Graduate General Relativity

Following Misner, Wheeler and Thorne.

Foundations of General Relativity

• All laws of physics can be expressed geometrically
• Local reference frames are flat

Basic Definitions and Math Things

• Events: Things that happen at a particular point in spacetime. Does not depend on the underlying spacetime structure (re: could be flat or curved)
• Curves: A curve is some path through spacetime $P(\lambda)$. Once again, this is independent of the notion of any underlying spacetime (although a metric is needed to ascribe a length to the curve)
• Coordinates: How you label events in spacetime. There is no unique way of assigning coordinates
  • Written as $x_{\mu}$, where $\mu$ ranges from 0 to 3
  • A coordinate transformation is a set of 4 equations. Each one determines how the new coordinate depends on the old 4 coordinates
  • Coordinate singularities can occur (think of the north pole of the earth on latitude and longitude)
    • These can be dealt with via multiple patches of coordinates
• Vectors: The seperation between two events in spacetime. In flat spacetime, this is the difference between the two coordinates assigned to each event. This falls apart in curved spacetime, but is an good approximation for arbitrarily close events
  • These transform under coordinate transformations as: $\epsilon^{\beta} = \frac{\partial x^{\alpha}}{\partial x^{\beta}} \epsilon^{\beta}$
    • Follows from Taylor expansion
  • Can reduce the notion of a vector from one that requires 2 events to one that requires 1
    • Imagine a parameterized line between the two events: $P(\lambda) = A + \lambda(B-A)$, where $0\leq \lambda \leq 1$
    • Taking the derivative w.r.t. $\lambda$ evaluated at $\lambda=0$ gives P(1)-P(0). This construction $\frac{dP}{d\lambda}_{\lambda=0}$ is a 1 point object called the tangent vector
• Summation Notation: $\epsilon_{\alpha}\gamma^{\alpha} = \Sigma_{\alpha=0}^{3} \epsilon_{\alpha}\gamma^{\alpha}$
• The metric tensor:
  • a machine for computing scalar products of vectors. It takes in two vectors and spits out a scalar
  • It’s symmetric in it’s arguments: $g(u,v) = g(v,u)$
  • It’s linear w.r.t. it’s arguments: $g(a\vec{u}+b\vec{v},\vec{w}) = a g(\vec{u},\vec{w}) + b g(\vec{v},\vec{w})$
  • If you know the basis vectors of the frame ($\vec{e_{\alpha}}$), then you can calculate it’s output for any input (follows from linearity)
    • Define the metric coefficients to be $g(e_{\alpha}, e_{\beta}) = \vec{e_{\alpha}} \cdot \vec{e}_{\beta} $
    • The scalar product then becomes: $u^{\alpha}v^{\beta}g_{\alpha\beta}$
  • In special relativity, this is called $\eta_{\alpha\beta}$, which is diagonal with time being -1 and the space coordinate being 1
• 1-forms: The 3rd class of geometric objects
  • Think of the 4-momentum vector $p_{alpha} = m\mu_{\alpha}$
  • The de Broglie wavelength gives an alternative interpretation for momentum. If you diffract the wave on a lattice and observe the diffraction pattern, you can then map surfaces of equal phase. This series of surfaces are a one-form
  • If you run a vector through this one form from ove event to another, then you can calculate the phase difference between the point via $<\tilde{k},\vec{v}>$
  • You can think of the 1-form as the local form of these equal phase surfaces (re: the best linear approximation of the phase $\phi$ near an event)
  • More mathematically, you can think of a 1 form as a linear, real-valued funtion of vectors (re: something that maps vectors to scalars)
  • the set of all 1-forms at a given event is a “vector space”
  • For each vector $\vec{p}$, the is a unique 1-form defined by $\vec{p}\cdot \vec{v} <\tilde{p},\vec{v}>$ which holds for all $\vec{v}$
    • In words: the projection of v onto the 4 momentum equals the number of surfaces the vector v pieces of the 1-form
  • The gradient of a scalar $\mathbf{d}f$ is the simplest example of a 1-form
    • The more familiar vector notion of the gradient is the unique vector associated with the 1-form notion of the gradient
    • Define $\partial_{\vec{v}} = (\frac{d}{d\lambda}) $ at $\lambda=0$ along the curve $P(\lambda)-P(0) = \lambda v$ as the directional derivative
    • The gradient describes the first order changes in f in the neighborhood of $P_{0}$: $f(P) = f(P_{0}) <\mathbf{d}f, P-P_{0}>$
    • The gradient is related to the directional derivative. Apply the directional derivative to a scalar function f: $\partial_{\vec{v}} f = <\mathbf{d}f, \frac{dP}{d \lambda} = <\mathbf{d}f, \vec{v}>$
• Tensors: The generalization of these lower dimensional objects. They can have multiple indices which transform under the Poincare group
  • Think of tensors as multilinear maps which take in an appropriate number of one-forms and vectors and then outputs as scalar
  • The order in which you insert vectors/1-forms matters
  • If one knows the value of a tensor in 1 frame for some basis vectors/1-forms, then you can transform to any other rest frame (just transform each of vectors/1-forms with their appropriate matrices, then tack those onto the tensor to get the transformation law)
  • We define the “rank” of the tensor as the number of indices used to describe it
  • Tensor Operations (for example’s sake, we will deal with a 4-tensor):
    • Gradient: $\nabla S(u,v,w,\epsilon)$ is $\partial_{\epsilon} S(\vec{u},\vec{v},\vec{w})$ with u,v and w fixed
      • This is roughly the difference of S(u,v,w) from the tip of $\epsilon$ to the tail of $\epsilon$
      • In component notation: $\nabla S(u,v,w,\epsilon) = (\frac{\partial S_{\alpha\beta\gamma}}{\partial x^{\alpha}} \epsilon^{\delta}) u^{\alpha}v^{\beta}w^{\gamma} = S_{\alpha\beta\gamma,\delta} u^{\alpha}v^{\beta}w^{\gamma}\epsilon^{\delta}$
      • This increases the rank of the tensor by 1
    • Contraction: You can reduce the rank of a tensor by 2 via contraction
      • $M(u,v) = \Sigma_{a=0}^{3} R(e_{a}, u, w^{a}, v) $
      • You need to have one index up and one index down prior to summing them
    • Divergence: You take the gradient of a tensor, and then contract it with one of the original slots
      • Hence $\nabla \cdot S = S^{\alpha}_{\beta\gamma,\alpha}$
        • If you take the divergence of S on the first slot
    • Tranpose: interchange two slots for each other
    • Symmetrization and Anti-symmetrization
      • A tensor is symmetric if any transposition of it’s indices yields the same tensor
      • A tensor is antisymmetric if swapping indicies causes the tensor to reverse sign
    • Wedge product: a way to construct a completely antisymmetric tensor from a set of vectors and 1-forms
      • Define the “bivector” $u \wedge v = u\otimes v - v\otimes u$
      • Define the “2-form” $\alpha \wedge \beta = \alpha \otimes \beta - \beta \otimes \alpha$
      • If a vector is a linear combination of other vectors in the set, the wedge product between all the vectors is 0
      • You can define a generalized p-form as a completely anti-symmetric tensor of rank p, which you need to normalize by $p!$. Wheather there is a - or + is determined by the Levi-Civiti symbol
      • Wedge products obey distributive and addition, and commute with scalar multiplication and addition
      • If you are commuting wedge products (say that you have a p-form a, and a q-form b):
        • $\alpha \wedge \beta = (-1)^{pq} \beta \wedge \alpha$
    • Duals: For vectors, antisymmetric rank 2 tensors, and antisymmetric rank 3 tensors:
      • $*J_{\alpha\beta\gamma} = J^{\mu}\epsilon_{\mu\alpha\beta\gamma}$
      • $*F_{\alpha\beta} = \frac{1}{2} F^{\mu\nu}\epsilon_{\mu\nu\alpha\beta}$
      • $*B_{\alpha} = \frac{1}{3!} B^{\lambda\mu\nu}\epsilon_{\lambda\mu\nu\alpha}$
      • More generally, the dual is denoted as Hodge star (*) and describes the one to one correspondence between a p form and an n-p form
        • Requires specifying the metric for this to make sense
        • $*(\epsilon^{i_1}\wedge … \wedge \epsilon^{i_p}) = \frac{1}{(n-p)!}{\epsilon^{i_1…i_{p}}}_{j{p+1}… j{n}}\epsilon^{j{p+1}}\wedge … \epsilon^{jn}$
          • Uses the metric to raise the first p indices of $\epsilon_{1…n} = \sqrt{-g} \tilde{\epsilon}_{1…n}$ where $\tilde{e}$ is the Levi Civiti symbol
            • The $\sqrt{g}$ is needed to make it transform like a tensor
• Reference frames: a system to relate two coordinate systems to each other
  • Intertial reference frames are the preferred class since they require the minimal set of forces needed to describe motion. Non-interital frames tack on additional forces (re: forces that arise from accelerating w.r.t. the inertial one)

