1

Lecture 11 Seeing Near Black Holes

Ref. , “Exploring Black Holes” Chapter 5, Project B

I Various Measures of the Speed of Light in Curved Spacetime

We could now go through and discuss exactly how massive objects move about in the Schwarzschildgeometry. This is the content of chapter 4 of “Exploring Black Holes”. In particular, we are rather close tobeing able to derive the general relativistic prediction of the advance of the perihelion of Mercury (Project Cof “Exploring Black Holes”). However we are going to skip such discussions and proceed to the discussionof the motion of light in Schwarzschild geometry.

To begin, we will examine the motion of light from different perspectives (reference frames).

A) Free floating observer.B) Far away observer (bookkeeper in the text).C) Shell observer, (kept at a fixed distance from the black hole).

A Free floating frame

Since the free float observer is in a (local) IRF, the speed of light will be observed to be c withinthis frame. (For measurements that span great distances, a different value may result).

B Far away frame

We have already discussed this case previously but the result is easily obtained from the metric.Restricting to the spatial two-plane (θ=π/2), we have,

22 2 2 2S

S

R dr0 1 c dt r dRr 1r

= − − − ϕ −

. I.1

There are two cases to consider, a light ray moving radially and a light ray moving tangentially (maintainingthe same distance form the origin).

-Light ray moving radially (dφ = 0),22

2 2 2 2 2S S

S

S

R Rdr0 1 c dt 0 dr 1 c dtRr r1r

Rdr c 1 , light moving radially, far away observer.dt r

= − − − → = − −

= ± −

-Light ray moving tangentially (dr = 0)2 2 2 2 2 2 2 2S S

S

R R0 1 c dt 0 r d r d 1 c dtr r

Rdr c 1 , light moving tangentially, far away observer.dt r

= − − − ϕ → ϕ = −

ϕ= ± −

Notice that the far away observer measures that the speed of light slows as it approaches thehorizon. Is this a problem? (light moving less than c?). NO, for the far away observer never directlymeasures the light. The measurements are performed close to Rs and relayed to the far away observer whoadjusts for her own coordinates. Also notice that the radial and tangential velocities are different.

C Shell frame

2

When the shell observer measures the speed of a passing light ray in a small nearby region, thelocal metric is the same as for flat spacetime and always measures it to be c. However if a measurement farfrom the shell observer were to be made, a value different from c would result.

In general, whenever a measurement is made nearby, in a small region of an observer, the speed oflight is c. If extended measurements are made (such as the far away observer ‘measuring’ a ray near a blackhole) then c may seem to be different than c.

II Fish in Lake Analogy

To try to understand how these different views of the motion of light can be consistent, we return tothe fish in the lake analogy. To explore the different perspectives we will introduce a little more structure.

Consider a large sea, which has a sinkhole in the center. Water flows in from a very large distancefrom the hole. This sea is populated with a large number of fish, which always swim at a very fast rate. Atthe edge of the sea is a cannery, which launches a large number of fishing boats. These boats spacethemselves out about the sinkhole to catch the fish. The boats must use their engines to maintain a setposition with respect to the hole (or dock). The closer to the hole, the more power is needed. There is a pointat which no boat can maintain their position and if they approach any closer they will wind up at PointSingularity, crashing on the rocks. The fish always swim as fast as the fastest boat. They too can notapproach any closer to the hole than the horizon and be free, any closer and they crash onto Pt. Singularity.The manager of the cannery sits on the dock and tries to keep tabs of his fleet. There are no radios and it is afoggy day (the manager can not see any of the boats). In order to communicate with his fleet special“homing fish” are used. These well-trained fish are kept on the boats and, when released, always swim back

to the dock.

To send supplies to the boats, the manager drops, from rest, a buoy from the dock. As the buoypasses each boat supplies are removed and a homing fish is released to notify the manager that the drop wassuccessful. Since it is foggy out, the only way to tell about the progress of the buoy is via the homing fish.

