Title: Part IA / Dynamics and Relativity
Tags: Systems_Of_Particles Forces Special_Relativity Central_Forces Newton’s_Laws Non_Inertial_Frames

Q: Definition. Particle. An object of [...]. The only information you have to specify is its [...]
A: insignificant size; position.

Q: Definition. The position, velocity and acceleration of a particle is defined. by \[[...]}}\]
A: \mathbf{x} = \mathbf{x}(t), \mathbf{v} \equiv \mathbf{x}, \mathbf{a} \equiv \ddot{\mathbf{x

Q: Definition. Inertial Frame
A: A frame which in which a free particle has a constant velocity.

Q: Definition. Galilean Relativity Principle. A frame related to an inertial frame by a Galilean transformation, is also an inertial frame, and all laws of physics remain the same.
A: Translations: \(\mathbf{x}' = \mathbf{x} + a\) for constant a. There is no special point in the Universe
Rotations: \(\mathbf{x}' = R\mathbf{x}\) for an orthogonal matrix. There is no special direction in the Universe
Boosts: \(\mathbf{x}' = \mathbf{x} + \mathbf{v}t\), for constant velocity v. There is no special velocity in the Universe
We can show easily that \(\ddot{\mathbf{x}}' = 0 \iff \ddot{\mathbf{x}} = 0\)
Sometimes we also include \(t' = t + t_0\), this is a time translations, so clocks tick at the same rate in inertial frames called absolute time

Q: Definition. Newton's 1st Law.
A: Left alone, a particle moves with constant velocity
Inertial frames exist

Q: Definition. Newton's 2nd Law.
A: In an inertial frame \(\dot{\mathbf{p}} = \mathbf{F}\).
Where the momentum \(\mathbf{p} = m\dot{\mathbf{x}}\), where m is the inertial mass.
Normally this mass in constant so \(\dot{\mathbf{p}} = m\ddot{\mathbf{x}}\)

Q: Definition. Potential energy V(x).
A: \[F(x) = - \nabla V\]
Defined up to an additive constant since we integrate both sides
\[ \nabla V = (\frac{\partial V}{\partial x_1}, \frac{\partial V}{\partial x_2}, \frac{\partial V}{\partial x_2})\]

Q: Definition. Conservative Forces.
A: Ones which have a conserved energy: \[E = \frac{1}{2}m\dot{x}^2 + V(x)\]
Can differentiate and check that its 0 as \(\frac{\partial V}{\partial x_i} = -F_i\)
If and only if \(\mathbf{F} = -\nabla V\)

Q: Definition. Work done.
A: \[W = \int_C \mathbf{F} \cdot d\mathbf{x} = \int_{t_1}^{t_2}\mathbf{F} \cdot \frac{d\mathbf{x}}{dt}dt\]
Integrated is the power \(P = \mathbf{F \cdot x}\)

Q: Claim. The total work done is independent of the trajectory C for a conservative force.
A: \[W = \int_{t_1}^{t_2}\mathbf{F \cdot \dot{x}}dt = m\int_{t_1}^{t_2} \mathbf{\ddot{x} \cdot \dot{x}}dt = T(t_2) - T(t_1) = V(t_1) - V(t_2)\]
The last line only follows because energy is conserved so \(V(t_1) + T(t_1) = V(t_2) + T(t_2)\)

Q: Definition. Lorentz force. Force experienced by a particle with electric charge q.
A: \[\mathbf{F} = q \left ( \mathbf{E(x) + \dot{x} \times B(x)}\right)\]
with \(\bf E\) being the electric field and \(\bf B\) being the magnetic field
If we assume that the electric field is static then we can write it as \[\mathbf{E} = -\nabla \phi\] where \(\phi(\mathbf{x})\) is the electric potential

Q: Claim. The energy \(E = \frac{1}{2}m\mathbf{\dot{x} \cdot \dot{x}} + q\phi(\mathbf{x})\) is conserved.
A: \[\dot{E} = m\mathbf{\dot{x} \cdot \ddot{x}} + q\nabla\phi \cdot \mathbf{\dot{x}} = \mathbf{\dot{x}} \cdot (\mathbf{F} + q\nabla \phi) = q\mathbf{\dot{x} \cdot(\dot{x} \times B)} = 0\]

