February 21, 2019 – Physicspupil

February 21, 2019October 24, 2022

201902210854 Short Review I (Electric Current)

Definition.

An electric current is a flow of electric charge (through a conductor). We express electric current $I$ by the total electric charge $Q$ flowing through a cross-sectional area per unit time $t$ .

$\displaystyle{I=\frac{Q}{t}}$

One unit current is of one ampere ( $1\,\mathrm{A}$ ).

$\displaystyle{[\mathrm{current}]=\frac{[\mathrm{charge}]}{[\mathrm{time}]}}$ , so $1\,\mathrm{A}=1\,\mathrm{C\,s^{-1}}$ .

Concept Test

An electric current can result from
1. the movement of atoms.
2. the movement of electrons.
3. the simultaneous movement of positive charges and electrons.
  1. I only
  2. III only
  3. I and II only
  4. II and III only
A direct current (d.c.) of flows through a wire. How much charge passes through the wire in 30 minutes? And how many electrons flow through it?
1. $24\,\mathrm{C}$ ; $3.84\times 10^{-18}$ electrons
2. $24\,\mathrm{C}$ ; $1.5\times 10^{20}$ electrons
3. $1440\,\mathrm{C}$ ; $2.304\times 10^{-16}$ electrons
4. $1440\,\mathrm{C}$ ; $9\times 10^{21}$ electrons
Suppose, on average, passengers arrive at Central station from Chai Wan station by MTR every hour. The journey takes up minutes, with average speed . The average train frequency is one every minutes. Which of the following statements is/are correct?
1. The total number of passengers on this journey by MTR at any instant is $1900$ .
2. $3800$ passengers arriving in Central per hour is analogous to a current passing through a cross-sectional area.
3. Average train speed $70\,\mathrm{km/h}$ is analogous to the current in a conducting wire segment.
4. The average train frequency is analogous to the current flowing through a circuit.
1. I and II only
2. I and III only
3. II and III only
4. III and IV only

Answers:

Explanation:

(I) is wrong because atoms are neutral and do not carry charge. Movement of atoms is not a flow of charge, viz. current. (II) is correct because electrons are charge-carriers. (III) is correct. An example is electrolytes with positive ions and negative ions as charge-carriers.
Charges passing through the wire $Q=It=0.8\times 30\times 60=1440\,\mathrm{C}$ (Caution: use SI-unit). An electron $e$ has $1.6\times 10^{-19}\,\mathrm{C}$ of negative charge. So $1\,\mathrm{C}$ of charge consists in $\displaystyle{\frac{1}{1.6\times 10^{-19}}}=6.25\times 10^{18}$ electrons. Thus $1440\,\mathrm{C}$ corresponds to $1440\times 6.25\times 10^{18}=9\times 10^{21}$ electrons. Remark: Options A and C do not make sense, how can an electron be split into pieces?
Draw an analogy like this: (i) trains $\sim$ charge carriers; (ii) passengers $\sim$ charge; and (iii) railway $\sim$ circuit. (2) is correct because current $I=\displaystyle{\frac{Q}{t}}\sim \displaystyle{\frac{3800\mathrm{\,passengers}}{1\,\mathrm{hr}}}$ . (I) is correct because the total number of passengers on board a train at any instant is $3800\times$ time needed for one journey $=3800\times 0.5=1900$ . (III) is wrong because the average train speed $\sim$ the flow of charge carriers $\neq$ the flow of charge. (IV) is wrong because average train frequency $\sim$ density of charge carriers (i.e., electrical conductivity) $\neq$ the current.

Current Direction

By definition, current is the flow of charges (e.g., carried by electrons).

(Charge carriers can be positive, e.g., holes in semi-conductors, positive ions in electrolytes.)

In nature, current is due to the flow of negative electrons from the negative (-ve) terminal to the positive (+ve) terminal of a power source.

An old convention, which is wrong, lasts to date:

Conventional electric current is a flow of positive charge from the +ve terminal to the -ve terminal of a battery.

Conventional current direction is opposite to the direction of electron flow.

External_circuit_direction_of_potential_current_electrons

Measuring Current

By ammeter (also by a current sensor with data-logger, or by galvanometer, which is used for large current, large voltage, and any resistance, like a multimeter), connected in series to a component in a circuit. An ideal ammeter should have zero resistance.

Current in Series and Parallel Circuits

In a series circuit, the current is the same at all points.

