Astrophysics

Following Misner, Wheeler and Thorne. Foundations of General Relativity All laws of physics can be expressed geometrically Local reference frames are flat gravitational mass is equivalent to inertial mass All reference frames are equivalent (re: there is no preferred reference frame) Minimal coupling: work in SR to find physical laws , then go to general frame to include gravity Basic Definitions and Math Things Take naturalized units (c=G=1) when working with GR 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 Postulates: All inertial frames are equivalent There is a maximum speed that information can travel at (happens to be c) 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 $\begin{pmatrix}\gamma & -\beta \gamma & 0 & 0\\ -\beta \gamma & \gamma &0 &0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ Apply this matrix to some vector (t,x,y,z) to get the transformations to a primed system moving at some velocity v w.r.t. the unprimed Can derive by noting that: Time dialates: ($\Delta t’ = \gamma \Delta t$) where ’ is the moving frame Lengths along the motion contract ($\Delta x’ = \frac{1}{\gamma}\Delta x$ Lengths perpendicular to the motion are unchanged 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 $\alpha$ Spacetime Interval The interval is defined as $ds^{2} = -dt^{2} +dx^{2}$. This is invariant under Lorentz transformation This of this as a frame-invariant way of stating the distance between two spacetime events This can also be written as $ds^{2} = \eta_{ab} dx^{a} dx^{b}$ The interval is also invariant under spatial rotations (can think of boosts as rotations in t-x plane) The interval has 3 cases: timelike (<0), spacelike (>0) or null (=0) Timelike implies a causal lightcone structure Spacelike involves a non-causal structure Nulllike implies moving at light speed SR Mechanics Define $\tau$ to be the proper time to be the time experienced by an observer when at rest (ie. $d\tau^{2} = -dt^{2} +dx^{2}$ This proper time will be less than or equal to every other frame (b/c time dialation) $\tau$ is Lorentz invariant Suppose that you parameterize your interval 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 ) The magnitude of $v^{2}$ dictates the angle that the 4-velocity makes with the temporal axis This transforms like the coordinate $x^{a}$ since $\tau$ is invariant The norm of u is -1 U is tangent to the particle’s path From the above, we can construct a 4-momentum for massive objects: $p^{a} = mu^{a} = (\gamma m, \gamma m \vec{v})^{a} = (E, \vec{p})^{a}$ The norm of the 4-momentum is $-m^{2}$ For a massless particle, we demand that $p^{a} = (\frac{h}{\lambda}, \frac{h}{\lambda} \vec{e})^{a}$ $\lambda$ is the deBroglie wavelength $\vec{e}$ points along the direction of propagation The four-force is defined as $\vec{F} \frac{d\vec{p}}{dt}$ where t is the coordinate time and p is the relativistic 3 momentum $\vec{p} = \gamma m \vec{v}$ The power is $\vec{v}\cdot\vec{F}$, which can be seen by differentiating $E^{2} = \vec{p}^{2}+m^{2}$ w.r.t. time Flux Densities The relativistic flux of some substance q is given by $j^{a} = (\rho, \vec{J})$ $\rho = frac{dq}{d^{3}\vec{r}}$ $j^{i} = \frac{dx^{i} dq}{d^{4}x}$ $d^{4}x$ is Lorentz invariant ($\det(\Lambda)=1$ is the volume element) If q is conserved, then we have $\partial_{a} j^{a} = 0$ 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 ($\nabla\cdot\frac{F_{g}}{m} = -4\pi \rho$ )to see that $\kappa = 8\pi$ Hence we have $G_{\mu\nu} = 8\pi T_{\mu\nu}$ EFE Applications Thermodynamics Assume that we are dealing with equillibrium thermo of a perfect fluid with fixed chemical compositions We can charaterize the system via a set of potentials (n, $\rho$, P, T, s ,$\mu$) Potentials are frame independent functions (since they are scalars) However, we can evaluate the potentials in the fluid’s rest frame The potentials, evaulated in the rest frame, are: n: baryon number density: number of baryons per unit 3d volume of the rest frame. Anti-baryons are negative $\rho$: density of the total mass-energy rest frame P : isotropic pressure in rest frame T: temperature s: entropy per baryon in rest frame $\mu$: chemical potential of baryons We are assuming: the chemical composition is fixed by the baryon density and entropy per baryon We are assuming that chemical reaction rates are two slow to matter, or that they are so fast that equillibrium is reached We assume baryon conservation: $\nabla \cdot S = 0$ where $S = nu^{\mu}$ $\frac{d}{d\tau}(nV) = 0$ The 2nd law states entropy can be generated, but not destroyed Assume that entropy very slowly flows in and out of a fluid element $ $\frac{ds}{d\tau} \geq 0$ The 1st law states: In words: The differential of the energy in a volume element containing a fixed number A of baryons equals the negative pressure times the volume differential plus the temperature times the entropy differential $d(\frac{\rho A}{n}) = -P d(\frac{A}{n}) + T d(As) \rightarrow d\rho = \frac{\rho+P}{n} dn +nT ds$ “d” is a exterior derivative Can think of $\rho = \rho(n,s)$. We can then think of the first law as a total differential to state: $\frac{\rho+P}{n} = \frac{\partial \rho}{\partial n}_{s}$ and similar for entropy The two partials must obey the Maxwell relationship $(\frac{\partial P}{\partial s}) = n^{2} \frac{\partial T}{\partial n}_{s}$) for consistency The chemical potential is defined as follows: Take a sample of the simple fluid in a fixed thermodynamic state (fix n and s) Take a much smaller sample containing $\delta A$ baryons in the same state as the larger sample (fix n and s) Inject the smaller sample into the larger sample, holding the volume fixed during the injection process The total mass energy, plus the work required to perform the injection is $\mu \delta A$ $\mu \delta A = \rho(\frac{\delta A}{n})+ P \frac{\delta A}{n} = \frac{\partial \rho}{\partial n}_{s}$ Hydrodynamics is just enforcing that $\nabla \cdot T=0$ E&M The are unmodified from SR! Just change your metric. Recall that $F^{0j} = E^{j}$ and $F^{jk} = \epsilon^{jkl} B^{l}$ We have that: $F^{\alpha\beta}_{;\beta} = 4\pi J^{\alpha}$ $F_{\alpha\beta;\gamma} + F_{\beta\gamma;\alpha} + F_{\gamma \alpha; \beta} = 0$ $ma^{\alpha} = F^{\alpha\beta}q u_{\beta}$ Geometric Optics Define the following length scales : $\lambda$ is the wavelength of the wave involved L is the length over which the wave parameters vary R is the radiius of curvature through which the waves propagate Geometric optics is the regime where $\lambda «L$ and $\lambda « R$ Hence, we can assume that waves are locally plane waves The main results from these assumptions are that: Light rays are null geodesics The polarization vector is perpendicular to the rays and the parallel-propagated along the rays The amplitude is governed by an adiabatic invariant (ie. Number of photons is conserved)

