GENERAL RELATIVITY



GENERAL RELATIVITY

Some basic concepts

Event: an extent-less occurrence, specified by coordinates xμ =(x1,x2,x3,x4)

(Comment: we don’t specify here how they are to be measured; similar to generalized coordinates q; in classical mechanics, and contrary to the dogma of operationalism. No restriction to 3 space and 1 time coordinate – could have 4 time coordinates.)

Space-time: Totality of all events (real of imagined)

Particle: a moving point, endowed with mass. Traces out a path in space-time xμ= xμ(u), called a world-line, the parameter u specifying a point or the world line.

Given a fixed event on a world line, we assume we can divide all other events on the world line with?

Past: all events happening earlier

Future: all events happening later

Null cone

We retain the belief that a particle can’t be in two places at the same time or (??), some pairs of events can’t be lined by the world-line of any particles

For any event O, all other events can be divided into 2 cones:

Evens which can be linked to O via a particle world line

Can’t

These tow cones occupy distinct “volumes” in space-time, separated by a “surface” called the null cone. Schematically

A happens after O on any world-line ?? O

B before

Definition

Given two neighboring events:

Displacement in d xμ“timelike” of P and Q can be linked by a particle’s world line

“null” photon’s

“space-like” otherwise

Note: null displacement from O in previous diagram lies on surface of null cone, hence name “null cone”

Postulate I(a)

Space-time is Riemanian, ie, there exists a metric tensor gμν(=gνμ) whose components are functions of the coordinate xК.

Definition: The “interval” ds between xμ and xμ+dxμ is given by

ds2=gμνdxμdxν

Postulate I(b)

For a time-like displacement, gμνdxμdxν>0, and [pic]is the elapsed time between xμ and xμ+dxμ along the world time joining them, ie in the frame in which the two events occur at the same places

[Comment: ds/c is measured by a standard clock. It is the “proper-time” between P and Q, and denoted by dτ]

For a null displacement gμνdxμdxν=0, i.e., ds = 0

For a space-like displacement, gμνdxμdxν>0, ie, ds is imaginary

Note

1. ds = 0, divides space-like from time-like displacement. Hence ds = 0 for any two neighboring points on the null cone.

2. Exp. Verification of I(a) and I(b)A: Take fixed event P, and average for 10 observers to move along 10 different world lines starting from P. Each move from P to a neighbouring point and measures ds/c, since g has 10 independent components, ds2= gμνdxμdxν, gives us 10 ? and 10 unknowns. Solve for gμν(at P). Now get 11th observer to move from D to an arbitrary neighbouring point Q. Measure ds, and compare with calculated volume[pic].

Geodesics

Euclidean 3-specs ??

dl2 = dx2 + dy2 + dz2

Given two fixed points A and B

[pic]=0

Is the equation of a straight line.

Ie, gives shortest distance between A < B

Definition: Be analogy, a geodesics is a curve in space-time whose “length” has a stationary value with regard to small variations, end points being fixed

[pic]=0

Now any curve between A and B may be describe parametrically by xμ = xμ (u), where u is a parameter

0=[pic]=[pic]

Where

L=[pic]=L(xμ,[pic])

Now have analysis analogous to deriving Lagrange’s equations from Hamilton’s principles:

[pic]

Since from for arbitrary variations between A and B

[pic]=0

(analogue of Lagrange’ equations)

Now [pic]

Where ( ), κ means [pic]( )

And [pic]

[pic]

Now let u = s, provided ds ( 0, ie we are not dealing with a null geodesics

Then L = 1 and [pic]

ie. [pic]

But [pic]

[pic] [pic]

×gλκ and use gλκgμκ=δλμ =>

[pic]

Where [pic] are Christoffel symbols

Special Case – Null Geodesics

A null geodesic is a path for which the initial ds between neighboring points is zero.

Here ds = 0, so can’t use s as parameter in xμ = xμ (s),

Procedure:

gμν dxμ dxν = 0, since ds = 0, and xμ = xμ (u), u (s.

