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IZhO 2022, Теория — Задача 3. Problem 3. Corpuscular interpretation of light pressure

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XVIII International Zhautykov Olympiad/Theoretical Competition · 16 февруари 2022 г. · 10 т.

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The duration of the theoretical competition is 4 hours. There are three problems.
You can use your own calculator for numerical calculations.
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Условие

Introduction

Electromagnetic waves, reflected from the interfaces between media or absorbed by them, exert mechanical pressure, whose corpuscular interpretation is the subject of this problem. The main postulate of the corpuscular theory of electromagnetic radiation states that electromagnetic radiation, in particular light, is a beam of particles called photons with an energy determined by Planck's formula. In what follows, take the speed of light be equal to $c = 2.98\cdot10^{8}\ \mathrm{m/s}$ .

3.1 Let a parallel beam of light with intensity $I_0$ fall on a flat surface at an angle $\varphi$ with the normal, and the coefficient of reflection from the surface is $R = I_r/I_0$ , where $I_r$ stands for the intensity of the reflected light. Find the pressure of light $p_s$ exerted on the surface, provided that the reflection coefficient does not depend on the angle of incidence.

3.2 Calculate the pressure $p_s$ of the solar radiation, whose intensity is equal to $I_s = 1400\ \mathrm{W/m^{2}}$ . Assume that light is incident perpendicular to: (a) the completely absorbing earth's surface; b) the completely reflective (mirror) surface.

3.3 A ball with the radius $R = 1.00\ \mathrm{m}$ is illuminated by a wide parallel beam of sunlight with intensity $I_s = 1400\ \mathrm{W/m^{2}}$ . One half of the ball is made of a material that completely reflects light, while the other half absorbs it completely. Both halves are symmetrically illuminated by the beam. Calculate the moment of light pressure forces acting on the ball about its axis of symmetry, which is perpendicular to the beam and lies in the plane dividing the ball into the mirror and absorbing halves.

Space station with the mirror sail

The space station rests far from the planets at a distance of $R_0 = 5.00\cdot10^{7}\ \mathrm{km}$ from the Sun, held by a solar sail (fully reflective mirror) oriented perpendicular to the sun's rays. At certain time moment, the station's engines turn on for a short time and give it an initial speed $V_0 = 9.00\ \mathrm{m/s}$ in the direction from the Sun. Completely ignore the influence of the solar wind, which is a beam of ionized particles, mostly protons, helium nuclei and some others. Consider it known that the Earth moves around the Sun in a circular orbit of the radius $r_E = 1.50\cdot10^{8}\ \mathrm{km}$ with the speed of $V_E = 30.0\ \mathrm{km/s}$ .

3.4 Calculate the maximum distance $R$ from the Sun to the station.

3.5 Determine the minimum value of the speed $V_0 = V_{min}$ , at which the station will be able to fly away from the Sun

Poynting-Robertson effect

In astrophysics, the physical process is well known, such that in the solar system solid dust particles slowly fall onto the Sun along a spiral trajectory whose shape is very close to circular. Let us consider a similar spherical dust particle of radius $a = 1.00\ \mathrm{mm}$ and density $\rho = 3.00\cdot10^{3}\ \mathrm{kg/m^{3}}$ , which rotates around the Sun at such a distance from it that light with intensity $I_s = 1400\ \mathrm{W/m^{2}}$ falls on the dust particle. Under given conditions, the radial component of the pressure of light on the particle can be neglected in comparison with the gravity of the Sun. Consider that dust the particle completely absorbs the incident radiation.

3.6 Evaluate the characteristic time $\tau$ , during which the distance to the Sun $R$ decreases by the relative value $\mu = \frac{\delta R}{R} = 10^{-4}$ .

Laser tweezer

In 2018, the Nobel Prize in physics was awarded to A. Ashkin for the creation of "laser tweezer", a device that allows one to hold and move transparent microscopic objects with the help of light. In one of the devices of such a "laser tweezer", a parallel beam of light from a laser passes through a converging lens $L$ and hits a microparticle $M$ , which can also be considered a converging lens. Point $F$ is the common focus of $L$ and $M$ (see figure 3.1 below). The light intensity in the beam is $I = 1.00\ \mu W/sm^{2}$ , the beam radius is $R = 1.00\ \mathrm{cm}$ , the focal length of the lens $L$ is $F = 10.0\ \mathrm{cm}$ . Ignore completely the absorption and reflection of light.

3.7 Calculate the force acting on the microparticle.

To create a force acting on the particle in the transverse direction of the beam, the left half of the lens $L$ is covered by a diaphragm (see figure 3.2 below).