Scales Black holes are described by 3 parameters: Mass, spin, and charge In practice, astrophysical black holes have no charge, since they are embedded in a plasma which “shorts out” the hole We can define some characteristic scales of black holes $R_{s} = \frac{2GM}{c^{2}}$ is the Schwartzchild radius $\Delta t_{LC} = \frac{R_{s}}{c}$ is the light crossing time (roughly the time needed to cross the black hole, up to a scale factor) How bright can be a black hole be? We would want to convert all of the black hole’s mass into radiation in the relavent time scale $L_{max} = \frac{Mc^{2}}{\frac{2GM}{c^{3}}} \approx \frac{c^{5}}{G} \approx 10^{59} \frac{erg}{s}$ For reference, supernovae are around $10^{52}$ What is the luminosity of the observable universe (excluding the blackholes)? This is Fermi estimation problem: Find the mass of the sum, calculate the mass to energy conversion, and calculate the time needed for a photon to escape the center The escape time is roughly a billion years, which gives a luminosity of around $10^{37}$ There are roughly $10^{10}$ stars in the universe. Assume all stars are like the sun There are also roughly $10^{10}$ galaxies in the universe Combining these gives that the luminosity of the universe is $10^{53}$ The point of this is that gravity is very efficient at converting mass to energy If the black hole is moving close to the speed of light towards you (pileup of photons), then the observed luminosity can exceed this threshold Gamma ray bursts (GRBs) are the prime example of this boosted luminosity Eddington Limit Imagine that you have a star which is gravitationally bound and emits radiation. These two forces oppose each other: gravity attracts to the center, while radiation pressure pushes the star boundary outwards Larger stars have more mass, but they also tend to produce more radiation pressure The Eddington limit is when these two forces balance each other Define $\dot{m}$ as the accretion rate (gas falling onto a compact object or star) The luminosity is $L = \frac{GM\dot{m}}{r}$, where $\frac{M}{r}$ is the compactness The associated Eddington luminosity is an upper bound on the brightness of these objects (there are exceptions though) Let’s look at the simplest case: We have a fully ionized Hydrogen gas (plasma) accreting isotropically onto a star/ compact object Electrons getting accelerated around protons produces radiation, which creates the outward pressure Thomson Scattering Imagine a EM wave propagating along the $\hat{z}$ direction. This will oscillate an electron in a direction transverse to z (fix this direction to be x) This oscillating generates the radiation Newton’s 2nd Law, and letting $E(z,t) = exp(i\omega t) \hat{x}$, the resulting motion is $x(t) = A exp(i\omega t)$ where $A = \frac{qE(z)}{m\omega^{2}}$ The cross section scales like $A^{2}$, which means that the cross section of the proton is roughly 1 million times smaller than the electron $\frac{\sigma_{Tp}}{\sigma_{Te}} = \frac{m_{e}^{2}}{m_{p}^{2}}$ The radiation force is defined as the energy flux times the cross section divided by the speed of light $F = \frac{L}{4\pi r^{2}} \rightarrow f_{rad} = \frac{\sigma_{T} L }{4\pi r^{2} c}$ Setting the gravitational force (ignoring the electron mass) equal to the radiative force (ignoring the proton cross section) gives the target luminosity (re: Eddington Luminoosity) as $L_{edd} = \frac{4\pi c G M m_{p}}{\sigma_{T}}$ Setting the mass scale to that of the sun, we get that $L_{Edd} = 1.3E38 (\frac{M}{M_{*}}) \frac{ergs}{s}$ For a black hole, $L_{edd} = 1.5E45 M_{7} \frac{erg}{s}$ where $M_{7} = \frac{M}{1E7 M_{*}}$ If we assume that the object is a perfect blackbody, we can define an effective temperature $L = 4\pi R^{2} \sigma_{B} T_{eff}^{4}$ Variability The acceleration is $\frac{GM}{r^2}$. Suppose that the radiation pressure suddenly vanished. How long would it take for the star to collapse? Naively, one would say that $\frac{1}{2} a t_{dyn}^{2} = R \rightarrow t_{dyn} = \sqrt{\frac{2R^{3}}{GM}}$ Assuming that force at star radius is the same for all smaller radius This is roughly $t_{dyn} = \frac{1}{\sqrt{G\rho}}$ $t_{dyn}$ sets the timescale for how quickly the luminosity of the star can change Sense of Scale The above outlines the rough energies, wavelengths, frequencies and temperatures for each band of radiation Atmospheric Attenuation Our atomsphere absorbs different amounts of radiation depending on the frequency This change is absorption is described by the opacity of the sky The above sets constraints on where expeiments/telescopes can be (radio and visible can be ground based, but other bands might demand high altitudes or be in space) Intensities Define $I_{\nu} = \frac{dE}{dA dt d\nu d\Omega}$ as the frequency dependent brightness of a source This is constant along rays in free space Imagine that you have 2 detectors which are coaxial with seperation R and area $dA_{1}$ and $dA_{2}$ respectively The energy change across each is $dE_{1} = I_{\nu_{1}} dA_{1} dt d\Omega_{1} d\nu_{1}$. Same for the 2nd detector Solid angle is related to area via $d\Omega_{1} = \frac{dA_{2}}{R^{2}}$ The frequency and time crossing are the same Use conservation of energy to see that the brightness is conserved Define the flux as $F_{\nu} = \int I_{\nu} \cos \theta d\Omega$ cosine modulation to account for when you don’t look at a source head on You can recover the inverse square law by integrating over a small solid angle Define the pressure as $P_{\nu} = \frac{1}{c} \int I_{\nu} \cos^{2}(\theta) d\Omega$ Additional cosine arises from projecting onto the area of interest Define the energy density $u_{\nu}$ as $dE = u_{\nu}(\Omega) dV d\Omega d\nu$, or in terms of the brightness: $u_{\nu} = \frac{I_{\nu}}{c}$ Define the mean intensity as $J_{\nu} = \frac{1}{4\pi} \int I_{\nu} d\Omega$ Particle acceleration Cosmic Rays There exists cosmic rays which are protons of energies 1E21 eV The LHC boosts protons to 1E13 eV When they hit the atmosphere, they create showers of high energy particles, which produce Cherenkov radiation cones in the air There is a hierachy of densities in our Universe: On earth, the density of air is roughly 1E20 per cc In the ISM, there is a density of 1E0 particles per cc In the IGM, there is a density of 1E-6 per cc In the ISM and IGM, the mean free path $\lambda = \frac{1}{n \sigma}$ is very long, which lets particles get accelerated for a long time before getting scattered Fermi Acceleration There are these objects called molecular clouds These are objects in the ISM which form in slightly over-dense regions They are called “star nurseries” Point being, despite being made up of a bunch of particles, collectively the clouds have some average velocity in some direction There are a lot of these molecular clouds scattered around. Fermi realized that a particle could bounce off of these clouds and potentially gain these massive energies This hinges on the fact that the velocity distribution of the clouds is assymetric A non-relativistic calculation of this phenomena: Let $\Delta$ be the distance from the particle to the nearest cloud Let $v$ and $V$ be the velocities of the particle and the cloud respectively Let’s calculate the probability of a collision The time until a head-on collision is $\tau_{u} = \frac{\Delta}{v+V}$ The velocities point in opposite directions The time until a catch-up collision is $\tau_{c} = \frac{\Delta}{v-V}$ The velocities point in same direction The probability of a head-on collision is $\frac{\frac{1}{\tau_{u}}}{\frac{1}{\tau_{u}}+\frac{1}{\tau_{c}}} = \frac{v+V}{2v}$ The probability of a catch-up collision is $P_{c} = \frac{v-V}{2v}$ What is the change in energy for each case? Assume that the mass of the cloud is much larger than the particle. This is just an elastic collision For a head-on collision: velocity after the collision is $v+2V$, hence $\Delta E_{h} = \frac{m}{2} (v+2V)^{2} -\frac{m}{2} v^{2}$ For a catch-up collision: velocity after the collision is $-v+2V$ What is the expected change is energy? Its $<\Delta E > = \Sigma_{i} P_{i} E_{i}$ Doing that calculation out gives $\frac{<\Delta E >}{E} = 4 (\frac{V}{v})^{2}$ This is a 2nd order process (b/c of the -2), which means it’s pretty inefficient This Fermi acceleration is a general concept. There are “converging magnetic flows” where the “clouds” are always moving towards you. Hence you get a more efficient energy gain Let $\frac{dE}{dt} \propto \nu < \Delta E>$, where $\nu$ is the frequency of collisions For the molecular clouds, we see that $\frac{dE}{dt} = \alpha E$ for some characteristic $\alpha$ of the ISM Let’s think of this problem in phase space, where our axes are E and x $\frac{d N}{d t} = -\frac{\partial \phi_{s}}{\partial x} - \frac{\partial \phi_{e}}{\partial E} + \frac{\partial N}{\partial t}$ Think Stokes Theorem, where the flow on the boundary and the production of particles determines the change in the number of particles in some volume The spatial flux is defined by $\phi_{s} = -D \frac{\partial N}{\partial x}$, where D is some diffusion constant The energy flux is $\phi_{e} = \frac{N \Delta E}{\Delta t}$ We define some timescale $\tau$, which dictates how quickly particles leave the system (for instance, the Larmor radius grows so large so as to leave the system). This then gives that $\frac{\partial N}{\partial t} \approx -\frac{N}{\tau}$ For molecular clouds, D=0, and the system is in equillibrium, so $\frac{dN}{dt} = 0$ Solving for N yields $N(E) \approx N_{0} exp(-(1+\frac{1}{\alpha}{\tau}))$ where $\alpha = 4 \nu (\frac{V}{v})^{2}$ In log-log space, this is a straight line (re: a power law) MHD (Magneto-hydrodynamics) We have Maxwell’s equations $\nabla \cdot E = 4\pi \rho_{e}$ $\nabla \cdot B = 0$ $\nabla \times E = -\frac{1}{c} \frac{\partial B}{\partial t}$ $\nabla \times B = \frac{4\pi}{c} J \frac{1}{c} \frac{\partial E}{\partial t}$ From Maxwell, we can derive charge conservation: $\frac{\partial \rho_{e}}{\partial t} = \nabla \cdot J$ We also make the ideal assumption (re: no resistivity) This gives that there is no Lorentz force in the plasma (re: if there was a force, then due to no resistivity, the charges would rearrange themselves and short out the force) Hence we have $\vec{E} \approx - \frac{\vec{v}}{c} \times \vec{B}$ We add on the additional constraint of conservation of mass: $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{v}) = 0$ Writing down momentum conservation gives: $\rho \frac{d\vec{v}}{dt} = \rho \frac{\partial \vec{v}}{\partial t} + \rho (\vec{v} \cdot \nabla) \vec{v} = -\frac{\vec{B}}{4\pi} (\nabla \times \vec{B} - \frac{1}{c} \frac{\partial \vec{E}}{\partial t})$ In non-relativistic MHD, we can drop the $\frac{\partial E}{\partial t}$ term Suppose that you have some B field along the z axis, which experiences some sinusoidal peturbation in the x axis: $\vec{B} = B_{0} \hat{z} + B_{A} exp(ikz- i\omega t) \hat{x}$ This could arise via neutron star quakes, or acretion disk turbulances These quakes have some length scale $\lambda$, which modules the perturbation by $\sin(k_{y} y)$ The E field which gets created is $\vec{E} = \frac{ic}{\omega}(ikB_{a} \sin(k_{y}y) \hat{y} - k_{y} B_{A} \cos (k_{y} y) \hat{z}) exp(ikz-i\omega t)$ So this peturbation creates an E field along the magnetic field line, which can accelerate particles to relativistic speeds This speed up is reduced due to synchrotron radiation Acretion Disks We have gas around a object with a strong gravitational field Gravity can accelrate the charges, causing energy loses via radiation; angular momentum is preserved though Radially, The balance of centrifugal acceleration and gravity causes the gas to gather into a disk Vertically, the gravity gets balanced by the radiation pressure Thin Disks A thin disk is defined as being thin in the z-axis (due to being cold and having low pressure) The radial pressure gradient is negligible A thicker disk is hotter (hence larger pressure gradients) How does all the gas fall into the black hole when the portions of the disk have angular momentum? There is a net flux of angular momentum out of the disk Imagine a ring of gas of radial thickness dR at radius R. The angular momentum on the inside is $R\Omega(R)$, where $\Omega(R) = \frac{v_{\phi}}{r}$, and on the outside is $(R+dR)\Omega(R+dR)$ We imagine that there is some particle exchange from inside the disk to the outside, and vice versa The torque due to the outflow is $\tau = \dot{M_{1}} (R+dR) R \Omega(R)$ The torque due to the inflow is $\tau = \dot{M_{2}} R(R+dR) R \Omega(R+dR)$ In short, the particles keep their old angular momentum as they cross the boundary The inflow and outflow mass rate are the same ($\dot{M_{1}} = 2\pi R \rho(R) H \tilde{v}$). H is the height of the disk. $\rho(z)$ is the density. $\tilde{v}$ is the average radial velocity of the disk Hence, the net torque is $\tau = \dot{M_{1}} R(R+dR)(\Omega(R)-\Omega(R+dR)) = 2\pi R H \tilde{v} \rho(z) R(R+dR)(dR \Omega’(R))$ The kinematic viscosity is defined as $\nu = dR \tilde{v}$ (units of length squared over time) Hence: $\tau = -2\pi \nu \epsilon R^{3} \Omega'$ There is a characteristic length and velocity scales in the problem (height of disk and speed of sound). You can encode this in the kinematic viscosity via $\nu = \alpha c_{s} H$, where $0 \leq \alpha \leq 1$ Due to the viscosity of the gas, the friction generated causes the gas to radiate like a blackbody The mass flow of a ring is given by $R\frac{\partial \Epsilon}{\partial t} + \frac{\partial}{\partial R}(R\Epsilon v_{R}) = 0$ Imagine the mass flow at the inner and outer radii, then take the limit as dR goes to 0 $\Epsilon$ is the mass density of the ring, $v_{r}$ is the radial velocity Similarly, there is a flow of angular momentum: $R \frac{\partial}{\partial t}(\Epsilon R^{2} \Omega) + \frac{\partial}{\partial R}(R\Epsilon v_{r} R^{2}\Omega) = -\frac{1}{2\pi} \frac{\partial \tau_{out}}{\partial R}$ Same story: Take the difference between the inner and outer disks, and also add on the torques at the inner and outer rings Since the rings are Keplerian, we know that $\Omega = \frac{v}{R} \propto R^{-\frac{3}{2}}$ We can combine mass and angular momentum conservation, along with the Keplerian ring assumption to calculate the radial velocity: $v_{r} = -\frac{3}{\Epsilon R^{\frac{1}{2}}} \frac{\partial}{\partial R}(v \Epsilon R^{\frac{1}{2}})$ We can define $\dot{m} = 2\pi R \Epsilon (-v_{R})$ Thick Disks If the cooling is inefficient, or the mass accretion rate is too large, then disk has a tendency to become thick There are a lot of different regimes with funny names: ADAF: advection-dominated accretion flow RIAF: raidatively inefficient accretion flow CDAF: convection dominated accretion flow ADIOS: advection dominated inflow/outflow solution You get like a quadropole movement in the flow, where one axis is an inflow while the other axis is an outflow Magnetic fields become more important, and the ideal MHD approximation breaks down Analytic Assume that the angular velocity is now a function of z and R ($\Omega(z,R)$) For simplicity, assume that the radial velocity is much smaller than the angular velocity. the vertical velocity is also small Force balance gives $\frac{1}{\rho}\nabla P = -\nabla \Phi +\Omega^{2}\vec{R} = \vec{g_{eff}}$ To simplify the analysis, let’s examine isobaric surfaces $\nabla P = 0$ To further simplify the analysis, pretend that the z