April 2020 – Physicspupil

April 24, 2020January 12, 2023

202004241907 Homework 1 (Q2)

(a) State the Mean Value Theorem and the Taylor’s Theorem.

(b) Find an $O(h^4)$ and an $O(h^5)$ approximation to $\cos h$ . Compare those approximate values to the actual value $0.9950042$ when $h=0.1$ . Correct answers in this question to seven decimal places.

Solution.

(a)

Theorem. (Mean-Value Theorem) Suppose $f\in C[a,b]$ , and $f'(x)$ exists on $(a,b)$ . For every $x\in [a,b]$ and some $x_0\in [a,b]$ ,

$f(x)=f(x_0)+f'(\eta (x))(x-x_0)$

where $\eta (x)$ is between $x_0$ and $x$ .

Theorem. (Taylor’s Theorem) Suppose $f\in C^n[a,b]$ , and $f^{(n+1)}$ exists on $(a,b)$ . For every $x\in (a,b)$ and some $x_0\in [a,b]$ , there exists a number $\eta (x)$ between $x_0$ and $x$ with $f(x)=P_n(x)+R_n(x)$ , where

$P_n(x)=f(x_0)+f'(x_0)(x-x_0)+ \displaystyle{\frac{f''(x_0)}{2!}}(x-x_0)^2+\cdots + \displaystyle{\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n}$ , and

$R_n(x)=\displaystyle{\frac{f^{(n+1)}(\eta (x))}{(n+1)!}}(x-x_0)^{n+1}$ .

(b) Let $f(x)=\cos x$ . Then we have $f'(x)=-\sin x$ , $f''(x)=-\cos x$ , $f'''(x)=\sin x$ , $f^{(4)}(x)=\cos x$ , $f^{(5)}(x)=-\sin x$ , etc. By Taylor expansion of $\cos x$ at $x_0$ , we have

$\cos x=\cos x_0+(-\sin x_0)(x-x_0)+\displaystyle{\frac{-\cos x_0}{2!}}(x-x_0)^2+\displaystyle{\frac{\sin x_0}{3!}}(x-x_0)^3+\displaystyle{\frac{\cos x_0}{4!}}(x-x_0)^4+\displaystyle{\frac{-\sin x_0}{5!}}(x-x_0)^5+\cdots$

Now letting $x_0=0$ , we obtain further that

$\cos x=1-\displaystyle{\frac{x^2}{2!}}+\displaystyle{\frac{x^4}{4!}}+\cdots =\displaystyle{\sum_{n=0}^\infty \frac{(-1)^n}{(2n)!}x^{2n}}$

Thus, the $O(h^4)$ approximation to $\cos h$ is in the form of

$\cos h=1-\displaystyle{\frac{h^2}{2!}}+O(h^4)$

where $O(h^4)$ is the higher-order terms $\displaystyle{\sum_{n=2}^\infty \frac{(-1)^n}{(2n)!}h^{2n}}$ .

And the $O(h^5)$ approximation to $\cos h$ is in the form of

$\cos h=1-\displaystyle{\frac{h^2}{2!}}+\displaystyle{\frac{h^4}{4!}}+O(h^5)$

where $O(h^5)$ is the higher-order terms $\displaystyle{\sum_{n=3}^\infty \frac{(-1)^n}{(2n)!}h^{2n}}$ .

Then, when $h=0.1$ , $\cos h$ in the $O(h^4)$ approximation would be

$\cos 0.1=1-\displaystyle{\frac{0.1^2}{2!}}=0.995$ ;

whereas in the $O(h^5)$ approximation,

$\cos 0.1=1-\displaystyle{\frac{0.1^2}{2!}}+\displaystyle{\frac{0.1^4}{4!}}=0.995004166\approx 0.9950042\enspace \textrm{(corr. to 7 d.p.)}$ .

Given the actual value $\cos h=0.9950042$ , the $O(h^4)$ approximation has an absolute error of $|0.9950042-0.995|=0.0000042$ and a relative error of $\displaystyle{\frac{|0.9950042-0.995|}{|0.9950042|}}=4.2210877\times 10^{-6}$ . And the $O(h^5)$ approximation has an absolute error of $|0.9950042-0.9950042|=0$ and thus a relative error of zero.

Remark. The $0$ error is owing to the correction of 7 decimal places in both the direct computation of $\cos 0.1$ and its $O(h^5)$ approximation.

