General Relativity | Lecture 6
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UniverseNegationBrewster's angleFACTS (newspaper)MassRoll formingVideoPlane (tool)FlatcarStellar atmosphereParticleCartridge (firearms)Finger protocolMechanicForceMobile phoneNoise figureWatchAngle of attackTrajectoryDirect currentTypesettingQuality (business)Gravitational singularityBook coverNorthrop Grumman B-2 SpiritGeokoronaCosmic distance ladderTemperaturabhängiger WiderstandFunkgerätRubber stampMaterialArbeitszylinderShort circuitTARGET2Coulomb's lawMAN-Miller Druckmaschinen GmbHNewtonsche FlüssigkeitOrbitCentrifugeAzimuth thrusterCentrifugal forceClassical mechanicsMitsubishi A6M ZeroLecture/Conference
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OrbitAmplitudeRoots-type superchargerCartridge (firearms)Angle of attackGeokoronaEnergiesparmodusCombined cycleBird vocalizationNoise figureDrehmasseGirl (band)MinerYearAccelerationStagecoachGas turbineMassRadioactive decayHeatElectronic componentVideotapeSpare partRadial engineSatelliteSource (album)ForceDirect currentMail (armour)Angular velocityMitsubishi A6M ZeroFACTS (newspaper)SoundLambda baryonMembrane potentialMechanicLecture/Conference
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Thermodynamic equilibriumLimiterRoots-type superchargerCartridge (firearms)Hydrogen atomNoise figureStock (firearms)Particle physicsPhotonOrbitHalo (optical phenomenon)Radial velocityAngular velocityParticleHomogeneous isotropic turbulencePhotographyMissileYearMinuteGentlemanFinger protocolRadial engineMassEffects unitFunkgerätRulerMechanical fanSpeed of lightDrehmasseFood storageNanotechnologyCosmic distance ladderBand gapArtillery batterySpare partHot workingAlternatorLambda baryonLecture/Conference
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PhotonThermodynamic equilibriumPhotocopierOrbitSpiral galaxyYearNyquist stability criterionCosmic distance ladderRainPhotographyParticleMeasuring cupGeokoronaDestroyerAudio frequencyAngle of attackMaterialRoots-type superchargerDirect currentEffects unitNorthrop Grumman B-2 SpiritGravitational singularityCombined cycleOncotic pressureTool bitCogenerationBlack holeMitsubishi A6M ZeroPhotosphereDifferential (mechanical device)Lecture/Conference
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Spare partCylinder blockFunkgerätRing (jewellery)Ship breakingNoise figureThermodynamic equilibriumMaskRail transport operationsAtomhülleFACTS (newspaper)ElektronenbeugungDVD playerCartridge (firearms)Active laser mediumEnergiesparmodusPaintWhiteAnalog signalLightDirect currentCasting defectParticlePhotonLeistungsanpassungOrbitYearGentlemanStarNegationPattern (sewing)Black holeStandard cellMassHot workingCatadioptric systemBicycle lightingLiterary fragmentAngle of attackRadial velocityFirearmCosmic microwave background radiationGeokoronaLine-of-sight propagationNewtonsche FlüssigkeitCosmic distance ladderClockGreen politicsLecture/Conference
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Analog signalCapital shipAM-Herculis-SternBird vocalizationCosmic distance ladderAngle of attackDrehmasseBirefringenceClassical mechanicsNegationAccelerationInversion <Meteorologie>TrajectoryCut (gems)Focus (optics)Runic calendarCartridge (firearms)Tape recorderProgrammable Array LogicDeck (ship)Differential (mechanical device)WatchMagic (cryptography)Car seatGasbohrlochScrewdriverSpaceportRailroad carWatercraft rowingRolling (metalworking)Foot (unit)KopfstützeOrbital periodYearLecture/Conference
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Black holeCosmic distance ladderSpantKickstandMassAccelerationInversion <Meteorologie>Veränderlicher SternReference workElektronentheorieFocus (optics)Direct currentNear field communicationJukosNoise figureBird vocalizationFlatcarMonthMicroscopeWatercraft rowingTurningWater vaporMagic (cryptography)DrehmasseTowingNetztransformatorToolSunriseGround (electricity)Lecture/Conference
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Cell (biology)Orbital periodGravitational singularityApparent magnitudeToolParticleHourWednesdayVideoYachtDrehmasseGentlemanCar seatBauxitbergbauKey (engineering)Filing (metalworking)Speed of lightYearDayCrystal habitNoise figureHalo (optical phenomenon)ClockAtmosphere of EarthBlack holeBending (metalworking)WeatherClassical mechanicsPlane (tool)Direct currentMitsubishi A6M ZeroAlpha particleNegationLecture/Conference
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AutumnAutomobile platformWatchGravitational singularityStock (firearms)Direct currentOrbitYachtBlack holeFoot (unit)Will-o'-the-wispVideoTrajectoryBird vocalizationFinger protocolClockTypesettingSteelYearQuarkFACTS (newspaper)Hot workingMarble sculptureFirearmLeadGround stationSeries and parallel circuitsToolA Large Ion Collider ExperimentMusical ensembleMeasurementVehicle armourAccelerationAM-Herculis-SternTauApparent magnitudeMitsubishi A6M ZeroYardstickLecture/Conference
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MassClockFunkgerätKosmischer StaubSpeed of lightDirect currentAngle of attackBlackLightPhotonFlashlightAlcohol proofWolkengattungSupernovaTrajectoryQuantumTape recorderAudio frequencyHyperbelnavigationBlack holeA Large Ion Collider ExperimentSignal (electrical engineering)Gravitational singularityContinuous trackWavelengthSpread spectrumDrehmasseFACTS (newspaper)SensorStandard cellSurfingAccelerationSpaceflightWeightCylinder headManned mission to MarsGroup delay and phase delaySizingFirearmAngeregter ZustandHot workingMicrowaveGentlemanDayBurst (band)LeadBird vocalizationCrystal structureNear field communicationBombOrbital periodRoll formingGameMint-made errorsNoise figureLecture/Conference
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StarPhotonSource (album)Cosmic distance ladderRadial engineBlack holeDayStandard cellGangBrickyardBook designDirect currentLightElectronic mediaFirearmCartridge (firearms)Hot isostatic pressingYearLimiterArray data structureAtomhülleAutomated teller machineContinuous trackOrbitAutumnHot workingRail profileMaterialDefecationHydraulic accumulatorSunriseKopfstützeRainAngle of attackGeokoronaSpeed of lightMassWeekKosmischer StaubAmplitude-shift keying
Transcript: English(auto-generated)
00:05
Stanford University. OK, let's begin. We were in the midst of discussing the properties of the Schwarzschild black hole.
00:23
We really barely began. So let me just review quickly. The Schwarzschild solution, first of all, is an idealized solution.
00:41
It's idealized in the sense that the Newtonian force law, F, where are we, this is broken, F equals MMG over R squared.
01:04
That's an idealization. It's an idealization under the assumption that the mass creating the gravitational field is a point at the center. A point mass creates a gravitational field like that.
01:21
If the mass is spread out, then, of course, we don't really believe this to be the case in the interior of where there's mass, in the interior of the earth. This is not correct, but it is true in the exterior. It is true out beyond the surface of the earth, at least
01:41
if we ignore the atmosphere and the other forms of gravitating mass. And so it's an idealization for a point mass. In Newtonian mechanics, you never really ever reach a point mass. Real materials have a certain stiffness. You can squeeze on them, but you can never squeeze them
02:03
to an infinitely small radius. They just have a certain resistance to being squeezed. And gravity is only so strong. Gravity is only so strong, even though the gravitational
02:21
field appears to get very, very large at short distances, the nature of materials is always such that if you squeezed it down to arbitrarily small distances, it would spring back. Not so in the general theory of relativity. We're going to see an example tonight of a way in
02:41
which that's true that if matter gets too close to the singularity, it gets sucked in in a way that doesn't happen in Newtonian gravity. Newtonian gravity, for example, in Newtonian gravity or in Newtonian physics in general, if you have a force
03:04
center like this and you put, let's say, an idealized force center right at the center there, and you put in a particle which is moving straight toward the force center, and with infinite precision, yes, indeed, it
03:22
will hit the center. But even if you have the littlest deviation away from ideal perfect perfection in the trajectory of this particle, it will not hit the center and it will swing around it.
03:43
Effectively, there's a kind of force repelling it from the center. You know what that force is called? Centrifugal force. Centrifugal force, if your aim is not perfect, the particle will have a little bit of angular momentum.
04:02
And because of that angular momentum, there will effectively be centrifugal force and the centrifugal force will keep it from going through the center. So in Newtonian mechanics, even though in Newtonian gravity, even though the center pulls very hard when
04:21
you get close to it, nevertheless, it is infinitely unlikely for an infalling point particle to actually hit the center. Not so in the general theory of relativity. General theory of relativity, gravity is even stronger.
04:43
It's so strong that it overwhelms the centrifugal barrier and pulls anything into the center. In Newtonian mechanics, wouldn't that also be because the particle is infinitely small? It would be almost impossible to hit it?
05:01
No. Well, this is true. This is true. But in the general theory of relativity, you do hit it. Yeah. You do hit it. Okay. Let's write down the Schwarzschild metric.
05:21
The Schwarzschild metric, we did not derive. I just wrote it. And we're going to analyze it. Then later we're going to come back and I'll tell you what equations the Schwarzschild metric is a solution to. It's a solution to Einstein's field equations, but let's
05:46
take it as a target. This is something later we will want to derive. For now, it's a specific metric, a specific geometry whose properties we're going to analyze. All right. So it's described by a proper time, a proper time
06:04
interval, and I've written it down before. It's 1 minus 2m, the mass of the object times the gravitational constant divided by the radial coordinate.
