IPhO 2023 · Theory problem 2

Neutron Stars

We discuss the stability of large nuclei and estimate the mass of neutron stars theoretically and experimentally.

Part A. Mass and stability of nuclei (2.5 points)

The rest-energy of a nucleus \(m(Z,N)c^2\) consisting of \(Z\) protons and \(N\) neutrons is smaller than the sum of rest-energies of protons and neutrons, hereafter called nucleons, by the binding energy \(B(Z,N)\), where \(c\) is the speed of light in vacuum. Ignoring minor corrections, we can approximate the binding energy consisting of the volume term with \(a_V\), the surface term with \(a_S\), the Coulomb energy term with \(a_C\), and the symmetry energy term with \(a_{\mathrm{sym}}\) in the following way.

\[ m(Z,N)c^2=Am_Nc^2-B(Z,N),\qquad B(Z,N)=a_VA-a_SA^{2/3}-a_C\frac{Z^2}{A^{1/3}} -a_{\mathrm{sym}}\frac{(N-Z)^2}{A}, \tag{1} \]

where \(A=Z+N\) is the mass number and \(m_N\) is the nucleon mass. In the calculation, use \(a_V\approx15.8\ \mathrm{MeV}\), \(a_S\approx17.8\ \mathrm{MeV}\), \(a_C\approx0.711\ \mathrm{MeV}\), and \(a_{\mathrm{sym}}\approx23.7\ \mathrm{MeV}\) (\(\mathrm{MeV}=10^6\) electron volts).

A.1 — 0.9pt

Under the condition of \(Z=N\), determine \(A\) for maximizing the binding energy per nucleon, \(B/A\).

A.2 — 0.9pt

Under the condition of fixed \(A\), the atomic number of the most stable nucleus \(Z^*\) is determined by maximizing \(B(Z,A-Z)\). For \(A=197\), calculate \(Z^*\) using Eq. (1).

A.3 — 0.7pt

A nucleus having large \(A\) breaks up into lighter nuclei through fission in order to minimize the total rest-mass energy. For simplicity, we consider one of multiple ways to break a nucleus with \((Z,N)\) into two equal nuclei, which occurs when the following energy relation holds,

\[ m(Z,N)c^2>2m(Z/2,N/2)c^2. \]

When this relation is written as

\[ Z^2/A>C_{\mathrm{fission}}\frac{a_S}{a_C}, \]

obtain \(C_{\mathrm{fission}}\) up to two significant digits.

Part B. Neutron star as a gigantic nucleus (1.5 points)

For large nuclei with a large enough mass number \(A>A_c\) with a threshold \(A_c\), these nuclei stay stable against nuclear fission because of the sufficiently large binding energy due to gravity.

B.1 — 1.5pt

We assume that \(N=A\) and \(Z=0\) is realized for sufficiently large \(A\) and Eq. (1) continues to hold with the addition of the gravitational binding energy. The binding energy due to gravity is

\[ B_{\mathrm{grav}}=\frac35\frac{GM^2}{R}, \]

where \(M=m_NA\) and \(R=R_0A^{1/3}\) with \(R_0\simeq1.1\times10^{-15}\ \mathrm{m}=1.1\ \mathrm{fm}\) are the mass and the radius of the nucleus, respectively.

For \(B_{\mathrm{grav}}=a_{\mathrm{grav}}A^{5/3}\), obtain \(a_{\mathrm{grav}}\) in the MeV unit up to the first significant digit. Then, ignoring the surface term, estimate \(A_c\) up to the first significant digit. In the calculation, use \(m_Nc^2\simeq939\ \mathrm{MeV}\) and \(G=\hbar c/M_P^2\) where \(M_Pc^2\simeq1.22\times10^{22}\ \mathrm{MeV}\) and \(\hbar c\simeq197\ \mathrm{MeV}\cdot\mathrm{fm}\).