3.8 Calculate the force acting on the microparticle in the transverse direction of the beam.

A parallel beam of width 2R falls on lens L; rays converge to focus F, cross there, and pass through microparticle M which again collimates them downwards. — Figure 3.1

Same arrangement as Figure 3.1 but the left half of lens L is covered by a dark diaphragm; only the right half of the beam converges through focus F onto microparticle M. — Figure 3.2

Решение

Problem 3. Corpuscular interpretation of light pressure (10.0 points)

Introduction

3.1 Let the concentration of photons with the energy $\varepsilon$ in the incident radiation be equal to $n$ , then the wave intensity is determined by the relation

I_0 = c\varepsilon n, \qquad (1)

where $c$ stands for the speed of light.

The number of photons $\Delta N$ falling on the area element $\Delta S$ at the angle $\varphi$ per unit of time is written as

\Delta N = cn\Delta t\Delta S\cos\varphi. \qquad (2)

The number of absorbed photons per unit of time is found as follows

\Delta N_a = (1-R)\Delta N, \qquad (3)

whereas the number of reflected ones

\Delta N_r = R\Delta N. \qquad (4)

The normal component of the momentum, transferred by one photon to the area element upon absorption, is equal to

\Delta p_a = \frac{\varepsilon}{c}\cos\varphi, \qquad (5)

and the same value at reflection is put down as

\Delta p_r = 2\,\frac{\varepsilon}{c}\cos\varphi. \qquad (6)

The total momentum transferred to the area element is determined by the expression

\Delta p = \Delta N_a\Delta p_a + \Delta N_r\Delta p_r, \qquad (7)

and the pressure sought is calculated by the formula

p_s = \frac{\Delta p}{\Delta S\Delta t} = \frac{I_0}{c}(1+R)\cos^2\varphi. \qquad (8)

3.2 At normal incidence $\varphi = 0$ and at complete absorption $R = 0$ , we obtain

p_s = \frac{I_s}{c} = 4.70\cdot 10^{-6}\ \mathrm{Pa}. \qquad (9)

and, accordingly, at total reflection $R=1$

p_s = \frac{2I_s}{c} = 9.40\cdot 10^{-6}\ \mathrm{Pa}. \qquad (10)

3.3 Consider a section of the spherical surface perpendicular to the incident light flux. For the mirror part of the surface, which completely reflects light, the mechanical torque is equal to zero, since the transmitted momentum is directed strictly along the radius of the sphere.

Cross-section of a sphere of radius R perpendicular to the incident light flux; the radius R, the distance x from the center to a strip of width dx, and a dashed vertical axis through the center are shown.

Let us consider a strip in the section located from the center of the sphere at distances from $x$ to $x+dx$ . The selected part of the completely absorbing surface has the area

dS = 2\sqrt{R^2 - x^2}\,dx, \qquad (11)

and the number of absorbed photons per unit time is equal to

\Delta N_a = \frac{I_s}{\varepsilon}dS, \qquad (12)

each of which has the momentum

\Delta p_a = \frac{\varepsilon}{c}. \qquad (13)

The force shoulder is

l = x, \qquad (14)

therefore, the torque of forces acting on the selected area is obtained as

dM = \Delta N_a\Delta p_a l = \frac{2I_s}{c}\sqrt{R^2-x^2}\,x\,dx, \qquad (15)

and the total torque of forces is determined by the integral

M = \int\limits_0^R dM = \frac{2I_s R^3}{3c} = 3.13\cdot 10^{-6}\ \mathrm{N}\cdot\mathrm{m}. \qquad (16)

Space station with the mirror sail

3.4 At the initial rest point of the station of mass $m$ with the sail of area $S$ , located at the distance $R_0$ from the Sun of mass $M_S$ , the gravitational force is exactly balanced by the light pressure force, which leads to the equation

G\,\frac{M_S m}{R_0^2} = \frac{2n_0\varepsilon}{c}\,S, \qquad (17)

where $G$ refers to the gravitational constant, $n_0$ is the concentration of photons of solar radiation with energy $\varepsilon$ at the location of the station.

Due to the spherically symmetric expansion, the photon concentration changes with the distance $r$ from the Sun according to the law

n(r) = n_0\left(\frac{R_0}{r}\right)^2. \qquad (18)

The initial momentum of photons before the collision with the sail is written as

p_0 = \frac{\varepsilon}{c}, \qquad (19)

whereas the final one constitutes

p = \frac{\varepsilon}{c}\,\frac{c-V}{c+V}. \qquad (20)

This relationship is easily obtained from the kinematics and is actually the formula for the Doppler effect. In addition, the momentum of a photon after reflection from the sail mirror can be easily obtained from the laws of conservation of momentum and energy by considering an absolutely elastic collision of a photon with a moving massive mirror.