[pic]

[pic]= 0 where L= [pic]

Analysis same as before except no 1/2L factor in [pic]and [pic] gives eventually

[pic]

Properties of Christoffed Symbols

[ μν, κ] ’ [νμ, κ] by inspection

[ μν, κ] + [κν, μ] ’ 1/2 ( gμκ,ν + gνκ,μ − gμν,,κ) + 1/2( gκμ,ν + gνμ,κ − gκν,μ)

’ gμκ,ν

1. Γμν τ ’ Γνμ τ by inspection

[pic]

From gμκ,ν’ [μ ν, κ] + [κ ν, μ], we deduce

2. gμκ,ν ’ gμνΓμν τ + gμνΓνμ τ

But bear in mind: Christoffel symbols are NOT tensors (see later), but simply functions of the metric tensors and its derivatives

Postulate II

Free particles (one on which no real forces act) follow time-like geodesics through space-time. Photons follow null geodesics

Hence

Particles: Photons:

Equation of motion is Equation of motion is

[pic] [pic]

with particular integral with particular integral

[pic][pic]= 1 [pic]= 0

Solution xμ(s) requires ?? on xμ,[pic]. Solution xμ(s) requires ?? on xμ,[pic].

Notes:

1. Motivation for making this postulate: In Newtonian mechanics, free particles move in straight lines. In space-time with geodesics metric, the analogue of a straight line is a geodesic.

2. Postulate that photons follow null geodesics means that the null cone introduced earlier is a surface determined by the world line of photons converging on O and diverging from O.

Special Relativity

We now confirm the validity of the above postulate for those frames of reference (the initial frames) in which special relativity holds (locally)

Cartesian spatial coordinates: x, y, z

Define: x’ = x, x2 = y, x3 = z, x4 = ct

Postulate I(a) – Existence of metric tensor

c2dt2 – (dx2 + dy2 + dz2) is invariant under transformation between initial frames

Write ds2 = c2dt2 – (dx2 + dy2 + dz2) = (dx4)2 - (dx’)2 - (dx2)2 - (dx3)2 = gμνdxμdxν

Where [pic]

Note: we assume that gμν(xμ) = (-1, -1, -1, +1) for all xκ, even though we know the this will be so only ?? a limited region of space-times

Postulate I(b) – Meaning of [pic]

In that frame of reference for which dx = dy = dz = 0, ie in frame in which the two events occur at the same place.

Postulate II – Geodesics

Since gμν= constant, all Γμν τ are zero

For particles, [pic]( xλ = αλs + βλ, where αλ, βλare constants

Velocity of particles in x-directions is

[pic]

Similarly vy = const, vz = const: Particles move with constant velocity

Particular integral is [pic]= 1

[pic] Where v2 = vx2 + vy2 + vz2

This is the time-dilation formula

[Automatically, [pic] where [pic]= proper time integral]

For photons, [pic] ( photons have constant velocity

Particular integral [pic]= 0 => [pic]

ie. v=c

Hence postulate correct. But note severe restriction to metric frames. In a non-initial frame, even with certain coordinates, we are NOT going to find [pic], and hence free particles are NOT going to move with constant velocity.

Riemannian Coordinates (sometimes call geodesic coordinates)

Aim: starting from an arbitrary coordinate system, to find a coordinate system in which SR holds locally.

O is a fixed point with coordinates xoλ

xoλ OP is an arbitrary time-like geodesic whose equation is

[pic] or [pic]

Where dτ=ds/c is proper time

Suppose we measure s and therefore τ from O, then xλ= xλ(τ)along geodesic, is the proper frame.

Expanding about O,

[pic]

Define [pic], the velocity vector

[pic]

But [pic]

[pic]

[pic]

[pic]

(ν ((σ in second term)

[pic]

Now define new coordinates x’λ = ν0λτ, so that O has coordinates x’λ = 0.

Then xλ = x0λ + x’λ − 1/2[Γμν τ]0 x’μ x’ν + 1/6[- Γμν,ρ τ + 2Γμσ λΓνρ σ]0 x’μ x’ν x’ρ + ...

These equations now serves as a transformation equation from xλ to x’λ , originally along geodesic OP, but now valid for ANY geodesic since τ has been eliminated. These Riemannian Coordinates x’λ , are defined up to the radius of convergence of the infinite space on the RHS, ie for all space-time point in the vicinity of O.

Now in the Riemannian coordinate system, geodesic equations is

[pic]

But x’λ = v0λτ, ’’> [pic]

[pic]

At O, where is dx’μ is arbitrary, Γμσ λ ’ 0

It follows that g’μκ,ν ’ at O, (see Postulate of Christoffed symbols)

Significance: [pic]

In vicinity of O is the main feature of SR

Hence deduce that coordinate system x’λ is one in which SR holds locally.