profile goes like $\pm R \tan \alpha $ The potential takes the form $\Phi = -\frac{GM}{(R^{2}+z^{2})^{\frac{3}{2}}}$ Assume that z « R Doing the algebra and calculus gives $\Omega^{2}(R) = \frac{GM \cos \alpha}{R^{3}}$ This reduces to a Keplerian disk in the small $\alpha$ limit Otherwise, the angular velocity is sub-Keplerian The radiation force becomes $\vec{F_{rad}} = -\frac{c}{\kappa \rho} \nabla P$ Integrating the force over the surface area of the disk gives you the luminosity: $L = \frac{4\pi c GM}{\kappa} -\frac{c}{\kappa} \int \nabla \cdot \Omega^{2} dV$ Doing the integral, you get $L = L_{edd}(1+\sin \alpha \frac{r_{2}}{r_{1}})$ This can be super-Eddington! This can explain how these super-massive black holes can be created in a Hubble time Acretion Columns Imagine that the central accretor has some non-zero magnetic dipole moment. The B field for a non-zero dipole is $\vec{B} = \frac{3\hat{n}(\hat{n}\cdot\hat{m})-\vec{m}}{r^{3}}$ In our case, $\hat{m}$ is along the rotation axis of the acretor, and $\hat{n}$ denotes the direction we are measuring the field at As a function of radius, assuming we are in the plane, we see that $B(R) = B_{0}(\frac{R_*}{R})^{3}$ The pressure of the gas is given by $\frac{\rho GM}{R}$ The magnetic pressure is given by $P = \frac{B^{2}}{2\mu_{0}}$ The magnetic pressure has dynamical consequences once these pressures are comparable to each other The radius where this interchange happens defines this transition on behavior The gas, once below this radius, will flow along the magnetic field lines towards the poles of the acretor. This forms an accretion column MHD We can describe our system as a function of pressure, velocity, and density at all x,y,z,t. These are not all conserved. We can construct a set of conserved variables from these: the total energy density (kinetic plus thermal), the momentum density and the density itself There are two pictures of flows that are useful The Eulerian description imagines that you fix a box in space. You then look at the flow of stuff through the faces of the box and any stuff which gets generated in the box The Lagrangian description follows a particular section of stuff around as a function of time and space With conservation of mass, we see that $\frac{\partial \rho}{\partial t} + \nabla \cdot (\rho \vec{v}) = 0$ Can derive by looking at the mass flow in a little box is size $\delta V = \delta x \delta y \delta z$ With momentum conservation, we find that: $\frac{\partial}{\partial t}(\rho \vec{v}) + \nabla \cdot (\rho \vec{v}\vec{v}) = -\nabla \cdot \sigma + \vec{F_{b}}$ $\vec{v}\vec{v}$ and $\sigma$ are 3x3 matrices (re:tensors) You can split the stress tensor into diagonal pressure components and off-diagonal shear components You can dot the above with $\vec{v}$ to get an power conservation law The 1st law of thermodynamics states: $\frac{du}{dt} = T\frac{ds}{dt} + \frac{P}{\rho^{2}}\frac{d\rho}{dt}$ S is the entropy per unit mass Bondi Accretion This describes the accretion of cold gas onto a central mass where the flow is primary in the radial direction The interstellar medium (ISM) is such a cold gas Far away from the center accretor, the thermal energy density is given by $\rho c_{s}^{2}$. Define some radius $R_{a}$ where the thermal energy density equals the gravitational energy density Another relavent radii is $r_{s}$ (the sonic radius), where in infalling gas speed equals the sound speed Imagine that you have some steady radial flow: From mass conservation, we know that $\frac{1}{r^{2}} \frac{d}{dr}(r^{2}\rho v) = 0$ The mass accretion rate is then defined as $\dot{M} = 4\pi r^{2} \rho (-v)$ (minus sign since the gas is always moving inwards) From the momentum conservation, we have that $+\frac{\nabla P}{\rho}+ \frac{GM}{r^{3}} + v \frac{dv}{dr} = 0$ We assume an equation of state $P = D \rho^{\Gamma}$, which is representative of some ideal gas equation of state (or if $\Gamma=1$, it’s an isotherm) If we define $c_{s}^{2} = \frac{dP}{d\rho}$, do some chain rule to change $\frac{dP}{d r} = \frac{\partial P}{\partial \rho}\frac{d \rho}{dr}$, and incorporate the mass continuity equation, we see that: $\frac{1}{2}(1-\frac{c_{s}^{2}}{v^{2}}) \frac{d}{dr}(v^{2}) = -\frac{GM}{r^{2}}(1-\frac{2c_{s}^{2}r}{GM})$ This describes a transition from subsonic to supersonic speed Radii which this occurs at is when $v(r_{s}) = c_{s}(r_{s})$, which translates to $r_{s} =frac{GM}{2c_{s}^{2}(r_{s})$ We know the sound speed and the density at infinity. We need to relate these parameters to the values at $r_{s}$ Look at energy conservation: $\frac{1}{2}v^{2} + \int \frac{dP}{\rho} - \frac{GM}{r} = constant$ Assuming that $\Gamma \neq 1$, we know that v=0 at $r=\infty$. This let’s you write the sound speed as a function of the boundary conditions. Remembering that $v = c_{s}$ at $r_{s}$, we see that: $c_{s}(r_{s}) = c_{s}(\infty) (\frac{2}{5-3r})^{\frac{1}{2}}$ $\rho(r_{s}) = \rho(\infty) (\frac{c_{s}(r_{s})}{c_{s}(\infty)})^{\frac{2}{\Gamma-1}}$ Roche Lobes Most massive stars are in binary systems (~70%) We define the ratio between the two masses as $q = \frac{M_{2}}{M_{1}}$, where $M_{2} \leq M_{1}$ In an inertial frame, the potential is just the sum due to each mass In a co-rotating frame, you can peg the masses in place, and add in some additional potential to model this rotation For simplicity, we will only assume that the centrifugal force is relavent. Euler and Coreolis forces are small The centrifugal acceleration is $-\omega\ \times \omega \times r$, which means the potential is $\Phi(r) = -\frac{1}{2}(\vec{\omega}\times \vec{r})^{2}$ (just take the derivative w.r.t. r! to get back the force!) The relavent lengths of the system are: The binary separation a: related to the period via Kepler’s Law: $4\pi^{2}a^{3} = G(M_{1}+M_{2})P_{orb}^{2}$ $R_{i}$ the average radii of each mass $b_{i}$ the distance from the mass to the L1 Lagrange point The equipotentials of this system in the co-rotating frame looks like two lobes which meet at the Lagrange point L1 Some heutristics for how these parameters evolve are: $\frac{R_{2}}{a} 0.38+0.2 \log q$ $\frac{b_{1}}{a} = 0.5 -0.277 \log q$ Jets Imagine you have an electron moving through a constant magnetic field. Align your coordinate system along the electrons velocity, with the z axis orthogonal in the direction of the B field There is some pitch angle $\alpha$ which the electron makes with the B field the gyration radius is given by $r_{g} = \frac{v\sin\alpha}{\omega_{g}}$ where $\omega_{g} = \frac{eB}{m_{e}c}$ ( follows from Lorentz force law) The radiated power in the electron’s rest frame is then $P = \frac{2e^{2}}{3c^{3}}\omega_{g}^{2} v \sin^{2} \alpha$ (Larmor formula) The power emitted by the electron in some other frame is $P’ = \frac{4}{3} \sigma_{T} c \beta^{2} \gamma^{2} u_{B}$ $\sigma_{T}$ is the Thompson cross section (need Klein-Nishina if quantum is important) $U_{B} = \frac{B^{2}}{8\pi}$ You also need to average over the pitch angle: $\int 2\pi \sin^{3} \alpha d\alpha = \frac{8\pi}{3}$ Since energy and time both have the same transformation, then the above powers are the same This means that the energy gets relativistically beamed forwards As the electron whips around, the beam very quickly enters and leaves the line of sight of an observer Hence, the time we observe the beam is short, which means the frequency content is very large (hence why radio pulsars are a thing…) The width of the cone is roughly $\frac{1}{\gamma}$ The actual measured power depends on the velocity distribution of electrons