Q: Definition. Electric potential and Coulomb Force.
A: \[\mathbf{E} = -\nabla \left ( \frac{Q}{4\pi \epsilon_0 r}\right ) = \frac{Q}{4\pi \varepsilon_0}\frac{\mathbf{\dot{r}}}{r^2}\]

Q: Definition. Friction.
A: Friction is a macroscopic force that acts between an object and the medium it is travelling through
1)Doesn't conserve energy
2)Irreversible, energy always lost

Q: Definition. Linear and Quadratic Drag.
A: Linear: \(\mathbf{F} = -\gamma \mathbf{v}\)
Quadratic: \(\mathbf{F} = -\gamma |\mathbf{v}|\mathbf{v}\)
Where \(\gamma\) is the coefficient of friction.

Q: Definition. What are the coefficients of friction for linear and quadratic drag
A: Linear: \(\gamma = 6\pi\eta L\), where \(\eta\) is the viscosity of the fluid and L is the radius of the object.
Quadratic: \(F \propto \rho A v^2\) where \(\rho\) is the density of the medium and A is the surface area.

Q: Definition. Terminal velocity with quadratic drag
A: Equation of motion is \[m \frac{dv}{dt} = -mg + \gamma v^2\]
Solving this gives \(v = -\sqrt{\frac{mg}{\gamma}}\tanh \left( \sqrt{\frac{\gamma g}{m}}t \right)\)
So as \(t \to \infty\), \(v \to -\sqrt{\frac{mg}{\gamma}}\)

Q: Definition. Central Force Potential.
A: Such a system obeys the equation of motion \[m \mathbf{\ddot{x}} = -\nabla V(r) \]
where the potential depends only on \(r = |\mathbf{x}|\)

Q: Definition. Angular Momentum.
A: \[\mathbf{L} = m\mathbf{x \times \dot{x}}\]

Q: Definition. Torque.
A: \[\bf \tau = x \times F\]

Q: Claim. Angular momentum is conserved in a central potential.
A: \[\frac{d\mathbf{L}}{dt} = m\frac{d}{dt}(\mathbf{x \times \dot{x}})  = m\mathbf{\dot{x} \times \dot{x}} + m \mathbf{x} \times \mathbf{\ddot{x}}= \mathbf{x \times F} = 0\]
Since \(\mathbf{F} = -\nabla V(r)\) and \(\nabla V\) is parallel to \(\mathbf{x}\)

Q: Definition. Equation for orbit of object about a star and energy of the object.
A: \[r = \frac{r_0}{e \cos \theta + 1}\]
where \(r_0 = \frac{l^2}{k}\) and \(e = \frac{Al^2}{k}\)
and \(V(r) = -\frac{km}{r}\)
Then we can write \[E = \frac{1}{2}m\dot{r}^2 + \frac{ml^2}{2r^2} - \frac{km}{r} = \frac{1}{2}m\left( \frac{dr}{d\theta}\right)^2 \frac{l^2}{r^4} + \frac{ml^2}{2r^2} - \frac{km}{r}\] which we can simplify by writing things in terms of \(\theta\) and then back again to get \[E = \frac{mk^2}{2l^2}(e^2 - 1)\]

Q: Definition. Kepler's Laws of Planetary Motion.
A: 1: Each planet moves in an ellipse, with the sun at one focus.
2: The line between the planet and the sun sweeps out equal areas in equal times.
3: The period of the orbit is proportional to \(\text{radius}^{\frac{3}{2}}\)

Q: Proof. Derive Kepler's third law.
A: Use dimensional analysis.
\[T = cR^Ak^B\] where \(k = GM\)
\[[T] = [L]^{A-3B}[T]^{2B}\] which results in \(T = c \frac{R^{\frac{3}{2}}}{k^{\frac{1}{2}}}\).
Can get the dimensionless constant by considering the area swept out by a single period.

Q: Proof. Derive Kepler's 2nd Law.
A: Follows from conservation of angular momentum.
\[\delta A = \frac{1}{2}r^2 \delta \theta \implies \frac{dA}{dt} = \frac{1}{2}r^2 \dot{\theta} = \frac{l}{2}\]
which we know is constant.