However, note that g’μν will not necessarily be diagonal with diagonal elements equal to (-1, -1, -1, +1). To achieve this, a further transformation of coordinates is necessary which DIAGONALISES the metric tensor. Don’t want to go into details here, but main task is to solve the equation

Equation [pic]

For the unknown [pic] (16 unknowns, 10 equations ( solutions contains 6 arbitrary parameters – because initial frames are not unique. These correspond to 3 components of velocity and 3 parameters (eg Euler angles) specifying orientations of coordinate axes.

Diagonalisation of gμν (see Roberston & Noew ?? p214)

Given: arbitrary coordinate system xμ, with metric tensor gμν and transformation to new coordinate x’μ

[pic]

Can we find a coordinate system x’μ, such that [pic] at a particular event O? now gμν(0) are a set of know numbers, the unknowns are [pic],16 quantities in all. There are 16 equations to determine them, but since gμνis symmetric, only 10 are independent [ α = 1, β = 2 give same equations as α = 2, β = 1]

Hence can solve for [pic], in terms of 6 arbitrary parameters.

Hence can calculate [pic] from[pic]

Can now calculate any tensor components at O in the x’ρ coordinate system. In fact, can be shown that the above procedure gives real solutions for[pic], only if of diagonal elements have same signs and fourth has opposite sign, ie with no lack of generality can take [pic] at O.

Ultimate reason for this: space-time must be separable into past, present and future with pas and future linked by present. See Spring ? the Special Theory ? p17.

Suppose now gμν,τ ’ 0 at O, ie xμ coordinate system already Riemannian. Then if we take [pic]is LINEAR transformation, then immediately follows that g’αβ,τ ’ 0, also, at O.

????? in fact that ds2 must be appreciate to [pic] ??????

Hence procedure is to

1. Find Riemannian coordinate system starting from arbitrary coordinate system, then,

2. Diagonalise in terms of 6 arbitrary parameters.

Reasons for appearance of 6 arbitrary parameters: spatial frames are not unique starting from any one initial frame, most general metrical frame with Cartesian coordinate requires 6 parameters to specify it: eg 3 Euler angles to establish rotation of axes and 3 components of velocity

Covariant Differentiation

Suppose Tμ is a tensor. The partial derivative Tμ,τ is NOT a tensor

For [pic]

[pic]

Tμ,τ would be a tensor only if 2nd term on RHS = 0, not correct generally. [though correct in SR where we restrict ourselves to transformations between initial frames]

The following derivatives, call covariant derivatives are tensors:

1. Tμ;τ = Tμ,τ + Γμστ Tσ

2. Tμ;τ = Tμ,τ − Γσμτ Tσ

Proof of 2 [pic]

[pic]

[pic]

[pic][pic]

[pic]

Proof of 1 (Illustrate different methods)

Tμ,τ is a tensor when coordinate transformations are between those in which SR holds

At any event O, define a tensor Tμ;τ such that, in any locally Riemannian coordinate system based on O,

T’μ;τ = T’μ,τ

Since T’μ;τ is required to be a tensor

[pic]

=[pic]

[pic]

[pic]

[pic]

[pic]

Now xλ - x0λ = x’λ − 1/2[Γμν τ]0 x’μ x’ν + ...

x’λ ’ xλ - x0λ + 1/2[Γμν τ]0 (xμ - x0μ ) (xν - x0ν ) + ...

At O, [pic] , [pic]

[pic]Tμ;τ = Tμ,τ + Γμστ Tσ, all quantities evaluated at O. But O was any event, so result must be true generally.

Extensions to tensors of higher (and lower) order

Second orders

Tμν;τ = Tμν,τ + Γμστ Tσν + Γνστ Tμν

Tμν;τ = Tμν,τ − Γσμτ Tσν − Γσντ Tμσ

Tμν;τ = Tμν,τ + Γμστ Tσν + Γσντ Tμσ

Zero’th order

T;τ = T,τ

Extension to higher orders is obvious. Note that the covariant derivative of Tμν for fixed μ,ν ?? not just the Tμν component but other components. Note also that at the origne of any Riemannian coordinate system all Christoffed symbol ?? so: τ reduces to τ ?? in all cases

Theorems

1. gμν,τ ’ 0 Proof

gμν,τ ’ gμν,τ − Γσμτ gσν − Γσντ gνσ ’ 0, by one of the relations for Christoffed symbols

2. δμν;τ = 0 Proof

δμν;τ = δμν,τ + Γμστ δσν − Γσντ δμν ’ Γμντ − Γμντ = 0

3. (X + Y);τ = X;τ + Y;τ for any tensors X, Y of same type. Proof obvious

4. (XY);τ = XY;τ + X;τ Y for any tensor X and Y

Proof: at an arbitrary event O, choose Riemannian Coordinates.

