# Oh Yeah # Dat Boi # Dis Boi   Given a vector function, I show that $\vec{F}(x,y,z,t)$, I show that $d\vec{F} = (d\vec{r} \cdot \nabla) \vec{F} + \frac{\partial \vec{F}}{\partial t}$  by writing $d\vec{F}$ in terms of independent variations in the $x, y, z$ and $t$ direction. I write $d\vec{F}$ as a sum of four increments, one purely in the $x$ direction, the $y$ direction, the $z$ direction and the $t$ direction as follows:

\$latex dF(x,y,z,t)

# Conservation of Total Angular Momentum Proof

For this post, I want to prove that in the absence of external forces, the total angular momentum of an N-particle system is conserved.

I start with $\vec{P} = \displaystyle \sum_{\alpha} \vec{p}_{\alpha}$ which is the total momentum of an N-particle system. Now I can vectorially multiple the total momentum by $\vec{r}$ which is the position vector measured from the same origin $O$ for each particle. This will give me the total angular momentum of the system which can be written as $\vec{L} = \displaystyle \sum_{\alpha = 1}^{N} \vec{\ell}_{\alpha} = \displaystyle\sum_{\alpha = 1}^{N} \vec{r}_{\alpha} \times \vec{p}_{\alpha}$. After differentiating with respect to $t$, I obtain $\dot{\vec{L}} = \displaystyle \sum_{\alpha} \dot{\vec{\ell}}_{\alpha} = \displaystyle \sum_{\alpha} \frac{d}{dt} (\vec{r} \times \vec{p}) = \displaystyle \sum_{\alpha} ( \dot{\vec{r}} \times \vec{p}) + (\vec{r} \times \dot{\vec{p}})$. In the first cross product, I can substitute $\vec{p}$ with $m\dot{\vec{r}}$, and since the cross product of any two parallel vectors is zero, the first term becomes zero. This leaves implies that my sum becomes $\displaystyle\sum_{\alpha}\vec{r}_{\alpha} \times \vec{F}_{\alpha}$. Now, I can rewrite the net force on particle $\alpha$ as $\vec{F}_{\alpha} = \displaystyle \sum_{\beta \neq \alpha} \vec{F}_{\alpha \beta}$, where $\vec{F}_{\alpha \beta}$ represents the force exerted on particle $\alpha$ by particle $\beta$. Now I can make a substitution for $\vec{F}_{\alpha}$ to give me $\dot{\vec{L}} = \displaystyle \sum_{\alpha} \displaystyle \sum_{\beta \neq \alpha} \vec{r}_{\alpha} \times \vec{F}_{\alpha \beta}$

…I will finish the rest of this at some point. I seem to have misplaced the book.

# Matrix Proof

I want show that $e^{(At)} = Se^{(\Lambda t)} S^{-1}$, where $S = (v_{1} v_{2})$, $A$ is a $2 \times 2$ matrix, and $\Lambda = \begin{pmatrix} \lambda_{1} & 0 \\0 & \lambda_{2} \end{pmatrix}$

I start by writing the middle sum in summation notation which gives me $Se^{(\Lambda t)}S^{-1} = S( \displaystyle\sum_{k = 0}^{\infty} \frac{1}{k!}(\Lambda t)^{k})S^{-1}$. Now I can use the identity $S^{-1}AS = \Lambda$ which will then give me $S(\displaystyle\sum_{k = 0}^{\infty} \frac{1}{k!}(S^{-1}AS)^{k}t^{k})(S^{-1})$. After pulling terms out of the sum, I will get $SS^{-1} (\displaystyle \sum_{k = 0}^{\infty} \frac{t^{k}A^{k}}{k!})SS^{-1}$. The $SS^{-1}$ terms create an identity matrix and the middle sum is equivalent to $e^{At}$ as shown below.

# Matrix Proof

Given the matrix equation $\dot{u} = Au$, where $u =$ $\begin{pmatrix} x \\ y \end{pmatrix}$ and $A$ is a $2 \times 2$ matrix, I want to show that $\dot{u} = e^{At}u_{0}$ is the solution where $e^{At} = I + At + \frac{1}{2!}A^{2}t^{2} + \frac{1}{3!}A^{3}t^{3} + ...$ and $I$ is the identity matrix.

