General Relativity
Lecture notes from Phys415

Niklas Karlsen
University of Victoria, Canada

May 4, 2003

Contents

• Space-time
  • Inertial reference frames
  • Events, world lines
  
  
• The Lorentz transformation
  • Transformation of length and time
  • Transformation of speed
  
  
• Relativistic mechanics
  • Energy and momentum
  
  
• Introduction to general relativity
  • Gravity
  • The principle of equivalence
  
  
• Tensor Calculus
  • Higher rank tensors
  • Symmetry
  • Addition & subtraction
  • Contraction
  • Quotient law
  • Christoffel symbols
  • Parallell propagation
  • The Riemann tensor
  • Some useful tensor-relations
  • The Einstein field equations
  
  
• Applications of GR
  • Empty space
  • Gravity around the sun
  
  
• Tests of GR
  • Mercurys precession
  • The gravitational red-shift
  • 3rd test
  • New 4th test of GR
  
  
• Gravitational Collapse & Black Holes
• The Universe
  • Static universe
  • Non-static Models, ($\Lambda = 0$)
  
  
• Bibliography
• About this document ...

Space-time

Inertial reference frames

These are ones in which a free body moves with constant velocity. Newtons laws hold in inertial reference frames. Note: In newtonian physics gravitational effects proceed instantaneously, but there is a speed limit, c, in nature!

Newtons theory of relativity: physics is the same in all reference frames.
Einstens principle of relativity: newton + c

NB: The speed limit c is the same inn all inertial frames.

Events, world lines

$S_{12}$ is called the ``space-time interval''. It's invariant:

$$ds^2 = c^2dt^2 - dl^2$$ (1)

Real ``spacetime-intervals'' are called time-like (they are separated in time.) Imaginary ``spacetime-intervals'' are called space-like (they are separated in space.) AND they are not causally linked. See fig. 1

Figure 1: The spacetime of Minkowski.

The Lorentz transformation

Consider,

$$x_{0'} = 0 + ct'\sinh{\phi}$$ (2)

$$(ct)_{0'} = 0 + ct'\cosh{\phi}$$ (3)

where the primed variables refer to a primed reference system K' moving with a velocity V away from K, our restframe. The K' frame has its origin at the origin of K at t=t'=0. Now we divide eq.  2 with eq.  3 to get:

$$\frac{x_{0'}}{(ct)_{0'}} = \frac{V}{c} = \frac{\sinh{\phi}}{\cosh{\phi}}$$ (4)

If we square eq.  4 and use that ${\sinh^2{\phi}} - {\cosh^2{\phi}} = 1$ we get:

$$\sinh{\phi} = \pm \sqrt{ \frac{ \frac{V^2}{c^2} }{ 1 - \frac{V^2}{c^2} } }$$ (5)

$$\cosh{\phi} = \pm \sqrt{ \frac{ 1 }{ 1 - \frac{V^2}{c^2} } }$$ (6)

If we substitue this back into eq.  2 and eq.  3 for the general situation, we get the Lorentz transformations:

$$x = \frac{ x' + Vt' }{ 1 - \frac{V^2}{c^2} }$$ (7)

$$y = y'$$ (8)

$$z = z'$$ (9)

$$ct = \frac{ ct' + \frac{ Vx' }{ c } }{ 1 - \frac{V^2}{c^2} }$$ (10)

Transformation of length and time

$$L_0 = \frac{ L }{ 1 - \frac{V^2}{c^2} }$$

$$T_0 = T\sqrt{ 1 - \frac{V^2}{c^2} }$$

Transformation of speed

$$v_x = \frac{ v_x' + V }{ 1 + \frac{Vv_x'}{c^2}}$$

$$v_y = v_y'\frac{ \sqrt{ 1 - \frac{V^2}{c^2}} }{ 1 + \frac{Vv_x'}{c^2}}$$

$$v_z = v_z'\frac{ \sqrt{ 1 - \frac{V^2}{c^2}} }{ 1 + \frac{Vv_x'}{c^2}}$$

Relativistic mechanics

Great law of physics:

For every physical system, there is an invariant integral, S (=action), such that when the system changes configuration, $1\rightarrow2$, it does so by extremizing this action.

$$\delta S = 0$$ (11)

When S is written as a time integral, then it is called the Lagrangian, L.

$$S = \int{Ldt}$$ (12)

In special relativity, for a free particle, the invariant is ds.

$$S = K\int{ds}$$ (13)

We can write eq.  1 as:

$$ds^2 = c^2dt^2(1 - \frac{V^2}{c^2})$$ (14)

We see then that from eq.  12, eq.  13 and eq.  14 that the Lagrangian is:

$$L = Kc\sqrt{ 1 - \frac{V^2}{c^2}}$$

What is K? We know that when $v \ll c$ the laws should return to the classical form. This gives us the final solution for the Lagrangian:

$$L = -mc^2\sqrt{1 - \frac{V^2}{c^2}}$$ (15)

Energy and momentum

$$\vec{p} = m_0\vec{v}$$

where $m_0$ is the relativistic mass. If we set c=1 we have :

$$E^2 = \vec{p}^2 + m^2$$

$$\frac{E\vec{v}}{c^2} = \vec{p}$$

Introduction to general relativity

Gravity

Consider the strength of gravity

• Does gravity play a role in atomic physics? If we calculate the ratio of the gravitational force to the electromagnetic force we get that .
• For the nucleus we get .
• In the cohesion of matter and bulk? Consider the energy of the interatomic bond; find it by melting the material, remember . For sulphur $T_{melt} = 385$K. Then J, while gravity has a strength of $10^{-50}$J for $r = 10^{-10}$m.

Gravity matters on the large scale.

• Strong and weak interactions go to zero very quickly with distance. But gravity is a long range force.
• But the electrostatic force is also $\frac{1}{r^2}$, but matter in bulk tends to be neutral; and gravity only has positive mass.

The principle of equivalence

A gravitational field is locally equivalent to a accelerated reference frame.

``Local'' because differences in gravitational attraction can be detected on large scales. When we are in an inertial reference frame, we have no gravity, then the spacetime interval is:

$$ds^2 = c^2dt^2 - dl^2$$ (16)

Now we make phony gravity by accelerating our reference frame, we move into a rotating reference frame with the coordinate transformation:

$$\begin{eqnarray*} x = x' \cos \omega t + y' \sin \omega t \\ y = -x' \sin \omega t + y' \cos \omega t \\ t = t' \end{eqnarray*}$$

Then the spacetime interval is:

$$ds^2 = c^2dt'^2(1 - \frac{\omega^2}{c^2}(x'^2 + y'^2)) - dx'^2 - dy'^2 - dz'^2 + 2\omega y' dx' dt' - 2\omega x' dy' dt'$$

which in short form, can be written:

$$\sum_{i,k=0}^zg_{ik}dx^idx^k$$ (17)

Einstein:

Since phoney gravity is characterized by metric functions $g_{ik}$, since by the equivalence principle, so characterize real gravity by $g_{ik}$ as well.

Tensor Calculus

If we want to do a transformation of coordinates from $dx, dy, dz \to dr, d\theta, d\phi$ we use:

$$\begin{eqnarray*} x = r\sin\theta\cos\phi \\ y = r\sin\theta\sin\phi \\ z = r\cos\theta \end{eqnarray*}$$

Generally, if we want to transform from a set of old coordinates $x^i$ to $\bar{x}^i$ we can write:

$$d\bar{x}^i = \sum_{k=0}^3\frac{\partial\bar{x}^i}{\partial x^k}dx^k$$ (18)

We will call any set of functions $A^0, A^1, A^2$ and $A^3$ the components of a contravariant vector, if the coordinattransformation goes as in eq.  18. We write a contravariant vector transformation as:

$$\bar{A}^i = \frac{\partial\bar{x}^i}{\partial x^k} A^k$$ (19)

Consider, $\nabla\varphi$, and change the coordinate system to $\bar{x}$:

$$\frac{\partial\varphi}{\partial\bar{x}^k} = \frac{\partial\... ... {\partial x^1}\frac{\partial x^1}{\partial\bar{x}^k} + \ldots$$

or

$$\frac{\partial\varphi}{\partial\bar{x}^k} = \frac{\partial\varphi} {\partial x^i}\frac{\partial x^i}{\partial\bar{x}^k}$$ (20)

This is a covariant vector transformation. Note, the partials are reversed: in eq.  19

$$\frac{\partial\bar{x}^i}{\partial x^k}$$

and in eq.  20

$$\frac{\partial x^i}{\partial\bar{x}^k}$$

Definition 5.0.1   Any set of 4 functions $B_i$, which under a coordinate transformation changes just like in eq.  20, is called a covariant vector.