In a parallel circuit, the sum of currents passing through each branch is equal to the current in the main circuit.

February 21, 2019January 27, 2023

201902210410 Short Review I (Electric Charge)

Electrostatics is the study of charges at rest.

Two kinds of charge

1. There are two kinds of charge—the positive charge and the negative charge.
2. Like charges repel; unlike charges attract.
3. Since neutrons have no charge, it is said to be neutral.
4. When an object is neither positively nor negatively charged, it is also said to be (electrically) neutral.

How to charge an object?

By (a) rubbing/friction, by (b) inducing charges, or by (c) contact/sharing, there is always a transfer of free charges.

(Capacitors, made up of conductors, can be charged by Extra-High Tension (EHT) power supply.)

Where do charges come from?

Only when an object gains electrons it becomes negatively charged; only when an object loses electrons it becomes positively charged.

Conservation of charge: Charge cannot be created or destroyed.

Unit of charge

The unit is coulomb, written as $\mathrm{C}$ . The charges of a proton and of an electron are $+e=1.6\times 10^{-19}\mathrm{C}$ and $-e=-1.6\times 10^{-19}\mathrm{C}$ respectively.

For your information:

The Triboelectric Series (shortened)

It is the relative position of two materials in this series that determines which one receives electrons and which one donates. The one nearer to the positive end will be positively charged when being rubbed with the other less near, which will then be negatively charged.

$(-)$ Negative End of Series: Silicon rubber $<$ Teflon $<$ Polyethylene $<$ Saran $<$ Orlon $<$ Synthetic rubber $<$ Brass and silver $<$ Nickel and copper $<$ Hard rubber $<$ Sealing wax $<$ Amber $<$ Wood $<$ Steel $<$ Cotton $<$ Paper $<$ Aluminium $<$ Silk $<$ Lead $<$ Wool $<$ Nylon $<$ Glass $<$ Acetate $<$ Asbestos $<$ Human Hands : $(+)$ Positive End of Series

Concept Test

Which of the following statements is correct?
1. Given that glass rod is positively charged after it is rubbed with a piece of silk, it is also positively charged after it is rubbed with any other materials.
2. An object could be negatively charged by transferring away some of its protons.
3. After rubbing two different materials together, each material is charged with the same quantity of charge but with opposite sign.
4. In the process of inducing charges, an object’s net charge either increases or decreases.
Which of the following statements is/are correct?
1. 1. Each carbonate ion, $\mathrm{CO_3\,^{2-}}$ , has more electrons than protons.
  2. Sodium ion $\mathrm{Na^{+}}$ has one positive coulomb of charge.
  3. Chloride ion $\mathrm{Cl^{-}}$ has $-1.6\times 10^{-19}\mathrm{C}$ of charge.

1. I only.
2. I and II only.
3. I and III only.
4. All of the above.

Answers:

Explanation:

A is wrong: for instance, the glass rod would become negatively charged when it is rubbed with asbestos. Please refer to The Triboelectric Series. B is wrong because the protons are always bound with the neutrons in the nucleus. Only free electrons could be transferred. C is correct because only as such could the charge be conserved. D is wrong because net charge of an object must be unchanged, unless earthing occurs, which will be discussed later on.
I is correct. The superscript $(2-)$ indicates it has two extra electrons than protons. II is wrong, for the superscript $(+)$ means one extra proton than electron so that the charge should be $+e$ , or $+1.6\times 10^{-19}\mathrm{C}$ . By the same reason, III is correct.

Conductors and Insulators

Conductors allow free charges (e.g., electrons, electrolytes) to flow through them easily; insulators do not.

(Not all conductors are metal, e.g., electrolyte is not a metal but a conductor.)

An isolated and charged conductor always has its charges distributed on its surface; the charge density is higher on curved edges and cusps than on flat and smooth surface.

Friction_Induce_Contact_Earthing_Illustration

The above chart illustrates the three processes of charging and the process of earthing. In doing experiments, and also exercises, we often use a combination of some of these four techniques, so make sure you understand and remember them.

Coulomb’s Law

$F=k\times \displaystyle{\frac{Q_1Q_2}{r^2}}$

where $k=\displaystyle{\frac{1}{4\pi \epsilon_0}}$ is the proportionality constant/Coulomb constant.

$F = \displaystyle{\frac{Q_1Q_2}{4\pi \epsilon_0r^2}}$

where $\epsilon_0=8.85\times 10^{-12}\mathrm{C^2N^{-1}m^{-2}}$ is the permittivity of free space.

