Physicspupil – Page 33

February 25, 2021August 18, 2025

202102250456 Remarks on “Conceptual Change”

$\textrm{\Huge{\S}}$ 0 Foreword

This article is sheer homework submitted for earning a grade. And it is not intended to survive academic rigor that it may have any contribution to the current literature.

$\textrm{\Huge{\S}}$ 1 Introduction

May the author introduce his readers to the history of conceptual change approach.

The term “conceptual change” originated from The Structure of Scientific Revolutions published by Thomas Kuhn in 1962. Kuhn, first a theoretical physicist and later a scholar in the history of science and philosophy, recorded the scientific revolutionary change that took place in physics at that time—from Newtonian mechanics to quantum mechanics. It was nevertheless the tension beneath that surface that loaned its name— the tension between the Received View $\dagger$ and the Theory Reduction $\ddagger$ (Vosniadou, 2007, pp. 1–2). The idea of “conceptual change” and “paradigm shift” transitioned from the philosophy of science into the science education circle equally appropriately, as did the idea of “gestalt switch” and Piaget’s stages of cognitive development from the study of psychology (Lillian Hoddeson, 2007, pp. 26–27).

(The Received View)

"... [t]he attempts by logical positivists and logical empiricists to treat scientific theories as sets of axioms that could be formulated in mathematical logic."

(The Theory Reduction)

"According to theory reduction, a theory that enjoys a high degree of confirmation cannot ever be disconfirmed, but can only be expanded to a theory with a wider scope, or absorbed into a more inclusive and comprehensive theory (see Suppe, 1977)."

Criticisms brought about to ”conceptual change approach” in teaching science are in many aspects, just to list a few:

“… [C]onceptual change is a slow and gradual process and not a dramatic, gestalt shift that happens over a short period of time.” (Caravita & Hallden, 1994; Vosniadou, 2003; Hatano & Inagaki, 1994; as cited in Vosniadou, 2007, pg. 3)
“… [S]cience learning does not require the replacement of ‘incorrect’ with ‘correct’ conceptions, but the ability on the part of the learner to take different points of view and understand when different conceptions are appropriate depending on the context of use.” (e.g., Pozo, Gomez, & Sanz, 1999; Spada, 1994; as cited in Vosniadou, 2007, pg. 3)
“… [C]ognitive conflict is not a successful instructional strategy for producing conceptual change, as students tend to patch up local inconsistencies in a superficial way.” (Chinn & Brewer, 1993; Smith, diSessa, & Roschelle, 1993; Vosniadou, 1999; as cited in Vosniadou, 2007, pg. 3)

Hoddeson (2007, pg.27) recalled how his PhD teacher Kuhn studied the “conceptual change” as in “Werner Heisenberg’s intellectual route to his formulation of quantum mechanics”. It can be seen that the term “conceptual change” was originally intended to:

$\checkmark$ : make reference to the leap of philosophy of science from the Received View to the Theory Reduction.

$\times$ : it is an intrinsic end, $\textrm{\scriptsize{NOT}}$ an instrumental end, i.e., this approach is descriptive of how science, in particular physics, evolves, but not prescriptive of how science education, inclusive of high school physics, should be attained or designed.

Von Aufschnaiter and Rogge (2010), arguing both theoretically with literature and empirically with their studies on pupils’ conceptual development in physics, posited that pupils are typically lacking in any explanatory conceptual understanding of the content knowledge, and a focus on missing conceptions is much more promising than on misconceptions.

The author couldn’t agree more with all their conclusions, and one of which he cannot overemphasize but ought to restate here is their lately approach—the importance of inclusion of pupils’ misconceptions to teaching is attached to the design of instruction rather than making them an explicit discussion and contrasting them with scientific concepts.

$\textrm{\Huge{\S}}$ 2 Research questions

Andrew A. diSessa (2017, pg. 10) defined “clinical interviewing” to be the technical version of “just talking with people”, and explained how clinical interviewing is connected to instruction. Namely, there are three ways of input.

Input to conjectures and expectations

“Knowing how people think within their zone of felt competence provides important, general and often very specific conjectures about how they can learn from instruction.”

Input to design

“… [S]uch research affects the very goals of instruction (what and when topics should be taught, and how they should be construed), in addition to instructional strategies.”

Input to observation

The clinical interviewer is able to observe “[b]oth the productive and sometimes less productive roles of pre-instructional knowledge”.

Research question

To what extent the argument proposed by Von Aufschnaiter and Rogge, namely, that missing conceptions should be focused prior to misconceptions, is confirmed by a clinical interview with a high school physics pupil?

$\textrm{\Huge{\S}}$ 3 Methodology

One interview was held with a Form 5 pupil, which lasted for eight minutes. A transcript of the interview is in the appendix.

The interviewee had not been taught uniform circular motion before. And the author found it good to have a closed-book talk on this topic with him.

During the interview, there is not any rigid interview protocol, as long as any digression from the subject might still maintain a good understanding between the interviewer and the interviewee.

Delimitation.

Notwithstanding that the interview questions should reveal equally likely misconceptions and missing conceptions, the interviewer must have confirmation bias towards a greater likelihood of missing conceptions.

