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IZhO 2018, Теория — Задача 3. Optics of moving media

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XIV International Zhautykov Olympiad/Theoretical Competition · 12 януари 2018 г. · 10 т.

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THEORETICAL COMPETITION

January 12, 2018

The time available for the theoretical competition is 4 hours. There are three questions.
Use only the pen provided.
You can use your own calculator for numerical calculations. If you don’t have one, please ask for it from Olympiad organizers.
You are provided with Writing sheet and additional paper. You can use the additional paper for drafts of your solutions but these papers will not be checked. Your final solutions which will be evaluated should be on the Writing sheets. Please use as little text as possible. You should mostly use equations, numbers, figures and plots.
Use only the front side of Writing sheets**.** Write only inside the boxed area.
Begin each question on a separate sheet of paper.
Fill the boxes at the top of each sheet of paper with your country (Country), your student code (Student Code), the question number (Question Number), the progressive number of each sheet (Page Number), and the total number of Writing sheets used (Total Number of Pages). If you use some blank Writing sheets for notes that you do not wish to be evaluated, put a large X across the entire sheet and do not include it in your numbering.
At the end of the exam, arrange all sheets for each problem in the following order:
- Used Writing sheets in order;
- The sheets you do not wish to be evaluated
- Unused sheets and the printed question.
Place the papers inside the envelope and leave everything on your desk. You are not allowed to take any paper out of the room.

SOLUTIONS TO THE PROBLEMS OF THE THEORETICAL COMPETITION

Attention. Points in grading are not divided!

Условие

Part 1. 4-dimensional vectors

Consider two inertial reference frames $S$ and $S'$ of which the second one moves at a speed $V$ relative to the first one as shown in the figure on the right. It is assumed that the origins $O$ and $O'$ coincide at the initial time moment $t = t' = 0$ by the clocks of both reference frames. It is known that the Lorentz transformation of space-time coordinates $(x', y', z', ct')$ of any event in the frame $S'$ into the space-time coordinates $(x, y, z, ct)$ of the same event in the frame $S$ have the following form

x = \frac{x' + (V/c)ct'}{\sqrt{1 - V^2/c^2}}, \qquad y = y', \qquad z = z', \qquad ct = \frac{ct' + (V/c)x'}{\sqrt{1 - V^2/c^2}},

where $c = 2.9979\ \mathrm{m/s}$ is the speed of light.

Two reference frames: S with origin O and axes X, Y, Z, and S′ with origin O′ and axes X′, Y′, Z′, moving along the common X, X′ axis with velocity V.

In the formulas of the Lorentz transformation the spatial coordinates and the time are intentionally brought to the same dimension as they together form components of the so-called 4-vector. It is known that all the components of 4-vectors are transformed in the same way at the transition from one inertial frame to the other. In particular, the momentum and the energy constitute the components of the 4-vector.

Suppose that an object moves in the reference frame $S$ such that it has the total energy $E$ and the momentum projections $p_x, p_y, p_z$ on the coordinate axes $OX$ , $OY$ and $OZ$ , respectively.

1.1 Write down the transformation of energy and momentum of the object from the reference frame $S$ to the reference frame $S'$ .

Suppose an object of the rest mass $m$ moves in the reference frame $S$ such that it has the total energy $E$ and the momentum $p$ . When converting its energy and momentum from one reference frame to another the value $E^2 - p^2c^2 = inv$ remains invariant.

1.2 Express $inv$ in terms of $m$ and $c$ .

Part 2. Doppler effect and light aberration

Let a plane electromagnetic wave (EMW) propagate in $XY$ -plane of the reference frame $S$ so that it has the frequency $\omega$ and makes the angle $\varphi$ with the axis $OX$ .

2.1 Find the frequency $\omega'$ of EMW registered by an observer in the reference frame $S'$ .

2.2 Find the angle $\varphi'$ that EMW makes with the axis $O'X'$ in the reference frame $S'$ .

Astronomical observations have shown that the position of a newly discovered massive star $X$ on the celestial sphere (i.e. relative to very distant objects) does not remain constant throughout the year. It moves along an ellipse with axial ratio 0.900. Ecliptic latitude of a star is the angle between the direction to the star and the ecliptic plane, which can be assumed coincident with the plane of the Earth orbit around the Sun.

