Geometry of Curved spaces

Note we have carefully avoided saying what we mean by a curved space

If you take the example of the 2-D curved surface of the Earth, this is embedded in a 3-D space. Hence If a massive body curves space, it almost implies extra dimensions. In fact we can carry out tests to decide if we live in a "normal" 3-D space (Euclidean) e.g.
parallel lines
angles of a triangle add up to 1800 
α + β + γ = 180
Specifically, consider a sphere: circles don't satisfy 

C = 2πa
In fact, 
C = 2πx = 2πRsin(a/R) = 2πa(1-a2/6R2)

i.e. by measuring the circ. of large circles, we can determine if we live in a curved space. These are experiments that we can almost do. (Gauss tried the 2nd!). We can measure the curvature k ≡ 1/R2; (a physicist would say we are measuring distant-dependence corrections to π!)

Can also have negative curvature K<0.

How do we know analytically if we are in a curved space?
e.g. a cylinder will satisfy (most) Euclidean geometry. We can "unwrap" a cylinder into a flat surface, we can't unwrap a sphere without distorting it.
Constant metric implies a flat space, but the opp. doesn't hold: e.g. Cartesians $$ \color{red}{ (x,y) \Rightarrow g_{\mu \nu } = \left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & 1 \\ \end{array}} \right]} $$ describes same space as polar coords $$ \color{red}{ (r,\varphi ) \Rightarrow g_{\mu \nu } = \left[ {\begin{array}{*{20}c} 1 & 0 \\ 0 & {r^2 } \\ \end{array}} \right]} $$ If we go to cylinder, then r=R is fixed, and the distance $$ \color{red}{ \Delta s^2 = \Delta z^2 + R^2 \Delta \varphi ^2 = \Delta x^2 + \Delta y^2 :x = z,y = R\varphi } $$
Note that on a cylinder we can have closed and open geodesics, in flat plane can only have open geodesics

Compare with distance on a sphere: ,$$ \color{red}{ \Delta s^2 = R^2 \Delta \theta ^2 + R^2 \sin ^2 \left( \theta \right)\Delta \varphi ^2 \Rightarrow g_{\mu \nu } = \left[ {\begin{array}{*{20}c} {R^2 } & 0 \\ 0 & {R^2 \sin ^2 \left( \theta \right)} \\ \end{array}} \right]} $$ How do we know we cannot flatten this? Gauss' tells us how to measure curvature in frame-indep fashion. $$ \color{red}{ K = \frac{1}{{2g_{11} g_{22} }}\left[ { - \frac{{\partial g_{11} ^2 }}{{\partial \left( {x^2 } \right)^2 }} - \frac{{\partial g_{22} ^2 }}{{\partial \left( {x^1 } \right)^2 }} + \frac{1}{{2g_{11} }}\left( {\frac{{\partial g_{11} }}{{\partial x^1 }}\frac{{\partial g_{22} }}{{\partial x^1 }} + \left( {\frac{{\partial g_{11} }}{{\partial x^2 }}} \right)^2 } \right) + \frac{1}{{2g_{22} }}\left( {\frac{{\partial g_{11} }}{{\partial x^2 }}\frac{{\partial g_{22} }}{{\partial x^2 }} + \left( {\frac{{\partial g_{22} }}{{\partial x^1 }}} \right)^2 } \right)} \right]} $$ (no, it isn't obvious! Gauss called it the "Theorem Egregium"). Note this is the 2-D case: in general we will have a curvature tensor. Rijkl: much of GR can be done with this (e.g. a single black hole but not a rotating BH and definitely not colliding BH's!). Very important for what we are going to do later is 3-D constant curvature solution (hyper-spherical geometry). Generalise fixed r relation $$ \color{red}{ \Delta s^2 = r^2 \Delta \theta ^2 + r^2 \sin ^2 \left( \theta \right)\Delta \varphi ^2 } $$ to 3-D: but now we expect some r dependence: $$ \color{red}{ \Delta s^2 = f\left( r \right)\Delta r^2 + r^2 \Delta \theta ^2 + r^2 \sin ^2 \left( \theta \right)\Delta \varphi ^2 } $$
Can only interpret this in "slices"
Result can't depend on φ or θ, so simplify by taking θ =π/2; Then $$ \color{red}{ g_{\mu \nu } = \left[ {\begin{array}{*{20}c} {f\left( r \right)} & 0 \\ 0 & {r^2 } \\ \end{array}} \right]} $$ Gauss gives us $$ \color{red}{ K = \frac{1}{{2rf^2 }}\frac{{df}}{{dr}} \Rightarrow f = \frac{1}{{C - Kr^2 }}} $$ which can easily be solved: C is const. of integration: obviously f(r) = 1 for a flat surface. Hence $$ \color{red}{ \Delta s^2 = \frac{{\Delta r^2 }}{{1 - Kr^2 }} + r^2 \Delta \theta ^2 + r^2 \sin ^2 \left( \theta \right)\Delta \varphi ^2 } $$
We can find the area and radius of an r-sphere:$$ \color{red}{ a\left( r \right) = \int_0^a {ds} = \int_0^r {\frac{{dr}}{{\left( {1 - Kr^2 } \right)^{1/2} }}} = \frac{1}{{\sqrt K }}\sin ^{ - 1} \left( {r\sqrt K } \right)} $$
How do we interpret this?