One single interview alone can demonstrate but that at most one interviewee is able to get along with a pre-lesson discussion on expertise and argumentation. And without following up the interviewee’s learning progress, it cannot give evidence to whether such an axiomatic approach as concerns classical physics is a better alternative to the prevailing conceptual change approach.

Findings of interviews with more interviewees and on more diverse and deeper physics topics are in order for presenting themselves to the current literature.

$\textrm{\Huge{\S}}$ 4 Findings and discussion

Misconception 1: Impetus theory—there exists a force acting upon an object moving with constant velocity

Referring to the transcript from line 25 to 32. The interviewer prompted the interviewee to recall Newton’s first law of motion. The interviewee self-corrected his statement later on though, it might not be an over-extrapolation that he held the impetus theory which is common to physics beginners and laymen.

Such misconception is an instance of ontologically inappropriate miscategorisation of concepts.

Misconception 2: Vector vs. scalar—inability to distinguish speed and velocity.

Referring to the transcript from line 40 to 65. The interviewee was asked whether or not he could differentiate between speed and velocity. Although the reply being affirmative, it can be inferred that he had not mastered the notion of vectors, and could not tell speed and velocity apart.

When debriefed after the interview, he said that he had never studied vectors in mathematics, as the teaching was not then compulsory. What he could tell was by reciting his bookwork, that speed is the rate of change of distance, and that velocity the rate of change of displacement.

Missing conception 1: Centripetal force can be non-contact force.

Referring to the transcript from line 68 to 79. It can be extrapolated that a beginner, when first introduced to the centripetal force acting upon an object in circular motion, might be inclined to think there must be some tension between the centre and the orbiter, and dismiss the fact that centripetal force can account for non-contact force, let alone the fact that it is not a force a priori, but a posteriori the net force.

Missing conception 2: Force vs. pseudo force—distinguishing centripetal and centrifugal forces by inertial or non-inertial frame of reference.

Referring to the transcript from line 97 to 118. It can be observed, that in the scenario when a bus was taking a turn, the interviewee maintained that he should term the force centripetal, despite the fact that its pulling the passenger away from his seat should naturally be termed centrifugal. As an aside, the answer shall depend on which frame—the ground’s or the bus’—is taken reference to, and at this stage it is all too demanding of one high school pupil to tell the difference between inertial and non-inertial frames of reference.

It is worth mentioning, however, that the interviewee admitted that he opted for the term “centripetal” because he had had his supposition that “centrifugal force is a pseudo-force”, based on what disseminated from popular science.

This echoes with that of Ivarsson, Schoultz, and Säljö‘s sociocultural perspective, that human cognition is an interactive process in sociocultural context rather than a purely cognitve one, and “prior knowledge is neither an obstacle nor a prerequisite for conceptual change” (Mayer, 2002, pg. 106).

Misconception 3: The motion of an object is always in the same direction of the net force which it is applied.

Referring to the transcript from line 130 to 152. The interviewer asked the interviewee to consider the situation where there is a centripetal force acting on the satellite by the Earth but the satellite will not clash into the Earth. One reading between the lines could identify there might be one assumption held by the interviewee, that “the motion of an object is the same as that of its net force it experiences”.

The interviewee had previously been taught about electrostatics, and he should have rejected the wrong assumption above by giving a counterexample: the path of a charged particle in an E-field is not necessarily the same as the path of field lines.

Missing conception 3: Relating pre-conceptions to form new ones—from projectile motion to circular motion

Referring to the transcript from line 18 to 20. The interviewee was asked whether he could deduce from definition of uniform circular motion that there must exist a net force acting upon it, still less defining it the centripetal force. It was not until in halfway of the interview that he could understand the axiomatic deduction that come up to its existence.

Referring to the transcript from line 147 to 160. The interviewee could not account for the reason that in the presence of a centripetal force an orbiter and the centre are able to keep their distance. The failure might be mostly due to misconception 3, and partly attributed to an over-compartmentalisation of interrelated physics concepts. That is to say, should he relate an orbiter as a projectile with some initial velocity and a downward constant acceleration perpendicular to its motion, he might be given a clue.

$\textrm{\Huge{\S}}$ 5 Conclusions and implications on teaching

Conceptual change approach has had a long standing reputation for three decades in not only tertiary science teacher education, but teaching high school science also. Care and caution must be taken by educators, to cite Mayer (2002, pg. 127), that it is “… [n]ot a simple process of deletion or replacement of p-prims (i.e., phenomenological primitives), as in contrasting views of conceptual change, but rather a complex process of integration and reorganization”.

The author contends that a pre-instructional talk, chat, or, informal discussion, can help the teacher to design his instruction. Judging misconceptions is with the benefit of hindsight, as searching for missing conceptions is with the benefit of the doubt. Focusing on missing conceptions is indeed more promising than on misconceptions, at least in physics.

Last but not least, as food for thought, if what taught in high school physics is not only classical, but also non-classical (quantum), we have an excuse not to spare the big words—paradigm shift, gestalt switch, conceptual change, and what not. Otherwise, in the senior secondary school physics curriculum, as one in Asian countries, should we not found our education upon normal science, in Kuhn’s terms?