06:22
This is just twice the potential energy of a unit mass particle moving at distance r. That's times dt squared and minus 1 over the same thing, 1 minus 2mg over r, the r squared.
06:46
And notice that when you pass r equals 2mg, as you come in toward the center, this changes sign. But so does this. So there will always be one coordinate with a positive sign
07:03
here and three other coordinates with a negative sign. Let's write in the other three coordinates. The other three coordinates would be r squared. And I've expressed this as d omega squared. The omega squared, let's write it over here, the omega squared is just the metric of a unit sphere.
07:23
It's just the metric of a sphere in polar, in azimuthal and polar angle. So it's just d theta squared plus cosine of theta squared d
07:40
phi squared. We'll just call it that. Sometimes we may only be interested in one of these angles. For example, we're going to be interested in the problem of an orbiting particle. If the orbiting is around the equator like that, then we can
08:03
forget phi altogether. No, we can forget, sorry, we can forget, we only need one angle. We only need one angle. We don't need two angles. And we can just call it d phi squared. Incidentally, cosine theta at the equator is just one.
08:20
And so at the equator, this would just be d phi squared. Okay, so that's this metric here. Let's draw a picture of it. Let's draw a picture of it. First, let's draw the time axis. The time axis I will draw vertically. There's the time axis. Horizontal is space.
08:41
We only have two directions of space that I can draw. So you have to think of a flat plane intersecting time equals zero. But of course, it's really a three-dimensional plane. But I'm not going to try to draw that. And then at r equals two mg, something is happening. This of course, the center here could be thought of as r equals zero.
09:02
So this is r equals zero. R equals zero at the center. And we would do exactly the same thing if we were talking about a Newtonian point particle. We would just place the particle at the center. Then somewhere as r at r equals two mg, there's
09:24
something happening or not happening. Some funniness in any case in the mathematical structure of it. It's like a cylinder surrounding r equals zero. It's called the Schwarzschild radius.
09:42
And it's called the horizon of the black hole. We'll understand why as we go along. This is the horizon of the black hole or it's the Schwarzschild radius. Out beyond that, very far away, let's look at what's going on very far away. Very far away, one over two mg, that's negligible if we
10:01
go to very large distances. One over r becomes very small. So this just becomes dt squared. This just becomes minus dr squared minus r squared d omega squared. That's just flat space. That's just flat space in polar coordinates. Flat space time. So far away, it's just flat space time.
10:21
But as we move into our r equals two mg, something happens and the geometry is characteristically different. Now, it is true that something looks bad at r equals two mg. First of all, the coefficient of dt becomes zero. But even worse, the coefficient of dr squared
10:42
becomes infinite. So something's blowing up, but we're going to see that nothing of any real physical significance is happening at that point. On the other hand, at r equals zero, at r equals zero, the
11:01
denominator here becomes zero. One minus two mg over r becomes infinite. And you might ask, is something real going on at r equals zero? Indeed, something real is going on at r equals zero. And what's going on at r equals zero is the curvature
11:22
is becoming infinite. That's the thing that's going on. Nothing special. No large curvature is happening. No enormous deviation from flat space is happening at r equals two mg. But at r equals zero, all hell is breaking loose.
11:40
If we were to have calculated the curvature tensor to find out just how curved this geometry is, we would find all of its components become infinite right at r equals zero. There's another way to describe that. Curvature means tidal forces. Anything falling in when it would hit r equals zero would
12:04
experience infinitely strong tidal forces and be torn to pieces. It would be stretched radially, squeezed in the angular direction an infinite amount. So r equals zero is a real place.
12:20
In other words, a real serious disaster. r equals two mg, nothing special. Okay, we talked a little bit about radial infalling things. I'll just remind you of what the conclusion was.
12:43
I don't need to derive this again because we're going to see exactly how it works in terms of pictures. We're not going to need the equations to see how it works that it takes an infinite time for something falling in to pass r equals two mg.
13:01
That means in the coordinates that I've drawn here, time, surfaces of constant time are horizontal. The time axis is vertical. If something were falling in, it would look like this. It would fall in and in, it would start far away,
13:26
fall toward the black hole or the horizon or whatever it is, and it would accelerate for a while. I won't try to draw it accelerating. I'm not good enough for that. But as it got closer and closer to the horizon,
13:41
it would asymptote, getting closer and closer, but never quite getting there. If this object had a length, let's say it was a meter stick falling in the long way, if it were a meter stick falling in the long way, the front end of the meter stick
14:01
might look like that. The back end of the meter stick would do exactly the same thing. It would fall in, and it too would never cross the horizon. And so the only way to make sense out of this is to say at least in these coordinates, the gap in r between the front
14:21
end of the meter stick and the back end of the meter stick shrinks to zero. This is actually a form of Lorentz contraction. As something falls in, even though it appears to be slowing down, its momentum is increasing and its Lorentz contracting. I also think about a clock.
14:43
Let's imagine that this meter stick was not just a meter stick, but it was a clock. How could you build a clock out of a meter stick? Well, you could put two mirrors at either end of the meter stick and allow a light beam to go back and forth and back and forth, and you simply count the number of pulses of light beam that go back and forth.
15:05
That would be one way of making a clock out of this. But clocks, as they move, however they are moving, they don't tick off this kind of time.
15:20
They tick off proper time. That's the meaning of the proper time. The proper time is the ticking of a moving clock. How many ticks the moving clock makes from some starting point. This time over here, that's called coordinate time. It's just the coordinate that we're using to describe
15:42
the entire system. Notice that if R is very far away, supposing we're way out here, not moving at all, not moving in, not moving out,
16:01
then d tau is equal to dt, at least to a high approximation, because one minus two mg over R is just one. So very far away, proper time and ordinary time are the same. Very far away, this factor is just basically one, and d
16:22
tau and dt are the same thing. If we're not moving, so the dr and the omega are zero, then a clock moving very far away will really just measure dt. But as we get closer and closer in, for a given amount of
16:41
dt, d tau gets smaller and smaller. So for example, here's a given amount of time transpiring. Time, I mean little t. A given amount of ordinary time, coordinate time
17:00
transpires between this surface and that surface over here. How much proper time? As you get closer and closer, the proper time that transpires along this trajectory here gets smaller and smaller. So while it appears that the amount of t that it takes to
17:25
fall in through the black hole is infinite, we could ask how much tau evolves by the time an infalling object eventually arrives at the horizon.
17:41
And that is finite. Again, we're going to see this in pictures. We're going to see this without doing any mathematics, without doing any calculation. We'll see it just from simple diagrams. But here we can see why this might be true. Even though the time dt or the time interval t is getting
18:01
longer and longer because tau is shrunken by comparison with t, there's a chance in any case that the amount of time along one of these trajectories here is finite until you cross the boundary. Now, somebody falling in with a clock, with or without a
18:23
clock, ordinary human being, do they experience, and by experience I mean experience in the usual sense, do they experience an infinite amount of time before they get to the horizon or a finite amount? The answer is they experience the proper time.
18:42
Your heart is a clock, your brain is a clock. Every timekeeper that you have will experience the same shift or relative size scale for the beating of the clock, the tau, and we can say the beating of clocks very, very
19:02
far away. All right, so it would be interesting to know whether the amount of proper time that it takes to fall is finite or infinite, and we could solve equations, we could solve equations, we can calculate, we can do integrals, but
19:22
fortunately we don't really have to. The pictures will take care of it. Before I get to those pictures, I want to show you how to solve another problem that does not have to do with somebody falling into the black hole, but has to do with
19:42
another question, orbiting the black hole. Orbiting the black hole means, well, it's a little hard to draw. Orbiting the black hole means corkscrew going around as
20:04
time goes forward, but let's take the case of a circular orbit. The circular orbit means that if we projected the orbit onto space without worrying about time, that the orbit would just be a circular orbit.
20:23
So if we were looking down at this thing, if we were looking from the future down at the center, we would see R equals zero, we would see some sort of circle corresponding to the Schwarzschild radius, R equals 2mg, R
20:49
we would see an orbiting particle going around, and I'll take the case of a circular orbit.
21:02
Circular orbits are, of course, particularly simple. Can we figure out what the circular orbits are? Yes, we can, but I'm going to show you just one specific kind of circular orbit. The particular kind of circular orbit we're going to talk about is the orbit of a light ray. As you might expect, and as you probably know, that a light
21:26
ray is, again, let's draw another picture of this, but now let's draw it so that the horizon is small. I haven't changed the size of the horizon, I'm just looking at it from a larger distance. A light ray goes by.
