IPhO 2024 · Theory problem 3

Black Widow Pulsar

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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.

Figure 1

Figure 1 - (a) The falling gases of the companion star onto the neutron star - (b) The companion of the neutron star PSR J2215+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.

A-1 Write \(\varphi(x,y)\) in terms of \(M_1\), \(M_2\), \(G\), and \(a\). 1.0 pt
A-2 Assuming \(M_1>M_2\), plot the function \(\varphi(x,0)\) qualitatively. 0.7 pt

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.

A-3 Find the numerical value of \(\dfrac{x_0}{a}\), up to two significant figures. You may use the calculator. 0.5 pt

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.

A-4 Calculate the rate of change of \(a\) and \(P\) in terms of \(\beta\), \(M_1\), \(M_2\), \(G\), and \(a\). 0.6 pt

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.

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). 1.0 pt

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 \(M_\odot=1.98\times10^{30}\ \mathrm{kg}\) is the mass of the Sun. The rotational period is \(P=4.14\ \mathrm{hr}\), and the Stefan-Boltzmann constant is \(\sigma=5.67\times10^{-8}\ \mathrm{W/m^2K^4}\), and the gravitational constant is \(G=6.67\times10^{-11}\ \mathrm{m^3/kg\,s^2}\).

Assume that the mass flow rate to the neutron star is \(\beta=\dot M_{\mathrm{NS}}=9\times10^{-10}\,M_\odot\,\mathrm{yr}^{-1}\).

A-6 Calculate the temperature of the disc at the radius \(r=\dfrac{a}{10}\) in kelvins. 0.5 pt

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.

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 pt

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.

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 pt
B-2 Derive a (differential) equation for determining \(\rho(r)\) at equilibrium, and write it in the following form: \(\dfrac{d}{dr}\left[h_1(\rho,r)\dfrac{d\rho}{dr}\right]+h_2(r)\rho=0\). Find the functions \(h_1\) and \(h_2\). 0.6 pt
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 pt
B-4 Rewrite the (differential) equation of task B-2 in the following form: \(\dfrac{d}{dx}\left[A_1(u,x)\dfrac{du}{dx}\right]+A_2(x)u(x)=0,\) where \(x=\dfrac{r}{r_0}\) and \(u=\dfrac{\rho}{\rho_c}\). Find the functions \(A_1(u,x)\) and \(A_2(x)\). 0.3 pt
B-5 For \(\gamma=2\) one finds \(u(x)=\dfrac{f(x)}{x}\). Determine \(f(x)\). 0.6 pt

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

Figure 2

Figure 2 - The plot of \(\dfrac{du}{dx}\)

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 pt

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 \(\varepsilon(t)\ll1\).

B-7 Find \(\tilde\rho\) and \(\tilde g\) in terms of \(\rho\) and \(g\) to the first order in \(\varepsilon\). 0.9 pt
B-8 Using Newton's equation of motion for the spherical layer with the equilibrium radius of \(r\) find \(\dfrac{d^2\tilde r}{dt^2}\) in terms of \(\tilde g\), \(\tilde\rho\), \(K\), \(\gamma\), and \(\dfrac{\partial\tilde\rho}{\partial\tilde r}\) (By \(\dfrac{\partial\tilde\rho}{\partial\tilde r}\) we mean derivative of \(\tilde\rho\) with respect to \(\tilde r\) at constant \(t\).) 0.6 pt
B-9 Obtain \(\dfrac{d^2\varepsilon}{dt^2}\) 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 pt

Official English text of the problem.

Answers

Write \(M=M_1+M_2\), and let the stellar positions be \(x_1=-aM_2/M\) and \(x_2=aM_1/M\).

  • A-1 \[ \boxed{ \varphi(x,y)= -\frac{GM_1}{\sqrt{(x+aM_2/M)^2+y^2}} -\frac{GM_2}{\sqrt{(x-aM_1/M)^2+y^2}} -\frac{GM}{2a^3}(x^2+y^2) } \] up to an additive constant.

