April 2021 – Page 2 – Physicspupil

The Schrödinger equation for the delta-function barrier reads

$\displaystyle{-\frac{\hbar^2}{2m}}\dfrac{\mathrm{d}^2 \varphi}{\mathrm{d}x^2}+\alpha \delta (x)\varphi =E\varphi$

It yields both bound states ( $E<0$ ) and scattering states ( $E>0$ ). For the time being we consider only scattering states.

For $x<0$ , $V=\alpha \delta (x)=0$ .

The Schrödinger equation reads

$\displaystyle{\dfrac{\mathrm{d}^2\varphi}{\mathrm{d}x^2}=-\frac{2mE}{\hbar^2}\varphi=-k^2\varphi}$ ,

where $\displaystyle{k\stackrel{\textrm{def}}{=} \frac{\sqrt{2mE}}{\hbar}}$ is real and positive.

The general solution being

$\varphi (x)=Ae^{ikx}+Be^{-ikx}$ ,

and this time we cannot rule out either term, since neither of them blows up. Similarly, for $x>0$ ,

$\varphi (x)=Fe^{ikx}+Ge^{-ikx}$

The continuity of $\varphi (x)$ at $x=0$ requires that

$F+G=A+B$

The derivatives are

$\mathrm{d}\varphi /\mathrm{d}x=ik(Fe^{ikx}-Ge^{-ikx})$ for $x>0$ ;

$\mathrm{d}\varphi /\mathrm{d}x=ik(Ae^{ikx}-Be^{-ikx})$ for $x<0$ .

Here we define

$\displaystyle{\Delta (\mathrm{d}\varphi /\mathrm{d}x)=\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg|_{0^+}-\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg|_{0^-}}$

Hence, $\Delta (d\varphi /dx)=ik(F-G-A+B)$ .

By the second boundary condition, we have

$\displaystyle{\Delta \bigg(\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg)=\frac{2m}{\hbar^2}\lim_{\epsilon \to 0}\int_{-\epsilon}^{+\epsilon}V(x)\varphi (x)\,\mathrm{d}x}$ ,

which in turn gives

$\displaystyle{\Delta \bigg(\frac{\mathrm{d}\varphi}{\mathrm{d}x}\bigg)=\frac{2m\alpha}{\hbar^2}\varphi (0)=\frac{2m\alpha}{\hbar^2}(A+B)}$ .

Combining the results of first derivative continuity and second boundary condition, we write

$\displaystyle{ik(F-G-A+B)=\frac{2m\alpha}{\hbar^2}(A+B)}$ ,

Calculations:

Suppose the wave is coming from the left so that there will be no wave scattering from the right, i.e., $G=0$ .

Recall that $F+G=F+0=F=A+B$ by the continuity condition.

Calculations:

$\begin{aligned} F& =A+B\\ & = A-\frac{i\beta}{1+i\beta}A\\ & = \frac{1}{1+i\beta}A\\ \end{aligned}$

Reflection coefficient $R$ :

$\begin{aligned} R & \equiv \frac{|B|^2}{|A|^2}\\ & = \bigg|-\frac{i\beta}{1+i\beta}\bigg|^2\\ & = \frac{|i\beta|^2}{|1+i\beta|^2}\\ & = \frac{(\sqrt{\beta^2})^2}{(\sqrt{1^2+\beta^2})^2}\\ & = \frac{\beta^2}{1+\beta^2}. \end{aligned}$

We could find the transmission coefficient $T$ in two ways:

(1) By the formula $\displaystyle{T\equiv \frac{|F|^2}{|A|^2}}$ ; or
(2) By the fact that the sum of reflection coefficient and transmission coefficient has to be unity,

i.e., $R+T=1$ .

Calculations:

By (2):

$\begin{aligned} T & = 1-R\\ & = 1-\frac{\beta^2}{1+\beta^2}\\ & = \frac{1}{1+\beta^2} \end{aligned}$

Notice that $R$ and $T$ are functions of $\beta$ , hence:

$\displaystyle{R=\frac{1}{1+(2\hbar E/m\alpha^2)}}$ ,

$\displaystyle{T=\frac{1}{1+(m\alpha^2/2\hbar^2E)}}$ .

Readers should verify this result.

Compare this delta-function barrier with the delta-function trap, we notice that the reflection coefficient and transmission coefficient are each identical respectively in two cases.

April 14, 2021July 25, 2022

202104142057 Sidenote on “deriving momentum operator”

$\begin{aligned} \langle x\rangle &:= \int_{-\infty}^{+\infty}x|\Psi (x,t)|^2\,\mathrm{d}x.\\ \frac{\mathrm{d}\langle x\rangle}{\mathrm{d}t} & = \int_{-\infty}^{+\infty}x\frac{\partial}{\partial t}|\Psi (x,t)|^2\,\mathrm{d}x.\\ & = \int_{-\infty}^{+\infty}x\frac{i\hbar}{2m}\Bigr(\Psi^*\frac{\partial^2\Psi}{\partial x^2}-\frac{\partial^2\Psi^*}{\partial x^2}\Psi \Bigr)\,\mathrm{d}x\\ & = \frac{i\hbar}{2m}\int_{-\infty}^{+\infty}(x)\Bigr(\Psi^*\frac{\partial^2\Psi}{\partial x^2}+ \frac{\partial \Psi \partial \Psi^*}{\partial x^2}-\frac{\partial^2\Psi^*}{\partial x^2}\Psi -\frac{\partial \Psi \partial \Psi^*}{\partial x^2}\Bigr)\,\mathrm{d}x\\ & =\frac{i\hbar}{2m}\int_{-\infty}^{+\infty}(x)\Bigr[\frac{\partial}{\partial x}\Bigr(\Psi^*\frac{\partial \Psi}{\partial x}\Bigr)- \frac{\partial}{\partial x}\Bigr(\Psi \frac{\partial \Psi^*}{\partial x}\Bigr)\Bigr]\,\mathrm{d}x\\ &= \frac{i\hbar}{2m}\int_{-\infty}^{+\infty}\frac{\partial}{\partial x}(x)\Bigr(\Psi^*\frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^*}{\partial x}\Bigr)\,\mathrm{d}x\\ &= -\frac{i\hbar}{2m}\int_{-\infty}^{+\infty}\Bigr(\Psi^*\frac{\partial \Psi}{\partial x}-\Psi \frac{\partial \Psi^*}{\partial x}\Bigr)\,\mathrm{d}x\\ &= -\frac{i\hbar}{2m}\Bigr[\int_{-\infty}^{+\infty}\Bigr(\Psi^*\frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr)-\int_{-\infty}^{+\infty}\Bigr(\Psi \frac{\partial \Psi^*}{\partial x}\Bigr)\,\mathrm{d}x\Bigr]\\ &= -\frac{i\hbar}{2m}\Bigr[\Bigr(\Psi^*\Psi \Bigr|^{+\infty}_{-\infty}-\int_{-\infty}^{+\infty}\Psi \frac{\partial \Psi^*}{\partial x}\,\mathrm{d}x\Bigr)-\Bigr(\Psi \Psi^* \Bigr|^{+\infty}_{-\infty}-\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr) \Bigr]\\ & = -\frac{i\hbar}{2m}\Bigr( 0-\int_{-\infty}^{+\infty}\Psi \frac{\partial \Psi^*}{\partial x}\,\mathrm{d}x-0+\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr) \end{aligned}$

