### Create a StudySoup account

#### Be part of our community, it's free to join!

Already have a StudySoup account? Login here

# TOP SURFACESINTERFACES EXP PHYS 591

Virginia Commonwealth University

GPA 3.81

### View Full Document

## 28

## 0

## Popular in Course

## Popular in Physics 2

This 16 page Class Notes was uploaded by Dr. Nia O'Kon on Wednesday October 28, 2015. The Class Notes belongs to PHYS 591 at Virginia Commonwealth University taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/230668/phys-591-virginia-commonwealth-university in Physics 2 at Virginia Commonwealth University.

## Similar to PHYS 591 at Virginia Commonwealth University

## Reviews for TOP SURFACESINTERFACES EXP

### What is Karma?

#### Karma is the currency of StudySoup.

#### You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!

Date Created: 10/28/15

1 Problems with the Schwarzschild Metric The spacetime outside of a nonerotating star or planet or whatever of total mass m is described by the metric tensor d527lt1727mgtdt2 T d7 mm 1 7 T This metric solves the vacuum Einstein equations and according to Birkhoff s Theorem is the only spherically symmetric metric that does It obviously has a problem at T 2m One metric coef cient blows up while another goes to zero Consider a clock that is holding its position at a constant value of the radial coordinate If it does this for an interval At of coordinate time then the time elapsed on the clock will be AT At 1 7 72m V T For T near 2m very little proper time will elapse on the clock even though a great deal of coordinate time elapses This result tells us that the t constant hypersurfaces are failing to advance in time near T 2m During the early days of general relativity the problem with the Schwarze schild solution was regarded mostly as a curiosity of little consequence because for normal astronomical objects such as the sun and the earth the critical value of the radius is extremely small For an object with the mass of the earth the critical radius is about a centimeter For the sun it is a kilometer The metric inside an ordinary star is not given by the Schwarzchild vacuum solution and is quite regular everywhere For the external vacuum spacetime to extend to the critical radius the sun would have to be compressed to a radius of a kiloe meter Until 1939 Oppenheimer and Volkoff Oppenheimer and Snyder no astrophysicist seriously believed that a mass equal to that of the sun could be compressed into an object only a kilometer in radius Not too many believed it after that either From a mathematical point of view the coordinateeindependent nature of spacetime geometry was not well understood during the early days so it was some time before someone considered the possibility that the T 2m singularity was simply the result of bad coordinates One indication that this might be the case is that none of the scalar curvature invariants such as R RPUWRUWquot and invariant ratios of lightlike components of curvatures become in nite at T 2m 2 The Kruskal Extension 21 Orthogonal Surface Metric Once it is realized that the T 2m singularity is really just a problem with the coordinates it is fairly easy to x The angle coordinates 950 are xed by the spherical symmetry group so nothing could be wrong with them That leaves the coordinates Tt and the metric 2 2 2 71 2 27 ds7fdt f dr f177 on the timelike two7surfaces orthogonal to the group orbits We need to nd new coordinates in which this metric looks more regular An obvious geometrical feature of this two dimensional metric is the light cone On any such surface it is always possible to nd coordinates U V such that curves at constant U are lightlike and so are curves at constant V For example in two dimensional Minkowski spacetime one can take U tT V t 7 7 and obtain the metric in the form 2d52 7 d In general so long as the metric takes the form 2152 7ltIgtdUdV then an interval with either AU 0 or AV 0 will be lightlike Thus we seek coordinates U V suc that 710sz f ldTZ 7ltIgtdUdV for some function CI Once we nd such coordinates we can construct space and time coordinates u 7 v U in which the light cones look exactly like the ones in Minkowski spacetime If there is any coordinate system in which this metric tensor is regular then this has to be the one 22 Conditions on Advanced and Retarded Time Coordi nates Represent partial derivatives with respect to tr by subscripts so that dU UTdr Utdt dV Wdr Vdt and therefore 710sz f ldTZ 41gt Um Utdt Wdr mt or PUnVnf7 PUTVTf 1 UTVt Uth 0 Divide all of these equations by UTVT and obtain the results U Vt 2 Vt Ut if 7 i i 0 UT m f m UT which can be solved for the two ratios in the form U V E 7 f7 VT 7 if Thus if the coordinates exist then they must satisfy these two conditions Cone versely the argument can be reversed to show that solving these two conditions is enough to put the spacetime metric into the desired form where the conformal factor CI can be found from the equation ltIgtUtVt f In more explicit form the conditions to be solved are 8U7 2mgt8U aviilt 2mgtal 5 E E 7 8T T 23 Solving the Conditions Tortoise Coordinate To see what to do next recall what these conditions would look like for null coordinates in Minkowski spacetime 8U 7 8U 8V 7 8V 8t 7 8r 8t 7 8r which would yield the result that U depends only on t T and V depends only on t 7 T To get the conditions into this form we need a new radial