Then ;τ −> ,τ at O.

(XY);τ = (XY), τ = X,τY + XY, τ ’ X;τY + XY; τ

Hence is true in the Riemannian system. But is a tensor equations, so is true in ALL coordinate systems. (at O). Since O is arbitrary, is true everywhere.

5. gμν;τ ’ 0 Proof:

gμν gνσ ’ δμσ

gμν;τ gνσ + gμν gνσ;τ ’ δμσ;τ

( (

0 0

( gμν;τ ’ 0

6. Tμν;τ ’ ( gμσ Tσν);τ ’ gμσ Tσν;τ

Order of raising or lowering a suffix, and covariant differentiation, is irrelevant.

The Curvature Tensor

Extension to second order derivatives:

Partial derivatives:

Covariant derivatives

Tμ;στ = (Tμ;σ);τ [pic]

Now Tμ;τ = Tμ,σ - Γαμτ Tσ is a second order covariant tensor

(Tμ;σ);τ ’ (Tμ;σ),τ - Γβμτ Tβ;σ - Γβστ Tμ;β

Tμ;στ = [Tμ.στ] − Γαμσ,τTα − ΓαμσTα,τ − ΓβμτTβ,σ − ΓβμτΓαβσTα − ΓβστTμ,β + ΓβστΓαμβTα

If calculate Tμ;στ - Tμ;τσ, terms bracketed dropout, ie terms symmetric in σ & τ.

Leave

Tμ;στ - Tμ;τσ = RαμστTα

Where

Rαμστ = Γαμτ,σ − Γαμσ,τ + Γβμτ Γαβσ − Γβμσ Γαβτ

Obviously Rαμστ is a tensor of the type indicated. If is called the (mixed) curvature tensor, or Riemannian-Christoffed tensor. Its importance is that it is the next most complicated tensor in space-time, after gμν ?? itself, that is a function on of gμν and its derivatives (note: depends on gμν and 1st and 2nd derivatives of gμν)

Properties of Rαμστ and the fully covariant form Rαμστ ’ gαλRλμστ

1. Rαμστ = - Rαμτσ , by inspection

2. Rαμστ + Rαστμ + Rατμσ ’ 0 (proof is direct and simple)

3. Rαμστ = - Rαμτσ (from 1 together with definition of Rαμστ)

4. Rαμστ = - Rματσ

5. Rαμστ = - Rσταμ proved in problem sheet. No! too ??

6. Rαμστ + Rαστμ + Rατμσ = 0 (from 2 together with definition of Rαμστ)

Taking these into account can be shown that of the 256 components of Rαμστ, only 20 are independent.

Bianchi Identity

Rαμστ;ρ + Rαστρ;μ + Rαμρσ;τ = 0 (Note:1024 equations)

Proof: Given an arbitrary event O, we select Riemannian coordinates with O as origin. Then all Christoffed symbols are zero (thought derivatives not necessarily zero) at O and ; ρ −> ,ρ.

At O, Rαμστ;ρ = Rαμστ,ρ = Γαμτ,σρ − Γαμσ,τρ

Rαμστ;ρ + Rαστρ;μ + Rαμρσ;τ =

Γαμτ,σρ − Γαμσ,τρ + Γαμρ,τσ − Γαμτ,ρσ + Γαμσ,ρτ − Γαμρ,στ ’ 0

Hence equation is true (at O) in Riemannian coordinate system. But is a tensor equation: If true in one coordinate system, true in all (at O). Also O was arbitrary, hence true everywhere.

Flatness

Definition: A space is flat if it is possible to find coordinate for which gμν is diagonal, with each diagonal element = ±1. Otherwise, it is curved.

Example:

Euclidean plane: dl2 = gμνdxμ dxν = dx2 + dy2 flat

Euclidean 3-space: dl2 = dx2 + dy2 + dz2 flat

Cylindrical surface: dl2 = dz2 + a2dφ2 = dz2 + [d( a φ)]2 flat

Spherical surface: dl2 = a2dθ2 + a2sin2θdφ2 Can’t find any transformation which would make dl2 = (dx’)2 ± (dx2)2 curved.

