An introduction to general relativity.

Logistics

  • Location: 214 Pupin

  • Time: 4-5:30 PM

  • Textbooks: “Gravity” (Hartle)

  • Office Hours: 3:30-5 pm on Mondays (Pupin 914)

  • Grading Scheme:

    • Problem Sets: 45% (11 total)
      • Assigned Thursdays, and due the following week @ 4:10
      • Lowest grade dropped
    • Midterm: March 7 (25%)
      • On March 7th
    • Final (30%)

Conventions

  • Time is t or $\frac{x^{0}}{c}$

  • Space coordinates: $x^{1},x^{2},x^{3}$

    • position 3-vector: $\vec{x}$

  • Position 4-vector: $(ct,\vec{x})$

  • Metric: $(-,+,+,+)$

    • $ds^{2} = -dt^{2}+(dx^{1})^{2}+(dx^{2})^{2}+(dx^{3})^{2}$

  • Index notation:

    • Roman labels (i,j,k,…) span from 1 to 3, and are spatial coordinates
      • so $\vec{x} = x^{i}$
    • Greek labels ($\mu,\nu,\alpha…$) span from 0 to 3 and are space-time coordinates

  • Einstein Summation notation:

    • Whenever we have a pair of repeated indices of differing heights, we sum over them
      • $\vec{x}\cdot \vec{y} = x_{i}y^{i} = \Sigma_{i=1}^{3} \delta_{ij} x^{i}y^{j}$
      • Can write Minkowski metric as $ds^{2} = dx_{\mu}dx^{\mu} = \eta_{\mu \nu}dx^{\mu}dx^{\nu} = \Sigma_{\mu=0}^{3}\Sigma_{\nu=0}^{3} \eta_{\mu \nu}dx^{\mu}dx^{\nu}$
    • Rules:
      • Indices appear , at most, twice in a given term
        • $x_{\alpha}y^{\alpha}z_{\alpha}$: ill-defined, since order of contraction can’t be determined
      • Equations must have the same set of free indices on both sides
        • free indices are indices which are not being contracted
      • Paired indices should always appear with one index upstairs and one index downstairs
      • A trick: the labels of repeated indices can be changed to whatever variable you want

Newtonian Gravity

  • $\vec{F} =\frac{GMm}{r^{2}}\vec{r}$

    • G = 6.67E-11 $\frac{m^{3}}{kg\cdot s^{2}}$

  • Can define the gravitational potential as $\Phi(\vec{r}) = \frac{-Gm_{a}}{r_{AB}} = \frac{-Gm_{a}}{|\vec{r_{a}}-\vec{r_{b}}|}$

    • This implies a Poisson equation for gravity: $\nabla^{2}\Phi = 4\pi G \rho$

The Two Body Problem

  • Write down the Lagrangian of the system where you utilize the reduced mass to reduce the two-body problem to a one body problem with reduced mass ($\mu = \frac{Mm}{M+m}$): $\mathcal{L} = \frac{\mu\dot{r}^{2}}{2}+\frac{mr^{2}\phi^{2}}{2}-U(r)$

    • where $U(r) = \frac{GmM}{r}$

  • Do the Euler-Lagrange equations, realize that that that from Noether’s theorem that $l = \mu r^{2}\phi$ is conserved, make the substitution $u = \frac{1}{r}$, you get the following equation:

    • $\frac{d^{2}u}{d\phi^{2}}+u = \frac{G\mu^{2}M}{l^{2}}$
      • Solutions are of the form: $u(\phi) = (\frac{G\mu^{2}M}{l^{2}})(1+\epsilon \cos(\phi-\phi_{0}))$
        • $\epsilon = \sqrt{1+\frac{2El^{2}}{G\mu^{2}M}}$
      • This describes various conic sections:
        • $\epsilon < 1$ means that $E0$ which implies an ellipse
        • $\epsilon=0$ implies that $E = \frac{-G\mu^{2}M}{2l^{2}}$ which moves in a circle
        • $\epsilon > 1$ implies that $E>0$ and moves like a hyperbola
        • $\epsilon = 1$ implies that $E=0$ which moves like a parabola
    • The effective potential energy then becomes $V_{eff}(r) = \frac{-Gm}{r}+\frac{l^{2}}{2\mu r^{2}}$

Kepler’s Second Law

  • Equal areas are swept out in equal times

  • The area swept is given by the base of the triangle($rd\phi$) multiplied by the height ($\frac{r}{2}$). Taking time derivative, you can use constancy of angular momentum to prove the 2nd Law

Kepler’s Third Law

  • the total area is given by $\int dA = \int \frac{r^{2}}{2} d\phi = \int_{0}^{\tau} (\frac{dA}{dt}) dt = \frac{l\tau}{3\mu} = \pi ab \rightarrow \tau = \frac{2\mu ab \pi}{l}$. Can use relationship between semimajor and semiminor axes, can show that $\tau \propto a^{\frac{3}{2}}$

From Kepler To Newton

  • You can run the above argument in reverse
    • We know that the gravitational acceleration is equal to the centripetal acceleration $g = \omega^{2} r = (\frac{2\pi }{\tau})^{2}r$
    • Plug Kepler’s 3rd Law for $\tau$, let $a=r$ yields that $g(r) = r^{-2}$

The Equivalence Principle

  • “Mass” appears in both Newton’s 2nd law, as well as Newton’s law of gravitation

    • Empirically, these two masses are equivalent (Eotvos experiment)

  • This implies that there always exists some accelerating reference frame where the observer will not feel a gravitational field

    • ie. There is no local experiment that you can to to determine if you are falling

  • More formally:

    • all uncharged, freely falling test particles and massive gravitating objects follow the same trajectories
    • in all freely-falling frams, the same S.R. laws of physics hold locally, regardless of position or velocity

Gravitational Redshift

  • Imagine two observers: Observer A is located height h above the Earth’s surface, while Observer B is located on the Earth’s surface. Every time interval $\delta t_{a}$, A sends out a light pulse, while $\delta t_{b}$ is the time between when B receives the pulses

  • Can thing of the same situation, were we are instead in an accelerating reference moving upward with acceleration g

  • Calculating the time differences under weak gravity (ie. that $gt « c$)

  • $\Delta t_{A} = \Delta t_{b}(1+\frac{gh}{c^{2}})$

    • Since $\phi_{B}-\phi_{A} = -gh$, we see that $t_{b}^{-1} \approx t_{a}^{-1}(1-\frac{\phi_{b}\phi_{a}}{c^{2}})$
      • Clocks tick slower the deeper you are in a potential

Light falls

  • Imagine a light beam being shot through a box accelerating upwards

    • From an inertial frame, the light beam moves in a straight line
    • From the accelerating frame, the light appears to fall to a lower height

  • By the equivalence principle, this means that light gets bent in the presence of gravity

Differential Geometry

Definitions/Basics

  • Events: a point in spacetime

    • Events exist, regardless of how you describe them (ie. independent of the coordinates used)

  • Worldline: the path through spacetime (ie. a set of events)

  • Coordinates: 4-tuples (ie. a set of 4 numbers) that label events

  • The laws of nature must be written in a coordinate independent form

Crash Course

  • flat Euclidean space has a well-defined notino of physical distance between two points, regardless of coordinates