Black Holes

Chap 5 in Berry tells you how to write down and use the metric for a isolated massive body

First step is curvature due to a massive body $$ \color{red}{ K\left( r \right) = - \frac{{GM}}{{c^2 r^3 }}} $$ (Dimensional argument, but also can show this reduces to Newton's laws.

Gravitational Red-shift

A ball thrown up near the earth's surface will lose energy.
Again can get this via equivalence principle: light emitted from floor of elevator with frequency νE, hits ceiling after a time t = h/c. during this time, lab (elevator) has accelerated to a speed u = gt, so the light will have a red-shift $$ \color{red}{ z = \frac{u}{c} = \frac{{gh}}{{c^2 }} \Rightarrow \frac{{GM}}{{c^2 r}}:\lambda _o = \lambda _E \left( {1 - \frac{{GM}}{{c^2 r}}} \right)} $$
(Confirmed by Rebka-Pound using Mossbauer techniques in 1960. An atom is a good clock:

This is another consequence of the equivalence principle: confirmed in numerous experiments over the last 40 years. Implies that clocks run slow in gravitational fields $$ \color{red}{ t' = \frac{t}{{\left( {1 - \frac{{2GM}}{{c^2 r}}} \right)^{1/2} }}} $$ Then Schwarzchild metric $$ \color{red}{ g_{\mu \nu } = \left[ {\begin{array}{*{20}c} {1 - \frac{{2GM}}{{c^2 r}}} & 0 & 0 & 0 \\ 0 & { - \frac{1}{{c^2 \left( {1 - \frac{{2GM}}{{c^2 r}}} \right)}}} & 0 & 0 \\ 0 & 0 & { - \frac{{r^2 }}{{c^2 }}} & 0 \\ 0 & 0 & 0 & { - \frac{{r^2 \sin ^2 \left( \theta \right)}}{{c^2 }}} \\ \end{array}} \right]} $$ (can find g11 by variant of last argument: to find g00 need to have the metric independent of time even though. clocks run slow. Schwarzchild radius R = 2GM/c2 is where metric becomes singular.

in particular, if 

r = 2GM
     c2
then t = ∞: i.e. if I watch a clock at the Schwarzchild radius, it appears to have stopped. From this can get all the classic tests of GR (see Clifford Will).

Gravitational force

gets changed $$ \color{red}{ F = \frac{{GMm}}{{r^2 }} \Rightarrow \frac{{GMm}}{{r^2 }} - \frac{{GMJ^2 }}{{c^2 r^3 }}} $$

What does this look like? As long as speeds are small, exactly the same as Newton (Ha!), but if velocities are "large" then the force gets changed

Fact that orbits are closed is "coincidence" not true for any potentials except r2, 1/r and 1/r2

Hence get "rosette" orbits.

Much less dramatic in practice: perihelion (closest approach to sun) of Mercury advances by 43" arc/century

And light gets does bent: this is a very large cluster of galaxies, which acts as a very large (and rather bad!) lens. It produces several images of a much more distant galaxy

A final consequence:

gravitational waves

  • Vibrating charge radiates E.M. waves
  • Vibrating mass radiates grav. waves

Differences: that

  • Gravitational force between 2 electrons ~ 10-42 electric force Radiation is quadrupole, not dipole, which also means it is still weaker
  • Quadrupole nature means that grav. radiation cannot be produced by monopole or dipole system: e.g. supernova collapse (which has plenty of energy) is probably symmetric, so no radiation

    • Hence (well, more or less hence!) it requires a large amount of mass to produce a grav. wave, and a large amount to see one: e.g need to detect motions of << atomic radius in a one ton sapphire crystal.

    History:
    1. Joseph Weber claimed to have seen grav waves in 1970.
    2. LIGO (Laser interferometer) turned on in 2004: will see coalescing binary systems
    3. LISA (space interferometer will detect grav. radiation from anything in local supercluster

    Hulse and Taylor: Binary Pulsar PSR1913+16 discovered 1974. Like all pulsars, emits very regular radio pulse every 59 ms. (Frequency is 16.940 539 184 253 Hz: i.e. is better known than atomic clocks)

    This consists of two neutron stars, in orbit 106 km in radius, with period of hours. Change in frequency allows orbit to be calculated exactly, and can measure..

    Rate of precession = 4.22662 0/yr (i.e. 30,000x that of Mercury)

    and that pulsar is losing energy, by gravitational radiation (mass~1.4 M0, and accns are large)

    Decrease of the orbital period P (about 7h 45 min) of the binary pulsar PSR B1913+16, measured by the successive shifts T(t) of the crossing times at periastron; the continuous curve corresponds to $$ \color{red}{ T(t) = \frac{{t^2 }}{{2P}}\frac{{dP}}{{dt}}} $$ given by the general relativity (reaction to the gravitational waves emission)

    Reference: Taylor J.H. 1993, Testing relativistic gravity with binary and millisecond pulsars, in General Relativity and Gravitation 1992, eds. R.J. Gleiser, C.N. Kozameh, O.M. Moreschi. Institute of Physics Publishing (Bristol).

    Hence 1993 Nobel Prize

    So we conclude that Relativity (Special and General) works because

    But we are doing Cosmology!