Classical Mechanics

Logistics Principle of “Least” Action Symmetries Gauge Invariance Example Reduce to Quadratures Hamiltonian Mechanics Phase Portraits Poisson Brackets Properties Jacobi Proof Sketch Poisson’s theorem Canonical Transformations Why We Care Liouville’s Theorem Canonical Invariants Entropy Tangent Infinitesimal CTs Hamiltonian Noether’s Theorem Hamilton Jacobi Equation Action Variables Multiperiodic motion Planetary motion Liouville’s Integrability Theorem Cannonical Peturbation Theory Adiabatic Motion Quantum Mach Zender Hilbert Space Linear Algebra Review Time Evolution Heisenberg Formulation Compatible Observables Symmetries Uncertainty Relationship Phase Space Interpretation Entanglement And Mixed States Mixed States Von Neumann Entropy Partial Trace (“Tracing Out”) Infinite Spaces Translation Operator Scattering Probability Current Delta Function Path Integral Free Particle Saddle Point Approximation Path Integral Proof WKB Logistics Matthew Klebon [email protected] Office Hours: Grading: TBD TA: [email protected]. Room 943 Principle of “Least” Action N particles in 3D There are 6N coordinates We want to find $\vec{q}(t)$ of the system where q represents the coordinates in position for each particle $q = {x_{0},x_{1},… x_{n}}$ As long as the equations of motions are 2nd order, there exists a unique solution for the system Let’s examine a simple system: the double pendulum Naively, one would expect 6 equations of motions for the 6 coordinates. However, there are 4 constrains of the system $z_{1} = z_{2} = 0$ $x_{1}^{2}+y_{1}^{2} = l_{1}^{2}$ $x_{2}^{2}+y_{2}^{2} = l_{2}^{2}$ With these constraints, you can reduce the system to a function of two coordinates: $\theta_{1}$ and $\theta_{2}$ holonomic: when the constraints of a system are solely of a position $f_{k}(q_{1}…q_{n}) = 0$ Not all constraints are holonomic: Rolling without slipping, inequalities etc. To incorporate these constraints, you can either use Lagrange multipliers, or use the constraints to eliminate degrees of freedom Assuming no dissipative forces, you can describe the system by a Lagrangian $\mathcal{L(q, \dot{q}, t)}$ This is typically $\mathcal{L} = T-V$ In polar coordinates, we have that $\dot{r}^{2}+(r\dot{\theta})^{2}$ In spherical coordinates, we have that $\dot{r}^{2}+(r\dot{\theta}\sin \phi)^{2} + (r\dot{\phi})^{2}$ From this Lagrangian, you can define something called the action of the path: $S(q_{i}(t)) = \int_{t_{1}}^{t_{2}} \mathcal{L}(q(t), \dot{q(t)}, t) dt$ This is a functional (ie. it’s arguments are other functions) Units of Plank’s constant and angular momentum You want to find the stationary points of the action function as a function of time. Pretend that the endpoints of the path are fixed at two points in time: $q(t_{1}) == q(t_{2})$ This implies that $\delta q_{i}(t_{1}) = \delta q_{i}(t_{2}) = 0$ Taking the total derivative of the action, use integration by parts, utilize the boundary conditions and rearrange to get $\delta S = \int_{t_{1}}^{t_{2}} dt (\frac{\partial L}{\partial q_{1}}-\frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{q_{i}}})\delta q_{i}) = 0$ This must hold for all possible $q_{i}$, so if we imagine placing delta functions everywhere, this implies that the bracketed term must be 0 these are the Euler-Lagrange equations: $\frac{\partial L}{\partial q_{1}}-\frac{\partial}{\partial t}(\frac{\partial L}{\partial \dot{q_{i}}}) = 0$ Sometimes, it’s