  • A-2 There are three strictly concave branches, each with one maximum. The potential tends to \(-\infty\) on both sides of \(x_1,x_2\), and as \(x\to\pm\infty\). With the usual labels, the maxima occur in the order \[ x_{L_3}<x_1<0<x_{L_1}<x_2<x_{L_2}, \] and their heights satisfy \[ \varphi(L_1)<\varphi(L_2)<\varphi(L_3)<0. \]

  • A-3 \[ \boxed{\frac{x_0}{a}=0.36}. \]

  • A-4 \[ \boxed{\dot a=2\beta a\left(\frac1{M_2}-\frac1{M_1}\right)}, \qquad \boxed{\dot P=6\pi\beta\sqrt{\frac{a^3}{GM}} \left(\frac1{M_2}-\frac1{M_1}\right)}. \]

  • A-5 Using local orbital-energy release and radiation from both faces, \[ \boxed{T(r)=\left(\frac{GM_1\beta}{8\pi\sigma r^3}\right)^{1/4}}. \]

  • A-6 \[ \boxed{T(a/10)\simeq8.9\times10^3\ \mathrm K}. \]

  • A-7 The escape-speed threshold is \[ \boxed{v'_{\mathrm{crit}}=\sqrt{\frac{2G(M_1'+M_2)}a}}, \] with a bound orbit requiring \(v'<v'_{\mathrm{crit}}\). For isotropic ejection, \[ \boxed{M_1'>\frac{M_1-M_2}{2}}. \] Equality in either condition is the marginal, parabolic escape case.

  • B-1 \[ \boxed{g(r)\simeq-\frac{4\pi}{3}G\rho_c r}. \]

  • B-2 One convenient normalization is \[ \boxed{h_1(\rho,r)=K\gamma r^2\rho^{\gamma-2}}, \qquad \boxed{h_2(r)=4\pi G r^2}. \]

  • B-3 \[ \boxed{l=-\frac12,\quad m=\frac12,\quad n=-1}, \qquad \boxed{r_0=\sqrt{\frac{p_c}{G\rho_c^2}}}. \]

  • B-4 \[ \boxed{A_1(u,x)=\gamma x^2u^{\gamma-2}}, \qquad \boxed{A_2(x)=4\pi x^2}. \]

  • B-5 \[ \boxed{f(x)=\frac{\sin(\sqrt{2\pi}\,x)}{\sqrt{2\pi}}}. \]

  • B-6 \[ \boxed{\gamma\simeq1.67}. \]

  • B-7 \[ \boxed{\tilde\rho=\rho(1-3\varepsilon)+O(\varepsilon^2)}, \qquad \boxed{\tilde g=g(1-2\varepsilon)+O(\varepsilon^2)}. \]

  • B-8 \[ \boxed{\frac{d^2\tilde r}{dt^2} =\tilde g-K\gamma\tilde\rho^{\gamma-2} \frac{\partial\tilde\rho}{\partial\tilde r}}. \]

  • B-9 \[ \boxed{\ddot\varepsilon+ \frac{4\pi G\rho_c}{3}(3\gamma-4)\varepsilon=0}. \] The critical value is \(\boxed{\gamma_{\mathrm{crit}}=4/3}\), with restoring oscillations for \(\gamma>4/3\) and marginal behavior at equality. Their angular frequency is \[ \boxed{\omega=\sqrt{\frac{4\pi G\rho_c}{3}(3\gamma-4)}}. \]

Solution

A-1

Let \(M=M_1+M_2\). The center-of-mass condition and the separation give \[ M_1x_1+M_2x_2=0,\qquad x_2-x_1=a, \] so \[ x_1=-\frac{M_2}{M}a,\qquad x_2=\frac{M_1}{M}a. \]

If the binary’s angular velocity is \(\Omega\), Newton’s equation for either star gives \[ \Omega^2=\frac{GM}{a^3}. \]

The gravitational potential per unit test mass is the sum of the two point-mass potentials. In the uniformly rotating frame, the outward centrifugal acceleration is generated by the potential \[ \varphi_{\mathrm{cf}}=-\frac12\Omega^2(x^2+y^2). \] Consequently, \[ \boxed{ \varphi(x,y)= -\frac{GM_1}{\sqrt{(x+aM_2/M)^2+y^2}} -\frac{GM_2}{\sqrt{(x-aM_1/M)^2+y^2}} -\frac{GM}{2a^3}(x^2+y^2) }. \]

A-2

On the \(x\)-axis, \[ \varphi(x,0)= -\frac{GM_1}{|x-x_1|} -\frac{GM_2}{|x-x_2|} -\frac12\Omega^2x^2. \]

A precise description of the sketch is as follows:

  • There are downward vertical singularities at \(x=x_1\) and \(x=x_2\): the potential approaches \(-\infty\) from either side of each star.
  • At large positive or negative \(x\), \[ \varphi(x,0)\sim-\frac12\Omega^2x^2, \] so both outer ends descend to \(-\infty\).
  • Each of the three intervals separated by the stars contains exactly one maximum.

