gives a circumference for a circle as equal to

r k ≠ 0

A = 4πr2 k ≠ 0 r

k dr dr/ 1 −kr2− −−−−−

√

k > 0x, y, z w

x, y, zw, x, y, z w+dw, x+dx, y+dy, z+dz

d = d +d +d +d .s2 w2 x2 y2 z2 (2.7)

R

+ + + = .w2 x2 y2 z2 R2 (2.8)

+ + =x2 y2 z2 R2

x, y, z r, θ,ϕ

d = d +d + (d + θd )s2 w2 r2 r2 θ2 sin2 ϕ2 (2.9)

+ = .w2 r2 R2 (2.10)

r

w

2wdw+2rdr = 0 r dr r2 2rdr wdw R d = 0R2 dw2

w2 r2 R2

d = + (d + θd ).s2dr2

1 − /r2 R2r2 θ2 sin2 ϕ2 (2.11)

= 1/kR2



by definition. We defined the angular coordinate , in the 2D case, so that the distance from to is equalto . Since the circumference is this means and are the same point.

3. We first asserted that a 2D homogeneous and isotropic space with coordinates constructed as above has the followingdistance rule: , where is some constant with units of inverse area. We also asserted that inthe 3D case the distance rule in spherical coordinates (with an analogous construction procedure that we did not explicitlydescribe) is .

4. Partly to demonstrate that these are indeed homogeneous and isotropic spaces, we then demonstrated how such a spacecould be constructed, for the case of , by embedding it in a Euclidean space with one additional dimension. We wereable to visualize such an embedding for the 2D case. Although we could not visualize the embedding in the 3D case, we atleast mathematically treated its embedding in a 4D Euclidean space to derive the distance rule.

We also introduced techniques for doing geometrical calculations. We introduced paths through a space described by use of anindependent parameter; e.g., with ranging from 0 to 1. If these are Cartesian coordinates in a 2-dimensional Euclidean space then this describes a straight line from to . We calculated distances along suchparameterized paths by calculating from one endpoint of the path to another. Continuing with ourexample, that's length = . We also reminded you of a result from thecalculus of variations, that you have seen in your study of classical mechanics, and applied it to such length integrals, in orderto derive the differential equations obeyed by extreme paths, which are usually (but not always) the shortest paths.

We used these calculational techniques to calculate circumferences and radiuses of circles in the 2D case. With ourexploration, in this way, of homogeneous and isotropic spaces, our fervent desire is that your mind is now freed somewhatfrom its Euclidean intuitions about the nature of space. If you can imagine that the area interior to a circle can be larger than

, where and is the circumference, or the volume interior to a sphere can be larger than where and is the surface area, then you are now better prepared to contemplate the expansion of space.

The expansion of space happens over time. In the next two chapters we thus turn our attention to the geometry of spacetime.We'll find in Chapter 4 that spatial distances are no longer invariant under changes of coordinate systems; you may have beenexposed to this before via the phenomenon of Lorentz contraction. We'll therefore be forced to alter how we relate coordinatedifferences between infinitesimally separated pairs of points to things we can actually measure. In particular, for this purpose,we will introduce the ``invariant distance'' that presumably you have studied in special relativity.

HOMEWORK ProblemsFor the homework problems we will be considering a two-dimensional space labeled with coordinates and with invariantdistance given by

In the following, assume unless otherwise specified.

How long would a path be that stretches from to ? Call the length and express it as a function of and .Assume that . Unlike in the exercises, do not use a Taylor series approximation to .

Consider the set of points all at with all values of . This set is a circle since it is the set of all points located aparticular distance away from another point ( ). What is the circumference of the circle? First find thecircumference as a function of and then use your result from the previous problem to express it as a function of and .

2πr ϕ r,ϕ r,ϕ+dϕrdϕ 2πr r,ϕ r,ϕ+2π

d = d /(1 −k ) + ds2 r2 r2 r2 ϕ2 k

d = d /(1 −k ) + (d + θd )s2 r2 r2 r2 θ2 sin2 ϕ2

k > 0

x(λ) = λ, y(λ) = λ λ(0, 0) (1, 1)

∫ ds = ∫ dλ(ds/dλ)

dλ = dλ =∫ 10 (dx/dλ +(dy/dλ)2 )2

− −−−−−−−−−−−−−−−√ ∫ 10 1 +1

− −−−√ 2

–√

πr2 r ≡ C/(2π) C 4/3πr3

= A/(4π)r2 A

r ϕ

d = + ds2dr2

1 −kr2r2 ϕ2

k > 0

Problem 2.1

r = 0 r = r1 ℓ r1 k


If you were a two-dimensional creature, and could travel around this space with measuring tape, describe in a fewsentences at least one way for measuring the value of .