April 24, 2020October 24, 2022

202004241649 Problem 2, Ch. 1 Sec. 1

Prove the following equalities:

(a) $|ab|=|a|\cdot |b|$ ;

(b) $|a|^2=a^2$ ;

(c) $\displaystyle{\bigg|\frac{a}{b}\bigg|=\frac{|a|}{|b|}}$ (where $b\neq 0$ );

(d) $\sqrt{a^2}=|a|$ .

Proof.

(a) (proof by cases)

i. When $a,b\geqslant 0$ :

$|ab|=ab=|a||b|$ .

ii. When $a,b< 0$ :

$|ab|=ab=(-a)(-b)=|a||b|$ .

iii. When $a\geqslant 0$ and $b<0$ :

$|ab|=-ab=a(-b)=|a||b|$ .

iv. When $a<0$ and $b\geqslant 0$ :

$|ab|=-ab=(-a)b=|a||b|$ .

QED

(b) (proof by induction)

By making a stronger claim

$P(n)$ : For any positive integer $n$ , $|a^n|=|a|^n$ .

Proof.

The trivial cases $n=0$ and $n=1$ are evident.

Consider the case $n=2$ ,

$\begin{aligned} |a^2| & = |a||a| \qquad \textrm{(by equality (a))} \\ & = |a|^2 \end{aligned}$

$P(2)$ is true.

Assume now that $P(n)$ is true,

$\begin{aligned} P(n+1): \qquad |a^{n+1}|& = |a^n\cdot a| \\ & = |a^n||a| \qquad \textrm{(by equality (a))}\\ & = |a|^n|a| \qquad \textrm{(by the assumption }P(n)\textrm{ is true)} \\ & = |a|^{n+1} \end{aligned}$

it can be seen that $P(n+1)$ is also true.

From the fact that $P(2)$ is true and by the principle of mathematical induction, $P(n)$ is true for all positive integers $n$ .

It follows that $|a|^2=|a^2|=a^2$ holds.

QED

(c) (direct proof)

$\because |a|=\displaystyle{\bigg| \frac{a}{b}\cdot b\bigg|} =\bigg|\displaystyle{\frac{a}{b}}\bigg| |b|$ ,

where the second equality sign is due to equality (a),

$\therefore \displaystyle{\bigg| \frac{a}{b} \bigg| = \frac{|a|}{|b|}}$ .

QED

(d) (proof by definition)

The absolute value of a real number $a$ , denoted by $|a|$ , is defined by

$|a|= \begin{cases} x & \textrm{if } x\geqslant 0, \\ -x & \textrm{if }x<0 \end{cases}$

For any non-negative real number $a$ , the symbol $\sqrt{a}$ denotes the non-negative square root of $a$ .

QED

April 24, 2020July 25, 2022

202004241533 Problem 1, Ch. 1 Sec. 1

Prove that if $a$ and $b$ are real numbers then

$||a|-|b||\leqslant |a-b|\leqslant |a|+|b|$ .

Proof.

As $|a|$ is non-negative and $-|a|$ non-positive, one has

Eq. (1):

$-|a|\leqslant a\leqslant |a|$

Eq. (2):

$-|b|\leqslant b\leqslant |b|$

Combining Eq. (1) and Eq. (2),

$-(|a|+|b|)\leqslant a+b\leqslant |a|+|b|$ ,

or, Eq. (3): (the triangle inequality)

$|a+b|\leqslant |a|+|b|$ .

Applying the triangle inequality to $|a-b|$ , one gets

Eq. (4):

$|a-b|=|a+(-b)|\leqslant |a|+|-b|=|a|+|b|$ ,

or, Eq. (4)’:

$|a-b|\leqslant |a|+|b|$ .

Applying the triangle inequality to $|a|=|(a-b)+b|$ , one gets

Eq. (5):

$|a|=|(a-b)+b|\leqslant |a-b|+|b|$ ,

or, Eq. (5)’:

$|a|-|b|\leqslant |a-b|$ .

Applying the triangle inequality to $|b|=|(b-a)+a|$ , one gets

Eq. (6):

$|b|=|(b-a)+a|\leqslant |b-a|+|a|$ ,

Roughwork.

or, Eq. (6)’:

$-|a-b| \leqslant |a|-|b|$

Combining Eq. (5)’ and Eq. (6)’:

$-|a-b| \leqslant |a|-|b| \leqslant |a-b|$ ,

or, Eq. (7):

$||a|-|b||\leqslant |a-b|$ .