$\begin{aligned} \frac{\mathrm{d}\langle x\rangle}{\,\mathrm{d}t} & = -\frac{i\hbar}{2m}\Bigr[-\Bigr(\Psi \Psi^* \Bigr|^{+\infty}_{-\infty}-\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr)+\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr]\\ &= -\frac{i\hbar}{2m}\Bigr(2 \int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\Bigr)\\ &= -\frac{i\hbar}{m}\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\\ &= \frac{\hbar}{im}\int_{-\infty}^{+\infty}\Psi^* \frac{\partial \Psi}{\partial x}\,\mathrm{d}x\\ \langle p\rangle & := m\frac{\mathrm{d}\langle x\rangle}{\,\mathrm{d}t}\\ & = \int_{-\infty}^{+\infty}\Psi^* \Bigr(\frac{\hbar}{i}\frac{\partial}{\partial x}\Bigr)\Psi \,\mathrm{d}x \end{aligned}$

April 14, 2021October 24, 2022

202104161839 Homework 1 (Q2)

At time $t=0$ a particle is represented by the wave function

$\Psi (x) = \begin{cases} Ax/a, & \textrm{if }0\leq x\leq a \\ A\big( (x-a)^2+1\big), &\textrm{if }a\leq x\leq 2a \\ A(a^2+1)\displaystyle{\frac{3a-x}{a}},&\textrm{if }2a\leq x\leq 3a\\ 0, & \textrm{otherwise} \end{cases}$

where $A$ , $x$ are constants.

(a) Normalize $\Psi$ (that is, find $A$ in terms of $a$ ).
(b) Sketch $\Psi (x,0)$ as a function of $x$ .
(c) Where is the particle most likely to be found, at $t=0$ ?
(d) What is the probability of finding the particle to the left of $a$ .
(e) What is the expectation value of $x$ ?

(a)

$\begin{aligned} 1 & = \int_{-\infty}^{+\infty}|\Psi |^2\,\mathrm{d}x \\ & = \int_{0}^{a}\frac{|A|^2x^2}{a^2}\,\mathrm{d}x + \int_{a}^{2a}|A|^2[(x-a)+1]^2\,\mathrm{d}x+\int_{2a}^{3a}|A|^2(a+1)^2\bigg( \frac{3a-x}{a}\bigg)^2\,\mathrm{d}x \\ & = \frac{|A|^2}{a^2}\bigg[\frac{x^3}{3}\bigg]\bigg|^a_{0} + |A|^2\bigg[\frac{\big( (x-a)+1\big)^3}{3}\bigg]\bigg|_{1}^{a+1} + |A|^2\bigg(\frac{a+1}{a}\bigg)^2\bigg[\frac{(x-3a)^3}{3}\bigg]\bigg|_{-a}^{0} \\ & = \frac{|A|^2a}{3} + |A|^2\bigg(\frac{(2)^3}{3} - \frac{(2-a)^3}{3} \bigg) + |A|^2\bigg( \frac{a+1}{a} \bigg)^2\bigg( \frac{(-3a)^3}{3} - \frac{(-4a)^3}{3} \bigg) \\ \dots \enspace & \textrm{ (to be continued) }\dots \end{aligned}$

Roughwork.

Simplifying the second term,

$\begin{aligned} &\quad |A|^2\bigg(\frac{(2)^3}{3} - \frac{(2-a)^3}{3} \bigg) \\ & = |A|^2\bigg[ \bigg(\frac{8}{3}\bigg) -\bigg( \frac{8-12a+6a^2-a^3}{3}\bigg) \bigg] \\ & = |A|^2\bigg( \frac{4}{3}a - 2a^2 + \frac{a^3}{3}\bigg) \end{aligned}$

Simplifying the third term,

$\begin{aligned} & \quad |A|^2\bigg( \frac{a+1}{a} \bigg)^2\bigg( \frac{(-3a)^3}{3} - \frac{(-4a)^3}{3} \bigg) \\ & = |A|^2\bigg( 1+\frac{2}{a}+\frac{1}{a^2}\bigg) \bigg( \frac{-27a^3}{3} + \frac{64a^3}{3} \bigg)\\ & = |A|^2\bigg( 1+\frac{2}{a}+\frac{1}{a^2}\bigg) \bigg( \frac{37a^3}{3}\bigg) \\ & = |A|^2\bigg( \frac{37}{3}a^3+\frac{74}{3}a^2 + \frac{37}{3}a \bigg) \end{aligned}$

$\begin{aligned} \dots\enspace &\textrm{ (continue) }\dots \\ 1 & = |A|^2 \bigg[ \bigg( \frac{1}{3} + \frac{4}{3} + \frac{37}{3} \bigg) a + \bigg( -2+\frac{74}{3} \bigg) a^2 + \bigg( \frac{1}{3}+\frac{37}{3} \bigg) a^3 \bigg] \\ 1 & = |A|^2 \bigg( 14a + \frac{68}{3}a^2 + \frac{38}{3}a^3 \bigg) \\ A & = \frac{1}{\sqrt{14a+\frac{68}{3}a^2+\frac{38}{3}a^3}} \end{aligned}$

(b)

(c)

at $x=2a$ where $|\Psi |^2$ attains its maximum value.