Following Baumann’s Cosmology. Conventions We are assuming a (- + + + ) signature for the metric. Greek letters for spacetime indices, and Latin letters for spatial indices overdots for physical time derivatives and primes for conformal time derivatives Qualitative Overview Meters are a cumbersome unit to work with at universe scales. We use lightyears and parsecs instead 1 ly is roughly 9.5E15 m 1 parsec (pc) is defined as the distance at which the radius of the Earth’s orbit around the Sun subtends an angle of one arcsecond (re: $\frac{1}{3600}$ of a degree) In light years, 1 parsec is around 3.26 lightyears For a sense of scale: The nearest star (Proxima Centauri) is 4.2 lightyears away Our Galaxy, the Milky Way, is around 30 kpc across The nearest galaxy (Andromeda) is around 2 million light years away (2 Mpc) The local group (~ 50 nearby galaxies) is around 10 Mpc in diameter The local supercluster (Laniakea) is 500 Mpc The observable universe is around 14 Gpc or 46.5 billion lightyears This is ostensibly larger than the age of the universe (13.8 lightyears) due to expansion The consituents of the universe are: Ordinary matter: This is us. It’s 5% Dark matter: This is the vast majority of the matter contents Major experimental evidence for this is increased rotational speeds of hydrogen gas in the outer reaches of galaxies (Vera Rubin) Only can be explained if galaxies embedded in halos of dark matter Another major piece is lensing of the CMB is in tension with measurements of element abundances without non-baryonic dark matter Also, the anisotropies of the CMB wouldn’t manifest as they do without dark matter Dark Energy: This is the biggest component, and is roughly 70% of the universe’s energy content Some negative pressure associated with the universe’s expansion Big Bang In the beginning, it was hot and dense, with all particle species in roughly equal abundances At some point, it expanded rapidly, and cooled off in the process We can map out the various phases by converting these temperatures to energies: EW phase transition (100 GeV): the electroweak symmetry is broken, allowing EM and the weak force to become distinct entities and particles gain mass via the Higgs As the temperature drops below the mass of particle species, particle antiparticle annihilation begins while pair production becomes inefficient QCD phase transition (150 MeV): the remaining quarks condense to hadrons At the grand old age of 1, the neutrinos decouple from equillibrium due to the expansion of the universe creating the cosmic neutrino background ($c\nu B$) Big Bang Nucleosynthesis (BBN) occurs around 1 minute in, creating elements lighter than lithium-7 370,000 years later, the universe cooled enough to allow recombination to occur. This allowed the photons to stream through the universe (CMB) There are two periods which we don’t know exactly when they occured dark matter production: lots of theories here (WIMPs, axions etc.) baryogenesis: The symmetry breaking between matter and antimatter (1 part in $10^{10}$) to give the matter to photon ratio observed Structure formation The first stars (Population III stars) are boorn 100 millions years after the Big Bang They were massive, and died off rapidly. The UV light emitted heated up the surrounding gas, leading to reionization They could have created the supermassive black holes They created the heavier elements in the universe The first galaxies formed around 1 billion years after the Big Bang The distribution of galaxies is not uniform (on small scales): there are spatial correlations Inflation We observe some objects in the universe to be older than the universe’s age Inflation helps solve this: if the distances between objects suddenly increases in a short time frame, you can explain these apparent causality violations This period would need to happen before the Big Bang In 1 billionth of a trillionth of a trillionth of a second, the universe doubled in size about 80 times The Expanding Universe GR Concept Refresher Spacetime is defined by some metric: $ds^{2} = \Sigma_{\mu,\nu}^{3} g_{\mu\nu}dx^{\mu}dx^{\nu}$ $ds^{2}$ is invariant to all observers In Minkowski space, this reduces to $ds^{2} = -c^{2}dt^{2} + \delta_{ij} dx^{i} dx^{j}$ the metric is used to convert between contravariant vectors ($A_{\mu}$) and covariant vectors ($A_{\mu}$) You can contract a convariant with a contravariant vector via the metric to produce an invariant scalar FRW Metric On large enough scales, the distribution of mass is isotropic (same in all directions) and homogeneous (same at every point in space) The metric then takes the form $ds^{2} = -c^{2} dt^{2} + a^{2}(t) d\vec{l}^{2}$ homogeneous and isotropic spacetimes must have constant intrinsic curvature This means that we can write $d\vec{l}^{2} = dx^{2}+ k \frac{(x \cdot dx)^{2}}{R_{0}^{2}-kx^{2}}$ $R_{0}$ denotes the intrinsic curvature of the space In spherical coordinates, we have $dl^{2} = \frac{dr^{2}}{1-k\frac{r^{2}}{R_{0}^{2}}}+ r^{2} d\Omega^{2}$ k can be 0 for flat space, 1 for spherical, and -1 for hyperbolic The RW metric becomes $ds^{2} = -c^{2} dt^{2} + a^{2}(t)(\frac{dr^{2}}{1-k\frac{r^{2}}{R_{0}^{2}}}+ r^{2} d\Omega^{2})$ The line element has a rescaling symmetry $a \rightarrow \lambda a$, $ r\rightarrow r/ \lambda$ and $R_{0} \rightarrow R_{0}/\lambda$ This allows $a(t_{0}) = 1$ (re: the scale factor at the present time) as well as $R_{0} = R(t_{0})$ to be interpreted as the physical curvature today we call r a comoving coordinate, while $r_{phys} = a(t) r$ is the physical coordinate The physical velocity is just the time derivative of $r_{phys}$ $v_{phys} = H r_{phys} + v_{pec}$ We define $H = \frac{\dot{a}}{a}$ to be the Hubble parameter $v_{pec}$ to be the velocity of a comoving observer (re: someone who follows the Hubble flow) We can make the change of coordinates $d\chi = \frac{dr}{\sqrt{1-\frac{kr^{2}}{R_{0}^{2}}}}$ The FRW metric becomes: $ds^{2} = -c^{2} dt^{2} + a^{2}(t) (d\chi^{2}+ S_{k}(x)^{2} d\Omega^{2})$ where $S_{k}(\chi)$ is $\chi$ for k=0, $R_{0} \sin (\frac{\chi}{R_{0}})$, and $R_{0} \sinh(\frac{\chi}{R_{0}})$ The t in this FRW metric is the coordinate time Another useful time variable is the conformal time $d\eta = \frac{dt}{a(t)}$ Kinematics Define a free particle lagrangian in an arbitrary coordinate system as $\mathbf{L} = \frac{m}{2} g_{ij}(x^{k}) \dot{x}^{i}\dot{x}^{j}$ Substitute into the Euler-Lagrange equution to get $\frac{d^{2} x^{i}}{dt^{2}} = - \Gamma_{ab}^{i} \frac{dx^{a}}{dt}\frac{dx^{b}}{dt}$ We define $\Gamma_{ab}^{i} = \frac{1}{2} g^{ij} (\partial_{a} g_{jb} + \partial_{b} g_{ja} + \partial_{j} g_{ab})$ These are the Christoffel symbols This is easily extended to 4 dimensions. Simply replace the latin indices for greek ones (re: add time equations in) The derivaion for massless particles require some care though If you define the 4-momentum as $P^{\mu} = m \frac{dx^{\mu}}{d\tau}$, the geodesic equation becomes $P^{\alpha}(\partial_{\alpha}P^{\mu} + \Gamma_{\alpha \beta}^{\mu} P^{\beta}) = 0$ The bracketed term is the covariant derivative, denoted as $\nabla_{\alpha} P^{\mu}$ Substituting in the FRW metric to the geodesic equation tells you how the 4-momentum depends on the scale factor Looking at $\mu=0$, we see that: For massless particles, we have $\frac{1}{E} \frac{dE}{dt} = - \frac{\dot{a}}{a}$ This implies an energy scalinig of $E = \frac{1}{a}$ For massive particles, we have that the momentum scales like $\frac{1}{a}$ Redshift Since spacetime is getting stretched out, the wavelength of light gets stretched as well. This increase in observed wavelength is called redshifting Since energy of photons scales like $\frac{1}{a}$, the wavelength is proportional to a. Hence $\lambda_{0} = \frac{a(t_{0})}{a(t_{1})} \lambda_{1}$ redshift is defined as the fractional shift in wavelength: $z = \frac{\lambda_{0}-\lambda_{1}}{\lambda_{1}}$ Using $a(t_{0})=1$ for today’s scale factor, we have that $1+z = \frac{1}{a(t_{1})}$ Conventionally, we label events in the universe’s history by their redshift: the surface of last scattering happens at z=1100 and the first galaxies formed around z= 10 For nearby sources (z< 1), we can expand the scale factor around $t_{0}$ and define $H_{0} = \frac{\dot{a_{0}}}{a(t_{0})}$ as Hubble’s constant to get that $a(t_{1}) = 1+ (t_{1}-t_{0})H_{0}$ For close objects, $\Delta t$ is simply $frac{d}{c}$, which means that redshift increases linearly with distance Alternatively, the recession speed $v = cz$ becomes $v \approx H_{0}d$ The above is called the Hubble-Lemaitre law It’s convention to describe Hubble constant measurements as $H_{0} = 100 h \frac{km}{s Mpc}$ since the measurements originally have very large uncertainties h is $0.73 \pm 0.01$ from supernovae measurements while h is $0.674 \pm 0.005$ from CMB measurements There is a statistically significant difference between the two, giving rise to the “Hubble tension” Distances These are hard to measure. The metric distance is unobservable, and the physical distance assumes a fixed time To make measurements, we need definitions which take into account the expansion and the finite travel time of light Luminosity Distance We can use “standard candles” (re: objects of known intrinsic brightness) and compare against their observed brightness to calculate distances We have the Cepheids, which have a periodic brightness. This period is correlated with the intrinsic brightness of a star More explicitly, the brightness of the star versus the logarithm of the period shows a linear relationship So if you can measure the distance to a Cepheid via parallax, you can extrapolate the distances to the rest of the Cepheids The next rung up on the cosmic distance ladder are the Type 1a supernovae These happen when a white drawf accretes too much matter from a companion star and explode They are so bright that they outshine the stars in their local group These explosions occur at a precise moment (re: when the white drawf exceeds the Chandrasekhar limit) and hence have a fixed brightness To generalize, suppose that we have a source of known luminosity L (re: energy per unit time) There is some observed flux (re: energy per unit time per unit area) from with L can be inferred Suppose that the source is at redshift z. The comoving distance is $\chi(z) = c\int_{0}^{z} \frac{dz}{H(z)}$ The flux in a static space is $F = \frac{L}{4 \pi \chi^{2}}$ since the source is isotropic the expansion complicates this for 3 reasons: the radius of the sphere expands so that the area becomes $4\pi a^{2}(t_0) d_{M}^{2}$ The arrival rate of the photons is smaller than the rate at which they are emitted by a factor of $\frac{1}{1+z}$ The energy of the photons get red shifted, which contributes another factor of $\frac{1}{1+z}$ The end result is that $F = \frac{L}{4\pi d_{M}^{2} (1+z)^{2}} = \frac{L}{4\pi d_{L}^{2}}$ where $d_{L}$ is the luminosity distance $d_{L} = (1+z) d_{M}(z)$ where $d_{M}$ is the metric distance Angular Diameter Distance Standard rulers are objects of know physical size An example is the typical size of the hot and cold spots in the CMB from theory Once again, assume a static space. The angular size of an object that is of known length D at some comoving distance $\chi$ away is $\delta \theta = \frac{D}{\chi}$ With expansion, the formula becomes $\delta \theta = \frac{D}{a(t_{1}) d_{M}} = \frac{D}{d_{A}}$ $d_{A}(z) = \frac{d_{M}}{1+z}$ relates the angular diameter distance to the metric distance Dynamics To calculate the scale factor, we need to solve the Einstein equation: $G_{\mu\nu} = \frac{8\pi G}{c^{4}} T_{\mu\nu}$ G is the Einstein tensor (curvature measurement) which T is the energy momentum tensor First, let’s look at number density $N^{\mu}$ $\mu=0$ is the number density of particles $\mu=i$ component is the flux of the particles in direction $x^{i}$ What constraints must $N^{\mu}$ obey to satisfy homogeneity and isotropy? Isotropy implies that $N^{i} = 0$ while homogeneity implies $N^{0} = c n(t)$ In a comoving frame, we have $N^{\mu} = n U^{\mu}$ Particle number should be conserved, which implies $\partial_{\mu} N^{\mu} = 0$ In curved spacetime, we have $\nabla_{\mu} N^{\mu} = 0$ $\nabla_{\mu} A^{\nu} = \partial_{\mu}A^{\nu} + \Gamma_{\mu\lambda}^{\nu} A^{\lambda}$ $\nabla_{\mu} B_{\nu} = \partial_{\mu}B_{\nu} - \Gamma_{\mu\nu}^{\lambda} B_{\lambda}$ So we have $\partial_{\mu} N^{\mu} = -\Lambda_{\mu \lambda}^{\mu} N^{\lambda)$ In the rest frame, we have $ \frac{\dot{n}}{n} = -3 \frac{\dot{a}}{a}$ This implies that $n(t) \propto a^{-3}$ In an analagous way to the number density, we can constrain the form of $T_{\mu\nu}$ We can break the tensor into 3 components: an energy density (scalar), a vector (momentum density/energy flux. Same values because of symmetry), and a smaller 3x3 stress tensor Isotropy forces the vector components to be 0 Homogeneity requires the energy density to be indpendent of position, but dependent on time Isotropy around the origin constrains the mean value of the stress tensor to be proportional to $\delta_{ij}$. Homogeneity requires that the proportionality constant be only a function of time All of that allows us to write down the perfect fluid stress-energy tensor: $T_{\mu\nu} = (\rho+\frac{P}{c^{2}})U_{\mu}U_{\nu}+P g_{\mu\nu}$ Local energy and momentum conservation imply $\nabla_{\mu} T^{\mu}{\nu} = 0$ Expand out in terms of Christoffel symbols and examine the $\nu=0$ term The $T^{i}_{0}$ term vanishes by isotropy In what remains, $\Gamma_{\mu 0}^{\lambda}$ is 0 unless $\lambda$ and $\mu$ are the same spatial indices, which implies $\Gamma_{i0}^{i} = \frac{3}{c} \frac{\dot{a}}{a}$ From the continuity equation, we arrive at: $\dot{\rho} + 3 \frac{\dot{a}}{a} (\rho + \frac{P}{c^{2}}) = 0$ Since the time translation is broken in an expanding space, global energy conservation doesn’t hold Most cosmological fluids are parametered as $P = w(\rho c^{2})$ This implies $\rho \propto a^{-3(1+w)}$ Matter is a fluid whose pressure is much smaller than it’s energy density. w=0, which implies $\rho \propto a^{-3}$ This could be baryons or dark matter Radiation is anything which obeys $P = \frac{1}{3} \rho c^{2}$. So $w=\frac{1}{3}$ and $\rho \propto a^{-4}$ This can be very light particles, photons, neutrinos, gravitons Dark energy has negative pressure, which implies $\rho \propto a^{0}$ In detail: EFE is $\G^{\mu\nu} + \Lambda g^{\mu\nu} = 8\pi G T^{\mu\nu}$. If you take $\nabla_{\mu}$, the $\Lambda$ term doesn’t contribute to the constraint Hence, you can define an additional contribution to the energy tensor $T_{de}^{\mu\nu} = -\frac{\Lambda}{8\pi G} g^{\mu\nu}$ without messing up the constraints. Plugging into EFE gives the $P = -\rho$ This iis the standard