Today in Astronomy 102 black holes and how

Today in Astronomy 102: black holes and how to prevent them q The Schwarzschild singularity and the sizes of black holes. q Degeneracy pressure: a quantum-mechanical effect that might stop matter from collapsing to form a black hole, when gas pressure or material strength aren’t enough. 9 October 2001 Astronomy 102, Fall 2001 1

Karl Schwarzschild’s Work In 1916 Schwarzschild read Einstein’s paper on general relativity. He was interested in the physics of stars, and had a lot of spare time between battles on the Russian front, so he solved Einstein’s field equation for the region outside a massive spherical object. His solution had many interesting features, including q prediction of space warping in strong gravity, and invention of embedding diagrams to visualize it. q verification gravitational time dilation, just as Einstein had pictured it q prediction of black holes, though this was not recognized at the time. 9 October 2001 Astronomy 102, Fall 2001 Karl Schwarzschild 2

Einstein’s field equation The field equation is the ultimate mathematical expression of Einstein’s general theory of relativity. Astronomy 102 version: “Spacetime, with its curvature, tells masses how to move; masses tell spacetime how to curve. ” Physics 413 - Astronomy 554 version: é 2 l é ¶x h ¶ 2 x a ¶x s ¶ 2 x a ¶ 2 g ml ¶ 2 g mk ù ¶ 2 g lk ¶x h ¶ 2 x a ¶x s ¶ 2 x a ù 1 ê ¶ gl ú + g hs ê + - a ú a l a m k k a m l 2 ê ¶x k ¶x m ¶x k ¶x l ¶x m ¶x l ú ¶x ¶x ¶x l ¶x ¶x ¶x úû êë ¶x ¶x ¶x l ¶x ¶x ¶x ë û m æ é ¶ 2 gl h é ¶ h ¶ 2 xa ¶ s ¶ 2 xa 2 x a ¶ s ¶ 2 x a ùö ¶ 2 g lm ù ¶ 2 gl ¶ ¶ 1 1ê x x l ç ú + g hs ê ú÷ - g mk - m - a 2 m a l a m m a l ç ¶x l ¶x m ú ê ¶x ¶x ¶x l ¶x ¶x m ¶x ú÷ø 2 ¶x ¶ ¶ x x l m è 2 êë ¶x ¶x m ¶x ¶x l ë û û = -8 G åp n 9 October 2001 nm dx nk 3 d (x - xn ) dt Astronomy 102, Fall 2001 3

What you get when you solve the field equation In case you’re interested (i. e. not on the exam): q The solution to the field equation is a function called the metric tensor. This function tells how much distance or time is involved for unit displacements in the spacetime coordinates. q Accordingly, the metric tensor is related to the absolute interval. Each different solution to the field equation corresponds to a different absolute interval. The absolute interval corresponding to Schwarzschild’s metric turns out to be given (in spherical coordinates) by 9 October 2001 Astronomy 102, Fall 2001 4

Calculated from Schwarzschild’s solution for an object with M=c 2/2 G. (From lecture on 13 September) Distances between the circles: 1. 185 1. 074 1. 057 1. 106 1. 065 1. 087 1. 135 1. 300 All the distances would be 1 if space weren’t warped. y x 9 October 2001 Astronomy 102, Fall 2001 5

One way to visualize warped space: “hyperspace” (Also from lecture on 13 September) To connect these circles with segments of these “too long” lengths, one can consider them to be offset from one another along some imaginary dimension that is perpendicular to x and y but is not z. (If it were z, the circles wouldn’t appear to lie in a plane!). Such addi-tional dimensions comprise hyperspace. 1. 057 1. 065 1. 074 1. 087 1. 106 1. 185 1. 135 1. 300 y x Hyperspace 9 October 2001 Astronomy 102, Fall 2001 6

Schwarzschild’s view of the equatorial plane of a star Figure from Thorne, Black holes and time warps (equator) Center of star 9 October 2001 Astronomy 102, Fall 2001 7