$\textrm{\Huge{\S}}$ 6 Reference

diSessa, A. A. (2017). Knowledge in pieces: an evolving framework for understanding knowing and learning. In Amin, T. G., & Levrini, O. (Eds.). (2017). Converging perspectives on conceptual change: Mapping an emerging paradigm in the learning sciences. Routledge.

Hoddeson, L. (2007). In the wake of Thomas Kuhn’s theory of scientific revolutions: The perspective of an historian of science. In Vosniadou, S., Baltas, A., & Vamvakoussi, X. (Eds.). (2007). Re-framing the conceptual change approach in learning and instruction. Elsevier Ltd.

Mayer, R. E. (2002). Understanding conceptual change: a commentary. In Limón, M. & Mason, L. (Eds.). (2002). Reconsidering conceptual change: issues in theory and practice. Kluwer Academic Publishers

Von Aufschnaiter, C., & Rogge, C. (2010). Misconceptions or missing conceptions?. Eurasia Journal of Mathematics, Science & Technology Education, 6(1).

Vosniadou, S. (2007). The conceptual change approach and its re-framing. In Vosniadou, S., Baltas, A., & Vamvakoussi, X. (Eds.). (2007). Re-framing the conceptual change approach in learning and instruction. Elsevier Ltd.

$\textrm{\Huge{\S}}$ 7 Appendix

Please turn this page over and see Transcript of Interview attached thereafter.

(Submitted on Saturday, 4 May, 2019)

February 19, 2021October 24, 2022

202102190301 Homework 1 (Q3)

Find the analytic functions $w(z)=u(x,y)+\mathrm{i}v(x,y)$ if

(a) $u(x,y)=x^3-3xy^2$ ;

(b) $v(x,y)=e^{-y}\sin x$ .

Attempts.

(a)

$\displaystyle{\frac{\partial u}{\partial x}}=3x^2-3y^2$ .

From (first of the Cauchy-Riemann equations)

$\displaystyle{\frac{\partial u}{\partial x}}=\displaystyle{\frac{\partial v}{\partial y}}$ ,

we have

$v=\displaystyle{\int} (3x^2-3y^2)\,\mathrm{d}y+g(x)$

where $g$ is a function independent of $y$ . Then

$v=3x^2y-y^3+g(x)$ .

Now

$\displaystyle{\frac{\partial v}{\partial x}}=6xy+g'(x)$ ;

$\displaystyle{\frac{\partial u}{\partial y}}=-6xy$

From (second of the Cauchy-Riemann equations)

$\displaystyle{\frac{\partial u}{\partial y}}=-\displaystyle{\frac{\partial v}{\partial x}}$ ,

we have

$-6xy=-6xy-g'(x)$

i.e.,

$g'(x)=0$ ;

$g(x)=C$ .

Thus $v(x,y)=3x^2y-y^3+C$ for some constant $C\in\mathbb{R}$ .

Part (b) is left to the reader.

February 19, 2021October 24, 2022

202102190222 Homework 1 (Q2)

Show whether or not the function (a) $f(z)=\mathrm{Re}(z)=x$ ; (b) $f(z)=z^2$ is analytic.

Setup.

Theorem 1. Suppose $f:\Omega \rightarrow \mathbb{C}$ is analytic. Then, writing

$f(z)=f(x+\mathrm{i}y)=u(x,y)+\mathrm{i}v(x,y)$

with $u(x,y)=\mathrm{Re}(f(z))$ and $v(x,y)=\mathrm{Im}(f(z))$ , we have the Cauchy-Riemann equations

$\begin{cases} \displaystyle{\frac{\partial u}{\partial x}} & = \displaystyle{\frac{\partial v}{\partial y}} \\ \displaystyle{\frac{\partial u}{\partial y}} & = \displaystyle{-\frac{\partial v}{\partial x}} \end{cases}$

everywhere on $\Omega$ .

Theorem 2. If $f:\Omega\rightarrow \mathbb{C}$ is of class $\mathcal{C}^1$ , $f=u+\mathrm{i}v$ . Then $(u,v)$ satisfies the Cauchy-Riemann equations if and only if $f$ is analytic.

(a) Now that

$u(x,y)=x$ and $v(x,y)=0$ ,

the partial derivatives are as follows

$\begin{aligned} \frac{\partial u}{\partial x} & = 1 & \qquad \frac{\partial u}{\partial y} = & 0 \\ \frac{\partial v}{\partial y} & = 0 & \qquad -\frac{\partial v}{\partial x} = & 0 \\ \end{aligned}$

Since

$1=\displaystyle{\frac{\partial u}{\partial x}}\neq \displaystyle{\frac{\partial v}{\partial y}}=0$ ,

the Cauchy-Riemann equations are not satisfied, and $f$ is $\textrm{\scriptsize{NOT}}$ analytic.

(b)

$\begin{aligned} f(z) & = z^2 \\ f(x+\mathrm{i}y) & = (x+\mathrm{i}y)^2 \\ & = (x^2-y^2)+\mathrm{i}(2xy) \\ & = u(x,y)+\mathrm{i}v(x,y) \end{aligned}$

where $u(x,y)=x^2-y^2$ and $v(x,y)=2xy$ .