Thus, the change in the momentum of the photon is transferred to the mirror and is equal to

\Delta p = p - p_0 = \frac{2\varepsilon}{c+V}, \qquad (21)

and the number of photons falling per unit time $\Delta t$ on the sail is derived as

\frac{\Delta N}{\Delta t} = n(r)S(c-V). \qquad (22)

Hence, the force acting on the station due to the solar radiation is determined by the expression

f = \Delta p\,\frac{\Delta N}{\Delta t} = 2n_0\varepsilon S\left(\frac{R_0}{r}\right)^2\frac{c-V}{c+V} = G\,\frac{M_S m}{r^2}\,\frac{c-V}{c+V}. \qquad (23)

The station is also subject to the force of gravitational attraction from the Sun

f_g = G\,\frac{M_S m}{r^2}. \qquad (24)

which means that the motion of the station in the radial direction is described by Newton's second law in the form

m\frac{dV}{dt} = f - f_g = -2G\,\frac{M_S m}{r^2}\,\frac{V}{c+V}. \qquad (25)

Bearing in mind that for a small displacement

dr = Vdt, \qquad (26)

we obtain from expression (25) the differential equation

(c+V)dV = -2GM_S\,\frac{dr}{r^2}, \qquad (27)

which is easily integrated and, if the station stops, gives rise to

cV_0 + \frac{1}{2}V_0^2 = 2GM_S\left(\frac{1}{R_0} - \frac{1}{R}\right). \qquad (28)

Solving equation (28), we find the distance sought as

R = \frac{R_0}{1 - \dfrac{\left(cV_0 + \frac{1}{2}V_0^2\right)R_0}{2GM_S}}, \qquad (29)

which, under the condition of the Earth's orbital motion

GM_S = V_E^2 r_E, \qquad (30)

as well as the relation $V \ll c$ , yields the final answer of the form

R = \frac{R_0}{1 - \dfrac{cV_0 R_0}{2V_E^2 r_E}} = 9.93\cdot 10^{10}\ \mathrm{m}. \qquad (31)

3.5 It follows from formula (31) that the station is able to fly away to infinity $R \to \infty$ only if the denominator of the expression becomes zero, which results in

V_{\min} = \frac{2V_E^{2}r_E}{cR_0} = 18.1 m/s . (32)

Poynting-Robertson effect

3.6 The mass of the dust particle is determined by the expression

m = \rho \frac{4}{3}\pi a^{3}, (33)

and its cross-sectional area is

S = \pi a^{2}. (34)

Let us determine the effective force acting on the particle as a result of light absorption. To reduce it to the pressure of light, let us move to the frame of reference associated with the dust particle. In this frame of reference, the particle is affected by the pressure of light, calculated by formula (9), but its direction does not coincide with the radial one due to the aberration of light, namely, it makes a small angle $V/c$ with it. Thus, in the tangential direction of the particle trajectory, a force appears due to the absorption of photons, equal to

F = -V \frac{I_S}{c^{2}}S , (35)

which creates a torque about the center of attraction found as

M = -FR. (36)

Since the trajectory of the dust particle is almost circular, its velocity can be written as

V = \sqrt{\frac{GM_S}{R}}, (37)

and the angular momentum relative to the attracting center

L = mVR . (38)

Collecting equations (33)-(38) together, we write

\frac{dL}{dt} = M , (39)

whence we finally find the time sought in the following form

\tau = \frac{2\mu\rho a c^{2}}{3I_S} = 1.27\cdot 10^{8} s . (40)

At the derivation, change in the intensity of solar radiation with distance is neglected, since the radius of the orbit decreases only slightly and the corresponding corrections are of higher order of smallness.

Note: A consistent explanation of the Poynting-Robertson effect is based on the following interpretation. In the reference frame associated with the particle, it absorbs the solar radiation, which propagates at a small angle to the radial direction, and then reradiates the accumulated energy isotropically in all directions. In the reference frame associated with the Sun, the primary radiation of the Sun propagates in the radial direction, and the reradiation of the particle itself is no longer isotropic. In the first case, the appearance of the braking force moment is explained by the aberration of solar radiation, whereas in the second case, by the Doppler effect for the reradiation of the particle itself.

Laser tweezer

3.7 Let us calculate the force acting on the first converging lens, which is equal to the total change in the momentum of photons incident on the lens per unit time. Obviously, the momentum changes due to the refraction of light in the glass, since its direction changes, but not the module.