Combining Eq. (4)’ and Eq. (7), one obtains readily

$||a|-|b||\leqslant |a-b|\leqslant |a|+|b|$ .

QED

April 24, 2020February 5, 2023

202004240713 Homework 1 (Q3)

Suppose $G$ be a finite group of even order. Prove that there exists an element $a\in G$ such that $a\neq e$ and $a^2=e$ .

Attempts.

Try negating the statement by the following:

$\forall\, a\in G$ , either $a=e$ or $a^2\neq e$ ,

then looking for contradiction to the assumption that $G$ should be a finite group of even order.

CASE 1: Would it be possible that $a=e$ ?

Given so, the inverse and the identity of $G$ would be $e$ . And the set $G$ would have contained $e$ only, i.e., the singleton set $G=\{ e\}$ . This contradicts with the assumption that its order be even.

CASE 2: If for all elements in the group $G$ , their order cannot ever be $2$ .

If $a^2\neq e$ , then $a\neq a^{-1}$ for all $a\in G$ . It would then become an ill-posited negation, because the identity element $e$ is one of all $a\in G$ , and both statements $e^2\neq e$ and $e\neq e^{-1}$ are abhorrent to the basic axioms of a group. So perhaps to my discretion $e$ should be precluded from $a$ ‘s, and understood as one ( $e=e^{-1}$ ) member apart from the other members.

Proceeding to observe that if $a\neq a^{-1}$ for all $a\in G$ , I may then construct a pair of $a_i$ and $a_i^{-1}$ for $i\in \{ 1,2,\dots ,n:n\in\mathbb{Z}^+\}$ , provided from the definition of a group that the inverse of each element must exist and, as it follows, that it must be unique. That is, if $a_5=a_{27}^{-1}$ , I may relabel $a_{27}$ as $a_5^{-1}$ . By such a construction I can guarantee that all elements, except the identity $e$ , are now in pairs.

One therefore, by counting in total how many elements there are in group $G$ , will get an odd number $(2n+1)$ , the $2n$ counted from the pairs, and the single $1$ the identity $e$ itself alone counts.

This contradicts with the assumption that $|G|$ be of even parity.

From the negation of statement arises contradiction, the original statement is therefore proven by contradiction.

Remark.

I could not prove otherwise than loosely to so naive myself. Not until I had survived mathematical rigor.

April 24, 2020October 24, 2022

202004240630 Homework 1 (Q2)

Call $G$ a group and $e$ the identity of $G$ . If the order of $G$ is $2$ , then $a^2=e$ for all $a\in G$ and $G$ is abelian (Prove it yourself). Show that

$G$ is abelian if $a^2=e$ for all $a\in G$ .

Give an example of such a group whose order is greater than $2$ .

Solution.

Recall that a group $G$ is said to be abelian if $xy=yx$ for all $x,y\in G$ .

For any $a,b\in G$ , there is $ab=(bb^{-1})ab(a^{-1}a)$ , because a group satisfies, first, the inverse axiom $xx^{-1}=x^{-1}x=e$ under the notation $x^{-1}$ being the inverse of $x$ , and secondly, the identity axiom $ex=xe=x$ .

Thence by (associativity axiom) $x*(y*z)=(x*y)*z$ I may change brackets and obtain as follows:

$\begin{aligned} ab & =(bb^{-1})ab(a^{-1}a) \\ & =b(b^{-1}a)(ba^{-1})a \\ & =b(b^{-1}a)(b^{-1}a)^{-1}a \end{aligned}$

Provided that $a^2=e$ for all $a\in G$ , one may deduce $a=a^{-1}$ , $b=b^{-1}$ , etc. Thus,

$ab=b(b^{-1}a)(b^{-1}a)a$

In addition $(b^{-1}a)(b^{-1}a)=(b^{-1}a)^2=e$ , hence $ab=ba$ , and $G$ abelian.

The product group $\mathbb{Z}_2\times \mathbb{Z}_2$ is an example of one abelian group whose order is greater than $2$ and the order of whose elements is $2$ .