Q: Proposition. Rutherford Scaterring.
A: The potential is \(V = \frac{qQ}{4\pi \epsilon_0 r}\)
We know that b = b' since angular momentum is conserved.
We can write \(\phi = \pi - 2\alpha\)
Also \(\cos \alpha = \frac{1}{e}\)
We can equate this with our equation for initial energy to get \(E = \frac{1}{2}mv^2 = \frac{mk^2}{2l^2}(e^2 - 1) = \frac{mk^2}{2l^2}\tan^2 \alpha \)
Then replace \(l = bv\) to get \[\phi = 2\tan^{-1}\left( \frac{|k|}{bv^2}\right)\]

Q: Definition. Momentum of System.
A: \[\mathbf{R} = \frac{1}{M} \sum_{i=1}^N m_i\mathbf{x}_i\]
\[\mathbf{P} = M\mathbf{\dot{R}}\]

Q: Definition. Rate of change of momentum of centre of mass.
A: \[\mathbf{\dot{P}} = \sum_i \mathbf{\dot{p}}_i = \sum_i \left ( \mathbf{F}_i^{ext} + \sum_{j \neq i}F_{ij}\right ) = \sum_i \mathbf{F}_i^{ext} \sum_{i < j}(F_{ij} + F_{ji}) = \sum_i \mathbf{F}_i^{ext} \]
So it only depends on the external forces

Q: Definition. Angular Momentum of System of Particles. How does it change w.r.t time?
A: \[\mathbf{L} = \sum_i \mathbf{x_i} \times \mathbf{p_i}\]
\[\frac{d\mathbf{L}}{dt} = \sum_i \mathbf{x_i \times \dot{p}_i} = \sum_i \mathbf{x_i \times \left ( F_i^{ext} + \sum_{j\neq i} F_{ij}\right )} = \mathbf{\tau + \sum_i \sum_{j \neq i} x_i \times F_{ij}}\]
This simplifies to \[\frac{d\mathbf{L}}{dt} = \sum_{i<j}\mathbf{(x_i - x_j) \times F_{ij}}\]
So when the force due to Newton's Third law is parallel to the line joining the two particles we get \[\frac{d\mathbf{L}}{dt} = \mathbf{\tau}\] where \(\mathbf{\tau} = \sum_i \mathbf{x_i \times F_i^{ext}}\) is the total external torque.

Q: Definition. Total Energy of System of Particles.
A: \[T = \frac{1}{2}\sum_i m_i \mathbf{\dot{x}_i \cdot \dot{x}_i}\]
Rewrite position of particle as \(\mathbf{x_i = R + y_i}\)
So we write \[T = \frac{1}{2}\sum_i m_i(\mathbf{\dot{R} + \dot{y}_i})^2 = \frac{1}{2}M \mathbf{\dot{R}}^2 + \frac{1}{2}\sum_i m_i \mathbf{\dot{y}}_i^2\]
So the kinetic energy splits up into the kinetic energy of the centre of mass, together with the kinetic energy of the particles moving around the centre of mass.

Q: Proof. Why is the two body problem really a one body problem?
A: Write the positions of the particles in terms of their separation from the centre of mass.
\[\mathbf{\ddot{r}} = \mathbf{\ddot{x}_1 - \ddot{x}_2} = \frac{1}{m_1}\mathbf{F_{12}} - \frac{1}{m_2}\mathbf{F_{21}} = \frac{m_1 + m_2}{m_1m_2}\mathbf{F_{12}}\]

Q: Definition. Angular Velocity.
A: \[\mathbf{\omega} = \dot{\theta}\mathbf{\hat{z}}\] is the angular velocity.
Then \[\mathbf{\dot{x}} = \mathbf{\omega \times x}\]
Speed is \[v = |\mathbf{\dot{x}}| = r\omega \sin \phi = d\omega\]
where d is the perpendicular distance to the axis of rotatiom.
Also then we can calculate the kinetic energy \[T = \frac{1}{2}m \mathbf{\dot{x} \cdot \dot{x}} = \frac{1}{2}m(\omega \times \mathbf{x})\cdot (\omega \cdot \mathbf{x}) = \frac{1}{2}md^2\omega^2\]