The magnitude of Coulomb force is given above. The direction of the force on each point charge is pointing away from each other when the two point charges $Q_1$ and $Q_2$ are of the same sign, whereas pointing towards each other when of opposite sign.

(Point charge, as well as point mass, is an ideal model of particle. When the size of two charge carriers is exceedingly less than their separating distance, we treat them as point charges. For example, the radius of an electron is approximated to $10^{-15}$ order of magnitude, much less than the radius of an atom $10^{-10}$ .)

Concept Test

1. A positively charged glass rod is brought near (without contact) an isolated aluminium rod which is electrically neutral.
  
  Which of the following statements is correct?
  1. The glass rod induces a net negative charge on the aluminium rod.
  2. The electrons are transferred from the aluminium rod to the glass rod.
  3. The net electrostatic force acting on the aluminium rod is zero.
  4. The net electrostatic force acting on the aluminium rod is non-zero.
2. Sally wants to charge a metallic conductor by friction. After rubbing it with a tablecloth for 5 minutes, she still finds the conductor uncharged. Which of the following statements could best explain the failure?
  1. Conductors can never be charged by friction.
  2. Sally shall continue the rubbing for a longer time.
  3. The charge obtained by the conductor during rubbing flows away through Sally’s body.
  4. The air surrounding the rubbed surface of the conductor is ionized such that the charges acquired flow away through air.
3. Two identical spherical conductors of radius are separated by a distance of between their origins. Each of them contains the same amount of charges . But one is positively charged; another is negatively charged. Which of the following about electrostatic force is true?
  1. $F=\displaystyle{k\frac{(Q)(-Q)}{(5r)^2}}$
  2. $F=\displaystyle{k\frac{(Q)(-Q)}{(3r)^2}}$
  3. None of the above.

Answers:

Explanation:

Unless there is earthing on the aluminium rod, its net charge is conserved. A is wrong. Unless being in contact, the electrons are never transferred from the aluminium rod to the glass rod in mid-air. B is wrong. This aluminium rod would be induced negative (opposite) charges on its front side near the glass rod and positive (like) charges on the back side away from the glass rod. Because the aluminium’s front side is closer to the glass rod than its back side, and electrostatic force $\displaystyle{\varpropto \frac{1}{r^2}}$ , we know the force of attraction is greater than that of repulsion. So there is a non-zero net force acting on the aluminium rod. C is wrong and D is correct.
When Sally’s hand touches the metallic conductor, the conductor is earthed. So the charge acquired by friction is discharged first to her body and then to the ground. Thus, to charge a metallic conductor by friction, it must be held with an insulator.
The Coulomb’s force law does not apply here because we cannot treat the two spherical conductors as point charges, for two reasons: (i) the charge distribution of both conductors is uneven—the charges are denser at the surfaces close to each other, and (ii) the separating distance $5r$ is only five times the radius $r$ of the spherical conductors, they are in same order of magnitude.

February 21, 2019January 19, 2023

201902210304 Exercise 6.3.1

Fill in the gap between equations (6.18) and (6.19).

Roughwork.

Equation (6.16):

$g^{\mu\alpha}g_{\alpha\nu}=\delta^\mu_{\enspace\nu}$ and $g'^{\mu\beta}g'_{\beta\nu}=\delta^\mu_{\enspace\nu}$

Equation (6.9):

$g'_{\mu\nu}=\displaystyle{\frac{\partial x^\alpha}{\partial x'^\mu}} \displaystyle{\frac{\partial x^\beta}{\partial x'^\nu}}g_{\alpha\beta}$

Equation (6.11):

$g'^{\mu\nu}= \displaystyle{\frac{\partial x'^\mu}{\partial x^\alpha}} \displaystyle{\frac{\partial x'^\nu}{\partial x^\beta}}g^{\alpha\beta}$