April 24, 2020October 24, 2022

202004240606 Homework 1 (Q1)

Let $G=\{ a,b,c,d\}$ and the Cayley table of $G$ be

$\begin{aligned} * \quad | &\textrm{} \quad a &\textrm{}\quad b \quad &\textrm{}\quad c &\textrm{}\quad d \\ \hline a \quad | &\quad a &\quad b \quad &\quad c &\quad d \\ b \quad | &\quad b &\quad a \quad &\quad c &\quad d \\ c \quad | &\quad c &\quad b \quad &\quad a &\quad d \\ d \quad | &\quad d &\quad d \quad &\quad b &\quad c \end{aligned}$

Is $(G,*)$ a group?

Attempts.

I recall that a group $(G,*)$ is a set $G$ altogether with an operation $*:G\times G\rightarrow G$ , $(x,y)\mapsto x*y$ such that the following axioms be satisfied:

i. (Associativity) $(x*y)*z=x*(y*z)$ ;

ii. (Identity) There exists $e\in G$ such that $e*x=x*e=x$ for all $x\in G$ ; and

iii. (Inverse) For each $a\in G$ , there exists $b\in G$ such that $a*b=b*a=e$ .

Let me check them one by one.

i. For $x,y,z\in \{ a,b,c,d \}$ , it suffices to check $4\times 4\times 4=64$ operations. And I found that the following operation failed axiom (i):

$d*(c*b)=d*b=d$ but $(d*c)*b=b*b=a$ .

That said, $G$ under the operation $*$ is not a group.

ii. By inspection, the identity of $G$ exists, and that is $a$ because $a*x=x*a=x$ for all $x\in \{ a,b,c,d\}$ . Axiom (ii) is thus satisfied.

iii. There exists $d\in G$ , such that for all $x^i\in\{ a,b,c,d\}=G$ ,

$d*x^i=x^i*d\neq a$ .

The claim above is validated by simply checking the identity found to be $a$ in part ii. cannot be found in the entries, and by the fact that $a$ is unique.

Axioms i. and iii. having been failed, I may conclude that $G$ under the operation $*$ cannot form a group.

April 24, 2020October 24, 2022

202004240535 Homework 1 (Q4)

What is the effective temperature of a neutron star with radius $10^6\,\mathrm{cm}$ which is detected to emit an energy flux $F=9\times 10^{-15}\,\mathrm{erg\, s^{-1}\, cm^{-2}}$ ? The distance to this neutron star is $1\,\mathrm{kpc}$ .

Setup.

Recall that the effective temperature $T_{\textrm{eff}}$ of a blackbody defines the frequency-integrated intensity of radiation at the source:

$\sigma_{\textrm{SB}}T^4_{\textrm{eff}}=F=\displaystyle{\int}I_\nu\cos\theta\,\mathrm{d}\nu\,\mathrm{d}\Omega$

where $\sigma_{\textrm{SB}}=5.67\times 10^{-5}\,\mathrm{erg\cdot cm^{-2}\, K^{-4}}$ is the Stefan-Boltzmann constant.

Solution. (guesstimate)

The luminosity $L$ of a source is related to the energy flux $F$ detected at some distance $d$ away from the source by

$F=\displaystyle{\frac{L}{4\pi d^2}}$ ,

from which one can infer the luminosity of the neutron star in question is

$\begin{aligned} L & =9\times 10^{-15}\times 4\pi (3.085677581\times 10^{21})^2 \\ & = 1.076845664\times 10^{30}\enspace \mathrm{erg\, s^{-1}} \end{aligned}$

where $d=1\,\mathrm{kpc}=3.085677581\times 10^{21}\,\mathrm{cm}$ is given.

Since $L=4\pi R^2\sigma_{\textrm{SB}}T_{\textrm{eff}}^4$ ,

$\begin{aligned} T^4_{\textrm{eff}} & =L/4\pi R^2\sigma_{\textrm{SB}} \\ T_{\textrm{eff}} & = \bigg(\frac{L}{4\pi R^2\sigma_{\textrm{SB}}}\bigg)^{1/4} \\ & = \bigg(\frac{1.076845664\times 10^{30}}{4\pi \times (10^6)^2\times 5.67\times 10^{-5}}\bigg)^{1/4} \\ T_{\textrm{eff}} & = 197169.6816\enspace \mathrm{K} \end{aligned}$

April 24, 2020October 24, 2022

202004240521 Homework 1 (Q5)

Estimate the apparent magnitude of Mercury and Venus when the Earth-planet-Sun forms a right angle. (Assuming the planet reflects 50% of solar light.)