    • There exists a special set of coordinates where calculating the distance is the simplest (ie. Cartesian): $dist(A,B)^{2} = \Sigma_{i=1}^{3}\Sigma_{j=1}^{3}\delta_{ij}(x_{A}^{i}-x_{B}^{i})(x_{A}^{j}-x_{B}^{j}) = \delta_{ij}(x_{A}^{i}-x_{B}^{i})(x_{A}^{j}-x_{B}^{j})$

  • For Curved space, distance is harder to define

    • On very small scales, we can draw infinitesimal line segments, and then sum up those distances
    • The metric of a curved space coverts between coordinate distance and physical distance
      • While you can have many different values for coordinate distance, the all yields the same physical distance after being ground up by the metric

Example

  • Consider a flat space. The infinitesimal distance between two points is $dl^{2} = dx^{2}+dy^{2}$. In polar coordinates, we have that $dr^{2}+r^{2}d\theta^{2}$

    • In terms of the metric: $dl^{2} = \Sigma_{i,j=1}^{3} g_{ij} dx^{i}dx^{j} = g_{ij}dx^{i}dx^{j}$
      • For cartesian coordinates: $g_{ij} = \begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}$
      • For polar coordinates: $g_{ij} = \begin{pmatrix}1 & 0\\ 0 & r^{2}\end{pmatrix}$

  • To derive the polar metric we need to:

    • Utilize the invariance of $dl^{2}$
    • The chain rule: $dx^{i} = \frac{\partial x^{i}}{\partial x^{j}}dx^{j}$

  • Let unprimed be cartesian and primed be polar

  • $dl^{2} = g_{ij}dx^{i}dx^{j} = g_{‘i’j}dx^{‘i}dx^{‘j}$

    • $g_{ij}dx^{i}dx^{j} = g_{ij} (\frac{\partial x^{i}}{\partial x^{m}})\frac{\partial x^{j}}{\partial x^{n}}dx^{m}dx^{n}$
      • Can relabel repeated indices to whatever you want: $g_{ij}dx^{i}dx^{j} = g_{mn} (\frac{\partial x^{m}}{\partial x^{i}})\frac{\partial x^{n}}{\partial x^{j}}dx^{i}dx^{j}$

  • Comparing to the initial equation yields that $g_{ij}^{’} = g_{mn}\frac{\partial x^{m}}{\partial x^{i}}\frac{\partial x^{n}}{\partial x^{j}}$

    • This is a general formula
    • To do calculations, it is useful to write down the Jacobian matrix as a reference

  • For cartesian to polar: $g_{mn} = \delta_{ij}$, $x^{1} = x, x^{‘1} = r,x^{2} = y, x^{‘2} = \theta$

    • $x^{1} = x^{‘1}\cos x^{‘2}$
    • $x^{2} = x^{‘1}\sin x^{‘2}$
    • Jacobian is then $\frac{\partial x^{i}}{\partial x^{‘j}} = \begin{pmatrix} \cos(x^{‘2}) & -x^{‘1}\sin(x^{‘2}) \\ \sin(x^{‘2}) & x^{‘1}\cos(x^{‘2}) \end{pmatrix}$

  • As a mechanical example, we see that $g_{11} = g_{mn}\frac{\partial x^{m}}{\partial x^{1}}\frac{\partial x^{n}}{\partial x^{1}}$

    • Only the 11 and the 22 term contribute

  • Similarly: $g_{22} = g_{mn}\frac{\partial x^{m}}{\partial x^{2}}\frac{\partial x^{n}}{\partial x^{2}}$

2-Sphere

  • How do you find the line element on the surface of a sphere (a 2D object)

  • We embed the 2-sphere into 3D Euclidean space, and then induce the metric from the ambient 3D space

    • $dl^{2} = R^{2}(d\theta^{2}+\sin^{2}\theta d\phi^{2})$
    • In Cartesian coordinates: $dl^{2} = dx^{2}+dy^{2}+\frac{(xdx+ydy)^{2}}{R^{2}-(x^{2}+y^{2})}$

  • Consider a neighborhood around the north pole (0,0,R)

    • $x = \epsilon, y = \delta, Z = R$ where $\epsilon,\delta « R$
    • The metric around the north pole is then $dl^{2} + dx^{2}+dy^{2}$
      • The metric looks Euclidean around the North Pole
      • Since the north pole is not a special point, we can rotate the north pole to any other point and repeat the analsis. Hence, at every point on the sphere, local space looks Euclidean

  • The Certesian coordinate system fails at the equation (ie. z=0), so we need another set of coordinates to describe the equation

Special Relativity

  • Postulates

    • Assume laws of physics are the same in all intertial reference frames
    • The speed of light is a universal constant, independent of reference frame

  • Empirically, we know that the laws of physics are

    • homogeneous: no special position
    • isotropic: no special direction
    • Stationary: no special time

  • Can derive the Minkowski metric from these assumptions: $ds^{2} = -(c dt)^{2}+dx^{2}+dy^{2}+dz^{2}$

    • $ds^{2}<0$ is called timelike. This means you can always find a reference frame with $dx^{i}=0$
      • ie. the events overlap in space, but not in time
    • $ds^{2}=0$
      • Null or lightlike interval. Only light can have these trajectories
      • You can also find frames where $dx^{0} >0$ and $dx^{0}<0$. Hence, there is no absolute notion of “before”, “after” or “simultaneous” for spacelike seperated events
      • $ds^{2}>0$ can always find a frame where events are simultaneous (ie. at the same time)

  • As a result of the homogenous, isotropic, and stationary constraint, we demand that coordinates be related to each other via linear transformations

    • $x^{’} = \Lambda_{\nu}^{\mu}x^{\nu}+a^{\mu}$
      • a is some constant vector

  • From 2nd postulate, we have that $\eta_{\nu\mu}\Delta_{x}^{\mu}\Delta_{x}^{\nu} = 0 = \eta_{\nu\mu}\Delta_{x’}^{\mu}\Delta_{x’}^{\nu}$

    • Plugging in linear transformation law (remembering that the deltas eliminate the constant offset vector) yields that
    • $\Lambda_{\mu}^{\alpha}\eta_{\alpha\beta}\Lambda_{\nu}^{\beta} = \eta_{\mu\nu}$
    • Can move around orders of entries freely in index notation (don’t have to worry about stuff commuting)

  • For Minkowski space, we know that the $\Lambda$ matrices take the following form (derived from Lorentz transformations):

    • ${\Lambda^{\mu}}_{\nu} = \begin{pmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$
    • $\gamma = \frac{1}{\sqrt{1-\beta^{2}}}$
    • $\beta = \frac{v}{c}$

Minkowski Spacetime Diagrams

  • Draw orthogonal $x^{0}$ and $x^{1}$ axes just like in cartesian coordinates

  • For the primed coordinate system, draw $x^{‘0}$ and $x^{‘1}$ at some angle offset to the unprimed coordinates

  • Lines parallel to each axis correspond to worldines which have a constant value for that coordinate

Metric

  • Can convert to any metric from another one if you know the coordinate transformation