easier to plug in $q+\delta q$ directly into the action and reduce up to first order terms You can add a total derivative $\frac{d}{dt}f(q,t)$ (note lack of dependence on $\dot{q}$) to a Lagrangian and leave the equations of motion unchanged Simply do variational principle with additional term The converse is not necessarily true The principle of least action is invariant under coordinate changes Symmetries When there is a symmetry, there is a conserved quantity (Noether’s Theorem) For instance, if the Lagrangian is time independent (ie. $L(q,\dot{q},t) = L(q,\dot{q})$), then via the Euler Lagrange equations, energy conservation follows Taking the total derivative of the Lagrangian w.r.t. time, applying the time-independent constraint, and then rearranging yields that that the following is conserved. This is the energy (usually)! $\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}\dot{q}-L) = 0$ A trivial symmetry is when $\frac{\partial L}{\partial q} = 0$. This is called a cyclic coordinate Another example is $L(q_{i}+\delta q_{i}) = L(q_{i})$. If q is a position, then the associated momentum is also conserved Yet another is conservation of angular momentum: $L = \frac{1}{2} m \dot{r}^{2}- V(|\vec{r}|)$ $\frac{\partial \vec{r}}{\partial \theta_{i}} = \hat{n_{i} \times \vec{r}} = \epsilon_{jkl} n_{k} r_{l}$ Can reduce this to $P_{\theta_{i}} = \epsilon_{ilj} r_{l} (m\dot{r}_{j})$, which is just $\vec{r}\times \vec{p}$ This can be done by utilizing the fact that $\hat{n_{ik}} = \delta_{ik}$ that the Levi-Civita symbol is invariant under cyclic permuations of it’s indicies the definition that $p_{\theta} = \frac{\partial L}{\partial \dot{r_{j}}}\frac{\partial \dot{r_{j}}}{\partial \dot{\theta_{i}}}$ More formally, a transformation $q_{i} => f(\epsilon, \bar{q}); f_{i}(0) = q_{i}$ (where $\epsilon$ is the implicit variable and $\bar{q}$ represents all of the coordinates) is a symmetry of the system if the Lagrangian is unchanged (ie. $\delta L = \frac{d}{dt}(\frac{\partial L}{\partial q_{i}} \delta q_{i}) = 0$) If the symmetry holds, then the conserved quantity obeys the following equation $\frac{\partial L}{\partial q_{i}}\delta q_{i} = \epsilon \frac{\partial}{\partial q_{i}}\frac{\partial f_{i}}{\partial \epsilon}|_{\epsilon = 0}$ Gauge Invariance We know that $L’ = L + \frac{d\Phi}{dt}$ holds in general for some scalar field (gauge invariance) This modifies Noether’s therom: $\frac{d}{dt}(\frac{\partial L}{\partial q_{i}}\dot{q_{i}}-\Phi) = 0$ Hence $\frac{\partial L}{\partial q_{i}}\dot{q_{i}}-\Phi$ is conserved Example Let’s couple a charge particle with some fields: $L = \frac{1}{2} m\vec{\dot{x}^{2}}-qV+\frac{q}{c}\vec{A}\cdot \vec{x}$ There is a gauge invariance from the transformation $\vec{A} => \vec{A}+c\nabla \Lambda$ and $V => V - \frac{\partial}{\partial t} \Lambda$ Plugging in this transformation yields that $L’ = L + \frac{\partial}{\partial t}(q\Lambda)+\frac{\partial}{\partial x_{i}}(q\Lambda) \vec{\dot{x_{i}}} = L + \frac{d}{dt} \Phi$ Reduce to Quadratures Instead of solving a PDE to find equations of motion, you can solve a different PDE in terms of conserved quantities For instance: $L = \frac{1}{2}m\dot{x}^{2} -V(x)$ has $h = \frac{1}{2}m\dot{x}^{2} +V(x)$ as a conserved quantity You could solve $m\ddot{x} = 9\frac{\partial V}{\partial x}$ You could also solve $(\frac{dx}{dt})^{2} = \frac{2}{m} (h-V(x))$ Hamiltonian Mechanics The Legendre transform of the Lagrangian