$\begin{aligned} g'^{\mu\nu} g_{\mu\nu} & = \bigg( \displaystyle{\frac{\partial x'^\mu}{\partial x^\alpha}} \displaystyle{\frac{\partial x'^\nu}{\partial x^\beta}} g^{\alpha\beta} \bigg) \bigg( \displaystyle{\frac{\partial x^\gamma}{\partial x'^\mu}} \displaystyle{\frac{\partial x^\delta}{\partial x'^\nu}} g_{\gamma\delta} \bigg) \\ & = \bigg( \displaystyle{\frac{\partial x^\gamma}{\partial x^\alpha}} \displaystyle{\frac{\partial x^\delta}{\partial x^\beta}} \bigg) g^{\alpha\beta}g_{\gamma\delta} \\ & = \delta^\gamma_{\enspace \alpha} \delta^\delta_{\enspace \beta} g^{\alpha\beta} g_{\gamma\delta} \\ & = g^{\alpha\beta} g_{\alpha\beta} \end{aligned}$

February 21, 2019January 19, 2023

201902210301 Exercise 6.2.1-6.2.3

Again consider polar coordinates on a flat plane. The transformation equations between polar coordinates $r$ , $\theta$ (the primed coordinate system) and Cartesian coordinates $x$ , $y$ (the unprimed coordinate system) are given by equation 6.15. Consider also the vector $\mathbf{v}$ whose components are $v^x=1$ and $v^y=0$ .

Lowering an index as given by Equation (6.5):

$A_\mu \equiv g_{\mu\nu}A^\nu$ .

In Cartesian coordinate system the metric tensor is

$\mathbf{g}=\begin{bmatrix} g_{xx} & g_{xy} \\ g_{yx} & g_{yy} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Thus

$v_x\equiv g_{x\nu}v^\nu = g_{xx}v^x+g_{xy}v^y = (1)(1) + (0)(0)=1$

$v_y\equiv g_{y\nu}v^\nu = g_{yx}v^x+g_{yy}v^y = (0)(1) + (1)(0)=0$

The definition of covector is given by Equation (6.2):

$B'_\nu = \displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}B_{\mu}$

So,

$v_r=\displaystyle{\frac{\partial x}{\partial r}}v_x + \displaystyle{\frac{\partial y}{\partial r}}v_y = (\cos\theta )(1) + (\sin\theta )(0) = \cos\theta$

$v_\theta =\displaystyle{\frac{\partial x}{\partial \theta}}v_x + \displaystyle{\frac{\partial y}{\partial \theta}}v_y = (-r\sin\theta )(1) + (r\cos\theta )(0) = -r\sin\theta$

The metric tensor for the polar coordinate basis is given by Equation (5.19):

$g_{\mu\nu}\equiv \mathbf{e}_{\mu}\cdot\mathbf{e}_\nu = \begin{bmatrix} \mathbf{e}_{r}\cdot\mathbf{e}_r & \mathbf{e}_{r}\cdot\mathbf{e}_\theta \\ \mathbf{e}_{\theta}\cdot\mathbf{e}_r & \mathbf{e}_{\theta}\cdot\mathbf{e}_\theta \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & r^2 \\ \end{bmatrix}$

Exercise 6.2.3.

One can show that in the polar coordinate system, $v^r=\cos\theta$ and $v^\theta =-\sin\theta /r$ (see Problem P6.1). Show that $v'^\mu v'_{\mu}=1$ . Does this make sense?

Equation (6.5):

$A_{\mu} \equiv g_{\mu\nu}A^\nu$

Hence

$\begin{aligned} v_r & \equiv g_{r\nu}v^\nu \\ \cos\theta & = g_{rr}v^r + g_{r\theta} v^\theta \\ \cos\theta & = (1)(v^r) + (0)(v^\theta ) \\ v^r & = \cos\theta \end{aligned}$

$\begin{aligned} v_\theta & \equiv g_{\theta\nu}v^\nu \\ -r\sin\theta & = g_{\theta r}v^r + g_{\theta\theta} v^\theta \\ -r\sin\theta & = (0)(\cos\theta ) + (r^2)(v^\theta) \\ v^\theta & = \displaystyle{\frac{-\sin\theta }{r}} \\ \end{aligned}$

are checked.

$\begin{aligned} v'^\mu v'_{\mu} & = v^r v_r + v^\theta v_{\theta} \\ & = (\cos\theta )(\cos\theta ) + (\displaystyle{\frac{-\sin\theta}{r}})(-r\sin\theta ) \\ & = \cos^2\theta + \sin^2\theta \\ & = 1 \end{aligned}$ .

Remark. Invariant norm.