To establish the last point, away from the masses, \[ \frac{d^2\varphi}{dx^2} =-\frac{2GM_1}{|x-x_1|^3} -\frac{2GM_2}{|x-x_2|^3} -\Omega^2<0. \] Thus each branch is strictly concave. Its derivative changes from positive to negative between its two ends, giving one maximum.

Denote the left, inner, and right maxima by \(L_3,L_1,L_2\), respectively. Since \(M_1>M_2\), the heavier star is closer to the origin, and \[ \left.\frac{d\varphi}{dx}\right|_{x=0} =\frac{GM_1}{|x_1|^2}-\frac{GM_2}{x_2^2}>0. \] Therefore the inner maximum is to the right of the origin: \[ x_{L_3}<x_1<0<x_{L_1}<x_2<x_{L_2}. \]

The left outer maximum is highest, the right outer maximum is next, and the inner maximum is lowest: \[ \varphi(L_1)<\varphi(L_2)<\varphi(L_3). \]

With the additive constant chosen in A-1, the whole curve is negative and has no horizontal-axis intercept. Its vertical-axis intercept is \[ \boxed{\varphi(0,0) =-\frac{GM}{a}\left(\frac{M_1}{M_2}+\frac{M_2}{M_1}\right)}. \]

A-3

The gas first passes toward \(M_1\) through the inner equilibrium point \(L_1\), where the effective force vanishes. This point is a maximum of \(\varphi(x,0)\) along the line joining the stars and a saddle of the two-dimensional effective potential.

Here \[ M_1=\frac34M,\qquad M_2=\frac14M, \] so \[ x_1=-\frac a4,\qquad x_2=\frac{3a}{4}. \]

Writing \(z=x_0/a\), the condition \(\partial\varphi/\partial x=0\) between the stars becomes \[ \frac{3/4}{(z+1/4)^2} -\frac{1/4}{(3/4-z)^2} -z=0. \]

The root in \(-1/4<z<3/4\) is \[ z=0.360743\ldots. \] Thus, to two significant figures, \[ \boxed{\frac{x_0}{a}=0.36}. \]

A-4

Treat the transfer as conservative: all mass lost by \(M_2\) reaches \(M_1\), and there is no external torque. In the point-mass model, spin and angular momentum temporarily stored in the stream are neglected. Hence \[ \dot M_1=\beta,\qquad \dot M_2=-\beta,\qquad \dot M=0, \] and the orbital angular momentum is conserved.

For a circular binary, \[ L=\frac{M_1M_2}{M}a^2\Omega =M_1M_2\sqrt{\frac{Ga}{M}}. \] Taking its logarithmic derivative, \[ 0=\frac{\dot L}{L} =\frac{\beta}{M_1}-\frac{\beta}{M_2} +\frac12\frac{\dot a}{a}. \] Therefore \[ \boxed{\dot a =2\beta a\left(\frac1{M_2}-\frac1{M_1}\right)}. \]

Kepler’s relation gives \[ P=2\pi\sqrt{\frac{a^3}{GM}}. \] Since \(M\) is constant, \[ \frac{\dot P}{P}=\frac32\frac{\dot a}{a} =3\beta\left(\frac1{M_2}-\frac1{M_1}\right). \] Thus \[ \boxed{\dot P =6\pi\beta\sqrt{\frac{a^3}{GM}} \left(\frac1{M_2}-\frac1{M_1}\right)}. \]

Both the separation and the period increase when mass flows from the lighter star to the heavier one.

A-5

Use a thin disk with nearly circular Keplerian orbits, negligible radial kinetic energy, and local radiation of the orbital mechanical energy lost while matter crosses an annulus.

