Physicspupil – Page 34

November 11, 2020July 25, 2022

202011110657 Sidenote of Separation of Variables

Two-particle system: $\displaystyle{i\hbar\frac{\partial \Psi}{\partial t}=H\Psi}$

in which $\displaystyle{H=-\frac{\hbar^2}{2m_1}\nabla_1^2-\frac{\hbar^2}{2m_2}\nabla_2^2+V(\mathbf{r_1},\mathbf{r_2},t)}$ .

Assuming that the potential is time-independent.

Let $\Psi (\mathbf{r}_1,\mathbf{r}_2,t)=\psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t)$ ,

$\begin{aligned} & i\hbar \frac{\partial \psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t)}{\partial t} \\ = & -\frac{\hbar^2}{2m_1}\nabla_1^2\bigg( \psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t)\bigg) -\frac{\hbar^2}{2m_2}\nabla_2^2\bigg( \psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t) \bigg) \\ &\qquad\enspace +V(\mathbf{r_1},\mathbf{r_2})\psi (\mathbf{r}_1,\mathbf{r}_2)\varphi (t) \\ & \\ & \textrm{\scriptsize OR} \\ & \\ & i\hbar\Big[ \frac{\partial \psi (\mathbf{r}_1,\mathbf{r}_2)}{\partial t}\varphi (t)+\psi (\mathbf{r}_1,\mathbf{r}_2)\frac{\partial \varphi (t)}{\partial t}\Big]\\ = & -\frac{\hbar^2}{2m_1}\varphi (t)\nabla_1^2\psi (\mathbf{r}_1,\mathbf{r}_2)-\frac{\hbar^2}{2m_2}\varphi (t)\nabla_2^2\psi (\mathbf{r}_1,\mathbf{r}_2) +V\psi \varphi \\ \end{aligned}$

November 11, 2020July 25, 2022

202011110613 Sidenote of Dummy Variables

Q: What is a dummy variable?

A: $f(x)=x^2$ and $g(y)=y^2$ are dummy variables because they describe the same pattern.

The indefinite integrals $\int x\,\mathrm{d}x=x^2/2+C$ and $\int y\,\mathrm{d}y=y^2/2+C$ are $\textrm{\scriptsize \textbf{NOT}}$ dummy variables because they are functionals admitting of different functions.

But if we put an upper and a lower limit to make it a definite integral,

i.e.,

$\begin{aligned} \int_0^1x\,\mathrm{d}x=\bigg[ \frac{x^2}{2} \bigg]_0^1=\frac{1}{2} \\ \int_0^1 y\,\mathrm{d}y=\bigg[ \frac{y^2}{2} \bigg]_0^1=\frac{1}{2} \\ \end{aligned}$ ,

they are dummy variables as the structure preserves the value.

November 9, 2020October 24, 2022

202011091423 Exercise 1 (Q2)

If $f(x)=(x-1)(x+5)$ , find the values of $f(2)$ , $f(1)$ , $f(0)$ , $f(a+1)$ , $f\bigg(\displaystyle{\frac{1}{a}}\bigg)$ , $f(-5)$ .

Solution.

$\begin{aligned} f(2) & = (2-1)(2+5)=7 \\ f(1) & = (1-1)(1+5)=0 \\ f(0) & = (0-1)(0+5)=-5 \\ f(a+1) & = ((a+1)-1)((a+1)+5) = a(a+6) \\ f\bigg( \frac{1}{a}\bigg) & = \Bigg( \bigg( \frac{1}{a}\bigg) -1\Bigg) \Bigg( \bigg( \frac{1}{a}\bigg) +5 \Bigg) \\ f(-5) & = (-5-1)(-5+5) = 0 \end{aligned}$

On reflection.

Suppose you are given an unknown quadratic function $f(x;a_0,a_1,a_2)=a_0+a_1x+a_2x^2$ such that

$\begin{aligned} 0 & \mapsto -5 \\ 1 & \mapsto 0 \\ 2 & \mapsto 7 \\ \end{aligned}$

and you are asked to make polynomial interpolation.

$\begin{cases} a_0 + a_1 (0) + a_2(0)^2 = f(0) = -5 \\ a_0 + a_1 (1) + a_2(1)^2 = f(1) = 0 \\ a_0 + a_1 (2) + a_2(2)^2 = f(2) = 7 \\ \end{cases}$
$\begin{cases} a_0=-5 \\ a_0 + a_1 + a_2 = 0\\ a_0 +2a_1 + 4a_2 = 7\\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 + a_2 = 5 \\ a_1 + 2a_2 = 6 \\ \end{cases}$
$\begin{cases} a_0 = -5 \\ a_1 = 4 \\ a_2 = 1 \\ \end{cases}$

And then is the interpolated polynomial $f(x)=x^2+4x-5\qquad \textrm{\scriptsize OR}\qquad (x+5)(x-1)$ .