    • $g_{\mu\nu}(x’) = g_{\alpha\beta}(x) \frac{\partial x^{\alpha}}{\partial x^{\mu}}\frac{\partial x^{\beta}}{\partial x^{\nu}}$

  • Can define the proper time as $d\tau = \frac{1}{c}\sqrt{-ds^{2}} = \frac{1}{c}\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}$

    • $\tau = \int_{A}^{B} d\tau$

  • Can define instantaneous coordinate 3-velocity as $v^{i} = \frac{dx^{i}}{dx^{0}}*c = \frac{dx^{i}}{dt}$

    • Can express proper time in terms of this velocity
    • Recall that $\frac{dx^{0}}{dt} = c$
    • $g_{0i} = g_{i0}$
    • $d\tau = \sqrt{-g_{00}-2g_{0i}(\frac{v^{i}}{c})-g_{ij}(\frac{v^{i}}{c})(\frac{v^{j}}{c})}dt$

Useful Tricks

  • $\frac{\partial x^{\mu}}{\partial x^{\nu}} = \delta^{\mu}_{\nu}$

  • $\delta^{\mu}_{\nu} = \frac{\partial x^{\mu}}{\partial x^{’\alpha}}\frac{\partial x^{’\alpha}}{\partial x^{\nu}}$

4-Vectors

4-velocity

  • We can represent 4-velocity

    • $u^{\mu} = \frac{dx^{\mu}}{d\tau} = \frac{dt}{d\tau}\frac{dx^{\mu}}{dt} = (c,\vec{v})$
      • In an inertial reference frame, $\frac{dt}{d\tau} = \frac{dt}{\frac{dt}{\gamma}}$, which means that $u^{\mu} = \frac{1}{\sqrt{1=\beta^{2}}}(c,\vec{v})$
      • In a different coordinate system:
        • $u^{’\mu} = \frac{dx^{’\mu}}{d\tau} = \frac{dx^{’\mu}}{dx^{\nu}}\frac{dx^{\nu}}{d\tau} = \frac{dx^{’\mu}}{dx^{\nu}}u^{\nu}$
        • For an intertial reference frame: $u^{’\mu} = \Lambda^{\mu}_{\nu} u ^{\nu}$
      • The norm of the vector is given by:
        • $u_{\mu}u^{\mu} = g_{\mu\nu}u^{\mu}u^{\nu} = g_{\mu\nu}\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt} = \frac{d}{dt^{2}}(s^{2}) = -c^{2}\frac{d\tau^{2}}{d\tau^{2}} = -c^{2}$
        • So the norm is a constant time-like vector

  • $V^{’\mu} = \frac{\partial x^{’\mu}}{\partial x^{\nu}} V^{\nu}$

  • $V^{’\mu} = \Lambda^{\mu}_{\nu} V^{\nu}$

  • If a law is written in purely in terms of 4-vectors, then Lorentz invariance is automatically satisfied

  • In general, 4-vectors transform as follows:

    • $V_{obs}^{0} = \frac{-1}{c}g_{\mu\nu}u^{\mu}V^{\nu}$

  • u is the 4 velocity vector

Dual Vectors

  • Define the dual vector $V_{\mu}$ of $V^{\mu}$ as

    • $V_{\mu} = g_{\mu\nu}V^{\nu}$
    • Also called covariant vector or a one form
    • FYI: a vector is also called the contravariant vector

  • Can imagine the dual vector as an object which can always be contracted with a 4 vector to obtain a scalar quantity

  • Dual Vectors transform under coordinate changes as follows:

    • $V_{\mu}^{’} = g’_{\mu\nu} V^{’\nu} = $
    • $g’_{\mu\nu}\frac{\partial x^{’\nu}}{\partial x^{\alpha}}V^{\alpha} =$
    • $g_{\beta\gamma}\frac{\partial x^{\beta}}{\partial x^{’\mu}}\frac{\partial x^{\gamma}}{\partial x^{’\nu}}\frac{\partial x^{\nu}}{\partial x^{\alpha}}V^{\alpha} = $
    • $g_{\beta\gamma}\frac{\partial x^{\beta}}{\partial x^{’\mu}}\delta^{\gamma}_{\alpha}V^{\alpha} = $
    • $V_{\mu}’ = \frac{\partial x^{\beta}}{\partial x^{’\mu}} V_{\beta}$

4-momentum

  • $p^{\mu} = mu^{\mu} = (\frac{E}{c},\vec{p})$

  • $p_{\mu}p^{\mu} = -m^{2}c^{2}$

    • For light, the 4-momentum magnitude is 0

Inverse Metric

  • Let $g_{\mu\nu}$ be the metric associated with $x^{\mu}$

  • Define $g^{\mu\nu}$ such that $g_{\mu\beta}g^{\mu\alpha} = \delta^{\alpha}_{\beta}$

  • The inverse metric transforms like

    • $g^{’\alpha\beta} = \frac{\partial x^{’\alpha}}{\partial x^{\mu}}\frac{\partial x^{’\beta}}{\partial x^{\mu}}g^{\mu\nu}$

Variational Approach

  • For SR, we want to maximize the proper time of along a particle’s worldline

    • $\tau = \frac{1}{c}\int_{A}^{B}\sqrt{-ds^{2}}$
    • Assume events are time-like seperated

  • $\tau = \frac{1}{c}\int_{0}^{1} d\lambda \sqrt{-\nu_{\mu\nu}}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}$

    • $\lambda$ is some monotonically increasing parameter that ranges from 0 to 1 that parameterizes all the possible curves
    • Define $\dot{x^{\mu}} = \frac{dx^\mu}{d\lambda}$
    • Can then write this in terms of a Lagrangian: $\tau = \frac{1}{c}\int_{0}^{1}L(x^{\mu},\dot{x^\mu},\lambda)$
      • $L = \sqrt{-\eta_{\mu\nu}\dot{x^\mu}\dot{x^\nu}}$
    • To maximize this, we get Euler-Lagrange Equations:
      • $\frac{d}{d\lambda}(\frac{\partial L}{\partial \dot{x^\mu}})-\frac{\partial L}{\partial x^{\mu}} = 0$
      • Solving this for our La rgrangian, we find that we recover the equation of motion for an unconstrained particle (ie. 4-velocity is constant)

Moving to GR

  • For non-intertial frames, we don’t expect free particles to move in straight lines

  • To derive how a free particle moves, examine the proper time (since the change w.r.t. proper time is the same in all frames):

    • $\frac{d^{2}x^{\mu}}{d\tau^{2}} = 0$
    • Via Chain Rule: $\frac{d^2x^{\mu}}{d\tau^{2}} = \frac{dx’^{\sigma}}{d\tau} \frac{\partial ^{2} x^{\mu}}{\partial x’^{\nu}\partial x’^{\sigma}}$
    • Substituting in, multiplying by $\frac{\partial x’^{\alpha}}{\partial x^{\mu}}$, recongnize the Kronecker delta in the expression to eventually get that
    • $\frac{\partial ^{2}x’^{\alpha}}{\partial d\tau^{2}}+\Gamma_{\nu\sigma}^{\alpha}\frac{\partial x’^{\sigma}}{\partial \tau}\frac{\partial x’^{\nu}}{\partial \tau} = 0$
      • Called Geodesic equations
      • $\Gamma_{\nu\sigma}^{\alpha} = \frac{\partial x’^{\alpha}}{\partial x^{\mu}}\frac{\partial^{2}x^{\mu}}{\partial x’^{\sigma}\partial x^{\nu}}$
        • Called the Christoffel symbols
        • NOT A TENSOR
        • Symmetric in $\nu$ and $\sigma$
        • Vanishes in Minkowski space
        • Can think of these symbols as encoding how the basis vectors transform as you move around in some non-Euclidean space

  • What if you are tracking a massless particle? then $d\tau$ is 0. You simply replace $\tau$ with $\lambda$ where $\lambda$ is some affine parameter that monotonically increases along the path

Christoffel Symbols

  • I’m being kind of lazy an only describing how to do this manipulate, along with checkpoint equations to guide you

  • Can you write the Christoffel Symbols in terms of the metric?