Part C. Neutron star in a binary system (6.0 points)

Some neutron stars are pulsars regularly emitting electromagnetic waves, which we call "light" for simplicity here, at a constant period. Neutron stars often make binary systems with a White Dwarf. Let us consider the star configuration shown in Fig. 1, where a light pulse from a neutron star N to the Earth E passes near a White Dwarf W of the binary system. Measuring these pulses influenced by the star's gravity leads to an accurate estimation of the mass of W as explained below, resulting in the estimation of the mass of N.

Figure 1

Fig. 1: Configurations with the \(x\)-axis along the line connecting N and E. (a) for \(x_N<0\) and (b) for \(x_N>0\).

C.1 — 1.0pt

As shown in the figure below, under the constant gravitational acceleration \(g\) we place two levels I and II with the height difference \(\Delta h(>0)\). Set the identical clocks at I, II, and \(F\), the free-falling system, denoted by clock-I, clock-II, and clock-\(F\), respectively.

Figure 2

Set-up of the thought experiment.

We assume that an observer sits with clock-\(F\), and initially \(F\) is placed at the same height as that of clock-I and its velocity is zero. Since the clocks are identical, they register equal time intervals, \(\Delta\tau_F=\Delta\tau_{\mathrm{I}}\). Then, we let \(F\) fall freely, and work in the frame of \(F\), which is considered to be inertial. In this frame, clock-II passes by clock-\(F\) with velocity \(v\), so that the time dilation of clock-II can be determined by the Lorentz transformation. When time \(\Delta\tau_{\mathrm{I}}\) elapses on clock-\(F\), time \(\Delta\tau_{\mathrm{II}}\) elapses on clock-II.

Determine \(\Delta\tau_{\mathrm{II}}\) in terms of \(\Delta\tau_{\mathrm{I}}\) up to the first order in \(\Delta\phi/c^2\), where \(\Delta\phi=g\Delta h\) is a difference of the gravitational potential, i.e., the gravitational potential energy per unit mass.

C.2 — 1.8pt

Under the gravitational potential \(\phi\), time delays change the effective speed of light, \(c_{\mathrm{eff}}\), observed at the infinity, though the local speed of light is \(c\). When \(\phi(r=\infty)=0\), \(c_{\mathrm{eff}}\) can be given up to the first order in \(\phi/c^2\) as

\[ c_{\mathrm{eff}}\approx\left(1+\frac{2\phi}{c^2}\right)c \]

including the effect of space distortion, which was not featured in C.1. We note that the light path can be approximated as a straight line.

As shown in Fig. 1 (a), we take the \(x\)-axis along the light path from the neutron star N to the Earth E and place \(x=0\) at the point where the White Dwarf W is the closest to the light path. Let \(x_N(<0)\) be the \(x\)-coordinate of N, \(x_E(>0)\) be that of E, and \(d\) be the distance between W and the light path.

Estimate the changes of the arrival time \(\Delta t\) of the light from N to E caused by the White Dwarf with mass \(M_{\mathrm{WD}}\) and evaluate the answer in a simple form disregarding higher order terms of the following small quantities: \(d/|x_N|\ll1\), \(d/x_E\ll1\), and \(GM_{\mathrm{WD}}/(c^2d)\ll1\). If necessary, use the following formula.

\[ \int\frac{dx}{\sqrt{x^2+d^2}} =\frac12\log\left( \frac{\sqrt{x^2+d^2}+x}{\sqrt{x^2+d^2}-x} \right)+C. \]

(\(\log\) is the natural logarithm)

C.3 — 1.8pt

As shown below, in a binary star system N and W are assumed to be moving in circular orbits with zero eccentricity around the center of mass \(G\) on the orbit plane. Let \(\varepsilon\) be the orbital inclination angle measured from the orbit plane to the line directed toward E from \(G\), and let \(L\) be the length between N and W and \(M_{\mathrm{WD}}\) be the mass of the White Dwarf. In the following, we assume \(\varepsilon\ll1\).

Figure 3

Binary star system.