2.3 Find the ecliptic latitude $\delta$ of the star $X$ and evaluate it in arc degrees.

Observation of the emission spectrum of the star $X$ has shown that all frequency in its spectrum are shifted to the red. The relative frequency deviation is $(\Delta\omega/\omega)_0 = 9.9945 \cdot 10^{-3}$ . It has independently been established that the recession velocity of the star $X$ from the Sun is equal to $v_x = \frac{1}{100}c$ .

2.4 Find and evaluate the escape velocity $v_{II}$ at the surface of the star $X$ .

Part 3. Light in a moving medium

Consider the same two reference frames as in Part 1. Let an object move in the plane $X'Y'$ of the reference frame $S'$ such that the projections of its velocity on the axis $O'X'$ and $O'Y'$ are equal $u_x'$ and $u_y'$ , respectively.

3.1 Find the projections of the object velocity $u_x$ on the axis $OX$ and $u_y$ on the axis $OY$ in the reference frame $S$ .

Consider a water flow moving at the speed of $V$ relative to the bottom of the vessel. A plane EMW falls onto the water surface to make the angle $\alpha$ to the normal in the laboratory reference frame. A detector is fixed at the bottom of the vessel. The refractive index of water is known to be $n$ .

A ray falls on the surface of a water layer at angle α to the normal and is refracted at angle β towards a detector on the bottom; the water flows with velocity V; labels Bottom and Detector.

If the water velocity $V \ll c$ , the expression for the sine of the angle $\beta$ at which the detector registers the direction of EMW propagation, takes the form:

\sin\beta = A_1 + B_1 V.

3.2 Find $A_1, B_1$ and express them in terms of $\alpha$ and $n$ .

If the water velocity $V \ll c$ , the expression for velocity $v_m$ of the light propagation in the laboratory reference frame is found as

v_m = A_2 + B_2 V.

3.3 Find $A_2, B_2$ and express them in terms of $\beta, n, c$ .

In 1860 A. Fizeau set up the following experiment. A monochromatic beam with the wavelength $\lambda$ comes out of the source $A$ and falls onto the semitransparent plate $B$ at which it is divided into two coherent beams. The first beam, being reflected from the plate $B$ , goes along the way $BKDEB$ ( $K, D$ and $E$ are the mirrors), whereas the second beam, passing through the plate $B$ , goes along the way $BEDKB$ . The first beam, when returning to the plate $B$ , is partially reflected from it and reaches the interferometer $F$ . The second beam, when returning to the plate $B$ , partially passes through it and reaches the interferometer $F$ . Both interfering beams travels the same distance in the laboratory reference frame, including section $BE$ and $KD$ in which the water flows with the velocity $v$ . The total distance covered by each beam in water in the laboratory reference frame is equal to $2L$ .

Fizeau’s set-up: source A, semitransparent plate B, mirrors K, D, E and interferometer F; water flows with velocity v through the U-shaped tube sections BE and KD; wavy arrows show the two counter-propagating beams.

3.4 Find the number of bands $\Delta N$ the interference pattern is shifted by when the liquid velocity changes from 0 to $v$ and express it in terms of $L, n, v, c$ , and $\lambda$ .

In his own experiment A. Fizeau obtained $\Delta N = 0.230$ at $L = 1.49\ \mathrm{m}$ , $v = 7.06\ \mathrm{m/s}$ and $\lambda = 536\ \mathrm{nm}$ .

3.5 Evaluate refractive index of water $n$ in Fizeau’s experiment.

Mathematical hint

You may need to know the following approximate equality: $(1 + x)^{\alpha} \approx 1 + \alpha x$ , at $x \ll 1$ .

Отговори

Решение

Part 1. 4-dimensional vectors (1,4 points)

1.1 [0,8 points] To bring the momentum and the energy to the same unit it is sufficient to divide the energy by the speed of light or to multiply the momentum by the speed of light. Moreover, by virtue of the principle of relativity, it is necessary to make the substitution $V \rightarrow -V$ . Thus, one gets

p_x' = \frac{p_x - (V/c)(E/c)}{\sqrt{1 - V^2/c^2}}, \qquad (1)

p_y' = p_y, \qquad (2)

p_z' = p_z, \qquad (3)

E'/c = \frac{E/c - (V/c) p_x}{\sqrt{1 - V^2/c^2}}. \qquad (4)