SR: ds2 = -dx2 - dy2 - dz2 + c2dt2 ?? flat

In a curved space, geometry is intrinsically different from that is a flat space, eg sum of angles of a ?? Dearth >< 1800, straight lines (geodesic) always intersect etc.

Note that curvature is intrinsic to a space – not to be elucidated ?? by companying the space with another in which it is embedded.

Modification of definition for G.R.

Definition

Space-time is flat at a point O if it is possible to find coordinates for which [pic] everywhere in a finite region about O. Otherwise it is

curved

Theorem

If Rαμστ = 0, is obviously a necessary condition. For if gμν’ diag(±1, ±1, …) everywhere, on one coordinate system, then gμν.σ ’ 0, Γαμσ ’ 0, Γαμσ,τ ’ 0 ’’> Rαμστ = 0 in that coordinate system. But this is a tensor equation. Must be true in all coordinate system.

Proof of sufficiency: needs concept of absolute differentiation – tends to be lengthy. See Springer & ??. No proof given here.

Connection between space-time Rαμστ and matter

Arbitrary event O. Select Riemannian coordinate system with O as origin, so gμν,σ ’ 0, Γαμτ = 0 at O in this coordinate system. Two cases

a. O is far from all matter. The locally – initial frame is initial one large region of space-time.

Ie gμν,σ ≈ 0 over large region near O

Γαμτ ≈ 0 ………………………..

Γαμτ,σ ≈ 0 at O. Hence Rαμστ ≈ 0 at O, ie space-time is “locally” flat

b. O is near large gravitating bodies

a. gμν,σ ≈ 0 in small region near O

b. Γαμτ ≈ 0 ………………………..

c. Γαμτ,σ ( 0 at O.

d. Rαμστ (0 at O. Hence space-time not “locally flat”, is curved.

This is an empirical observation. In setting up the theory, would like this to emerge as a consequence of a postulate; ie would like to postulate an equation of form

F(Rαμστ) = Some tensor density?? distribution of matter in universe

??

Note

Rαμστ = 0 is a tensor equation – true in all frames if true in one.

Hence Rαμστ =0 means NO GRAVITY, an absolute property.

Rαμστ (0 means gravitational field present, an absolute property, even though locally some of its effects may be nullified by going to the locally initial frame. This is relevant to some “strange” forms of the principle of Equivalence eg “ No experiment. can distinguish a gravitational field from its effects of acceleration”

- not true as it stands since gravitational field lines can verge, enable lines ? from uniform acceleration.

Matter

Most successfully approach: treat as non-rigid, continuous medium, ie fluid

(rigid body not easy to assimilate into SR or GR due to finite spread of propagation of signals)

Assume matter forms a perfect fluid, ie the only stress is an internal isotropic pressure – no sheet stress. (The assumption is good for stars and for mater on scale for which GR rive results significantly different from Newtonian gravitational theory)

Newtonian theory of a perfect fluid

In an inertial frame let

ρ(r, t) = density of fluid

P(r, t) = pressure ……..

v(r, t) = velocity

Assume no sources, mass conservation gives [pic]

Or using Cartesian cords, (x, y, z)=(x1,x2,x3) , [pic]

Assume no external forces; then Euler’s equations (see separate sheet derivations)

?? conservation of momentum [pic] i=1,2,3

This means have 4 equations, but 5 unknowns, so need more information for explicit solutions

ρ , = const (incompressible fluid)

ρ, ρ(P, T) (equation of state) + information on entropy

see McVittie p44, 45

Euler’s equation for a perfect fluid with no external forces

Consider an infinitesimal volume of the fluid contained within an imaginary box with sides δx, δy, δz as shown. We assume no external forces and no viscosity. Let P(r, t) , ρ(r, t) and v(r, t) be the pressure, the density and the velocity of the fluid.