Consider two vectors $A^i$ and $B_i$ and form $\bar{A}^i\bar{B}_i$:

$$\begin{eqnarray*} \bar{A}^i\bar{B}_i = A^k\frac{\partial\bar{x}^i}{\partial x^k... ...partial x^l}{\partial\bar{x}^i} \\ = A^kB_l\delta^l_k = A^kB_k \end{eqnarray*}$$

So $\bar{A}^i\bar{B}_i$ is an invariant, or a scalar.

Higher rank tensors

Consider taking $A^i$ and $B^k$ and evaluating

$$A^iB^k = \left( \begin{array}{ccc} A^0B^0 & A^0B^1 & \ld... ... \vdots & \vdots & \ddots \end{array} \right) \equiv C^{ik}$$

How does $C^{ik}$ transform?

$$\begin{eqnarray*} \bar{A}^i = A^l \frac{\partial\bar{x}^i}{\partial x^l} \\ \bar{B}^k = B^m \frac{\partial\bar{x}^k}{\partial x^m} \end{eqnarray*}$$

So

$$\bar{C}^{ik} = \bar{A}^i\bar{B}^k = A^lB^m\frac{\partial\ba... ...r{x}^i} {\partial x^l} \frac{\partial\bar{x}^k}{\partial x^m}$$

Call $C^{ik}$ a second rank contravariant tensor. It is a set of 16 functions that transform as above. Similarly, if we have two covariant vectors $D_i$ and $E_k$, you can write an outer product $D_iE_k$, which is a set of 16 functions that transform as the products of $D_i$ and $E_k$, i.e. let $D_iE_k = F_{ik}$. This transforms as:

$$\bar{F}_{ik} = F_{lm}\frac{\partial x^l}{\partial\bar{x}^i} \frac{\partial x^m}{\partial\bar{x}^k}$$

This is the transformation law for a covariant tensor of rank 2 ¹.

Consider a covariant vector $A_i$ and a contravariant vector $B^k$, and form:

$$A_iB^k = G_i^{\phantom{1}k}$$

This transforms as:

$$\bar{G}_i^{\phantom{1}k} = G_l^{\phantom{1}m}\frac{\partial x^l} {\partial\bar{x}^i}\frac{\partial\bar{x}^k}{\partial x^m}$$

This is a mixed second rank tensor.

$D^{ik}_{\phantom{ik}lmn}$ is a 5th rank tensor. It transforms as:

$$\bar{D}^{ik}_{\phantom{ik}lmn} = D^{ab}_{\phantom{ab}cde} ... ...^d}{\partial\bar{x}^m} \frac{\partial x^e}{\partial\bar{x}^n}$$

Symmetry

In general $C^{ik} \ne C^{ki}$. But suppose,

$$C^{ik} = C^{ki}$$

Then $C^{ik}$ is called symmetric².

Proof:

Suppose $C^{ik} = C^{ki}$ in the $x^l$-coordinate system. Then change to the new coordinate system $\bar{x}^m$. In the new coordinate system we get:

$$\bar{C}^{rs} = C^{ik}\frac{\partial\bar{x}^r}{\partial x^i} \frac{\partial\bar{x}^s}{\partial x^k}$$

We can change the dummy summation indices, such that:

$$= C^{ki}\frac{\partial\bar{x}^r}{\partial x^k} \frac{\partial\bar{x}^s}{\partial x^i}$$

Now we use the symmetry property, eq.  5.2:

$$\begin{eqnarray*} = C^{ik}\frac{\partial\bar{x}^r}{\partial x^k} \frac{\parti... ...^i} \frac{\partial\bar{x}^r}{\partial x^k} \\ = \bar{C}^{sr} \end{eqnarray*}$$

So

$$\bar{C}^{rs} = \bar{C}^{sr}$$

We have shown what we wanted to prove.

Similarly, $C^{ik} = -C^{ki}$, is called antisymmetric, or skew-symmetric.

Addition & subtraction

Consider two tensors $A^{ik}, B_l$ and make the sum . But this is no good, since they have different laws of transformation. But, say is fine. Consider

$$\begin{eqnarray*} A_{ik} = \frac{1}{2} A_{ik} + \frac{1}{2} A_{ik} = \frac{1}... ... + A_{ki}) + \frac{1}{2}(A_{ik} - A_{ki}) = A_{(ik)} + A_{[ki]} \end{eqnarray*}$$

That is, a sum of a symmetric and an antisymmetric part.

Contraction

Consider a $3^rd$ rank tensor:

$$\bar{A}^{rs}_{\phantom{rs}t} = A^{ik}_{\phantom{ik}l} \fra... ...r{x}^s}{\partial x^k} \frac{\partial x^l}{\partial\bar{x}^t}$$

Put s=t and sum:

$$\begin{eqnarray*} \bar{A}^{rs}_{\phantom{rs}s} = A^{ik}_{\phantom{ik}l} \frac{... ... = A^{ik}_{\phantom{ik}k}\frac{\partial\bar{x}^r}{\partial x^i} \end{eqnarray*}$$

This is a contravariant vector. Contraction reduces rank by 2.

Quotient law

Let $A^{ik}$ be a tensor. And suppose , where $C_l$ is a good covariant vector. You do not know what $B_{ikl}$ is.

Quotient law 5.5.1   $B_{ikl}$ is what you think it is

Proof:

$$\bar{A}^{ik}\bar{B}_{ikl} = \bar{C}_l$$

left hand side:

$$A^{rs}\frac{\partial\bar{x}^i}{\partial x^r} \frac{\partial\bar{x}^k}{\partial x^s}\bar{B}_{ikl} =$$

right hand side:

$$\begin{eqnarray*} C_t\frac{\partial x^t}{\partial\bar{x}^l} \\ = A^{rs}B_{rst}\frac{\partial x^t}{\partial\bar{x}^l} \end{eqnarray*}$$

So

$$A^{rs} \left[ \bar{B}_{ikl}\frac{\partial\bar{x}^i}{\partia... ... - B_{rst}\frac{\partial x^t} {\partial\bar{x}^l} \right] = 0$$

Since $A^{rs}$ is arbitrary, we could have picked $A^{12} \ne 0$ and let all the other components equal 0. To satisfy the equation, we have to have the bracket equal zero. Muliply the bracket with $\frac{\partial\bar{x}^l}{\partial x^m}$.

$$\begin{eqnarray*} \bar{B}_{ikl}\frac{\partial\bar{x}^i}{\partial x^r} \frac{\p... ...\bar{x}^l}{\partial x^m} \\ = B_{rst}\delta^t_m \\ = B_{rsm} \end{eqnarray*}$$

$B_{rsm}$ is a $3^{rd}$ rank covariant tensor. So $B_{ikl}$ is a good $3^{rd}$ rank covariant tensor. As we wanted to show.

Application

Recall $ds^2 = g_{ik}dx^idx^k$. The first term is a scalar, and $dx^idx^k$ is a good contravariant symmetric $2^{nd}$ rank tensor. According to the quotient law $g_{ik} + g_{ki}$ is a good tensor. Consider $g_{ik}dx^idx^k = g_{(ik)}dx^idx^k + g_{[ik]}dx^idx^k$. Let

$$g_{[ik]} \equiv A_{ik} = \left( \begin{array}{cccc} 0 & ... ... 0 & \ldots \\ \vdots & \vdots & \ddots \end{array} \right)$$

If we calculate the product $A_{ik}dx^idx^k$ we get 0. Since $g_{[ik]}$ contributes nothing to $ds^2$, we might as well take $g_{ik}$ to be symmetric.

Now we introduce the conjugate metric tensor, $g^{il}$, defined by

$$g_{ik}g^{il} \equiv \delta_k^l$$

From $A^k$ we form a new vector $A_i$, . We call this the associate vector of $A^k$. Similarly, from $B_i$ form $B^k$, .

From classical physics, form a good 4-vector (4-velocity). We do the coordinate transformation:

$$\frac{d\bar{x}^k}{ds} = \frac{dx^i}{ds} \frac{\partial\bar{x}^k} {\partial x^i}$$

This is the transformation of 4-velocity. To find the acceleration we take $\frac{d}{ds}$ of the 4-velocity:

$$\frac{d^2\bar{x}^k}{ds^2} = \frac{d^2 {x}^i}{ds^2}\frac{\pa... ...artial^2\bar{x}^k}{\partial x^i \partial x^l}\frac{dx^l}{ds})$$

Because, of the second term $\frac{d^2\bar{x}^k}{ds^2}$ is not a good vector. Consider the partial derivatives of $A^i$.

$$\frac{\partial A^i}{\partial x^l} = \frac{\partial\bar{A}^k... ...^k \partial\bar{x}^m} \frac{\partial\bar{x}^m} {\partial x^l}$$ (21)

Here also, the second term spoils $\frac{\partial A^i}{\partial x^l}$ from being a good $2^nd$ rank mixed tensor.