3

What do the various observers witness?Dock observer: Near the dock the manager observes the fish to arrive at regular intervals since the

sea is calm nearby. Each fish which is released closer to the sinkhole must fight against the current whichgrows stronger as on nears. Thus, if the boats were regularly spaced, the arrival of fish at the dock wouldtake longer. For the fish released nearest the horizon the current is so strong that it would take many years toreach the dock. And if a fish were released right at the warning sign, it would take an infinite amount of timeto reach the dock. Since the fish arrival indicates when the fish were released, the manager concludes thatthe buoy initially speeds up as it floats down but then slows to a stop as it nears the horizon. Also, knowinghow fast the swim fish nearby, the dock observer would calculate that the fish swim slower near thesinkhole.

Boat Observers: The buoy is viewed to pass faster and faster the closer the boat is to the horizon.Each boat only observes it to pass its boat. Once past, the boat observer could view the homing fish pass ontheir way to the dock. Each boat would observe the time between each fish passing to grow in time. Thus theboat observer would see the buoy pass by and then slow down as it nears the horizon. Once it is past theboat, the view is very much like that of the dock observer. Note that close to the horizon the buoy would betraveling very fast but then immediately slows to a crawl and stops at the horizon. In fact right at the horizonthe buoy is traveling at the speed of the fish!

Buoy Observer: For a person riding on the buoy the sea appears calm since it is floating with thecurrent. Thus in the local vicinity of the buoy all fish appear to swim at the fastest rate.

II.2 Trajectories of Light Near a Black Hole

We can extend this analogy to attempt to explain the nature of null geodesics (light rays) near ablack hole. We do this now and later will go through the quantitative analysis.

Examine a boat a short distance from the warningsign (horizon). If this captain releases fish in all directions it isclear that many of them will fall into the sinkhole, exactly howmany depends upon how far from the sign and at what anglethe fish swim. A boat very close to the sign would only have asmall fraction make it out to the dock, and , of course, at thesign no fish will make it out. There will be one boat for whichfish released perpendicular to the radial line from the hole tothe boat (in the tangential direction) will just make it out (infact they will end up back at the boat). Any fish released at anangle closer to the sinkhole will end up in the hole. We callthis the “critical boat” for reasons that will be clear later.

What we would like to do is to derive arelation that predicts how far from the hole the criticalboat sits and for what angles fish will end up in thehole. Carrying this back to the black hole, the criticalboat corresponds to the radius at which circular orbitsof light are allowed.

Once past the sign post all fish and boatsproceed to Point Singularity. They can not climb upthe torrential flow of water. All objects eventually endup on the Point. Also, no signal can be sent to theboats above the signpost, since no fish can travel thatfast. Signals can be sent to a boat within but not out. It

is not possible to have a stationary boat within the horizon so it is not reasonable to talk about suchobservers within (there are no shell observers within the horizon).

4

II.3 Spinning Sinkhole – Spinning Black Holes.

Before moving on to the quantitative analysis weextend this analogy even further to a case which we can notgo into fully in this course. This is the case of the spinningblack hole. The analogy is amended now to have the seaspiral in as it flows inward. In this case it may seem obviousthat the light cone (angles at which fish can just make it out)tip in the direction of the spiral. Once beyond the signpostfish (and boats) must spiral in, they can not follow a straightline to Point Singularity. This is true for the black holescenario, once within the horizon, motion must be inward butspiraling down. A direct geodesic does not exist.

We can get a little more mileage out of this analogy by examining boats near the horizon. For aspinning hole it turns out the signpost is located closer to Point Singularity than for the non-spinning case.However there is another interesting point. If the spiral is strong enough (has enough angular momentum)there will be a point before reaching the signpost at which it is no longer to have a boat at rest. The boatmust spiral but is not committed to falling into the hole. This is the point where the outward lightcone(fishcone?) tips over so much that a stationary boat is not allowable. This point is called the static limit inblack hole terms – it is the point beyond which static observers can not exist. The region between the staticlimit and the signpost (horizon) is called the ergosphere. It is still possible to escape to infinity, but not bytraveling only in the radial direction. Returning to the black hole picture, a spinning black hole is notspherically symmetric. The picture discussed above would translate to the two spatial plane perpendicular tothe axis of rotation. For other angles the static limit is smaller than at the equator. And at the pole (axis ofrotation) the static limit meets the horizon.