We observe light pulses from N on E far away from N. The light path to E varies with time depending on the configuration of N and W. The delay in the time interval of arriving pulses on E has the maximum value \(\Delta t_{\max}\) for \(x_N\simeq-L\) and the minimum value \(\Delta t_{\min}\) for \(x_N\simeq L\) (see Fig. 1 (b) for the configuration).

Calculate \(\Delta t_{\max}-\Delta t_{\min}\) in a simple form disregarding higher order terms of small quantities as done in C.2. We note that the delays due to gravity from stellar objects other than W are assumed to cancel out in \(\Delta t_{\max}-\Delta t_{\min}\).

C.4 — 0.8pt

The below figure shows the observed time delays as a function of the orbital phase \(\varphi\) for the binary star system with \(L\approx6\times10^6\ \mathrm{km}\) and \(\cos\varepsilon\approx0.99989\). Estimate \(M_{\mathrm{WD}}\) in terms of the solar mass \(M_\odot\) and show the results for \(M_{\mathrm{WD}}/M_\odot\) up to the first significant digit. Here the approximate relation, \(GM_\odot/c^3\approx5\ \mu\mathrm{s}\), can be used.

Figure 4

Observed time delays \(\Delta t\) as a function of the orbital phase \(\varphi\) (see the figure in C.3) to locate N and W on the orbits.

C.5 — 0.4pt

In the binary system of neutron stars, two stars release energy and angular momentum by emitting gravitational waves and eventually collide to merge. For simplicity, let us consider only a circular motion with the radius \(R\) and the angular velocity \(\omega\) and then \(\omega=\chi R^p\) holds with the constant \(\chi\) depending on neither \(\omega\) nor \(R\) if relativistic effects are ignored. Determine the value for \(p\).

C.6 — 0.2pt

The amplitude of the emitted gravitational wave from the binary system in C.5 is proportional to \(R^2\omega^2\). Figure below qualitatively shows four different temporal profiles of the observed gravitational waves before the two-star collision. Select the most appropriate profile from (a) to (d).

Figure 5

Observed data profiles of gravitational waves.

Official English text of the problem.

Answers

  • A.1: \(\displaystyle A=\frac{2a_S}{a_C}\approx50.\)
  • A.2: \(\displaystyle Z^*=\frac{A}{2+\frac{a_C}{2a_{\mathrm{sym}}}A^{2/3}}=78.55\ldots\), giving the integer atomic number \(\boxed{79}\).
  • A.3: \(\displaystyle C_{\mathrm{fission}}=\frac{2^{1/3}-1}{1-2^{-2/3}}\approx\boxed{0.70}.\)
  • B.1: \(\displaystyle a_{\mathrm{grav}}\approx\boxed{6\times10^{-37}\ \mathrm{MeV}},\qquad A_c\approx\boxed{4\times10^{55}}\), using the positive-total-binding-energy criterion discussed below.
  • C.1: \(\displaystyle \boxed{\Delta\tau_{\mathrm{II}}\approx\Delta\tau_{\mathrm{I}}\left(1-\frac{\Delta\phi}{c^2}\right)}.\)
  • C.2: \(\displaystyle \boxed{\Delta t\approx\frac{2GM_{\mathrm{WD}}}{c^3}\ln\left(\frac{4x_E|x_N|}{d^2}\right)}.\)
  • C.3: \(\displaystyle \boxed{\Delta t_{\max}-\Delta t_{\min}\approx\frac{2GM_{\mathrm{WD}}}{c^3}\ln\left(\frac{1+\cos\varepsilon}{1-\cos\varepsilon}\right)\approx\frac{4GM_{\mathrm{WD}}}{c^3}\ln\left(\frac{2}{\varepsilon}\right)}.\)
  • C.4: \(\displaystyle \boxed{\frac{M_{\mathrm{WD}}}{M_\odot}\approx0.5}\), or \(M_{\mathrm{WD}}\approx0.5M_\odot\).
  • C.5: \(\displaystyle \boxed{p=-\frac32}.\)
  • C.6: \(\boxed{\text{(b)}}\): both amplitude and frequency increase as the stars approach collision.