The equation of motion in the x direction for the fluid considered is

[pic]

i.e. [pic]

Where d/dt is a derivative following the motion of the fluid. Generalizing this to three dimensions, we have

[pic]

From the mass conservation equation

[pic]

We get [pic]

And so [pic]

or [pic]

i.e. [pic]

SR Theory of a Perfect Fluid

Here [pic] with x’ = x, x2 = y, x3 = z, x4 = ct

Note that[pic] also

Definition:

4 velocity of fluid[pic]

now[pic]

where v = (vx, vy, vz) is the velocity of the fluid

In 4-vector notation, Uμ = γ(v) (v, c)

Now define 2 invariants

[pic]is density and pressure in frame in which fluid is locally at rest

Local reference from in which fluid is at rest

Then define [pic]

Energy-momentum tensor

Properties of Tμν:

1. Tμν ’ Tνμ (obvious)

2. In frame in which fluid is locally stationary,

3. Tμν,ν ’ 0 − expresses conservation of mass-momentum

Assuming no external forces, for proof, see Schwarz, Intro to SpRel, p180, 186

Reduction of Tμν,ν ’ 0 to Newtonian conservation equations for small velocities

Assume v1 0

All components of Gμν and Tμν are zero except for diagonal ones. Also G22 = G33 and T22 = T33. Hence only 3 different non-trivial equations, and only 2 of these are independent since Gμν and Tμν are divergenceless ??. Hence we look only at the μ ’ ν ’ 1 and μ ’ ν ’ 4 equations.

μ ’ ν ’ 4 [pic] (1)

μ ’ ν ’ 1 [pic]

Subtracting (1) from the equation gives

[pic] (2)

Now (1) can be integrated

i.e. [pic]

Note: the arbitrary constant is essentially contained in the lower limit of integration. We have taken this to be zero, to avoid a singularity in e-λ at r = 0.

Now define

[pic]

Then

[pic] [pic] (3)

[Remark: would surely have expected the mass appearing has to be the total proper mass of the source, ie

[pic] since radial element of distance is[pic]

However, gravitational potential energy lowers mass

[pic]

To held to find ν, we use the conservation law Tμν;ν = 0 or Tμν;μ = 0

Now[pic]

and Tμν;μ = Tμν,μ + Γμσμ Tσν − Γσνμ Tμσ

Only ν = 1 gives a non-trivial equations

0. = Tμ1,μ + Γμσμ Tσ1 − Γσ1μ Tμσ

1. ’ T11,1 + [Γ111 + Γ122 + Γ133 + Γ144]T11 −[Γ111 T11+ Γ112 T21+ Γ113 T31+ Γ114 T41]

i.e. [pic]

To proceed further, we have to know the equation of state, ie the relation between p and ρ

Simplest assumption: ρ ’ constant (incompressible fluid)

[then our previous equations are

[pic] [pic]

and (3) became

[pic] [pic]

Integrating (4), -v/2 = ln(P + ρc2) + const

Or (P + ρc2) = ke−ν/2, r < a where k = const

We find k from the boundary conditions, knowing that

eν = 1 – 2m/r r > a

At r = a, we require eν to be continuous and P = 0.

[pic]

We now combine equations (5) and (2) to eliminate P:

[pic]

[pic]

A partial integral of this equation is [pic]

The corresponding homogenous equation is

[pic]

[pic]

[pic] where b2=const

General solution of inhomogeneous equations is

[pic]

But [pic] at r = a , from the ??? solution

[pic] [pic]

Hence from (5) we find

[pic] [pic]

Notes

1. when m/a > 0, (ie v real) or P >> 0, we just have

[pic] for all [pic]

∴[pic]1 ∴[pic]

So an incompressible spherical source must have its Schwarzschild radius inside the sphere itself

3. Inside source, [pic]

Hence embedding arguments give

[pic] z(0)=0

i.e. [pic]

Since this is an equation of circle centered on r = 0,z=[pic]we deduce that embedding surface for r < a is spherical.

Experimental Tests of GR [TUTORIAL MATERIAL]

A. Planetary Motion

Assuming can approximate a planet as a particle (ie ignore any distance ?? to Schwarzschild metric round sun due to the planet itself), motion of planet follows a time-like geodesic:

[pic] xλ ’ r, θ, φ, ct

([pic]

From (2) we note that a solution exists for which θ ’ π/2

[Similarly, (3) has a solution, φ = const] These are orbits in a plane passing through the origin, ie motion is confined to a plane

Without any lack of generality we’ll take θ ’ π/2.

