Appendix

KaTeX Promotion

In case you are confused about how I manage to write Math Equations in a website… Press Here

Appendix B

• Used in my post about HKDSE Physics

Proposition 1.2.1: $$r^2 \dot{\theta} \equiv h=\frac{

\underline{L}

}{m}$$

Proof

Going back to Planar Polar Coordinates:

\[\hat{r}=\cos (\theta) \hat{x}+\sin (\theta) \hat{y}, \quad \hat{\theta}=-\sin (\theta) \hat{x}+\cos (\theta) \hat{y} .\]

By Chain Rule,

\[\dot{\hat{r}}=\dot{\theta}[-\sin (\theta) \hat{x}+\cos (\theta) \hat{y}]=\dot{\theta} \hat{\theta}, \quad \dot{\hat{\theta}}=\dot{\theta}[-\cos (\theta) \hat{x}-\sin (\theta) \hat{y}]=-\dot{\theta} \hat{r} .\]

And

\[\underline{\dot{r}}=\frac{d}{d t}(r \hat{r})=\dot{r} \hat{r}+r \dot{\hat{r}}=\dot{r} \hat{r}+r \dot{\theta} \hat{\theta}\]

Further Differentiation yields:

\[\begin{aligned} \underline{\ddot{r}}=\frac{d}{d t}(\dot{r} \hat{r}+r \dot{\theta} \hat{\theta}) & =\ddot{r} \hat{r}+\dot{r} \dot{\hat{r}}+\dot{r} \dot{\theta} \hat{\theta}+r \ddot{\theta} \hat{\theta}+r \dot{\theta} \dot{\hat{\theta}} \\ & =\ddot{r} \hat{r}+\dot{r} \dot{\theta} \hat{\theta}+\dot{r} \dot{\theta} \hat{\theta}+r \ddot{\theta} \hat{\theta}-r \dot{\theta} \dot{\theta} \hat{r} \\ & =\left(\ddot{r}-r \dot{\theta}^2\right) \hat{r}+(2 \dot{r} \dot{\theta}+r \ddot{\theta}) \hat{\theta} \end{aligned}\]

Since the force is a central force, we have:

\[m\left(\ddot{r}-r \dot{\theta}^2\right) \hat{r}+m(2 \dot{r} \dot{\theta}+r \ddot{\theta}) \hat{\theta}=F_r(r) \hat{r} .\]

We can quickly identify two equations:

\[\begin{aligned} \ddot{r}-r \dot{\theta}^2 & =F_r(r) / m \\ 2 \dot{r} \dot{\theta}+r \ddot{\theta} & =0 . \end{aligned}\]

From the second equation

\[\frac{1}{r} \frac{d}{d t}\left(r^2 \dot{\theta}\right)=0 \quad \Longrightarrow \quad r^2 \dot{\theta} \equiv h=\text { const }\] \[\underline{L}=\underline{r} \times \underline{p}=m r \hat{r} \times \underline{\dot{r}}=m r \hat{r} \times(\dot{r} \hat{r}+r \dot{\theta} \hat{\theta})=m r \hat{r} \times r \dot{\theta} \hat{\theta}=m r^2 \dot{\theta} \hat{n} \equiv m h \hat{n}\]

Hence:

\[r^2 \dot{\theta} \equiv h=\frac{|\underline{L}|}{m}\] \[\blacksquare\]

Appendix C

Dimensional Analysis is one of the most helpful knowledge in Physics. You can immediately tell if an answer is correct or not using it. For example if we accidentally squared a term that contains dimensions, we can almost immediately identify the mistake.

N.B.: Two objects with different units cannot be added

Proposition 1.3: \(\sin x\) doesn’t have any units

Proof: Consider the series expansion of \(\sin x\), i.e.

\[\sin x=x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\cdots=\sum_{k=0}^n \frac{(-1)^k x^{2 k+1}}{(2 k+1) !}\]

First note that \(x\) cannot have any units, otherwise \(x-\frac{x^3}{3 !}+\frac{x^5}{5 !}-\cdots\) won’t have any physical sense, as discussed above.

Therefore \(\frac{x^3}{3 !}\) and \(\frac{x^5}{5 !}\) will have the same units as \(x\), i.e. no units, so as the preceding terms. Hence \(\sum_{k=0}^n \frac{(-1)^k x^{2 k+1}}{(2 k+1) !}\) have no units, and hence \(\sin x\) has no units \(\blacksquare\)

Note that dimensional analysis cannot ensure that your answer is correct; rather, it occasionally picks up errors, but this is already extremely useful.

Appendix D

• Used in my post about Cosets and Lagrange’s Theoremc

Lemma

Claim: Let \(G\) be a group. If \(g, h \in G\), then

There exists (Existence) a unique element (Uniqueness) \(k \in G\) such that \(k * g=h\)

Proof: (Existence) Let \(k:=h * g^{-1}\). Then \(k * g=\left(h * g^{-1}\right) * g=h *\left(g^{-1} * g\right)=h * e=h,\) Where we have used associativity of group axioms. Hence \(k\) exists \(\blacksquare\)

(Uniqueness) Now suppose that \(k^{\prime} * g=h\). Then \(k=h * g^{-1}=\left(k^{\prime} * g\right) * g^{-1}=k^{\prime} *\left(g * g^{-1}\right)=k^{\prime} * e=k^{\prime}\)

hence \(k\) is unique \(\blacksquare\)

Therefore there exists a unique element \(k \in G\) such that \(k * g=h\) \(\blacksquare\)

Appendix E

Ellispes are a form of conic sections.

Incomplete!