21:44
We know from Einstein from a number of different places that the light ray will bend when it goes by the massive object. I did show you at one point that the equivalence
22:00
principle, the accelerating elevator, does tell you that a light ray moving across the elevator, as the elevator is accelerating, will bend its trajectory. It's the same thing here, gravity bends trajectories, and in fact, obviously, as you move in closer and closer, that
22:22
light ray will get bent more and more. Why? Because gravity becomes stronger. As you move in toward the Schwarzschild radius, it really whips around there, and at some point, it's not the horizon, it's further out than the horizon, there exists
22:44
trajectories which just go right around, circular trajectories which just go right around the gravitating object. Now, this can only happen, this will not happen near the surface of the earth. You'd have to squeeze the earth down to practically become
23:03
a black hole, you'd have to squeeze it way, way down past the Schwarzschild radius, incidentally, an object with the mass of the earth would have a Schwarzschild radius of about a centimeter. So, in other words, the earth would not do anything like this unless it was squeezed down to the diameter of a
23:22
peanut. So, not to worry, you're not going to see light rays circling around the earth, but nevertheless, it's very interesting to try to see if we can understand the orbiting light rays. Now, this is a little bit calculationally intensive here,
23:42
and I'm going to show you the set of things that you do to calculate the circular orbits. We're going to do it by using the rules of classical mechanics, things that we've already set up in previous quarters, and part of the reason I'm doing it this way, the reason I'm doing it this way is because it's the
24:02
only easy way to do it, and even it is not so easy. But it is worth the point that the things we learned before about classical mechanics, for example, are not disconnected from what we're doing now. We're going to use classical mechanics and some
24:22
conservation laws, not much more than conservation laws, just a little bit more, the basic principle of stability that things, the basic principle of equilibrium, that equilibrium happens at extremal points or at stationary
24:41
points of potential energy. We're going to use that, and we're going to use the mechanics of a object orbiting, but not an object, a Newtonian object, but an object moving with this metric over here. Okay, so we talked about this last time, I'm going to go
25:00
back over it, but not at great length, and then show you the calculation. We start with the action. Whenever you're working out equations of motion for anything, it is almost always easiest to start with the action principle. The action principle, from the action principle, derive a
25:22
Lagrangian, and from the Lagrangian, derive the Euler-Lagrange equations of motion, or some other trick. And we're not going to do quite that. We're going to use conservation laws. All right, so again, we start with this metric here, and the rule for the action is the action of a moving
25:47
particle. The action of a moving particle is a parameter in front of it. The parameter is, of course, the mass of a particle. This is necessary to give action. What are the units of action?
26:00
Anybody know? What's the units of Lagrangian? Units of the Lagrangian are energy. Units of action are energy times time. All right, in order to give it the right units, there's got to be a mass out in front. It's going to be energy times time. Mass is energy.
26:20
I'm not going to fool around with the speed of light. We'll leave the speed of light equal to one tonight. Units of action are mass times time, energy times time. So here's your energy, and what's the rest of it? The rest of it is the proper time of the orbit between
26:41
one point and another point. In other words, the integral dt, d tau, along the trajectory from an initial point to a final point. Okay, d tau is the square root of what appears here.
27:01
So, let's rewrite d tau, square root of, I'm going to give, in order to not keep writing this thing over and over, and this thing over and over, I'm going to give them some new names.
27:24
We're going to call this object f, script f, and one over it, I will call script g. The reason I'm using script is because in order to not
27:40
write this thing equal this thing, I didn't want to use just ordinary g, so I called this one script g, but then I said, well, I better call this one script f. Okay, so this is script f and script g. It's script f times dt squared minus script g times dr
28:04
squared, and then finally minus r squared. Now, for d omega squared, we're talking about a particle moving on a trajectory. Let's assume that the trajectory is a circle
28:20
parameterized by a single angle. It could be the angle around the equator, and so I'll just call it minus r squared d theta squared. Maybe it's d phi squared, it doesn't matter whether it's the polar angle or the azimuthal angle, I'll just call
28:42
it d theta squared. That's for a trajectory moving around, let's call it the equator of the sphere. So, in other words, this angle here is theta, r squared d theta squared. Now, what we do with this, always the same thing, divide
29:02
by dt squared inside the square root and then multiply by dt outside the square root. Same thing. So, we're going to get rid of the dt squared here. This dr squared is going to become dr by dt squared, so that's just called r dot squared, the radial component of
29:22
the velocity squared. This becomes d theta by dt squared, which is just theta dot squared, the angular velocity, this is the angular velocity, but now I have to put back dt. I remove dt squared, but from under the square root, and
29:43
so I make that up by multiplying by dt here. F and g are functions of r. For the moment, I won't specify them, but let's just write them down and do remember that g is one over f. As it happens, g is one over f.
30:04
For a circular orbit, r dot is equal to zero. For a circular orbit, this will get a little bit simpler because for a circular orbit, r dot is equal to zero, but we don't want to quite go there yet. All right, so what do we read off the Lagrangian? The Lagrangian is just the thing inside the integral.
30:26
The Lagrangian, let's call it script L, is minus m, contains the factor minus m, times the square root f minus g, whoops, not s, but g, r dot squared minus r squared
30:42
theta dot squared. Now, what do I want from this? I want two things. There are two conservation laws for this problem. The conservation laws are energy. Energy is always a conservation law. Any other conservation law that you can think of for
31:02
orbiting a satellite? Angular momentum. Angular momentum. Angular momentum is just the momentum conjugate to theta. Remember the definition of a momentum in Lagrangian mechanics.
31:21
If we have a coordinate called Q, any coordinate called Q, the momentum associated with Q is just equal to the derivative of the Lagrangian with respect to Q dot. We've seen that over and over again. Here it is again. All right, so the angular momentum, unfortunately, angular
31:44
momentum is also called L, but I'll call it not script L. The angular momentum is just equal to the derivative of the Lagrangian with respect to theta dot. We're just applying this equation when Q is equal to
32:03
theta, and that is equal to m over the square root. Now, I'm not going to continuously write the square root over and over again. I'm just going to call it square root without writing anything in here. But when I write square root, you keep in mind that it's
32:23
the square root of what's in here times the derivative of what's inside with respect to theta dot, and that just gives us an R squared theta dot. That's the angular momentum, and the importance of the angular momentum is that it's constant.
32:42
The importance of the angular momentum is that it's constant. It's also proportional to the mass. I'm going to write it as the mass times something I'll call Greek L or lambda.
33:00
I'm going to factor out the mass, and what's left over, I will just call lambda. What is lambda? Lambda is everything in here with the exception of the mass. So we could write this, I'll just divide by L over, I'll divide by M, remove M from here, and remove M from here.
33:25
So here we have an expression for a quantity which is conserved. It's conserved because the angular momentum is conserved, and it's a function of theta dot and R dot.
33:41
For a circular orbit, it simplifies a little bit because R dot is equal to zero, but let's not do the circular orbit yet. Let's calculate the, this is, well okay, this L is the momentum associated with angle. There's another momentum, and the other momentum is
34:02
associated with R, the other, there are two coordinates in this problem, R and theta, and to find PR, we have to differentiate the Lagrangian with respect to R dot. That's also equal to some, let's see, what is that? That's M script G R dot, I think divided by the same
34:25
square root, same square root, I think that's right, yeah. So we have R dot, but R dot is not conserved. R dot is not conserved. There are forces, radial forces, and when there are radial
34:43
forces, there are accelerations along the radial axis. If there are accelerations along the radial axis, PR is not conserved, but what is conserved is the energy. So I'm going to write over here the general expression for the energy. The general expression for the energy is the good old
35:00
Hamiltonian. The classical Hamiltonian, I'll remind you what it is, it's given by the sum over the various coordinates of the momentum times the velocity associated with that
35:20
coordinate, in other words, the time derivative, sum over the coordinates, in this case there are only two coordinates, and then you subtract off the Lagrangian. I bet you all remember that. But the point is, there is always an energy, it's always
35:42
given by a certain construction, you can go back in your notes later and check that this is the definition of the energy, and it is always conserved. It's conserved as long as the Lagrangian does not explicitly depend on time, there's no explicit time dependence in here, and so the energy is conserved.
36:02
All you have to do now is take, where is P? Here's PR, multiply it by R dot, PL, P theta is just L, multiply it by theta dot, add them together and subtract the Lagrangian. It sounds like a horror, but it's not so bad.
36:20
What you get is pretty simple, H is script F of R, times the mass, divided by good old square root.
36:41
We will not need PR any further, we only needed it to calculate the energy, and this is of course is equal to the energy, and it is conserved, it does not change with time. All right, so let's just put it under here, where I really want it, the energy, script F of R, M, and then
37:07
square root, and now I'll actually write in what goes into the square root here. But, I'm now at the stage now where I'm going to use the fact that we're interested in circular orbits. For circular orbits, R dot is equal to zero, and so
37:23
for a circular orbit, the energy is F minus, no, no minus G, R dot, no R dot, minus R squared theta dot squared. All right, so this is an expression for the energy that involves theta dot squared.
37:45
It's partly kinetic energy, partly kinetic energy due to the motion of theta. There's no kinetic energy due to the motion of R, because R is not changing for a circular orbit. No R velocity, just theta velocity. This is a combination of potential energy, it has R
38:02
dependence, and it has theta and it has velocity dependence. And it's conserved. That's the energy, and what about the angular momentum here? The angular momentum is also conserved and it has the same denominator, F minus R squared theta dot squared.
38:24
Okay, what do you do with this? I'll tell you what you do with it. I'll tell you what the prescription is. The first thing you do is you solve this equation, and I'll let you do it at home. You solve this equation for theta dot, and you find theta
38:40
dot as a function of R, and this is usually called the reduced angular momentum, the angular momentum divided by the mass. You find the angular velocity as a function of angular, sorry, yes, you find the angular velocity as a function of
39:03
the reduced angular momentum. And that's not hard to do, but there's no point in doing it. It's very easy to do. I won't do it, but then when you get it, you plug it back into this equation, into the energy equation. And the result is that you have the energy as a function of R and the reduced angular momentum.
39:23
So, I'll tell you what you get. Now, physics always consists of this alternation between principles and having fun figuring out the principles and what you should do and then the boring work of doing the
39:42
little calculations. And then back to principles again, and I will sort of shortcut the boring calculational part. We solve for theta dot here, and then we plug into the energy, and I'll write down what we get.