Then (3) ([pic]

i.e. [pic]

i.e. [pic],where h=const.(??? of angular momentum)

(4) ([pic]

[pic]

[pic] where k=const ( ??? of energy)

Instead of taking (1), we use the particle integral

[pic]

(setting [pic])

[Eq (1) is now obtained by differentiating with regard to r, then substituting for [pic]]

Now [pic](obviously we take the exterior solution)

[pic]

[pic]

Now let 1/r=u [pic]

[pic]

Physical Significance of k

[pic] now [pic]

[pic]

For 2m/r = 1 ( unbound states

Imaginary when kc < 1 ( bound states

b. V -> c as r -> 2m, provided source is a black hole (regardless of value of [pic])

Solutions of orbit equation:

Classically, we would write KE + PE = const

i.e. [pic] M=mass of source

i.e. [pic] m=GM/r

Also [pic] (analogue momentum constant) (note h has meaning analogue to previous h)

Now use [pic]

=> [pic]

Now let r = 1/u , [pic] => [pic]

Rel. equation has extra factor 3mu2 on RHS

Solution of classical equation [pic] e=const

For e < 1, this is an ellipse, with eccentricity e and re ? angle ω. Obviously r has min value when φ – ω ’ 0, 2π, 4π … ??

Actually e is related to the const k - easy to show that

[pic]

So e < 1 implies kc < 1 and vice versa.

Solution of GR equ

Assume 3mu2 r/c > r1, m/r2 negligible

[pic] r1 ~ radius of sun

But note complications due to v/c term – often will be of same magnitude as m/r1. If taken v = 0, get [pic] (red shift)

Best experiment on redshift was by

Brault (1962) – measured sodium D1 line emitted by sun – agreement to 5% (see note on this in Misner at all, p1058-9)

White dwarf stars have large m/r value, but difficult to estimate independent values of mass to radius

Altogether, whole field is rather messy.

2. Earth-bound (Laboratory) Experiments

Let r1 – r2 = l, l 0 blue shift

L < 0 read shift

Redshift experiment of highest precision was Pound + Rebka (196)) who used 144 Kev γ-ray, see Misner P1056, McVittie p100, with “observer” observing the radiation, using the Mossbauer effect. Distance l used was 22.5m, moving source was alternatively placed ?? above & below observer, so that[pic]

Theoretical value = 4.93 x 10-15

Experimental value = (5.13 ± 0.51) x 10-15

Alternative Derivation of ν2/ν1 = (1 + gl/c2) by Principle of Equivalence

(Argument first given by Einstein, 1911)

Higher ray frequency

Tube accelerates to right with acceleration g with regard to metrical frame. Initially velocity of tube > zero. Light ray of frequency emitted as shown.

Tube to reach other end is ~ l/c

But then, tube has velocity gl/c

By SR Doppler effect, observed frequency is ν2 ∼ ν1 (1 + velocity/c)

ν2/ν1 = (1 + gl/c2)

Second Alternative Proof of ν2/ν1 = (1 + gl/c2)

Photon, frequency ν1, has effective mass given by hν1 = m*c2, m* effective mass. Following through l, gains energy m*gl = hν1/c2 gl

hν2 = hν1 + hν1gl /c2

ν2/ν1 = (1 + gl/c2)

The Hafele – Keating ? Experiment

Two identical clocks, A & B, one kept at rest in a lab, the other transported around the earth at height h (above sea-level) and speed v (relative to surface).

Synchronize clocks at start and compare again when 2nd clock returns.

For simplicity assume travel around equator, so θ ’ π/2. During journey dr = dθ = 0. Let R = radius of earth, ω = angular rotation of earth.

Note: lab clock as well as flying clock is moving with respect to the underlying system of reference coordinate to (r, θ, φ, t)

[Assume rotation of earth, being slow, has no effect on the Schwarzschild solution]

Lab Clock: element of proper time dτ given by

[pic]

Now r = R , [pic], so

[pic]

Where TA = agreed time for lab clock A

t = agreed time according to ‘coordinate clocks”

Note:

The factor -2m/r within the [pic] referes to the slowing of clocks placed in a gravitational field, as previously discussed

The factor [pic] refers to the SR result that moving clocks run slow

Traveling Clock

By similar argument to above

[pic]

Here r = R + h , and [pic] [by “relevant angular velocity”]

∴ [pic] [wh ~ 1ms-1, neglible if ν)

∴ [pic] = elapsed time ? to traveling clock

Fractional ?? increase in round-the-world journey time is

[pic]

In 1st order , where g=GM/R2

For h = 104, = + 300 ms-1, RHS = -1.0 x 10-12 east-bound

h = 104, = - 300 ms-1, RHS = 2.1x 10-12 west-bound

If fractional accuracy of modern atomic clock is ~ 10-13

1971: Hafele – Heating took atomic clocks round the world on economic ? flights – not exactly equatorial, so they ? modified

Theory Experiment

Results for TB – TA : West-bound 275 ±21 273 ± 7

Eest-bound -40 ±23 -59 ±10

Radar Echoes from Planets

(Shapiro Delay)

Radar signal from earth to Venus & return. Measure time for round-trip. This round-trip time measures slightly if the signal passes close to the sun.