Consider all the rays passing through the ring on the lens, located from its center at distances from $r$ to $r+dr$ .

Ring element on the lens for problem 3.7: ring of radius r and width dr at the top, a refracted ray making a small angle θ with the dashed vertical axis, and the common focus F at the bottom.

The area of this ring is written as

dS = 2\pi r dr . (41)

The change in the longitudinal momentum of photons passing through the given ring per unit time is equal to

dp_{\parallel} = \frac{I}{c}(1 - \cos\theta)dS , (42)

where the angle of refraction is found as folows

\sin\theta = \frac{r}{F}, (43)

since all rays converge at the focus of the lens.

Integrating the resulting expression over the entire surface of the lens, we obtain

f_{\parallel} = \int\limits_0^{R} dp_{\parallel} = \frac{\pi I}{c}\left(R^{2} - \frac{2}{3F}\left[F^{3} - \left(F^{2} - R^{2}\right)^{3/2}\right]\right) \approx \frac{\pi IR^{4}}{4cF^{2}} = 2.64\cdot 10^{-17} N . (44)

Since the foci of the lens $L$ and the particle $M$ coincide, when leaving the "lens-particle" system, the light beam propagates again parallel to the optical axis, and, therefore, as a result of refraction on the particle $M$ , the photon momentum is restored. Consequently, the force acting on the particle $M$ is equal in magnitude to $f_{\parallel}$ , but is directed towards the converging lens. This force draws the particle into the laser radiation field. This is the principle of operation of the "laser tweezer".

3.8 Consider all the rays passing through the element of the semiring on the lens, located from its center at distances from $r$ to $r+dr$ , and also cut off by azimuth angles from $\beta$ to $\beta + d\beta$ . The area of this semicircle element is derived as

dS = r dr d\beta . (45)

The change in the transverse momentum of photons passing through the given ring per unit time is equal to

dp_{\perp} = \frac{I}{c}\sin\theta\sin\beta dS , (46)

and integration over the entire surface of the half of the lens, taking into account formula (43), leads to the expression

f_{\perp} = \int dp_{\perp} = \frac{I}{cF}\int\limits_0^{\pi}\int\limits_0^{R} r^{2}dr\sin\beta d\beta = \frac{2IR^{3}}{3cF} = 2.24\cdot 10^{-16} N . (47)