February 21, 2019January 19, 2023

201902210253 Exercise 6.1.1-6.1.3

Consider polar coordinates on a flat plane. The transformation equations between the polar coordinates $r$ , $\theta$ (the primed coordinate system) and cartesian coordinates $x$ , $y$ (the unprimed coordinate system) are

Equation (6.15a):

$x=r\cos\theta$ , $y=r\sin\theta$

Equation (6.15b):

$r=\sqrt{x^2+y^2}$ , $\theta =\tan^{-1}\bigg( \displaystyle{\frac{y}{x}} \bigg)$

Consider also the scalar function $\varPsi =bxy=br^2\cos\theta\sin\theta$ .

Calculate the four transformation partials $\displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}$ for the transformations given above:

$\displaystyle{\frac{\partial x}{\partial r}}=\cos\theta$ , $\displaystyle{\frac{\partial x}{\partial \theta}}=-r\sin\theta$ , $\displaystyle{\frac{\partial y}{\partial r}}=\sin\theta$ , $\displaystyle{\frac{\partial y}{\partial \theta}}=r\cos\theta$ .

It can be shown that the gradient of $\varPsi$ in cartesian coordinate system is

$\partial_x\varPsi =by$ and $\partial_y\varPsi =bx$

and that of $\varPsi$ in polar coordinate system is

$\partial_r\varPsi =2br\cos\theta\sin\theta$ and $\partial_\theta\varPsi =br^2[\cos^2\theta -\sin^2\theta]$ .

To make practice of the covector transformation rule:

Equation (6.3):

$\partial_\mu\varPsi \equiv \displaystyle{\frac{\partial \varPsi}{\partial x^\mu}}$

Equation (6.4):

$\partial'_\nu \equiv \displaystyle{\frac{\partial \varPsi}{\partial x'^\nu}} = \displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}\displaystyle{\frac{\partial \varPsi}{\partial x^\mu}} = \displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}\partial_\mu \varPsi$

Thus,

$\begin{aligned} &\quad\enspace \displaystyle{\frac{\partial x}{\partial r}}\partial_x\varPsi + \displaystyle{\frac{\partial y}{\partial r}}\partial_y\varPsi \\ & = (\cos\theta )(by) + (\sin\theta )(bx) \\ & = (\cos\theta )(br\sin\theta ) + (\sin\theta )(br\cos\theta) \\ & = 2br\cos\theta\sin\theta \\ & = \partial_r \varPsi \end{aligned}$

and

$\begin{aligned} &\quad\enspace \displaystyle{\frac{\partial x}{\partial \theta}}\partial_x\varPsi + \displaystyle{\frac{\partial y}{\partial \theta}}\partial_y\varPsi\\ & = (-r\sin\theta )(by) + (r\cos\theta )(bx) \\ & = (-r\sin\theta )(br\sin\theta ) + (r\cos\theta )(br\cos\theta ) \\ & = br^2[\cos^2\theta -\sin^2\theta ] \\ & = \partial_\theta \varPsi \end{aligned}$

February 21, 2019January 19, 2023

201902210247 Exercise 5.5.1

Check the cases where $\alpha =t$ and $\beta =x$ , and where $\alpha =\beta =x$ .

Roughwork.

This exercise is to check the transformation law for the metric tensor of flat spacetime, given by Equation (5.16):

$\eta'_{\alpha\beta}=\displaystyle{\frac{\partial x^\mu}{\partial x'^\alpha}}\displaystyle{\frac{\partial x^\nu}{\partial x'^\beta}} \eta_{\mu\nu}=\eta_{\mu\nu}(\varLambda^{-1})^\mu_{\enspace \alpha}(\varLambda^{-1})^\nu_{\enspace \beta}=\eta_{\alpha\beta}$

(where the second equality is by Equation (5.13), and the third equality by Equation 4.19)

The textbook has already had a checking for $\eta'_{tt}=\eta_{tt}$ through Equations (5.29) and (5.30). I will do the remaining.