Setup.

Recall the apparent magnitude $m_b$ , the Sun’s luminosity being the reference, is given by the formula

$m_b=-0.23+5\log D-2.5\log (L/L_{\odot})$ ,

where $D$ is the distance (in parsec) from the source to the Earth, $L$ the luminosity of the source, and $L_\odot =3.8\times 10^{33}\,\mathrm{erg\, s^{-1}}$ the solar luminosity.

CASE I of Mercury

According to http://coolcosmos.ipac.caltech.edu/ask/25-How-far-is-Mercury-from-Earth-, the average distance between Mercury and Earth is 0.0000024954 pc. From Wikipedia on Mercury (planet), the mean radius of Mercury $R_{\textrm{Mercury}}$ is $2439.7\pm 1\,\mathrm{km}$ .

The luminosity of Mercury $L_{\textrm{Mercury}}$ due to reflection of sunlight, is related to the solar luminosity $L_\odot =3.8\times 10^{33}\,\mathrm{erg\, s^{-1}}$ by

$L_{\textrm{Mercury}}=\displaystyle{\frac{\pi R^2_{\textrm{Mercury}}}{4\pi D^2}}L_\odot$ ,

where the Sun-Earth distance $D$ is one astronomical unit $1\,\mathrm{AU}=1.5\times 10^{13}\,\mathrm{cm}$ .

Thence one computes

$\begin{aligned} L_{\textrm{Mercury}} & = \displaystyle{\frac{\pi (2.4397\times 10^8)^2}{4\pi (1.5\times 10^{13})^2}}\times 3.8\times 10^{33} \\ & = 2.513124127\times 10^{23}\enspace \mathrm{erg\, s^{-1}} \end{aligned}$

Thus, the apparent magnitude $m_{b,\textrm{mercury}}$ of Mercury is

$\begin{aligned} m_{b,\textrm{mercury}} & =-0.23+5\log (0.0000024954)-2.5\log (6.613484544\times 10^{-11})\\ & = -2.795375004 \end{aligned}$

CASE II of Venus

According to http://coolcosmos.ipac.caltech.edu/ask/52-How-far-away-is-Venus-from-Earth-, the average distance between Venus and Earth is 0.0000012963 pc. From Wikipedia on Venus, the mean radius of Venus $R_{\textrm{Venus}}$ is $6051.8\pm 1\,\mathrm{km}$ .

The luminosity of Venus $L_{\textrm{Venus}}$ due to reflection of sunlight, is related to the solar luminosity $L_\odot =3.8\times 10^{33}\,\mathrm{erg\, s^{-1}}$ by

$L_{\textrm{Venus}}=\displaystyle{\frac{\pi R^2_{\textrm{Venus}}}{4\pi D^2}}L_\odot$ ,

where the Sun-Earth distance $D$ is one astronomical unit $1\,\mathrm{AU}=1.5\times 10^{13}\,\mathrm{cm}$ .

Thence one computes

$\begin{aligned} L_{\textrm{Venus}} & = \displaystyle{\frac{\pi (6.0518\times 10^8)^2}{4\pi (1.5\times 10^{13})^2}}\times 3.8\times 10^{33} \\ & = 1.546358626\times 10^{24}\enspace \mathrm{erg\, s^{-1}} \end{aligned}$

Thus, the apparent magnitude $m_{b,\textrm{venus}}$ of Venus is

$\begin{aligned} m_{b,\textrm{venus}} & =-0.23+5\log (0.0000012963)-2.5\log (4.069364804\times 10^{-10})\\ & = -6.190288956 \end{aligned}$

Correction.

Without heed to the 50% reduction of solar light reflection, this problem need be redone.

April 24, 2020October 24, 2022

202004240406 Homework 1 (Q6)

Derive the Planck function $B_\nu$ .

HINTS: You can make the following steps.

(1) Show that the density of states for photons is given by

$g(E)=\displaystyle{\frac{\mathrm{d}N}{\mathrm{d}E}=\frac{8\pi VE^2}{h^3c^3}}$ ,

(2) the photon density per unit frequency is given by

$\displaystyle{\frac{\mathrm{d}n}{\mathrm{d}\nu}}=\displaystyle{\frac{8\pi \nu^2}{c^3(e^{h\nu /kT+1})}}$ , and

(3) the definition of the specific intensity $B_\nu =I_\nu$ is the energy flux per unit frequency per unit solid angle.