is the Hamiltonian Graphically, you can take the slope of the initial function, and project that onto the y-axis. In symbols $v \frac{\partial L}{\partial v}-L = H(p)$ where v is the x axis coordinate and p is the y axis coordinate This assumes that $\frac{dL}{dv} = p(v)$ We assume that p is invertible (ie. p(v) implies v(p)) $L’’(v) \neq 0$ This one allows H to be a well defined function With the above in mind, we define the Hamiltonian as $H(q_{i},p_{i},t) = \Sigma_{i=1}^{3} p_{i}\dot{q_{i}}-L(q_{i},\dot{q_{i}},t)$ $p_{i} = \frac{\partial L}{\partial \dot{q_{i}}}$ Taking the total derivative of the Hamiltonian, applying the equation $p_{i} =\frac{\partial L}{\partial \dot{q_{i}}}$ and applying the EL equations yields $H = \dot{q}dp_{i}-\dot{p_{i}}dq_{i}-\frac{\partial L}{\partial t} dt$ $\frac{\partial H}{\partial q_{i}} = -\dot{p_{i}}$ $\frac{\partial H}{\partial p_{i}} = \dot{q_{i}}$ $\frac{\partial H}{\partial t} = \frac{\partial L}{\partial t}$ You solve double the number of equations, but they are 1st order now Phase Portraits Can plot momentum and position of a particle. Closed loops are called librations Unbounded paths are called rotations The boundary between the two is called the separatrix Poisson Brackets Suppose that you have some observable $f(q_{i}, p_{i},t)$ Utilizing the total time derivative and Hamilton’s equations, you show $\frac{df}{dt} = [f,H]+ \frac{\partial f}{\partial t}$ $[A,B] = \frac{\partial A}{\partial q_{i}}\frac{\partial B}{\partial p_{i}}- \frac{\partial A}{\partial p_{i}}\frac{\partial B}{\partial q_{i}}$ This is the Poisson bracket Properties $[\alpha f_{1} + \beta f_{2}, g] = [\alpha f_{1}, g] + [ \beta f_{2}, g]$ $[f,g] = -[g,f]$ $[f,[g,h]]+[h,[f,g]]+[g,[h,f]] = 0$ (Jacobi identity) $[f,gh]= [f,g]h + g[f,h]$ (Leibniz’s rule) Any object which satisfies the above properties is called a Lie Algebra. ...

Notes for Mechanics class for Spring 2023. Chapter 1 Main Concepts Scalars, Vectors and Matrices Derivatives of Position Chapter 2 Main Concepts Newton’s Laws Definitions Conservation Laws Chapter 3 Main Concepts Chapter 4 Main Concepts Nonlinear Oscillations Phase diagrams Plane Pendulum Chapter 5 Main Concepts Chapter 6 Main Concepts Chapter 7 Main Concepts Chapter 8 Main Concepts Chapter 9 Main Concepts Chapter 10 Main Concepts Chapter 11 Main Concepts Euler Equations Force Field Chapter 1 Main Concepts Scalars, Vectors and Matrices A scalar is some quantity that is invariant under any coordinate transformation A vector is an object in $\mathbb{R^{n}}$ that is invariant under rotations Vectors are closed under addition and scalar multiplication The Dot product is defined as $\vec{A}\cdot \vec{B} = \Sigma_{i} A_{i}B_{i}$ or equivalently $\vec{A}\cdot \vec{B} = |A| |B| \cos(\vec{A},\vec{B})$ $|A| = \sqrt{\Sigma_{i} A_{i}^{2}}$ This is invariant under rotation (since it is a scalar) Abelian The Cross Product is defined as $C = A \times B$ where $C_{i} = \Sigma_{j,k} \epsilon_{ijk}A_{j}B_{k}$ whre $\epsilon_{ijk}$ is the Levi-Civata tensor (0 if indices are the same, 1 if an even permutation of 1,2,3 and -1 if an odd permutation of 1,2,3) Alternatively $|C| = |A||B|\sin\theta$ Non-Abelian $A\times(B\times C) = (A\cdot C)B-(A\cdot B)C$ $A\cdot (B\times C) = B\cdot (C\times A) = C \cdot (A\times B) = ABC$ Matrices are objects in $\mathbb{R^{n\times m}}$ A simple 2D rotation is given by (can be derived by rotating the primed coordinate system