1.2 [0,6 points] In any inertial frame of reference the expression for the momentum is written as

p = \frac{mv}{\sqrt{1 - v^2/c^2}}, \qquad (5)

and the expression for the total energy has the form

E = \frac{mc^2}{\sqrt{1 - v^2/c^2}}. \qquad (6)

This implies that the invariant sought is equal to

inv = E^2 - p^2 c^2 = m^2 c^4. \qquad (7)

Part 2. Doppler effect and light aberration (4,6 points)

2.1 [1,0 points] Since the rest mass of photons is zero, it follows from (7) that the momentum and energy of a photon are related as follows

p = \frac{E}{c}. \qquad (8)

It is known that the photon energy is given by the Planck formula as

E = \hbar\omega. \qquad (9)

The photon momentum projections on the coordinate axes are written as

p_x = p\cos\varphi, \qquad (10)

p_y = p\sin\varphi, \qquad (11)

and on substituting into (B1.4), one finds

\omega' = \omega\frac{1 - V\cos\varphi/c}{\sqrt{1 - V^2/c^2}}. \qquad (12)

This is the well known formula for the relativistic Doppler effect.

2.2 [0,4 points] It follows from (2), 8) and (9) that

\frac{\hbar\omega'}{c}\sin\varphi' = \frac{\hbar\omega}{c}\sin\varphi. \qquad (13)

Using (12), it is merely found that

\sin\varphi' = \frac{\sqrt{1 - V^2/c^2}\sin\varphi}{1 - V\cos\varphi/c}. \qquad (14)

Expression (14) is a classical formula for the light aberration.

2.3 [1,0 points] The position of the star on the celestial sphere varies throughout the year due to the orbital motion of the Earth around the Sun and the aberration of light which is schematically shown in the figure on the right. Since the speed of Earth's orbital motion is much less than the speed of light, it follows from (14) that the aberration angle is equal to

\delta\varphi = \varphi' - \varphi \approx \frac{V}{c}\sin\varphi, \qquad (15)

where $\varphi$ denotes the angle between $V$ and the direction towards the star.

The Earth orbit around the Sun drawn as an ellipse with points A, B, C, D and a celestial hemisphere; a star on the hemisphere, the angle δ between the direction Sun–star and the orbital plane is marked.

The figure shows that the angle $\varphi$ varies periodically from a minimum value $\delta$ at the point $D$ , reaches the value of $\pi/2$ at the point $B$ , has a maximum value of $\pi - \delta$ at point $C$ , and finally becomes equal to $\pi/2$ at point $A$ . Hence, one can infer that the star apparent position on the celestial sphere moves along an ellipse with angular dimensions of the semi-axes

a_1 = \frac{V}{c} \qquad (16)

and

a_2 = \frac{V}{c}\sin\delta. \qquad (17)

It is found from the given data that

\delta = \arcsin\left(\frac{a_2}{a_1}\right) = 64.2^{\circ}. \qquad (18)

2.4 [2,2 points] According to formula (12) for the Doppler effect the relative frequency shift at $\varphi = 0$ is found to be

\left(\frac{\Delta\omega}{\omega}\right)_D = 1 - \sqrt{\frac{1 - v_X/c}{1 + v_X/c}} \approx 9.95 \times 10^{-3}. \qquad (19)

This shows that the Doppler effect cannot fully explain the red shift in the spectrum of the star. It is natural to assume that when the light leaves the surface of the star the photon frequency decreases due to the gravitational redshift.

The gravitational mass is found from the principle of equivalence as

m_{ph} = \frac{\hbar\omega}{c^2}, \qquad (20)

and the gravity force, acting on the photon at a distance $r$ from the star, is equal, according to the Newton law, to

F = G\frac{m_{ph} M}{r^2}. \qquad (21)

The energy conservation law for the motion of the photon can be written as

\hbar d\omega = -F dr. \qquad (22)

Thus,

\frac{d\omega}{\omega} = -\frac{GM}{c^2}\frac{dr}{r^2}. \qquad (23)

On integrating (B4.5) in the range of the stellar radius $R$ до $\infty$ leads to the following equation

\ln\left(\frac{\omega}{\omega_0}\right) = -\frac{GM}{c^2 R}, \qquad (24)

where $\omega_0$ and $\omega$ stand for the frequencies of the photon on the stellar surface and at infinite distance from it, respectively.