Special Relativity

• Suppose that you parameterize your line via the proper time $\tau$
  • $\tau$ is the 4-magnitude of your vector (Think of it as the frame where no velocity is happening)
  • With a set of 4 orthonormal basis vectors $\vec{e_{\alpha}}$, you can construct the 4-velocity u as $\vec{u} = u^{0}\vec{e_{0}} + u^{1}\vec{e_{1}} + u^{2}\vec{e_{2}} + u^{3}\vec{e_{3}}$
    • $u^{0} = \frac{dt}{d\tau} = \frac{1}{\sqrt{1-v^{2}}}$
    • $u^{j} = \frac{dx^{j}}{d\tau} = \frac{v^{j}}{\sqrt{1-v^{2}}}$
    • $v^{2}$ is the ordinary 3d velocity, and $v^{j}$ is that paricular coordinate of the velocity (doesn’t need to be Cartesian )

Lorentz Transformations

• Define some set of matrices $\Lambda_{\alpha}^{\beta}$ which transforms your coordinates from one frame to another
  • $x^{a} = \Lambda_{b}^{a} x^{b}$ and $x^{b} = \Lambda_{a}^{b} x^{a}$
• Going from a primed frame to an unprimed frame and back is the identity. Hence the Lorentz transform must be inverses of each other: $\Lambda_{\beta}^{\alpha} \Lambda_{\gamma}^{\beta} = \delta^{\alpha}_{\gamma}$
• The Lorentz transforms define how the basis vectors and 1 forms transforms via the Einstein summation notation
• In matrix notation, suppose that your frame is moving along the x axis. Then your Lorentz matrix takes the form $\begin{pmatrix}\gamma & -\beta \gamma & 0 & 0\\ -\beta \gamma & \gamma &0 &0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$
• This can be framed in hyperbolic geometry by making the associations $\beta = \tanh (\alpha)$, $\sinh (\alpha) = \beta \gamma$ and $\cosh (\alpha) = \gamma$
  • This lets you easily add boosts together by simply adding the $\alpha$