III Trajectories of Light

We now examine quantitatively the behavior of light near a black hole. We return to our expressionderived in a previous lecture relating the energy and angular momentum for an object moving freely near ablack hole.

2 r 2 22 2

2s S

E (p ) Lm cR R r(1 ) (1 )r r

= − −− −

The case we want to discuss now is for a light ray traveling in a circular orbit about the black hole. At whatradius is this allowed? Examining our qualitative analysis, it appears there should be some point at whichthis can occur. To find where this point lies, we set m=0 and dr =0 in the above expression. (Setting dr = 0 isequivalent to setting pr = 0, there is no momentum in the r-direction). Thus, we begin with,

2 2

2s

E L0R r(1 )r

= −−

. III.1

This expression gives us a relation between the orbit radius, the energy, and the angular momentum. Sincethe energy and the angular momentum are constants of the motion for an object freely moving, we can solvethe expression in terms of L/E, which is itself a constant of the motion. (The expression L/E for light is equalto the impact parameter b, the perpendicular distance of minimum approach for an initial trajectory atinfinity). Expressing the above as an equation in r, we have,

22 s

2

2 23 2 2

S2 2

R L0 E (1 )r r

L Lr c r c R 0E E

= − −

− + =

5

This expression gives us a relation between r0 (the radius of the circular orbit of light) and L/E. The ratio /E= b has the dimensions of a radius, thus we express it in terms of the natural scale factor, RS, (L/E) =b = yRS.To scale the radius, we define r = xRS. To get a useful quantitative form we set the following,

22

S s 0 S2 2

2GM LR , yR , r xRc E

= = =

.

Substituting this into the expression above we have the following equation (in terms of the dimensionlessparameters x and y),

3 3 3 3 3S S S

3

3

x R xyR yR 0 x xy y 0

x (1 x)y 0xy

x 1

− + = → − + =

+ − =

=−

This last expression gives us a relation between r0 and L/E.2 3

2S 0 S

L x R , r xRE x 1

= = − In fact this expression tells us the distance of minimum approach, r0, given an impact parameter b = L/E.However this expression is not limited to the circular orbit of light we are interested in. This is a generalequation of an object free falling in (or orbiting) a black hole.

Our goal now is to find an even more specific expression for the radius of a circular orbit of light. Ifr = r0 then the expression (III.1) is satisfied. If r differs ever so slightly from r0 then (III.1) is no longersatisfied. In order to satisfy the general expression,

2 r 2 2

2s S

E (p ) L0R R r(1 ) (1 )r r

= − −− −

III.2

pr must be different than zero, meaning that dr is different than zero and the light ray has a radial component.For a circular orbit there can be no radial component. Thus, if we change r by an infinitesimal amount, ε, weexpect the expression to still be satisfied to lowest order in ε. (For those of you familiar with calculus we aresimply taking the derivative of the expression with respect to r to find the extrema). Thus we consider thesolution discussed above, r0, and an infinitesimal deviation from it ε, r = r0 + ε. We insert this into theequation and ignore terms of ε2 and higher (since these are very very small). Following this we have,

23 2 2 20 0 S 0 2

3 2 2 3 2 2S 0 0 S

2 2 2 2 20 0 0 0 S

23 2 2 2 2 3 2 2 20 0 0 0 0 0 S

Lr b r b R 0 , r = r , bE

r b r b R (r ) b (r ) b R 0

(r )(r 2 r ) b r b b R 0

(r 2 r r ) (r 2 r ) b r b b R 0

− + = + ε ≡

− + → + ε − + ε + =

+ ε + ε + ε − − ε + =

+ ε + ε + ε + ε + ε + − − ε + =

In this last expression we ignore any terms having ε2 or higher since these are very small compared to the ε1

terms. This gives,23 2 2 2 2

0 0 0 0 S2 2 3 2 20 0 0 S

r 2 r r b r b b R 0

(3r b ) (r b r b R ) 0

+ ε + ε − − ε + =

ε − + − + =Notice that the last term in parenthesis is simply the original expression, which automatically vanishes. Thuswe are led to the expression,

2 20 0

b3r b or r3

= = .