Theory: light signal satisfies ds = 0. Choose coordinates so that motion is in the plane θ ’ π/2, then

[pic]

Or [pic]

Now,

[pic]

At point of closes approach, r = r0 say , [pic]

[pic]

So [pic]

∴

[pic]

Coordinate time for light signal to travel from r0 to r is

[pic]

If electromagnetic signals traveled in straight lines, all m/r & m/r0 terms would vanish and the result would be

[pic]

Which is obviously correct. Now we expand integrand to 1st order in m/r and m/r0!

[pic]

[pic]

Which eventually simplifies (exercise) to

[pic]

1st term in the integrand is the classical term, so

[pic]

Now

[pic]

[pic]

Now signal was to travel from earth at (r = rE) to Venus (at r = rv) and return. The overall time delay (compared with the time when the sun is not close to path) is

[pic]

But rE, rV >> r0

[pic]

Now rE ~ 150 x106 km, rV ~ 108 x106 km, m = 1.48 km

So if radar signal just grazes sun, r = R = 6.96 x105 km

( Δt = 2.5 x 10-4 s.

Experiment carried out by Shapiro et all in 1970 – effect confirmed with uncertainty of 3%. Hence the “Shapiro Delay”

See also Nature (2001) 412, 158 – Observation on binary pulsar which demonstrates same effect

Black Holes

Final stage in evolutions of some stars, ie those with mass >~ 3 solar masses.

Can also have super-massive black holes believed to exist at the centre of every galaxy – with masses ~ 109 solar mass.

“Active” black holes devour nearby matter – quasars. Also have inactive ones.

Consider a radial journey into a black hole, by a free particle, θ = const, φ = const, so the geodesic equation (see above) gives the non-trivial equs

[pic] and [pic]

i.e [pic]

Hence

[pic]

Take the – sign for approaching a black hole, and near r = 2m

Let r = 2m + ε where ε is small

[pic]

So as r -> 2m, t ->∞

ie the object taken on infinite coordinate time to reach the Schwarzschild radius!

Velocity as r -> 2m?

We had [pic], so v -> c as r -> 2m

Frequency shift as r -> 2m?

[pic] For v outwards

Let invward bound observer, at position r (decreasing) send periodic signal to distant observer at r2 ≈∞

[pic]

As r -> 2m, v -> c, so [pic]

All frequencies received by distant observer -> O ,so signals gradually fade out

Time of journey o/c to traveler?

Traveller measures ds/c

Now [pic] at r(2m.

Time taken to reach Schwarzschild radius r = 2m is finite; the exact value is

[pic]

Fate of objects inside r = 2m?

[pic] “exterior” solution

Impossible for any object for r < 2m to be stationary, since dr = dθ = dφ = 0, implies ds2 < 0, contradicting the requirement that particles should ? only time-like ? [and can’t avoid them by invoking the “interior” solutions, since ev must be continours] in space-time

All free particles must continue their journey towards r = 0 – the big crunch, ? density become infinite (Penrose, Howting ? 1965 – 70) and GR breaks down.

Can show that it takes about 10-5 s for a particle to travel from r = 2m to r = 0, for a typical black hole.

Summary:

A non-rotatory black hole consists of a singularity at r = 0 where density =∞ , surrounded by a Schwarzschild horizon at r = 2m.

Any non-rotatory star which suffies gravitational collapses to a black hole must end up as perfectly spherical black hole, no matter what its initial shape and initial structure.

There are also rotatory black holes (Kerr black holes) – ultimate fate of certain rotatory stars – which are not spherically symmetric but exhibit on equatorial bulge (like the earth)

D.

-----------------------

x, y, z

dl

x + dx, y + dy, z + dz

A

B

dl

O

P

τ

xoλ

xλ

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