Hence, the frequency of the photon at infinite distance from the star is obtained as

\omega = \omega_0 \exp\left(-\frac{GM}{c^2 R}\right) = \omega_0 \exp\left(-\frac{v_{II}^2}{2c^2}\right), \qquad (25)

where the escape velocity is determined by the classical expression

v_{II} = \sqrt{\frac{2GM}{R}}. \qquad (26)

Combining formulas (19) and (25) yields

\left(\frac{\Delta\omega}{\omega}\right)_0 = 1 - \exp\left(-\frac{v_{II}^2}{2c^2}\right)\sqrt{\frac{1 - v_X/c}{1 + v_X/c}}. \qquad (27)

On substituting numerical values, one gets

v_{II} = \sqrt{2\ln\left(\frac{\sqrt{\frac{1 - v_X/c}{1 + v_X/c}}}{1 - \left(\frac{\Delta\omega}{\omega}\right)_0}\right)}\,c = 2.83 \cdot 10^6\ \mathrm{m/s}. \qquad (28)

Part C. Light in a moving medium (4,0 points)

3.1 [1,1 points] By definition, the projections of the object velocity in the reference frame $S'$ are defined as expressions

u_x' = \frac{dx'}{dt'}, \qquad (29)

u_y' = \frac{dy'}{dt'}. \qquad (30)

The same projections in the reference frame $S$ are given by

u_x = \frac{dx}{dt}, \qquad (31)

u_y = \frac{dy}{dt}. \qquad (32)

The Lorentz transformations can be rewritten in the form of finite differences as

dx = \frac{dx' + V dt'}{\sqrt{1 - V^2/c^2}}, \qquad (33)

dy = dy', \qquad (34)

dt = \frac{dt' + dx' V/c^2}{\sqrt{1 - V^2/c^2}}. \qquad (35)

On dividing term by term the left and right hand sides of (33)-(35) and using (29)-(32) yields

u_x = \frac{u_x' + V}{1 + \frac{u_x' V}{c^2}}, \qquad (36)

u_y = \frac{\sqrt{1 - V^2/c^2}}{1 + \frac{u_x' V}{c^2}} u_y'. \qquad (37)

3.2 [1,4 points] Let us sit in the reference frame associated with the water. According to formula (14) the light aberration appears in this reference frame, whereby making the angle $\alpha'$ of incidence of a plane wave on the water surface equal to

\cos\alpha' = \frac{\sqrt{1 - V^2/c^2}\cos\alpha}{1 - V\sin\alpha/c} \approx \cos\alpha\,(1 + V\sin\alpha/c) \quad \text{or}

\sin\alpha' = \frac{\sin\alpha - V/c}{1 - V\sin\alpha/c} \approx \sin\alpha - V\cos\alpha^2/c. \qquad (38)

In the reference frame associated with the water flow, the refraction law has a usual form

\sin\alpha' = n\sin\beta', \qquad (39)

and the speed of light propagation is

v_{ph} = \frac{c}{n}. \qquad (40)

Going back to the laboratory reference frame with the aid of (36) and (37) one finds

v_m \sin\beta = \frac{v_{ph}\sin\beta' + V}{1 + \frac{v_{ph} V \sin\beta'}{c^2}} \approx v_{ph}\sin\beta' + V, \qquad (41)

v_m \cos\beta = \frac{\sqrt{1 - V^2/c^2}}{1 + \frac{v_{ph} V \sin\beta'}{c^2}} v_{ph}\cos\beta' \approx v_{ph}\cos\beta'. \qquad (42)

Using (38)-(42), it is finally obtained that

\sin\beta \approx \frac{1}{n}\sin\alpha - \frac{n^2 + \cos 2\alpha}{n}\frac{V}{c}, \qquad (43)

making

A_1 = \frac{1}{n}\sin\alpha, \qquad (44)

B_1 = -\frac{n^2 + \cos 2\alpha}{n}. \qquad (45)

3.3 [0,4 points] Again using (38)-(42) yields

v_m \approx \frac{c}{n} + V\left(1 - \frac{1}{n^2}\right)\sin\beta. \qquad (46)

whereby

A_2 = \frac{c}{n}, \qquad (47)