November 5, 2020October 24, 2022

202011051527 Exercise 1 (Q1)

If $f(x)=2x^2-4x+1$ , find the values of $f(1)$ , $f(0)$ , $f(2)$ , $f(-2)$ , $f(a)$ , $f(x+\delta x)$ .

Solution.

Given $f(x)=2x^2-4x+1$ .

$\begin{aligned} f(1) & =2(1)^2-4(1)+1=-1 \\ f(0) & = 2(0)^2 - 4(0) +1 = 1 \\ f(2) & = 2(2)^2-4(2)+1 =1 \\ f(-2) & = 2(-2)^2-4(-2)+1=17\\ f(a) & = 2a^2 - 4a +1 \\ f(x+\delta x) & = 2(x+\delta x)^2 - 4 (x+\delta x) +1 \end{aligned}$

This exercise is done.

On reflection.

Suppose you are given the following conditions:

$\begin{aligned} x_0 = 0 & \qquad f(x_0) = 1 \\ x_1 = 1 &\qquad f(x_1) = -1 \\ x_2 = 2 &\qquad f(x_2) =1 \end{aligned}$

and you are asked to interpolate by Lagrange polynomials over the range $[0,2]$ .

$\begin{aligned} \mathcal{L}(x) & = (1)\bigg( \displaystyle{\frac{x-1}{0-1}} \bigg)\bigg( \displaystyle{\frac{x-2}{0-2}} \bigg) + (-1)\bigg( \displaystyle{\frac{x-0}{1-0}} \bigg) \bigg( \displaystyle{\frac{x-2}{1-2}} \bigg) + (1)\bigg( \displaystyle{\frac{x-0}{2-0}} \bigg) \bigg( \displaystyle{\frac{x-1}{2-1}} \bigg) \\ & = \displaystyle{\frac{(x-1)(x-2)}{2}} + x(x-2) + \displaystyle{\frac{x(x-1)}{2}} \\ & = \displaystyle{\frac{(x-1)(x-2)+2x(x-2)+x(x-1)}{2}} \\ & = \displaystyle{\frac{x^2-3x+2+2x^2-4x+x^2-x}{2}} \\ & = \displaystyle{\frac{4x^2-8x+2}{2}} \\ & = 2x^2-4x+1\\ \end{aligned}$

The interpolating polynomial $\mathcal{L}(x)$ checks with the original function $f(x)$ .

October 26, 2020November 7, 2023

202010260718 Exercises 2.1 (Q1)

(Sketch of a proof)

$\displaystyle{\frac{\mathrm{d}^2u^{i}}{\mathrm{d}t^2}} + \Gamma_{jk}^{i}\displaystyle{\frac{\mathrm{d}u^{j}}{\mathrm{d}t}}\displaystyle{\frac{\mathrm{d}u^{k}}{\mathrm{d}t}} = h(s) \displaystyle{\frac{\mathrm{d}u^{i}}{\mathrm{d}t}}$

where $h(s)=-\displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}\bigg( \frac{\mathrm{d}t}{\mathrm{d}s}\bigg)^{-2}}$

will reduce to Eq. (2.11):

$\displaystyle{\frac{\mathrm{d}^2u^{i}}{\mathrm{d}t^2}} + \Gamma_{jk}^{i}\displaystyle{\frac{\mathrm{d}u^{j}}{\mathrm{d}t}}\displaystyle{\frac{\mathrm{d}u^{k}}{\mathrm{d}t}}=0$

if and only if $t=As+B$ .

That is to say,

$h(s) \displaystyle{\frac{\mathrm{d}u^{i}}{\mathrm{d}t}}=0$ if and only if $t=As+B$ .

(if-part) Assume $t=As+B$ , then $\displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}}=0$ . From $h(s)= - \big( 0\big) \bigg( \displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}} \bigg)^{-2}=0$ , we have $h(s)\displaystyle{\frac{\mathrm{d}u^{i}}{\mathrm{d}t}}=0$ .

(only-if part). Assume $h(s)\displaystyle{\frac{\mathrm{d}u^i}{\mathrm{d}t}}=0$ , then some one of the following should be true:

i. $\displaystyle{\frac{\mathrm{d}^2t}{\mathrm{d}s^2}}=0$ ;

ii. $\bigg(\displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}}\bigg)^{-2}=0$ ;

iii. $\displaystyle{\frac{\mathrm{d}u^i}{\mathrm{d}t}}=0$ .