Schwarzschild’s solution to the Einstein field equations Results: the curvature of spacetime outside a massive star. q For a given, fixed star mass M, he examined how space and time are curved if the star is made smaller and smaller in size. q Singularity: if the star is made smaller than a certain critical size, the gravitational redshift of light (time dilation, remember) predicted by his solution was infinite! (See next slide. ) 9 October 2001 Astronomy 102, Fall 2001 8

Figure from Thorne, Black holes and time warps 9 October 2001 Astronomy 102, Fall 2001 9

Implications of “Schwarzschild’s singularity” q If a star is made too small in circumference for a given mass, nothing can escape from it, not even light. • This would be a black hole, and the critical size is the size of the black hole’s horizon. q This is similar to an 18 th century idea: “dark stars” (Michell, Laplace. . . ); that if light were subject to gravitational force, there could be stars from which light could not escape. • The critical size of Schwarzschild’s singularity turns out to be the same as that for the 18 th century dark star. 9 October 2001 Astronomy 102, Fall 2001 10

The Schwarzschild singularity According to Schwarzschild’s solution to Einstein’s field equation for spherical objects, the gravitational redshift becomes infinite (i. e. time appears to a distant observer to stop) if an object having mass M is confined within a sphere of circumference CS, given by 4 p. GM Schwarzschild circumference c 2 where G =6. 67 x 10 -8 cm 3/(gm sec 2) is Newton’s gravitational constant, and c = 2. 9979 x 1010 cm/sec is, as usual, the speed of light (and = 3. 14159. . ). CS = You need to understand this formula. 9 October 2001 Astronomy 102, Fall 2001 11

The Schwarzschild singularity (continued) Any object with mass M, and circumference smaller than CS, would not be able to send light (or anything else) to an outside observer -- that is, it would be a black hole. The sphere with this critical circumference the Schwarzschild singularity itself - is what we have been calling the event horizon, or simply the horizon, of the black hole. 9 October 2001 Astronomy 102, Fall 2001 CS 12

Examples: calculation using the horizon (Schwarzschild) circumference Example 1: what is the horizon circumference of a black hole? (Compare to the discussion of the black hole Hades, pg. 29 in Thorne) 9 October 2001 Astronomy 102, Fall 2001 13

Examples: calculation using the horizon (Schwarzschild) circumference, continued Example 2: what is the horizon circumference of a black hole with the same mass as the Earth (6. 0 x 1027 gm)? CS= = 4 p. GM c 2 3 cm 4 ´ 3. 14 ´ 6. 67 ´ 10 -8 2 s gm 2 cm æ ö ç 3. 00 ´ 10 10 ÷ è s ø = 5. 6 cm (!!) 9 October 2001 Astronomy 102, Fall 2001 ´ 6. 0 ´ 10 27 gm 14

Examples: calculation using the horizon (Schwarzschild) circumference, continued Example 3: what is the mass of a black hole that has a horizon circumference equal to that of the Earth (4. 0 x 109 cm)? First, rearrange the formula: CS = 4 p. GM c 2 c 2 CS = 2 4 p. G c CS c 2 =M 4 p. G 9 October 2001 Astronomy 102, Fall 2001 15

Examples: calculation using the horizon (Schwarzschild) circumference, continued Then, put in the numbers: (!!) 9 October 2001 Astronomy 102, Fall 2001 16

Mid-lecture Break q Exam #1 is still having its grades recorded. It’ll be available tomorrow in recitation, and Thursday in lecture. Einstein and his violin 9 October 2001 Astronomy 102, Fall 2001 17

Singularities in physics, math and astronomy A formula is called singular if, when one puts the numbers into it in a calculation, the result is infinity, or is not well defined. The particular combination of numbers is called the singularity. Singularities often arise in the formulas of physics and astronomy. They usually indicate either: q invalid approximations -- not all of the necessary physical laws have been accounted for in the formula (no big deal), or q that the singularity is not realizable (also no big deal), or q that a mathematical error was made in obtaining the formula (just plain wrong). 9 October 2001 Astronomy 102, Fall 2001 18

