GradGeneralRelativity

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
• 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:
      • $*J_{\alpha\beta\gamma} = J^{\mu}\epsilon_{\mu\alpha\beta\gamma}$
      • $*F_{\alpha\beta} = \frac{1}{2} F^{\mu\nu}\epsilon_{\mu\nu\alpha\beta}$
      • $*B_{\alpha} = \frac{1}{3!} B^{\lambda\mu\nu}\epsilon_{\lambda\mu\nu\alpha}$
• 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 Formulation of E&M

Exterior Calculus

• 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 $d\phi = \frac{1}{f!} \frac{\partial \phi_{\alpha_{1}…\alpha_{f}}}{\partial x^{\alpha_{0}}} dx^{\alpha_{0}}\wedge … dx^{\alpha_{f}}$
• 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 $g_{\mu\nu,\alpha}(P_{0}) = 0$