  • First, what is $\frac{\partial g_{\mu\nu}}{\partial x’^{\lambda}}$

    • This metric is in terms of primed coordinates. Convert from cartesian coordinates (ie. using Minkowski space) to the actual metric
    • $\frac{\partial g_{\mu\nu}}{\partial x’^{\lambda}} = \frac{\partial}{\partial x’^{\lambda}}(\eta_{\mu\nu}\frac{\partial x^{\alpha}}{\partial x’^{\mu}}\frac{\partial x^{\beta}}{\partial x’^{\nu}})$
    • Expanding out derivitive, and staring really hard at the mixed partials, you can realize that the mixed partials with some expression containing the Christoffel symbols
      • $\Gamma^{\alpha}_{\mu\sigma}\frac{\partial \gamma}{\partial x’^{\alpha}} = \frac{\partial x^{\gamma}}{\partial x’^{\nu}\partial x’^{\sigma}}$
    • Substituting Christoffel symbols in, and once again staring really hard, you realize that you can use the definition of the metric $g_{\rho\nu} = \frac{\partial x^{\alpha}}{\partial x’^{\rho}}\frac{\partial \beta}{\partial x’^{\nu}}\eta_{\alpha\beta}$ to simplify things
    • So $\frac{\partial g_{\mu\nu}}{\partial x’^{\lambda}} = g_{\mu\nu}\Gamma_{\lambda \mu}^{\rho}+g_{\sigma\mu} \Gamma^{\sigma}_{\lambda \nu}$

  • To deal with the last equation, we permutate the free indices (ie. $\lambda \rightarrow \mu \rightarrow \sigma$) to generate a new valid equation. We permutate again to generate a 3rd equation, and then we add and subtract them.

  • Denote the unpermutated equation as 1, the first permutated equation as 2, and the 2nd permutated equation as 3.

    • If we add 2 and 3 and subtract 1, and once again stare really hard, we find that
    • $\frac{g_{\lambda\mu}}{\partial x’^{\mu}}+\frac{g_{\lambda\sigma}}{\partial x’^{\mu}}-\frac{g_{\nu\mu}}{\partial x’^{\mu}} = \Gamma_{\mu\nu}^{\sigma} g_{\sigma \lambda}+\Gamma_{\mu\nu}^{\sigma}g_{\sigma\lambda} = 2\Gamma_{\mu\nu}^{\sigma} g_{\sigma\lambda}$

  • Multiplying by $g^{\alpha\lambda}$ and dividing by 2 yields that

    • $\Gamma_{\mu\nu}^{\alpha} = \frac{1}{2}g^{\alpha\lambda}(\frac{\partial g_{\lambda\mu}}{\partial x’^{\nu}}+\frac{\partial g_{\lambda\nu}}{\partial x’^{\mu}}-\frac{\partial g_{\nu\mu}}{\partial x’^{\lambda}})$
    • What this means: we can write the Christoffel symbols in terms of the metric
    • So, given some arbitrary coordinate system $x’^{\mu}$ with some associated metric, we can directly compute the Christoffel symbols, and hence, find the equation of motion for a free particle in some arbitrary non-inertial coordinate system

We haven’t done GR yet. Let’s do that

  • From now on, c is 1

    • This makes the units of time and length the same

  • To extend to GR, note these two important facts

    • You cannot define a globally valid inertial frame. There only exists an inertial frame locally
    • Since the Christoffel symbols are proportional to the first partial of the metric, and since the Christoffel symbols vanish for inertial frames, then we expect that locally, the metric looks like
      • $g_{\mu\nu} = \eta_{\mu\nu}$
      • That the first partials of the metric vanishes

  • Combining the above, re realize that we expect that locally, we expect that g must deviate locally from the Minkowski metric on order 2

    • $g_{\mu\nu} \approx \eta_{\mu\nu}+\frac{\partial g_{\mu\nu}}{\partial x^{\alpha}x^{\beta}}|_{a^{\alpha}}(x^{\alpha}-a^{\alpha})(x^{\beta}-a^{\beta})$

Qausi-Stationary

  • Assume that $g_{\mu\nu} = \eta_{\mu\nu}+h_{\mu\nu}$
    • $|h_{\mu\nu}| « 1$
    • $\eta_{\mu\nu}$ is a constant matrix
  • Quasi stationary means that $|\frac{\partial h_{\mu\nu}}{\partial t}| « |\frac{\partial h_{\mu\nu}}{\partial x^{i}}|$
    • For when the metric does not change rapidly with time
  • Under Quasi stationary term, the geodesic equation collapses to:
    • $\frac{d^{2}x^{\mu}}{d\tau^{2}} = \Gamma_{00}^{\mu}\frac{dx^{0}}{d\tau}\frac{dx^{0}}{d\tau} = 0$
      • Follows from $|\frac{dx^{i}}{d\tau}| « |\frac{dt}{d\tau}|$ (ie. non-relativistic assumption)
    • Evaluating Christophel symbol: $\Gamma_{00}^{\mu} = \frac{1}{2}(\eta^{\mu\alpha}-h^{\mu\alpha})(\frac{\partial h_{\alpha 0}}{\partial x_{0}}+ \frac{\partial h_{ 0\alpha}}{\partial x_{0}}-\frac{\partial h_{00}}{\partial x^{\alpha}} )$
      • Since time portion much smaller than spatial components, only $\frac{\partial h_{00}}{\partial x^{\alpha}}$ survives
      • Can also drop time portion of sum
    • Hence $\Gamma_{00}^{\mu} \approx \frac{1}{2} \eta^{\mu i}(-\frac{\partial h_{00}}{\partial x^{i}})$
      • $\Gamma_{00}^{0} \approx 0$
      • $\Gamma_{00}^{i} \approx -\frac{1}{2} \eta^{ij} \frac{\partial h_{00}}{\partial x^{j}} $
    • $\frac{dx^{\mu}}{d\tau^{2}}-\frac{1}{2}\eta^{ij} \frac{\partial h_{00}}{\partial x^{j}}(\frac{dx^{0}}{d\tau})^{2}=0$
      • $\frac{dx^{0}}{d\tau} = 1$
    • Defining $\Phi(x^{\alpha}) = \frac{-1}{2}h_{00}(x^{\alpha})$
      • In real units, we have that $\frac{-2\Phi}{c^{2}} h_{00}$
      • From the assumptions, potential is only spatial dependent. It is currently written more generally
      • Substituting this into the geodesic equation yields Newton’s 2nd law
      • Substituting into metric yields:
        • $g_{00} = -(1+2\Phi)$
        • $g_{ij} = (1-2\Phi)\delta_{ij}$
        • Called the weak, quasi-static metric

Curved or not Curved

  • It is possible to transform the quasi-static metric into one that looks like the Minkowski metric (ie. the realization of the equivalence principle)

  • In general, given some metric, how do we know if the space is curved or not?