Singularities in physics, math and astronomy (continued) Example of a classical physics law with a singularity: Newton’s law of gravitation. r GMm F= 2 r F m M q r is the distance between the centers of the two spherical masses. A spherical mass exerts force as if its mass is concentrated at its center. q Clearly, if r were zero, the force would be infinite! This formula will not appear on homework or exams. It is used only because it is a good example of a singularity. 9 October 2001 Astronomy 102, Fall 2001 19

Singularities in physics, math and astronomy (continued) This singularity is not realized, however: q the mass really isn’t concentrated at a point. q a spherical shell of matter does not exert a net gravitational force on a mass inside it. m r. M 9 October 2001 Consider mass m inside mass M: outer (yellow) matter’s forces on m cancel out, and only inner (green) exerts a force. As m gets closer to the center (r 0), the force gets smaller, not larger. No singularity! Astronomy 102, Fall 2001 20

Reaction to the Schwarzschild singularity q Schwarzschild’s solution to the Einstein field equation was demonstrated to be correct - the singularity is not the result of a math error. q Thus most physicists and astronomers assumed that the singularity would not be physically realizable (just like the singularity in Newton’s law of gravitation) or that accounting for other physical effects would remove it. q Einstein (1939) eventually tried to prove this in a general relativistic calculation of stable (non-collapsing or exploding) stars of size equal to the Schwarzschild circumference. q He found that this would require infinite gas pressure, or particle speed greater than the speed of light, both of which are impossible. 9 October 2001 Astronomy 102, Fall 2001 21

Reaction to the Schwarzschild singularity (continued) q Einstein’s results show that a stable object with a singularity cannot exist. q From this he concluded (incorrectly) that this meant the singularity could not exist in nature. q Einstein’s calculation was correct, but the correct inference from the result is that gas pressure cannot support the weight of stars similar in size to the Schwarzschild circumference. q If nothing stronger than gas pressure holds them up, such stars will collapse to form black holes -- the singularity is real. • Stronger than gas pressure: degeneracy pressure. 9 October 2001 Astronomy 102, Fall 2001 22

Degeneracy pressure This involves a concept from quantum mechanics called the waveparticle duality: q All elementary particles from which matter and energy are made (including light, electrons, protons, neutrons. . . ) have simultaneously the properties of particles and waves. q Which property they display depends upon the situation they’re in. Degeneracy pressure consists of a powerful resistance to compression that’s exhibited by the elementary constituents of matter when these particles are confined to spaces small enough to reveal their wave properties. In more detail…. 9 October 2001 Astronomy 102, Fall 2001 23

Particles and waves Particles exist only at a point in space. Waves extend over a region of space. Electric Field, for instance Location in space 9 October 2001 Astronomy 102, Fall 2001 24

Light can be either a particle or a wave Particle example: the photoelectric effect -- the 1905 explanation of which, in these terms, won Einstein the 1921 Nobel Prize in physics. Light (in the form of photons) Electrons Metal slab 9 October 2001 Astronomy 102, Fall 2001 25

Light can be either a particle or a wave (continued) Wave example: the Doppler effect. Lasers Observer sees: To observer V (close to c) 9 October 2001 Astronomy 102, Fall 2001 26

Electrons can be particles or waves Particle example: collisions between free electrons are “elastic” (they behave like billiard balls). Wave example: electrons confined to atoms behave like waves. Nucleus Atom 9 October 2001 Electron “cloud” (extends over space as a wave does) Astronomy 102, Fall 2001 27

How to evoke the wave properties of matter All the elementary constituents of matter have both wave and particle properties. If a subatomic particle (like an electron, proton or neutron) is confined to a very small space, it acts like a wave rather than a particle. How small a space? q The size of an atom, in the case of electrons (about 10 -8 cm in diameter). q A much smaller space for protons and neutrons (about 10 -11 cm diameter). q Generally, the more massive a particle is, the smaller the confinement space required to make it exhibit wave properties. 9 October 2001 Astronomy 102, Fall 2001 28