cosmological model of $\Lambda$ CDM Einstein Tensor for FRW metric The Einstein tensor is defined as $G_{\mu\nu} = R_{\mu\nu}-\frac{1}{2}R g_{\mu\nu}$ $R_{\mu\nu}$ is the Ricci tensor and $R = g^{\mu\nu}R_{\mu\nu}$ is the Ricci scalar You can write the Ricci tensor in terms of the Christoffel symbols: $\partial_{\lambda} \Gamma_{\mu\nu}^{\lambda} - \partial_{\nu} \Gamma_{\mu\lambda}^{\lambda} + \Gamma_{\lambda\rho}^{\lambda} \Gamma_{\mu\nu}^{\rho} - \Gamma_{\mu\lambda}^{\rho}\Gamma_{\nu\rho}^{\lambda}$ $R_{0i} = R_{i0} = 0$ because of isotropy (re: vector component) $R_{00} = -\frac{3}{c^{2}} \frac{\ddot{a}}{a}$ $R_{ij} = \frac{1}{c^{2}} (\frac{\ddot{a}}{a} + 2 (\frac{\dot{a}}{a})^{2} + 2 \frac{kc^{2}}{a^{2} R_{0}^{2}}) g_{ij}$ The author doesn’t directly calculate this. Instead, he computes $R_{ij}$ around x=0, then argues that since this is a tensorial reslationship, the equation holds everywhere The spatial metric is $\gamma_{ij} = \delta_{ij} + \frac{k x_{i} x_{j}}{R_{0}^{2}-k(x_{k}x^{k})}$ They expand to 2nd order since you need derivatives of $\gamma_{ij}$, and $\gamma_{ij}(x=0) = \delta_{ij}$ doesn’t have this information $\gamma_{jk}^{i} = \frac{k}{R_{0}^{2}} x^{i} \delta_{jk}$ The Ricci scalar is then $R = \frac{6}{c^{2}}(\frac{\ddot{a}}{a} + (\frac{\dot{a}}{a})^{2} +\frac{kc^{2}}{a^{2} R_{0}^{2}})$ Friedmann Equations Plugging in the FRW metric into the Einstein equations gives the Friedmann Equations The time-time component gives: $(\frac{\dot{a}}{a})^{2} = \frac{8\pi G}{3} \rho - \frac{kc^{2}}{a^{2}R_{0}^{2}}$ This is often written in terms of the hubble parameter $H = \frac{\dot{a}}{a}$ The first Friedmann equation can also be written as $\frac{H^{2}}{H_{0}^{2}} = \Omega_{r} a^{-4} + \Omega_{m} a^{-3} \Omega_{k} a^{-2} + \Omega_{\Lambda}$ Define $\Omega_{i} = \frac{\rho_{i}}{\rho_{crit}}$ where $\rho_{crit} = \frac{3H_{0}^{2}}{8 \pi G}$ os the density of a flat universe evaluated at today $\Omega_{\Lambda} = -\frac{kc^{2}}{(R_{0}H_{0})^{2}}$ If you evaluate at $t_{0}$, you get that $1 = \Omega_{r} + \Omega_{m} + \Omega_{\Lambda} + \Omega_{k}$ The spatical component(s) yield: $\frac{\ddot{a}}{a} = -\frac{4\pi G}{3}(\rho+\frac{3P}{c^{2}})$ The 2nd Friedmann equation is also called the Raychaudhuri equation It can also be derived by taking the time derivative of the 1st equation and applying continuity for $\dot{\rho}$ Explicating use $\nabla_{\mu} T_{\mu\nu} = 0$ Exact Solutions of the Friedmann Equations In general, you need numerics to solve the Friedmann equation Single Component Universes Assume a flat universe with just a single component You can parameterize the specific component by its associated w factor Namely assume that $\frac{P}{\rho} = w $ We have $\frac{d \ln a}{dt} \approx H_{0} \sqrt{\Omega_{i}} a^{-\frac{3}{2}(1+w_{i})}$ Transforming to conformal time, we find $a(\eta) \propto \eta^{\frac{2}{1+3 w_{i}}}$ The special case of $w_{i} = 1/3$ is a universe dominated by spatial curvature A pure radiation universe scales like $a(\eta) = 2 H_{0} \sqrt{\Omega_{\gamma}} \eta$ A pure matter universe is a called a Einstein-de Sitter universe (ie. w=0, since the pressure is much much smaller than the rest mass) This is a good approximation for long matter-dominated periods in the universe Using a Matter dominated universe implies that the Hubble constant is $\frac{2}{3} \frac{1}{t_{0}}$. This gives a universe age of 9 billion years, which is the famous age problem $a(\eta) = \frac{1}{4} H_{0}^{2} \Omega_{m} \eta^{2}$ for very late times A pure curvature universe is only permitted if k=-1. This is called a Milne universe A pure curvature universe is only permitted if k=-1. This is called a Milne universe If $w=-1$, then we have a de Sitter space, which is a good approximation in the far past and far future $a(t) \propto exp(\sqrt{\frac{\Lambda}{3}}t)$ If we have $\Lambda<0$ and $k=-1$, it’s called an Anti-de Sitter spacetime All of these solutions (baring de Sitter space) have a singularity at t=0. This singularity (which goes like $\rho \propto t^{-2}$ is a generic feature of all Big Bang cosmologies (Hawking and Penrose) This assumes a strong energy condition ($\rho c^{2} + 3P \geq 0 $ which implies that $\ddot{a} \leq 0$ at all times Gravity is attractive, so it slows down the expansion Our Universe There are a handful of parameters that parameterize our universe First off is the temperature of the cosmic microwave background (CMB), which is given as $T_{0} = 2.7255 \pm 0.0006 K$ The energy density of the photons is given by $\Omega_{\gamma} = 5.4E-5$ The energy density of the neutrinos is 68% that of the photons This is currently being constrained by $0.0012 < \Omega_{\nu} < 0.003$, where the lower limit comes from neutrino oscillation experiments and the upper bound comes from cosmology All of these measurements came from COBE The anisotropy of the CMB places constrains on the curvature energy density ($|\Omega_{k}| < 0.005$ This implies that curvature is a very negligable part of the universe’s energy contents (and more so in the past) We have the baryon density, which we can infer from the abundances of light chemical elements produced in the Big Bang. This coems to around $\Omega_{b} \approox 0.05$ The dark matter density is $\Omega_{c} \approx 0.27$ Dark energy, responsible for the expansion of the universe, makes up $\Omega_{\Lambda} \approx 0.68$ of the energy content of the universe Integrating the Friedmann equation yields: $H_{0} t = =\int_{0}^{a} \frac{da}{\sqrt{\Omega_{r}a^{-2}+\Omega_{m}a^{-1}+\Omega_{\Lambda}a^{2}+\Omega_{k}}}$ We can infer the age of the universe (a=1) to be $t_{0} \approx 13.8$ GYrs We can evaluate the time scales at which matter-radiation and matter-dark energy equality occur at: $t_{mr} \approx 50000$ yrs $t_{m\Lambda} \approx 10.2$ Gyrs The coincidence problem: Why did dark energy come to dominate the expansion so close to the present time? The Hot Big Bang Natural Units We define natural units, which set the speed of light and $\hbar$ to unity We also define the reduced Planck mass as $M_{PI} = \sqrt{\frac{\hbar c}{8\pi G}}$ The Boltzmann constant is also typically set to unity, equating temperature and mass Boltzmann Hierarchy From Liouville’s Theorem, we know that the phase space density doesn’t change over time: $\frac{d}{dt}(\Delta^{3}x \Delta^{3} p)=0$ ...

Logistics Conventions Newtonian Gravity The Two Body Problem Kepler’s Second Law Kepler’s Third Law From Kepler To Newton The Equivalence Principle Gravitational Redshift Light falls Differential Geometry Definitions/Basics Crash Course Example 2-Sphere Special Relativity Minkowski Spacetime Diagrams Metric Useful Tricks 4-Vectors 4-velocity Dual Vectors 4-momentum Inverse Metric Variational Approach Moving to GR Christoffel Symbols We haven’t done GR yet. Let’s do that Qausi-Stationary Curved or not Curved Tensors General Covariance Useful facts Scalar Field Transformation Vector Field Transformation Dual Vector Transformation Stress-Energy Tensor $T^{\mu\nu}$ Dust Covariant Derivatives Properties of Covariant Derivatives Tidal Forces of Curvature Newtonian GR version Directional Covariant Derivative Reimannian tensor Reimannian Tensor properties Parallel Transport Stress Energy Tensor Conservation EM Force Law Einstein’s Equation What the Heck is $\kappa$? Cosmological Constant Killing Vector Schwarzschild Solution Potential Redshift Lensing An introduction to general relativity. ...

