Съдържание

IPhO 2024, Теория — Задача 3. Black Widow Pulsar

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от IPhO 2024

Black Widow Pulsar

Условие

A significant number of the observed stars are binaries. One or both of the stars may be neutron stars rotating with a high angular velocity and emitting electromagnetic waves; such stars are called pulsars. Sometimes a companion star is an expansive mass of gas that gradually falls down onto the neutron star and causes its mass to increase (Figure 1-a). In this way, a neutron star gradually swallows up a portion of the mass of its companion star. For this reason, the neutron star has been compared to a black widow (or redback spider), a female spider which eats its mate after mating. The heating of the gas falling down onto the black widow generates radiation which can be observed. The heaviest neutron stars often are black widows and they serve as natural laboratories for testing fundamental physics. Figure 1-b shows the picture of the companion of the neutron star PSR J2215+5135, taken by the 3.4-meter optical telescope of the Iranian National Observatory. No neutron star can be seen in this image and the observed light is due to its companion.

(a) Artist's impression of a neutron star with an accretion disk pulling gas from a large yellow-red companion star. (b) Telescope image (green/blue field) with several bright spots, one marked with a red circle, showing the companion of PSR J2215+5135. — Figure 1 - (a) The falling gases of the companion star onto the neutron star - (b) The companion of the neutron star PSRJ2215+5135

A. A Binary System

Consider a simple model in which the black widow and its companion star, are represented by two point masses $M_1$ and $M_2$ moving on a circular orbit around their center of mass. To investigate the dynamics of this system, consider a rotating coordinate system in which the two bodies are stationary. Take the center of mass to be the origin of the coordinate system. Assume that the two point masses lie on the $x$ -axis on both sides of the origin at a distance $a$ from each other, and that $M_1$ lies on the negative $x$ -axis. At an arbitrary point $(x, y)$ in the plane of motion, the effective potential $\varphi(x, y)$ for a unit test mass is the sum of the gravitational potentials of the two point masses plus the centrifugal potential.

Suppose (just for task A-3) $M_2 = M_1/3$ and assume that $M_2$ is surrounded by a rarefied gas of very low density. The mass of this gas is insignificant and we ignore its gravitational effects. If the size of this gas envelope becomes greater than a specific limit, the gas will overflow onto $M_1$ . Suppose the overflow occurs through $x = x_0$ on the $x$ -axis.

Take the rotational period of the stars around their center of mass to be $P$ . Assume that mass flows from $M_2$ to $M_1$ at a very small rate of $dM_1/dt = \beta$ . This rate is so small that in each period of rotation, the distance between the two stars can be assumed to be constant. However, after a long period of time, the distance between the two stars changes, while the motion remains circular.

The gas separated from $M_2$ forms a disk rotating around $M_1$ and heats up due to friction (Figure 1-a). As the gas loses energy, it spirals inward toward $M_1$ and finally falls onto it. In the steady state, the mass flows at the constant rate of $\beta$ , from $M_2$ to the disc and from the disc onto $M_1$ . At the same time, the heated disk emits thermal radiation as a blackbody. This disk forms very close to the neutron star so the gravitational pull of the $M_2$ star can be ignored for the analysis of the disk's motion. Also, ignore the heat capacity of the gas.

In the binary system PSR J2215+5135, the mass of the neutron star is $M_{\mathrm{NS}} = 2.27\,M_\odot$ and the mass of its companion star is $M_{\mathrm{S}} = 0.33\,M_\odot$ , where $M_\odot = 1.98\times 10^{30}\ \mathrm{kg}$ is the mass of the Sun. The rotational period is $P = 4.14\ \mathrm{hr}$ , and the Stefan-Boltzmann constant is $\sigma = 5.67\times 10^{-8}\ \mathrm{W/m^2K^4}$ , and the gravitational constant is $G = 6.67\times 10^{-11}\ \mathrm{m^3/kgs^2}$ . Assume that the mass flow rate to the neutron star is $\beta = \dot{M}_{\mathrm{NS}} = 9\times 10^{-10}\,M_\odot\,\mathrm{yr}^{-1}$ .

Assume that after a sudden explosion, the $M_1$ star ejects a part of its mass out of the binary system at a very high speed, and its mass becomes $M_1'$ . Take the magnitude of the velocity of $M_1'$ relative to $M_2$ to be $v'$ after the explosion.