B_2 = \left(1 - \frac{1}{n^2}\right)\sin\beta. \qquad (48)

3.4 [0,9 points] When the light propagates in the direction of the water flow, the angle $\beta$ in formula (48) should be taken $\pi/2$ and the corresponding speed is found as

v_+ = \frac{c}{n} + V\left(1 - \frac{1}{n^2}\right), \qquad (49)

when the light propagates in the direction opposite to the water flow, the corresponding speed is obtained as

v_- = \frac{c}{n} - V\left(1 - \frac{1}{n^2}\right). \qquad (50)

Since the total path covered by the two light beams in water is $2L$ , the difference in their propagation time $\Delta t$ is equal to

\Delta t = \frac{2L}{v_-} - \frac{2L}{v_+} \approx \frac{4Lv(n^2 - 1)}{c^2}, \qquad (51)

and the corresponding path difference is derived as follows

\Delta l = c\Delta t = \frac{4Lv(n^2 - 1)}{c}. \qquad (52)

Thus, the interference pattern is shifted by the number of bands equal to

\Delta N = \frac{\Delta l}{\lambda} = \frac{4Lv(n^2 - 1)}{c\lambda}. \qquad (53)

3.5 [0,2 points] Using formula (53) the water refraction index is found to be

n = \sqrt{1 + \frac{c\lambda\Delta N}{4Lv}} = 1.37. \qquad (54)