Then,

$\displaystyle{\frac{\partial u}{\partial x}} = 2x$ , $\displaystyle{\frac{\partial u}{\partial y}} = -2y$ , $\displaystyle{\frac{\partial v}{\partial y}} = 2x$ , $\displaystyle{-\frac{\partial v}{\partial x}} = -2y$

and the Cauchy-Riemann equations are satisfied. $f$ is thus analytic.

February 19, 2021October 24, 2022

202102190159 Homework 1 (Q1)

(a) Find the reciprocal of $x+\mathrm{i}y$ , working entirely in the Cartesian representation.

(b) Repeat part (a), working in polar form but expressing the final result in Cartesian form.

Solution.

(a)

$\begin{aligned} \frac{1}{x+\mathrm{i}y} & = \bigg( \frac{1}{x+\mathrm{i}y}\bigg) \bigg( \frac{x-\mathrm{i}y}{x-\mathrm{i}y} \bigg) \\ & = \frac{x-\mathrm{i}y}{x^2-(\mathrm{i}y)^2} \\ & = \frac{x-\mathrm{i}y}{x^2+y^2} \\ & = \bigg( \frac{x}{x^2+y^2} \bigg) - \mathrm{i}\bigg( \frac{y}{x^2+y^2} \bigg) \end{aligned}$

(b)

Let

$x+\mathrm{i}y\stackrel{\mathrm{def}}{=}z=re^{\mathrm{i}\varphi}$

where $r=\sqrt{x^2+y^2}$ and $\varphi = \arctan\bigg( \displaystyle{\frac{y}{x}} \bigg)$ .

Then, the reciprocal of $z=x+\mathrm{i}y$ is

$\begin{aligned} \frac{1}{z} & = \frac{1}{re^{\mathrm{i}\varphi}} \\ & = \frac{1}{r}\, e^{\mathrm{i}(-\varphi )} \\ & = \frac{1}{r}\, \mathrm{cis}(-\varphi )\\ & = \frac{1}{r}\big(\cos (-\varphi)+\mathrm{i}\sin (-\varphi )\big)\\ & = \frac{1}{r}(\cos\varphi -\mathrm{i}\sin\varphi ) \\ & = \frac{1}{\sqrt{x^2+y^2}}\bigg[ \frac{x}{\sqrt{x^2+y^2}} - \mathrm{i}\Big( \frac{y}{\sqrt{x^2+y^2}} \Big) \bigg] \\ & = \bigg( \frac{x}{x^2+y^2} \bigg) - \mathrm{i}\bigg( \frac{y}{x^2+y^2} \bigg) \end{aligned}$

January 14, 2021October 24, 2022

202101141547 Homework 1 (Q5)

Let $G=\mathbb{Z}_2$ .

(a) Describe all the elements in $\mathrm{Perm}(G)$ and the maps $l_a$ for $a\in \mathbb{Z}_2$ . Is $l$ surjective? (Use the notation $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ to denote the map sending from $0$ to $1$ and from $1$ to $0$ .)

(b) Repeat part (a) for the group $G=\mathbb{Z}_3$ .

Setup.

$\textrm{Perm}(X)$ is the set of all bijective maps from $X$ to itself. And the group $(\textrm{Perm}(X),\circ )$ is the group where $\circ$ is the composition of maps.

$l:G\rightarrow \textrm{Perm}(G),\enspace a\mapsto l_a\qquad \forall\, a\in G$

is the group embedding of $G$ into $\textrm{Perm}(G)$ , where

$l_a:G\rightarrow G,\enspace g\mapsto ag\qquad g\in G$

is the left translation of $G$ .

(a)

Thus, the elements of $\textrm{Perm}(\mathbb{Z}_2)$ are altogether the following bijections:

$\begin{aligned} & \\ [0] & \mapsto [0] \\ [1] & \mapsto [1] \end{aligned}$

and

$\begin{aligned} & \\ [0] & \mapsto [1] \\ [1] & \mapsto [0] \end{aligned}$ ,

or equivalently, in matrices,

$\begin{pmatrix} 0 & 0 \\ 1 & 1 \\ \end{pmatrix}$ and $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$

The left translation $l_0$ is

$l_0:\mathbb{Z}_2\rightarrow\mathbb{Z}_2,\enspace g\mapsto 0g\qquad g\in\{ 0,1\}= \mathbb{Z}_2$

But $[0]+[0]=[0]\quad (\mathrm{mod}\,2)$ and $[0]+[1]=[1]\quad (\mathrm{mod}\,2)$ ,
hence in matrix $l_0$ is $\begin{pmatrix} 0 & 0 \\ 1 & 1 \\ \end{pmatrix}$ .

The left translation $l_1$ is

$l_1:\mathbb{Z}_2\rightarrow\mathbb{Z}_2,\enspace g\mapsto 1g\qquad g=\{ 0,1\}\in \mathbb{Z}_2$

But $[1]+[0]=[1]\quad (\mathrm{mod}\,2)$ and $[1]+[1]=[0]\quad (\mathrm{mod}\,2)$ ,
hence in matrix $l_1$ is $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ .

It is hence checked that the range of the left translations $l_0$ , $l_1$ is inside (or, more precisely, equal to) the codomain $\textrm{Perm}(\mathbb{Z}_2)$

Then, I can say $l:\mathbb{Z}_2\rightarrow\mathrm{Perm}(\mathbb{Z}_2)$ , the mappings of which are explicitly the following two:

$0\mapsto \begin{pmatrix} 0 & 0 \\ 1 & 1 \\ \end{pmatrix}$ and $1\mapsto \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$ ;

by the definition $l(a): a\mapsto l_a\qquad \forall\, a\in\{ [0],[1]\}=\mathbb{Z}_2$ .

Recall that a function is said to be surjective if the range of function constitutes its whole codomain. Thus, $l$ is surjective.

(b)

Let $G=\mathbb{Z}_3(=\{ [0],[1],[2] \})$ .

All the one-to-one mappings are listed as follows:

$\textrm{Perm}(\mathbb{Z}_3)=\Bigg\{ \begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \end{pmatrix},\begin{pmatrix} 0 & 0 \\ 1 & 2 \\ 2 & 1 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ 2 & 2 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 2 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 0 \\ 2 & 1 \\ \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 1 \\ 2 & 0 \\ \end{pmatrix}\Bigg\}$ .

First, the left translation $l_0$ is

$l_0:\mathbb{Z}_3\rightarrow\mathbb{Z}_3,\enspace g\mapsto 0g\qquad g\in\{ 0,1,2\}=\mathbb{Z}_3$

Since $[0]+[0]=[0]\quad (\mathrm{mod}\,3)$ , $[0]+[1]=[1]\quad (\mathrm{mod}\,3)$ , and $[0]+[2]=[2]\quad (\mathrm{mod}\,3)$ , I may write

$l_0=\begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \end{pmatrix}$ .

Secondly, the left translation $l_1$ is

$l_1:\mathbb{Z}_3\rightarrow\mathbb{Z}_3,\enspace g\mapsto 1g\qquad g\in\{ 0,1,2\}=\mathbb{Z}_3$

Since $[1]+[0]=[1]\quad (\mathrm{mod}\,3)$ , $[1]+[1]=[2]\quad (\mathrm{mod}\,3)$ , and $[1]+[2]=[0]\quad (\mathrm{mod}\,3)$ , I may write

$l_1=\begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 2 & 0 \\ \end{pmatrix}$ .

Thirdly and lastly, the left translation $l_2$ is

$l_2:\mathbb{Z}_3\rightarrow\mathbb{Z}_3,\enspace g\mapsto 2g\qquad g\in\{ 0,1,2\}=\mathbb{Z}_3$

Since $[2]+[0]=[2]\quad (\mathrm{mod}\,3)$ , $[2]+[1]=[0]\quad (\mathrm{mod}\,3)$ , and $[2]+[2]=[1]\quad (\mathrm{mod}\,3)$ , I may write

$l_2=\begin{pmatrix} 0 & 2 \\ 1 & 0 \\ 2 & 1 \\ \end{pmatrix}$ .

To sum up what is up till now, $l_a=\Bigg\{ \begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 2 & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 2 \\ 1 & 0 \\ 2 & 1 \\ \end{pmatrix}\Bigg\}$

Then, I can say $l:\mathbb{Z}_3\rightarrow\mathrm{Perm}(\mathbb{Z}_3)$ , the mappings of which are explicitly the following three:

$0\mapsto \begin{pmatrix} 0 & 0 \\ 1 & 1 \\ 2 & 2 \\ \end{pmatrix}$ , $1\mapsto \begin{pmatrix} 0 & 1 \\ 1 & 2 \\ 2 & 0\\ \end{pmatrix}$ , and $2\mapsto\begin{pmatrix} 0 & 2 \\ 1 & 0 \\ 2 & 1 \\ \end{pmatrix}$ .

by the definition $l(a): a\mapsto l_a\qquad \forall\, a\in\{ [0],[1],[2]\}=\mathbb{Z}_2$ .

$l$ in the case of $\mathbb{Z}_3$ is $\textrm{\scriptsize NOT}$ surjective because in the codomain $\mathbb{Z}_3$ , the following three elements

$\begin{pmatrix} 0 & 0 \\ 1 & 2 \\ 2 & 1 \\ \end{pmatrix}$ , $\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ 2 & 2 \\ \end{pmatrix}$ , and $\begin{pmatrix} 0 & 2 \\ 1 & 1 \\ 2 & 0 \\ \end{pmatrix}$

are not inside the range of $l$ .

January 14, 2021May 4, 2023

202101141538 Homework 1 (Q4)

$\phi :G_1\rightarrow G_2$ is said to be a group embedding if it is an injective group homomorphism.

For a group $G$ and $a\in G$ , we define the left translation of $G$ by $a$

$l_a:G\rightarrow G$ , $g\mapsto ag,\qquad g\in G$ .

Thus $l$ is a mapping from a value into a function of the value.

I would like to first check that $l$ is a homomorphism,

i.e., $l(a_1a_2)=l(a_1)l(a_2)$ .