(d)

$\begin{aligned} P(x\leqslant a) & =\int_{0}^{a}|\Psi |^2\,\mathrm{d}x \\ & = \int_{0}^{a}\bigg|\frac{Ax}{a}\bigg|^2\,\mathrm{d}x \\ & = |A|^2\frac{a^2}{3} \end{aligned}$

(e)

$\begin{aligned} \langle x\rangle & = \int_{0}^{3a}x|\Psi |^2\,\mathrm{d}x \\ & = |A|^2\bigg( \int_{0}^{a}\frac{x^3}{a^2}\,\mathrm{d}x + \int_{a}^{2a}[(x-a)^2+1]^2x\,\mathrm{d}x + \int_{2a}^{3a}(a^2+1)^2\bigg(\frac{3a-x}{a}\bigg)^2 x\,\mathrm{d}x \bigg) \\ \dots\enspace & \textrm{ (to be continued) }\dots \end{aligned}$

April 8, 2021October 31, 2022

202104080551 Solution to 1973-AL-AMATH-I-5

A particle of mass $m$ moving in a straight line in a medium with a velocity $v$ is subjected to a force $mk(v^3+a^2v)$ in the direction opposite to the velocity, where $k$ and $a$ are constants. Show that, at any subsequent time $t$ , the velocity $v$ is related to the initial velocity $v_0$ by

$\displaystyle{\frac{v}{(v^2+a^2)^{1/2}}=\frac{v_0}{(v_0^2+a^2)^{1/2}}e^{-a^2kt}}$ .

Show that for any value of the initial velocity, no matter how large, the particle will never travel a distance greater than $\pi /(2ka)$ .

Background.

Newton's second law:

$\begin{aligned} F = -mk(v^3+a^2v) & = ma \\ -mk(\dot{x}^3+a^2\dot{x}) & = m\ddot{x} \\ \frac{\mathrm{d}\dot{x}}{\mathrm{d}t} & = -k(\dot{x}^3+a^2\dot{x}) \\ \frac{\mathrm{d}\dot{x}}{\dot{x}^3+a^2\dot{x}} & = -k\,\mathrm{d}t \\ \dots \textrm{integrating from }& (v=v_0,\, t=0)\textrm{ to }(v=v,\, t=t) \\ \int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} & = \int_{0}^{t}-k\,\mathrm{d}t \end{aligned}$

(to be continued)

Roughwork.

Let . Then,

$\begin{aligned} 1 & \equiv A(u^2+a^2)+Bu^2+Cu \\ 1 & \equiv (A+B)u^2+Cu+Aa^2 \\ A+B & = 0 \\ C & = 0 \\ A & = \frac{1}{a^2} \\ \Rightarrow B & = -\frac{1}{a^2} \end{aligned}$
Thus,
$\displaystyle{\int \frac{1}{u(u^2+a^2)}\,\mathrm{d}u = \int \bigg(\frac{1}{a^2u} - \frac{u}{a^2(u^2+a^2)}\bigg) \,\mathrm{d}u}$

(continue)

$\begin{aligned} & \quad \int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} \\ & = \int_{v_0}^{v} \bigg(\frac{1}{a^2\dot{x}} - \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\bigg) \,\mathrm{d}\dot{x} \\ & = \int_{v_0}^v\frac{1}{a^2\dot{x}}\,\mathrm{d}\dot{x} - \int_{v_0}^v \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} \\ \end{aligned}$

The first term is

$\begin{aligned} \int_{v_0}^v\frac{1}{a^2\dot{x}}\,\mathrm{d}\dot{x} & = \bigg[ \frac{\ln \dot{x}}{a^2} \bigg]\bigg|_{v_0}^{v} \\ & = \bigg( \frac{\ln v}{a^2} \bigg) - \bigg( \frac{\ln v_0}{a^2} \bigg) \\ & = \frac{1}{a^2}\ln \bigg( \frac{v}{v_0} \bigg) \end{aligned}$

and the second term is

$\begin{aligned} &\quad -\int_{v_0}^v \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x}\\ \dots & \textrm{ let } \dot{x}=a\tan\theta \enspace\dots \\ \dots & \textrm{ let } \theta = \tan^{-1}\bigg(\frac{v}{a}\bigg) \textrm{ and } \theta_0 = \tan^{-1}\bigg(\frac{v_0}{a}\bigg) \\ & = -\int_{\theta_0}^{\theta} \frac{a\tan\theta}{a^2(a^2\tan^2\theta +a^2)}\cdot (a\sec^2\theta)\,\mathrm{d}\theta \\ & = -\int_{\theta_0}^{\theta} \frac{\tan\theta}{a(a^2\sec^2\theta)}\cdot (a\sec^2\theta )\,\mathrm{d}\theta \\ & = -\int_{\theta_0}^{\theta} \frac{\tan\theta}{a^2}\,\mathrm{d}\theta \\ & = -\bigg[ -\frac{\ln |\cos\theta |}{a^2} \bigg]\bigg|_{\theta_0}^{\theta} \\ & = \frac{1}{a^2} \bigg[\ln \bigg| \frac{a}{\sqrt{\dot{x}^2+a^2}} \bigg|\bigg]\bigg|_{v_0}^{v} \\ & = \frac{1}{a^2} \bigg( \ln\bigg| \frac{a}{\sqrt{v^2+a^2}} \bigg| -\ln\bigg| \frac{a}{\sqrt{v_0^2+a^2}} \bigg| \bigg) \\ & = \frac{1}{a^2} \ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} \bigg) \\ \end{aligned}$

Back to an earlier line

$\displaystyle{\int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} = \int_{0}^{t}-k\,\mathrm{d}t}$

we have

$\begin{aligned} \frac{1}{a^2}\ln\bigg( \frac{v}{v_0} \bigg) + \frac{1}{a^2}\ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}}\bigg) & = -kt \\ \ln\bigg( \frac{v}{v_0} \bigg) + \ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}}\bigg) & = -a^2kt \\ \ln\bigg( \frac{v}{v_0}\frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} \bigg) & =-a^2kt \\ \frac{v}{v_0}\frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} & = e^{-a^2kt} \\ \end{aligned}$

or,

$\boxed{\displaystyle{\frac{v}{\sqrt{v^2+a^2}}=\frac{v_0}{\sqrt{v_0^2+a^2}}e^{-a^2kt}}}$

as desired.