Situation ii. implies that $\displaystyle{\frac{\mathrm{d}s}{\mathrm{d}t}}=0$ , which is impossible for $\displaystyle{\frac{\mathrm{d}t}{\mathrm{d}s}}\neq \infty$ and $t=f(s)$ cannot have a point at infinity.

Situation iii. is impossible because it is only for some, but not any, $i$ ‘s in spherical coordinates (i.e., $r$ , $\theta$ , $\phi$ ), that $u^i=0$ . It is also for some $i$ ‘s that $\displaystyle{\frac{\mathrm{d}u^i}{\mathrm{d}t}}=0$ . As $i$ ‘s are to be chosen arbitrarily, the equality cannot hold.

As the second and the third were ruled out, the first situation is what that could be left possible.

The proof is as yet incomplete. It remains to be shown that

$\displaystyle{\iint\bigg(\frac{\mathrm{d}^2t}{\mathrm{d}s^2}\bigg)\,\mathrm{d}s\,\mathrm{d}s}=0 \qquad \Longrightarrow \qquad t=As+B$ .

October 23, 2020October 20, 2022

202010230206 Sidenote of Clopen

Observe that both $\emptyset$ and $X$ are open and closed in $X$ , i.e., clopen.

Proof. Pastime.

Remark. ( $S$ ‘s in several symbols )

$S$ is a set of points.
$S^{0}$ , the interior of set $S$ , contains all interior points.
$S'$ , the derived set of set $S$ , contains all accumulation/cluster/limit points.
$\bar{S}$ , the closure of set $S$ , contains all adherent points.
$\partial S$ (also denoted by $b(S)$ or $S^{b}$ ), the boundary of set $S$ , contains all boundary points.

October 23, 2020October 20, 2022

202010230028 Problem 2.4.11

Let $(X,d)$ be a metric space, $a\in X$ and $0<r<r'$ .

Prove that the set

$\{ x\in X:\enspace r<d(x,a)<r' \}$

is open in $(X,d)$ .

Setup.

Understand the definition given to each of the following:

First, what is meant by whether a set is open or not in some metric space?

Definition. (open set) Let $(X,d)$ be a metric space. A set $G\subset X$ is said to be an open set if it is a neighborhood of each of its points. (Equivalently, a set $G\subset X$ is said to be an open set
if for each $x\in G$ , there exists an $r>0$ such that $S_r(x)\subset G$ .)

_{Please refer to pg. 20, Jain and Ahmad’s Metric Spaces.}

Second, what is referred to as a neighborhood of some point(s)?

Definition. (neighborhood) Let $(X,d)$ be a metric space and $x\in X$ . A set $N\subset X$ is said to be a neighborhood (nbd) of $x$
if there exists an open sphere centred at $x$ and contained in $N$ ,
i.e., if $S_r(x)\subset N$ for some $r>0$ .

_{Please refer to pg. 19, Jain and Ahmad’s Metric Spaces.}

Third, what is an open sphere?

Definition. (open sphere) Let $(X,d)$ be a metric space. Let $x\in X$ and $r>0$ be a real number. The open sphere with centre $x$ and radius $r$ , denoted by $S_r(x)$ , the subset of $X$ given by $S_r(x)=\{ y\in X:\enspace d(x,y)<r \}$ N.b. An open sphere is always non-empty since it contains its centre at least.

_{Please refer to pg. 16, Jain and Ahmad’s Metric Spaces.}

October 22, 2020October 24, 2022

202010220700 Problem 1, Chapter 1.1

The Fahrenheit temperature scale is defined so that ice melts at $32^\circ \textrm{F}$ and water boils at $212^\circ \textrm{F}$ .

(a) Derive the formulas for converting from Fahrenheit to Celsius and back.

(b) What is absolute zero on the Fahrenheit scale?

Solution.

Given that the melting point is $32^\circ\textrm{F}$ and the boiling point $212^\circ \textrm{F}$ .

Therefore $1\,\textrm{C}^\circ =\displaystyle{\frac{212-32}{100}}\textrm{F}^\circ = 1.8\,\textrm{F}^\circ$ .

Then $F=32+1.8C$ and $C=\displaystyle{\frac{F-32}{1.8}}$ .

Absolute zero on Fahrenheit scale is $F=32+1.8(-273)=-459.4^\circ\textrm{F}$ .

October 22, 2020October 24, 2022

202010220448 Exercises 2.1A (Q1)

Suppose $|S|=19$ , $|T|=11$ and $|S\cap T|=8$ . Find $|S\cup T|$ and $|S\backslash T|$ .

Hint. (Verbal translation)

You are given that the number of elements in set $S$ is $19$ ,the number of elements in set $T$ is $11$ , and the number of elements in the intersection of set $S$ and set $T$ is $8$ .