coordinate Tquot suchthat 1 2m 8 7 8 T 8T 7 BTquot From the chain rule for partial derivatives 878V 8 E73 sowehave 1 2m 8Tquot 8 7 8 7 araw aw or 8V7 1 arii m which can be rewritten as follows BTquot T 2m 7771 7 8T T72m T72m and integrated directly T 2mlan 72ml This coordinate has been called the tortoise coordinate after Zeno s paradox and the way that the coordinate creeps up on the value T m which is achieved only as Tquot 7gt 700 For large values of T the new radius coordinate becomes quite similar to T In terms of the tortoise coordinate we have the conditions 83 av 81 8V 8 aw a aw which simply mean UUtT VVt7T We still have two functions U V of a single variable to choose One potential dif culty needs to be anticipated at this point The tortoise coordinate Tquot is not continuous across the boundary at T 2m When we get the nal form of the metric we need to be sure that it is regular there 24 Regularizing the Metric Now all we have to do is nd the conformal factor CI and see if there is any way to choose the functions U V so that the singularity at T 2m goes away Use primes to denote derivatives of functions of a single variable and write the condition that is to be solved for CI CPUnVnf U7 U037U0 tiat T 7 T U V 12 T T For this to work the conformal factor must not vanish at T 2m or we will still have a singularity there Thus the functions U V must be chosen so that they produce a factor of T 7 2m This can be arranged because U U i T UtT2mlan 72ml VVt7T Vt7T72mlan72ml depend explicitly on T 7 2m To pull out the overall factor of T 7 2m that we need choose the functions to be exponentials Uac AeJr Vac Be The constants A and B will be chosen later The condition to be solved for the conformal factor CI is then tw 1 kw 7 T 72m 1 7ltIgtABie 4m 76 4m 4m 4m T or AB 2 L T 7 m 71677151 f or AB 2 T1mlan71ml T 7 m 7 I m 7 167712 e 1 T 71652 e 1nh72ml T 12m or 2 iqDABlT 72ml 16m 67 T 7 m T Now we see why the constants A and B have been carried along They can each be either 1 or 71 but their de nitions rnust switch as the T m boundary is crossed so that 7AB lT 7 2771 T 7 2m which yields the nal result 16m2 L I 5 1 T and puts the orbit7orthogonal 27metric into the form 2 15 7 T e dUdV which no longer has any trace of irregularity at T 2m It remains singular at T 07 however 25 Understanding Kruskal s Coordinates The lightlike Kruskal coordinates U and V are related to the usual Schwarzschild coordinates y U Ae w V Be so that UV ABeEL ABe U mlan Zml ABe r 7 ml or UV 7e T 7 2m and A 5 Ae EN 01 All of these relationships have been derived for the region of spacetime where the Schwarzchild coordinates make sense Within this region we have U gt 0 V lt 0 with the T 2m singularity happening at U V 0 For the de nitions of U and V to work in this region we need A LB 71 Since T gt 2m in this region we con rm that the relation 7AB lT 7 27ml T 7 2m works there Now switch to the space and time coordinates de ned by U v u V v 7 u for which the spacetime metric takes the form 2152 16er 711122 du2 In these coordinates the surfaces of constant T are given by v2 7u2 7e T 72m which implicitly de nes T uv Similarly the constant t surfaces are given by v7u A 7ezm vu B The region within which we have derived the new coordinate system cor7 responds to a wedge of the full uv plane vgtu vgt7u or equivalently u gt lvl V The boundary of this region at u lvl corresponds to T 2m Within this region surfaces of constant T gt 2m correspond to hyperbolas Surfaces of constant time correspond to straight lines through the origin with t 700 at u vandtooatuv V x9 I K fl lJ quot 4 l t 0 55 Tquot K x HTquot 22 qu U The nature of the coordinate problem at 7 2m can now be seen very clearly The Schwarzchild time coordinate t fails to advance at that point 26 Extending the Solution Now consider the other quadrants of the Kruskal plane In the upper quadrant U gt 0 V gt 0 so that the constants A and B can each be chosen to be 1 The sign of 7 2m is negative so we con rm that AB 7 Zml 7 2m in this region as required Similarly the required signs work out in each of the other quadrants Since nothing goes wrong with the spacetime metric at the 7 2m boundaries of the coordinates that we are using we can simply use the whole solution for all values of u v The only place where something still goes wrong is at 7 0 where the timelike 2 surface metric is still regular but the group orbits have zero area A map of the full solution in Kruskal coordinates is called the Kruskal di agram The region between the two hyperbolas is the maximal extension of the Schwarzschild metric The Schwarzschild metric itself corresponds to the right hand quadrant of the map Singl To use this diagram note that light rays move along 450 lines so that the allowed world lines of particles always move upward at less than 450 to the vertical The diagonal lines in the picture correspond to r 2m Light signals from the right hand quadrant of the picture can reach the outside universe However if an ob ject s world line carries it across the r 2m surface then any light signals that it might emit will hit the singularity and never get out It is important to remember that the Kruskal solution is the maximal ex tension of the vacuum solution Most physical situations will use only part of it as an exterior solution which is matched to an interior solution with matter sources For a normal stable object such as the Earth or the