Christoffel symbols

Define Christoffel symbol of the first kind.

$$[ij,k]\equiv \frac{1}{2}(\frac{\partial g_{ik}}{\partial x^j}... ..._{jk}}{\partial x^i} - \frac{\partial g_{ij}} {\partial x^k})$$ (22)

Define Christoffel symbol of second kind.

$$\left[ \begin{array}{c} l \\ ij \end{array} \right] = \Gamma_{ij}^l = g^{kl}[ij,k]$$ (23)

How do these change when we change from one coordinate system to another? It can be shown that eq.  22 transforms as

$$\overline{[lm,n]} = [ij,k]\frac{\partial x^i}{\partial\bar{... ...al\bar{x}^n} \frac{\partial^2x^j}{\partial\bar{x}^l\bar{x}^m}$$

and eq.  23 transforms as

$$\bar{\Gamma}_{lm}^p = \Gamma_{ij}^s\frac{\partial x^i}{\par... ...}^l \partial\bar{x}^m} \frac{\partial\bar{x}^p}{\partial x^s}$$

We solve this equation to get the second derivate by itself. We start by multiplying both sides with $\frac{\partial x^k} {\partial\bar{x}^p}$:

$$\bar{\Gamma}_{lm}^p\frac{\partial x^k}{\partial\bar{x}^p} - \Gamm... ...} \frac{\partial\bar{x}^p}{\partial x^s}\frac{\partial x^k} {\partial\bar{x}^p}$$ (24)

$$\frac{\partial^2x^s}{\partial\bar{x}^l \partial\bar{x}^m} = \bar{... ...^k\frac{\partial x^i}{\partial\bar{x}^l} \frac{\partial x^j}{\partial\bar{x}^m}$$ (25)

We put this into eq.  21, and do some change of dummy summation indices:

$$\begin{eqnarray*} \frac{\partial A^i}{\partial x^l} = \frac{\partial\bar{A}^k... ...}{\partial\bar{x}^p} \frac{\partial\bar{x}^m}{\partial x^l} \\ \end{eqnarray*}$$

$$\frac{\partial A^i}{\partial x^l} + \Gamma_{cl}^iA^c = (\frac{\pa... ...p)\frac{\partial x^i}{\partial\bar{x}^p} \frac{\partial\bar{x}^m}{\partial x^l}$$ (26)

Eq.  26 is a good transformation of a $2^{nd}$ rank mixed tensor. Also is a good contravariant vector.

A free particle in special relativity moves with no acceleration, $\frac{d^2x^i}{ds^2} = 0$. To see how this is related to another coordinate system we do the transformation:

$$\begin{eqnarray*} \frac{d^2\bar{x}^k}{ds^2} + \bar{\Gamma}_{lm}^k\frac{d\bar{x}... ...\footnotemark ) \frac{\partial\bar{x}^k}{\partial x^i} \\ = 0 \end{eqnarray*}$$

So in a new coordinate system in SR, the equation of motion is

$$\frac{d^2\bar{x}^k}{ds^2} + \bar{\Gamma}_{lm}^k\frac{d\bar{x}^l}{ds} \frac{d\bar{x}^m}{ds} = 0$$ (27)

Since Einstein recognized that locally, gravity is like an accelerated reference frame, this is also the equation of motion of a free particle in GR, i.e real gravity as well.

Some remarks

$$\begin{eqnarray*}[ij,k]= [ji,k] \\ \Gamma_{ij}^l = \Gamma_{ji}^l \end{eqnarray*}$$

We now introduce a writing convention. We write for the covariant derivative of $A^i$ (we use a comma whenever we do the partial derivative). is the covariant derivative of a covariant vector.

Find the covariant derivative of the metric tensor.

$$\begin{eqnarray*} g_{ik;l} = \frac{\partial g_{ik}}{\partial x^l} - \Gamma_{i... ..._{il,k}) - \frac{1}{2}(g_{ik,l} + g_{il,k} - g_{kl,i}) \\ = 0 \end{eqnarray*}$$

So .

Properties

1. 
2. 

What about covariant derivative of a scalar? Consider $\frac{\partial f}{\partial x^i}$.

$$\frac{\partial f}{\partial\bar{x}^k} = \frac{\partial f}{\partial x^i}\frac{\partial x^i}{\partial\bar{x}^k}$$

Since $\frac{\partial f}{\partial x^i}$ is allready a good covariant vector, define $f_{;k} \equiv \frac{\partial f}{\partial x^i}$.

Consider , called the intrinsic derivative, and here it is a good $2^{nd}$ rank tensor.
eg.

$$B_{\phantom{i};k}^{i}\frac{dx^k}{ds} = \frac{D}{ds}B^i$$

We work it out:

$$\begin{eqnarray*} B_{\phantom{i};k}^i = \frac{\partial B^i}{\partial x^k} - \G... ...c{D}{ds}B^i = \frac{dB^i}{ds} - \Gamma_{kl}^iB^l\frac{dx^k}{ds} \end{eqnarray*}$$

Wonderful application: Let $B^i$ be $\frac{dx^i}{ds}$.

$$\frac{D}{ds}u^i = \frac{du^i}{ds} + \Gamma_{lk}^iu^lu^k$$

This is good acceleration, and equals zero for a free particle.

Parallell propagation

In cartesian coordinates the parallell propagation of a vector is expressed as . We go to a new coordinate system $\bar{x}^k$. Then in $\bar{x}^k$ coordinates the vector is . Consider

$$\frac{d\bar{A}^k}{ds} = \frac{dA^i}{ds}\frac{\partial\bar{x... ...artial x^i} + A^i \frac{d^2\bar{x}^k}{dx^ldx^i}\frac{dx^l}{ds}$$

We replace in eq.  25, and think about what $\Gamma_{bc}^a$ is in the original coordinate system, i.e. flat space. Then we get

$$\frac{D\bar{A}^k}{ds} = 0$$

Properties

1. If a vector is taken for a parallell walk, the length stays the same.
   Proof
   $$\begin{eqnarray*} \frac{d}{ds}A^2 = \frac{d}{ds}g_{ik}A^iA^k = \frac{D}{ds}g_{i... ...{ds} = 0 \\ \Rightarrow A \ne 0 \Rightarrow \frac{dA}{ds} = 0 \end{eqnarray*}$$
   
   
   
   
   

2. Also the angle between two vectors that are parallelly propagated stays fixed.
   
   $$A^iB_i = AB\cos\theta$$
   
   

In 3 dimensional flat space we can describe the space time intervall in polar coordinate like so,

$$ds^2 = dr^2 + r^2d\theta^2 + r^2\sin^2\theta d\varphi^2$$

Take r=constant=1. Then we get

$$ds^2 = d\theta^2 + sin^2\theta d\varphi^2$$

This is 2 dimensional curved space. It is not euclid! Lets work here. Let . The metric tensor then is

$$\begin{eqnarray*} g_{11} = 1, g_{22} = \sin^2x^1 \\ g_{12} = g_{21} = 0 \\ ... ...{2\sin^2x^1}(2\sin x^1\cos x^1) \ = \cot x^1 = \Gamma_{21}^2 \end{eqnarray*}$$

To summarize, the only gammas that are not zero are

$$\begin{eqnarray*} \Gamma_{21}^2 = \Gamma_{12}^2 = \cot x^1 \\ \Gamma_{22}^1 = -\sin x^1\cos x^1 \end{eqnarray*}$$

Let's take a vector for a parallell walk in this 2-space.

$$\frac{DA^i}{ds} = 0 = \frac{dA^i}{ds} + \Gamma_{kl}^iA^k\frac{dx^l}{ds}$$

We let $\theta = constant$ for the path we choose. We then get

$$\begin{eqnarray*} 0 = \frac{dA^1}{ds} + \Gamma_{kl}^1A^k\frac{dx^l}{ds} \\ 0 ... ... 0 = \frac{dA^2}{d\phi} + \cot\alpha A^1 \end{array} \right] \end{eqnarray*}$$

We need to decouple the two equations,

$$\begin{eqnarray*} 0 = \frac{d^2A^2}{d\phi^2} + \cot\alpha\frac{dA^1}{d\phi} \ ... ...a\cos\alpha A^2 \\ = \frac{d^2A^2}{d\phi^2} + \cos^2\alpha A^2 \end{eqnarray*}$$

This has the solution

$$A^2 = C\sin(\cos\alpha\phi) + D\cos(\cos\alpha\phi)$$

Let $A^i = (1,0)$ at $\phi = 0$.