around the orign by some angle and adding up lengths) $x_1^{’} = x_1 \cos \theta +x_2 \sin \theta$ $x_2^{’} = -x_1 \sin \theta +x_2 \cos \theta$ In general $x_{i} = \Sigma_{j=1}^{3} \lambda_{ji} x_{j}^{’}$ where the $\lambda$ are the directional cosines defined by $\lambda_{ij} = \cos(x_{i}^{’},x_j)$ ie. the cosine of the angle b/w the axes The inverse transform is such that $\Sigma_{i}\lambda_{ij}\lambda_{ik} =\delta_{jk}$ For orthogonal (ie. rotation matrices), we have that $\lambda^{T} = \lambda^{-1}$ and that det($\lambda = \pm 1$), where the -1 refers to a changing of handedness Derivatives of Position In standard Cartesian coordinates, we have that position is given by $\vec{r} = a \hat{x}+b \hat{y}+c \hat{z}$ In any non-Cartesian coordinates, the basis vectors themselves change as a function of time. For polar coordinates, we have that $\dot{\hat{r}} = \dot{\theta}\hat{\theta}$ $\dot{\hat{\theta}} = -\dot{\theta}\hat{r}$ Hence, the velocity and acceleration in polar coordinates are $\vec{v} = \dot{r}\hat{r}+r\dot{\theta}\hat{\theta}$ $\vec{a} = (\ddot{r}-r\theta^2)\hat{r}+(r\ddot{\theta}^{2}+\dot{r}\dot{\theta})\hat{\theta}$ We can define $\omega = \dot{\theta}$ and define $\vec{v} = \omega \times \vec{r}$ Chapter 2 Main Concepts Newton’s Laws A body remains at rest or in uniform motion unless acted upon by a force A body acted upon by a force moves in such a manner that the time rate of change of momentum equals the force If two bodies exert forces on each other, these forces are equal in magnitude and opposite in direction These laws are only valid in inertial frames of reference (ie. frames where objects move in straight lines when subject to no external force). If you are in a non-inertial frame, you need to introduce forces that explain this deviation from straight lines Much of Mechanics comes down to solving the equation $\vec{m\ddot{r}} = \vec{F(\vec{r},\vec{v},t)}$ Definitions Momentum p is given by $\vec{p} = m\vec{v}$ Force F is given by $\frac{d\vec{p}}{dt} = \frac{d}{dt}(m\vec{v})$ inertial mass (the mass of F=ma) and gravitational mass (the mass in the law of gravitation) are the same Conservation Laws Momentum is conserved when the net force on a system is zero (ie $\dot{p} = 0$) Define angular momentum as $\vec{L} = \vec{r}\times \vec{p} = I\omega$ and torque as $\vec{T} = \vec{r}\times \vec{F} = I\alpha$ $\frac{d\vec{L}}{dt} = \vec{T}$ Hence, when the net torque is 0, angular momentum is conserved Define Work as $\int \vec{F}\cdot \vec{dr}$, kinetic energy as $\frac{mv^2}{2}$. Define the potential U as $\vec{F} = -\nabla U$ The work is given by the change in kinetic energy The total energy E is T+U is conserved as long as the force is not time-dependent From the potential energy, we can determine the stability of a solution at equilibrium (ie when $\frac{dU}{dx} = 0$) if $\frac{d^{2}U}{d^{2}x} >0$ we are stable, if $\frac{d^{2}U}{d^{2}x} <0$, we are unstable Chapter 3 Main Concepts $\vec{F} = -kx$ where $K = -\frac{dF}{dx}$ is called Hooke’s Law For an SHO, we have that $\ddot{x}+\omega_0^2x = 0$ where $\omega_0 = \sqrt{\frac{k}{m}}$ $\omega_0 =2\pi \nu_0$ $\nu_0 = \frac{1}{\tau_0}$ For a damped oscillator with a linear friction term of the form $-b\dot{x}$, we have that $\ddot{x}+2\beta\dot{x}+\omega_{0}^{2}x = 0$ where $\beta = \frac{b}{2m}$ The general solution has the form of $x(t) = e^{-\beta t}(A_{1}exp(t\sqrt{\beta^{2}-\omega_{0}^{2}})+A_{2}exp(-t\sqrt{\beta^{2}-\omega_{0}^{2}}))$ $\omega_{0}^{2} > \beta^{2}$ is underdamping function oscillates, but dies off over time $\omega_{0}^{2} = \beta^{2}$ is critial damping you much change the form of x(t) to becomes $x(t) = (A+Bt)e^{-\beta t} $ This system returns to zero faster than any other system $\omega_{0}^{2} < \beta^{2}$ is over damping Sytem returns to zero slower than critically damped For a forced damped oscillator with a linear friction term, we have that $\ddot{x}+2\beta\dot{x}+\omega_{0}^{2}x = A \cos (\omega t)$ The particular solution takes the form of $D \cos(\omega t-\delta)$ where you need to solve for what D and $\delta$ are by comparing the coefficient of the sin and cosine terms after differentiating You also need the complementary solution from solving the regular damped oscillator When $\omega = \omega_{R}$ where $\omega_{R} = \sqrt{\omega_{0}^{2}-2\beta^2}$, you get resonance, which is the frequency that produces the maximum amplitude response The quality factor Q is defined to be $\frac{\omega_{R}}{2\beta}$. Q is large when little damping occurs For any periodic driving term ($F(t_\tau) = F(t)$), we can see that $F(t) =\frac{a_0}{2}+\Sigma_{n=1}^{\infty}(a_{n}\cos (n\omega t)+b_{n}\sin (n\omega t))$ where $a_{n} = \frac{2}{\tau}\int_{0}^{\tau} F(t’)\cos (n\omega t’) dt'$ $b_{n} = \frac{2}{\tau}\int_{0}^{\tau} F(t’)\sin (n\omega t’) dt'$ Chapter 4 Main Concepts A differential equation is non-linear if it contains powers of $\dot{x}$ and x higher than linear, or if it contains cross terms xy, or if it contains non-linear functions of x Deterministic chaos refers to the motion of a system whose time evolution is sensitivty to the initial conditions Nonlinear Oscillations When the deviation from linearity of an oscillator is symmetric (ie. the force at -x equals the force at x), then the new force becomes $F(x) = -kx+\epsilon x^3$, which has a corresponding potential of $U(x) = \frac{kx^2}{2}-\frac{\epsilon x^2}{4}$ $\epsilon > 0$ is called soft since the new force is less than the linear term alone $\epsilon <0$ is called hard For the antisymmetric case, we have that $F(x) = -kx+\lambda x^3$ with $U(x) = \frac{kx^2}{2}-\frac{\lambda x^3}{3}$ Phase diagrams From energy considerations, we know that $\dot{x} \propto \sqrt{E-U(x)}$. Hence, if U(x) is known, we can construct the phase diagram easily For systems with damping, computers can construct phase paths quite easily from a given set of initial conditions Stable orbits look like ellipses or inwards spirals Unstable orbits spiral outward Plane Pendulum Closed orbits are stable orbits The unit cell refers to the region between $-\pi$ and $\pi$, since this region is repeated The period of the system is energy dependent FOr energies larger than $E_{0}$, the motion corresponds to the pendulum executing complete revolutions around its support axis Chapter 5 Main Concepts $\vec{F} = \frac{-GmM}{r^2}\hat{r}$ $\vec{F} = -Gm \int \frac{\rho(r’)}{r^2} e_{r} dv'$ $r’$ location of mass density w.r.t. origin and r distance b/w mass density and test mass Potential $V = -G \int \frac{pho}{r} dv'$ Potential energy is $U = mV$ an the force is $F -\nabla U$ $V = \frac{-GM}{r}$ for a point mass and when outside a sphere with spherically symmetric mass distribution $\nabla^2 V = 4 \pi G \rho$ Chapter 6 Main Concepts $\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y} = 0$ f is a function of the form $f(x,y,y’)$ whose extrema you are