It remains to be shown that for any value of the initial velocity, no matter how large, the particle will never travel a distance greater than $\pi /(2ka)$ .

(to be continued)

April 6, 2021October 14, 2023

202104060103 Solution to 1973-AL-AMATH-I-2

When a pan of mass $m$ is put on top of a vertical spring scale of negligible weight, the downward displacement of the spring is observed to be $d$ . A ball of mass $M$ is now dropped from a height $h$ above the pan onto the pan. Suppose that the coefficient of restitution is equal to zero.

i. Show that the velocity of the pan immediately after the impact is

$\displaystyle{\frac{M\sqrt{2gh}}{M+m}}$ .

ii. From energy considerations, determine the maximum deflection of the pan after the ball has been dropped onto it.

Note: Force in spring scale is proportional to displacement.

Solution. (bad, if not wrong)

i.

Background.

1. If the coefficient of restitution is zero, the collision is perfectly inelastic.

2. Recall Hooke’s law:

$\begin{aligned} \mathbf{F_s} & =-k\mathbf{x} \\ F & = kx \\ U(x) & =\displaystyle{\frac{1}{2}}kx^2 \\ \end{aligned}$

Let $t$ be the time it takes for the ball to free-fall a vertical distance $h$ .

$\begin{aligned} s & =ut+\frac{1}{2}at^2 \\ \dots & \enspace\textrm{(take downward to be positive) }\dots \\ h & =(0)t+\frac{1}{2}(g)t^2 \\ t & =\sqrt{\frac{2h}{g}} \end{aligned}$

The velocity of the ball right before the impact is given by

$\begin{aligned} v & =u+at \\ \dots & \enspace\textrm{(take downward to be positive) }\dots \\ & = (0) + (g)\cdot\bigg(\sqrt{\frac{2h}{g}}\bigg) \\ & = \sqrt{2gh} \end{aligned}$

By conservation of linear momentum,

$\begin{aligned} M\cdot \sqrt{2gh} & = (M+m)v \\ v & = \frac{M\sqrt{2gh}}{M+m} \end{aligned}$

is the velocity of the ball and the pan immediately after collision.

ii.

Setup.

From Hooke’s law, the spring force is given by $\mathbf{F}_s=-k\mathbf{x}$ , or $mg=kd$ . The spring constant, characteristic of the spring, is $k=mg/d$ . Let $x$ be the magnitude of downward displacement of the spring from its relaxed position in reaching the final state of equilibrium. Then,
$\begin{aligned} F & = kx \\ (M+m)g & = \bigg( \frac{mg}{d} \bigg) x \\ x & = \frac{(M+m)d}{m} \end{aligned}$

By the law of conservation of energy,

$\begin{aligned} \Delta\textrm{Mechanical Energy} & = 0 \\ \Delta\textrm{KE} + \Delta\textrm{PE} & = 0 \\ \Big( \textrm{KE}_{\textrm{final}} - \textrm{KE}_{\textrm{initial}}\Big) + \Big( \textrm{PE}_{\textrm{final}} - \textrm{PE}_{\textrm{initial}}\Big) & = 0 \end{aligned}$

$\begin{aligned} \textrm{KE}_{\textrm{initial}} & = \frac{1}{2}(M+m)\bigg( \frac{M\sqrt{2gh}}{M+m} \bigg)^2 \\ & = \frac{M^2gh}{M+m} \\ \textrm{KE}_{\textrm{final}} & = 0 \\ \textrm{PE}_{\textrm{initial}} & = \frac{1}{2}kd^2\\ \textrm{PE}_{\textrm{final}} & = \frac{1}{2}k(x+d)^2 - (M+m)gx \end{aligned}$

$\begin{aligned} & \textrm{KE}_{\textrm{final}}-\textrm{KE}_{\textrm{initial}} \\ & = 0 - \frac{M^2gh}{M+m} \\ & = - \frac{M^2gh}{M+m} \\ & \\ & \textrm{PE}_{\textrm{final}}-\textrm{PE}_{\textrm{initial}} \\ = & \frac{1}{2}k(x+d)^2-(M+m)gx - \frac{1}{2}kd^2 \\ = & \frac{1}{2}kx^2+kxd - (M+m)gx \\ = & \frac{1}{2}\bigg( \frac{mg}{d} \bigg) x^2 + \bigg( \frac{mg}{d} \bigg) xd - (M+m)gx \\ = & \bigg( \frac{mg}{2d}\bigg) x^2 - Mgx \end{aligned}$

Thus,

$\begin{aligned} 0 & = -\frac{M^2gh}{M+m} + \frac{mgx^2}{2d} - Mgx \\ 0 & = \bigg( \frac{m}{2d} \bigg) x^2 + (-M)x + \bigg( -\frac{M^2h}{M+m} \bigg) \\ x & = \frac{-(-M)\pm\sqrt{(-M)^2-4(\frac{m}{2d})(-\frac{M^2h}{M+m})}}{2(\frac{m}{2d})} \\ x & = \frac{Md}{m} \pm \frac{Md}{m} \sqrt{1+\frac{2hm}{(M+m)d}} \end{aligned}$

$\therefore$ The maximum deflection of the pan after the ball has been dropped onto it is

$\boxed{x = \displaystyle{\frac{Md}{m} \pm \frac{Md}{m} \sqrt{1+\frac{2hm}{(M+m)d}}}}$ .