E&M Fields

• We want to write the Lorentz Force law in some frame-invariant way
• The standard form: $\frac{\vec{p}}{dt} = e(\vec{E} + \vec{v}\times\vec{B})$
  • This depends on a lot of frame dependent quantities (3-momentum, the time of a specific frame, E and B fields etc.)
• The frame-invariant form should only depend on frame invariant quantities. Swap the 3-momentum with the 4 momentum, and swap t for $\tau$. We want to find some machine $\mathbf{F}(u^{\alpha})$(re: tensor) which takes in the 4-velocity and outputs the appropriate derivative of the 4 momentum divided by the charge
  • In index notation: $\frac{dp^{\alpha}}{d\tau} = e F^{\alpha}_{\beta}u^{\beta}$
• You can calculate the spatial parts of F via comparing against the standard form of the Lorentz force law
• You can calculate the time components as $\frac{dp^{0}}{d\tau} = \frac{1}{\sqrt{1-\vec{v}^{2}}} \frac{dE}{dt} = e \vec{E}\cdot\vec{u}$
• The end result is that the force law generalizes to $\frac{dp^{\mu}}{d\tau} = e F^{\mu}_{\nu} u^{\nu}$
  • The transformation of the fields under Lorentz boost can easily be seen by applying a Lorentz transformation on the Faraday tensor
• People typically take about the “covariant components” of F, which can be found by lowering an index: $F_{\alpha\beta} = \eta_{\alpha\gamma}F^{\gamma}_{\beta}$
• Maxwell’s equations can also be written in terms of the Faraday tensor:
  • Define the 4-current J where the time component is $\rho$ and the spatial components are the current density
  • We have that $\mathbf{d}F = F_{\alpha \beta, \gamma} + F_{\beta \gamma, \alpha} + F_{\gamma \alpha, \beta} = 0$ (re: The exterior derivative of the tensor vanishes) where $F_{a,b} = \frac{\partial}{\partial x^{b}} A_{a}$
    • This encode $\nabla \cdot \vec{B} = 0$ and $\frac{\partial B}{\partial t}+\nabla \times E = 0$
  • We have that $\mathbf{d*} F = 4\pi * J \rightarrow F^{\alpha\beta}_{,\beta} = 4\pi J^{\alpha}$ (or the exterior derivative of the dual of F is proportional to the dual of J)
    • This encodes $\nabla \cdot E = 4\pi \rho$ and $\frac{\partial E}{\partial t}-\nabla\times b = -4\pi J$

Exterior Calculus

Exterior Calculus Formulation of E&M

• Recall that the Faraday tensor is $F = F_{\alpha\beta} \mathbf{d}x^{\alpha} \otimes \mathbf{d}x^{\beta}$
• The Faraday tensor can instead be written in terms of wedge products: $F = \frac{1}{2} F_{\alpha\beta} dx^{\alpha} \wedge dx^{\beta}$
  • $dx^{\alpha} \wedge dx^{\beta} = dx^{\alpha} \otimes dx^{\beta} -dx^{\beta} dx^{\alpha}$
  • Any antisymmetric, second-rank tensor can be expanded like this
  • Observe that $B_{2}-E^{2} = \frac{1}{2} F_{\alpha \beta}F^{\alpha\beta}$ and that $\vec{E}\cdot \vec{B} = \frac{1}{4} F_{\alpha\beta}* F^{\alpha\beta}$
  • The 4-density current is defined as $J^{\mu} = e \int \delta^{4}(x^{nu}-a^{\nu}(a)) \frac{da^{\mu}}{d\alpha} d\alpha$
• Can think of a simple 2-form as a “honeycomb” structure with circulation in each of the cells. You can generate this simple 2-form by intersecting two one forms together
  • In general, you can’t think of general 2-forms like this. You need to think of a general 2-form as a sum of simple 2-forms
  • general 2-forms may or may not simplify to a simple 2-form
• Combinatorics dictates the number of 2-forms you can construct from a set of 1 forms (for 4 basis 1-forms, you have 6 unique basis 2-forms)
• Just like how 1-forms can convert vectors to numbers via the number of surfaces pierced by the vector, 2 forms convert 2 vectors to a number by the number of cells spliced by the parallelogram generated by the wedge product
• Hence, a general f-form can be written as $\phi = \frac{1}{f!} \phi_{\alpha_{1}\alpha_{2},…\alpha_{f}} dx^{\alpha_{1}} \wedge … dx^{\alpha_{f}}$
• The exterior derivative of a general f-form $d\phi = \frac{1}{f!} \frac{\partial \phi_{\alpha_{1}…\alpha_{f}}}{\partial x^{\alpha_{0}}} dx^{\alpha_{0}}\wedge … dx^{\alpha_{f}}$
  • Notationally, this is a boldface d
• We can write $F = \mathbf{d} A$ to automatically satisfy $\mathbf{d} F = 0$
  • In alternative notation: $F_{\alpha\beta} = \partial_{\alpha}A_{\beta}-\partial_{\beta}A_{\alpha}$
• You can use the Lorentz gauge ($\frac{\partial A^{\nu}}{\partial x^{\nu}} = 0$) and the following definition of the 4-density current $J^{\mu} = e\int \delta^{4}(x^{\nu}-a^{\nu}(\alpha))\dot{a^{\mu}}(\alpha) d\alpha$ to get:
  • $\square A_{\mu} = - 4\pi J_{\mu}$
  • $\square A = (-\frac{\partial^{2} \phi}{\partial t^{2}} + \nabla^{2} )A $

Stress-Energy Tensor and Conservation Laws

• Spacetime has a river of 4-momentum, which consists of particles moving along their worldlines
• We can quantify the flow of this river via the stress energy tensor $T_{\mu\nu}$
• Define the volume 1-form to be $\Sigma_{\mu} = \epsilon_{\mu\alpha\beta\gamma}A^{\alpha}B^{\beta}C^{\gamma}$
  • A,B, and C are the sides of the parallelpiped
• $T(…,\Sigma)$ is the momentum flowing through the box
  • $T^{00}$ is the energy density
  • $T^{j0}$ is the momentum flux is the 4-momentum per unit volume
  • $T^{0k}$ is the energy flux
  • $T^{jk}$ iis the stress tensor (re jth component of force produced by fields at x^{k}$
• Define the number-flux vector $S_{A} = N_{A} \vec{u}_{A}$
  • This is a 4 vector whose time component is the Lorentz contracted number density, and whose spatial components is the flux of particles

Perfect Fluids

• We have a box of particles whose velocities are distributed isotropically
• The off-diagonal elements are zero, the on diagonal elements are just the rest-energy density and a pressure along each axis
• In index notation: $T_{\alpha\beta} = \rho u_{\alpha}u_{\beta} + P (\eta_{\alpha\beta}+ u_{\alpha} u_{\beta})$
• In geometric language: $T = P \mathbf{g} + (\rho+P) \mathbf{u} \otimes \mathbf{u}$