  • Suppose that we have some arbitrary coordinate system $c^{\mu}$ with an associated metric $g_{\mu\nu}$

  • Converting to some primed system, we get that

    • $g_{\mu\nu}’ = \frac{\partial x^{\alpha}}{\partial x^{\mu}}\frac{\partial x^{\beta}}{\partial x^{\nu}} g_{\alpha\beta}$
    • This is 10 independent equations. (In general, it is $\frac{N(N+1)}{2}$ independent equations)
    • Since we have 16 entries to play with, we can easily satisfy these constraints

  • We want the metric to locally look like $\eta_{\mu\nu}$. (ie. the first derivative of the metric should be 0)

    • Taking the derivative w.r.t $\frac{\partial}{\partial x’^{\gamma}}$ of both sides, collecting like terms after relabelling indices, and demanding that this equals 0, we find that we must examine the constraints on $\frac{\partial x^{\alpha}}{\partial x'^{\mu}x'^{\gamma}}$
      • Since this is symmetric in the bottom indices, and there are 4 possibilities for $\alpha$, we see that there are 40 free entries
      • From $\frac{\partial g_{\mu\nu}}{\partial x'^{\gamma}}$, we have 40 constraints, which is just satisfied by the number of free entries
    • The above computation can be extended to higher order derivatives. It turns out, you cannot make the 2nd order derivatives and higher vanish due to being overconstrained. For 2nd order, you are missing 20 numbers to statisfy 20 missing constraints

Tensors

  • Building off of above, we can encode this missing information as the Reimannian tensor to encode this

  • We need some scaffolding to understand what that means though

General Covariance

  • Simple statement: The laws of physics should be able to be written in a form that is valid in any coordinate system

Useful facts

  • $J = \frac{\partial x^{\mu}}{\partial x^{\nu}}$ (transformation matrix)

  • $\frac{\partial x^{\mu}}{\partial \bar{x}^{\nu}}\frac{\partial \bar{x}^{\alpha}}{\partial x^{\nu}} = \delta^{\alpha}_{\nu}$

  • $\frac{\partial x’^{\mu}}{\partial \bar{x}^{\nu}} = \frac{\partial x’^{\mu}}{\partial \bar{x}^{\alpha}}\frac{\partial \bar{x}^{\alpha}}{\partial \bar{x}^{\nu}}$

Scalar Field Transformation

  • Scalar fields are unchanged by a change of coordinates

Vector Field Transformation

  • Consider some curve $g^{\alpha}(\lambda)$ in $x^{\mu}$ coordinates

  • Define the tangent field $T^{\alpha} = \frac{dy^{\alpha}}{d\lambda}$

  • Converting to a different coordinate system yields that $\bar{T}^{\alpha} = \frac{\partial \bar{x}^{\alpha}}{\partial x^{\mu}}T^{\mu}$

    • This is exactly how 4-vectors transform
    • Vectors that transform like this are called covariant vectors

Dual Vector Transformation

  • The canonical example of a dual is the gradient

    • Suppose that we have some function $f(x^{\mu})$. define the gradient as
      • $F_{\alpha} = \frac{\partial f}{\partial x^{\alpha}}$
      • From chain rule, we see that $\bar{F_{\alpha}} = \frac{\partial x^{\beta}}{\partial \bar{x}^{\alpha}} F_{\beta}$
      • This transformation is general
    • Things that transform like a gradient are called contravariant

  • Contracting a contravariant vector with a covariant vector yields a scalar (ie. an invariant under coordinate transformation)

    • Can also think of these as linear maps from 4 vectors to scalars

  • The above can easily be generalized to objects with an arbitrary number of indices. Their transformation law immediately generalizes from above.

    • To be succient: $\bar{T} = AT$. Assume an n,m tensor. To fill out A as follows: A starts T 1
      • For each upper index, add a $\frac{\partial \bar{x}^{\alpha’}}{\partial x^{\alpha}}$
      • For each lower index, add a $\frac{\partial x^{\alpha}}{\partial \bar{x}^{\alpha’}}$

  • Not everything with indices is a tensor (like the Christoffel symbols)

Stress-Energy Tensor $T^{\mu\nu}$

  • A symmetric (2,0) tensor which generalizes the mass in Newtonian gravity

  • $T^{\mu\nu} = (\rho+P)U^{\nu}U^{\mu}+Pg^{\mu\nu}$

    • $U^{\nu}$ is the 4 velocity
    • Assumes a perfect fluid (a fluid that is perfectly characterized by it’s rest-frame energy $\rho$ and it’s rest frame pressure P)
    • Let’s define that
      • Going back to SR
      • $\rho = \Sigma\frac{E+m}{V}$
        • energy density, including rest mass
        • In rest frame: $\rho_{0} = \frac{\Sigma_{i} m_{i}}{V_{0}}$
        • Transforming to another frame moving at v yields that $m \rightararow \gamma m$ and $V\rightarrow \frac{V}{\gamma}$
        • Hence $\rho’ \rightarrow \gamma^{2} \rho_{0}$
        • So the energy density alone is NOT invariant
      • Hence, we need a full tensor to get some invariant quantity. On an element by element basis, we have that
        • $T^{00}$ flux of $p^{0}$ (energy) in the $x^{0}$ (time) direction $\rho$
        • $T^{0i}$ flux of $p^{i}$ along the $x^{0}$ direction: momentum density
        • $T^{ij}$ denotes some forces
          • off diagonal are shearing forces
          • diagonal are pressures in x,y,z direction

Dust

  • Define $N^{\mu} = nU^{\mu}$ where n is the rest-frame number density (ie. the number of particles in a given volume)

  • The stress energy tensor of dust is then $T^{\mu\nu} = \rho U^{\mu}U^{\nu}$

  • Generalizing to an arbitrary local frame, we get that $T^{\mu\nu} = (\rho+P)U^{\mu}U^{\nu}+P\eta_{\mu\nu}$

  • To generalize to GR, we cacn just replace $\eta_{\mu\nu}$ with $g_{\mu\nu}$

Covariant Derivatives

  • How does the derivatie of a tensor transform like?