$$\begin{eqnarray*} \Rightarrow 0 = D \\ A^2 = C\sin(\cos\alpha\phi) \\ 0 = C\cos(\cos\alpha\phi)\cos\alpha + \cot\alpha A^1 \end{eqnarray*}$$

$$\begin{eqnarray*} 0 = C\cos\alpha\cos 0 + \cot\alpha \\ \Rightarrow C = - \fr... ...\cos(\cos\alpha\phi)\cos \alpha \\ A^1 = \cos(\cos\alpha\phi) \end{eqnarray*}$$

Lets do one orbit:

$$\begin{eqnarray*} A^1 = \cos(\cos\alpha 2\pi) \\ A^2 = -\frac{1}{\sin\alpha}(... ...orall \alpha / \{ \frac{\pi}{2}\} \mid (\alpha \in [0,2\pi)) \end{eqnarray*}$$

But for $\alpha = \frac{\pi}{2}$ we get $A^1 = 1$ and $A^2 = 0$. That it is, we get back our original vector. This path is called the geodesic. The equator curve is a geodesic (the arc of a great circle). The extremal(minimal) distance curve.

$$\frac{D}{ds}u^i = 0$$

Also in GR, this is the equation of the path of a free particle.

$$\frac{Du^i}{ds} = \frac{du^i}{ds} + \Gamma_{kl}^iu^k\frac{dx^l}{ds} = 0$$

or

$$\frac{d^2x^i}{ds^2} + \Gamma_{kl}^i\frac{dx^k}{ds}{dx^l}{ds} = 0$$

The Riemann tensor

We find the change in the components of a vector $A^i$ under a parallell displacement around an infinitesimal closed path, defined by four curves of the two parameter set . Since we have parallell transport,

$$\frac{DA^i}{ds} = 0 = \frac{dA^i}{ds} + \Gamma_{jk}^iA^j\frac{dx^k}{ds}$$

so

$$dA^i = -\Gamma_{jk}^iA^jdx^k$$

Net change in $A^i$ is

$$(dA^i)_{net} = \oint_c dA^i = -\oint_c\Gamma_{jk}^iA^jdx$$

On $c_1 + c_3$, v=constant so $dx^k = \frac{\partial x^k}{\partial u}du$. So

$$\begin{eqnarray*} dA^i \mid_{c_1 + c_3} = -\int_{c_1}\Gamma Adx -\int_{c_3}\Gam... ...ta v}A_{v + \Delta v}^j\frac{\partial x^k} {\partial u}\Delta u \end{eqnarray*}$$

So

$$dA^i\mid_{c_1 + c_3} = \frac{\partial}{\partial v}[\Gamma_{jk}^iA^j] \Delta v \frac{\partial x^k}{\partial u}\Delta u$$

Similarly, on $c_2 + c_4$

$$dA^i\mid_{c_2 + c_4} = -\Gamma_{jk}^i\mid_{u + \Delta u}A_{... ...mma_{jk}^i\mid_uA_u^j \frac{\partial x^k}{\partial v}\Delta v$$

So

$$dA_{net}^i = [\frac{\partial}{\partial v}[\Gamma_{jk}^iA^j]... ...mma_{jk}^iA^j]\frac{\partial x^k}{\partial v}]\Delta u\Delta v$$

Now , etc. So

$$\begin{eqnarray*} dA_{net}^i = \Delta u\Delta v[\Gamma_{jk,l}^i\frac{\partial x... ...\frac{\partial x^t}{\partial u}\frac{\partial x^k}{\partial v}] \end{eqnarray*}$$

or

$$dA_{net}^i = \Delta A^i = A^j \Delta u\Delta v\frac{\partia... ...^i - \Gamma_{rk}^i\Gamma_{lj}^r + \Gamma_{rl}^i\Gamma_{kj}^r]$$

Also is a vector, $A^j, \frac{\partial x^l}{\partial v}, \frac{\partial x^k} {\partial x^k}{\partial u}$ are vectors. So

$$R^i_{\phantom{i}jkl} \equiv \Gamma_{jl,k}^i - \Gamma_{jk,l}^i + \Gamma_{rk}^i\Gamma_{jl}^r - \Gamma_{rl}^i\Gamma_{jk}^r$$ (28)

is a tensor, called the Riemann tensor.

$$\Rightarrow dA^i = -dudvA^jR^i_{\phantom{i}jkl}\frac{\partial x^l}{\partial v}\frac{\partial x^k}{\partial u}$$

The Riemann tensor is key to curvature, and curvature is gravity. If we multiply the Riemann tensor 28 through by $g_{im}$, we get

$$g_{im}R^i_{\phantom{i}jkl} = R_{mjkl}$$

called the curvature tensor.

$$\Rightarrow R_{rjnp} = \frac{1}{2}(g_{rp,jn} + g_{jn.rp} - g_{rn,jp} - g_{jp,rn} + g^{ts}[[jn,s][rp,t] - [jp,s][rn,t]]$$

Properties

1. Consider switching $r \leftrightarrow j$
   
   $$R_{rjnp} = -R_{jrnp}$$
   
   

2. Consider switching $n \leftrightarrow p$
   
   $$R_{rjnp} = -R_{rjpn}$$
   
   

3. For this, make the proof easy by going to the pole of a geodesic coordinate system. There all the Christoffel symbols vanish. Do the proof there.
   
   $$R_{rjnp} + R_{rpjn} + R_{rnpj} = 0$$
   
   

   Proof:

   $$\begin{eqnarray*} = \frac{1}{2}(g_{rp,jn} + g_{jn,rp} - g_{rn,jp} - g_{jp,rn})\... ...ac{1}{2}(g_{rn,pj} + g_{pj,rn} - g_{rj,pn} - g_{pn,rj})\\ = 0 \end{eqnarray*}$$
   

   
   

   
   

4. Consider switching $rj \leftrightarrow np$
   
   $$R_{rjnp} = R_{nprj}$$
   
   

The tensor $R_{rjnp}$ has 256 components, but due to the 4 symmetries, the number of components is $\frac{N^2(N^2 - 1)}{12}$. For N=4 we get 20 components! And for N=2 we have 1 independent component!!

Some useful tensor-relations

Ricci tensor

$$R^i_{\phantom{i}jkl}$$

We contract the Riemann tensor by setting k=i

$$\Rightarrow R^i_{\phantom{i}jil} \equiv R_{jl}$$ (29)

Ricci scalar

$$R^{ij}_{\phantom{ij}kl} \rightarrow R^{ij}_{\phantom{ij}il} \rightarrow R^{il}_{\phantom{il}il} \equiv R$$ (30)

This is called the Ricci scalar.

Bianchi identities

$$R^n_{\phantom{n}ikl;m} + R^n_{\phantom{n}mk;l} + R^n_{\phantom{n}lm;k} = 0$$ (31)

To prove it, go to the pole of a geodesic coordinate system, where all the $\Gamma = 0$, but not $\Gamma_,$. Then the sum is:

$$\begin{eqnarray*} \Gamma^n_{\phantom{n}il,km} - \Gamma^n_{\phantom{n}ik,lm} + ... ...mma^n_{\phantom{n}ik,ml} - \Gamma^n_{\phantom{n}il,mk} \\ = 0 \end{eqnarray*}$$

Einstein tensor

$$\begin{eqnarray*} R_{mjnp;r} + R_{mjrn;p} + R_{mjpr;n} = 0 \mid g^{mp}g^{jr}\ ... ..._{;n} - R^{p}_{\phantom{p}n;p} = 0 \\ -2R^r_{n;r} + R_{;n} = 0 \end{eqnarray*}$$

Let

$$G^r_{\phantom{r}n} = R^r_{\phantom{r}n} - \frac{1}{2}\delta^r_nR$$ (32)

be the Einstein tensor. Consider

$$\begin{eqnarray*} G^r_{\phantom{r}n;r} = R^r_{n;r} - \frac{1}{2}\delta^r_nR_{;r} \\ = R^r_{\phantom{r}n;r} - \frac{1}{2}R_{;n} \\ = 0 \end{eqnarray*}$$

THE COVARIANT DIVERGENCE OF THE EINSTEIN TENSOR EQUALS ZERO.