Q: Definition. Moment of Inertia.
A: We write the kinetic energy for a rigid body as \[T = \frac{1}{2}\sum_i m_i \mathbf{\dot{x}_i \cdot \dot{x}_i} = \frac{1}{2}I \omega^2\]where \[I \equiv \sum_{i=1}^N m_id_i^2\]is the moment of inertia. The moment of inertia is to rotations, what mass is to translations. Bigger I, more energy you need to supply to the body to make it spin

Q: Definition. Angular Momentum of a Rigid Body.
A: For a rigid body all particles rotate with the same angular velocity \[\mathbf{\dot{x}_i} = \omega \times \mathbf{x_i}\]Then \[\mathbf{L} = \sum_{i} m \mathbf{x}_i \times \mathbf{\dot{x}_i} = \sum_im_i\mathbf{x_i} \times (\omega \times \mathbf{x_i})\] If we write \(\boldsymbol{\omega} = \omega \mathbf{\hat{n}}\), then the magnitude of the angular momentum in the direction of \(\boldsymbol{\omega}\) is \[\mathbf{L} \cdot \mathbf{\hat{n}} = \omega \sum_i m_i (\mathbf{x_i \times (\hat{n} \times x_i)}) \cdot \mathbf{\hat{n}} = \omega \sum_i m_i(\mathbf{x_i \times \hat{n}}) \cdot (\mathbf{x_i} \times \mathbf{\hat{n}}) = I\omega\]

Q: Definition. Torque of Rigid Body.
A: Torque causes change in angular momentum and is in the same direction as the angular velocity.
We know that \(\mathbf{L \cdot \hat{n}} = I\omega\) and \(\tau= \tau \mathbf{\hat{n}}\), so \[I\dot{\omega} = \tau\]

Q: Proposition. Calculating the Moment of Inertia.
A: We can treat rigid bodies as continuous objects. Replace the discrete particle masses \(m_i\) with continuous density distribution \(\rho(\mathbf{x})\). The total mass the body is then \[M = \int \rho(\mathbf{x}) dV\]and the moment of inertia is \[I = \int \rho(\mathbf{x})x_{\perp}^2 \; dV = \int \rho(\mathbf{x}) (x\sin\phi)^2 dV\]where \(x_{\perp} = x \sin \phi\) is the perpendicular distance from the point \(\mathbf{x}\) to the axis of rotation.

Q: Proposition. Calculate the moment of inertia for a sphere with the axis of rotation passing through the centre.
A: We get a distance \(r\sin\theta\) from the axis of rotation, and the Jacobian factor \(r^2 \sin \theta\), so the moment of inertia is \[I = \int_0^a \int_0^{\pi}\int_0^{2\pi} \rho(r \sin \theta)^2r^2 \sin \theta \; dr\; d\theta\; d\phi = \frac{8}{15}\pi \rho a^5 = \frac{2}{5}Ma^2\]Since we know that \(M =\frac{4}{3}\pi \rho a^3\)

Q: Theorem. Parallel Axis Theorem. A rigid body has mass M and moment of inertia \(I_{CoM}\) about an axis which passes through its centre of mass. Let I be the moment of interim about a parallel axis that lies a distance h away. Then [...]
This is the parallel axis theorem
A: \[I = I_{CoM} + Mh^2\]
Back Extra: Pick an origin that lies on the second axis and label the unit vector along with axis as \(\mathbf{\hat{n}}\). From here the position to any particle can be written as \[\mathbf{r_i = R + y_i}\] where \(\bf R\) is the centre of mass position and \(\mathbf{y_i}\) are constrained to obey \(\sum_i m_i\mathbf{y_i} \). WE can then write the moment of inertia as \[I = \sum_i m_i (\mathbf{\hat{n} \times r_i}) \cdot (\mathbf{\hat{n} \times r_i})\]
Simplifying this leads to the result.