  • Let’s first look at how a dual vector transform’s like

    • $\frac{\partial F_{\alpha}}{\partial x^{\beta}} = \frac{\partial^{2}f}{\partial x^{\alpha}\partial x^{\beta}}$
    • $F_{\alpha} = \frac{\partial f}{\partial x^{\alpha}}$ where f is a scalar field
    • What you get is a normal tensor transformation, plus a term that involves the Christoffel symbol. Hence, dua vectors DON’T transform in a tensorial matter
    • As an aside, this implies the Hessian is NOT a tensor

  • As such, let us define a new type of derivative called the covariant derivative. Acting on a covariant tensor

    • $\nabla_{\beta}F_{\alpha} = \partial_{\beta}F_{\alpha}-\Gamma_{\alpha\beta}^{\mu}F_{\mu}$

  • Similarly, the covariant derivative acting on a contravariant tensor yields:

    • $\nabla_{\beta}F^{\alpha} = \partial_{\beta}F^{\alpha}+\Gamma_{\alpha\beta}^{\mu}F_{\mu}$

  • The above immediately generalizes to arbitrary rank tensors

Properties of Covariant Derivatives

  • Obeys linearity

    • $\nabla_{\alpha}(aT+bS) = a\nabla_{\alpha}T+b\nabla_{\alpha}S$
    • Assumes that the tensors have the same rank and the same index structure (ie. all indices match up)

  • Normal product rule

    • $\nabla_{\alpha}(TS) = S\nabla_{\alpha}T+T\nabla_{\alpha}S$
    • This holds for arbitrary tensors

  • The covariant derivative commutes with contraction

    • $\nabla_{\beta}(T_{\alpha}^{\alpha}) = (\nabla T)^{\alpha}_{\alpha\beta}$
      • Expanded out yields: $\partial_{\alpha} T_{\beta}^{\alpha}+\Gamma_{\alpha\sigma}^{\alpha}T_{\beta}^{\sigma}-\Gamma_{\alpha\beta}^{\sigma}T_{\alpha}^{\sigma}$

  • Metric compatible. ie.

    • $\nabla_{\alpha}g_{\mu\nu} = 0$
    • This also implies that $\nabla_{\alpha}g^{\mu\nu} = 0$

  • We assume that covariant derivatives commute (assumes torsion-free field)

  • Aside: Short hand for partials is ,a, and shorthand for covariant derivative is ;a

Tidal Forces of Curvature

  • In order to measure tidal fores, we need at least 2 particles

Newtonian

  • In Newtonian, we have that $\vec{a} = -\nabla\phi \rightarrow \frac{dx^{2}}{dt^{2}} = -\delta_{ij}\partial_{j}\Phi$

    • $\partial_{j} = \frac{\partial}{\partial x^{j}}$

  • Consider a second particle some distance $\vec{b(t)}$ away from the first particle. The equation of motion becomes

    • $\frac{d^{2}b}{dt^{2}} = -\delta^{ij}b^{k}(\partial_{k}\partial_{j}\phi)$
    • The tidal tensor is then $\frac{\partial^{2} \Phi}{\partial x^{k} \partial x^{j}}$
      • The trace of this tensor shows up in Poisson’s equation
        • $\nabla^{2} \Phi = \delta^{ij}\partial_{j}\partial_{k}\Phi =4\pi G\rho$

GR version

  • Consider two geodesic. One is defined by some 4 vector $x^{\mu}(\tau)$, the other by $(x^{\mu}+b^{\mu})(\tau)$ where the b is the deviation from the first particle

Directional Covariant Derivative

  • Define the directional covariant derivative ($V^\mu$) of $b^\mu$ along the first geodesic: $V^{\mu} = \frac{Db^{\mu}}{D\tau} = \frac{dx^{\nu}}{d\tau}\nabla_{\nu} B^{\mu} = UB^{\nu}_{;\nu}$

    • Writing this out: $V^{\mu} = \frac{db^{\mu}}{d\tau}+\Lambda_{\sigma\nu}^{\mu}U^{\nu}b^{\sigma}$
    • Can interpret this as the relative velocity of the geodesics

  • Can also define a relative acceleration:

    • $A^{\mu} = \frac{dV^{\mu}}{d\tau}+\Lambda_{\sigma\nu}^{\mu}U^{\nu}V^{\sigma}$

  • Writing down the fundamental steps

    • Plugging in definition of $V^{\mu}$ into relative acceleration, and expand out (very messy)
    • Use geodesic equation to substitute derivative of 4 vectors: $\frac{dU^{\beta}}{d\tau} = -\Lambda_{\mu\nu}^{\beta}U^{\mu}U^{\nu}$
    • Make the observation that $\frac{dx^{\delta}}{d\tau} \Lambda_{\beta\gamma}^{\alpha} = U^{\delta}\partial_{\delta}\Lambda_{\beta\sigma}^{\alpha}$
    • After a bit of algebra, you can see that you can factor out a factor of $U^{\mu}U^{\epsilon}B^{\gamma}$ from everything. Do that.
    • Now, in parallel, write down the geodesic equation for the second path, and Taylor expand to first order w.r.t. $B^{\mu}$.
      • Can plug in geodesic equation of 1st path to simplify things
      • Eventually, you get $\frac{d^{2}B^{\alpha}}{d\tau^{2}}+2\Gamma_{\beta\gamma}^{\alpha}U^{\beta}\frac{dB^{\gamma}}{d\tau} = -U^{\beta}U^{\gamma}B^{\delta} \partial_{\delta}\Gamma_{\beta\gamma}^{\alpha}$
    • Plugging in the above into the relative acceleration equation yields that
      • $A^{\alpha} = -R_{\beta\gamma\sigma}^{\alpha}U^{\beta}U^{\gamma}B^{\gamma}$
        • where $R_{\beta\gamma\sigma}^{\alpha} = -\partial_{\delta}\Gamma_{\beta\gamma}^{\alpha}+\partial_{\gamma}\Gamma_{\beta\delta}^{\alpha}+\Gamma_{\epsilon\gamma}^{\alpha}\Gamma_{\beta\delta}^{\epsilon}-\Gamma_{\beta\epsilon}^{\alpha}\Gamma_{\delta\gamma}^{\epsilon}$ is the Reimann tensor (ie. the curvature tensor)

Reimannian tensor

  • An important case: what is the metric if the space is locally Minkowski? The first derivatives vanish, the 0th derivative is constant, and the 2nd derivatives are non-zero

  • Eventually, can show that $R_{\nu\eta\lambda}^{\sigma} = \frac{1}{2}\eta^{\sigma\delta}(\partial_{\mu}g_{\mu\nu} \partial_{\nu}+\partial_{\lambda}\partial_{\delta}g_{\nu\mu})$ (ie. the 2nd derivatives of metric is the Reimannian tensor)

Reimannian Tensor properties

  • $(\nabla_{\mu}\nabla_{\nu}-\nabla_{\nu}\nabla_{\mu})V^{\sigma} = R_{\lambda \nu\nu}^{\sigma} V^{\lambda}$

  • This is a rank 4 tensor

  • Can be written in terms of the metric

  • The total number of independent entries are $\frac{N^{2}(N^{2}-1)}{12}$

  • Can be used to determine if a given metric describes a curved space

  • Symmetries

    • $R_{\alpha\beta\gamma\delta} = -R_{\beta\alpha\gamma\delta}$
    • $R_{\alpha\beta\gamma\delta} = -R_{\alpha\beta\delta\gamma}$
    • $R_{\alpha\beta\gamma\delta} = -R_{\beta\alpha\delta\gamma}$
    • $R_{\alpha\beta\gamma\delta} = R_{\gamma\delta\alpha\beta}$
    • $\nabla_{\lambda}R_{\mu\nu\rho\sigma}+\nabla_{\nu}R_{\lambda\mu\rho\sigma}+\nabla_{\mu}R_{\nu\lambda\rho\sigma}=0$
      • Bianchi identity