B. Analysis of the Stability of a Star

In this part we study the stability of a single star. Consider a star containing a specific kind of matter with the equation of state $p = K\rho^{\gamma}$ where $K$ and $\gamma$ are constants. Let $p(r)$ and $\rho(r)$ be the pressure and density at a distance $r$ from the center of the star, respectively. The pressure and density at the center of the star are $p_c$ and $\rho_c$ , respectively. In all tasks of the part B, take all outward vectors to be positive.

Assume that for a particular star $\frac{du}{dx}$ , as a function of $x$ , is given by the curve given in Figure 2.

Graph of du/dx versus x: a parabola-like curve starting at 0 at x=0, reaching a minimum of about -1.0 at x≈0.65, and rising back to about -0.2 at x≈1.5; axis ticks from 0.2 to 1.4 and du/dx from -0.2 to -1.0. — Figure 2 - The plot of $\frac{du}{dx}$

To analyze the stability of the system, we assume that the star deviates slightly from its equilibrium state: we assume that the spherical shell, which was in equilibrium at radius $r$ , now has a radius $\tilde{r}$ , similarly the parameters $g$ , $p$ , and $\rho$ have changed to $\tilde{g}$ , $\tilde{p}$ , and $\tilde{\rho}$ respectively. For convenience, we shall only consider small $r$ 's near the center of the star, for which we can assume that $\tilde{r} = r(1 + \varepsilon(t))$ , where $\varepsilon(t) \ll 1$ .

A-1 Write $\varphi(x, y)$ in terms of $M_1$ , $M_2$ , $G$ , and $a$ . [1 т.]

A-2 Assuming $M_1 > M_2$ , plot the function $\varphi(x, 0)$ qualitatively. [0,7 т.]

A-3 Find the numerical value of $\frac{x_0}{a}$ , up to two significant figures. You may use the calculator. [0,5 т.]

A-4 Calculate the rate of change of $a$ and $P$ in terms of $\beta$ , $M_1$ , $M_2$ , $G$ , and $a$ . [0,6 т.]

A-5 Determine the temperature of the disc at distance $r$ from the center of the star $M_1$ in terms of $\beta$ , $M_1$ , $G$ , and $\sigma$ (Stefan-Boltzmann constant). [1 т.]

A-6 Calculate the temperature of the disc at the radius $r = \frac{a}{10}$ in kelvins. [0,5 т.]

A-7 Determine the maximum value of $v'$ , in terms of $M_1'$ , $M_2$ , $G$ , and $a$ , that allows the new binary system to stay bounded. Assuming that the explosion is isotropic, what is the minimum value of $M_1'$ for the binary system to remain bounded? [0,7 т.]

B-1 Determine the gravitational acceleration $g(r)$ near the center of the star in terms of $r$ and the constants $G$ and $\rho_c$ . [0,2 т.]

B-2 Derive a (differential) equation for determining $\rho(r)$ at equilibrium, and write it in the following form: $\frac{d}{dr}[\,h_1(\rho, r)\,\frac{d\rho}{dr}\,] + h_2(r)\rho = 0$ . Find the functions $h_1$ and $h_2$ . [0,6 т.]

B-3 Construct a quantity $r_0$ of the form $r_0 = G^l p_c{}^m \rho_c{}^n$ with the dimension of length. [0,4 т.]

B-4 Rewrite the (differential) equation of task B-2 in the following form:

\frac{d}{dx}[\,A_1(u, x)\,\frac{du}{dx}\,] + A_2(x)\,u(x) = 0,

where $x = \frac{r}{r_0}$ and $u = \frac{\rho}{\rho_c}$ . Find the functions $A_1(u, x)$ and $A_2(x)$ . [0,3 т.]

B-5 For $\gamma = 2$ one finds $u(x) = \frac{f(x)}{x}$ . Determine $f(x)$ . [0,6 т.]

B-6 Use the behavior of the curve in Figure 2, in the vicinity of the point $x = 0$ , to find $\gamma$ up to 3 significant figures. Use the given ruler if necessary. [0,8 т.]

B-7 Find $\tilde{p}$ and $\tilde{g}$ in terms of $\rho$ and $g$ to the first order in $\varepsilon$ . [0,9 т.]