You are asked:

What is the number of elements in the union of set $S$ and set $T$ ?What is the number of elements in the relative complement $S\backslash T$ of set $T$ with respect to set $S$ ?

Definition. The relative complement of $T$ with respect to $S$ is the set

$S\backslash T=\{ x\, |\enspace x\in S\textrm{ and }x\notin T\}$ .

Can you try drawing a Venn diagram?

Attempts.

(constructive)

Let

$S=\{ a,\, b,\, c,\, d,\, e,\, f,\, g,\, h,\, i,\, j,\, k,\, l,\, m,\, n,\, o,\, p,\, q,\, r,\, s\}$ ,

and also

$T=\{ l,\, m,\, n,\, o,\, p,\, q,\, r,\, s,\, t,\, u,\, v\}$ ,

so that $S\cap T=\{ l,\, m,\, n,\, o,\, p,\, q,\, r,\, s \}$ .

The union $S\cup T$ of set $S$ and set $T$ must as follows be:

$S\cup T=\{ a,\, b,\, c,\, d,\, e,\, f,\, g,\, h,\, i,\, j,\, k,\, l,\, m,\, n,\, o,\, p,\, q,\, r,\, s,\, t,\, u,\, v\}$ ,

such that $|S\cup T|=22$ .

The relative complement of set $T$ w.r.t. set $S$ is

$S\backslash T=\{ a,\, b,\, c,\, d,\, e,\, f,\, g,\, h,\, i,\, j,\, k\}$

and the number of its elements is

$|S\backslash T| = 11$ .

(analytic)

By observation of the Venn diagram,

you are writing out

$|S\cup T|=|S|+|T|-|S\cap T|$ ,

keeping in mind that

$|S\cap T|=|S|+|T|-|S\cup T|$

shall answer another question of a different subject.

October 13, 2020October 24, 2022

202010132314 Homework 1 (Q1)

The radius $r$ of a right circular cylinder is decreasing at a rate of $12\,\mathrm{cm\, s^{-1}}$ , while its height $h$ is decreasing at a rate of $25\,\mathrm{cm\, min^{-1}}$ . How is the volume changing when $r=180\,\mathrm{cm}$ and $h=500\,\mathrm{cm}$ ? Is the volume increasing or decreasing?

Solution.

The volume of a right circular cylinder is calculated by the formula

$V=\pi r^2h$ .

The volume $V$ (a dependent variable) of a cylinder varies with its radius $r$ and height $h$ (both independent variables). The change of volume, simply put it, is a derivative of volume $V$ with respect to time $t$ :

$\displaystyle{\frac{\mathrm{d}V}{\mathrm{d}t}}$ .

Differentiate $V(r,h,t)$ wrt. time $t$ :

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} V(r,h,t) & = \frac{\mathrm{d}}{\mathrm{d}t} (\pi r^2 h) \\ & = \pi \frac{\mathrm{d}}{\mathrm{d}t} (r^2h) \\ & = \pi \bigg[ r^2 \frac{\mathrm{d}}{\mathrm{d}t}(h) + h \frac{\mathrm{d}}{\mathrm{d}t} (r^2) \bigg] \\ & = \pi \bigg[ \big( r(t) \big)^2 \frac{\mathrm{d}}{\mathrm{d}t}\big( h(t) \big) + \big( h(t)\big) \bigg( \frac{\mathrm{d}}{\mathrm{d}t}\big( r(t) \big)^2 \bigg) \bigg] \\ & = \pi (r^2h' +2hrr' ) \\ & = \pi r^2h' + 2\pi hrr' \\ \end{aligned}$

Given $r=180\,\mathrm{cm}$ , $h=500\,\mathrm{cm}$ , $r'=-12\,\mathrm{cm\, s^{-1}}=-720\,\mathrm{cm\, min^{-1}}$ , and $h'=-25\,\mathrm{cm\, min^{-1}}$ , you would have it.

In some formalism of partial derivatives,

$\begin{aligned} \frac{\mathrm{d}V(r,h)}{\mathrm{d}t} & = \frac{\partial V}{\partial r}\frac{\mathrm{d}r}{\mathrm{d}t} + \frac{\partial V}{\partial h}\frac{\mathrm{d}h}{\mathrm{d}t} \\ & = (2\pi hr)\frac{\mathrm{d}r}{\mathrm{d}t} + (\pi r^2)\frac{\mathrm{d}h}{\mathrm{d}t}\\ & = 2\pi hrr' +\pi r^2h' \\ \end{aligned}$

you could have it also.

Afterthought.

It just so happens that there are two lines of attack, by taking total/ordinary derivatives and by taking partial derivatives. Is here anyhow the difference? Is there anything the matter?