$\begin{aligned} \textrm{LHS} & = l(a_1a_2)\\ & =l_{a_1a_2}(g)\qquad\qquad (g\in G)\\ & = a_1a_2g\qquad\qquad (a_1,a_2\in G)\\ & = l_{a_1}(l_{a_2}(g))\\ & =(l_{a_1}\circ l_{a_2})(g)\qquad\qquad (\circ\,\textrm{as the operation in }\textrm{Perm}(G))\\ & = (l_{a_1}l_{a_2})(g)\\ & = l(a_1)l(a_2)\\ & = \textrm{RHS}\\ \end{aligned}$

$\therefore$ $l$ is a homomorphism.

Secondly, I would like to check that $l$ is injective.

If $l(a_1)\neq l(a_2)$ , or $l_{a_1}(g)\neq l_{a_2}(g)$ , or, $a_1g\neq a_2g$ for $a_1,a_2, g\in G$ , then $a_1gg^{-1}\neq a_2gg^{-1}$ guaranteed that the inverse of $g$ always exists in $G$ . It follows that $a_1\neq a_2$ .

$\therefore$ $l$ is injective.

In conclusion, $l$ is a group embedding.

January 14, 2021May 20, 2024

202101141527 Homework 2 (Q3)

Let $G$ be a group and $e\neq a\in G$ in which $e$ is the identity of $G$ . Suppose $\mathrm{ord}(a)=n$ .

i. If $a^h=e$ , show that $n\big| h$ .

ii. Evaluate the order of $a^m$ where $m$ is an integer.

iii. Show that $|H|=n/(m,n)$ if $G=\langle a\rangle$ and $H=\langle a^m\rangle$ where $m\in\mathbb{N}$ .

iv. If G is cyclic of order $n$ , show that for any $d\big| n$ , there exists a subgroup of order $d$ in $G$ . Could $G$ have more than one subgroup of order $d$ ? (Clearly the answer is no if $d=1$ or $n$ .) Justify with a proof (if you think yes) or give a counterexample (if you think no).

Attempts.

i. By Division Algorithm, $h=qn+r$ for some $q\in\mathbb{Z}$ and $0\le r<n$ . Then one may write $a^r=a^{h-qn}=a^h(a^n)^{-q}=e$ . It can be seen that $r$ must be $0$ lest $a^r=e$ where $1\le r<n$ contradicts to the assumption that $n=\mathrm{ord}(a)$ where $n$ is the smallest positive integer such that $a^n=e$ . Thus $h=qn$ , and $n|h$ .

ii. First, $(a^m)^{\frac{n}{(n,m)}}=a^{\frac{mn}{(n,m)}}=(a^n)^{\frac{m}{(n,m)}}=e^{\frac{m}{(n,m)}}=e$ . Secondly, if $k\in\mathbb{N}$ such that $(a^m)^k=e$ , from the result of part i. previously, it follows that $\frac{n}{(n,m)}|k$ because if $a|bc$ and $(a,b)=1$ , then $a|c$ (here $a=\frac{n}{(n,m)}$ ), $b=\frac{m}{(n,m)}$ , and $c=k$ . Such a $k$ will always be greater than or equal to $\frac{n}{(n,m)}$ . Thus $\mathrm{ord}(a^m)=\frac{n}{(n,m)}$ .

iii. If $H=\langle a^m\rangle$ then $|H|=\mathrm{ord}(a^m)\stackrel{\textrm{by (ii)}}{=}\frac{n}{(n,m)}$ .

iv. Let $G=\langle a\rangle$ as it is cyclic. Let $m=n/d$ where $d|n$ is given in the problem. The order of the cyclic subgroup $\langle a^m\rangle$ of $G$ is then $\frac{n}{(n,m)}=\frac{n}{m}=d$ , in view of the result in part iii. We have constructed a subgroup of order $d$ in $\langle a\rangle$ where $d|n$ .

We wish to prove that it is unique: If $H$ is a subgroup of $G=\langle a\rangle$ and $|H|=d$ , then $H=\langle a^{\frac{n}{d}}\rangle$ . Note that $H=\langle a^m\rangle$ and $\frac{n}{(n,m)}=d$ . Consider the subgroup $\langle a^{\frac{n}{d}}\rangle$ . Because $\frac{n}{d}=(n,m)$ divides $m$ , we have $a^m\in \langle a^{\frac{n}{d}}\rangle$ and thus $H\subset \langle a^{\frac{n}{d}}\rangle$ . From the result of part iii. it follows that $|\langle a^{\frac{n}{d}}\rangle | =\frac{n}{(n,\frac{n}{d})}=d$ . Thus $H=\langle a^{\frac{n}{d}}\rangle$ is unique.

January 14, 2021October 24, 2022

202101141501 Homework 2 (Q1)

Let $H$ be a subset of the group $G$ . Show that $H$ is a subgroup of $G$ if and only if $H\neq \emptyset$ and $gh^{-1}\in H$ whenever $g,h\in H$ .

Attempts.

(only-if part) If $H$ is a subgroup of $G$ , then $H$ must contain the identity $e$ and both $x^{-1}$ and $xy$ is inside $H$ whenever $x,y\in H$ . Thus $H\neq \emptyset$ and for any $g,h\in H$ , $h^{-1}$ is inside $H$ . Thus $gh^{-1}\in H$ .