Maxwell Stress Energy Tensor

• $T^{\mu\nu} = F^{\mu\alpha}F_{\alpha}^{\nu} - \frac{1}{4}\eta^{\mu\nu} F_{\alpha\beta}F^{\alpha\beta}$

Symmetry of Stress Energy Tensor

• We know that $T^{0j} = T^{j0}$ (re: momentum density equals energy flux) due to the equivalence of mass and energy
• For the stress tensor portion, make the following argument:
  • WLOG, examine the torque around the z-axis for a cube of side L
  • $\tau^{z} = -T^{yx} L^{2} \frac{L}{2} + T^{yx} L^{2} (-\frac{L}{2}) - (-T^{xy}L^{2}) \frac{L}{2} - T^{xy}L^{2}(-\frac{L}{2})$
    • In words, $\tau^{z}$ is the y component of the force on the +x face times the lever arm on the +x face. plus the y-component of force on the -x face times the lever arm to the -x face minus the x component of force on the +y face times the lever arm to +y face times the lever arm to the +y face minus the x-component of force on the -y face times the lever arm to the -y face
  • $\tau^{z} = (T^{xy}-T^{yx})L^{3}$
• The moment of inertia on the cube is $T^{00} L^{3} * L^{2}$
• Torque decreases as $L^{3}$ and moment of inertia decreases as $L^{5}$, hence the torque could create an arbitrarily high angular acceleration, which is absurd
  • The paradox is avoided if $T^{xy} = T^{yx}$
  • The same arguments hold for the other axes

Conservation of 4-Momentum

• We know that $\oint_{\partial V} T^{\mu\alpha} d^{3}\Sigma_{a} = \int_{V} T^{\mu\alpha}_{,\alpha} dV$
  • In words, the flux of 4-momentum outwards across a closed 3-surface must vanish
• Since the volume we choose is arbitrary, we know that $T^{\mu\alpha}_{,\alpha} = 0$

Conservation of Angular Mommentum

• We can define a conserved angular momentum $J^{\alpha\beta}$ from the symmetry of the stress energy tensor
• Define some event A to be your origin which your angular momentum is defined around
  • $x^{\alpha}(A) = a^{\alpha}$
• $J^{\alpha\beta\gamma} = (x^{\alpha}-a^{\alpha})T^{\beta\gamma}-(x^{\beta}-a^{\beta})T^{\alpha\gamma}$
  • $x^{\alpha}-a^{\alpha}$ is the vector separation of the field point x from the origin a
  • $J^{\alpha\beta\gamma}_{,\gamma} = 0$ from the stress-energy tensor symmetry
  • From the vanishing divergence, we know that $\oint_{\partial V}J_{,\gamma}^{\alpha\beta\gamma} d^{3} \Sigma_{\gamma}= 0$
  • We can see that the spacelike surface of a constant time t is the standard angular momentum
    • $J^{\alpha \beta} = \int J^{\alpha\beta0}dx dy dz$

Accelerated Observers

• SR works for accelerating observers. Just imagine an interpolation of constant velocity frames which you transition between
• Define $a^{\alpha} = \frac{du^{\alpha}}{d\tau}$
  • From $u^{2} = -1$, we can see that $0 = a_{\alpha}u^{\alpha}$
• In the rest frame of the passenger (re: instantaneous inertial frame), we have that $a^{\alpha} = (0, \vec{a})$
  • This frame is the co-moving frame

Fermi-Walker Tetrad

• This comoving frame by construction cannot be global, because you run into paradoxes
  • If you confine this frame locally (Fermi-Walker Tetrad), then you are fine
• This tetrad is defined as follows:
  • Define a set of 4 basis vectors $e_{\alpha}$ (the subscript denotes vectors,not components of one vector!)
    • Let the $e_{0}$ vector be aligned with the 4-velocity. Define all the other vectors such that they are orthogonal and nonrotating
    • The orthogonal condition can be encoded as $e_{\mu}\cdot e_{\nu} = \eta_{\mu}{\nu}$
  • For the non-rotating condition, we know that the basis vectors at two successive instants must be related by a Lorentz transformation
    • Since the unit 4-velocity has a constant magnitude, then accelerations therefore imply some “rotation” of the 4-velocity
    • Lorentz boosts can be thought of as “rotations” in the x-t plane
    • Hence, “non-rotating” means that there is no additional boosts or rotations other than the bare minimum required to move the 4-velocity