Q: Proposition. Motion of Rigid Bodies.  If the centre of mass follows some trajectory \(\mathbf{R}(t)\), together with a rotation about the centre of mass. What is the kinetic energy of the rigid body?
A: Write \(\mathbf{r_i - R + y_i} \implies \mathbf{\dot{r}_i = \dot{R} + \dot{y}_i}\). If the body rotates with angular velocity \(\boldsymbol{\omega}\) around the centre of mass, we have \(\mathbf{\dot{y}_i} = \boldsymbol{\omega}\times \mathbf{y_i}\), which means that we can write \[\mathbf{\dot{r}_i} = \mathbf{\dot{R}} + \boldsymbol{\omega} \times (\mathbf{r_i - R})\].
Then we know that \[ T = \frac{1}{2}M \mathbf{\dot{R} \cdot \dot{R}} + \frac{1}{2}\sum_i m_i \mathbf{\dot{y}_i \cdot \dot{y}_i} = \frac{1}{2}M \mathbf{\dot{R} \cdot \dot{R}} + \frac{1}{2}I\omega^2 \]

Q: Claim. In a rotating frame what are the velocity and acceleration?
A: Write \(\mathbf{\dot{e}_i' = \boldsymbol{\omega} \times e_i'}\). Then \(\mathbf{r} = r_i'\mathbf{e}_i'\).
So \(\mathbf{\dot{r}} = \dot{r}_i' \mathbf{e}_i' + r_i'\mathbf{\dot{e}}_i' = \dot{r}_i'\mathbf{e}_i' + \boldsymbol{\omega} \times \mathbf{r}\).
Then \(\mathbf{\ddot{r}} = \ddot{r}_i' \mathbf{e}_i' + 2\dot{r}_i' \boldsymbol{\omega} \times \mathbf{e}_i' + \boldsymbol{\dot{\omega}} \times \mathbf{r} + r_i' \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{e}_i')\)

Q: Definition. Fictitious Forces.
A: Using the fact that \(m\left( \frac{d^2 \mathbf{r}}{dt^2}_S\right ) = \mathbf{F}\) in the inertial frame S.
Then for the rotating frame S' it is \[m\left( \frac{d^2 \mathbf{r}}{dt^2}\right )_{S'} = \mathbf{F} - 2m\boldsymbol{\omega} \times \left ( \frac{d\mathbf{r}}{dt}\right )_{S'} - m\boldsymbol{\dot{\omega}} \times \mathbf{r} - m\boldsymbol{\omega} \times (\boldsymbol{\omega} \times \mathbf{r})\]
First term is the Coriolis Force.
Second term is the Euler force.
Third term is the centrifugal force.

Q: Definition. Lorentz Transformations. For two inertial frames S and S' moving with relative speed v, how are their co-ordinates related?
A: \[x' = \gamma(x - vt)\]\[t' = \gamma(t - \frac{v}{c^2}x)\]\[\gamma = \sqrt{\frac{1}{1 - \frac{v^2}{c^2}}}\]

Q: Claim. Addition of Velocities. A particle moves with constant velocity u' in frame S', which in turn moves with constant velocity v with respect to frame S. What is the velocity of u of the particle as seen in S?
A: Newtonian answer: u = u' + v
Special relatavistic answer: \(u = \frac{x}{t} = \frac{\gamma(x' + vt')}{\gamma(t' + vx'/c^2)} = \frac{u'+v}{1+u'v/c^2}\)

Q: Definition. Invariant Interval. Consier a spacetime with just a single spacial coordinate, x. In frame S, two events \(P_1\) and \(P_2\) have coordinates \((ct_1, x_1)\) and \((ct_2, x_2)\). The events are seperated by \(\Delta t = t_1 - t_2\) in time and \(\Delta x = x_1 - x_2\) in space. We define the invariant interval \(\Delta s^2\) as a measure of the distance between these two points: \[[...]\]
A: \Delta s^2 = c^2 \Delta t^2 - \Delta x^2
Back Extra: In a frame S' we have, \[\Delta s^2 = \gamma^2(c\Delta t' + \frac{v\Delta x'}{c})^2 - \gamma^2(\Delta x' + v\Delta t')^2 = \gamma^2(c^2 - v^2)\Delta t'^2 - \gamma^2(1-\frac{v^2}{c^2})\Delta x'^2 = c^2 \Delta t'^2 - \Delta x'^2\]So different inertial frames disagree about time passed and the spatial seperation, but agree on \((\Delta s)^2\)
We can write this as \[(\Delta s)^2 = (c \Delta t \; \; \; \Delta x) \left ( \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right)\left (\begin{matrix} c\Delta t \\ \Delta x \end{matrix}\right )\]Where the matrix is the Minkowski metric