40:07
The energy is equal to the square root of F times another square root, and the other square root is R squared lambda squared, reduced angular momentum squared, plus R to the
40:25
fourth, all divided by R squared, times M, times M. You always have M there.
40:44
And that's it. That's it. There's nothing too interesting about it. How do you find a circular orbit? What you do is, this does not depend on R dot. We're now interested only on the radial aspects of the motion.
41:01
We want to know at what radius the equilibrium position for the circular orbit is. The equilibrium position depends on the angular momentum. For an ordinary, okay good, for an ordinary Newtonian problem, is the angular momentum bigger or smaller the
41:21
further away you go for an orbit? It's not obvious. It's not obvious. I'm going to leave it to you to work out whether it's bigger or smaller as you go further away. The best constant angular frequency, let's say, right? No, no, no, no.
41:40
The further away you go, the slower the angular frequency, but remember, and the smaller the velocity, but remember that when you construct angular momentum, you have to multiply by R, velocity times R, so it's not clear which way it goes. If you know a little about the hydrogen atom, you can guess. Yeah?
42:01
Yeah, probably would be. No doubt. Do I have R squared? Yeah, yeah, yeah. It would certainly be, but I'm going to leave it that way anyway. I'm going to leave it that way for the moment because I want this way. But now I want to go to the limit of a photon. I'm interested now in the bizarre question of whether a
42:23
photon can orbit the geometry. Can a photon ever get in close enough that the speed of light is not enough to eject it from the circular orbit? So we're going to look for a circular orbit and the rule
42:44
is to minimize the energy as a function of R for fixed angular momentum. OK, but now there's something going on. Photons have zero mass. So we've got a bit of a funny problem here.
43:02
Photons have zero mass. Let's go back. Let's remember the angular momentum was M times lambda. Photons have zero mass, so it looks like they have zero energy. But they don't have zero energy. A moving photon, a high energy photon will knock your
43:20
teeth out if it hits you. Why? Because it has momentum and energy. OK, so photons do have energy. As M goes to zero, something had better get big to cancel that M. But they also have angular momentum, or momentum and angular momentum. A photon, if it's moving around in orbit or even if it's
43:44
just going past the star, does have angular momentum. So what happens as M goes to zero? The answer is that the thing that you hold fixed is the angular momentum. If you're interested in a photon of a given angular momentum, you hold the angular momentum fixed, you let
44:01
the mass go to zero, but obviously that's going to make lambda get infinite. In other words, the right way to take the limit for studying a photon of a fixed energy and a fixed angular momentum is to take the limit that the mass goes to zero,
44:20
lambda becomes infinite in such a way that the product of M times lambda stays fixed. The angular momentum stays fixed. All right, now let's just look at this formula. What does happen when lambda gets very big? Another way to say it is lambda is the angular momentum per unit mass.
44:42
If the mass goes to zero, lambda will get big if we keep the angular momentum fixed. All right, so here's an expression. Lambda is going to be the biggest thing in sight. M is going to go to zero, so let's analyze it.
45:00
If lambda is going to infinity for any value of R, lambda squared times R squared will be eventually bigger than R to the fourth and much bigger. Eventually, R squared times lambda squared will be bigger than R to the fourth and much bigger, so we can simplify
45:20
this and just say what's under the square root here is just R squared lambda squared and forget the R to the fourth because it's negligible when lambda gets very big. Well, that's just R times lambda. I don't need to write such a fancy expression.
45:40
It's just R times lambda times M, but lambda times M is just the angular momentum. All right, so what's in the numerator here, what's in the square root in any case times M, the square root times M is just the radial distance times the angular
46:06
momentum in the limit, in the limit that we pass to a massless particle holding the angular momentum fixed. This just becomes lambda times M, which is L, all times
46:20
square root of R squared. Then we have this thing over here and we have this thing, square root of F. We're going to put that back in in a minute and figure out what that is. And finally, we have a one over R squared downstairs, which means altogether one over R. One over R, L, and that's
46:45
the energy. What was this? What were we calculating? We were calculating the energy of the orbit. It's a function only, it contains L, but it only contains L as a multiplicative factor. All of the interesting stuff is a function of R. It's a
47:04
function of R and where is the equilibrium, where is the equilibrium? The equilibrium is when the energy is stationary. You can think of this as a kind of potential energy. It doesn't depend on any velocities. All the velocities have been eliminated out of this
47:22
problem. The radial velocity has been eliminated out because it has no radial velocity. The angular velocity was eliminated by substituting, by solving for the angular velocity in terms of the angular momentum. Here we have the angular momentum and we have a function
47:41
of R. Where is the orbit? The orbit, the photon orbit will be where the function of R is stationary, either maximum or minimum. So that's all we have to find. We have to find where and what value of R the
48:01
combination square root of F over R achieves either a maximum or a minimum value. Okay, let's now go back and remember what F of R is. F of R is just the Schwarzschild factor here, just this one minus two MG over R. So it's square root of
48:27
one minus two MG over R all divided by R. Let's imagine plotting this function. Let's imagine plotting the function and trying to see what
48:43
the minimum, what it looks like, where its minimum is and sorry, it is not a minimum as we will see. It is not a minimum. Okay, first go very far away. When you go very far away, two MG over R is much
49:03
smaller than one. It's negligible compared to one. And so when you're very far away, this is just approximately one over R. In other words, it just falls
49:21
off like one over R. But what happens as you move inward? As you move inward, what are we moving into? We're going to move inward to R. The center, the origin here is not R equals zero. It's R equals two MG, the horizon.
49:41
We're moving in toward the horizon and at R equals two MG, what happens to the numerator here? It becomes zero. The denominator is not zero at that point and so the whole energy function becomes zero at the horizon. So it must look like this.
50:05
And that is what it looks like. Now how do you find the point at which it's maximum? How do you find the point at which it's maximum? You differentiate this thing and you set it equal to zero.
50:21
I will tell you the answer. The answer is, and notice that I left out L, didn't I? Yeah, but L just multiplies it. The properties of the maximum or minimum or whatever it is will not depend on L. Okay, so here we are.
50:42
Differentiate the square root thing divided by R, set the derivative equal to zero. Where is it? It's right over here. And that's the point R equals three MG. At three MG, there's a maximum to the energy.
51:00
Now if you have a potential energy that has a maximum, does that correspond to an equilibrium position? Yes. It corresponds to an equilibrium position but an unstable equilibrium. An unstable equilibrium means that if you put an object right at the equilibrium position, it will stay there,
51:20
but if you give it the slightest tap, it will fall off. The smaller the tap you give it or the smaller the imprecision in putting it right up at the top, the longer it will last at the top or near the top, but eventually it will roll down, pick up some speed and
51:40
fall off the top. We're talking about a circular orbit. We're talking about a circular orbit of a massless particle. We sent M to zero. So what we found is right at R equals three MG, a
52:00
massless particle, a photon can orbit the black hole. It's not at the Schwarzschild radius, it's one and a half times the Schwarzschild radius. That radius is called the photosphere or sometimes the photonsphere for obvious reasons. It's a sphere on which a photon can orbit.
52:22
You might expect that if you had a black hole and you went and you looked at that distance, you would find all kinds of photons orbiting around it. After all, any photon that gets started in that orbit will stay in that orbit. But of course, if the photon starts slightly away from the
52:45
top here, just infinitesimally away, if it starts out a little bit too far, it will go back out. That corresponds to the orbits of photons which are out
53:00
beyond the photonsphere and which are just the orbits which get pulled in. Whoops, I hit the photonsphere. I didn't want to hit the photonsphere. I wanted to be outside the photonsphere. If you're outside the photonsphere but just outside this photonsphere, the photon will go whip around and
53:25
then come out, not necessarily exactly in the opposite direction. The closer you get to the photonsphere, the more angle the photon will sweep out before it eventually goes back out. So in particular, if you're very, very close to the
53:43
photonsphere, a photon coming in can go around many, many times but then go back out. That corresponds to just being slightly off to the right here and slowly going outward.
54:01
Well, you can imagine coming in, not quite getting there and then going back out. What happens, however, if the photon is on that side? If the photon is on that side, it gets pulled right in. It can't withstand the gravitation if it's just slightly
54:22
inward of the photonsphere. Instead of orbiting, it may orbit a few times but it will spiral right in to the horizon and then, of course, eventually spiral into the singularity. Okay, so that's what we see from this little calculation.
54:43
Why did you keep angular momentum constant? Oh, what else should I do? I'm interested in a photon with a given angular momentum. That was the problem. Yeah. Well, every photon has some angular momentum so I might as well set it equal to what its angular momentum is.
55:01
Angular momentum is conserved so whatever it is, it will stay that way. Notice the orbit does not depend on the angular momentum. The angular momentum factored out. Angular momentum factored out. The orbit of the unstable equilibrium orbit is right at r
55:24
equals 3mg and it doesn't matter what the angular momentum is. So maybe I'm thinking classically but so here comes a photon and it has a certain linear momentum that's determined by the frequency basically. And depending on which radius it comes in, it's going to
55:43
have different angular momentum about the center, right, about the black hole. Right. But it could be, no, no, no. It could be at a fixed, we could be talking about a thing at a definite radius and it can have different angular momentum at that radius.