The circular-orbit condition is \[ v^2=\frac{GM_1}{r}. \] The gas’s mechanical energy per unit mass is consequently \[ e(r)=\frac12v^2-\frac{GM_1}{r} =-\frac{GM_1}{2r}. \]

Consider an annulus between \(r\) and \(r+dr\), where \(dr>0\). Matter enters through its outer edge and leaves through its inner edge. At mass flow rate \(\beta\), the released power is \[ dL=\beta\bigl[e(r+dr)-e(r)\bigr] =\frac{GM_1\beta}{2r^2}\,dr. \]

Both disk faces radiate. Their combined annular area is \[ dA=2(2\pi r\,dr)=4\pi r\,dr. \] Neglecting heat storage and equating the local energy release to blackbody radiation, \[ 4\pi r\,dr\,\sigma T^4 =\frac{GM_1\beta}{2r^2}\,dr. \] Hence \[ \boxed{T(r)=\left(\frac{GM_1\beta}{8\pi\sigma r^3}\right)^{1/4}}. \]

In particular, \(T\propto r^{-3/4}\).

A-6

The total mass and period are \[ M=(2.27+0.33)(1.98\times10^{30}) =5.148\times10^{30}\ \mathrm{kg}, \] \[ P=4.14(3600)=14904\ \mathrm s. \]

Kepler’s relation gives \[ a=\left(\frac{GMP^2}{4\pi^2}\right)^{1/3} =1.24548\times10^9\ \mathrm m. \] Therefore \[ r=\frac a{10}=1.24548\times10^8\ \mathrm m. \]

Using \(1\ \mathrm{yr}\simeq3.156\times10^7\ \mathrm s\), \[ \beta= \frac{9\times10^{-10}(1.98\times10^{30})} {3.156\times10^7} \simeq5.65\times10^{13}\ \mathrm{kg\,s^{-1}}. \]

With \(M_1=2.27(1.98\times10^{30})\ \mathrm{kg}\), \[ T= \left[ \frac{(6.67\times10^{-11}) (2.27\times1.98\times10^{30}) (5.65\times10^{13})} {8\pi(5.67\times10^{-8})(1.24548\times10^8)^3} \right]^{1/4}. \] This yields \[ T\simeq8.86\times10^3\ \mathrm K, \] or \[ \boxed{T\simeq8.9\times10^3\ \mathrm K}. \]

A-7

The explosion is sudden, so the separation immediately afterward remains \(a\). Once the rapidly ejected material has left, the relative orbital energy of the remaining stars is \[ E'=\frac12\mu'v'^2-\frac{GM_1'M_2}{a}, \qquad \mu'=\frac{M_1'M_2}{M_1'+M_2}. \]

A bound two-body orbit requires \(E'<0\). Thus \[ v'^2<\frac{2G(M_1'+M_2)}a, \] and the limiting speed is \[ \boxed{v'_{\mathrm{crit}} =\sqrt{\frac{2G(M_1'+M_2)}a}}. \] At equality the orbit is parabolic and is not bound.

If the ejection is isotropic in the exploding star’s rest frame, there is no recoil. The instantaneous velocities of both remaining stars are unchanged. Their relative speed therefore remains the original circular-orbit speed: \[ v'^2=\Omega^2a^2=\frac{G(M_1+M_2)}a. \]

Substitution into the binding condition gives \[ M_1+M_2<2(M_1'+M_2), \] or \[ \boxed{M_1'>\frac{M_1-M_2}{2}}. \]

Equivalently, less than half of the binary’s original total mass may be ejected: \[ M_1-M_1'<\frac{M_1+M_2}{2}. \] The stated minimum is a critical threshold; equality gives marginal escape.

B-1

By spherical symmetry, only the mass enclosed inside radius \(r\) contributes to the gravitational field there. Near the center, \(\rho\simeq\rho_c\), so \[ m(r)\simeq\frac{4\pi}{3}\rho_c r^3. \]

Taking outward acceleration as positive, \[ g(r)=-\frac{Gm(r)}{r^2} \simeq-\frac{4\pi}{3}G\rho_c r. \] Therefore \[ \boxed{g(r)\simeq-\frac{4\pi}{3}G\rho_c r}. \]

B-2

Hydrostatic equilibrium requires \[ \frac{dp}{dr}=\rho g. \] The enclosed mass satisfies \[ g=-\frac{Gm(r)}{r^2}, \qquad \frac{dm}{dr}=4\pi r^2\rho. \] Consequently, \[ \frac{d}{dr}(r^2g)=-4\pi G r^2\rho. \]

Using \(g=(1/\rho)\,dp/dr\), \[ \frac{d}{dr}\left(\frac{r^2}{\rho}\frac{dp}{dr}\right) +4\pi G r^2\rho=0. \]

From \(p=K\rho^\gamma\), \[ \frac{dp}{dr}=K\gamma\rho^{\gamma-1}\frac{d\rho}{dr}. \] The required density equation is therefore \[ \boxed{ \frac{d}{dr}\left( K\gamma r^2\rho^{\gamma-2}\frac{d\rho}{dr} \right)+4\pi G r^2\rho=0 }. \] Thus one valid choice is \[ \boxed{h_1=K\gamma r^2\rho^{\gamma-2}}, \qquad \boxed{h_2=4\pi G r^2}. \] Multiplying both functions by the same nonzero constant gives an equivalent equation.