Note: the energy and momentum of photons are related as

$E=h\nu =pc$ ,

the energy flux per unit frequency is given by

$h\nu \displaystyle{\frac{\mathrm{d}n}{\mathrm{d}\nu}}c$ ,

and total solid angle for isotropic emission is $4\pi$ .

Attempts.

In attempting to derive the Planck function, extensive reference was made to the following several sources found on the Internet:

i. G. B. Rybicki & A. P. Lightman. (1979). Radiative Processes in Astrophysics. John Wiley & Sons, Inc.

ii. https://edisciplinas.usp.br/pluginfile.php/48089/course/section/16461/qsp_chapter10-plank(DOT)pdf

iii. https://en.wikipedia(DOT)org/wiki/Planck\%27s_law#Derivation

iv. http://web.phys.ntnu.no/~stovneng/TFY4165_2013/BlackbodyRadiation(DOT)pdf

Here I follow Rybicki and Lightman’s derivation (pp. 20–22, 1979) direct, and have changed only some wordings.

Solution.

The wave vector of the photon of frequency $\nu$ propagating in direction $\mathbf{n}$ is $\mathbf{k}=(2\pi /\lambda )\,\mathbf{n}=(2\pi\nu /c)\,\mathbf{n}$ . For a photon gas in a box of dimension $(L_x,L_y,L_z)$ , the number of nodes of the standing wave is $n_i=k_iL_i/2\pi$ for each direction $i\in\{ x,y,z\}$ and for some wave number $k_i$ . For all directions, write the variation of the number of nodes with the wave number

$\Delta n_i=\displaystyle{\frac{L_i\Delta k_i}{2\pi}}$

The three-dimensional wave vector element $\Delta k_x\Delta k_y\Delta k_z\equiv \mathrm{d}^3k$ has the number of states to be

$\Delta N=\Delta n_x\Delta n_y\Delta n_z=\displaystyle{\frac{L_xL_yL_z\,\mathrm{d}^3k}{(2\pi )^3}}$

By the fact that $L_xL_yL_z=V$ is the volume of the box and the fact that photons have two independent polarizations (i.e., two states per wave vector $\mathbf{k}$ ), for each 3D-wave vector, the number of states for every unit volume is

$\displaystyle{\frac{\Delta N}{V\,\mathrm{d}^3k}}=2/(2\pi )^3$ .

Using solid angle as in spherical coordinates, rewrite

$\mathrm{d}^3k=k^2\,\mathrm{d}k\,\mathrm{d}\Omega =\displaystyle{\frac{(2\pi )^3\nu^2\,\mathrm{d}\nu\,\mathrm{d}\Omega}{c^3}}$ .

Then, the density of states, i.e., the number of states per solid angle per volume per frequency, is given by

$\rho_s\stackrel{\textrm{def}}{=}\displaystyle{\frac{\mathrm{d}N}{\mathrm{d}\Omega\,\mathrm{d}V\,\mathrm{d}\nu}}=\displaystyle{\frac{2\nu^2}{c^3}}$ .

As each state of $n$ photons each of energy $h\nu$ is of energy $E_n=nh\nu$ , and according to statistical mechanics the probability of a state of energy $E_n$ is proportional to $\mathrm{exp}(-\beta E_n)$ (where $\beta =1/kT$ and $k=\textrm{ the Boltzmann's constant}$ ), the average energy of each state is

$\overline{E}=\displaystyle{\frac{\sum_{n=0}^\infty E_n\mathrm{exp}(-\beta E_n)}{\sum_{n=0}^\infty \mathrm{exp}(-\beta E_n)}}=-\displaystyle{\frac{\mathrm{\partial}}{\mathrm{\partial}\beta}\ln \bigg( \sum_{n=0}^\infty \mathrm{exp}(-\beta E_n)\bigg)}$ .