B-8 Using Newton's equation of motion for the spherical layer with the equilibrium radius of $r$ find $\frac{d^2\tilde{r}}{dt^2}$ in terms of $\tilde{g}$ , $\tilde{\rho}$ , $K$ , $\gamma$ , and $\frac{\partial \tilde{p}}{\partial \tilde{r}}$ (By $\frac{\partial \tilde{p}}{\partial \tilde{r}}$ we mean derivative of $\tilde{p}$ with respect to $\tilde{r}$ at constant $t$ .) [0,6 т.]

B-9 Obtain $\frac{d^2\varepsilon}{dt^2}$ in terms of $\varepsilon$ and the constants given in the problem. Find the minimum value of $\gamma$ for a stable equilibrium, and find the oscillation's angular frequency of the star. [0,6 т.]

Решение

A. A Binary System

A-1. Assume $a_1$ and $a_2$ , are respectively, the distances of $M_1$ and $M_2$ from the center of mass:

\begin{cases} M_1 a_1 = M_2 a_2 \\ a_1 + a_2 = a \end{cases} \to\ a_1 = \frac{M_2}{M}a\ ,\ a_2 = \frac{M_1}{M}a\ :\ M = M_1 + M_2

In the rotating coordinate system, a centrifugal potential has to be added to the gravitational potential of the two masses:

U = -\frac{1}{2}\omega^2 r^2 \qquad \omega = \sqrt{\frac{GM}{a^3}}

\varphi(x,y) = -\frac{GM_1}{\sqrt{(x+a_1)^2+y^2}} - \frac{GM_2}{\sqrt{(x-a_2)^2+y^2}} - \frac{1}{2}\,\omega^2(x^2+y^2)

\varphi(x,y) = -\frac{GM_1}{\sqrt{\left(x+\frac{M_2}{M}a\right)^2+y^2}} - \frac{GM_2}{\sqrt{\left(x-\frac{M_1}{M}a\right)^2+y^2}} - \frac{1}{2}\frac{GM}{a^3}(x^2+y^2)

A-1 (1.0 pt)

\varphi(x,y) = -\frac{GM_1}{\sqrt{\left(x+\frac{M_2}{(M_1+M_2)}a\right)^2+y^2}} - \frac{GM_2}{\sqrt{\left(x-\frac{M_1}{(M_1+M_2)}a\right)^2+y^2}} - \frac{1}{2}\frac{G(M_1+M_2)}{a^3}(x^2+y^2)

A-2. We set $y = 0$ in the previous equation, and obtain:

\varphi(x,0) = -\frac{GM_1}{\left|x+\frac{M_2}{M}a\right|} - \frac{GM_2}{\left|x-\frac{M_1}{M}a\right|} - \frac{1}{2}\frac{GM}{a^3}\,x^2

We draw the diagram noting that:

The function has asymptotes at $x = -a_1$ and $x = a_2$ , and it tends to $-\infty$ at both sides of these asymptotes.
The function has three maxima which are called Lagrange points.
The function goes to $-\infty$ for $x \to \pm\infty$

A-2 (0.7 pt)

Graph of the effective potential φ(x,0) versus x: the curve tends to −∞ on both sides of vertical dashed asymptotes at x = −a₁ and x = a₂, has three maxima (Lagrange points) and goes to −∞ for x → ±∞.

A-3. Let $\bar{x} = x/a$ , and denote the Lagrange point in the middle (between $\bar{x} = 0$ and $\bar{x} = 0.75$ ) by $\bar{x}_0$ , we have $\frac{d\varphi}{d\bar{x}}(\bar{x}_0) = 0$ . Using the given ratios:

\varphi(\bar{x},0) = \frac{GM}{a}\left[-\frac{\frac{3}{4}}{\left(\bar{x}+\frac{1}{4}\right)} + \frac{\frac{1}{4}}{\left(\bar{x}-\frac{3}{4}\right)} - \frac{1}{2}\,\bar{x}^2\right]

Let $f(\bar{x}) = \frac{a}{GM}\frac{d\varphi}{d\bar{x}}$ , then we have to solve for $f(\bar{x}_0) = 0$ . We have $f(0) > 0$ and $f(0.5) < 0$ , so the answer lies between 0 and 0.5. For the midpoint, we have $f(\bar{x}_0 = 0.25) > 0$ so $0.25 < \bar{x}_0 < 0.5$ , so by trial and error:

\begin{cases} f(0) > 0 \\ f(0.5) < 0 \end{cases} \to f(0.25) > 0 \to 0.25 < \bar{x}_0 < 0.5 \to f(0.375) < 0 \to \cdots \to 0.358 < \bar{x}_0 < 0.361

\to f(0.360) > 0 \to 0.360 < \bar{x}_0 < 0.361\ \to\ \frac{x_0}{a} = \bar{x}_0 \approx 0.36

So, up to two significant figures the answer is 0.36.

