October 12, 2021 – Physicspupil

October 12, 2021October 24, 2022

202110121413 Exercise 8.5.B (Q21)

This exercise is related to Einstein’s famous law $E=mc^2$ . The relativistic momentum $p$ of a particle of mass $m$ moving at a speed $v$ along a straight line (say, the $x$ -axis) is

$\displaystyle{p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}}$ ,

where $c$ is the speed of light. The relativistic force on the particle along that line is

$\displaystyle{F=\frac{\mathrm{d}p}{\mathrm{d}t}}$ ,

which is the same formula as Newton’s Second Law of motion in classical mechanics. Assume that the particle starts at rest at position $x_1$ and ends at position $x_2$ along the $x$ -axis. The work done by the force $F$ on the particle is:

$\displaystyle{W=\int_{x_1}^{x_2}F\,\mathrm{d}x=\int_{x_1}^{x_2}\frac{\mathrm{d}p}{\mathrm{d}t}\,\mathrm{d}x}$

(a) Show that

$\displaystyle{\frac{\mathrm{d}p}{\mathrm{d}v}=\frac{m}{\bigg( \displaystyle{1-\frac{v^2}{c^2}}\bigg)^{3/2}}}$ .

(b) Use the Chain Rule formula

$\displaystyle{\frac{\mathrm{d}p}{\mathrm{d}t}=\frac{\mathrm{d}p}{\mathrm{d}v}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}}$

to show that

$\displaystyle{F\,\mathrm{d}x=v\frac{\mathrm{d}p}{\mathrm{d}v}\,\mathrm{d}v}$ .

(c) Use parts (a) and (b) to show that

$\displaystyle{W=\int_{0}^{v}\frac{\mathrm{d}p}{\mathrm{d}v}\, v\,\mathrm{d}v=\int_{0}^{v}\frac{mv}{\bigg(\displaystyle{ 1-\frac{v^2}{c^2}}\bigg)^{3/2}}\,\mathrm{d}v}$ .

(d) Use part (c) to show that

$\displaystyle{W=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2}$ .

(e) Define the relativistic kinetic energy $K$ of the particle to be $K=W$ , and define the total energy $E$ to be

$\displaystyle{E=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}}$ .

So by part (d), $K=E-mc^2$ . Show that

$E^2=p^2c^2+(mc^2)^2$ .

(Hint: Expand the right side of that equation.)

(f) What is $E$ when the particle is at rest?

Solution.

(a)

$\begin{aligned} \frac{\mathrm{d}p}{\mathrm{d}v} & = \frac{\mathrm{d}}{\mathrm{d}v}\bigg( \frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}\bigg) \\ & = \frac{\bigg(\sqrt{1-\displaystyle{\frac{v^2}{c^2}}}\bigg)(m)-(mv)\bigg(\displaystyle{\frac{1}{2}}\Big( 1-\displaystyle{\frac{v^2}{c^2}}\Big)^{-1/2}\Big( -\displaystyle{\frac{2v}{c^2}}\Big)\bigg)}{\bigg(\sqrt{1-\displaystyle{\frac{v^2}{c^2}}}\bigg)^2} \\ & = \frac{\bigg( 1-\displaystyle{\frac{v^2}{c^2}}\bigg) (m)-(mv)\bigg( \displaystyle{-\frac{v}{c^2}}\bigg)}{\bigg( 1-\displaystyle{\frac{v^2}{c^2}}\bigg)^{3/2}} \\ & = \frac{m}{\bigg( 1-\displaystyle{\frac{v^2}{c^2}}\bigg)^{3/2}} \end{aligned}$

(b)

$\begin{aligned} \int_{x_1}^{x_2}F\,\mathrm{d}x & = \int_{x_1}^{x_2}\frac{\mathrm{d}p}{\mathrm{d}t}\,\mathrm{d}x \\ \int_{x_1}^{x_2}F\,\mathrm{d}x & = \int_{x_1}^{x_2}\frac{\mathrm{d}p}{\mathrm{d}v}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}x \\ F\,\mathrm{d}x & = \frac{\mathrm{d}p}{\mathrm{d}v}\frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}\,\mathrm{d}x \\ & = v\,\frac{\mathrm{d}p}{\mathrm{d}v}\,\mathrm{d}v\\ \end{aligned}$

(c)

$\begin{aligned} W & =\int_{0}^{v}\frac{\mathrm{d}p}{\mathrm{d}v}\, v\,\mathrm{d}v \\ & = \int_{0}^{v}\Bigg( \frac{m}{\big( 1-\frac{v^2}{c^2}\big)^{3/2}}\Bigg) (v)\,\mathrm{d}v\\ & =\int_{0}^{v}\frac{mv}{\bigg(\displaystyle{ 1-\frac{v^2}{c^2}}\bigg)^{3/2}}\,\mathrm{d}v \\ \end{aligned}$ .