56:00
The different angular momentum will correspond, some of them will be orbits and some of them won't orbit, some of them will just go off. But we can fix both the radius and the angular momentum independently. Not if we want to have a circular orbit. Alright, if we want to have a circular orbit, then the radius and the angular momentum are connected but not at
56:21
the photon sphere. The photon sphere, all circular orbits or the circular orbits can have any angular momentum. That's what's curious. So not for photons, it seems like you don't have that freedom. You don't have what? The freedom to fix angular momentum and the radius and Well, you do. You do. You do. Well, no, the only circular orbit, wait, for the
56:43
only circular orbit. No, I mean even if there's no circular orbit. Yeah. No, you have the freedom if you don't worry about circular orbits. But for a photon, the only kind of circular orbit is the one at 3MG here. That's basically what we derived. Only kind of circular orbit and it's an unstable circular
57:02
orbit. Anything that's inside, any photon that is moving, let's say, along in an angular direction that's inside the photon sphere gets swept in by gravity.
57:21
And of course, anything that's moving slower than a photon has even more trouble staying outside. So anything which is inside the photon sphere moving in an angular direction, whether it's got a mass or it
57:41
doesn't have a mass, will always wind up falling in. Yeah. Why is it that the photon sphere occurs where the energy is the maximum? That's where you will find the stable orbit.
58:01
That's just the analog of saying that, not a stable orbit, an equilibrium. If you're looking, we're talking about a kind of equilibrium. What kind of equilibrium? An equilibrium with respect to the radial coordinate. So the equations of motion of the radial coordinate for
58:20
fixed angular momentum have a solution which is just to extremalize the energy. It's the same operation as when you say an equilibrium position corresponds to a minimum or maximum of the potential energy.
58:40
Potential energy is the part of the energy which doesn't depend on velocity. Here we're talking only about the radial motion, so we're talking about the part of the energy which doesn't depend on the radial motion. It doesn't depend on the radial velocity. Okay. Think about it, dwell on it a little bit, compare it with the corresponding Newtonian calculation.
59:01
You'll see all the pieces are the same as the Newtonian calculation of calculating, but the outcome is quite different. The outcome is quite different that in particular there are no circular orbits for anything, photons or otherwise. There are no circular orbits inside the photon sphere.
59:22
There are circular orbits for massive things outside the photon sphere, but for massless particles only the photon sphere. So that's just an example of working out a specific motion in the Schwarzschild geometry.
59:41
So what does this mean for what it would look like around the black hole? I mean, we talked about things looking like they're all plastered onto the event horizon, but you couldn't actually see anything like that, could you? Well, I mean, looking at the black hole from a distance? Let's talk about looking at the black hole.
01:00:00
of the black hole for the distance. Here's the black hole. You're out here looking at the black hole. Light that passes out here bends a little bit. So you think about what you see in the backdrop. You think about the backdrop, what's behind the black hole. That's what you see. You don't see the black hole.
01:00:21
The black hole emits no light. So what you see is the backdrop, the stars or whatever it is, Picasso painting behind, whatever it is, you see it. And of course, now what you see is it appears that
01:00:41
everything in the painting is shifted. You look and you see this point as if the light were coming from here. So everything gets shifted out a little bit away from the blue. But it gets worse than that.
01:01:02
As you look in, for example, there will be a light ray that goes from here to here and another light ray that goes from here to here. There may be a light ray that goes from there to there. And you'll see this point double. If it isn't this point that you see double, it will be
01:01:21
this point that you see double. Some point along here you will see double. In fact, it's not even that you'll see it double. You'll see it split in half. You'll see a mess. It's worse than that. No. The light ray can go this way. It can go this way. It can go behind the black hole.
01:01:41
And so what you actually see is a kind of ring around the black hole. You see a ring around the black hole. And it's even worse. There are light rays that can go around that can start over here and go like that. A black hole is a very bad thing to try to hide from your enemy. Well, not the black hole.
01:02:01
It's very hard to hide from your enemy by hiding behind the black hole. Your enemy is over here and you're over here. You may think you're hiding, but you're not because there are light rays which go around like that. But, of course, what you see is very, very complicated as
01:02:22
your line of sight moves closer and closer to the horizon because you'll see light rays which go around many, many times and it will fragment the background into just an impossibly complicated thing.
01:02:40
Yeah, Einstein rings are experimentally or observationally verified things. Is that the photon sphere or the horizon photon sphere? Well, I was thinking about the horizon over here. A light ray which is pointed outward can get out from just above the horizon.
01:03:00
It's the ones which are pointing tangentially this way which will fall in if they're inside. A light ray, not here, here. Here's the photon sphere over here. A light ray which is moving in the angular direction will not be able to escape from the photon sphere, but a light
01:03:23
ray which is radially outward will escape. And in particular, a light ray which is sort of partly radially, partly angular, and those will swing all the ways around and come out, for example, over here.
01:03:41
Right. So that makes looking at a black hole very disifying. You'll see a very complicated pattern.
01:04:12
I've never done it. It's a mess. It's a mess. Well, you know, with modern computers, you just throw the
01:04:22
equation on the computer and it'll spit out answers. You know, as I said. The Euler Lagrange for this particular circular motion, is that substantially harder than the conservation?
01:04:40
For which motion? For the circular orbit. No. It's not, it's, it's probably, it probably takes more steps than what I did. What would you do? I suppose what you would do, yeah, no, no, I realize, I know that. No, no. It's a fair question.
01:05:03
You write down Euler Lagrange equations and then you stare at them and you say, how the devil am I going to solve them? Oh, I use the conservation laws. Sometimes I have enough conservation laws that it completely determines the answer. Then you don't have to solve any differential equation.
01:05:21
And that's all we really did here. I could have written down the equations of motion and then observed that there were some conserved quantities and just used them. But the trick of starting with the action is always the most efficient way to do things.
01:05:41
Well, is it always? I think it is always. Maybe there's a case here and there where there's something easier, but not usually. Okay. Now we want to understand better what it is that's happening at this curious point.
01:06:02
It's not a point. It's kind of a shell around the black hole. Let's draw it again. Here's R equals zero. I don't know. What do you like? Green horizons or red horizons?
01:06:22
Red. Good. I do too. So there's a kind of shell around the black hole, the horizon, and something curious is going on there. Among other things, the sign of the coefficient of Dt squared changes as does the sign of the coefficient of Dr
01:06:43
squared. They both change. They keep fixed the number of negative signs and positive signs, but still, something curious is going on. We know that clocks, when viewed from far away, a clock
01:07:00
falling in seems to go slower and slower. That's that relationship between Dt and Dtau. So something is happening here. But on the other hand, I've also told you that in some sense nothing is happening. So how do we reconcile nothing with something?
01:07:40
Many things that happen in gravity can be understood by first understanding them for accelerated coordinate frames. That's the equivalence principle. As I've said, gravity is not really exactly the same as a uniformly accelerated coordinate system, but it's often a very good thing to see if we can see what's
01:08:02
happening for a uniformly accelerated coordinate system. So let's go back to that problem. The uniform acceleration in relativity is a little more complicated than it is in good old Newtonian physics.
01:08:24
We've already talked about it. And it corresponds to a coordinate system which is the analog of polar coordinates, but polar coordinates in
01:08:42
hyperbolic geometry. Hyperbolic geometry just means circles are replaced by hyperbolas. Okay, so I'll just very quickly remind you. Let's introduce, this is flat space now. This is plain old flat space. It's not a black hole.
01:09:01
It's just ordinary flat space-time. There may be some coordinates coming out of the blackboard. For the moment, I'm not going to be interested in them. What I am interested in is this coordinate, which I'll call capital X, and this coordinate here, which I will call capital T.
01:09:21
Time and space. And there may be two more coordinates which you can think of as coming out of the blackboard, but they're not going to play any big role in what I'm going to say for the moment. Okay, now what is the metric of this, oh, sorry. Next, we introduce hyperbolic coordinates, or we introduce
01:09:41
polar coordinates. Let me write them down for you. The definition is that X is equal to something which I'll call rho. Rho is distance away from this point over here, horizontally, rho times hyperbolic cosine of omega, and
01:10:00
omega is a kind of time variable that varies along a hyperbola. Along each hyperbola, omega varies. It's a kind of angle, but it's not really an angle. It's not periodic. It's a hyperbolic angle. It goes from minus infinity down here to plus infinity up
01:10:22
here. But this is the analog of X equals rho cosine theta for ordinary polar coordinates. What about T? T is equal to rho hyperbolic sine of omega.
01:10:43
What about X squared minus T squared? X squared minus T squared is equal to rho squared times cos squared minus sinh squared.
01:11:01
Cos squared minus sinh squared is one. So each one of these hyperbolas is a trajectory on which X squared minus T squared is constant, and that is what we call uniformly accelerated coordinates in special relativity. It's really just special relativity.
01:11:21
The metric of ordinary flat space is given by d tau squared equals rho squared d omega squared minus d rho squared. And that's the analog of r squared d theta squared plus
01:11:41
d r squared. This would be, metric would be, let's continue to call it rho squared. Rho squared d theta squared plus d rho squared would be the metric and polar coordinates for an ordinary geometry. Here it is for Minkowski space for special relativity.
01:12:06
The tau squared is rho squared, the omega squared minus the rho squared. Now I'm going to change some coordinates. Just boring, very, very boring coordinate changes for a moment.
01:12:21
I'm going to invent a new coordinate called xi. I hate making new notations, but we have no real choice. We have to study this theory in different coordinates, and I'm going to change coordinates a little bit. I'm going to take this rho squared here, and I'm going to call it four times the Greek letter xi.
01:12:50
Rho squared is four times xi. I'm going to call, let's write that down. Rho squared is equal to four times xi.
01:13:04
Similarly, well not similarly, but also we're going to write that omega is just t over two. No big deal there. It's just convenient to replace omega by t over two.