Sun only a part of the right hand quadrant is used and the r 2m surface never appears A col lapsing star uses more of the solution because its surface passes through r 2m and uncovers a bit of the upper quadrant quot n 39 lam If Even in that extreme case however the bottom quadrant and the left hand quadrant do not appear There is no known situation that requires the use of the bottom quadrant of the Kruskal extension Such a situation would be extremely awkward because signals could travel from the initial singularity to the outside universe Because no initial conditions can be defined at the singularity anything at all could come out of it and the entire universe would become completely unpredictable The final singularity in the Kruskal extension does not pose such a problem because it is cloaked by the r 2m event horizon A singularity that is not cloaked by a horizon in this way is called a naked singularity The Cosmic Censorship Conjecture asserts that nature always manages to avoid creating naked singularities 3 Black Holes 31 A little astrophysics Stars sbend nbsb bl blnen llyes lnslng hydxogen be make hellum In smallex stars snnln s am Sun blne lnsbn nennubn Lakes blnne only c blne nenbex blblne Stu Wlnen blne hydxogen lnel ums bub c blne nenuen blne nucleiquot ne bums bnbwnnd nd nnnses blne suxhce blblne Stu be swell up by pexhaps lnnm bl hundred 1 nedglnnb Fbmnxsnnblnelnydne evenlnnlly um bub nd blne xedrglzmb stage wlll be followed by ebbllng wlnme dwaxfsux lneld up by blne lnnb um elenlnens obey n pu clple um ln sun clnen becomes n excluslo my pm blcne sm s cnnbsblnene s n exbnndlng planetary nebnlnquot snnln s clne Hellx nebnl below The newboxn Wm dwufsux nnn be seen c clne nencen blclne expulde cloud Hnndneds bl newboxn wlnlce dwaxfs ne seen ln clne sky Bennnse clne expulde cloud dlsslbnms ln just lew cbbnsnnd yams unene axe nnnny more oldex Wm dwaxfstus clnnl have lost llnen nebnlne sms nnnnln maze masswe llnnn on sun 0 lbmgln nme elnbmce seues ofsuges bnc sull end up n nnl nbllnbse O en clney laws we nnnnln nnnss lb be stable Wm dwufs nd nbllnbse maul clney axe Jnsc lew kllbnnecens ln dl l long I support clne st 1 c gs exlsc All dune eleckons have been abmxbed lnlb nencnens lbmgn mvexse bec denny nd l cluslon pu clple le l ch l ln lds ch sed when clne sh nbllnbses lb snnln nnnllslne ls compauble lb llemne nnnlenx enexgy nelensed by clne sh duxmg lee bl snnln lugs nbnnc ofenexgy nnnses nnnnln bl clne sh lb be blown my ln supexnova exblbsbn Hene ls n mug e E E s E of supexnova chhc banned 1 1587 Evehchhuy11chc he leh enhe em he hexphhahhg cloud of 355W uphdly mum hemeh em he he core 1h some cases the hehcmh em dugs he mhgheue eld chmgh the neaxby hehulh acceleuhng phmelee ha geneuhng enough enexgy e hghc up che whole hehhh The Cu hehhh e h example of such eyeuem The he some may meme eherth 50 e 100 umeeche mes ofouxsun shah supexrma we ems are very n He he che use of em emhhe eheWh here demonstrates when stars of this sort nally run out of reworks the mass left behind can be too much for even a neutron star and the surface area of the star drops below 47T 2m2 The surface of the star is then inside the 7quot 2m limit of the Kruskal metric and all timelike curves lead toward smaller values of 7quot and thus smaller values for the surface area The star is then doomed to nal collapse no matter what physical processes go on inside it 32 Are black holes really out there Brie y Yes An isolated black hole would be impossible to nd since it is only a few miles across and of course it is black because it absorbs all light that hits it However matter that falls into a black hole releases enormous amounts of energy For example matter falling into a solar mass size hole would fall into a gravitational eld identical to that of the sun a distance of a half million miles closer to the center7 of the eld than the surface of the sun with the gravitational eld increasing the whole way Figure out the amount of work a given mass m could do if it is lowered gradually into a black hole at the end of a string The answer turns out to be meg A black hole has the potential to convert mass into energy with 100 ef ciency It is generally believed that the most violent processes in the universe are powered by matter accreting onto black holes When a black hole is consuming matter the matter usually starts out with some angular momentum so instead of going straight into the hole it forms 11 an accretion disk around the hole In such a disk the faster inner ring usually couples to the slower outer ring often through embedded magnetic elds and transfers the angular momentum outward until it is lost in a radial plume of ejected matter The gravitational eld of the black hole is not Newtonian and for matter that comes too close there are no longer any stable orbits As the inner ring of matter is slowed it dives into the hole The energy released by that matter s descent into the hole s gravitational eld is released as X rays Meanwhile the twisting magnetic eld generated by the rotating accretion disk causes some matter to escape along the rotation axis of the system and be focused into jets Such a jet can be seen in some detail in the pictures of the galaxy M87 below M07 Frum 20000 LightYears in 02 Light fear VLAE m yLA T rnrn VLA 00 cm 1lr39LBI 13 cm I 1uquotLBI13 cm Eradlt FIEEEI39 wan HF MEIL d hn Eira a ETEcslj and ml guaa 139quot The hE naJ Radius AEIrEInEimy ma nraiary iaa 130quotin 01 1ha H Hajivzlnal Ecianzaa Frau l39ldElfll39ll Fina label under pa la39li39s agraamani by quot fl l l L rail has Inn The black hole responsible for jets such as the one seen in M87 are thought to have masses that are millions of times the mass of our sun It has long been believed that a giant black hole of this sort exists at the centers of many galaxies including our own 4 Wormholes 41 The EinsteinRosen Bridge Suppose that we somehow have a spacetime that is the full Kruskal extension of the Schwartzchild vacuum solution Consider the space geometry at the instant of time symmetry de ned in Kruskal coordinates by v O Specializing the 12 implicit de nition of the Schwartzchild luminosity radius coordinate T to this case gives 136 T 72m For the dimensionless coordinates u T k 7 xQm 7 2771 we have the relation k2 6 s 7 1 which when plotted shows that the luminosity radius goes through a minimum value of T2m at oru The luminosity radius goes to in nity corresponding to being far from the source of the gravitational eld when u increases to in nity It does exactly the same thing when u decreases to minus in nity Thus we have two outside regions Recall that the luminosity coordinate T is de ned in terms of the area of the spherical group orbits Thus the two outside regions are connected by a sphere of minimum area called the wormhole throat To make a picture of this structure take the dimensionless luminosity radius 5 to be a radius given by s x 12 312 and plot the resulting circles as a function of the reduced Kruskal coordinate k The resulting surface is de ned by W71 Turn this into a parameterized plot For this to work you need to rst de ne a function K by k 6T71 T 1 KT E 7 1 and then form the coordinate vector T cos9Tsin9K T 13 Select ComputelPlot 3Dlrectangular with the cursor anywhere in this vector Click on the resulting picture and then on the box in the lower right corner in order to adjust the parameter ranges set to 7 gt 07 and 0 g 9 lt 27139 The resulting plot is Wormhole Throat If you are viewing this text with Scienti c Notebook or Scienti c Workplace7 you can use your mouse to rotate the picture and view it from different angles double click and drag 4 2 Wormholes An EinsteineRosen Bridge can be thought of as connecting two different unie verses7 each of which looks like at spaceetime far from the bridge lmagine cutting the bridge and looking at each of these two universes To make a pic ture7 de ne the function 1 7T7 7 gt 05 W 716 r g 05 and do a 3D plot of the surface twice 7 cos 97 sin 9 q 7 Because these two universes are so close to at spacetime at great distances7 we only have to bend them a little to t them together and have them both be the same universe with two widely separated holes like this Tcos977 sin97qlt Tcos97327 2sin29gtqlt Tcos93272sin29gt Now we have two more loose edges to take care of7 namely the two throats They tted together before7 so they will surely still match up now Identify the two throats so that anything going down into one comes out the other The result is an argument that Einstein s eld equations should permit a spacetime with a kind of builtein subway 7 that lets you go from one part of the universe to another without traversing the ordinary space and time between 15 43 Interstellar superhighways Brie y No Near the wormhole throat7 the spacetime geometry should be almost exactly the same as for an EinsteineRosen Bridge and will be described by the Kruskal extended vacuum solution For two ships passing through the wormhole7 we would draw a diagram like this The ship that makes it through without hitting the singularity and getting crushed to in nite density is moving on a fasterethan light trajectory The other ship has obviously suffered a failure of its ftl drive system and is in big trouble A fasterethanelight drive is even more problematic than wormholes7 so these things might not be all that useful A worse problem can also be seen in the diagram The initial singularity is uncovered a naked singularity and can send completely unpredictable things into the region of the throat If the initial singularity is covered by the surface of a collapsing star7 then that star will block the throat More complex sorts of wormholes are possible For example7 the gravitae tional eld of a rotating7 electrically charged star can be extended in much the same way as the eld of an uncharged7 nonerotating star In that case7 the singularities may be avoidable but naked singularities are still present and there is no known way to form these things 44 Then What good are they The fact that Einstein s theory admits solutions like the wormhole spacetimes indicates that the underlying topology of spacetime might not be simple It also raises the question of how that topology is determined In Einstein s the ory one proceeds by assuming an underlying manifold with a given topology The wormhole solutions and many other examples suggest that Einstein s eld equations can be solved for any underlying manifold The question then is hat determines the underlying manifo 7 That question has become far more serious now that an evereexpanding variety of underlying manifold structures is being tried in particle and quantume gravity theories7 from string theory 2 dimensions through Metheory with an 11 dimensional low energy limit Some theories have extra dimensions that are 16