(if-part) First, because $H\neq \emptyset$ , there exists some $x$ in $H$ such that $e=xx^{-1}\in H$ (by the condition $gh^{-1}\in H$ whenever $g,h\in H$ ). This proves the existence of the identity. Secondly, having proven $e$ is in $H$ , and noted that $x^{-1}=ex^{-1}\in H$ whenever $e,x\in H$ , one proves the existence of the inverse. Thirdly, $x,y\in H$ implies $xy\in H$ because $y\in H\Rightarrow y^{-1}\in H$ and also $xy=x(y^{-1})^{-1}$ . $H$ is therefore a subgroup of $G$ .

January 14, 2021October 24, 2022

202101141227 Exercise 1 (Q10)

If $f(x)=2x^2$ , find expressions for:

1. $f(x+h)$ .

2. $f(x+h)-f(x)$ .

3. $\displaystyle{\frac{f(x+h)-f(x)}{h}}$ .

Solution.

1.

$\begin{aligned} f(x+h) & = 2(x+h)^2 \\ & = 2(x^2+2hx+h^2) \\ & = 2x^2+4hx+2h^2 \end{aligned}$

2.

$\begin{aligned} f(x+h)-f(x) & = 2(x+h)^2 - 2(x)^2 \\ & = 2(x^2+2hx+h^2) - 2x^2 \\ & = 4hx+2h^2 \\ \end{aligned}$

3.

$\begin{aligned} \frac{f(x+h)-f(x)}{h} & = \frac{2(x+h)^2-2(x)^2}{h} \\ & = \frac{2(x^2+2hx+h^2)-2x^2}{h} \\ & = \frac{4hx+2h^2}{h} \\ & = 4x+2h \\ \end{aligned}$

Remark.

The first step to derivation from first principles, is calculate the slope of tangent

$\displaystyle{\frac{f(x+h)-f(x)}{h}}$ ,

and put the limit $h\to 0$ , i.e.,

$\begin{aligned} f'(x) & =\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}x}} \\ & = \displaystyle{\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}} \end{aligned}$

a right-hand limit.

Afterword.

$\begin{aligned} f(x) & =2x^2 \\ f'(x) & = \frac{\mathrm{d}f}{\mathrm{d}x} \\ & = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \\ & \stackrel{\textrm{(3)}}{=} \lim_{h\to 0} 4x+2h \\ & = 4x+2(0) \\ & = 4x \end{aligned}$

November 11, 2020July 25, 2022

202011110822 Short Review II (Heat and Internal Energy)

Internal Energy

Each matter has three states:

$\begin{tabular}{p{1cm}|p{4cm}|p{5cm}|p{4cm}} & Property & Molecular arrangement & Molecular movement \\ \hline \hline Solid & fixed volume and shape & closely packed and fixed in positions & vibrate about a fixed position \\ \hline Liquid & fixed volume, unfixed shape & closely packed and not fixed in positions & move freely \\ \hline Gas & unfixed volume and shape & far apart and not fixed in positions & move freely at high speed\\ \hline \end{tabular}$

Each object contains energy, measured in joules ( $\textrm{J}$ ).

Kinetic energy () of a body is the energy due to its motion.
- The average kinetic energy is the kinetic energy of each molecule.
- The total kinetic energy is the kinetic energy of all the molecules.
- $\mathrm{Total\,KE}=\mathrm{no.\,of\,molecules}\times \mathrm{average\,KE}$
Potential energy ( $\textrm{PE}$ ) of a body is the energy due to its position or configuration.
Internal energy () of a body is the sum of the (total molecular) kinetic energy (KE) and (total molecular) potential energy (PE) of all its molecules. It is the total energy stored in a body.
- $\mathrm{Internal\,energy}=\mathrm{Total\,KE}+\mathrm{Total\,PE}$

Temperature,

in common sense, is a measure of hotness;
in scientific sense, is a measure of the average kinetic energy of a body.

Heat $Q$ is the energy transferred from a hot object to a cold object as a result of temperature difference. Its unit is also joule ( $\textrm{J}$ ).

REMARK. It would be to speak of heat as a form of energy because by definition, heat is the energy transferred , or, the change in internal energy.

Power $P$ is the rate of transferring energy, and it is measured in watts ( $\textrm{W}$ ).

$\mathrm{Power}=\displaystyle{\frac{\mathrm{energy\,transferred}}{\mathrm{time}}}\qquad\qquad \bigg( P=\frac{Q}{t}\bigg)$

REMARK. If energy is transferred not only by heating (e.g. but also by doing work ), we may not assume and we may not write .

Specific Heat Capacity

Specific heat capacity $c$ of a substance is the energy $Q$ transferred by heating in order to raise the temperature of the substance of mass $m=1\,\mathrm{kg}$ by $\Delta T=1\,\mathrm{C^\circ }$ .

By making an experiment, we can find out the specific heat capacity $c$ of a substance of mass $m$ by measuring how much energy $Q$ is needed when we increase its temperature by an arbitrary degree $\Delta T$ :

$\displaystyle{c=\frac{Q}{m\Delta T}}$

$\textrm{Unit:}\quad \mathrm{J\,kg^{-1}\,C^{\circ -1}}$

Or, given the specific heat capacity $c$ and the mass $m$ of an unknown substance, and having decided to what degree $\Delta T$ we wish to raise its temperature, we can figure out how much energy $Q$ is needed:

$Q=mc\Delta T$

REMARK.