Done. Let’s overdo.

ii. (energy approach)

One can derive the equation of motion by considering energy conservation:

$\begin{aligned} \textrm{KE} & =\frac{1}{2}(M+m)\dot{x}^2 \\ \textrm{PE} & = \frac{1}{2}kx^2-(M+m)gx \\ 0 & = \frac{\mathrm{d}(\textrm{KE}+\textrm{PE})}{\mathrm{d}t} \\ 0 & = (M+m)\dot{x}\ddot{x} + kx\dot{x} - (m+M)g\dot{x} \\ 0 & = (M+m)\ddot{x} + kx - (m+M)g \\ \ddot{x} & = -\frac{k}{(M+m)}x + g \\ \end{aligned}$

or, by considering the Euler-Lagrange equation

where the Lagrangian $\mathcal{L}$ is defined

and then computing,

the Euler-Lagrange equation now reads

which agrees with the equation of motion derived previously.

ii. (force approach)

One can also derive the equation of motion by considering Newton’s second law:

$\begin{aligned} \dots\enspace \textrm{recall }& \textrm{Newton's second law }\dots \\ \textrm{Net }\mathbf{F} & = m\mathbf{a} \\ \dots\enspace \textrm{downward } & \textrm{taken to be positive \dots} \\ (M+m)\mathbf{g} - k\mathbf{x} & = (M+m)\ddot{\mathbf{x}} \\ \dots\enspace \textrm{consider } & \textrm{magnitude only }\dots \\ \ddot{x} & = -\frac{k}{M+m}x + g\\ \ddot{x} & = -\bigg( \sqrt{\frac{k}{M+m}} \bigg)^2x+g \\ \ddot{x} & = -\omega^2x+g\qquad \textrm{where }\omega =\sqrt{\frac{k}{M+m}} \\ \dots \textrm{ go back } & \textrm{to an earlier line }\dots \\ (M+m)g - kx & = (M+m)\ddot{x} \\ \dots \textrm{let }x_{\textrm{eff}}\textrm{ be} & \textrm{ the effective displacement} \\ x_{\textrm{eff}} & = x-\frac{(M+m)g}{k} \\ \textrm{then\qquad} (M+m)\ddot{x}_{\textrm{eff}} & = -kx_{\textrm{eff}} \\ \end{aligned}$

(to be continued)

Remark.

The effective displacement $x_{\textrm{eff}}$ is defined by

$x_{\textrm{eff}} = x-\displaystyle{\frac{(M+m)g}{k}}$ .

Substituting the spring constant $k$ with

$k=\displaystyle{\frac{mg}{d}}$ ,

$x_{\textrm{eff}}$ can be rewritten as

$x_{\textrm{eff}} = x-\displaystyle{\frac{(M+m)d}{m}}$ .

As seen in ii. Setup,

$\displaystyle{\frac{(M+m)d}{m}}$

is the equilibrium position.

$\therefore$ The effective displacement $x_{\textrm{eff}}$ is thus the distance extended or compressed with reference to the equilibrium position.

(continue)

$\begin{aligned} (M+m)\ddot{x}_{\textrm{eff}} & = -kx_{\textrm{eff}} \\ \ddot{x}_{\textrm{eff}} & = - \bigg( \sqrt{\frac{k}{M+m}} \bigg)^2x_{\textrm{eff}} \\ \ddot{x}_{\textrm{eff}} & =-\omega^2x_{\textrm{eff}} \qquad \textrm{where }\omega = \sqrt{\frac{k}{M+m}} \\ \dots\textrm{ Solving it, }& \dots\\ x_{\textrm{eff}}(t) & = c_1\cos (\omega t) + c_2\sin (\omega t) \end{aligned}$

The constant coefficients (i.e., $c_1$ , $c_2$ ) should be obtained from the initial and boundary conditions.

Let $t=0$ be the instant the ball hits on the pan.

When $t=0$ ,

$\begin{aligned} x_{\textrm{eff}}(0) & = d - \frac{(M+m)d}{m} \\ c_1(1) + c_2(0) & = \frac{md-(M+m)d}{m} \\ c_1 & = -\bigg( \frac{M}{m} \bigg) d \\ \end{aligned}$

As the velocity of the pan immediately after the impact
was found in part (i) to be

$\displaystyle{\frac{M\sqrt{2gh}}{M+m}}$ ,

$\begin{aligned} \dot{x}_{\textrm{eff}}(t) & = -c_1\sin (\omega t)+c_2\cos (\omega t) \\ \dot{x}_{\textrm{eff}}(0) & = -c_1(0) + c_2(1) = \frac{M\sqrt{2gh}}{M+m} \\ c_2 & = \frac{M\sqrt{2gh}}{M+m} \end{aligned}$

$\begin{aligned} \therefore x_{\textrm{eff}}(t) & = -\bigg( \frac{Md}{m} \bigg)\cos \Bigg( \bigg(\sqrt{\frac{k}{M+m}}\bigg) (t) \Bigg)\\ & \qquad + \bigg( \frac{M\sqrt{2gh}}{M+m}\bigg) \sin\Bigg( \bigg( \sqrt{\frac{k}{M+m}}\bigg) ( t)\Bigg) \\ \end{aligned}$

If there is no friction or none any other dissipation of energy, the spring will continue indefinitely and uninhibitedly its simple harmonic motion.

(to be continued)

Aside.

The angular frequency $\omega$ ( $=2\pi f$ ) is

$\omega =\displaystyle{\sqrt{\frac{k}{M+m}}}$ .

Hence the frequency $f$ is

$f=\displaystyle{\frac{\omega}{2\pi}}=\displaystyle{\frac{1}{2\pi}\sqrt{\frac{k}{M+m}}}$ ,

and the period $T$ is

$T=\displaystyle{\frac{1}{f}=2\pi\sqrt{\frac{M+m}{k}}}$ .

At time $t=\displaystyle{\frac{T}{2}}$ , the spring is at the equilibrium position (i.e., $x_{\textrm{eff}}(\frac{T}{2})$ ), whereas at time(s) $t=\displaystyle{\frac{T}{4},\, \frac{3T}{4}}$ , it reaches the extreme points (i.e., $x_{\textrm{eff}}(\frac{T}{4}),\, x_{\textrm{eff}}(\frac{3T}{4})$ ).

(continue)

Lemma.

The sinusoidal function

$x(t)=c_1\cos (\omega t)+c_2\sin (\omega t)$

can be written in the form

$x(t)=A\cos (\omega t-\varphi )$

where $A=\sqrt{c_1^2+c_2^2}$ and $\tan\varphi =\displaystyle{\frac{c_2}{c_1}}$ .

The amplitude (i.e., $A=\sqrt{c_1^2+c_2^2}$ above) is the maximum displacement of the spring from its equilibrium position.

$\begin{aligned} A & = \sqrt{c_1^2+c_2^2} \\ & = \sqrt{\bigg( -\frac{Md}{m} \bigg)^2 + \bigg( \frac{M\sqrt{2gh}}{M+m} \bigg)^2} \\ & = \sqrt{\frac{M^2d^2}{m^2} + \frac{2M^2gh}{(M+m)^2}} \\ \end{aligned}$

April 4, 2021October 24, 2022

202104160738 Solution to 1970-AL-PMATH-II-8

(a) Find the values of the constants $A$ , $B$ so that

$\displaystyle{\frac{-2x+4}{(x^2+1)(x-1)^2}=\frac{Ax+B}{x^2+1}-\frac{2}{x-1}+\frac{1}{(x-1)^2}}$ for all $x$ .

Hence or otherwise, find the indefinite integral

$\displaystyle{\int \frac{-2x+4}{(x^2+1)(x-1)^2}\,\mathrm{d}x}$ .

(b) Find the following indefinite integrals:

i. $\displaystyle{\int x^5e^{x^3}}\,\mathrm{d}x$ ;

ii. $\displaystyle{\int \frac{\mathrm{d}x}{\sqrt{(x-a)(b-x)}}}$ .

Solution.

(a)

$\begin{aligned} \textrm{RHS} & = \frac{(Ax+B)(x-1)^2-2(x^2+1)(x-1)+(x^2+1)}{(x^2+1)(x-1)^2} \\ & = \frac{[Ax^3+(B-2A)x^2+(A-2B)x+B]-[2(x^3-x^2+x-1)]+(x^2+1)}{(x^2+1)(x-1)^2} \\ & = \frac{(A-2)x^3+(B-2A+3)x^2+(A-2B-2)x+(B+3)}{(x^2+1)(x-1)^2} \\ & = \textrm{LHS} \end{aligned}$

Comparing coefficients,

$\begin{aligned} A-2 & = 0 \\ B-2A+3 & = 0 \\ A-2B-2 & = -2 \\ B+3 & = 4 \end{aligned}$

one gets $A=2$ and $B=1$ .

$\begin{aligned} & \int \frac{-2x+4}{(x^2+1)(x-1)^2} \,\mathrm{d}x \\ = & \int \bigg( \frac{2x+1}{x^2+1} - \frac{2}{x-1} + \frac{1}{(x-1)^2} \bigg) \,\mathrm{d}x \\ = & \int \frac{2x+1}{x^2+1}\,\mathrm{d}x - \int \frac{2}{x-1}\,\mathrm{d}x + \int \frac{1}{(x-1)^2}\,\mathrm{d}x \end{aligned}$

Evaluating term-by-term, the first term

$\begin{aligned} & \int \frac{2x+1}{x^2+1}\,\mathrm{d}x \\ \dots & \textrm{ Recall }1+\tan^2\theta = \sec^2\theta \enspace \dots \\ \dots & \textrm{ let }x=\tan\theta \enspace \dots \\ \dots & \textrm{ then }\mathrm{d}x = \sec^2\theta \,\mathrm{d}\theta \enspace \dots \\ = & \int \frac{2\tan\theta +1}{\tan^2\theta +1} \cdot \sec^2\theta \,\mathrm{d}\theta \\ = & \int \frac{2\tan\theta +1}{\sec^2\theta}\cdot \sec^2\theta\,\mathrm{d}\theta \\ = & \int (2\tan\theta +1)\,\mathrm{d}\theta \\ = & -2\ln |\cos\theta |+\theta +C_1 \\ = & 2\ln(\sqrt{x^2+1}) + \tan^{-1}x + C_1 \end{aligned}$

the second term

$\begin{aligned} & \int \frac{2}{x-1} \,\mathrm{d}x \\ & = \int \frac{2}{(x-1)}\,\mathrm{d}(x-1) \\ & = 2\ln (x-1) + C_2 \end{aligned}$

and the third term

$\begin{aligned} & \int\frac{1}{(x-1)^2}\,\mathrm{d}x \\ & = \int \frac{1}{(x-1)^2}\,\mathrm{d}(x-1) \\ & = -\frac{1}{(x-1)}+C_3 \end{aligned}$

Therefore

$\begin{aligned} & \displaystyle{\int \frac{-2x+4}{(x^2+1)(x-1)^2}\,\mathrm{d}x} \\ = & 2\ln \sqrt{x^2+1} + \tan^{-1}x - 2\ln (x-1) - \frac{1}{x-1} +C\\ = & \ln \Big( (x^2+1)/(x-1)^2 \Big) + \tan^{-1}x - \frac{1}{x-1} +C \end{aligned}$

(b)

i.

$\begin{aligned} & \int x^5e^{x^3}\,\mathrm{d}x \\ = & \int x^5e^{x^3}\cdot\frac{\mathrm{d}(x^3)}{3x^2} \\ = & \frac{1}{3}\int x^3e^{x^3}\,\mathrm{d}(x^3) \\ \dots\enspace & \textrm{let }u=x^3\enspace \dots \\ = & \frac{1}{3}\int ue^u\,\mathrm{d}u \end{aligned}$

(to be continued)

Recall integration by parts from the product rule of differentiation

$\begin{aligned} \frac{\mathrm{d}(uv)}{\mathrm{d}x} & = u\frac{\mathrm{d}v}{\mathrm{d}x} + v\frac{\mathrm{d}u}{\mathrm{d}x} \\ \int \frac{\mathrm{d}(uv)}{\mathrm{d}x}\,\mathrm{d}x & = \int u\bigg(\frac{\mathrm{d}v}{\mathrm{d}x}\bigg)\,\mathrm{d}x + \int v\bigg(\frac{\mathrm{d}u}{\mathrm{d}x}\bigg)\,\mathrm{d}x \\ \int \mathrm{d}(uv) & = \int u\,\mathrm{d}v + \int v\,\mathrm{d}u \\ \int u\,\mathrm{d}v & = uv - \int v\,\mathrm{d}u \end{aligned}$

(continue)

$\begin{aligned} = & \frac{1}{3} \int ue^u\,\mathrm{d}u \\ = & \frac{1}{3} \int u\,\mathrm{d}(e^u) \\ = & \frac{1}{3} \bigg( ue^u - \int e^u\,\mathrm{d}u \bigg) \\ = & \frac{1}{3}ue^u-\frac{1}{3}e^u + C \end{aligned}$

ii. It is left the reader.