  • Ricci tensor

    • The trace of the Reimann tensor over it’s first and third indices
    • ie $R_{\alpha\beta} = R_{\alpha\gamma\beta}^{\gamma} = g^{\gamma\gamma}R^{\gamma}_{\alpha\gamma\beta}$

  • Ricci scalar

    • Contract the ricci tensor again to get this.
    • $R = R_{\alpha}^{\alpha} = g^{\alpha\beta}R_{\alpha\beta}$

Parallel Transport

  • Consider some curve $y^{\alpha}(\lambda)$. Let the tangent vector be equal to $T = \frac{dy^{\alpha}}{d\lambda}$

  • Let the directional covariant derivative of the vector be be $\frac{DV^{\mu}}{D\lambda}=T^{\alpha}\nabla_{\alpha}V^{\mu}$

    • If the directional convariant derivative is 0, then the vector is parallel-transported along the curve

  • If you can’t parallel transport a vector around, then this is indicative of a underlying curved space

  • Can also derive Reimannian tensor via comparing parallel transport around opposing sides of parallelogram. End result of that is

    • $\delta V^{\mu} = R_{\alpha\rho\sigma}^{\mu} V^{\alpha}A^{\rho}B^{\sigma}$ where A and B and the 4 vector sides of the parallelogram

Stress Energy Tensor Conservation

  • In general: $\nabla_{\mu}T^{\mu\nu} = 0$
    • $\mu=0$ component is energy conservation
    • $\nu=i$ is Euler equation for perfect fluids (ie. momentum conservation in fluids)

EM Force Law

  • In GR, we find that $a^{\mu} = F^{\mu}_{\nu}u^{\nu}$

    • F is the Faraday tensor
    • In terms of the potential, you have: $F^{\mu\nu} = \partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}$

  • Can also describe current 4 vector: $J^{\mu} = (\rho,\vec{j})$

    • Conservation of charge is then $J^{\mu}_{,\mu} = 0$. Convert comma to semicolon to get GR form

  • Maxwell’s equations can then be described by: $\partial_{\gamma}F_{\alpha\beta}+F_{\alpha}F_{\beta\gamma}+\partial_{\beta}F_{\gamma\alpha} = J^{\mu}$

    • Assuming free space ($J^{\mu} =0$), you cover the Bianchi identity

Einstein’s Equation

  • Want to generalize Poisson’s equation: $\nabla^{2} 4\pi G\rho$

  • Since $T_{00} = \rho$, this suggests that $T_{\mu\nu}$ is the relativistic generalization

  • Similarly, we know that the LHS of Poisson is the trace of the tidal tensor. The tidal tensor generalizes to the Reimannian tensor. Hence, the most naive guess of the field equations is

    • $R_{\mu\nu} = KT_{\mu\nu}$
    • From conservation of energy/momentum, we know covariant derivative of RHS is 0. In general, however, $\nabla^{\mu}R_{\mu\nu} \neq 0$

  • We need to introduce something on the LHS to get 0

  • Start with Bianchi identity applied to Reimannian tensor:

    • $\nabla_{\lambda}R_{\mu\nu\rho\sigma}+\nabla_{\nu}R_{\lambda\mu\rho\sigma}+\nabla_{\mu}R_{\nu\lambda\rho\sigma} = 0 $
    • Apply $g^{\sigma\lambda}g^{\mu\rho}$, and make appropriate simplifications to get
      • $\nabla^{\sigma}R_{\nu\sigma}-\nabla_{\nu}R+ \nabla^{\rho}R_{\nu\rho} = 0$
      • Relabelling dummy indices and rearranging, we get: $\nabla^{\nu}R_{\mu\nu} = \frac{1}{2}\nabla_{\nu}R$
        • Since $R\neq 0 $ in general, this explains why the naive equation fails

  • Let’s define a new tensor that subtracts off the offending term. Letting this tensor be the LHS and stress-energy be the RHS yeilds

    • $G_{\mu\nu} = k T_{\mu\nu}$
      • $G_{\mu\nu} = R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R$ is the Einstein tensor
      • $\nabla^{\nu}G_{\mu\nu} = 0$

What the Heck is $\kappa$?

  • Only thing left to do is figure out what k is. We need this equation to reduce Newton’s law of gravitation. Let’s take this limit; doing so will enable use to figure out k

  • It can be show by multiplying by the raised metric that $-R=T^{\mu}_{\mu} \kappa = T\kappa$

  • This enables us to write the trace-reversed Einstein equation: $R_{\mu\nu} = \kappa(T_{\mu\nu}-\frac{1}{2}g_{\mu\nu}T)$

  • The Newtonian limit impies that $v« 1$ and $P « \rho$ (pressures small compared to mass energy)

    • $T_{00} = \rho$, and rest of tensor is 0
    • This implies $T_{\mu}^{\mu} \approx g^{00}T_{00} \approx \eta^{00}T_{00} =-\rho$
    • This also implies that $g_{\mu\nu} = \eta_{\mu\nu}+h_{\mu\nu}$
    • $g_{00} = -(1+2\phi(\vec{r}))$
      • $\phi « 1$

  • Plugging in the above quantities into trace reversed equation

    • $R_{00} = \kappa \rho(1-\frac{1}{2}) = \frac{\kappa\rho}{2}$
    • $R_{00} = \nabla^{2} \phi$ for quasi-static weak field
    • Comparing to Possion yields that $8\pi G = \kappa$
    • To recover physical units, $\kappa = \frac{8\pi G}{c^{4}}$

  • Highly nonlinear

Cosmological Constant

  • Einstein’s equations can be modifed to account for expansion via adding an additional term which does not violate the Bianchi identity:
    • $G^{\mu\nu} = 8\pi G (T^{\mu\nu}-\frac{\Lambda}{8\pi G} g^{\mu\nu})$

Killing Vector

  • Assume that the metric is independent of some coordinate

  • There esistes some 4-vector associated with this symmetry, which is called the Killing vector $X_{\mu}$

    • Might not be obvious in a particular coordinate system

  • All Killing vectors obey the Killing equation:

    • $\nabla_{\beta}X_{\alpha}-\nabla_{\alpha}X_{\beta} = 0$

  • Killing vectors also satisfy the equation $X_{\mu}u^{\mu} = constant$

  • Proof:

    • Recall the alternate form of the Geodesic equation $T^{\mu}\nabla_{\mu} T^{\alpha} = 0$
      • $T^{\alpha}$ is the tangent vector
      • We let $T^{\mu} = u^{\mu}$ (ie. a 4-velocity)
    • Since the metric is covariant compatible, can raise and lower indices inside covariant derivative with impunity. So we lower the $\alpha$
    • Expand out everything, can abuse symmetry of metric to see that terms get eliminated. You eventually get:
      • $\frac{du_{\alpha}}{d\tau} = \frac{1}{2}(u^{\mu}u^{\nu}\partial_{\alpha}g_{\mu\nu})$
      • if metric independent of a coordinate, then $\partial_{\alpha} g_{\mu\nu} = 0$
      • Since $u_{a} = X_{a}^{\mu} u_{\mu}$, then this implies the statement