Tensor Algebra

• With flat spacetime, a global Lorentz frame is impossible
• The basis vectors change as you move around spacetime
• Define your new metric to be $g_{\alpha\beta} = e_{\alpha}\cdot e_{\beta}$
  • $g_{\alpha\beta} g^{\beta\gamma} = \delta_{\alpha}^{\gamma}$
• Our Lorentz transformation matrices now become any arbitrary, nonsingular transformation matrices
  • $e_{\beta} = e_{\alpha}L^{\alpha}_{\beta}$
  • $p^{\beta} = L^{\beta}_{\alpha} p^{\beta}$
  • $L_{\alpha}^{\beta} L_{\gamma}^{\alpha} = \delta^{\beta}_{\gamma}$
• We can define coordinate bases: $e_{\alpha} = \frac{\partial P}{\partial x^{\alpha}}$
  • $L^{\alpha}_{\beta} = \frac{\partial x^{a}}{\partial x^{b}}$
• The Levi-Civita tensor has components which now depend on the bases:
  • $\epsilon_{\alpha \beta \gamma \delta} = \sqrt{-g} [\alpha\beta\gamma\delta]$
    • g is the determinant of $g_{\alpha\beta}$ $[…]$ means the antisymmetric symbol, where $0123 = 1$
• Each spacetime point resides in it’s own tangent plane to the manifold. You need to “parallel transport” a vector at another spacetime point
  • you can associate some tangent vector u with the corresponding directional derivative $\partial_{u}$ in the tangent plane
• Since a global Lorentz frame is impossible, the best we can do is demand that $g_{\alpha\beta}$ acts like $\eta_{\alpha\beta}$ in the neighborhood of some event $P_{0}$
  • Hence, we demand that $g_{\mu\nu,\alpha}(P_{0}) = 0$
• Define the covariant derivative $\nabla_{u} T$ along a curve $P(\lambda)$ whose tangent vector is $u = \frac{dP}{d\lambda}$
  • $lim_{\epsilon\rightarrow 0} \frac{T(P(\epsilon)-T(P(0))}{\epsilon}$ where you need to parallel transport $P(\epsilon)$ to $P(0)$
• Define the Christofel symbols as $\Gamma_{\beta\gamma}^{\alpha} = <\omega^{\alpha}, \nabla_{\gamma} e_{\beta}>$
  • A more convenient form of the symbols for a given basis in terms of the metric is: $\Gamma_{\mu\beta\gamma} = \frac{1}{2}(g_{\mu\beta,\gamma}+g_{\mu\gamma,\beta}-g_{\beta\gamma,\mu})$
• The geodesic equation is defined as parallel transporting the tangent vector onto itself: $\nabla_{u} u = 0$
  • In coordinates: $\frac{d^{2}x^{\alpha}}{d\lambda^{2}}+\Gamma_{\mu\gamma}^{\alpha} \frac{dx^{\mu}}{d\lambda}\frac{dx^{\gamma}}{d\lambda}=0$
• In component form, we can define the components of $\nabla_{u} T$ as $\frac{D T_{\alpha}^{\beta}}{d \lambda} = T_{\alpha;\gamma}^{\beta}u^{\gamma} = \frac{dT_{\alpha}^{\beta}}{d\lambda}+(\Gamma_{\mu\gamma}^{\beta} T_{\alpha}^{\mu}-\Gamma_{\alpha\gamma}^{\mu}T^{\beta}_{\mu}) \frac{dx^{\gamma}}{d\lambda}$

Differential Topology

• You don’t need a metric to describe curvature; it just makes calculation useful
• Many of our prior definitions still hold without a metric
  • Events are points in spacetime, irrespective of the underlying geometry
  • Curves are still paths through spacetime; without a metric, you can’t ascribe a proper length to the path though
    • The parameter $\lambda$ is arbitrary up to some origin and unit change (ie. the transformation $\lambda’ = a\lambda +b$ traces out the same path in spacetime
• Vectors in flat spacetime could be though of as the difference between two events
  • In curved coordinates, this doesn’t make sense
  • Vectors instead become the directional derivative along some curve: $u = \partial_{u} = \frac{d}{d\lambda}$ along the curve
  • The directional derivatives for a tangent space, which is isomorphic to the tangent vectors you normally think of
  • Another useful construction is to embed your manifold in some higher dimensional flat space, take some curve that is tangent to your point of interest, and take the limit as $\lambda=0$ to arrive at the tangent vector on the manifold
    • This is a bit of a hack, it that you shouldn’t need to refer to some higher dimensional space to extract properties of the manifold, but it works for computation
  • In the tangent space at event $P_{0}$, we can define a coordinate basis as $e_{\alpha} = \frac{\partial}{\partial x^{\alpha}}$
    • This let’s us write any vector as $v^{\nu}e_{(\nu)}$
    • If we are transforming between coordinate bases, then we know that $\frac{\partial}{\partial x^{\alpha}} = \frac{\partial x^{\beta}}{\partial x^{\alpha’}}\frac{\partial}{\partial x^{\beta}}$
    • In general, we have some non-singular transformation matrix $L^{\beta}_{\alpha}$
• One forms also reside in the tangent space
  • A set of one-forms $w^{\beta}$ obeys the identity $<w^{\beta},e_{\alpha}> \delta^{\beta}_{\alpha}$
    • This doesn’t require the metric to be interpretable! Just think of this as demanding that each $e_{\alpha}$ is parallel to the associated $w^{\alpha}$ and is perpendicular to every other one-form
  • In flat spacetime, we know that $<u,v> = u\cdot v$
• Without the metric, tensors can’t change their rank. They are the same otherwise
• Since vectors can be mapped to directional derivatives, one can ask of the order you act these operators on a function matter.
  • Define $[u,v] f = u(v(f))-v(u(f))$ as the commutator
  • In a coordinate basis (re: $u = \frac{\partial}{\partial x^{a}}$ and $v = \frac{\partial}{\partial x^{b}}$), then the commutator is 0 due to partial derivatives commuting
  • In general, this does not hold in arbitrary coordinate systems

Affine Geometry

• Remember that events and trajectories are independent of the underlying spacetime structure
• How you trace the curve (ie. the $\lambda$ parameter) is not unique
  • You can “start your clock” at some arbitrary point
  • You can change the units of your timescale (years to seconds for example)
  • Combining the above lets you define a more general affine parameter $\alpha = a\lambda +b$
• Given some initial event location, with some initial velocity, the worldline of the particle will be unique
  • Up to the affine reparameterization of course
• In order to define the length of a curve, you need a metric