This expression, coupled with the previous, gives us the radius at which circular orbits of light will occur.

6

3S

0 S S

3 32

Rb L 1 1 xr xR yRE x 13 3 3 3

1 x xx 3x 3(x-1)=x 2x=3x 1 x 13

3x=2

= = = = =−

→ = → = → →− −

Returning to our original variables we have,

0 s 2

3 3GMr R2 c

= = .

We have reached the result, at a radius of 3GM/c2 there can exist circular orbits of light. In addition, if we

find the value of S27b R2

= this tells us at what initial impact parameter the resulting light ray will orbit

the black hole. For impact parameters smaller than this value, the light ray terminates on the horizon. Seefigure (III 1). For impact parameters that are greater, the light rays may extend out to infinite distance.Figure (III 2): Consider a series of light sources which fire laser beams in tangential orbits near r0. For thelaser beam that is emitted exactly at r0, the beam will travel in a circle. (Colors in the image are merely todistinguish the beams).

FIGURE III 1 FIGURE III 2

This derivation of the radius of circularly orbiting light is much easier to obtain using calculus.From the expression (III.1) of the energy of light rays we take the derivative with respect to r and set it equalto zero to obtain the desired result. For those familiar with calculus this process is nothing but finding theextrema of a plot of the energy versus position. If we re-express equation III.1 and plot E2 as a function of rwe see a maximum at the radius r0 = 3GM/c2. For values of r different than r0 note that pr (or dr/dt) needs tobe nonzero in order to satisfy equation III.2 above – the light is changing radius.

IV Shell Observer’s Observations near the Black HoleWhat does a shell observer see when they look up at the sky when they are near a black hole?

Examining the fish analogy we see that for some angle the fish all end up in the horizon. Reversing this tellsus at which angles the fish emanating near the horizon reach the observer. Thus close to the horizon, the

7

image of the horizon will dominate the sky. The outward view of the sky will get compressed towards thezenith. The figure below shows light rays emanating from a shell.

Depending on where the shell is, one ray (shown in green in the figure) will be at the critical impactparameter, bcrit = 27

s2 R , and will just undergo a circular orbit. Any rays aimed more towards the horizon willfall in and rays aimed upwards tend to escape out to infinity. The angle θshell critical (not to be confused withthe spherical coordinate θ) measures the portion of the sky for which the horizon is viewed. We can do thissince an outgoing ray’s trajectory is the same as an ingoing one along the same path. Our goal now is toderive a relation for the angle θ shell critical, which will tell us what the sky looks like to a shell observer.

Care must be taken when working with the definitions of energy and angular momentum we haveused up to this point. These were expressed in terms of the proper time dτ. However, for a photon dτ = 0 soour definitions may become ill defined. To get around this we convert any necessary expressions involvingdτ to dt (the far away time) and then to dtshell (the time coordinate for the shell observer).

1/ 2s

1/ 2s

shell

proper time: d (= 0 for light)

Rfar away time: dt = 1- d

r

Rshell time: dt 1- dt

r

−

τ

τ

=

Consider a light ray passing a shell observer (at reduced radius r) at an angle θshell to the outwarddirection (see figure). We want to find the components of the light’svelocity vr (radial speed) and vt (the tangential speed) and want toexpress these in terms of L, E, and later b.

shellr t

shell shell

dr dv , v rdt dt

ϕ= = .

(vt has the factor of r because rdφ is the infinitesimal distance in the tangential direction).We begin with our general definitions of energy, angular momentum, and the metric (I.1).