April 3, 2021May 17, 2023

202104031305 Homework 1 (Q1)

Given two vectors $\mathbf{A}=-4\,\hat{\mathbf{x}}-2\,\hat{\mathbf{y}}+\hat{\mathbf{z}}$ and $\mathbf{B}=5\,\hat{\mathbf{x}}-3\,\hat{\mathbf{y}}-2\,\hat{\mathbf{z}}$ .

(a) Determine the angle $\theta$ between $\mathbf{A}$ and $\mathbf{B}$ .

(b) Determine the angle that the vector $\mathbf{A}$ makes with the $y$ -axis, and the angle that the vector $\mathbf{B}$ makes on the $yz$ -plane when it is projected on it.

(c) The vector $\mathbf{C}$ is the projection of vector $\mathbf{A}$ on the $xz$ -plane. Determine the unit vector along the direction of $\mathbf{C}\times \mathbf{A}$ .

(d) Express the vector $\mathbf{A}$ in terms of cylindrical coordinates, i.e., $\mathbf{A}=a_r\,\hat{\mathbf{r}}+a_\theta\,\hat{\boldsymbol{\theta}}+a_z\,\hat{\mathbf{z}}$ , with vectors $\hat{\mathbf{r}}$ , $\hat{\boldsymbol{\theta}}$ , and $\hat{\mathbf{z}}$ defined by the vector $\mathbf{A}$ . Determine the unit vectors $\hat{\mathbf{r}}$ , $\hat{\boldsymbol{\theta}}$ , and $\hat{\mathbf{z}}$ in terms of $\hat{\mathbf{x}}$ , $\hat{\mathbf{y}}$ , and $\hat{\mathbf{z}}$ , and the values of the coefficients $a_r$ , $a_\theta$ , and $a_z$ .

Solution.

(a) By the identity $\mathbf{A}\cdot\mathbf{B}=|\mathbf{A}||\mathbf{B}|\cos\theta$ where $\theta$ is the angle between vectors $\mathbf{A}$ and $\mathbf{B}$ ,

$\begin{aligned} \mathbf{A}\cdot\mathbf{B} & = (-4)(5) + (-2)(-3) + (1)(-2) = -16 \\ |\mathbf{A}| & = \sqrt{(-4)^2 + (-2)^2 + (1)^2 } = \sqrt{21} \\ |\mathbf{B}| & = \sqrt{(5)^2 + (-3)^2 + (-2)^2 } =\sqrt{38} \end{aligned}$

$\begin{aligned} \cos\theta & = \frac{\mathbf{A}\cdot\mathbf{B}}{|\mathbf{A}||\mathbf{B}|} \\ & = \frac{(-16)}{(\sqrt{21})(\sqrt{38})} \\ & = \frac{-16}{\sqrt{798}} \\ \theta & = \arccos \bigg( \frac{-16}{\sqrt{798}} \bigg) \\ & \Bigg[ \textrm{\scriptsize{OR}} \quad \cos^{-1}\bigg( \frac{-16}{\sqrt{798}} \bigg) \Bigg] \end{aligned}$

(b) Let the $y$ -axis be denoted by the vector $\mathbf{y}=(0,y,0)$ .

Then

$\begin{aligned} \mathbf{A}\cdot\mathbf{y} & = (-4)(0) + (-2)(y) + (1)(0) = -2y \\ |\mathbf{A}| & = \sqrt{(-4)^2+(-2)^2+(1)^2} = \sqrt{21}\\ |\mathbf{y}| & = \sqrt{(0)^2+(y)^2+(0)^2} = |y|\\ \end{aligned}$

The angle $\theta$ between vector $\mathbf{A}$ and $y$ -axis $\mathbf{y}$ is given by

$\begin{aligned} \theta & = \cos^{-1}\bigg( \frac{\mathbf{A}\cdot\mathbf{y}}{|\mathbf{A}||\mathbf{y}|} \bigg) \\ & = \cos^{-1}\bigg(\frac{-2y}{(\sqrt{21})(|y|)}\bigg) \\ & = \cos^{-1}\bigg(\frac{-2}{\sqrt{21}}\cdot \textrm{sgn}(y)\bigg) \\ \end{aligned}$

where the sign function $\textrm{sgn}(y)$ is defined below

$\textrm{sgn}(y) = \begin{cases} +1 & \textrm{for\enspace}y\geqslant 0 \\ -1 & \textrm{for\enspace}y<0 \end{cases}$

Let the $yz$ -plane be denoted by the plane of vectors $\mathbf{p}=(0,y,z)$ for any $y,z\in\mathbb{R}$ .

Then,

$\begin{aligned} \mathbf{B}\cdot\mathbf{y} & = (5,-3,-2)\cdot (0,y,z) \\ & = (5)(0)+(-3)(y)+(-2)(z) \\ & = -3y-2z \\ |\mathbf{B}| & = \sqrt{38} \\ |\mathbf{p}| & = \sqrt{y^2+z^2} \\ \theta & = \cos^{-1}\bigg( \frac{\mathbf{B}\cdot\mathbf{p}}{|\mathbf{B}||\mathbf{p}|} \bigg) \\ & = \cos^{-1}\bigg( \frac{-3y-2z}{\sqrt{(38)(y^2+z^2)}} \bigg) \end{aligned}$

Hold on, I smell a rat. Let’s go another way round. Find the angle $\phi$ between vector $\mathbf{B}$ and the $x$ -axis $\mathbf{x}=(x,0,0)$ .