B-3

The relevant dimensions are \[ [G]=L^3M^{-1}T^{-2},\qquad [p_c]=ML^{-1}T^{-2},\qquad [\rho_c]=ML^{-3}. \]

Requiring \(G^lp_c^m\rho_c^n\) to have dimension \(L\) gives \[ -l+m+n=0, \qquad 3l-m-3n=1, \qquad -2l-2m=0. \] Solving, \[ l=-\frac12,\qquad m=\frac12,\qquad n=-1. \] Hence \[ \boxed{r_0=\frac{\sqrt{p_c}}{\sqrt G\,\rho_c} =\sqrt{\frac{p_c}{G\rho_c^2}}}. \]

Since \(p_c=K\rho_c^\gamma\), it is also useful to write \[ r_0^2=\frac{K\rho_c^{\gamma-2}}G. \]

B-4

Substitute \[ r=r_0x,\qquad \rho=\rho_cu, \qquad \frac{d\rho}{dr}=\frac{\rho_c}{r_0}\frac{du}{dx} \] into the equation from B-2: \[ K\gamma\rho_c^{\gamma-1} \frac{d}{dx}\left(x^2u^{\gamma-2}\frac{du}{dx}\right) +4\pi G r_0^2\rho_c x^2u=0. \]

Because \(Gr_0^2\rho_c=K\rho_c^{\gamma-1}\), division by \(K\rho_c^{\gamma-1}\) gives \[ \boxed{ \frac{d}{dx}\left( \gamma x^2u^{\gamma-2}\frac{du}{dx} \right)+4\pi x^2u=0 }. \] Thus \[ \boxed{A_1(u,x)=\gamma x^2u^{\gamma-2}}, \qquad \boxed{A_2(x)=4\pi x^2}. \]

The central conditions are \[ u(0)=1,\qquad u'(0)=0, \] the latter following from regularity and spherical symmetry.

B-5

For \(\gamma=2\), the equation reduces to \[ \frac{d}{dx}(x^2u')+2\pi x^2u=0. \]

Let \(u=f/x\). Then \[ x^2u'=xf'-f, \qquad \frac{d}{dx}(x^2u')=xf''. \] For \(x>0\), \[ f''+2\pi f=0. \]

Writing \(k=\sqrt{2\pi}\), the general solution is \[ f=A\sin(kx)+B\cos(kx). \] Finiteness of \(u=f/x\) at the center requires \(B=0\). The normalization \(u(0)=1\) then gives \(Ak=1\). Therefore \[ \boxed{f(x)=\frac{\sin(\sqrt{2\pi}\,x)}{\sqrt{2\pi}}}, \] and \[ u(x)=\frac{\sin(\sqrt{2\pi}\,x)}{\sqrt{2\pi}\,x}. \]

The physical density profile ends at its first zero, \(x=\sqrt{\pi/2}\).

B-6

For this task, return to general \(\gamma\). Near the center, regularity permits the expansion \[ u(x)=1+cx^2+O(x^4), \qquad u'(x)=2cx+O(x^3). \]

Substitution into the dimensionless equilibrium equation gives, at order \(x^2\), \[ 6\gamma c+4\pi=0. \] Hence \[ u'(x)=-\frac{4\pi}{3\gamma}x+O(x^3). \]

If \(s\) is the slope of the plotted \(u'\)-versus-\(x\) curve at the origin, then \[ s=u''(0)=-\frac{4\pi}{3\gamma}, \qquad \gamma=-\frac{4\pi}{3s}. \]

The tangent at the origin in Figure 2 has slope approximately \[ s\simeq-2.51. \] Thus \[ \boxed{\gamma\simeq\frac{4\pi}{3(2.51)}\simeq1.67}. \] This is consistent with \(\gamma=5/3\).