The Einstein field equations

Consider

$$(\nabla^2 - \frac{1}{c^2}\frac{\partial^2}{\partial t^2}\phi) = 4\pi G\rho$$ (33)

$\rho c^2$ is the $T_0^0$ component of the energy-momentum tensor $T_i^k$. $\frac{T_0^{\alpha}}{c}$ is the linear momentum density. $T_{\alpha}^{\beta}$ expresses stresses or momentum flux density. Conservation of charge:

$$\nabla \vec{j} + \frac{\partial\rho}{\partial t} = 0$$

This can be rewritten as

$$\frac{\partial J^i}{\partial x^i} = 0$$

Conservation is expressed by the vanishing of the 4-divergence. Similarly, the conservation of energy and momentum if expressed by the vanishing of the energy-momentum tensor,

$$\frac{\partial T_i^k}{\partial x^k} = 0 \rightarrow T^{\phantom{i}k}_{i\phantom{k},k} = 0$$ (34)

For the energy and linear momentum in special relativity, this is expressed as follows in a ``bad'' coordinate system,

$$T^{\phantom{i}k}_{i\phantom{k};k} = 0$$ (35)

By the equivalence principle, use eq.  35 for energy and momentum conservation in the presence of gravity. Now we can rewrite eq.  33 as

$$? = \frac{4\pi G}{c^2}T_i^k$$

To make this a good tensor equation, we want the left hand side to carricature gravity, have $2^{nd}$ derivatives and be a $2^{nd}$ rank tensor with vanishing covariant divergence. So we use eq.  32, the Einstein tensor!

$$G_i^{\phantom{i}k} = \frac{8\pi G}{c^4}T_i^{\phantom{i}k}$$ (36)

These are the Einstein field equations.

Applications of GR

Empty space

In empty space the energy momentum tensor, $T_i^{\phantom{i}k} = 0$. Thus,

$$G_i^{\phantom{i}k} = 0$$

These are the vacuum Einstein equations. Let's rewrite them

$$R_{i}^{\phantom{i}k} - \frac{1}{2}\delta_i^kR = 0$$

Set i=k and sum,

$$\begin{eqnarray*} R_k^{\phantom{k}k} - \frac{1}{2}\delta_k^kR=0 \\ \Rightarro... ...i^{\phantom{i}k} - 0 = 0 \\ \Rightarrow R_i^{\phantom{i}k} = 0 \end{eqnarray*}$$

These are the Einstein vacuum equations expressed in terms of the Ricci tensor.

Gravity around the sun

Solve for Gravity around the sun. For spherical symmetry, we have

$$ds^2 = h(r,t)dr^2 + l(r,t)dt^2 + k(r,t)drdt + a(r,t)(d\theta^2 + \sin^2\theta d\varphi^2)$$

we do not have any cross terms include $\theta$ or $\varphi$, since we should get no change in $ds^2$ when $d\theta \rightarrow -d\theta$. Now, let $r \rightarrow r' = r'(r,t)$ and $t \rightarrow t' = t'(r,t)$. The structure of $ds^2$ stays the same. We have 2 degrees of freedom.

• use one of them to make $dr'dt' = 0$
• make the a coefficient equal $-r'^2$

These are the Schwarzschild choices. Rename $r' \rightarrow r$ and $t' \rightarrow t$.

$$ds^2 = e^{\nu (r,t)}dt^2 - e^{\lambda (r,t)}dr^2 - r^2(d\theta^2 + \sin^2\theta d\varphi^2)$$ (37)

We choose the functions in front of $dt^2$ and $dr^2$, to make the metric physical. Physical distance in r direction is no longer $dr$, but rather $e^{\frac{\lambda}{2}}dr$. Time is no longer $dt$, but rather $e^{\frac{\nu}{2}}dt$, t is not ``time''. Find $\lambda$ and $\nu$.

$$\begin{eqnarray*} 1 = r, 2 = \theta, 3 = \varphi, 0 = ct \\ g_{00} = e^{\nu},... ...tial\lambda}{\partial r} \equiv \frac{1}{2}\lambda_1 \\ \ldots \end{eqnarray*}$$

For the field around the sun ( $T_i^{\phantom{i}k} = 0$)

$$R_{i}^{\phantom{i}k} = 0$$

To find both inside and outside the sun use

$$R_i^{\phantom{i}k} - \frac{1}{2}\delta_i^kR = \frac{8\pi G}{c^4}T_i^{\phantom{i}k}$$

The solutions then become

$$\frac{8\pi G}{c^4}T_1^{\phantom{1}1} = -e^{-\lambda}(\frac{\nu_1}{r} + \frac{1}{r^2})\frac{1}{r^2}$$ (38)

$$\frac{8\pi G}{c^4}T_2^{\phantom{1}2} = \frac{8\pi G}{c^4}T_3^{\phantom{1}3}$$ (39)

$$\frac{8\pi G}{c^4}T_3^{\phantom{1}3} = -\frac{e^{-\lambda}}{2} (\... ...c{e^{-\nu}}{2}(\lambda_{00} + \frac{\lambda_0^2}{2} - \frac{\lambda_0\nu_0}{2})$$ (40)

$$\frac{8\pi G}{c^4}T_0^{\phantom{1}0} = -e^{-\lambda}(\frac{1}{r^2} - \frac{\lambda_1}{r}) + \frac{1}{r^2}$$ (41)

$$\frac{8\pi G}{c^4}T_0^{\phantom{1}1} = -e^{-\lambda}\frac{\lambda_0}{r}$$ (42)

where two indices below means $\frac{\partial^2}{\partial r^2}$. Outside of the sun $T_i^{\phantom{i}k} = 0$. Thus,

$$\begin{eqnarray*} % latex2html id marker 2327T_0^{\phantom{0}1} = 0 \Rightarr... ...rac{\lambda_1}{r}) = 0\\ \Rightarrow \nu_1 + \lambda_1 = 0 \\ \end{eqnarray*}$$

or

$$\begin{eqnarray*} \frac{\partial}{\partial r}[\nu + \lambda] = 0 \\ \Rightarr... ...lambda(r)}dt^2 - \ldots = e^{f(t)}dt^2e^{-\lambda (r)} - \ldots \end{eqnarray*}$$

Let $t \rightarrow \bar{t}$ such that $d\bar{t}^2 = e^{f(t)}dt^2$. Then eq. 37 becomes

$$ds^2 = e^{-\lambda (r)}d\bar{t}^2 - e^{\lambda (r)}dr^2 - r^2(d\theta^2 + \sin^2\theta d\varphi^2)$$

Birkhoffs theorem 1   Spherically symmetric empty space is intrinsically static.

This means that a spherically symmetric body, no matter how complicated, performing spherical pulsations, no matter how complicated (with $v<c$), maintains the same gravity picture outside, as if the mass stayed still. To solve for $\lambda = \lambda (r)$, use

$$e^{-\lambda}(\frac{1}{r^2} - \frac{\lambda_1}{r}) - \frac{1}{r^2} = 0$$ (43)

Try $e^{-\lambda} = 1 + \frac{k}{r}$.

Verify:

Take $\frac{\partial}{\partial r}: e^{-\lambda}(-\lambda_1) = -\frac{k}{r^2}$. Put this into eq. 43 and check:

$$(1 + \frac{k}{r})(\frac{1}{r^2} - \frac{1}{r}\frac{k}{r^2}e^{\lambda}) - \frac{1}{r^2} = 0$$

So $e^{-\lambda} = 1 + \frac{k}{r}$. (c=1)

Then we have

$$ds^2 = (1 + \frac{k}{r})dt^2 - \frac{dr^2}{1 + \frac{k}{r}} - r^2d\theta^2 - r^2\sin^2\theta d\varphi^2$$

Problem: get k In the Newtonian limit, we should get:

$$\begin{eqnarray*} \frac{d^2r}{dt^2} = -\nabla\varphi \\ \varphi = -\frac{GM}{r} \end{eqnarray*}$$

In GR

$$\frac{d^2x^i}{ds^2} + \Gamma_{kl}^i\frac{dx^k}{ds}\frac{dx^l}{ds} = 0$$

use this with

$$g_{ik} = \eta_{ik} + h_{ik}$$

where

$$\eta_{ik} = \left( \begin{array}{cccc} 1 & 0 & 0 & 0 \ ... ...0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{array} \right)$$

and $\mid h_{ik} \mid \ll 1$ for weak gravity. Use the comparison between this weak gravity in GR, with Newtonian gravity to get

$$k = -\frac{2GM}{c^2}$$

Then we have the Schwarzschild solution

$$ds^2 = (1 - \frac{2GM}{c^2r})c^2dt^2 - \frac{dr^2}{1 - \frac{2GM}{c^2r}} - r^2d^2\theta - r^2\sin^2\theta d\varphi$$ (44)

Tests of GR

Mercurys precession

Use the geodesic equation to see how the planets move.

$$\frac{Du^i}{ds} = \frac{d^2x^i}{ds} + \Gamma_{kl}^i\frac{dx^k}{ds}\frac{dx^l}{ds} = 0$$