8

1/ 22 s

2

12 2 2 2 2s s

R dtE mc 1r d

dL mrdR R0 1 c dt dr 1 r dr r

−

= − τ ϕ

=τ

= − − − − ϕ

Tangential speed vt:1/ 2 1/ 2

s st

shell shell

R Rd d dt d d dv r r 1 r 1 rdt dt dt r dt r d dt

ϕ ϕ ϕ ϕ τ = = = − = − τ In the last expression we write the second parenthesis in terms of L and the last parenthesis in terms of E.Our final expression must be free of m and τ.

1/ 2 1/ 2 1/ 22s s s

t

2s s

t

R R Rd d L mcv 1 r 1 1r d dt r mr E r

R RLc b v 1 c 1r rE r r

ϕ τ = − = − − τ

= − = −

Radial speed vr:

We begin with equation (I.1) and convert it to shell coordinates.2 2 2

2 2 2 2 2 2S S2 2

S S

2 2 2 2 2shell shell2 2S shell

2 2 2 2 2shell shellSshell

2 2S Sr

R Rdr dr 1 d0 1 c dt r d 1 c rR Rr r dt dt1 1r r

R dr dr dt 1 d dt0 1 c rRr dt dr dt dt dt1r

R R0 1 c v 1

r r

ϕ = − − − ϕ = − − − − −

ϕ = − − − −

= − − −

2 2 2 2St r t

Rv 1 v c v

r − − → = −

We have inserted the definition of vr and vt and converted dr and dt to drshell and dtshell. We see from thisexpression that the magnitude of the speed of light is equal to c, (i.e. 2 2 2

r tv v c+ = ).

So we now have a relation for the angle θshell,t s

shellv Rbsin 1c r r

θ = = −

.

The question we want to ask is what angle does the horizon subtend for shell observers near the black hole(what portion of the sky is filled with the image of the horizon). From our discussion above we see that thehorizon dominates more and more of the sky the closer the shell is to the horizon. The boundary betweenhorizon and sky is given by the trajectory that just orbits the black hole. This is given by the ray with the

critical impact parameter found before, crit s27b R2

= . Inserting this into the expression above gives us the

angle where the edge of the horizon is observed,

crit s s sshell critical

b R R R27sin 1 1r r 2 r r

θ = − = −

.

For various shell observers we have the following views,

Note that at r=3GM/c2 we get our previous result that light circles the black hole at this distance.

9

IV.1 What does the radially free falling observer see?To answer this question we need to relate the two reference frames. When the two frames are at the

same location we simply use special relativity to relate the observations (free falling is an IRF and so is theshell frame within a small neighborhood). If the free falling observer observes the shell to pass at speed vshell(this is equivalent to what the shell observer observes the free falling frame’s speed to be), we can use theaddition of velocity formula to find the speed of the light ray’s componentsas seen by the free falling observer (primed coordinates).

r shellr

r shell2

shell shellrshell

rfreefalling

r shell shell shell2

v vv ' v v1

cv vv cosv' c c ccos

v v cos vc 1 1cc

+=

+

+ θ += θ = =

θ+ +

We plot the portion of the sky taken up for the free falling observer whostarts from rest at an infinite distance away. To find the values recall that the speed of a free falling object,dropped from rest, as viewed by the shell observer is (previous lecture).

sshell

sshell

freefalls

shell

Rv cr

RcosrcosR1 cosr

=

θ +θ =

+ θ

By inserting the value of the critical angles found for the shell observer we can compare both referenceframe’s view at selected values of r. We can also plot other angles (colored dots) to see how the skybecomes deformed for the free falling observer.

Inside of the horizon we can assume that the relation for the shell angle still holds,1/ 22

srshell 2

Rv bcos 1 1c r r

θ = − = − − −

.

Insert this into the expression along with the shell speed to get,

10

( )2

1/ 2s s 1/ 23 22s s

freefall 1/ 221/ 2s s 3 / 2 2 2s

2 s

R Rb[1 1 ] r r R b r Rr rrcosR Rb R1 [1 1 ] r R r 1 br rr r

− − − + − − − + θ = = − − − − − −

where in the last expression we multiplied top and bottom by r3/2. This allows us to take the r 0 limit tofind what the free falling observer sees right before hitting the singularity.