trying to find Alternatively, we can say that $\frac{\partial f}{\partial x}-\frac{d}{dx}(f-y’\frac{\partial f}{\partial y’}) =0$. This is useful when f is not explicitly dependent on x, which implies that $f-y\frac{\partial f}{\partial y’} = const$ With the addition of constraints of the form $g(y,x) = 0$, we have that $\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y’}+\Sigma_{j} \lambda_{j}(x)\frac{\partial g_{j}}{\partial y_{j}}= 0$ Chapter 7 Main Concepts Define your lagrangian as $\mathcal{L} = T-U$. The Euler Lagrange equations then become $\frac{\partial L}{\partial x}-\frac{d}{dx}\frac{\partial L}{\partial \dot{x}} = 0$ These can be modified to include constraints of the form $g(q)=0$: $\mathcal{L} = \frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}+\Sigma_{j}\lambda_{j}(t)\frac{\partial g_{j}}{\partial q} = 0$ The force of constraint them becomes $F_{j} = \Sigma_{k} \lambda_{k} \frac{\partial f_{k}}{\partial q_{j}}$ You have a conserved quantity from Noether’s theorem whenever $\frac{\partial \mathcal{L}}{\partial q} = 0$ The Hamiltonian is defined as $(\Sigma_{j} p_{j}\dot{q}{j}) -L$ where $p{j} = \frac{\partial \mathcal{L}}{\partial \dot{q_{j}}}$ The equations of motion then becomes $\dot{q_{k}} = \frac{\partial H}{\partial p_{k}}$ and $-\dot{p_{k}} = \frac{\partial H}{\partial q_{k}}$ Chapter 8 Main Concepts The reduced mass is defined as $\mu = \frac{m_1 m_2}{m_1+m_2}$ For two-body central force problems, we can write down the Lagrangian as $\frac{\mu}{2}(\dot{r}r^2\dot{\theta}^2)-U(r)$ Via the Euler Lagrange equations and the coordinate transformation $r = \frac{1}{u}$ we find that $\frac{d^2u}{d\theta^2}+u = \frac{\mu}{l^2}\frac{1}{u^2}F(\frac{1}{u})$ where $l = \mu r^2\dot{\theta} = constant$ For a gravitational potential $\frac{-k}{r}$, the equation of motion is $\frac{\alpha}{r} = 1+\epsilon \cos\theta$ where $\alpha = \frac{l^2}{\mu k}$ and $\epsilon = \sqrt{1+\frac{2El^2}{\mu k^2}}$ The orbital period is related to the semi-major axis by the equation $\tau^2 = \frac{4\pi^2\mu}{k}a^3$ Chapter 9 Main Concepts $R = \frac{1}{M}\int \vec{r} dm $ is the center of mass (origin dependent) Equations of motion for composite objects can be reduced to following the point-like center of mass Elastic collisions have kinetic energy being conserved, while inelastic does not $\epsilon = \frac{|v_{2}-v_{1}|}{|u_{2}-u_{1}|}$ is coefficient of restitution. $\epsilon = 1$ is perfectly elastic Impulse is defined as $\int_{t_1}^{t_2} \vec{F} dt = \vec{P}$ b defines the impact parameter (the closest perpendicular distance of a particle approaching the target from infinity) The angular momentum then is $l = m_{1} u_{1} b$ we define the differential cross section as $\sigma(\theta)$ to be the number of interactions per target divided by the number of incidnet particles per unit area $\sigma(\theta) d\Omega = \frac{dN}{I}$ For an axial symmetry scattering problem, we have that $d\Omega = 2\pi \sin\theta d\theta$ $\Delta \Theta = \int_{r_{min}}^{r_{max}} \frac{\frac{l}{r^2} dr}{\sqrt{2\mu[E-U-(\frac{l^2}{2\mu r^2})]}}$ denotes the deflected angle For Rutherford scattering, we integrate from b to infinity $\sigma(\theta) = \frac{k^2}{(4T_0)^2 \sin^{4}(\theta /2)}$ Chapter 10 Main Concepts Let a vector $\vec{A}$ denote any general vector $(\frac{d\vec{A}}{dt}) = (\frac{d\vec{A}}{dt})_{rot} + \vec{\omega} \times \vec{A}$ The Force on an object in a rotating frame is given by $\vec{F}_{rot} = \vec{F}-m\ddot{R}-m\dot\omega\times \vec{r}- m\omega \times (\omega \times \vec{r}) - 2m \omega \times \vec{v}$ Chapter 11 Main Concepts $T_{rot} = \frac{1}{2}\Sigma_{\alpha} m_{\alpha} (\omega \times r)^2$ This can be simplified to $TT_{rot} = \frac{1}{2}\Sigma_{i,j} I_{ij}\omega_{i}\omega_{j}$ $I_{ij} = \Sigma_{a} m_{a} (\delta_{ij}\Sigma-{k} x_{a,k}^{2}-x_{\alpha,i}x_{\alpha,j})$ is called the moment of inertia tensor In the continuous case, we have that $I_{ij} = \int\rho(r)(\delta_{ij} \Sigma_{k} x_{k}^2 -x_{i}x_j)$ diagonal terms are the moments of inertia and the off-diagonal terms are the products of inertia Angular momentum now becomes $L_{i} = \Sigma_{j} I_{ij} \omega_{j}$ Problems become greatly simplified if the moment of inertia tensor is diagonal. Then angular momentum aligns with axis of rotation By solving the eigenvalue problem of the tensor, you can calculate these principal axes The eigenvalues then become the entries of the new moment of inertia tensor Suppose that you want to know what the moment of inertia tensor becomes upon taking a different point as your origin. You can use the following generalization of the parallel axis theorem The new $I_{ij}$ in terms of the original $J_{ij}$ where $\vec{a}$ denotes the displacement vector is $I_{ij} = J_{ij}-M(a^2\delta_{ij}-a_{i}a_{j})$ When the new axis is parallel to the old axis and you isolate yourself to a plane. We have that $I_{new} = I_{old}+Md^2$ It it occasionally more convenient to have equations of motion in the body frame (ie. axes that align with the principal axes). Made up of 3 Euler angles The first Euler angle is called $\phi$ and denotes a counter clockwise rotation around the z axis in the fixed frame. Matrix of the form $\begin{pmatrix} \cos \phi && \sin\phi && 0 \\ -\sin\phi && \cos\phi && 0 \\ 0 && 0 && 1 \end{pmatrix}$ The second Euler angle is called $\theta$ and is a counterclockwise rotation around the current x axis Matrix of the form $\begin{pmatrix} 1 && 0 && 0 \\ 0 && \cos\theta && \sin\theta \\ 0 && -\sin\theta && \cos\theta \end{pmatrix}$ The last Euler angle is called $\phi$ and is the counterclockwise rotation around the current z axis Matrix of the form $\begin{pmatrix} \cos \psi && \sin\psi && 0 \\ -\sin\psi && \cos\psi && 0 \\ 0 && 0 && 1 \end{pmatrix}$ Any vector can now be written in the body frame of reference $\vec{x_{new}} = \lambda_{\psi}\lambda_{\theta}\lambda_{\phi}x$ In addition, we can rewrite the rotation vector components in terms of $\phi$, $\theta$ and $\psi$ $\omega_1 = \dot{\phi}\sin\theta\sin\psi+\dot{\theta}\cos\psi$ $\omega_2 = \dot{\phi}\sin\theta\cos\psi-\dot{\theta}\sin\psi$ $\omega_1 = \dot{\phi}\cos\theta+\dot{\psi}$ Euler Equations Suppose that our Lagrangian is that of a free body (ie. V=0) and we only consider rotational kinetic energy We can express the Lagrangian as a function of $\psi$, $\phi$ and $\theta$. Taking these Euler angles as the generalized coordinates, one can show that $(I_{3}-I_{2})\omega_{2}\omega_{3}+I_{1}\dot{\omega_{1}} = 0$ $(I_{1}-I_{3})\omega_{3}\omega_{1}+I_{2}\dot{\omega_{2}} = 0$ $(I_{2}-I_{1})\omega_{1}\omega_{2}+I_{3}\dot{\omega_{3}} = 0$ Force Field One can calculate the torque on an object in a similar manner $(I_{3}-I_{2})\omega_{2}\omega_{3}+I_{1}\dot{\omega_{1}} = \tau_{1}$ $(I_{1}-I_{3})\omega_{3}\omega_{1}+I_{2}\dot{\omega_{2}} = \tau_{2}$ $(I_{2}-I_{1})\omega_{1}\omega_{2}+I_{3}\dot{\omega_{3}} = \tau_{3}$