The gammas are

$$\begin{eqnarray*} \Gamma_{11}^1 = \frac{\lambda_1}{2}, \Gamma_{10}^0 = \frac{\n... ...\frac{\lambda_0}{2}, \Gamma_{33}^1 = -r\sin^2\theta e^{-\lambda} \end{eqnarray*}$$

The geodesic equation then becomes, for i=2

$$\frac{d^2\theta}{ds^2} + \frac{2}{r}\frac{dr}{ds}\frac{d\theta}{ds} - \sin\theta\cos\theta (\frac{d\varphi}{ds})^2 = 0$$ (45)

i=3

$$\frac{d^2\varphi}{ds^2} + \frac{2}{r}\frac{dr}{ds}\frac{d\v... ...}{ds} + 2\cot\theta\frac{d\theta}{ds} \frac{d\varphi}{ds} = 0$$ (46)

i=0

$$\frac{d^2ct}{ds^2} - \lambda_1\frac{dr}{ds}{dct}{ds} = 0$$ (47)

We don't like the r equation. Use instead the metric itself.

$$1 = (1 - \frac{2m}{r})(\frac{dt}{ds})^2 - \frac{1}{1 - \fra... ...((\frac{d\theta}{ds})^2 + \sin^2\theta(\frac{d\varphi}{ds})^2)$$

Without loss of generality, let the planet move in the $\theta=\frac{\pi}{2}$ plane initially. So initially $\theta = \frac{\pi}{2} \Rightarrow \frac{d\theta}{ds} = 0$. Then eq. 45 becomes:

$$\begin{eqnarray*} \frac{d^2\theta}{ds^2} + \frac{2}{r}\frac{dr}{ds}\cdot 0 - 0\... ...s})^2 = 0 \\ \Rightarrow \frac{d^2\theta}{ds^2} = 0, initially \end{eqnarray*}$$

According to the mathematicians, we repeat by taking the derivatives of the $\theta$ equation to find:

$$\frac{d^n\theta}{ds^n} = 0$$

Which means that $\frac{d\theta}{ds}$ stays zero. Just as in classical mechanics, the planets move in a plane. Then eq. 46 becomes, after we have multiplied with $r^2$:

$$\begin{eqnarray*} r^2\frac{d^2\varphi}{ds^2} + 2r\frac{dr}{ds}\frac{d\varphi}{d... ... = 0 \\ \Rightarrow r^2\frac{d\varphi}{ds} = constant \equiv l \end{eqnarray*}$$

Which we recognize as the spin. Multiplying eq. 47 with $1 - \frac{2m}{r}$, it reduces to:

$$\begin{eqnarray*} \frac{d}{ds}([1 - \frac{2m}{r}]\frac{dt}{ds}) = 0 \\ \Rightarrow (1 - \frac{2m}{r})\frac{dt}{ds} = constant \equiv k \end{eqnarray*}$$

Substitute the l and k equations for $\frac{d\varphi}{ds}$ and $\frac{dt}{ds}$ back into the metric equation. Then after a little work, we get,

$$(\frac{dr}{ds})^2 = -\frac{l^2}{r^2}(1 - \frac{2m}{r}) - (1 - \frac{2m}{r}) + k^2$$ (48)

From the $\varphi$ equation,

$$(\frac{d\varphi}{ds})^2 = \frac{l^2}{r^4}$$

Divide eq. 48 by this and get,

$$(\frac{dr}{d\varphi})^2 = -r^2(1 - \frac{2m}{r}) - \frac{r^4}{l^2}(1 - \frac{2m}{r}) + \frac{k^2}{l^2}r^4$$

Let $r = \frac{1}{u}$ and derive by $\frac{d}{d\varphi}$ to get:

$$\frac{d^2}{d\varphi^2} + u = \frac{mG}{c^2l^2} + \frac{3mu^2G}{c^2}$$

For the planets, $\frac{m}{l^2} \ll 3mu^2$. So treat the correction as a perturbation of the classical orbit. The final answer is

$$u = \frac{1}{r} = \frac{m}{l^2}[1 + e\cos (\varphi - \varphi_0)]$$

for an unperturbed orbit. The second approximation gives us

$$\frac{1}{r} = \frac{m}{l^2}[1 + e\cos (\varphi - \varphi_0 - \Delta\varphi)]$$

where

$$\Delta\varphi = \frac{3m^2\varphi}{l^2}$$

For Mercury

$$(\Delta\varphi)_{\varphi=2\pi} \rightarrow 42.89'' / century$$

Observed value: $(42.6\pm1.0)'' / century$. This is one of the three classical tests of general relativity. The second one is:

The gravitational red-shift

$$ds^2 = (1 + \frac{2\varphi}{c^2})dt^2 - \frac{dr^2}{1 + \frac{2\varphi}{c^2}} - r^2d\Omega^2$$

Set $dr = d\theta = d\phi = 0$ to get proper time.

$$ds = \sqrt{1 + \frac{2\varphi}{c^2}}dt$$

This is the time read in the rest frame of the body. Since $\varphi$ varies with position, proper time also depends on position in the gravitational field. Let n waves be emitted at the suns position in time $(\Delta s)_s$ at frequency $\nu_s$. These waves are observed on earth in the time $(\Delta s)_e$ at the frequency $\nu_e$.

$$\begin{eqnarray*} n = \nu_s(\Delta s)_s = \nu_e (\Delta s)_e \\ \nu_s = \nu_e... ... \Rightarrow \frac{\Delta\nu}{\nu} = -\frac{\Delta\varphi}{c^2} \end{eqnarray*}$$

The gravitational potential at the earth is considerably smaller than at the sun $(\mid\varphi_e\mid \ll \mid\varphi_s\mid)$.

$$\Rightarrow \frac{\Delta \nu}{\nu} \approx \frac{\varphi_s}{c^2} = -\frac{GM}{r_sc^2}$$

for the photon emitting at the surface of the sun.

$$\begin{eqnarray*} \Rightarrow \frac{\Delta \nu}{\nu} < 0 \Rightarrow \nu_f - \... ... < 0 \Rightarrow \nu_f < \nu_i \Rightarrow \lambda_f > \lambda_i \end{eqnarray*}$$

So we see a red-shift of the light coming from the sun. Let's look at a different derivation. Consider the photon, and let it have an effective mass of $\lq\lq m_e$'' $= \frac{h\nu_e}{c^2}$ at the earth, and $\lq\lq m_s$'' $= \frac{h\nu_s}{c^2}$ at the sun. The total energy of the photon during it's existence is conserved, that is the total energy equals the intrinsic energy and the gravitational potential energy.

$$\begin{eqnarray*} h\nu_s + (\frac{h\nu_s}{c^2})\varphi_s = h\nu_e + (\frac{h\nu... ... \frac{\nu_e - \nu_s}{\nu_s} = \frac{\varphi_s - \varphi_e}{c^2} \end{eqnarray*}$$

So the second test is not really a test of GR.

Pound & Rebka

We can also perform this test in the gravity of the earth, which was done by Pound & Rebka [1][2]. Generally, the $\gamma$-rays emitted lead to line spectra that are not sufficiently sharp, due to Doppler line broadening, i.e. a nucleus recoils. But Mossbauer found that certain radioactive substances, like Fe$^{57}$, have the whole lattice recoil, and hence the line are very sharp. The atoms emit photons at the base of the earth, and they are gravitationally red-shifted when observed at top of a tower. So it will not get resonately absorbed by an iron sample above. To reestablish resident absorption, move the emitter upward to get a compensating Doppler blue shift. Measure the speed to get resident absorption back, and this gives the Doppler shift, and hence the gravitational red-shift. The calculated Doppler shift is

$$(\frac{\Delta\nu}{\nu}) = 4.92 \times 10^{-15}$$

And the experimental value is 0.997$\pm$0.008 of the predicted value.

3rd test

During an eclipse of the sun, it is possible to measure the position of stars close to the sun. The observed position will be slightly shifted to due to the sun, so if we compare to the position of the star when it is not close to the sun, we can measure the deflection. Leonard I. Schiff presents in [3] some mathematical considerations, showing that this test is really only a test of the equivalence principle, not the Einstein equations.

These are the three classical tests of general relativity.

New 4th test of GR

Irwin Shapiro has performed[4] radar bounces of some of the closer planets, and measured the retardation of the radio waves. The results were 1.02$\pm$0.05 within the GR prediction. Later[5] it has been done using spacecraft, giving more precise control of the radar beams reflections, and the results were 1.00$\pm$0.04.