A-3 (0.5 pt)

\frac{x_0}{a} = 0.36

A-4. The angular momentum of the system is:

J = \mu aV = \mu a^2\omega = \frac{M_1 M_2}{M}\,a^2\sqrt{\frac{GM}{a^3}} = \sqrt{\frac{GM_1^2M_2^2}{M}\,a},

where $\mu$ is the reduced mass and $V$ is the relative velocity of the two point masses. Taking the logarithm of both sides we'll have:

\ln J = \frac{1}{2}\left[\ln\frac{G}{M} + 2\ln M_1 + 2\ln M_2 + \ln a\right]

For slowly-varying quantities we'll obtain:

\frac{\dot{J}}{J} = \frac{\dot{M}_1}{M_1} + \frac{\dot{M}_2}{M_2} + \frac{1}{2}\frac{\dot{a}}{a}

because the total mass is a constant and $\dot{M}_1 + \dot{M}_2 = 0$ ; therefore:

\frac{\dot{a}}{a} = -2\frac{\dot{M}_1}{M_1}\left(1 - \frac{M_1}{M_2}\right) \quad\to\quad \dot{a} = -2\beta a\left(\frac{1}{M_1} - \frac{1}{M_2}\right)

For the period we'll have:

P = 2\pi\sqrt{\frac{a^3}{GM}}\ \to\ \frac{\dot{P}}{P} = \frac{3}{2}\frac{\dot{a}}{a} = -3\frac{\dot{M}_1}{M_1}\left(1 - \frac{M_1}{M_2}\right) \quad\to\quad \dot{P} = -6\pi\sqrt{\frac{a^3}{GM}}\,\beta\left(\frac{1}{M_1} - \frac{1}{M_2}\right)

A-4 (0.6 pt)

\dot{a} = -2\beta a\left(\frac{1}{M_1} - \frac{1}{M_2}\right)

\dot{P} = -6\pi\sqrt{\frac{a^3}{GM}}\,\beta\left(\frac{1}{M_1} - \frac{1}{M_2}\right)

A-5. In an infinitesimally thin ring with an inner radius of $r$ and an outer radius $r+dr$ , energy is leaving at a rate of $-\frac{GM_1\beta}{2r}$ and entering at a rate $-\frac{GM_1\beta}{2r} + \frac{GM_1\beta}{2r^2}\,dr$ . For the ring to stay in equilibrium, the excess energy of $\frac{GM_1\beta}{2r}\,dr$ per unit time must leave the system as radiation, so:

dP = \frac{GM_1\beta}{2r^2}\,dr = \sigma T^4\,2(2\pi r\,dr) = 4\pi\sigma T^4\,dr\ \to\ \ T = \left(\frac{GM_1\beta}{8\pi\sigma r^3}\right)^{\frac{1}{4}}

A-5 (1.0 pt)

T = \left(\frac{GM_1\beta}{8\pi\sigma r^3}\right)^{\frac{1}{4}}

A-6. From $P = 2\pi\sqrt{\frac{a^3}{GM}}$ we'll have:

a = \left[\frac{P^2 G(M_S + M_{NS})}{4\pi^2}\right]^{\frac{1}{3}}

Using the result of Part A.5, the temperature is:

T = \left(\frac{GM_{NS}\beta}{8\pi\sigma r^3}\right)^{\frac{1}{4}} = \left(\frac{500\pi\,M_{NS}\,\beta}{\sigma P^2(M_S + M_{NS})}\right)^{\frac{1}{4}} = 9\times 10^3 K

A-6 (0.5 pt)

T = 9\times 10^3\ K

A.7. For the system to remain bounded, the total mechanical energy of the system must be negative:

E' = \frac{1}{2}\mu' v'^2 - \frac{GM_1'M_2}{a} < 0 \to v' < \sqrt{\frac{2G(M_1' + M_2)}{a}}