### BOOM! Enjoy Your Free Notes!

We've added these Notes to your profile, click here to view them now.

### You're already Subscribed!

Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'

## Why people love StudySoup

#### "Knowing I can count on the Elite Notetaker in my class allows me to focus on what the professor is saying instead of just scribbling notes the whole time and falling behind."

#### "I used the money I made selling my notes & study guides to pay for spring break in Olympia, Washington...which was Sweet!"

#### "There's no way I would have passed my Organic Chemistry class this semester without the notes and study guides I got from StudySoup."

#### "It's a great way for students to improve their educational experience and it seemed like a product that everybody wants, so all the people participating are winning."

### Refund Policy

#### STUDYSOUP CANCELLATION POLICY

All subscriptions to StudySoup are paid in full at the time of subscribing. To change your credit card information or to cancel your subscription, go to "Edit Settings". All credit card information will be available there. If you should decide to cancel your subscription, it will continue to be valid until the next payment period, as all payments for the current period were made in advance. For special circumstances, please email support@studysoup.com

#### STUDYSOUP REFUND POLICY

StudySoup has more than 1 million course-specific study resources to help students study smarter. If you’re having trouble finding what you’re looking for, our customer support team can help you find what you need! Feel free to contact them here: support@studysoup.com

Recurring Subscriptions: If you have canceled your recurring subscription on the day of renewal and have not downloaded any documents, you may request a refund by submitting an email to support@studysoup.com

Satisfaction Guarantee: If you’re not satisfied with your subscription, you can contact us for further help. Contact must be made within 3 business days of your subscription purchase and your refund request will be subject for review.

Please Note: Refunds can never be provided more than 30 days after the initial purchase date regardless of your activity on the site.