B-7

Follow a particular material shell. Its mass is conserved as it moves from \(r\) to \[ \tilde r=r(1+\varepsilon). \] Since the same scale factor applies to neighboring central shells, \[ d\tilde r=(1+\varepsilon)\,dr. \]

Conservation of the mass in a thin shell gives \[ 4\pi r^2\rho\,dr =4\pi\tilde r^2\tilde\rho\,d\tilde r, \] so \[ \tilde\rho=\frac{\rho}{(1+\varepsilon)^3} =\rho(1-3\varepsilon)+O(\varepsilon^2). \] Therefore \[ \boxed{\tilde\rho\simeq\rho(1-3\varepsilon)}. \]

The mass enclosed by the moving shell is also conserved. Consequently, \[ \tilde g=-\frac{Gm(r)}{\tilde r^2} =\frac{g}{(1+\varepsilon)^2}, \] giving \[ \boxed{\tilde g\simeq g(1-2\varepsilon)}. \] Here \(g\) is negative, consistently with the prescribed outward-positive convention.

B-8

The outward pressure acceleration is \[ -\frac1{\tilde\rho} \frac{\partial\tilde p}{\partial\tilde r}. \] Adding gravity, Newton’s equation is \[ \frac{d^2\tilde r}{dt^2} =\tilde g-\frac1{\tilde\rho} \frac{\partial\tilde p}{\partial\tilde r}. \]

Using the same equation of state for the perturbed matter, \[ \tilde p=K\tilde\rho^\gamma, \] we obtain \[ \frac{\partial\tilde p}{\partial\tilde r} =K\gamma\tilde\rho^{\gamma-1} \frac{\partial\tilde\rho}{\partial\tilde r}. \] Hence \[ \boxed{ \frac{d^2\tilde r}{dt^2} =\tilde g-K\gamma\tilde\rho^{\gamma-2} \frac{\partial\tilde\rho}{\partial\tilde r} }. \]

B-9

At fixed time, \(\varepsilon\) is independent of the central shell label \(r\). Therefore \[ \tilde\rho(\tilde r,t) =(1+\varepsilon)^{-3} \rho\!\left(\frac{\tilde r}{1+\varepsilon}\right), \] and \[ \frac{\partial\tilde\rho}{\partial\tilde r} =(1+\varepsilon)^{-4}\frac{d\rho}{dr}. \]

The pressure term in B-8 consequently scales as \[ K\gamma\tilde\rho^{\gamma-2} \frac{\partial\tilde\rho}{\partial\tilde r} = K\gamma\rho^{\gamma-2}\frac{d\rho}{dr} (1+\varepsilon)^{2-3\gamma}. \]

In the unperturbed equilibrium, \[ K\gamma\rho^{\gamma-2}\frac{d\rho}{dr}=g. \] Together with \(\tilde g=g(1+\varepsilon)^{-2}\), this gives \[ \frac{d^2\tilde r}{dt^2} =g\left[(1+\varepsilon)^{-2} -(1+\varepsilon)^{2-3\gamma}\right]. \]

To first order, \[ (1+\varepsilon)^{-2}=1-2\varepsilon, \] \[ (1+\varepsilon)^{2-3\gamma} =1+(2-3\gamma)\varepsilon. \] Thus \[ \frac{d^2\tilde r}{dt^2}=g(3\gamma-4)\varepsilon. \]

Since \(r\) labels the original shell and is constant, \[ \frac{d^2\tilde r}{dt^2}=r\ddot\varepsilon. \] Using the near-center result \(g/r=-4\pi G\rho_c/3\), \[ \boxed{ \ddot\varepsilon =-\frac{4\pi G\rho_c}{3}(3\gamma-4)\varepsilon }. \]

For a restoring acceleration, the coefficient multiplying \(-\varepsilon\) must be positive. Hence \[ \gamma>\frac43. \] The critical value is therefore \[ \boxed{\gamma_{\mathrm{crit}}=\frac43}. \]

At equality there is no restoring force to this order; for \(\gamma<4/3\), a growing perturbation exists. For \(\gamma>4/3\), the motion is harmonic with \[ \boxed{\omega =\sqrt{\frac{4\pi G\rho_c}{3}(3\gamma-4)} =\sqrt{4\pi G\rho_c\left(\gamma-\frac43\right)}}. \]

The graph-derived value \(\gamma\simeq1.67\) lies in the stable range.

The solution was checked against the official solution.