Proof. ( $\eta'_{tx}=\eta_{tx}$ )

$\begin{aligned} \eta'_{tx} & = (\varLambda^{-1})^\mu_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{\mu\nu} \\ & = (\varLambda^{-1})^t_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{t\nu} + (\varLambda^{-1})^x_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{x\nu} + (\varLambda^{-1})^y_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{y\nu} \\ & \qquad\quad + (\varLambda^{-1})^z_{\enspace t}(\varLambda^{-1})^\nu_{\enspace x}\eta_{z\nu}\\ \end{aligned}$

But the metric tensor given by Equation (4.6):

$\begin{bmatrix} \eta_{tt} & \eta_{tx} & \eta_{ty} & \eta_{tz} \\ \eta_{xt} & \eta_{xx} &\eta_{xy} & \eta_{xz} \\ \eta_{yt} & \eta_{yx} & \eta_{yy} & \eta_{yz} \\ \eta_{zt} & \eta_{zx} & \eta_{zy} & \eta_{zz} \end{bmatrix}\equiv \begin{bmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

tells that

$\eta_{t\nu}$ is nonzero only when $\nu =t$ ;

$\eta_{x\nu}$ is nonzero only when $\nu =x$ ;

$\eta_{y\nu}$ is nonzero only when $\nu =y$ ;

$\eta_{z\nu}$ is nonzero only when $\nu =z$ .

So, the summation continues as follows:

$\begin{aligned} \eta'_{tx} & = (\varLambda^{-1})^t_{\enspace t}(\varLambda^{-1})^t_{\enspace x}\eta_{tt} + (\varLambda^{-1})^x_{\enspace t}(\varLambda^{-1})^x_{\enspace x}\eta_{xx} + (\varLambda^{-1})^y_{\enspace t}(\varLambda^{-1})^y_{\enspace x}\eta_{yy} \\ &\qquad\quad + (\varLambda^{-1})^z_{\enspace t}(\varLambda^{-1})^z_{\enspace x}\eta_{zz}\\ & = (\gamma )(\gamma\beta)(-1) + (\gamma\beta)(\gamma)(1) + (0)(0)(1) + (0)(0)(1) \\ & = -\gamma^2\beta + \gamma^2\beta \\ & = 0 \\ & \equiv \eta_{tx} \end{aligned}$

February 21, 2019January 19, 2023

201902210242 Exercise 5.4.1

Similarly, check that $\displaystyle{\frac{\partial x'^\mu}{\partial x^\nu}}=\varLambda^\mu_{\enspace\nu}$ when $\mu =x$ and $\nu =t$ and when $\mu =\nu =y$ .

Roughwork.

Equation (5.12):

$\begin{aligned} t' & = \gamma (t-\beta x) \\ x' & = \gamma (-\beta t+x) \\ y' & = y \\ z' & = z \end{aligned}$

and

$\begin{aligned} t & = \gamma (t'+\beta x') \\ x & = \gamma (\beta t'+x') \\ y & = y' \\ z & = z' \end{aligned}$

Equation (5.13a):

$\displaystyle{\frac{\partial x'^\mu}{\partial x^\nu}}=\varLambda^\mu_{\enspace\nu} =\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$

Proof.

$\displaystyle{\frac{\partial t'}{\partial t}}=\gamma$ , $\displaystyle{\frac{\partial t'}{\partial x}}=-\gamma\beta$ , $\displaystyle{\frac{\partial t'}{\partial y}}=0$ , $\displaystyle{\frac{\partial t'}{\partial z}}=0$

$\displaystyle{\frac{\partial x'}{\partial t}}=-\gamma\beta$ , $\displaystyle{\frac{\partial x'}{\partial x}}=\gamma$ , $\displaystyle{\frac{\partial x'}{\partial y}}=0$ , $\displaystyle{\frac{\partial x'}{\partial z}}=0$

$\displaystyle{\frac{\partial y'}{\partial t}}=0$ , $\displaystyle{\frac{\partial y'}{\partial x}}=0$ , $\displaystyle{\frac{\partial y'}{\partial y}}=1$ , $\displaystyle{\frac{\partial y'}{\partial z}}=0$

$\displaystyle{\frac{\partial z'}{\partial t}}=0$ , $\displaystyle{\frac{\partial z'}{\partial x}}=0$ , $\displaystyle{\frac{\partial z'}{\partial y}}=0$ , $\displaystyle{\frac{\partial z'}{\partial z}}=1$

February 21, 2019March 29, 2024

201902210108 Exercise 5.3.2

Calculate all eight partial derivatives $\displaystyle{\frac{\partial x'^\mu}{\partial x^\nu}}$ and $\displaystyle{\frac{\partial x^\mu}{\partial x'^\nu}}$ .

Roughwork.

Equation (5.23):

$p(x,y)=x$ , $q(x,y)=y-cx^2$ ;

Equation (5.24):

$x(p,q)=p$ , $y(p,q)=cp^2+q$ .