I start out by writing the above series in summation notation which gives me $\displaystyle\sum_{k = 0}^{\infty} \frac{t^{k}A^{k}}{k!}$. I can now take a time derivative of the sum to give me $\frac{d}{dt}e^{At} = \displaystyle\sum_{k = 0}^{\infty}\frac{kt^{k - 1}A^{k}}{k!}$. Since the first term of the series after differentiation is $0$, I can rewrite and reduce the sum to give me $\displaystyle\sum_{k = 1}^{\infty}\frac{t^{k - 1}A^{k}}{(k - 1)!}$. Now, I can pull out a single matrix term to give me $A \displaystyle\sum_{k = 1}^{\infty}\frac{t^{k - 1}A^{k - 1}}{(k - 1)!}$. I can now simplify the sum once again to give me $A \displaystyle\sum_{k = 0}^{\infty} \frac{t^{k}A^{k}}{k!}$. This is now equivalent to $Ae^{At}$. Assuming that $u = e^{At}u_{0}$  is the solution, I can differentiate it with respect to time to give me $\dot{u} = \frac{d}{dt}[e^{At}]u_{0}$. I just showed that $\frac{d}{dt} e^{At} = Ae^{At}$, which I can plug in to give me the equation $\dot{u} = Ae^{At}u_{0}$. Since $e^{At}u_{0} = u$, I am left with my original matrix equation $\dot{u} = Au$

# Mixture of States Proof

A particle of mass $m$ sits in a one-dimensional square well with infinitely high walls and width $L$. The particle is in a 50–50 mixture of states, half in the ground state and half in the first excited state. I want to show how to derive a formula for the complete, time-dependent wave-function of the particle.

I begin by the using the normalizing condition, that $\int|\Psi|^{2}dx=1$. This is because the probability of finding the particle somewhere in space must equal $1$ at all times. Because the particle is in a mixture of states, my wave function will take the form $\Psi = A(\Psi_{1}+\Psi_{2})$. Combining this with the normalizing condition, I get the equation $A^{2}\int(\Psi_{1}+\Psi_{2})(\Psi_{1}^{*}+\Psi_{2}^{*})dx = 1$. The individual wave-functions will take the forms $\Psi_{1} = \sqrt{\frac{2}{L}}\sin(\frac{\pi x}{L})e^{-i\omega_{1}t}$ and $\Psi_{2} = \sqrt{\frac{2}{L}}\sin(\frac{\pi x}{L})e^{-i\omega_{2}t}$. I can then plug these two equations in the the above normalizing condition for a particle in mixed states, convert all sine functions to cosine functions and cancel out like terms until I get down to the simple expression that $2A^{2}=1$ which implies that $A = \frac{1}{\sqrt{2}}$. Now I can use this constant to write down my mixed wave-function which will look as follows $\Psi = \frac{1}{\sqrt{L}}(\sin(\frac{\pi x}{L})e^{-\omega_{1}t} + \sin(\frac{2\pi s}{L})e^{-i\omega_{2}t})$. Now, I want to show that the probability of finding the particle between positions $\frac{1}{4}L$ and $\frac{1}{4}L+dx$, as measured from the left-hand side of the well, as a function of time, is $[\frac{3}{2}+\sqrt{2}\cos(\frac{3E_{1}t}{\hbar})]\frac{dx}{L}$. This is done by finding the probability amplitude which is the square modulus of the wave function. This will look as follows $|\Psi_{1}+\Psi_{2}|^{2} = (\Psi_{1}+\Psi_{2})(\Psi_{1}^{*}+\Psi_{2}^{*})$. After filling in each wave function and multiplying out each term I obtain $\frac{1}{L}(\sin(\frac{\pi x}{L}))^{2} + \frac{1}{L}(\sin(\frac{2\pi x}{L}))^{2} + \frac{1}{L}\sin(\frac{2\pi x}{L}) \sin(\frac{\pi x}{L}) ( e^{-i\omega_{1} t}e^{i\omega_{2} t}+e^{-i\omega_{2} t}e^{i\omega_{1} t})$. After converting the sine terms to cosines and converting the exponentials to trigonometric functions, I can plug in $x$ for $L$, which will give me $\frac{1}{4}L$ and $\frac{1}{4}L+dx$ which is what I wanted to show.

# Phase Velocity Proof

Using the relativistic energy equation $E^{2} = p^{2}c^{2}+m^{2}c^{4}$, I want to show that the resulting phase velocity for the de Broglie wave of an electron is greater than the speed of light.

I start by deriving an expression for Energy in terms of the phase velocity, which is the rate at which the phase of the wave propagates in space, and the momentum of the electron. Using the equation $v_{p}=f\lambda$, I can then make two substitutions namely that $f =\frac{E}{h}$ which gives the energy required or released when electrons change their energy levels and $\lambda=\frac{h}{p}$ which is the wavelength associated with a particle as postulated by Louis DeBroglie. After canceling like terms this gives me $E=v_{p}p$. Now I can plug this into my relativistic formula to obtain $v_{p}^{2}p^{2} = p^{2}c^{2} + m^{2}c^{4}$. Next I can make a substitution using the equation $p=mv_{p}$, and cancel out the masses. This leaves me with the bi-quadratic equation $y^{2}-c^{2}y-c^{4} = 0$ where $y=v_{p}^{2}$. This equation has only one real solution which is $\pm \sqrt\frac{c^{2}+\sqrt{5c^{4}}}{2}$. Taking the positive root, I obtain the velocity $3.8*10^{8} m/s$ which is greater than the speed of light.