Parallel Transport

• GR dictates that two test bodies, initially falling on parallel neighboring geodesics, get pushed towards each other by spacetime curvature. What does it mean for two geodesics to be parallel?
• Imagine that you have two points along some curve $r(\lambda)$. Two points along that curve A and B reside in their own tangent spaces. There is no way a priori to relate a vector in the tangent space of A to a vector in the tangent space of B
• What you can do is define some local coordinate system at each point along the curve which obeys Minkowski spacetime
  • Take the tangent vector in A that you want to compare with the tangent vector in B (both of them lying along the same curve)
  • As you move along the curve and transition between each coordinate system, keep the tangent vector unchanged as seen in each coordinate system
  • Once you arrive at B, compare the two vectors. If they are the same, then you are in a flat spacetime. Otherwise, you have curvature!
    • This follows from the equivalence principle (Minkowski spacetime is valid locally!)
• Schild’s ladder let’s you visualize this process:
  • Imagine that you have some curve through A and B (need not be a geodesic)
  • Take some tangent vector and propagate along that vector to some nearby point X
  • Define some nearby point R which lies along the curve from A to B
  • Take the geodesic from X to R with some affine parameterization $\lambda$. Define the point Y along XR at the point $\lambda_{Y} = \frac{1}{2}(\lambda_{x}+\lambda_{R})$
  • Take the geodesic from A to Y with some affine parameter $\lambda$. You need increase $\lambda$ by $\lambda_{Y}$. To arrive at Z, let $\lambda_{Z} = 2*\lambda_{X}$
  • The curve RP gives vector AX “parallel transported” along the curve AB
  • Rinse and repeat with R being the new A and P being the new X
  • In flat spacetime, AX and RP are parallel, and all vectors here are local, so this holds in curved spacetime (equivalence principle)
• Ask how rapidly a vector field $\vec{v}$ is changing along a curve with tangent vector $\vec{u} = \frac{d}{d\lambda}$
  • Call this $\nabla_{u} v$ (ie. the covariant derivative of v along u)
  • Construct this as follows:
    • Take the geodesic along the tangent vector $\vec{u}$ originating at some point defined at $\lambda_{0}$
    • Take the vector from the vector field defined at the point $\lambda = \lambda_{0}+\epsilon$ along the geodesic
    • Parallel transport v back to $\lambda_{0}$. Define this parallel transported vector as $\vec{w}(\lambda_{0}+\epsilon)$
    • Define $\nabla_{\vec{u}} \vec{v} = lim_{\epsilon\rightarrow 0} \frac{\vec{w}(\lambda_{0}+\epsilon)-\vec{v}(\lambda_{0})}{\epsilon}$
• Schild’s ladder let’s you pictorally show:
  • Symmetry: $\nabla_{u} v \nabla_{v} u = [u,v]$
  • chain rule: $\nabla_{u}(f \vec{v}) = f \nabla_{u} v+ v \partial_{u} f$
  • addition: $\nabla_{u}(v+w) = \nabla_{u} v + \nabla_{u} w$
    • From linearity: $\nabla_{a\vec{u}+b\vec{n}} \vec{v} = a \nabla_{u} v + b \nabla_{n} v$
      • a and b are some scalar functions (or numbers)
• Given the above, we can describe parallel transport as: $\frac{d\vec{v}}{d\lambda} = \nabla_{u} \vec{v} = 0$ implies that the vector field V is parallel transported along the vector $u = \frac{d}{d\lambda}$
• You can test if a curve is geodesic if it’s tangent vectors covariant derivative along itself vanishes: $\nabla_{u} u = 0$
• $\nabla$ is not a tensor!

Parallel Transport: Component Notation

• Since each point has it’s own tangent space, the basis vectors and dual basis vectors change from point to point
• To quantify how these basis change, use the covariant derivative. Define $\nabla_{e_{\beta}} = \nabla_{beta}$ to denote the covariant derivative along some basis vector $e_{\beta}$
  • This rate of change is a vector, so it can be expanded in the vector basis: $\nabla_{beta} e_{\alpha} = e_{\mu} \Gamma^{\mu}_{\alpha,beta}$
  • $\Lambda_{\alpha\beta}^{\mu} = (\omega^{\mu}, \nabla_{\beta} e_{\alpha})$
• Similarly, you can show that $\nabla_{\beta} \omega^{\nu} = -\Gamma^{\nu}_{\alpha\beta} = \omega^{\alpha}$
  • This implies that $(\nabla_{\beta} \omega^{\nu}, e_{\alpha}) = -\Gamma^{\nu}_{\alpha\beta}$
• This can be easily generalized to covariant derivatives of tensors:
  • You have the standard partial derivative part
  • You add terms of the form $S*\Lambda$
    • For an upper index, contract that upper index of S with the non-differentiating lower index in $\Lambda$
    • For a lower index, contract that index with the upper index of $\Lambda$, and preprend a minus sign
  • As an example, let $S_{\beta\gamma}^{\alpha}$. Denote $\nabla_{\delta} S_{\beta\gamma}^{\alpha} = S^{\alpha}_{\beta\gamma;\delta}$
    • $ S_{\beta\gamma;\delta}^{\alpha} = S_{\beta\gamma,\delta}^{\alpha} + S_{\beta\gamma}^{\mu} \Lambda_{\mu \delta}^{\alpha} - S_{\mu\gamma}^{\alpha} \Lambda_{\beta \delta}^{\mu} - S_{\beta\mu}^{\alpha} \Lambda_{\gamma \delta}^{\mu}$
    • Make sure that you contract on the non-differentiating index!