Heat capacity is denoted by a capital , defined . Heat capacity is related to specific heat capacity by or .

When an object is measured to be of a certain temperature, the unit we say degree Celsius or degree Fahrenheit. When an object has increased or decreased by a certain degree, the temperature change/difference we say is of certain Celsius degree(s) or Fahrenheit degree(s).

When two objects of different temperatures are in contact, energy is transferred from the hotter one to the colder one, until they reach the same temperature, in which there is no $\textrm{\scriptsize NET}$ flow of heat, i.e., thermal equilibrium.

Before the objects reach the thermal equilibrium, the energy lost by the hotter object will be gained by the colder object if there is no loss of energy to the surroundings

because the law of conservation of energy states that:

The total amount of energy in a closed system is conserved, i.e., it is always kept constant. Energy cannot be created or destroyed, although it can change from one form into another.

CONCEPT TEST (Internal Energy)

Which of the following statement(s) is/are correct?
1. Temperature is a measure of the internal energy of an object.
2. Heat is the average kinetic energy of an object.
3. Heat is the energy gained by an object as a result of heat flow.
  1. (I) only
  2. (III) only
  3. (I) and (II) only
  4. (II) and (III) only
Which of the following deductions must be correct?
1. Average KE $\uparrow$ $\Rightarrow$ Total KE $\uparrow$ .
2. Internal energy $\downarrow$ $\Rightarrow$ Total KE $\downarrow$ .
3. Total PE internal energy .
  1. (I) only
  2. (II) only
  3. (I) and (III) only
  4. (I), (II) and (III)
Which of the following statements is/are wrong?
1. Heat is a description of the stored energy in a body.
2. Temperature is a measure of the total KE of a body.
3. Total KE Average KE .
  1. (I) only
  2. (I) and (II) only
  3. (II) and (III) only
  4. (I), (II) and (III)
Which of the following is correct?
1. Molecules in random motion possess molecular kinetic energy.
2. The molecules of a gas are very far apart, so that they have less intermolecular potential energy than when they are in liquid or solid state.
3. The molecules of a liquid move at a faster speed than that of a gas.
4. The molecules of a solid vibrate about fixed positions, so that they have no molecular kinetic energy.

Answers:

Explanation:

(I) is wrong: temperature is a measure of the average kinetic energy. (II) is wrong: heat is the energy transferred from a hotter object to a colder object as a result of temperature difference. (III) is correct.
(I) is correct: total KE $=$ no. of molecules $\times$ average KE. (II) is wrong: a decrease in internal energy may be due to $\textrm{{\scriptsize EITHER}}$ a decrease in total KE, $\textrm{{\scriptsize OR}}$ a decrease in total PE. (III) is correct.
(I) and (II) are wrong by definition. (III) is wrong: total KE $\uparrow$ $\Rightarrow$ no. of molecules $\uparrow$ $\textrm{{\scriptsize OR}}$ average KE $\uparrow$ .
A is correct: molecular KE is related to the random motion of molecules. B is wrong: the molecules of a gas have the farthest separation between them, hence the greatest PE to overcome the attractive intermolecular forces. C is wrong: it should be the converse. D is wrong: when the molecules vibrate, there is molecular KE.

CONCEPT TEST (Specific Heat Capacity)

Which of the following statements is/are correct?
1. Heat capacity $C$ is a property of a substance.
2. Heat capacity $C$ depends on the mass of substance.
3. Specific heat capacity is a property of a substance.
  1. (I) and (II)
  2. (I) and (III)
  3. (II) and (III)
  4. (I), (II) and (III)
Which of the following statements is/are incorrect?
1. Specific heat capacity of water will be changed if it is mixed with salt.
2. Specific heat capacity $c$ can be negative.
3. Specific heat capacity of a substance is the energy that is released to lower the temperature of of the substance by .
  1. (II) only
  2. (III) only
  3. (I) and (II) only
  4. (II) and (III) only
Which of the following statements is incorrect?
1. When object $A$ is in thermal equilibrium with object $B$ , and $B$ is in thermal equilibrium with object $C$ , then $A$ and $C$ are in thermal equilibrium.
2. When two bodies are in thermal equilibrium, there is no heat flow between them.
3. Heat flows spontaneously from a hotter region to a colder region.
4. All liquid-in-glass thermometers have systematic errors because in reaching thermal equilibrium with the object to measure, the thermometer must have transferred some heat to/from the object.

Answers:

Explanation:

(I) is wrong; (II) correct: since heat capacity of a substance varies with its mass, it is not a (specific) property of the substance. (III) is correct.
(I) is correct: specific heat capacity depends on the material; when water is mixed with salt, the material is changed. (II) Specific heat capacity must be positive: as $\textrm{{\scriptsize EITHER}}$ $Q,\Delta T<0$ $\textrm{{\scriptsize OR}}$ $Q,\Delta T>0$ , so that $c=\displaystyle{\frac{Q}{m\Delta T}}>0$ . (III) is correct by definition.
B is incorrect because heat can be flowing from one body to another even though the net flow of heat is zero.