For an isotropic explosion, we would have $v' = v = \sqrt{\frac{GM}{a}}$ therefore:

\sqrt{\frac{G(M_1 + M_2)}{a}} < \sqrt{\frac{2G(M_1' + M_2)}{a}}

and:

\frac{M_1 - M_2}{2} < M_1'

A-7 (0.7 pt)

v'_{\max} = \sqrt{\frac{2G(M_1'+M_2)}{a}}

M_{1\min}' = \frac{M_1 - M_2}{2}

B. Analysis of the stability of a star

B-1. Using Newton's law of gravity:

g = -\frac{4\pi G\int_0^r r'^2\rho\,dr'}{r^2}\ \overset{\rho\cong\rho_c}{=}\ -\frac{4\pi G\rho_c r}{3}

B-1 (0.2 pt)

g = -\frac{4\pi G\rho_c r}{3}

B-2. Balance of forces for a differential element of volume with a surface area of $A$ and thickness $\Delta r$ between radii $r$ and $r+\Delta r$ is as follows:

\vec{F} = -\frac{GM(\vec{r})\rho}{r^2}\,A\,\Delta r - \Delta p\,A = 0

in which $M(r)$ is the mass of the part of the star confined within the radius $r$ . As $\Delta r$ is small, we can write:

\frac{G\rho}{r^2}\left(\int 4\pi r'^2\rho(r')dr'\right) = -\frac{dp(r)}{dr} = -K\gamma\rho^{\gamma-1}\frac{d\rho}{dr}

Multiplying both sides of the equation by $\frac{r^2}{4\pi G\rho}$ and taking the derivative once again, we get:

\frac{d}{dr}\left[r^2\rho^{\gamma-2}\frac{d\rho}{dr}\right] + \frac{4\pi G r^2}{K\gamma}\,\rho(r) = 0

B-2 (0.6 pt)

h_1(\rho,r) = r^2\rho^{\gamma-2}

h_2(r) = \frac{4\pi G r^2}{K\gamma}

B-3.

[\rho_c] = ML^{-3}\,,\quad [p_c] = ML^{-1}T^{-2}\,,\quad [G] = M^{-1}L^3T^{-2}

[G^l p_c^m \rho_c^n] = (M^{-1}L^3T^{-2})^l (ML^{-1}T^{-2})^m (ML^{-3})^n = L

\begin{cases} -l + n + m = 0 \\ 3l - 3n - m = 1 \\ -2l - 2m = 0 \end{cases} \to \begin{cases} l = -\frac{1}{2} \\ m = \frac{1}{2} \\ n = -1 \end{cases} \quad\to\quad r_0 = G^{-\frac{1}{2}}p_c^{\frac{1}{2}}\rho_c^{-1}

B-3 (0.4 pt)

r_0 = G^{-\frac{1}{2}}p_c^{\frac{1}{2}}\rho_c^{-1}

B-4.

\frac{K\gamma\rho_c^{\gamma-1}}{4\pi G r_0^2 x^2}\frac{d}{dx}\left[x^2 u^{\gamma-2}\frac{du}{dx}\right] = -\rho_c u(r)

\frac{K\gamma\rho_c^{\gamma-2}}{4\pi G r_0^2 x^2}\frac{d}{dx}\left[x^2 u^{\gamma-2}\frac{du}{dx}\right] = \frac{\gamma}{4\pi x^2}\frac{d}{dx}\left[x^2 u^{\gamma-2}\frac{du}{dx}\right] = -u

\frac{d}{dx}\left[x^2 u^{\gamma-2}\frac{du}{dx}\right] + \frac{4\pi x^2}{\gamma}u = 0

B-4 (0.3 pt)

A_1(u,x) = x^2 u^{\gamma-2}

A_2(x) = \frac{4\pi x^2}{\gamma}

B-5.

\gamma = 2\ \to\ \frac{d}{dx}\left[x^2\frac{du}{dx}\right] = -2\pi x^2 u(x) \to f''(x) = -2\pi f(x) \to\ f(x) = \frac{\sin(\sqrt{2\pi}\,x)}{\sqrt{2\pi}}

B-5 (0.6 pt)

f(x) = \frac{\sin(\sqrt{2\pi}x)}{\sqrt{2\pi}}

B.6.