Gravitational Collapse & Black Holes

We use the familiar Schwarzschild metric,

$$ds^2 = (1 - \frac{2m}{r})dt^2 - \frac{dr^2}{1 - \frac{2m}{r}} - r^2d\Omega^2$$ (49)

At r=2m, used to be called the Schwarzschild singularity, since then $g_{00} = 0, g_{11} = \infty$. Is really r=2m singular? Consider the transformation, as given by Novikov:

$$\begin{eqnarray*} c\tau = \pm ct\pm \int\frac{f(r)}{1 - \frac{2m}{r}}dr \\ cd... ...c{2m}{r})f(r)} \\ dR = cdt + \frac{dr}{(1 - \frac{2m}{r})f(r)} \end{eqnarray*}$$

Then consider

$$\begin{eqnarray*} c^2d\tau^2 - f^2dR^2 \\ = c^2dt^2 + \frac{f^2dr^2}{(1 - \fr... ... f^2dR^2}{1 - f^2} = c^2dt^2 - \frac{dr^2}{(1 - \frac{2m}{r})^2} \end{eqnarray*}$$

If we multiply this by $(1 - \frac{2m}{r})$ and subtract $r^2d\Omega^2$, we get back eq. 49. Choose $f^2 = \frac{2m}{r}$, then we get

$$ds^2 = c^2d\tau^2 - \frac{2m}{r}dR^2 - r^2d\Omega^2$$

At r=2m, no problem. At r=0, the metric is singular⁴! Consider

$$\begin{eqnarray*} R - c\tau = \int\frac{dr}{1 - \frac{2m}{r}}\sqrt{\frac{2m}{r}... ... \Rightarrow [(R - c\tau)\frac{3}{2}\sqrt{2m}]^{\frac{2}{3}} = r \end{eqnarray*}$$

So

$$ds^2 = c^2d\tau^2 - \frac{2m}{[(R - c\tau)\frac{3}{2}\sqrt{... ...^2 - [(R - c\tau)\frac{3}{2}\sqrt{2m}]^{\frac{4}{3}}d\Omega^2$$

Metric is now time dependent. For $r > 2m$, it is really the old story of actually being intrinsically static, but for $r < 2m$...watch out. Suppose you were at rest in this new coordinate system $\tau, R, \theta, \phi$. Then you are actually in physical free fall!

Proof:

Suppose you have any metric in the form:

$$ds^2 = c^2d\tau^2 + g_{\alpha\beta}dx^{\alpha}dx^{\beta}$$

called a synchronous metric. (note: $g_{00}=1 \& g_{0\alpha}=0$) Suppose you are at ``rest'' in this metric. Start with this assumption:

$$\frac{dx^{\alpha}}{ds} = 0$$

Then $ds^2 = c^2d\tau^2$

$$\begin{eqnarray*} \Rightarrow \frac{d\tau}{ds} = 1 \\ \frac{dx^{\alpha}}{ds} = 0 \end{eqnarray*}$$

Show that this velocity set satisfies

$$\begin{eqnarray*} \frac{du^i}{ds} + \Gamma_{kl}^iu^ku^l = 0 \\ \Rightarrow 0 ... ... = 0? \\ \frac{g^{im}}{2}(g_{m0,0} + g_{m0,0} - g_{00,m}) = 0? \end{eqnarray*}$$

$m = 0$ (from metric) $\Rightarrow g_{m0} = 0 \forall m\ne 0$, but $g_{00} = 1$, so $g_{00,0} = 0$. So being at rest in this coordinate system, really means that you are physically in free fall.

See fig. 2 for a plot of the radial⁵ trajectories near a black hole.

Figure 2: This figure describes the light cones of a free falling graduate student, into a black hole. The diagonal lines describes the trajectories of professors hovering at a constant distance from the black hole.

From the figure we see that when $R - c\tau = const. \rightarrow r = const.$. Now consider the trajectory for light. From SR, we remember that $ds^2 = 0$ for light. $\rightarrow ds^2 = 0 = c^2d\tau^2 - \frac{2m}{r}dR^2$. So for light

$$\begin{eqnarray*} \frac{c^2d\tau^2}{dR^2} = \frac{2m}{r} \\ \frac{cd\tau}{dR} = \pm\sqrt{\frac{2m}{r}} \end{eqnarray*}$$

Thus, we have found the slope of the light cones plotted in figure 2. Note: $\tau$ is the time read by the freely falling observer.

For someone falling into the black hole(R=const): r=0 for $\tau = \frac{R}{c}$= finite number. So for you falling into the black hole, it takes a finite amount of time to reach the big crunch, and less time to reach r=2m. But consider all this from the point of view of observers outside r=2m. Then,

$$\begin{eqnarray*} ds^2 = (1 - \frac{2GM}{c^2r})c^2dt^2 - \frac{dr^2}{1 - \frac{2GM}{c^2r}} - r^2d^2\theta - r^2\sin^2\theta d\varphi \end{eqnarray*}$$

for $r \gg 2m$, t is reading the time of outside observers. Radial geodesic solution gives

$$ct = \sqrt{1 - \frac{2m}{r_0}}\int_r^{r_0}\frac{dr}{(1 - \frac{2m}{r})(\frac{2m}{r} - \frac{2m}{r_0})^{\frac{1}{2}}}$$ (50)

What happens near r=2m?

$$\begin{eqnarray*} \sim \int\frac{dr}{1 - \frac{2m}{r}} = r + 2m\log (r - 2m) \\... ...- 2m) \\ \Rightarrow r - 2m \sim const\cdot e^{-\frac{ct}{2m}} \end{eqnarray*}$$

When $r \rightarrow 2m, t \rightarrow \infty$. Note: While it takes a finite time for the clocks of a free-fall observer to cross r=2m and hit r=0, from the point of view of the outside observers, r gets very close to 2m quickly, but never actually reaches r=2m.

How fast does a particle move, relative to a sequence of rest observers, as the particle approaches r=2m? ( $\theta = \varphi = const$)

$$\begin{eqnarray*} % latex2html id marker 2908v = \frac{(dl)_{proper}}{(dt)_{p... ...v^2 = \frac{(\frac{2m}{r} - \frac{2m}{r_0})}{1 - \frac{2m}{r_0}} \end{eqnarray*}$$

As $r \rightarrow 2m, v^2 \rightarrow (1 - \frac{2m}{r_0})\frac{1}{1 - \frac{2m}{r}} = 1$, i.e. $v \rightarrow c$, as $r \rightarrow 2m$. How long does it take to get a signal out as you approach r=2m?

$$\begin{eqnarray*} ds^2 = 0 = (1 - \frac{2m}{r})dt^2 - \frac{dr^2}{1 - \frac{2m}... ... - \frac{2m}{r})^2} \Rightarrow dt = \frac{dr}{1 - \frac{2m}{r}} \end{eqnarray*}$$

When $r \rightarrow r_0$, what is the time required?

$$\begin{eqnarray*} \int_t^{t_0} = \int_r^{r_0}\frac{dr}{1 - \frac{2m}{r}} = r + ... ...elta t = t_0 - t = (r_0 - r) + 2m\log\frac{r_0 - 2m}{r - 2m} \\ \end{eqnarray*}$$

As $r \rightarrow 2m$, $\Delta t \rightarrow r_0 - 2m + 2m\log\infty$. As $r \rightarrow 2m$ the frequency as seen, approaches 0...infinitely red-shifted.

The Universe

Figure 3: The static universe, radiation from a shell.

Let $\mu =$ the energy emitted per unit volume per unit time of a universe of chopped up stars. See figure 3. Consider the observer in 0, examining the shell at radius R, thickness dR, and the light coming to 0. the energy emitted per unit time from the shell. So intensity at 0

$$= \frac{4\pi R^2dR\mu}{R^2} = 4\pi\mu dR$$

For the contribution from the whole infinite universe, it is $\int_0^{\infty}4\pi\mu dR = \infty$. But the night sky is dark! (Olber's paradox)

Einstein: Yes, static universe. But beat Olber's paradox by making the universe also finite. An finite static model. To make it static, you have to resist the gravitational attraction. Change the Einstein equations:

$$R^{ik} - \frac{1}{2}g^{ik}R + \Lambda g^{ik} = \frac{8\pi G}{c^4}T^{ik}$$

where $\Lambda$ is the cosmological constant. Consider $[\Lambda g^{ik}]_{;k} = 0?$. Recall we showed $g_{ik;l} = 0$. So, OK.

Two objections:

1. We see a red-shift (universe is dynamic)
2. Static models are unstable

Adopt the ``cosmological principle'', at a given time, the universe has the same averaged appearance for all observers. We see an isotropic distribution of radiation, assume everyone does. Also, this implies that everybody sees the same density everywhere. Homogeneous, isotropic universe can be described by:

$$ds^2 = c^2dt^2 - R^2(t)[\frac{du^2 + u^2(d\theta^2 + \sin^2\theta d\phi^2)}{(1 + \frac{k}{4}u^2)^2}]$$

where $k = 0, +1 or -1$.

a)

k=0: Zero curvature. On a constant time-slice - geometry is flat. $\frac{C}{D} = \pi$. For a radius, r, equal to infinity, we get a curvature of 0. See figure 4. (Note: it is infinite, no boundaries.)