As a matter of routine, take dot product on $\mathbf{B}$ and $\mathbf{x}$ .

$\begin{aligned} \mathbf{B}\cdot\mathbf{x} & = (5,-3,-2)\cdot (x,0,0) \\ & = (5)(x)+(-3)(0)+(-2)(0) \\ & = 5x \\ |\mathbf{B}| & = \sqrt{38} \\ |\mathbf{x}| & = \sqrt{x^2} = |x| \\ \phi & = \cos^{-1}\bigg( \frac{\mathbf{B}\cdot\mathbf{x}}{|\mathbf{B}||\mathbf{x}|} \bigg) \\ & = \cos^{-1}\bigg( \frac{5x}{(\sqrt{38})(|x|)}\bigg) \\ & = \cos^{-1}\bigg( \frac{5}{\sqrt{38}}\cdot\textrm{sgn}(x) \bigg) \end{aligned}$

The angle that the vector $\mathbf{B}$ makes on the $yz$ -plane is thus $\theta =90^\circ - \phi$ .

Back to the original line of thought. I.e.,

$\begin{aligned} \theta & = \cos^{-1}\bigg( \frac{\mathbf{B}\cdot\mathbf{p}}{|\mathbf{B}||\mathbf{p}|} \bigg) \\ & = \cos^{-1}\bigg( \frac{-3y-2z}{\sqrt{(38)(y^2+z^2)}} \bigg) \end{aligned}$

When the vector $\mathbf{B}$ is projected onto the $yz$ -plane (i.e., $\mathbf{p}$ ), the angle of projection $\theta$ is defined the angle between $\mathbf{B}$ and its orthogonal projection $\textrm{proj}_{\mathbf{p}}\mathbf{B}$ on that plane. It is clear that $\textrm{proj}_{\mathbf{p}}\mathbf{B}=(0,-3,-2)$ .

The more careful should have I written

$\begin{aligned} \theta & = \cos^{-1}\bigg( \frac{\mathbf{B}\cdot\textrm{proj}_{\mathbf{p}}\mathbf{B}}{|\mathbf{B}||\textrm{proj}_{\mathbf{p}}\mathbf{B}|} \bigg) \\ & = \cos^{-1}\bigg( \frac{(5,-3,-2)\cdot (0,-3,-2)}{(\sqrt{(5)^2+(-3)^2+(-2)^2})(\sqrt{(0)^2+(-3)^2+(-2)^2})} \bigg) \\ & = \cos^{-1}\bigg( \frac{13}{(\sqrt{38})(\sqrt{13})} \bigg) \\ & = \cos^{-1}\bigg(\sqrt{\frac{13}{38}} \bigg) \end{aligned}$

As $\theta=90^\circ - \phi$ , or, $\theta +\phi =90^\circ$

$\cos (\theta +\phi )= \cos 90^\circ = 0$ .

One should check that

$\cos\theta\cos\phi - \sin\theta\sin\phi = 0$ .

By brute force

$\begin{aligned} \textrm{LHS} & = \cos\theta\cos\phi - \sin\theta\sin\phi \\ & = \bigg( \frac{\sqrt{13}}{\sqrt{38}}\bigg)\bigg( \frac{5}{\sqrt{38}}\bigg) - \bigg( \frac{\sqrt{(\sqrt{38})^2-(\sqrt{13})^2}}{\sqrt{38}}\bigg) \bigg( \frac{\sqrt{(\sqrt{38})^2-(5)^2}}{\sqrt{38}}\bigg) \\ & = 0 \\ & = \textrm{RHS} \end{aligned}$

(c) $\mathbf{C}=(-4,0,1)$

$\begin{aligned} \mathbf{C} \times \mathbf{A} & = \begin{bmatrix} \hat{\mathbf{x}} & \hat{\mathbf{y}} & \hat{\mathbf{z}} \\ -4 & 0 & 1 \\ -4 & -2 & 1 \end{bmatrix} \\ & = \begin{pmatrix} 0 & 1 \\ -2 & 1 \end{pmatrix} \hat{\mathbf{x}} - \begin{pmatrix} -4 & 1 \\ -4 & 1 \end{pmatrix} \hat{\mathbf{y}} + \begin{pmatrix} -4 & 0 \\ -4 & -2 \end{pmatrix} \hat{\mathbf{z}} \\ & = 2\,\hat{\mathbf{x}} + 8\,\hat{\mathbf{z}} \\ & = (2,0,8) \end{aligned}$

Thus,

$\begin{aligned} \hat{\mathbf{n}} & = \frac{\mathbf{C}\times\mathbf{A}}{|\mathbf{C}\times\mathbf{A}|} \\ & = \frac{(2,0,8)}{\sqrt{(2)^2+(0)^2+(8)^2}} \\ & = (\frac{2}{\sqrt{70}} , 0, \frac{8}{\sqrt{70}}) \end{aligned}$

(d)

$\begin{aligned} \hat{\mathbf{r}} & = \hat{\mathbf{x}}\cos\theta + \hat{\mathbf{y}}\sin\theta \\ \hat{\boldsymbol{\theta}} & = -\hat{\mathbf{x}}\sin\theta + \hat{\mathbf{y}}\cos\theta \\ \hat{\mathbf{z}} & = \hat{\mathbf{z}} \end{aligned}$

$\begin{aligned} \cos\theta & =\displaystyle{\frac{-4}{\sqrt{(-4)^2+(-2)^2}}=\frac{-4}{\sqrt{20}}} \\ \sin\theta & = \displaystyle{\frac{-2}{\sqrt{(-4)^2+(-2)^2}}=\frac{-2}{\sqrt{20}}} \end{aligned}$
$\begin{aligned} \hat{\mathbf{r}} & = -\frac{2\sqrt{5}}{5}\,\hat{\mathbf{x}} -\frac{\sqrt{5}}{5}\,\hat{\mathbf{y}}\\ \hat{\boldsymbol{\theta}} & = \frac{\sqrt{5}}{5}\,\hat{\mathbf{x}} -\frac{2\sqrt{5}}{5}\,\hat{\mathbf{y}} \\ \hat{\mathbf{z}} & = \hat{\mathbf{z}} \end{aligned}$