01:13:21
And now we do the metric. The omega squared, what's the omega squared? The omega squared is dt squared over four. Okay, so this is dt squared over four, that's the omega squared, dt squared over four, and now we have to
01:13:45
multiply that by rho squared. I'm talking about the first term here, and that's just four times xi, big deal. xi times dt squared. Okay, now what about minus d rho squared?
01:14:01
Let's take minus d rho squared. Here's rho squared. Let's calculate d rho squared, the differential of rho squared. Let's differentiate this equation. Two rho d rho is equal to four d xi.
01:14:23
That means that d rho is equal to, I think it's two d xi over rho. Do I have that right? I think I have it right.
01:14:43
Okay, so d rho squared is four d xi squared over rho squared, but now rho squared is four xi.
01:15:01
So let's put that in here, four xi. Four is cancel, and now maybe yes, maybe no, you'll notice a similarity with the Schwarzschild metric. In the Schwarzschild metric, you have a coefficient
01:15:22
function here times dt squared, and you have the inverse coefficient function here times dr squared. If you think of xi as being like r, and this has some family resemblance to this.
01:15:41
Notice another thing, that this changes sign when r gets in too close. xi changes sign when it becomes negative. Does xi being negative mean something? Yes, we will see that xi being negative does mean something.
01:16:00
So there is a similarity, but the basic similarity is a coefficient and another coefficient, which are inverses of each other, which we will see can meaningfully change sign as xi becomes negative. Okay, let's flesh this out a little more.
01:16:21
Let's flesh this out a little more. What's that? We're going to put it back. We'll put it back. It does matter, but for the moment, we're just concentrating on the radial and time-like directions. And then we'll put back the other one also.
01:16:42
But for the moment, let's just concentrate on these two guys over here. Okay, here's what I want to do. I want to focus and sort of zoom in on the place near r equals 2mg.
01:17:00
So blow it up, focus in on it, which is just another way of saying we're going to study this metric in the region where r is almost equal to 2mg, just to sort of blow it up and see what's going on there. Take a microscope and blow it up. Okay, let me first write it this way. r minus 2mg over r.
01:17:24
I haven't done anything. I've just written it as a numerator over a denominator. Now, I have to be careful. We're only going to be interested. Here's the horizon. Here's the horizon, and I'm interested in the vicinity very, very close to the horizon.
01:17:41
I want to explore that region very close to it, which means I'm not going to let r change very much. If I don't let r change very much, you would think I could just set it equal to 2mg. But then I would have a bit of a problem with the numerator here. I would have zero. I don't want to do it with a numerator.
01:18:00
But in the denominator, the denominator does not change much. It doesn't become zero when r equals 2mg. Nothing special happens to r when it becomes equal to 2mg. Implication is that in an approximation where you just hover near the horizon, you can set r equal to 2mg
01:18:22
without losing any accuracy very, very close to the horizon. It's like in Newton's equations or Newton's gravitational theory, if you're near the surface of the Earth, for many purposes you can set r equal to the radius of the
01:18:42
Earth. You have to be a little bit careful about things which go to zero and also things which can go to infinity, and this is something which can go to infinity over here. So let's look at this. This is r minus 2mg dt squared, and this is the inverse, 2mg over r minus 2mg, the r squared.
01:19:12
You know what I'm going to do just to simplify this? I don't want to carry along too much baggage. I'm going to set the mass of the black hole to be a
01:19:21
specific number. I'm going to set it equal to a number so that 2mg is just equal to one. That will simplify our lives a lot here. I can always do that. That's a choice of units. What are the units of 2mg? Units of length.
01:19:41
2mg is the Schwarzschild radius. There's units of length. For a particular black hole, I can choose units of length where the Schwarzschild radius is just equal to one. I can't do it for all black holes simultaneously, but if I'm interested in a particular black hole, I can do
01:20:02
that. And that makes this formula much simpler. Let's just do that. Let's set 2mg equal to one. It's r minus 1 dt squared, one over r minus one, the r squared. What's the next step?
01:20:24
What? The next step is to redefine r minus one and call it c. Just a change of variables. It's a change of variables which measures r relative to the horizon. Now the horizon is at r equals one.
01:20:42
r equals one equals horizon. So what we're going to do is we're going to change coordinates again so that our new coordinate is just measuring distance from the horizon. It's just measuring the deviation of r from its horizon
01:21:00
value. Let's call r minus one c. This one becomes one over c, the r squared, but the r and d c are the same thing. Let's see. What was the definition? C is equal to r minus one, d c is just equal to dr.
01:21:24
So this becomes d c squared. Exactly what we have here. What's going on? What this is telling us is that the vicinity of the horizon is in some mysterious way just flat space or approximately
01:21:43
flat space. Nothing dramatic is happening there because by a coordinate transformation, if by a coordinate transformation, I can change the metric so that it looks like this. It's saying that to an approximation, a good
01:22:00
approximation, the vicinity of the horizon is locally flat. Nothing very dramatic going on, no large curvature, more or less flat space. Now that's surprising, but it's true.
01:22:30
The uniformly accelerated coordinates, yes. But that doesn't mean that the space isn't flat. It's flat space in which there is a coordinate system
01:22:42
which is moving with uniform acceleration. Why would we be interested in a coordinate system for gravitating object which is mimicking uniform acceleration? Well, me standing here, I am experiencing an acceleration of
01:23:02
ten meters per second per second. I feel exactly the same things that I would if I were in space and the floor under me was accelerating upward with ten meters per second per second. In studying a gravitating object from the perspective of
01:23:21
somebody standing still, being supported, let's say above the horizon or wherever it is, you're effectively doing physics in a uniformly accelerated reference frame. What does physics feel like there? What does it do there? It does whatever the uniformly accelerated reference frame
01:23:42
does and here we see, we're just seeing exactly that. We're seeing the vicinity of the horizon of a black hole is simply a uniformly accelerated coordinate system, but the space itself, the space time itself is effectively flat.
01:24:02
Not much different than that. Hardly different from that. Now, let's talk about this inversion of space to time and
01:24:21
time to space. XE was equal to R minus one or if I put back MG, it would be related to R minus two MG. I set two MG equal to one, but what XE is is it's really
01:24:42
proportional to R minus two MG. XE can change sign. How does it change sign? By going from outside the horizon to inside the horizon. So, as I said before, yes it makes sense, at least in the black hole context here, to say that XE can change sign.
01:25:04
Does it make any sense in this flat space and if so, where is negative XE on this diagram? So, to answer that, we go back and we say, look, here's what XE was. XE was equal to rho squared over four.
01:25:23
The four is not so interesting. That's not the point here. Forget the four. It's not important. XE is rho squared over four. Does it mean anything for rho squared to change sign? How can rho squared change sign? Rho is a positive number and rho squared or rho is a real
01:25:42
number, it's square is positive. No, that's not quite right. The way you think about rho is it's the rho squared on the right hand side of this equation. Positive rho squared here means a hyperbola in this
01:26:03
quadrant. A hyperbola in this quadrant over here has a positive rho squared. It means that X squared is bigger than T squared. X squared is certainly bigger than T squared all along in this entire quadrant. X magnitude of X is bigger than the magnitude of T in this
01:26:23
whole quadrant here. That's why rho squared is positive. What does rho squared being negative mean? It means that T squared is bigger than X squared. It means you're up here. Let's forget the bottom and the left hand side. Let's just concentrate on this quadrant and this quadrant
01:26:41
for a moment. Rho squared being negative simply corresponds to the point in the upper quadrant here. That's what it means, rho squared being negative. So in going from XE positive to XE negative, you're simply passing from here to here.
01:27:01
Imagine now that we wanted to follow our way in. Yeah. We wanted to follow our way in till we got the XE equals zero and then where would we go from there?
01:27:22
Here's what we're doing. Yeah. Here's omega equals zero and we're following rho into the origin here at which point XE changes sign.
01:27:42
Where do we go from here as if we're following omega equals zero and letting XE vary? We don't go here. We go here. We jump into the upper plane.
01:28:02
So omega equals zero along here and omega equals zero here. Here, XE is greater than zero. Here, XE is less than zero. So what has happened is yes, a space-like coordinate,
01:28:21
namely XE, has jumped and become a time-like coordinate. Omega, which was this angular coordinate here, what happens to it? What happens if we vary omega in this quadrant? We're moving along here.
01:28:41
Also, a hyperboloid. What's that? Is it moving faster than light? Yes, but they're just coordinates. Coordinates can do anything you want. No physical system is moving faster than the speed of light.
01:29:00
Whether a system is moving faster than the speed of light is quite separate from whether a coordinate is moving faster than the speed of light. So these are just funny coordinates where XE goes from being a space coordinate to being a time coordinate and
01:29:23
omega goes from being time-like in here to being space-like in here. However, remember, all this was just flat space. There was nothing going on in the space. Somebody who jumps through here sees nothing special.
01:29:41
There's nothing happening. It's just peculiar coordinates that have some funny behavior as you go from one side to another. The same is true of the Schwarzschild coordinates. XE is R minus one or R minus two MG. When this changes sign, when you go from here to here,
01:30:00
you're doing exactly the same thing that you do going from here to here. Okay, let's go on a little bit and think about the black hole now. XE is R minus one. Varying R means varying R until you get to one, that's
01:30:22
the horizon, or two MG, and then continuing to vary R, you move up here. That's the nature of these coordinates. But eventually you'll get to R equals zero. Here R is decreasing, decreasing, decreasing, until it gets to two MG, and then it keeps decreasing until you get
01:30:43
to some point where R is equal to zero. When R is equal to zero, something nasty does happen. When R is equal to zero, something else is happening. Where is R equals zero? R equals zero is over here.