Q: Definition. Rapidity. \(\varphi\).
A: \[\gamma = \cosh \varphi\]So the Lorentz transformation \[\left ( \begin{matrix} \gamma & \frac{-\gamma v}{c} \\ \frac{-\gamma v}{c} & \gamma \end{matrix} \right ) = \left ( \begin{matrix} \cosh \varphi & -\sinh \varphi \\ -\sinh \varphi & \cosh \varphi\end{matrix} \right ) = \Lambda[\varphi]\]Two sequential Lorentz boosts: \[\Lambda[\varphi_1]\Lambda[\varphi_2] = \Lambda[\varphi_1 + \varphi_2]\]In terms of velocities: \[\Lambda[v_1]\Lambda[v_2] = \Lambda[\frac{v_1 + v_2}{1 + v_1v_2/c^2}]\]

Q: Definition. Inner products in terms of 4-vectors. We define the invariant distance between the origin and the point P as an inner product, written as \[[...]\]where \[\eta = \left( \begin{matrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1\end{matrix} \right )\]and \[X^{\mu} = (ct,x,y,z) \;\;\; \mu = 0,1,2,3\]
A: X \cdot X \equiv X^T \eta X = X^{\mu}\eta_{\mu \nu}X^{\nu}
Back Extra: Now \(X \cdot X = c^2t^2 - x^2 - y^2 - z^2\).
If \(X \cdot X > 0\): X called a timelike 4-vector
If \(X \cdot X < 0\): X called a spacelike 4-vector
If \(X \cdot X = 0\): X called lightlike or null 4-vector

Q: Definition. Lorentz Transformations in terms of 4-vectors. Define them to be the 4x4 matrices \(\Lambda\) such that \[[...]\]
A: X' = \Lambda X \text{ and } X' \cdot X' = X \cdot X \iff \Lambda^T \eta \Lambda = \eta
Back Extra: Lorentz transformations leave the inner product invariant.
We can solve to see that the possibilities are rotations and Lorentz boost in each of the 3 dimensions.
These form the Lorentz group: \(O(1,3)\)

Q: Definition. Proper Time. All inertial observers will agree on this parametrisation of a particle. \[[...]\]
A: \Delta \tau = \frac{\Delta s}{c}
Back Extra: Where \(\Delta s\) is the invariant interval

Q: Definition. 4-Velocity. Consider \[X = (ct(\tau), \mathbf{x}(\tau))\]\[[...]\]
A: U = \frac{dX}{d\tau} = \left( \begin{matrix} c \;\; dt/d\tau \\ d\mathbf{x}/d\tau\end{matrix} \right )
Back Extra: Also \[d\tau = \sqrt{dt^2 - \frac{d\mathbf{x}^2}{c^2}} = dt\sqrt{1 - \frac{1}{c^2}\left ( \frac{d\mathbf{x}}{dt}\right )^2} = dt\sqrt{1 - \frac{u^2}{c^2}}\]
From which we have \[\frac{dt}{d\tau} = \gamma\]

Q: Definition. 4-Momentum. How can we write it in terms of relativistic energy E and relativistic 3-momentum.
A: \[P = mU = \left ( \begin{matrix} m\gamma c \\ m \gamma \mathbf{u}\end{matrix} \right )\]
But we also know \[E = m \gamma c^2, \mathbf{p} = m \gamma \mathbf{u}\]
Where m is the rest mass.

Q: Proposition. Deriving mass energy equivalence relation.
A: In the rest frame of a particle, \(P = (mc,0,0,0)\), so \(P \cdot P = m^2 c^2\).
But the inner product is an invariant holding in any frame, so we have \(P\cdot P = \frac{E^2}{c^2} - \mathbf{p}^2\), so this gives us \[E^2 = \mathbf{p}^2c^2 + m^2c^4\]