Recall the formula for the sum of a geometric series:

$\displaystyle{\sum_{n=0}^\infty}\mathrm{exp}(-\beta E_n)=\displaystyle{\sum_{n=0}^\infty}\mathrm{exp}(-nh\nu\beta )=(1-\mathrm{exp}(-\beta h\nu ))^{-1}$ ,

one has therefore the result (in Bose-Einstein statistics)

$\overline{E}=\displaystyle{\frac{h\nu\,\mathrm{exp}(-\beta h\nu )}{1-\mathrm{exp}(-\beta h\nu )}=\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}}$ .

where it can be seen that the occupation number

$n_\nu =\bigg[ \mathrm{exp}\bigg( \displaystyle{\frac{h\nu}{kT}}\bigg) -1\bigg]^{-1}$

means the average number of photons in some frequency $\nu$ , as one energy $h\nu$ corresponds to one frequency $\nu$ .

The energy per solid angle per volume per frequency is the product of i. $\overline{E}$ , the average energy of each state, and ii. $\rho_s$ , the density of states.

I.e., $\overline{E}\cdot \rho_s = \bigg( \displaystyle{\frac{2\nu^2}{c^3}}\bigg) \displaystyle{\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}}$

Define the specific energy density $u_\nu$ the energy per unit volume per unit frequency range. Then the energy density per unit solid angle can be expressed as

$\begin{aligned} u_\nu (\Omega ) & =\displaystyle{\frac{\mathrm{d}E}{\mathrm{d}V\,\mathrm{d}\Omega\,\mathrm{d}\nu}} \\ \dots \textrm{arranging } & \textrm{and equating the previous two equations}\dots \\ u_\nu (\Omega )\,\mathrm{d}V\,\mathrm{d}\nu\,\mathrm{d}\Omega & =\bigg( \displaystyle{\frac{2\nu^2}{c^3}}\bigg) \displaystyle{\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}}\,\mathrm{d}V\,\mathrm{d}\nu\,\mathrm{d}\Omega \\ \dots \textrm{comparing} & \dots \\ u_\nu (\Omega ) & = \bigg( \displaystyle{\frac{2\nu^2}{c^3}}\bigg) \displaystyle{\frac{h\nu}{\mathrm{exp}(h\nu /kT)-1}} \end{aligned}$

Observe that the specific energy density $u_\nu$ and the specific intensity (or brightness) $I_\nu$ are related by

$u_\nu (\Omega )=\displaystyle{\frac{I_\nu}{c}}$ .

Now that the specific intensity, $I_\nu$ , is the same as the source function (of specific emission mechanism), $B_\nu$ .
The frequency distribution is said to be a blackbody form, i.e.,

The Planck function is expressed as

$\boxed{I_\nu =B_\nu (T)=\displaystyle{\frac{2h\nu^3}{c^2}}\bigg[ \mathrm{exp}\bigg( \displaystyle{\frac{h\nu}{kT}} \bigg) -1\bigg]^{-1}}$

(Units: $\mathrm{erg\cdot s^{-1}\cdot Hz^{-1}\cdot cm^{-2}\cdot ster^{-1}}$ )

April 23, 2020October 14, 2023

202004231606 Exercise 1, Section 1.1

Determine whether the vectors emanating from the origin and terminating at the following pair of points are parallel.

(a) $(3,1,2)$ and $(6,4,2)$

(b) $(-3,1,7)$ and $(9,-3,-21)$

(c) $(5,-6,7)$ and $(-5,6,-7)$

(d) $(2,0,-5)$ and $(5,0,-2)$

Background.

Two nonzero vectors $x$ and $y$ are called parallel if $y=tx$ for some nonzero real number $t$ . (Thus nonzero vectors having the same or opposite directions are parallel.)

_{Text on pg.3}

Solution.

(a) Let $x=(3,1,2)$ and $y=(6,4,2)$ . Apparently $\nexists\, t\in \mathbb{R}$ such that $y=tx$ . For otherwise $(6,4,2)=t(3,1,2)$ , the system of equations

$\begin{aligned} 6 & = 3t \\ 4 & = 1t \\ 2 & = 2t \\ \end{aligned}$

is inconsistent. They are not parallel.

(b) Let $x=(-3,1,7)$ and $y=(9,-3,-21)$ , then $y=-3x$ . The vectors $x$ and $y$ are in opposite direction and the magnitude of $y$ is three times that of $x$ . They are parallel.

(c) Let $x=(5,-6,7)$ and $y=(-5,6,-7)$ . Observe that they are in equal magnitude but in opposite direction, i.e., $y=-x$ . They are also parallel.