Spacetime Curvature

Relative Acceleration of Neighboring Geodesics

• Imagine a family of geodesics. You can distinguish between two geodesics by a continuous selector parameter n
• Hence, you can describe a point via two parameters:
  • $\lambda$ is your affine parameter for each geodesic
  • $n$ determines which geodesic you are on
• From this, we can define:
  • The tangent vector $\frac{\partial}{\partial \lambda}$
  • The separation vector $\frac{\partial}{\partial n}$ which denotes the separation between some fiducial geodesic at n and some typical nearby geodesic $n+\Delta n$
• geodesic deviation means that:
  • You parallel transport some seperation vector n along the fiducial geodesic
  • This tip of this vector traces out some cannonical trajectory
  • The actual trajectory of a test geodesic might deviate from this cannonical path
• Imagine that you displace the separation vector along the fiducial geodesic by $+\frac{\lambda}{2}$ and $-\frac{\lambda}{2}$
  • Calculate the covariant derivative at these two displacements (ie. $\nabla_{u} n$)
  • Move these derivative vectors back to $\lambda$ via parallel transport and subtract them after dividing through by $\delta \lambda$. The resulting 2nd covariant derivative is thus $\nabla_{u}\nabla_{u} n$
  • This is the “relative acceleration” between the geodesics
• Now imagine that you go around in a “square” loop:
  • You parallel transport the seperation vector by $\Delta \lambda$, the you parallel transport to an adjacent geodesic at $n+\Delta n$, parallel transport back along this new geodesic by $-\delta \lambda$, then move back to the start by parallel transporting by $-\Delta n$
  • This gives you that $\nabla_{u} \nabla_{u} n + [\nabla_{n},\nabla_{u}] u = 0$
    • This follows since when you “parallel transport”, you use the geodesic equation $\nabla_{u} a = 0$ where u is the tangent vector of the curve. We are also parallel transporting the initial tangent vector u.
• $[\nabla_{n},\nabla_{u}]$ is almost a valid curvature operator
  • The problem is that the above depends on the derivatives at the point of evaluation.
• The actual definition of the curvature tensor is $R(A,B) = [\nabla_{A}, \nabla_{B}]-\nabla_{[A,B]}$
  • $\nabla_{[A,B]}$ is the derivative along the vector $[A,B]$
• This gives the Riemann curvature tensor as $R(\sigma, C,A,B) = <\sigma, R(A,B)C>$
  • This is a type 1,3 tensor (takes in a one-form $\sigma$ and 3 vectors and spits out a number)
  • In a coordinate basis, we have $R_{\beta\gamma\delta}^{\alpha} = \frac{\partial \Gamma_{\beta\delta}^{\alpha}}{\partial x^{\gamma}} - \frac{\partial \Gamma_{\beta\gamma}^{\alpha}}{\partial x^{\delta}}+\Gamma_{\mu\gamma}^{\alpha}\Gamma_{\beta\delta}^{\mu}-\Gamma_{\mu\delta}^{\alpha}\Gamma_{\beta\gamma}^{\mu}$
  • A zero Reimann tensor implies a flat space (re: geodesics are straight lines)

Reimann Normal Coordinates

• In a curved spacetime, you can’t define a global coordinate system such that $\Lambda_{\beta\gamma}^{\alpha} = 0$ everywhere
• To define a local inertial system, use Riemann normal coordinates
  • Pick a point P in your manifold, and define an orthonormal basis at this point
  • Radiate all the geodesics out from point P. Each geodesic has a tangent vector v at point P associated to it
    • This overcounts by a lot
  • Suppose that you step $\lambda$ along a geodesic with tangent vector v
    • you reach the same point if you step $\frac{1}{2}\lambda$ along a geodesic with tangent vector v
    • This means that if we fix $\lambda$ and vary v, we can reach the same points as all possible geodesics
  • Find a tangent vector v at $P_{0}$ for which $P = \mathbf{P}(1,v)$
    • Expand v in terms of the basis $e_{a}$: $P = \mathbf{P}(1; x^{a}e_{a})$
• From the above construction, we can see that:
  • $e_{a}(\mathbf{P_{0}}) = \frac{\partial}{\partial x^{a}}_{P}$
  • $\Gamma_{\beta\gamma}^{\alpha}(P_{0}) = 0$
  • $\Gamma_{\beta\gamma,\mu}^{\alpha}(P_{0}) = -\frac{1}{3}(R_{\beta\gamma\mu}^{\alpha}+R_{\gamma\beta\mu}^{\alpha})$

Reimannian Geometry

• In order to define a notion of distance between points on a manifold, we define a metric
  • This is a rank 0-2 tensor that is symmetric in it’s indices
  • Explicitly, this is $g = g_{ij} dx^{i} \otimes dx^{j}$
• Define $g^{ab}$ to be the inverse
  • ie. $g^{\alpha\mu}g_{\mu\beta} = g_{\beta}^{\alpha} = \delta_{\beta}^{\alpha}$
• With the metric, you can define a one-to-one mapping between the cotangent and the tangent space
  • Use $g_{ab}$ to convert (1,0) tensors to (0,1) tensors
  • Use $g^{ab}$ to convert (0,1) tensors to (1,0) tensors
• To specialize to general relativity, we demand that at every point on the manifold, you need to be able to define an orthonormal basis such that $g_{ab} = e_{a}\cdot e_{b} = \eta_{ab}$
  • We also want the metric to be as “flat” as possible: $g_{ab,y}(P_{0}) = 0$
  • We won’t be able to set 2nd derivatives to 0 generically though
• In light of the above, you can also derive the geodesic equations from extermizing the action $S = \int (g_{\mu\nu}u^{\mu}u^{\nu})^{\frac{1}{2}} d\lambda$
  • You do a trick by reparameterizing the curve in terms of the differential of the action itself: $dS = (g_{\mu\nu}u^{\mu}u^{\nu})^{\frac{1}{2}} d\lambda$
  • You get back the geodesic equation via Eular-Lagrange
  • It’s useful to calculate Christoffel coefficients

Metric Induced Symmetries of Reimann Tensor

• $R_{\alpha\beta\gamma\delta}$ is antisymmetric between the last two indices
• It is symmetric under swapping adjacent pairs of indices: $R_{\alpha\beta\gamma\delta} = R_{\gamma\delta\alpha\beta}$
• $R_{\beta\gamma\delta}^{\alpha}$ satisfies a Bianchi identity: $R_{\beta\gamma\delta}^{\alpha}+R_{\gamma\delta\beta}^{\alpha}+R_{\delta\beta\gamma}^{\alpha} = 0$
  • You can notate this as $r_{[\beta\gamma\delta]}^{\alpha} = 0$
• Another Bianchi identity is $R_{\beta[\gamma\delta;\epsilon]}^{\alpha} = 0$
  • Reminder that ; denotes a covariant derivative w.r.t. the index to the right of it
• All the above implies that $R_{[\alpha\beta\gamma\delta]} = 0$