\frac{d^2u}{dx^2} + \frac{(\gamma-2)}{u}\left(\frac{du}{dx}\right)^2 + \frac{2}{x}\left(\frac{du}{dx}\right) + \frac{4\pi}{\gamma}\,u^{3-\gamma} = 0

u'(0) = 0\ ,\quad \lim_{x\to 0}\frac{u'(x)}{x} = u''(0)

u''(0) + 2u''(0) + \frac{4\pi}{\gamma} = 0 \quad\to\quad \gamma = -\frac{4\pi}{3u''(0)}

\gamma \sim [1.64, 1.70]

B-6 (0.8 pt)

\gamma = [1.64, 1.70]

B.7.

M(r) = \int_0^{\tilde r(r,t)} 4\pi r'^2\,\tilde\rho(r',t)\,dr' = \int_0^{r} 4\pi r'^2\rho(r')\,dr'

4\pi r^2\rho(r) = 4\pi\tilde r^2\,\tilde\rho(\tilde r, t)\,\frac{\partial\tilde r}{\partial r} \;\to\; \frac{\tilde\rho}{\rho} = \frac{r^2}{\tilde r^2}\left(\frac{\partial\tilde r}{\partial r}\right)^{-1} = (1+\epsilon)^{-3} \cong 1 - 3\epsilon

\frac{\tilde g}{g} = \frac{GM/\tilde r^2}{GM/r^2} = \frac{1/\tilde r^2}{1/r^2} = (1+\epsilon)^{-2} \cong 1 - 2\epsilon

B-7 (0.9 pt)

\tilde g \simeq g(1-2\epsilon)

\tilde\rho \simeq \rho(1-3\epsilon)

B-8. we have

\frac{\partial\tilde p}{\partial\tilde r} = \tilde\rho\left(\tilde g - \ddot{\tilde r}\right)

And

\tilde p = K\tilde\rho^{\gamma}

So:

\ddot{\tilde r} = \tilde g - \frac{\left(\frac{\partial\tilde p}{\partial\tilde r}\right)}{\tilde\rho} = \tilde g - K\gamma\tilde\rho^{\gamma-2}\frac{\partial\tilde\rho}{\partial\tilde r}

B-8 (0.6 pt)

\frac{d^2\tilde r}{dt^2} = \tilde g - K\gamma\tilde\rho^{\gamma-2}\frac{\partial\tilde\rho}{\partial\tilde r}

B.9. Using of the results in B.7 and B.8, we have:

\frac{d^2\tilde r}{dt^2} = \ddot{\tilde r} = \tilde g - K\gamma\tilde\rho^{\gamma-2}\frac{\partial\tilde\rho}{\partial\tilde r} = g(1-2\epsilon) - K\gamma\rho^{\gamma-2}\frac{\partial\rho}{\partial r}\left(\frac{(1-3\epsilon)^{\gamma-1}}{(1+\epsilon)}\right)

= g(1-2\epsilon) - K\gamma\rho^{\gamma-2}\frac{\partial\rho}{\partial r}\left(1 - 3(\gamma-1)\epsilon - \epsilon\right)

Equilibrium requires:

g - K\gamma\rho^{\gamma-2}\frac{\partial\rho}{\partial r} = 0 \;\Rightarrow\; K\gamma\rho^{\gamma-2}\frac{\partial\rho}{\partial r} = g

therefore:

\ddot{\tilde r} = r\ddot\epsilon = g(1-2\epsilon) - g\left(1 - 3(\gamma-1)\epsilon - \epsilon\right) = g(3\gamma - 4)\epsilon

and:

\ddot\epsilon = \frac{g}{r}(3\gamma-4)\epsilon

\ddot\epsilon = -\frac{4\pi G\rho_c}{3}(3\gamma-4)\epsilon

Stability requires that:

3\gamma - 4 > 0 \;\Rightarrow\; \gamma > \frac{4}{3}

and the angular velocity of the oscillations will be:

\omega = \sqrt{\frac{4\pi G\rho_c}{3}(3\gamma - 4)}

B-9 (0.6 pt)

\ddot\epsilon = -\frac{4\pi G\rho_c}{3}(3\gamma-4)\epsilon

\gamma_{\min} = \frac{4}{3}

\omega = \sqrt{\frac{4\pi G\rho_c}{3}(3\gamma - 4)}

Оригинал в Архива: IPhO_2024_Q3.pdf · официални решения: IPhO_2024_S3.pdf