01:31:07
You're behind the horizon, inside, interior to the horizon, this thing has changed sign, and guess what, the singularity at R equals zero is not a place, it's a time.
01:31:27
It's not a place, it's a time. It's not a single place. Here, many different places, all at the same, all at, well, they're not all at the same time, but this
01:31:41
surface R equals zero is not what you thought it was. It wasn't a vertical line like this, it was more horizontal and more space-like than time-like. So there's R equals zero. Pretty confusing, right?
01:32:02
Very confusing. Very confusing, but this is the way the black hole really is. You'll get used to it. You'll get used to it. If you think about it and you play with it, you doubt it? Well, let me get some simple questions. What if R becomes, on that chart, R could be negative just
01:32:24
by going past the geometrically, just past that curve, and you have two of the coordinates that, I don't know what their physical significance would be. The question is physically, what happens to a system? You say, you can, you can go, you meant me, behind
01:32:45
negative R. No, maybe you'd be up here, okay? Right. Maybe you'd be up here. But, the problem is you have to go through this. Is R equals zero still the center of the black hole? It is what it is.
01:33:03
You can think of it as the center of the black hole, but obviously it's a funny kind of center. Yeah, it's a time, not a place, if you like. That's why you can't avoid it.
01:33:22
You can't avoid the future. There's no way to escape from the future. There's a place. This thing is a place. I can avoid it. I can go around it. What I can't avoid is the future. That's why you can't avoid the singularity when you get in here, because no matter what you do, it's always in
01:33:44
your future and you will hit it. Just like the clock will run down, you'll hit it. So, the singularity is not avoidable the way it would be in Newtonian physics in which the center of coordinates is
01:34:01
a place. You can go around a place, you can't go around a time. So, there we are. That's the nature of the black hole singularity. What about the horizon? Well, the horizon is over here, but we're going to think about it a little bit differently eventually. We're going to think about this whole line as being a horizon.
01:34:22
But, let's erase some of this and yeah. On your diagram there, when you first drew it, you drew those two axes that actually aren't quite perpendicular. Wait, it's this and this? No, those are the ones you draw through second. The two that are out there that make an X, what are
01:34:42
those exactly? Those are the light cones. Those are the light light directions. Oh, okay. Got it. Those are the light light directions. Could it come in at a W, equal or minus some number? Yeah. A particle might start out over here, come in and pass
01:35:02
through there. Alright, now this figure explains something to us. This figure allows us to understand much better this idea that it takes an infinite amount of time for a thing to fall through the horizon, at least an infinite amount of coordinate time, but still it falls through in
01:35:21
its own timekeeping, it falls through in a finite time. So let's redraw it and see if we can understand that. Same picture, and now let's draw what the time slices
01:35:44
look like. These are the different times. Omega is like time. What was the relation between, where did I write it down? Yeah, in comparing this over here, yeah, what was the relation? The T, T was omega over two, or omega is T over two.
01:36:04
Right. So, time is simply this angular coordinate. Position is constant along these hyperbolas. Now, let's imagine somebody throws in, I don't know, Alice
01:36:24
throws Bob into the black hole. Alice stays outside, Bob goes in. Alice has coordinates which are just these polar type
01:36:47
coordinates, omega. Where? Here's omega equals zero, here's omega equals one, here's omega equals two, here's omega equals three, here's omega equals four, and oops, not very good,
01:37:02
omega equals four. Omega asymptotically goes to infinity when you get up to this light-like direction here. So, Alice, she says infinitely late time corresponds to
01:37:22
the surface over here. In fact, if she's looking back and watching Bob, she looks back, she looks back, she looks back, and it's not until an infinite amount of her time has transpired that she actually sees Bob go through the horizon. In other words, she never sees Bob go through the horizon.
01:37:41
According to her reckoning, it takes an infinite amount of omega before Bob falls through the horizon. That translates to an infinite amount of T in these coordinates. On the other hand, just looking at the diagram, you can see immediately that the amount of proper time between here and here is finite.
01:38:03
Just an ordinary trajectory, nothing special happening to it, it's ticking off time, there it is, a finite amount of time in order for this object to fall through the horizon, to fall through this surface here. It is the horizon, but we'll come back to that.
01:38:23
So, it's a coordinate property, a funny coordinate property, that Alice's reckoning is that Bob takes an infinite amount of time to fall through the horizon. Not so, says Bob, it only took a finite amount of time, I watched it in my clock, but the reason for the
01:38:41
discrepancy is this discrepancy between proper time and coordinate time. Alice uses coordinate time, Bob uses proper time. Yeah, Alice is staying outside the black hole. She's, like I am now, undergoing uniform acceleration.
01:39:04
She's standing on the floor above the horizon of the black hole, and so she's moving like this. If she were actually to watch, let's just imagine her looking back, she looks back, and at this instant over
01:39:20
here, where does she see Bob? She sees Bob back over here. She sees Bob over here. In other words, she has to look back in time, simply because it takes some time to get to her. A little bit later, she sees Bob over here. A little bit later, she sees Bob over here. She can wait forever, and she will not see Bob cross
01:39:42
that line. On the other hand, Bob crosses the line, as far as he's concerned. He doesn't see anything peculiar. Okay, let's look at it. Let's see how Bob sees Alice.
01:40:04
Well, before I answer that, let's just say one more thing. Alice obviously sees Bob slow down. Why does he slow down? Because it takes an infinite amount of time, or a very large amount of time from her perspective, to go from here to there.
01:40:21
To go from here to there, it takes a huge amount of time. Each heartbeat of Bob takes longer and longer from her point of view. So, Alice sees Bob slow down. What does Bob see? Sorry, what is that? Bob, what does Bob see?
01:40:42
All right, so let's draw that. Here's Bob. He just does that. Here's Alice out here, and Bob looks back. He sees Alice over here. He looks back. He sees Alice over here. He looks back right at this point where he crosses.
01:41:03
He sees nothing special. Even after he passes through, Alice can never see Bob at that point, but Bob has no trouble seeing Alice. So, Bob sees nothing special. He doesn't see Alice speed up. He doesn't see Alice shrink.
01:41:24
He does see Alice accelerating away from him. That's true. He sees Alice accelerating away from him, but that's all. He sees Alice, but at any point, he sees Alice perfectly
01:41:42
normal, except accelerating away from him. Alice, she's moving with uniform acceleration. That's because she's standing on the floor. She's standing on the floor at a fixed R. Remember, here we have...
01:42:03
She could be in orbit. She could be in orbit. Yeah. Or she could just be standing on a platform that's built above the black hole. Imagine a platform or suspended, suspended from a very, very distant object.
01:42:21
Whatever way is required to keep her just at a fixed height above the black hole, a fixed height above the black hole corresponds to a trajectory like that. Bob looks back. Nothing special happens to Alice when Bob crosses this point. On the other hand, from Alice's perspective, she never
01:42:43
sees Bob cross that point. It takes forever from her perspective. So, it's very asymmetric, and it is asymmetric. It's bizarre, too. When does Bob hit the... The singularity. All right. So, up here is a singularity someplace at R equals zero.
01:43:02
Bob hits the singularity right over here. By that point, the tidal forces will have become so large that Bob will no longer be with us. I mean, Bob will have been destroyed by the time he gets to here. That's why we don't think about trying to answer the
01:43:23
question of C when he gets, let's say, to negative R. By the time he gets to negative R, he has experienced infinitely strong tidal forces, and there is no more Bob. No, his proper time is tau.
01:43:43
Tau is proper time. That's what Bob measures along here. He's going parallel to T.
01:44:01
Oh, oh, in here, inside. Yeah. Inside, in this quadrant here, Bob would probably use R for time, or something like R for time. Inside here, R is varying from here to here, from R equals two mg here to R equals zero here, so his clocks
01:44:23
would have more to do with R than T, and T would be more like position in here. But as I said, he doesn't feel any funny thing happening where his clock starts to become a meter stick, and his meter stick becomes a clock.
01:44:42
Nothing like that. He just sails through happily. Now, I think it's apparent that this line over here is rather special. It's rather special in that anything that crosses behind
01:45:03
here, anything that crosses behind here cannot escape. We'll eventually remember in the coordinates that we're using, light moves with 45 degree angles in here. Light moves with 45 degree angles in these coordinates, and
01:45:23
so light cannot escape out here. It cannot escape out to here. All it can do is eventually hit the singularity, and anything that's moving slower than the speed of light will also hit the singularity. So anybody who finds himself in this quadrant in here is
01:45:44
doomed. Somebody out here has the possibility of escaping. Well, let's talk about light rays. If a light ray is moving radially outward here, it will escape. It will simply just keep going.
01:46:02
If it's pointing radially inward, it of course will, it's also doomed, it will hit the singularity, and there's everything in between. If the light ray is pointing outward, it may fall in, depending on whether it's beyond the photon radius or not.
01:46:21
But this is the picture that you want to get used to. If you want to understand and you want to be able to resolve funny paradoxes about who sees what in a black hole, go back to this picture. Remember that Alice's time on the outside, Alice who is far away, or moving with uniform acceleration, her clocks
01:46:45
record this hyperbolic angle as time. Bob's clocks record proper time, and it's that mismatch, it's that mismatch which is governing all of the peculiarities of motion near a black hole.