	Content	Points
3.1	Formula (1): $I_0 = c\varepsilon n$	0.1	0.8
	Formula (2): $\Delta N = cn\Delta t\,\Delta S\cos\varphi$	0.1
	Formula (3): $\Delta N_a = (1-R)\Delta N$	0.1
	Formula (4): $\Delta N_r = R\Delta N$	0.1
	Formula (5): $\Delta p_a = \dfrac{\varepsilon}{c}\cos\varphi$	0.1
	Formula (6): $\Delta p_r = 2\dfrac{\varepsilon}{c}\cos\varphi$	0.1
	Formula (7): $\Delta p = \Delta N_a\Delta p_a + \Delta N_r\Delta p_r$	0.1
	Formula (8): $p_s = \dfrac{I_0}{c}(1+R)\cos^{2}\varphi$	0.1
3.2	Formula (9): $p_s = \dfrac{I_s}{c}$	0.1	0.4
	Numerical value in formula (9): $p_s = 4.70\cdot 10^{-6}\ \mathrm{Pa}$	0.1
	Formula (10): $p_s = \dfrac{2I_s}{c}$	0.1
	Numerical value in formula (10): $p_s = 9.40\cdot 10^{-6}\ \mathrm{Pa}$	0.1
3.3	Moment of forces on the mirror part of the sphere $M = 0$	0.1	1.0
	Formula (11): $dS = 2\sqrt{R^{2}-x^{2}}\,dx$	0.1
	Formula (12): $\Delta N_a = \dfrac{I_s}{\varepsilon}dS$	0.1
	Formula (13): $\Delta p_a = \dfrac{\varepsilon}{c}$	0.1
	Formula (14): $l = x$	0.1
	Formula (15): $dM = \dfrac{2I_s}{c}\sqrt{R^{2}-x^{2}}\,dx$	0.1
	Formula (16): $M = \dfrac{2I_s R^{3}}{3c}$	0.2
	Numerical value in formula (16): $M = 3.13\cdot 10^{-6}\ \mathrm{N\cdot m}$	0.2
3.4	Formula (17): $G\,\dfrac{M_S m}{R_0^{2}} = \dfrac{2n_0\varepsilon}{c}S$	0.4	3.6
	Formula (18): $n(r) = n_0\left(\dfrac{R_0}{r}\right)^{2}$	0.2
	Formula (19): $p_0 = \dfrac{\varepsilon}{c}$	0.1
	Formula (20): $p = \dfrac{\varepsilon}{c}\,\dfrac{c-V}{c+V}$	0.2
	Formula (21): $\Delta p = p - p_0 = \dfrac{2\varepsilon}{c+V}$	0.1
	Formula (22): $\dfrac{\Delta N}{\Delta t} = n(r)S(c-V)$	0.2
	Formula (23): $f = G\,\dfrac{M_S m}{r^{2}}\,\dfrac{c-V}{c+V}$	0.2
	Formula (24): $f_g = G\,\dfrac{M_S m}{r^{2}}$	0.2
	Formula (25): $m\dfrac{dV}{dt} = -2G\,\dfrac{M_S m}{r^{2}}\,\dfrac{V}{c+V}$	0.4
	Formula (26): $dr = Vdt$	0.2
	Formula (27): $(c+V)dV = -2GM_S\,\dfrac{dr}{r^{2}}$	0.2
	Formula (28): $cV_0 + \dfrac{1}{2}V_0^{2} = 2GM_S\left(\frac{1}{R_0} - \frac{1}{R}\right)$	0.2
	Formula (29): $R = \dfrac{R_0}{1 - \dfrac{(cV_0 + \frac{1}{2}V_0^{2})R_0}{2GM_S}}$	0.2
	Formula (30): $GM_S = V_E^{2}r_E$	0.2
	Formula (31): $R = \dfrac{R_0}{1 - \dfrac{cV_0 R_0}{2V_E^{2}r_E}}$	0.4
	Numerical value in formula (31): $R = 9.93\cdot 10^{10}\ \mathrm{m}$	0.2
3.5	Formula (32): $V_{\min} = \dfrac{2V_E^{2}r_E}{cR_0}$	0.2	0.4
	Numerical value in formula (32): $V_{\min} = 18.1\ \mathrm{m/s}$	0.2
3.6	Formula (33): $m = \rho\,\dfrac{4}{3}\pi a^{3}$	0.1	2.0
	Formula (34): $S = \pi a^{2}$	0.1
	Formula (35): $F = -V\,\dfrac{I_s}{c^{2}}S$	0.4
	Formula (36): $M = FR$	0.2
	Formula (37): $V = \sqrt{\dfrac{GM_S}{R}}$	0.2
	Formula (38): $L = mVR$	0.2
	Formula (39): $\dfrac{dL}{dt} = M$	0.2
	Formula (40): $\tau = \dfrac{2\mu\rho a c^{2}}{3I_s}$	0.4
	Numerical value in formula (40): $\tau = 1.27\cdot 10^{8}\ \mathrm{s}$	0.2
3.7	Formula (41): $dS = 2\pi r dr$	0.1	1.0
	Formula (42): $dp_{\parallel} = \dfrac{I}{c}(1-\cos\theta)dS$	0.2
	Formula (43): $\sin\theta = \dfrac{r}{F}$	0.2
	Formula (44): $f_{\parallel} = \dfrac{\pi I}{c}\left(R^{2} - \dfrac{2}{3F}\left[F^{3} - \left(F^{2}-R^{2}\right)^{3/2}\right]\right) \approx \dfrac{\pi IR^{4}}{4cF^{2}}$	0.3
	Numerical value in formula (44): $f_{\parallel} = 2.64\cdot 10^{-17}\ \mathrm{N}$	0.2
3.8	Formula (45): $dS = r\,dr\,d\beta$	0.1	0.8
	Formula (46): $dp_{\perp} = \dfrac{I}{c}\sin\theta\sin\beta\,dS$	0.2
	Formula (47): $f_{\perp} = \dfrac{2IR^{3}}{3cF}$	0.3
	Numerical value in formula (47): $f_{\perp} = 2.24\cdot 10^{-16}\ \mathrm{N}$	0.2
Total			10.0

Mathematical hints for the theoretical problems

The following formulas may be useful:

\int x^n dx = \frac{x^{n+1}}{n+1} + C,

where $n \neq -1$ is a fixed number, $C$ refers to an arbitrary constant

\int \frac{dx}{x} = \ln \ln |x| + C,

where $C$ stands for an arbitrary constant

(1 + x)^{\gamma} \approx 1 + \gamma x + \frac{\gamma(\gamma-1)}{2}x^2,

for $|x| \ll 1$ and any value of $\gamma$

\ln \ln (1 + x) \approx x,

for $|x| \ll 1$

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