Part	Content	Points
1.1	Formula (1) $p_x' = \frac{p_x - (V/c)(E/c)}{\sqrt{1 - V^2/c^2}}$	0,2	0,8
	Formula (2) $p_y' = p_y$	0,2
	Formula (3) $p_z' = p_z$	0,2
	Formula (4) $E'/c = \frac{E/c - (V/c) p_x}{\sqrt{1 - V^2/c^2}}$	0,2
1.2	Formula (5) $p = \frac{mv}{\sqrt{1 - v^2/c^2}}$	0,2	0,6
	Formula (6) $E = \frac{mc^2}{\sqrt{1 - v^2/c^2}}$	0,2
	Formula (7) $inv = E^2 - p^2 c^2 = m^2 c^4$	0,2
2.1	Formula (8) $p = \frac{E}{c}$	0,2	1,0
	Formula (9) $E = h\omega$	0,2
	Formula (10) $p_x = p\cos\varphi$	0,2
	Formula (11) $p_y = p\sin\varphi$	0,2
	Formula (12) $\omega' = \omega\frac{1 - V\cos\varphi/c}{\sqrt{1 - V^2/c^2}}$	0,2
2.2	Formula (13) $\frac{h\omega'}{c}\sin\varphi' = \frac{h\omega}{c}\sin\varphi$	0,2	0,4
	Formula (14) $\sin\varphi' = \frac{\sqrt{1 - V^2/c^2}\sin\varphi}{1 - V\cos\varphi/c}$	0,2
2.3	Formula (15) $\delta\varphi = \varphi' - \varphi \approx \frac{V}{c}\sin\varphi$	0,2	1,0
	Formula (16) $a_1 = \frac{V}{c}$	0,2
	Formula (17) $a_2 = \frac{V}{c}\sin\delta$	0,2
	Formula (18) $\delta = \arcsin\left(\frac{a_2}{a_1}\right)$	0,2
	Numerical value $\delta = 64.2^{\circ}$	0,2
2.4	Formula (19) $\left(\frac{\Delta\omega}{\omega}\right)_D = 1 - \sqrt{\frac{1 - v_X/c}{1 + v_X/c}} \approx 9.95 \times 10^{-3}$	0,2	2,2
	Formula (20) $m_{ph} = \frac{\hbar\omega}{c^2}$	0,2
	Formula (21) $F = G\frac{m_{ph} M}{r^2}$	0,2
	Formula (22) $\hbar d\omega = -F dr$	0,2
	Formula (23) $\frac{d\omega}{\omega} = -\frac{GM}{c^2}\frac{dr}{r^2}$	0,2
	Formula (24) $\ln\left(\frac{\omega}{\omega_0}\right) = -\frac{GM}{c^2 R}$	0,2
	Formula (25) $\omega = \omega_0 \exp\left(-\frac{GM}{c^2 R}\right) = \omega_0 \exp\left(-\frac{v_{II}^2}{2c^2}\right)$	0,2
	Formula (26) $v_{II} = \sqrt{\frac{2GM}{R}}$	0,2
	Formula (27) $\left(\frac{\Delta\omega}{\omega}\right)_0 = \left[1 - \sqrt{\frac{1 - v_X/c}{1 + v_X/c}}\right]\exp\left(-\frac{v_{II}^2}{2c^2}\right)$	0,2
	Formula (28) $v_{II} = \sqrt{2}\ln\left(\frac{1 - \sqrt{\frac{1 - v_X/c}{1 + v_X/c}}}{\left(\frac{\Delta\omega}{\omega}\right)_0}\right) c$	0,2
	Numerical value $v_{II} = 7.108 \times 10^{-4} c = 21.31\ \mathrm{km/s}$	0,2
3.1	Formula (29) $u_x' = \frac{dx'}{dt'}$	0,1	1,1
	Formula (30) $u_y' = \frac{dy'}{dt'}$	0,1
	Formula (31) $u_x = \frac{dx}{dt}$	0,1
	Formula (32) $u_y = \frac{dy}{dt}$	0,1
	Formula (33) $dx = \frac{dx' + V dt'}{\sqrt{1 - V^2/c^2}}$	0,1
	Formula (34) $dy = dy'$	0,1
	Formula (35) $dt = \frac{dt' + dx' V/c^2}{\sqrt{1 - V^2/c^2}}$	0,1
	Formula (36) $u_x = \frac{u_x' + V}{1 + \frac{u_x' V}{c^2}}$	0,2
	Formula (37) $u_y = \frac{\sqrt{1 - V^2/c^2}}{1 + \frac{u_x' V}{c^2}} u_y'$	0,2
3.2	Formula (38) $\cos\alpha' = \frac{\sqrt{1 - V^2/c^2}\cos\alpha}{1 - V\sin\alpha/c} \approx \cos\alpha\,(1 + V\sin\alpha/c)$ or $\sin\alpha' = \frac{\sin\alpha - V/c}{1 - V\sin\alpha/c} \approx \sin\alpha - V\cos\alpha^2/c$	0,2	1,4
	Formula (39) $\sin\alpha' = n\sin\beta'$	0,2
	Formula (40) $v_{ph} = \frac{c}{n}$	0,2
	Formula (41) $v_m \sin\beta = \frac{v_{ph}\sin\beta' + V}{1 + \frac{v_{ph} V \sin\beta'}{c^2}} \approx v_{ph}\sin\beta' + V$	0,2
	Formula (42) $v_m \cos\beta = \frac{\sqrt{1 - V^2/c^2}}{1 + \frac{v_{ph} V \sin\beta'}{c^2}} v_{ph}\cos\beta' \approx v_{ph}\cos\beta'$	0,2
	Formula (44) $A_1 = \frac{1}{n}\sin\alpha$	0,2
	Formula (45) $B_1 = -\frac{n^2 + \cos 2\alpha}{n}$	0,2
3.3	Formula (47) $A_2 = \frac{c}{n}$	0,2	0,4
	Formula (48) $B_2 = \left(1 - \frac{1}{n^2}\right)\sin\beta$	0,2
3.4	Formula (49) $v_+ = \frac{c}{n} + V\left(1 - \frac{1}{n^2}\right)$	0,2	0,9
	Formula (50) $v_- = \frac{c}{n} - V\left(1 - \frac{1}{n^2}\right)$	0,2
	Formula (51) $\Delta t = \frac{2L}{v_-} - \frac{2L}{v_+} \approx \frac{4Lv(n^2 - 1)}{c^2}$	0,2
	Formula (52) $\Delta l = c\Delta t = \frac{4Lv(n^2 - 1)}{c}$	0,2
	Formula (53) $\Delta N = \frac{\Delta l}{\lambda} = \frac{4Lv(n^2 - 1)}{c\lambda}$	0,1
3.5	Formula (54) $n = \sqrt{1 + \frac{c\lambda\Delta N}{4Lv}}$	0,1	0,2
	Numerical value $n = \sqrt{1 + \frac{c\lambda\Delta N}{4Lv}} = 1.37$	0,1
Total			10,0

Оригинал в Архива: IZhO-2018-Theory_eng.pdf · официални решения: IZhO-2018-Theory_eng_sol.pdf