01:47:04
Okay, I think we are, yeah. So, with black hole forms, supernova or whatever it is, so what happens to all the matter from our perspective outside the black hole? I mean, is some of it trapped inside and some of it on the
01:47:22
surface? Let's talk about what Alice sees who stays outside the black hole. Alice, and we are speaking about classical black holes, we're not doing quantum mechanics, is it classical? Alice sees all of the matter that forms a black hole, not
01:47:41
falling into the horizon, but sort of a kind of sedimentary motion with everything just asymptotically getting closer and closer to the horizon, closer and closer and closer, flattening out and forming a kind of a sedimentary structure, a layered structure, which just asymptotically
01:48:04
migrates closer and closer to the horizon. And this is the matter that formed the black hole? That's the matter that formed the black hole, yeah. All the information is now smeared on the radius, so does it
01:48:22
mean that a new person, James, comes along much later, that person can now view the history of everything that's ever happened in terms of objects falling into the black Well, okay. So, how does James, James, is that him?
01:48:41
I'm not sure what happened to Charlie. Can we do Charlie? Charlie is okay? Okay. Charlie comes late and what does he see? All right. So, the first statement is to see anything, light has to
01:49:01
come from it. If Charlie can see Bob, it's because Bob is emitting light. Now, Charlie sees Bob's clocks slow down. As far as he's concerned, all of Bob's clocks are moving
01:49:24
slower and slower and slower, but that also means that Bob's light sources are emitting, let's suppose for the moment that Bob is sending out pulsed signals, one signal per microsecond.
01:49:42
And what Charlie sees is that first he sees signals coming out roughly at one per microsecond and a little bit later it's one per second. And somewhat later it's one every ten seconds. They get stretched out more and more and more. Why is that?
01:50:01
It is, again, that ratio of factors there. And as Bob gets closer and closer to the horizon, there's a bigger and bigger stretch between neighboring pulses. If we translate that into frequency of light, it's saying
01:50:21
that the frequency of light emitted by Bob. Bob thinks he has a flashlight that sends light at a particular frequency. He thinks his frequency is whatever is ten to the blah, blah, blah, cycles per second, but the light is
01:50:42
redshifted as far as Alice is concerned or as far as Charlie is concerned, the light is getting more and more progressively redshifted. So what Bob sends out as conventional optical light will
01:51:01
arrive at either Alice or Charlie, first as conventional light, then as infrared, then as microwaves, then as radio waves, then longer and longer radio waves, and of course, a given number of photons and radio waves is a
01:51:21
weaker and weaker signal, less and less energy. So the fact is that Charlie, after a long, long period of time, simply will not in any practical way see Bob. He'll see Bob's light having been shifted into the very, very, very far radio, which becomes harder and harder to
01:51:43
detect. So the answer is that Charlie will see nothing. But if he's capable of keeping track of these very, very faint, long, long wavelength radio waves, then he will just see Bob slow down.
01:52:04
When you say something becomes a time, like that the singularity becomes a time, you've studied black hole signals, how do you interpret that? No more or no less than this picture. No more or no less than this picture. What it means is that every trajectory in here that is
01:52:24
moving with slower than the speed of light will hit it, will get terminated at some point. That's what it means. That's all it means is that any trajectory starting from here cannot avoid it and will eventually hit it in some
01:52:43
specified amount of time along the trajectory. That's all. Well, let me say one other thing. There's a technical thing. It says that the surface is space-like. It says that the surface is space-like, that all
01:53:02
directions within the surface, within the surface, all those directions are space-like and there is no time-like direction within that surface.
01:53:32
As a massive object moves toward the horizon, the horizon
01:53:42
comes out to meet it. For that you have to solve Einstein's equations, but in fact that's what happens. The horizon comes out to meet it, meets it at some point, it blends in with the horizon and then this bulge in the
01:54:05
horizon spreads over the whole thing and increases the size of it, but still with the points of this thing coming in here always being on the outside. Suppose we have this huge mass in the middle of an
01:54:23
extremely large dust cloud and the dust is slowly collapsing in on the mass. The mass isn't big enough to form a black hole yet, but at some point it would reach that mass and begin to collapse. So it seems like from our perspective there is definitely
01:54:40
mass inside the black hole going all the way to the center, right? Somebody watching it from the outside will not see any of the mass pass through the horizon. Maybe it was there before the horizon formed is what I'm saying. There's no Schwarzschild radius when it starts. When it starts, yeah. Right. So we've got this big mass in the middle and all this
01:55:02
dust and finally enough mass accumulates to undergo the gravitational collapse and you get a Schwarzschild radius out there. But we've seen, we know the mass is at the center, so how do we now all of a sudden see it on the surface? We're going to do the problem of creating a black hole
01:55:21
by an in-falling shell of material and we're going to see exactly what happens with the horizon. Let me not try to get ahead of it right now. I think next time we will work out the picture that goes with an in-falling shell of matter, no black
01:55:41
hole to start with. When the in-falling shell passes its Schwarzschild radius, there will be a black hole there. The horizon actually begins before that. The horizon begins at the center, grows out to meet the in-falling shell. What happens, you know, all right, so here you have R
01:56:09
equals zero with no black hole yet, okay? Somebody throws in an in-falling shell of matter, it would come in, it would look like this. Here's the, can you see the shell of matter falling in?
01:56:26
Okay, the shell of matter is falling in, it gets to R equals zero right over there. But before it gets to the origin, at some earlier time, a
01:56:40
horizon starts growing from here and that horizon looks something like that. No light can ever escape from inside the horizon to outside because to do so it would have to go faster than the speed of light. So that means that Alice looking back will see a light ray
01:57:04
coming from here, from here, from here, from here, and if she tracks it all the ways back, it will come from here. She will never see, she will never see this shell passing through the horizon over here. She'll look back and she'll see the shell as she gets
01:57:21
later and later, she'll look back and she'll see that shell getting closer and closer to the horizon but never passing it. Well, we're going to do that problem, we're going to do that problem in detail. Fortunately, we don't have to do any equations, just pictures for that.
01:57:45
Becoming from where? 2GM and 3GM, yes.
01:58:03
Who has to be there? The source of the light. Yes. Well, if we see the source of the light in there, yes. If we see a source of light between 2GM and 3GM, we see a
01:58:22
source of light. No, we can't, can we? Didn't you say light can't escape if it's inside 3GM? No, it can escape if it's moving radially outward. No. Here's 2GM, that's the horizon, here's 3GM. If we discover a light ray over here that happens to be
01:58:44
moving, let's say, in that direction, we can be sure that it's going to fall in. If we see a light ray in that direction, no problem. That light ray will escape. If we see a light ray in that direction, we know it
01:59:03
won't escape. So at any given distance, you can divide the different directions into those which fall in and those which don't fall in. If you're at this point here and you consider a
01:59:20
tangential light ray, in other words, one moving with fixed radial distance, that one will fall in. Not so if it's out beyond the photon sphere. If it's out beyond the photon sphere, it'll move out. If it's inside the photon sphere, it'll fall in.
01:59:41
But as long as you're outside 2GM, then a radially outgoing light ray will get out. Okay? Yeah. Does that also mean that any light ray which comes out through that photon sphere has to originate in between
02:00:00
between that photon sphere and the horizon, and can't it come from outside the photon sphere? Can light cut through the photon sphere? No. No, I see what you're asking. Okay, so let's, I think you're asking, I think,
02:00:21
here's the horizon, here's the photon sphere, and I think you're asking whether a light ray can go in and come back out? No. Because if it went in and came back out, there would be some point at which it was moving tangentially, and once it's moving tangentially, it will fall in. So, no, a light ray cannot come in and go out of the
02:00:45
photon sphere. It's another property of the photon sphere. But, if, on the other hand, if somebody, it doesn't happen to be a happy person who has fallen through, his
02:01:02
motion is such that he knows that he's doomed, he can still send out a light ray radially outward that will get out. Not so when he passes the horizon. Once he passes the horizon, no way to communicate with the outside. The only limitation is that he will have to send the
02:01:23
light ray pretty close to radially out, or at least not too shallow an angle here. Yeah. If Bob and Alice were suspended, just outside, could they communicate? If what? If Bob and Alice were suspended, just outside, could
02:01:43
they communicate with each other? Yeah. Even inside the light? Even inside the light? Ah, ah, ah, oh, that's an interesting question. Okay. Yeah, I think if they're a little bit, just a little bit
02:02:00
inside, and they're suspended, as you say, suspended, just a little bit inside, a light ray which is just a little bit inside and which is moving in a tangential direction will go around many times before it falls in. So, yes, they can communicate. Now, okay, so there's an interesting question to which
02:02:25
I'm not exactly sure what the answer is. Supposing they were in diametrically opposed positions at some R, what are the limits on whether they can communicate?
02:02:40
I suspect there must be a distance beyond which they can't communicate. I'll work that out. It's not hard to work out, but I don't know the answer offhand. But as long as they're sufficiently close to this point here, yeah, they can, they can. But even in that case, they can always send out radially
02:03:02
out, which would then turn in and come back, right? Yeah, that's right. That's correct. They could have Charlie out here. Alice can send the message to Charlie. Charlie can take that message, move around to here perfectly nicely in orbit, and then send the message into Charlie. So they can communicate.
02:03:20
Even in the fear, they could actually just get an angle where it balances out. Yeah, yeah. They can still commit limitation on them communicating. The range goes down. It's only one special angle that can do it. Yeah, as they move closer and closer to the horizon,
02:03:42
yeah, I think they can still communicate. The way they communicate is Alice can send out almost radially outward, almost directly radially outward, and it doesn't even have to be a light ray. It could be, you know, a rock. She throws out a rock with enough velocity to almost escape, and it comes back and whack.
02:04:03
Yeah, so they can communicate. Okay, good. Till next week. For more, please visit us at stanford.edu.
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