Consider the embedding diagram in the chapter. Keep in mind that the 3-dimensional space, in which the 2-dimensionalsphere is embedded, is Euclidean. Use what you know about Euclidean geometry to show that the square of the length ofthe path between and , constrained to lie in the sphere, is indeed given by for theappropriate choice of as a function of the radius of the sphere . Also specify that function.

Don't let this calculation lead you astray conceptually. Although it's perfectly fine, and a useful tool for visualization, toconsider a two-dimensional space embedded in a three-dimensional Euclidean space, one can have curved two-dimensional spaces without there needing to be a third dimension. We see this mathematically in this chapter as we can dothings like calculate observables (lengths) without any reference to an additional (third) dimension.

Fun with the Schwarzschild solution. The spatial geometry outside of a central spherically symmetric mass distributioncan be labeled with coordinates such that

where is the "Schwarzschild radius" and is the total mass of the mass distribution. The azimuthal angle runs from 0 to . For the Earth, is about 9 mm.

Consider two concentric circles in the same plane with centers at the center of the mass distribution ( ). If thespatial geometry were Euclidean, the differences in their circumferences would be where is the radialdistance between the two circles. But the presence of the mass means the spatial geometry is not Euclidean, and insteadgiven by the Schwarzschild solution. If the two circles are at coordinate values and , show that, for much smallerthan or one instead gets . Hints: 1) Choose the circles so that it's just that'schanging, with fixed to and 2) Taylor expand to first order so = .

Note that if we take to be 42,000 km (about the distance to geostationary orbit from the center of the Earth) and tobe 6,000 km (the distance from center of Earth to the surface), the correction to the difference in the circumferences is

= 11 cm. A very small correction! The spatial geometry around Earth is very close to Euclidean.

Problem 2.3

k

Problem 2.4

r,ϕ r+dr,ϕ+dϕ + ddr2

1−kr2r2 ϕ2

k R

Problem 2.5

r, θ,ϕ

d = (1 − /r d + (d + θd )ℓ2 rs )−1 r2 r2 θ2 sin2 ϕ2

= 2GM/rs c2 M

ϕ 2π rs

r = 0ΔC = 2πΔℓ Δℓ

r1 r2 rsr1 r2 ΔC = 2πΔℓ −π ln( / )rs r2 r1 ϕ

θ π/2 (1 − /rrs )−1− −−−−−−−−−

√ 1 + 12rsr

r2 r1

π ln( / )rs r2 r1



3: Galilean RelativityWe now extend our discussion of spatial geometry to spacetime geometry. We begin with Galilean relativity, which we willthen generalize in the next section to Einstein (or Lorentz) relativity.

The notion of relativity of motion is not something new with Einstein. It's built into Newtonian mechanics as well. The idea isthat the results of experiments done in any inertial frame will be the same; i.e., one will not be able to determine by someexperimental means what frame is at "absolute rest" and which frames are moving -- all motion is relative. An inertial frame isone in which Newton's laws of motion are satisfied. These laws prescribe accelerations, not velocities, so a frame that ismoving at constant relative velocity with respect to an inertial frame must itself be an inertial frame.

Let's begin by considering a labeling of all points in space, at all points in time, with . Further, let's assume the spatialpart of this is a Cartesian coordinate system such that a ruler (at rest in this coordinate systems) with one end at andthe other at has length whose square is and the time labeling is such that aclock that goes from to will indicate that the time elapsed is . Further, we will assume that thisreference frame is inertial: Newton's laws of motion are satisfied in this frame.

Now imagine a ball that at is located at the origin of this frame and moving toward increasing values of at constantspeed so that . Let's attach another reference frame to this ball and call it the primed frame. In the primed frame theball is not moving, but the origin of the unprimed frame is moving toward decreasing values of at speed . The principle ofrelativity assures us that we can construct this primed frame so that it also has the same nice properties of the unprimed frame;i.e., the Pythagorean theorem applies for lengths, and the readings of stationary clocks make sense; i.e., a clock that goes from

to will indicate that the time elapsed is . The principle of relativity also leads us to theconclusion that if the unprimed frame is an inertial frame, the primed frame will be as well.

So far we have not mentioned Galileo. The above statements all follow from the principle of relativity, and our assumption thatspace is Euclidean. We specialize to Galilean relativity when we assume that the relationship between these two coordinatesystems is:

Because the relationship is time dependent (and because we are anticipating the generalization to Lorentz transformations), wehave explicitly included time in the transformation, even though time transforms trivially. We will call such a transformation a"Galilean Boost." In this case the above equations describe the relationship between a reference frame we denote withoutprimes (the unprimed reference frame) and a reference frame that we denote with primes (the primed reference frame), wherethe primed reference frame is moving relative to the unprimed reference frame with sp

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