( ) 1/ 23 2s s s

freefall 1/ 2r 0ss3 / 2 2 2s

ss

r r R b r R b R rcos limRRR b R 1r R r 1 b rr

→

− − − + − θ = ≈ ≈ − −− − −

.

Thus the observer sees all light rays come in from cosθfreefall = 0 or θfreefall = 90o. Inside the horizon the freefalling observer sees a black sky in front (corresponding to the singularity ahead), a black sky overhead, anda bright ring of light perpendicular to the direction of travel.

To the infalling observer the whole outward sky gets compressed into a ring near the boundary ofthe horizon. The sky upwards becomes dark. A bright ring perpendicular to the direction of travel isobserved as this observer passes the horizon and down to the singularity. To summarize we have thefollowing plot of what shell and free falling observers see.

11

Appendix A: Graphical Derivation of Circular Orbit

We begin with the energy equation for light moving in the Schwarzschild geometry.In this equation the constants of the motion are E (the energy) and L (the angular momentum) and R is theSchwarzschild radius.

= 0 − E2

− 1Rs

r

L2

r2

We want to plot this expression as a function of r to find possible radii of circular orbits. To do this, we re-express this as a plot of energy. Or more accurately, a plot of energy per unit angular momentum squared vs.radius. Thus rewrite the radius r terms of Rs , the Schwarzschild radius,

r = x Rs

In this expression x is a dimensionless number. Similarly, we rewrite the quantity L/E (which is equal to theimpact parameter, b, - a distance) in terms of Rs as follows,

L/E = y Rs

Rewriting these in this manner allows us to express the formula above in terms of dimensionless variableswhich we can plot easily.

= E2

L2

− 1Rs

rr2

Inserting our new expressions, this becomes,

= 1

y2 Rs2

− 11x

x2 Rs2

or,

= 1y2

− 1 1x

x2

12

Let's examine what this plot is telling us. Consider a light ray approaching the black hole from a largedistance away. Initially it has some energy (measured at infinity) and some angular momentum about theblack hole (if it is not directed exactly at the hole). We can define the impact parameter, b, as theperpendicular distance the ray would pass the black hole in the absence of curved space from its initialtrajectory. It was shown in lecture that the impact parameter is equal to b = L/E (for light, L/p for massiveparticles). Thus, our approaching ray has some energy per unit angular momentum (inverse b). The plotabove can be viewed as the potential energy graph for light trajectories (where potential energy isnormalized by the angular momentum - a constant). As you should be familiar with your course in mechanics, given a plot of potential energy versusdistance a particle of energy E can never reach regions for which E > PE. This would constitute a violationof energy conservation. Also, points where the energy curve meets the potential energy are points where E =PE or the kinetic energy is zero (it is at rest). With this in mind return to our plot and consider three initial light rays with different values ofE/L=1/b, as shown on the plot,

Ray 1: (E/L)2 = 0.1Ray 2: (E/L)2 = 4/27

Ray 3: (E/L)2 = 0.3

13

Note that ray 1 (with (E/L)2 = 0.1), if approaching from far away, does not have enough energy to penetratecloser than about r = 2.5 Rs . Or, put another way, its impact parameter is too great in order to penetrate to

the horizon. This means that the ray approaches but can not come any closer. The intersection gives theradius of closest approach for this ray. Ray 3 (with 1/b2 = 0.2) has a small enough impact parameter topenetrate to the center of the black hole. It is aimed more directly at the black hole. (It has enough energy toovercome the barrier). Ray 2 is the case where it just reaches the top of the curve. Any more it "bouncesaway" any less and it ends up in the horizon. Exactly at this point the light ray can undergo a circular orbit.(Figures III 1 & 2 above).

Appendix B Review of Energy in Classical Physics

In this appendix we review the classical conservation of energy.

Appendix C Angular Momentum and Impact Parameter