  • If $X^{\mu}$ satisfies the Killing equation, it is always possible to find a coordinate system where $\partial_{\alpha} g_{\mu\nu} = 0$

    • However, you can’t always find a coordinate system where all killing vectors satisfy the above equation

Schwarzschild Solution

  • We have a spherical distrubution of mass surrounded by a vacuum. What is the metric of this system

  • $G_{\mu\nu} = 0$ in the region outside the mass distribution

    • $R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu} = 0$
      • $R = -8\pi G Tr(T) = 0$ in vaccum (Tr(T) is 0 if T is 0)
    • Hence $R_{\mu\nu} = 0$

  • Make the following assumptions:

    • Static
      • Metric components are time independent
      • no time-space cross terms
    • Spherically symmetric
      • In Minkowski
        • $ds^{s} = dt^{2}+dr^{2}+r^{2}d\Omega^{2}$
          • $d\Omega^{2} = d\theta^{2}+\sin^{2}\theta d\phi^{2}$
      • Assume that the angular terms in the metric should look like the $d\Omega^{2}$ term in the metric (ignore $r{2}$ prefactor term. Thank can change)

  • Make the ansatz metric be

    • $ds^{2} = -e^{2\alpha(r)}dt^{2}+e^{2\beta(r)}dr^{2}+e^{2\gamma(r)}r^{2}d\Omega^{2}$
      • Arbitrary functions can’t be time dependent (static) or depend on angle( spherical symmetry)
      • Arbitrary functions in power of exponent since that prevents sign of trace from changing
    • Define $\tilde{r} = e^{\gamma(r)}r$
      • $d\tilde{r} = e^{\gamma(r)}dr+re^{\gamma(r)}d\gamma$
    • Making that substitution yields
      • $ds^{2}= -e^{2\alpha(r)}dt^{2}+(1+r\frac{d\gamma}{dr})^{-2}e^{2(\beta-\gamma)}\tilde{r}^{2}+\tilde{r}^{2}d\Omega^{2}$
      • Can rename $\tilde{r}$ to be r, and then redefine the prefactore of $dr^{2}$ to be some other arbitrary function called $e^{2\beta}$
        • $ds^{2}= -e^{2\alpha(r)}dt^{2}+e^{2\beta}\tilde{r}^{2}+\tilde{r}^{2}d\Omega^{2}$
          • This eliminates the the $\gamma$ constraint

  • Compute all the Christophel symbols now GLHF

    • The metric being diagonal helps eliminate a lot of terms

  • Calculate Reimann tensor, then calculate the Ricci tensor (lots of tedious algebra). You eventually get for the Ricci tensor that

    • $R_{tt} = e^{2(\alpha-\beta)}(\frac{\partial^{2}}{\partial r^{2}}+(\frac{\partial}{\partial r} \alpha)^{2}-(\frac{\partial}{\partial r} \alpha)\frac{\partial}{\partial r} \beta + \frac{2}{r}\partial_{r} \alpha)$
    • $R_{rr} = -\partial_{r}^{2}\alpha-(\partial_{r} \alpha)^{2} + \partial_{r}\alpha \partial_{r}\beta +\frac{2}{r}\partial_{r}\beta$
    • $R_{\theta\theta}= e^{-2\beta}(r(\partial_{r}\beta -\partial_{r} \alpha) -1)+1$
    • $R_{\phi\phi} = R_{\theta\theta}\sin^{2}\theta$

  • We demand that each of the entries in the Ricci tensor is 0

    • Combine $R_{tt}$ and $R_{rr}$, and the rearrange to get the following dif eq:
      • $\frac{2}{r}(\partial_{r}\alpha+\partial_{r}\beta) \rightarrow \alpha(r) = -\beta(r)+C$
        • Can set C=0 if you rescale time coordinate: $t \rightarrow e^{-C}t$
    • Solving $R_{\theta\theta} = 0$ yields
      • $e^{2\alpha}(-2r\partial_{r}\alpha -1) = -1$
        • $e^{2\alpha} = 1-\frac{R_{s}}{r} = e^{-2\beta}$

  • Hence, the metric becomes

    • $ds^{2} = -(1-\frac{R_{s}}{r})dt^{2}+(1-\frac{R_{s}}{r})^{-1}dr^{2} +r^{2}d\Omega^{2}$
    • Relating to Weak field metric, we find that $1-\frac{R_{s}}{r} = 1+2\Phi$, which implies that $R_{s} = \frac{2GM}{c^{2}}$
      • M here refers to the rest mass, plus the gravitational binding energy

  • Can use Einstein coordinates to remove singularity at $r = 2Gm$. Not done in class since it is really tedious (which is saying something)

Potential

  • $V(r) = \frac{\epsilon}{2}-\frac{GM}{r}\epsilon+\frac{l^{2}}{2r^{2}}-\frac{GMl^{2}}{r{3}}$
    • Derived from examining Killing Vectors of Schwarzchild metric
    • massless ($\epsilon=0$) only has unstable circular orbits, and will fly off to 0 or infinity if peturbed
    • massive has circular orbits of radius $r_{e} = \frac{l^{2}\pm l\sqrt{l^{2}-12(GM)^{2}}}{2GM}$
      • In the large angular momentum limit, this reduces to newtonian case $r_{e} = \frac{l^{2}}{GM}$

Redshift

  • Consider an observer moving at some 4 velocity $u^{\mu}$ who is stationary in th Schwarzchild coordinate system
    • $g_{\mu\nu}u^{\nu}u^{\mu} = -1$
    • $g_{00}(u^{0})^{2} = -1$
    • $u^{0} = (1-\frac{2Gm}{r})^{0.5}$
    • Consider some photon moving along a null geodesic.
      • $E_{\gamma} = h\nu$
      • In the observer coordinate system: $E_{ob} = -p_{\mu}u^{\mu} = -g_{00}p^0u^0 = \sqrt{1-\frac{2Gm}{r}}E_{\gamma}$
      • We know that $E = -\zeta_{\mu}\frac{dx^{\mu}}{d\lambda} = (1-\frac{2GM}{r})\frac{dt}{d\lambda}$
        • Hence: $\nu_{obs} = \frac{E}{h\sqrt{1-\frac{2GM}{r}}}$

Lensing

  • Find the equation of motion for a massless particle in Schwarzchild metric
    • $\epsilon = 0 = -g_{\mu\nu}u^{\mu}u^{\nu}$
    • Plug this into the potential of the Schwarzchild metric. The equation of motion becomes
      • $\frac{1}{2}(\frac{dr}{d\lambda})^{2}+\frac{l^{2}}{2r^{2}}-\frac{GMl^{2}}{r^{3}}=\frac{E^{2}}{2}$
        • Can solve this with $x = \frac{1}{r}$ substitution
        • $\frac{d^{2}x}{d\phi^{2}}+x = \frac{3G^{2}M^{2}}{l^{2}}x^{2}$
        • Can get approximate solution via peturbations: $x = x_{0}(\phi)+x_{1}(\phi)$
        • Final result is $\Delta \phi = \frac{4GM}{Rc^{2}}$