Then, the eight partial derivatives are

$\displaystyle{\frac{\partial p}{\partial x}}=1$ , $\displaystyle{\frac{\partial p}{\partial y}}=0$ , $\displaystyle{\frac{\partial q}{\partial x}}=-2cx$ , $\displaystyle{\frac{\partial q}{\partial y}}=1$ ,

$\displaystyle{\frac{\partial x}{\partial p}}=1$ , $\displaystyle{\frac{\partial x}{\partial q}}=0$ , $\displaystyle{\frac{\partial y}{\partial p}}=2cp$ , $\displaystyle{\frac{\partial y}{\partial q}}=1$ .

Equation (5.10):

$\mathrm{d}s^2=g_{pq}\,\mathrm{d}p\,\mathrm{d}q$ .

Equation (5.11):

$g'_{\mu\nu}=\displaystyle{\frac{\partial x^\alpha}{\partial x'^\mu}}\displaystyle{\frac{\partial x^\beta}{\partial x'^\nu}}\, g_{\alpha\beta}$

The metric tensor for the cartesian coordinates is given by equation (5.25):

$g_{\alpha\beta}=\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$

Using Equations (5.11) and (5.25):

$\begin{aligned} g_{pp} & =\displaystyle{\frac{\partial x^\alpha}{\partial p}}\displaystyle{\frac{\partial x^\beta}{\partial p}}g_{\alpha\beta} \\ & = \displaystyle{\frac{\partial x}{\partial p}}\displaystyle{\frac{\partial x}{\partial p}}g_{xx}+\displaystyle{\frac{\partial x}{\partial p}}\displaystyle{\frac{\partial y}{\partial p}}g_{xy}+\displaystyle{\frac{\partial y}{\partial p}}\displaystyle{\frac{\partial x}{\partial p}}g_{yx}+\displaystyle{\frac{\partial y}{\partial p}}\displaystyle{\frac{\partial y}{\partial p}}g_{yy}\\ & = (1)(1)(1) + (1)(2cp)(0) + (2cp)(1)(0) + (2cp)(2cp)(1) \\ &= 1+4c^2p^2 \end{aligned}$

$\begin{aligned} g_{pq} & =\displaystyle{\frac{\partial x^\alpha}{\partial p}}\displaystyle{\frac{\partial x^\beta}{\partial q}}g_{\alpha\beta} \\ & = \displaystyle{\frac{\partial x}{\partial p}}\displaystyle{\frac{\partial x}{\partial q}}g_{xx}+\displaystyle{\frac{\partial x}{\partial p}}\displaystyle{\frac{\partial y}{\partial q}}g_{xy}+\displaystyle{\frac{\partial y}{\partial p}}\displaystyle{\frac{\partial x}{\partial q}}g_{yx}+\displaystyle{\frac{\partial y}{\partial p}}\displaystyle{\frac{\partial y}{\partial q}}g_{yy}\\ & = (1)(0)(1) + (1)(1)(0) + (2cp)(0)(0) + (2cp)(1)(1) \\ &= 2cp \end{aligned}$

$\begin{aligned} g_{qp} & =\displaystyle{\frac{\partial x^\alpha}{\partial q}}\displaystyle{\frac{\partial x^\beta}{\partial p}}g_{\alpha\beta} \\ & = \displaystyle{\frac{\partial x}{\partial q}}\displaystyle{\frac{\partial x}{\partial p}}g_{xx}+\displaystyle{\frac{\partial x}{\partial q}}\displaystyle{\frac{\partial y}{\partial p}}g_{xy}+\displaystyle{\frac{\partial y}{\partial q}}\displaystyle{\frac{\partial x}{\partial p}}g_{yx}+\displaystyle{\frac{\partial y}{\partial q}}\displaystyle{\frac{\partial y}{\partial p}}g_{yy}\\ & = (0)(1)(1) + (0)(2cp)(0) + (2cp)(0)(0) + (1)(2cp)(1) \\ &= 2cp \end{aligned}$

$\begin{aligned} g_{qq} & =\displaystyle{\frac{\partial x^\alpha}{\partial q}}\displaystyle{\frac{\partial x^\beta}{\partial q}}g_{\alpha\beta} \\ & = \displaystyle{\frac{\partial x}{\partial q}}\displaystyle{\frac{\partial x}{\partial q}}g_{xx}+\displaystyle{\frac{\partial x}{\partial q}}\displaystyle{\frac{\partial y}{\partial q}}g_{xy}+\displaystyle{\frac{\partial y}{\partial q}}\displaystyle{\frac{\partial x}{\partial q}}g_{yx}+\displaystyle{\frac{\partial y}{\partial q}}\displaystyle{\frac{\partial y}{\partial q}}g_{yy}\\ & = (0)(0)(1) + (0)(1)(0) + (1)(0)(0) + (1)(1)(1) \\ &= 1 \end{aligned}$