Compilation of notes for Astrophysics and Cosmology class for Spring 2023. Special Relativity Spacetime Diagrams Worldline Inertial Frames Postulates of SR Deriving Time Dilation from diagrams Deriving Doppler Shift from Diagram Metrics Metric tensor (SR) Covariant versus contravariant vectors Metric in Spherical Coordinates 4-Momentum Example Classical Cosmology The Cosmological Principle Comoving Coordinates Hubble’s Law Wavelength relation to scale factor Luminosity Distance (L) Non-Euclidean Geometry Metric of a 2-Sphere Metric of the 3-sphere Metric of the Pseudosphere (hyperbolic geometry) Most general homogeneous isotropic metric What is $a(t)?$ Adiabatic expansion Friedmann Equations Summary Pure Matter Energy Density Pure Radiation Energy Density Dark Energy q Age of the Universe Einstein-de Sitter Universe Einstein Static Universe deSitter Universe Dark Energy versus Cosmological Constant Inconsistency of Hubble’s constant Distance Measures Comoving (radial) line of sight distance Transverse Comoving distance Transverse Angular Diameter Distance Luminosity Distance Horizon Distance Examples of Calculating Distances Using Type-I Supernovae to Extract Cosmological Parameters Distance Ladder Parallax Globular Cluster Fitting (Main Sequence Fitting) Cepheids (variable stars) Tully-Fisher Early Universe Physics Planck Mass Flatness Problem The Horizon Problem Solving the Horizon Problem Decoupling Time/Radiation Recombination Time Dark Matter Evidence Rotation curves Dark Baryon Problem Gravitational Lensing Particle Physics Theory Subatomic particles Force Mediators Feymann Diagrams Primary Vertices Propagators and Probability Amplitudes Why Antimatter? Klein-Gordan Equation Dirac Equation Building Particles Classical and Quantum Theory of the Scalar Field Quantitative Calculations Fermi’s Golden Rule (2 body) Calculating $T_{if}$ Directly Via Feynmann Diagrams Rules Cross section calculations Early universe $e^++e^-\rightarrow \mu^++\mu^-$ $e^++e^-\rightarrow \gamma+\gamma$ Thomson Scattering ($e+\gamma \rightarrow \gamma e$) $\mu^++e^-\rightarrow u^++e^-$ Higher Order Corrections to $ee\rightarrow ee$ Weak Interactions e $\nu$ scattering $e^+e^- \rightarrow \nu_e\bar{\nu_e}$ Discovery of Color $\beta$-decay of neutron $n\rightarrow pe^-\bar{v_e}$ Modern Fermi Theory Quark-Gluon Plasma GUTs Super-K (Proton Decay) Dark Matter Candidates Axions a Sterile neutrino $\chi$ Light DM (ALP or axion-like particles) Stat Mech of Early Universe Ultra-relativistic particles Family Degeneracy Current Temperature of CMB Reheating Neutrino Contribution to DM WIMP Miracle Inflationary cosmology revisited, and structure formation Energy Scales Structure Formation Transparency of the Universe The Cosmic Microwave Background (CMB) Observations of the CMB Measuring CMB Correlation Function CMB Acoustic Oscillations Primordial Black Holes and Hawking Radiation Application to Inflation Polarized CMB Neutrinos Super-K Detection Channel in Supe-K SNO CC Channel NC Channel ICECUBE Detection Channels Special Relativity Spacetime Diagrams Simplest visualization is 2D plot. One axis is ct, the other is x (ct is for consistent dimensions, could also do x/c to measure in units of time). The true diagram plots the vector $<ct,x,y,z>$ in a 4-dimensional manifold. Each one of these vectors is an event in spacetime Worldline Worldline denotes our trajectory through spacetime If you are stationary at x=0, then you get a vertical line at x = 0 If you walk to the right, then your x coordinate increases A light beam moves along the $45 \degree$ lines (either in 1st or 2nd quadrant) The faster you go, the closer you slope gets to unity (ie. light speed) The future light cone denotes the area above the x axis between the light worldlines The past light cone denotes the area below the x axis between the light worldlines Inertial Frames Consider two frames, one stationary, the other strapped to a rocket (the prime frame). Observer has their own clocks and rulers The events measured by the rocket are given by $<ct^{’},x^{’},y^{’},z^{’}>$, and the events in the lab are given by $<ct,x,y,z>$. The distinction is purely for bookkeeping. Assume that the rocket moves with velocity v relative to lab frame Draw the worldline of the rocket in the lab frame. This is the axis ct’ of the rocket frame This makes sense. The rocket observer is always at $x^{’}=0$ if they are not moving in their frame. So it makes sense that the worldline is the $ct^{’}$ axis To draw the x’ axis, take the angle between ct’ and ct axis. Transfer this axis to a positive angle from the x axis. This is the x’ axis Postulates of SR Light has a constant speed of c Light moves isotropically for all observers Imagine in the lab frame, you have a point source of light. Then $x = ct \rightarrow (ct)^{2} = x^{2}+y^{2}+z^{2}$ Similarly, in the rocket frame, you get $x^{’} = ct^{’} \rightarrow (ct^{’})^{2} = x’^{2}+y’^{2}+z’^{2}$ We can combine the above to get: $(ct^{’})^{2} - x’^{2}+y’^{2}+z’^{2} = (ct)^{2} - x^{2}+y^{2}+z^{2} = s^{2}$ $s$ is invariant between reference frames and holds true for any event, not just light. Called the interval and is used to define distance in spacetime can also be defined as $-s^{2} = x^{2}+y^{2}+z^{2}-c^{2}t^{2}$ Deriving Time Dilation from diagrams Suppose $\Delta x$ and $\Delta t$ displacements occur lab frame. This corresponds to a displacement of $\tau$ in rocket frame rocket: $(c\tau,0)$, lab: $(ct,x)$ The $s^{2}$ must be the same in both frames: $c^{2}t^{2}-x^{2} = c^{2}\tau^{2}$. Rearrange as needed to get time dilation Deriving Doppler Shift from Diagram The rocket is travelling at a velocity v. The rocket shots a light beam back to the lab frame (a 45 degree ray aimed at ct axis that starts at the end of the rocket trajectory) The rocket has displacements $c \Delta t$ and $\Delta x = \beta c \Delta t$. In the rocket frame, this displacement is $\tau$. The round trip time measured by the lab is: $\Delta t_{r} = c\Delta t + \Delta x = c\Delta t + \beta c \Delta t \rightarrow \Delta t_{r} = \Delta t (1+\beta)$ You add $\Delta x$ because this is the elasped time that the light takes to reach x=0 from the rocket Substitute $\Delta t$ with $\Tau$ using invariance of $s^{2}$ and simplify $\Delta t_{r} = \sqrt{\frac{1-\beta}{1+\beta}}$ $\tau \propto \frac{1}{\nu_{e}}$ and $\Delta t_{r} \propto \frac{1}{\nu_{r}}$, where $\nu_{e}$ is the emmision frequency and $\nu_{e}$ is the receiver frequency In terms of wavelength: $\lambda_{r} = \lambda_{e}\sqrt{\frac{1\pm \beta}{1\mp \beta}}$ (plus and minus depends on relative velocity direction) Metrics Metric tensor (SR) $ds^{2} = c^{2}dt^{2}-(dx^{2}+dy^{2}+dz^{2})$ ...