Figure 4: The universe depicted as an infinite plane

b)

k=+1: Positive curvature(ball). Finite universe. $\frac{C}{D} < \pi$. Here the curvature is $C = \frac{1}{r^2}$. See figure 5.

Figure 5: The universe depicted as a finite sphere

c)

k=-1: Negative curvature(saddle). $\frac{C}{D} > \pi$. The curvature is given by, $C = \frac{-1}{r_1} \frac{1}{r_2}$. See figure 6.

Figure 6: The universe depicted as an infinite saddle

Look at the light being sent to us from a distant observer.

$$\begin{eqnarray*} ds^2 = 0 = c^2dt^2 - R(t)^2[\frac{du^2}{(1 + \frac{k}{4}u^2)^... ...frac{cdt}{R(t)} = \int_u^0\frac{du}{1 + \frac{k}{4}u^2} \equiv l \end{eqnarray*}$$

Physical distance between E & O at time t is $R(t)l \equiv L$. Let the next pulse be emitted at $t_e + \Delta t_e$ and observed at $t_0 + \Delta t_0$.

$$\begin{eqnarray*} \int_{t_e + \Delta t_e}^{t_0 + \Delta t_0}\frac{cdt}{R(t)} = ... ...rac{cdt}{R(t)} = \int_{t_0}^{t_0 + \Delta t_0} \frac{cdt}{R(t)} \end{eqnarray*}$$

For $\Delta t$'s much less than the traveltime of light between E and 0, R(t) is approximately constant.

$$\begin{eqnarray*} \frac{c}{R(t_e)}\Delta t_e = \frac{c}{R(t_0)}\Delta t_0\\ \... ...er}} - 1 \\ \Rightarrow \frac{\Delta\lambda}{\lambda} \equiv Z \end{eqnarray*}$$

We see a redshift, i.e. $\lambda_0 > \lambda_e$.

$$\begin{eqnarray*} \Rightarrow Z > 0 \Rightarrow \frac{R_{now}}{R_{earlier}} - 1... ...frac{R_{now}}{R_{earlier}} > 1 \Rightarrow R_{now} > R_{earlier} \end{eqnarray*}$$

The universe is expanding. We will use GR to see how it is expanding. Use $G_i^{\phantom{i}k} = \frac{8\pi G}{c^4}T_i^{\phantom(i)k}$. Use approximation of the universe as a perfect fluid.

$$T_i^{\phantom{i}k} = \left( \begin{array}{cccc} \rho c^2... ...0 \\ 0 & 0 & -P & 0 \\ 0 & 0 & 0 & -P \end{array} \right)$$

For the universe NOW, $P \ll \rho c^2$, so $P \approx 0. T_i^{\phantom{i}k} \approx diag(\rho c^2, 0, 0, 0)$. Put this $T_i^{\phantom{i}k}$, with pressure for now, into the Einstein equations.

$$\begin{eqnarray*} \Rightarrow \frac{4\pi G}{c^2}(\rho + \frac{3P}{c^2}) = \Lamb... ...+ \frac{P}{c^2}) = \frac{k}{R^2} + (\frac{R'^2 - RR''}{c^2R^2}) \end{eqnarray*}$$

or

$$\frac{8\pi G}{c^2}\rho = -\Lambda + \frac{3k}{R^2} + \frac{3R'^2}{cR^2}$$ (51)

$$\frac{8\pi G}{c^2}\frac{P}{c^2} = \Lambda - \frac{k}{R^2} - \frac{2R''}{c^2R}$$ (52)

Static universe

$R' = 0, R = const$.

$$\Rightarrow \frac{4\pi G}{c^2}(\rho + \frac{3P}{c^2}) = \Lambda$$

Suppose $\Lambda = 0$. For $p > 0 \& P \ge 0 \Rightarrow l.h.s > 0 \& r.h.s = 0$. Contradiction.

Consider keeping $\Lambda$, and for static, show $\frac{k}{R^2} < \Lambda < \frac{3k}{R^2} \Rightarrow \Lambda \sim \frac{1}{R^2}$. Since $\Lambda$ is so small, it would not contradict the solar system tests of GR.

Non-static Models, ($\Lambda = 0$)

$$\frac{4\pi G}{c^2}(\rho + \frac{3P}{c^2}) = -\frac{3R''}{c^2R}$$

For $\rho > 0$ and $P \ge 0, l.h.s > 0 \Rightarrow R'' < 0 \rightarrow decellaration$. What is the solution?

$$\begin{eqnarray*} \frac{8\pi G}{c^2}\rho = \frac{3R'^2}{R^2c^2} + \frac{3k}{R^2... ... \frac{4\pi G}{c^2}(\rho + \frac{3P}{c^2}) = -\frac{3R''}{c^2R} \end{eqnarray*}$$

For NOW($P \approx 0$):

$$\begin{eqnarray*} \Rightarrow -\frac{6R''}{c^2R} = \frac{3k}{R^2} + \frac{3R'^2}{c^2R^2} \\ \Rightarrow 2RR'' + R'^2 + kc^2 = 0 \end{eqnarray*}$$

Solution is,

$$R'^2 = \frac{\alpha}{R} - kc^2$$

where $\alpha$ is a constant.

k=+1

$$\begin{eqnarray*}ct = \frac{\alpha}{2}(2\tau - \sin 2\tau) \\ R = \frac{\alpha}{2}(1 - \cos 2\tau) \end{eqnarray*}$$

This solution gives us a cycloid.

k=0

$$R = At^{\frac{1}{3}}$$

This solution gives us eternal expansion.

k=-1

$$\begin{eqnarray*}ct = \frac{\alpha}{2}(\sinh 2\tau - 2\tau) \\ R = \frac{\alpha}{2}(\cosh 2\tau - 1) \end{eqnarray*}$$

also monotonic expansion forever.

Which of these is the solution for our universe? Go back to eq. 51. Why is $(\frac{R'}{R})_{now}$ the Hubble parameter? Recall:

$$\begin{eqnarray*} L = R(t)l \\ \frac{dL}{dt} = V = R'(t)l = R'(t)\frac{L}{R(t... ...= (\frac{R'(t)}{R(t)})L \\ V = H_0\cdot D, (the Hubble law) \end{eqnarray*}$$

The Hubble parameter is $(\frac{R'(t)}{R(t)})_{now}$.

$$\frac{8\pi G}{c^2}\rho = \frac{3}{c^2}H_0^2 + \frac{3k}{R^2}$$

The solution for NOW, with $H_0 = \frac{1}{T_{age of universe}}$:

$$\rho - \frac{3}{8\pi G}\frac{1}{T^2} = \frac{3kc^2}{8\pi GR^2}$$

for k = $0, +1 \& -1$. Which one is it? If we insert the present age of the universe into $\frac{3}{8\pi G}\frac{1}{T^2}$ we get $\sim 2,1 \times 10^{-29} \frac{g}{cm^3}$. If

$$\begin{eqnarray*} \rho_{now} > 2,1 \times 10^{-29} \frac{g}{cm^3} \Rightarrow k... ...critical} = 2,1 \times 10^{-29} \frac{g}{cm^3} \Rightarrow k = 0 \end{eqnarray*}$$

From the visible matter in the universe we get $\rho_{visible} \sim 10^{-31} \frac{g}{cm^3}$. Is there enough dark matter to close the universe?

Bibliography

1

Pound & Rebka: Phys. Rev. Letters, Vol. 3, p. 357 & 459 (1960)

2

Pound & Snider: Phys. Rev. Letters, Vol. 13, p.539 (1964)

3

Leonard I. Schiff: Ameri. Jour. Phys., Vol. 28, p.340 (1960)

4

Irwin Shapiro: Phys. Rev. Letters, Vol. 13, p.789 (1964)

5

Irwin Shapiro: Phys. Rev. Letters, Vol. 26, p.1132 (1972)

About this document ...

General Relativity
Lecture notes from Phys415

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Footnotes

... 2¹

A vector is a tensor of rank 1. A scalar is a tensor of rank 0

... symmetric²

Symmetry is a covariant property, i.e. it is held in all coordinate systems

...$0$³

The $$ are 0 in cartesian coordinates

... singular⁴

This is the metric $ds^2$ in terms of coordinates both in the old system(r), and the new system(R, $\tau, \theta, \varphi$).

... radial⁵

$\theta = \varphi = 0$