(d)

$\begin{aligned} W & = \int_{0}^{v}\frac{mv}{\bigg(\displaystyle{1-\frac{v^2}{c^2}}\bigg)^{3/2}}\,\mathrm{d}v \\ \dots\enspace & \textrm{let }u=\frac{v}{c}\enspace\dots \\ \dots\enspace & \textrm{then }\mathrm{d}v=c\,\mathrm{d}u\enspace\dots \\ & = \int_{0}^{\frac{v}{c}}\frac{mc^2u}{(1-u^2)^{\frac{3}{2}}}\,\mathrm{d}u \\ \dots\enspace & \textrm{let }u=\sin\theta\enspace\dots \\ \dots\enspace & \textrm{then }\mathrm{d}u=\cos\theta\,\mathrm{d}\theta\enspace\dots \\ & = \int\frac{mc^2\sin\theta}{\cos^3\theta}\cdot\cos\theta\,\mathrm{d}\theta \\ & = \int\frac{mc^2\sin\theta}{\cos^2\theta}\,\mathrm{d}\theta \\ \dots\enspace & \textrm{let }w=\cos\theta\enspace\dots \\ \dots\enspace & \textrm{then }\mathrm{d}w=-\sin\theta\,\mathrm{d}\theta\enspace\dots \\ & \quad -\int \frac{mc^2}{w^2}\,\mathrm{d}w \\ & = \frac{mc^2}{w} \\ & = \frac{mc^2}{\cos\theta} \\ & = mc^2\sec\theta \\ & = \frac{mc^2}{\sqrt{1-u^2}} \\ & = \enspace\dots \\ & = \bigg[\frac{mc^2}{\sqrt{1-u^2}}\bigg]\bigg|^{\frac{v}{c}}_{0} \\ & = \frac{mc^2}{\sqrt{1-\displaystyle{\frac{v^2}{c^2}}}}-mc^2 \\ \end{aligned}$

Parts (e) and (f) are left to the readers.

October 12, 2021October 24, 2022

202110121137 Exercise 1.1.1

(a) Show that $x^2\in\langle x-y^2,xy\rangle$ in $k[x,y]$ ( $k$ any field).

(b) Show that $\langle x-y^2,xy,y^2\rangle = \langle x,y^2\rangle$ .

(c) Is $\langle x-y^2,xy\rangle = \langle x^2,xy\rangle$ ? Why or why not?

Definition.

Let $f_1,\dots ,f_s\in k[x_1,\dots ,x_n]$ . We let $\langle f_1,\dots ,f_s\rangle$ denote the collection $\langle f_1,\dots ,f_s\rangle = \{ p_1f_1+\cdots +p_sf_s:p_i\in k[x_1,\dots ,x_n]\}$ for $i= 1,\dots ,s$ .

(a)

$\begin{aligned} x^2 & = (x)(x-y^2) + (y)(xy) \\ & \in \langle x-y^2,xy\rangle \\ \end{aligned}$

(b)

First, we want to show

$\langle x-y^2,xy,y^2\rangle \in \langle x,y^2\rangle$ .

For any $p_1,p_2,p_3\in k[x,y]$ ,

$\begin{aligned} & \quad \langle x-y^2,xy,y^2\rangle \\ & = (p_1)(x-y^2)+(p_2)(xy)+(p_3)(y^2) \\ & = p_1x-p_1y^2+p_2xy+p_3y^2 \\ & = (p_1+p_2y)(x)+(p_3-p_1)(y^2) \\ & \in \langle x,y^2\rangle \\ \end{aligned}$

Next, we want to show

$\langle x,y^2\rangle\in\langle x-y^2,xy,y^2\rangle$ .

For any $p_1,p_2\in k[x,y]$ ,

$\begin{aligned} &\quad \langle x,y^2\rangle \\ & = (p_1)(x)+(p_2)(y^2) \\ & = (p_1)(x-y^2)+(0)(xy)+(p_1+p_2)(y^2) \\ & \in \langle x-y^2,xy,y^2\rangle \\ \end{aligned}$

All in all,

$\langle x-y^2,xy,y^2\rangle = \langle x,y^2\rangle$ .

(c) I guess $\textrm{\scriptsize{NOT}}$ .

October 12, 2021October 24, 2022

202110120936 Exercises 13.3.6 (Q43)

Graph each function by algebraically determining its key features. Then state the domain and range of the function.

$f(x)=x^2-7x+12$

Solution.

$\begin{aligned} f(x) & = x^2-7x+12 \\ & = (x-3)(x-4) \end{aligned}$

We see that $y=f(x)=0$ when $x=\{ 3,4\}$ .

i. $\therefore$ The $x$ -intercepts of $f(x)$ are thus $(3,0)$ and $(4,0)$ .

Plugging in $x=0$ will give the $y$ -intercept:

$\begin{aligned} y\textrm{-intercept}& = f(0) \\ & = (0)^2-7(0)+12 \\ & = 12 \end{aligned}$

ii. $\therefore$ The $y$ -intercept is thus $(0,12)$ .

Differentiating $y=f(x)$ with respect to $x$ ,

$\begin{aligned} & \quad f'(x) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f(x)\big) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(x^2-7x+12) \\ & = 2x-7 \\ \end{aligned}$

When $x=3.5$ and $f(3.5)=-0.25$ , the slope of $f(x)$ is zero, i.e., $f'(x)=0$ .

iii. $\therefore$ $(\frac{7}{2},-\frac{1}{4})$ is a turning point (/extreme point/vertex).

$\begin{aligned} y & =f(x) \\ & = x^2-7x+12 \\ & = x^2-7x+\bigg( \frac{7}{2}\bigg)^2 - \bigg( \frac{7}{2}\bigg)^2 + 12 \\ & = \bigg( x-\frac{7}{2}\bigg)^2 - \frac{1}{4}\\ \end{aligned}$

iv. $\therefore$ The axis of symmetry of the graph is $x=\frac{7}{2}$ .

Differentiating $y=f(x)$ twice with respect to $x$ ,

$\begin{aligned} &\quad f''(x)\\ & = \frac{\mathrm{d}}{\mathrm{d}x}\big( f'(x)\big) \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(2x-7) \\ & = 2 \quad (>0)\\ \end{aligned}$

We see that the slope is increasing with $x$ .

v. $\therefore$ The graph of $f(x)$ is concave upward (/convex downward).

vi. $\therefore$ The domain is $\mathbb{R}$ and the range $\{ y\in\mathbb{R}:y\ge -\frac{1}{4}\}$ .

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