Bianchi Identities

• From a physics POV, we demand the divergence of the stress energy tensor to be zero since this encodes a conservation law
  • Alternatively, we can say that the exterior derivative of T vanishes
  • What tensorlike feature of the geometry of spacetime can provide this conservation? (ie. What tensor derived from curvature objects has an exterior derivative of 0?)
  • If this exists, we can make it proportional to the stress-energy tensor
• The metric is a set of 10 potentials which can be combined into a single tensor
  • From the metric, you can define the curvature operator $R =\frac{1}{4} e_{\mu} \wedge e_{nu} R_{\alpha\beta}^{\mu\nu} dx^{\alpha} \wedge dx^{\beta}$
  • We know that R obeys two Bianchi Identities as listed above
  • The differential identity ( the 2nd one), is a manifestation of the general principle “the boundary of the boundary is zero”

Boundary of Boundary is 0

• Look at a 3D case. Define some tensor field at each point in space (for clarity, this can be a scalar field or a vector field)
• Imagine that you sum this quantity over all of the edges of the cube, where each edge has some orientation to it
  • There are no “exposed” edges . Namely, every edge has a corresponding edge on another face with the opposite orientation
  • Summing across edges gives you zero due to this pairing of opposite orientation edges
• The idea generalizes to higher dimensions.
  • Each volume
  • Has a set of oriented hypersurfaces
  • The “edges” of the hypersurfaces cancel out in pairs

Moment of Rotation

• Imagine a 3-volume. For simplicity, take it to be a cube with cartesian coordinates
• The rotation associated with the front face $\delta y \delta z$ is $e_{\lambda} \wedge e_{\mu} {R^{|\lambda\mu|}}_{yz} \delta y \delta z$
  • We are in Reimann normal coordinates, and we demand $\lambda < \mu$ as denoted by $|\lambda \mu|$
  • Follows from parallel transporting a test vector A around the surface edge and seeing how much it differs from the start
• The moment of rotation is taking $P_{cff}-P$ and $\wedge$ prior to the rotation
  • $P_{cff}$ is the center of the front face
  • P is some arbitrary point in space that you are comparing the moment against
    • The exact choice does not matter since the Bianchi identity will make it irrelevant only the back face is subtracted
  • The difference between the two is a well defined vector, but the actual locations of the two points are a bit nebulous
  • $P_{cff}$ and P need to be infinitesimally close though
• Do a similar calculation on the back face. We have that the difference between the two faces is $P_{cff} - P_{cbf} = \Delta x e_{x}$
• Sum across all 6 faces to get $e_{\nu} \wedge e_{\lambda} \wedge e_{\mu} {R^{|\lambda \mu|}}_{|\alpha\beta|} dx^{\nu}\wedge dx^{\alpha} \wedge dx^{\beta}$
  • This is evaluated on the cube $e_{x} \wedge e_{y} \wedge e_{z} \Delta x \Delta y \Delta z$
• Alternatively, we can write this as $dP \wedge R$
  • $dP = e_{\sigma} dx^{\sigma}$
  • $R = \frac{1}{4} e_{\mu} \wedge e_{\nu} {R^{\mu\nu}}_{\alpha\beta} dx^{\alpha}\wedge dx^{\beta}$
    • $dR =0$ is the 2nd Bianchi identity
• $dP \wedge R$ is a trivector. Taking the dual of this would just give a 1-form, which is equivalent and easier to work with
• In general, we can see that $*(e_{\nu} \wedge e_{\lambda} \wedge e_{\mu}) = {\epsilon_{\nu\lambda\mu}}^{\sigma} e_{\sigma}$
• We also se that $dx^{\nu}\wedge dx^{\alpha} \wedge dx^{\beta} = \epsilon^{\nu\alpha\beta\tau} d^{3} \Sigma_{\tau}$
  • $d^{3}\Sigma_{\tau}$ is the 3 volume element
• Use the above to rewrite the moments across the entire cube as : $*(dP \wedge R) = e_{\sigma} {X_{\nu}}^{\sigma\nu\tau} d^{3} \Sigma_{\tau}$
  • $X_{\nu} = (*R*)_{\nu}$
• ${X_{\nu}}^{\sigma\nu\tau}$ is the Einstein tensor
  • it relates the initial starting volume with the final moment
• Observe that $d*G = 0$ (follows from dd = 0 and that * and d commute)
  • Hence, *G is a candidate object whose exterior derivative vanishes
  • The Field equations use G as the source of curvature because of this

Einstein Field Equations

• The main idea is that the stress energy tensor should give rise to the geometry of the system and vice-versa.
• Written abstractly, we have that $G = \kappa T$ where T is the stress energy tensor, $\kappa$ is some proportionality constant, and G is a purely geometric tensor
• G must satisfy the following properties:
  • G should vanish when the spacetime is flat
  • G needs to purely be a function of the metric, and other tensors that can be contructed from the metric (like the Reimann tensor)
  • G needs to be linear in Reimann (simplest non-trivial assumption. Could include higher order terms)
  • G needs to be symmetric
  • G needs to have a vanishing divergence $\nabla \cdot G = 0$
• The only tensor which satisfies this is $G_{\mu\nu} = R_{\mu\nu}-\frac{R}{2}g_{\mu\nu}$
  • This can be though of as $G_{\mu\nu} = G_{\mu\alpha\nu}^{\alpha}$ where the rank (1,3) tensor is the double dual of the Reimann tensor
  • This can be proven by making the general ansatz $G = aR_{\alpha\beta}+bRg_{\alpha\beta}+\Lambda g_{\alpha\beta}$, then applying each condition you want to satisfy
• To calculate kappa, the above equation should reduce to the Newtonian limit
  • Assume that you are in a region of constant density
    • This implies that the $T_{00}$ component dominates all other components of the tensor. Also, the tensor is diagonal
  • In the weak field approximation, we have $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ where the magnitude of $h_{\mu\nu}$ is small
  • Examine the 00 component and grind through the algebra gives you a Poisson equation. Compare this against the actual Poisson equation to see that $\kappa = 8\pi$
• Hence we have $G_{\mu\nu} = 8\pi T_{\mu\nu}$

EFE Applications

Spinors

Regge Calculus