The metric tensor for $p$ – $q$ coordinate system is given by Equation (5.26):

$g'_{\mu\nu}=\begin{bmatrix} 1+4c^2p^2 & 2cp \\ 2cp & 1 \\ \end{bmatrix}$ .

February 21, 2019January 19, 2023

201902210102 Exercise 3.1.1

$\mathbf{A}=\begin{bmatrix} A'^t\\ A'^x\\ A'^y\\A'^z \end{bmatrix}=\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{bmatrix}\begin{bmatrix} A^t \\ A^x \\ A^y \\ A^z \end{bmatrix}$

$\mathbf{B}=\begin{bmatrix} B'^t\\ B'^x\\ B'^y\\B'^z \end{bmatrix}=\begin{bmatrix} \gamma & -\gamma\beta & 0 & 0 \\ -\gamma\beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\\end{bmatrix}\begin{bmatrix} B^t \\ B^x \\ B^y \\ B^z \end{bmatrix}$

are two four-vectors.

$\begin{aligned} A'^tB'^t & =(\gamma A^t-\gamma\beta A^x)(\gamma B^t-\gamma\beta B^x)\\ & = \gamma^2A^tB^t-\gamma^2\beta A^tB^x-\gamma^2\beta A^xB^t+\gamma^2\beta^2A^xB^x \end{aligned}$

$\begin{aligned} A'^xB'^x & = (-\gamma\beta A^t+\gamma A^x)(-\gamma\beta B^t+\gamma B^x)\\ & = \gamma^2\beta^2A^tB^t-\gamma^2\beta A^tB^x-\gamma^2\beta A^xB^t+\gamma^2A^xB^x\\ \end{aligned}$

$\begin{aligned} A'^yB'^y & = A^yB^y \end{aligned}$

$\begin{aligned} A'^zB'^z & = A^zB^z \end{aligned}$

$\begin{aligned} &\quad\enspace -A'^tB'^t+A'^xB'^x+A'^yB'^y+A'^zB'^z \\ & = (-\gamma^2+\gamma^2\beta^2)A^tB^t+(-\gamma^2\beta^2+\gamma^2)A^xB^x+A^yB^y+A^zB^z\\ & = -\gamma^2(1-\beta^2)A^tB^t+\gamma^2(1-\beta^2)A^xB^x+A^yB^y+A^zB^z\\ & = -\bigg( \frac{1}{\sqrt{1-\beta^2}} \bigg)^2(1-\beta^2)A^tB^t+\bigg( \frac{1}{\sqrt{1-\beta^2}} \bigg)^2(1-\beta^2)A^xB^x+A^yB^y+A^zB^z\\ & = -A^tB^t+A^xB^x+A^yB^y+A^zB^z \end{aligned}$

Scalar product of any two four-vectors is frame-independent.

February 21, 2019January 19, 2023

201902210054 Exercise 17 Chapter 2

It sometimes occurs that the generalized coordinates appear separately in the kinetic energy and the potential energy in such a manner that $T$ and $V$ may be written in the form

$T=\displaystyle{\sum_i}f_i(q_i)\dot{q_i}^2$ and $V=\displaystyle{\sum_iV_i(q_i)}$ .

Show that Lagrange’s equations then separate, and that the problem can always be reduced to quadratures.

Solution.

This solution is not mine. It was found on the Internet some years ago, to whose author(s) I lost references.

From the fact that

$\begin{aligned} \mathcal{L} & =T-V \\ & =\displaystyle{\sum_i f_i(q_i)\dot{q_i}^2-\sum_iV_i(q_i)}\\ & =\sum_i\bigg( f_i(q_i)\dot{q_i}^2-V_i(q_i)\bigg)\\ & =\sum_i(T_i-V_i)\\ & =\sum_iL_i\\\end{aligned}$ ,

the Lagrange’s equation can be separated into $i$ Lagrange’s equations.

Remark.

The solution is incomplete. It remains to be shown how the problem can always be reduced to quadratures.