Solution

A.1

For \(Z=N=A/2\), the symmetry term vanishes. The binding energy per nucleon becomes

\[ \frac BA=a_V-a_SA^{-1/3}-\frac{a_C}{4}A^{2/3}. \]

Treating \(A\) as continuous,

\[ \frac{d}{dA}\left(\frac BA\right) =\frac{a_S}{3}A^{-4/3}-\frac{a_C}{6}A^{-1/3} =\frac{2a_S-a_CA}{6A^{4/3}}. \]

This derivative is positive below \(A=2a_S/a_C\) and negative above it, so this stationary point is a maximum:

\[ A=\frac{2a_S}{a_C} =\frac{2(17.8)}{0.711} =50.07\ldots. \]

Thus

\[ \boxed{A\approx50}. \]

The nearest allowed even mass number, consistent with integer \(Z=N\), is \(A=50\).

A.2

At fixed \(A\), \(N=A-Z\), hence \(N-Z=A-2Z\). Only the Coulomb and symmetry terms depend on \(Z\):

\[ B(Z,A-Z)=\text{constant} -a_CZ^2A^{-1/3} -\frac{a_{\mathrm{sym}}}{A}(A-2Z)^2. \]

The maximizing condition is

\[ \frac{\partial B}{\partial Z} =-2a_CZA^{-1/3} +\frac{4a_{\mathrm{sym}}}{A}(A-2Z)=0. \]

Rearranging,

\[ 4a_{\mathrm{sym}}A =Z\left(8a_{\mathrm{sym}}+2a_CA^{2/3}\right), \]

so

\[ \boxed{ Z^*=\frac{A}{2+\dfrac{a_C}{2a_{\mathrm{sym}}}A^{2/3}} }. \]

It is a maximum because

\[ \frac{\partial^2 B}{\partial Z^2} =-2a_CA^{-1/3}-\frac{8a_{\mathrm{sym}}}{A}<0. \]

For \(A=197\),

\[ Z^*= \frac{197}{2+\dfrac{0.711}{2(23.7)}197^{2/3}} =78.553\ldots. \]

Since the dependence on \(Z\) is a concave quadratic, the optimal integer is the nearest integer:

\[ \boxed{Z^*=79}. \]

A.3

The total number of nucleons is conserved, so the free-nucleon rest-energy terms cancel. Fission is energetically favorable when

\[ 2B(Z/2,N/2)-B(Z,N)>0. \]

The volume and symmetry terms are unchanged in total. The daughters’ total surface term is

\[ -2a_S(A/2)^{2/3}=-a_S2^{1/3}A^{2/3}, \]

and their total Coulomb term is

\[ -2a_C\frac{(Z/2)^2}{(A/2)^{1/3}} =-a_C2^{-2/3}\frac{Z^2}{A^{1/3}}. \]

Therefore,

\[ 2B(Z/2,N/2)-B(Z,N) =-a_S(2^{1/3}-1)A^{2/3} +a_C(1-2^{-2/3})\frac{Z^2}{A^{1/3}}. \]

The condition for positive energy release becomes

\[ \frac{Z^2}{A}> \frac{2^{1/3}-1}{1-2^{-2/3}}\frac{a_S}{a_C}. \]

Hence

\[ \boxed{ C_{\mathrm{fission}} =\frac{2^{1/3}-1}{1-2^{-2/3}} =0.702414\ldots\approx0.70 }. \]

Physically, fission becomes favorable when the reduction in Coulomb repulsion exceeds the cost of the additional surface area.

B.1

Substituting the prescribed mass and radius into the gravitational binding energy gives

\[ B_{\mathrm{grav}} =\frac35\frac{G(m_NA)^2}{R_0A^{1/3}} =\frac{3Gm_N^2}{5R_0}A^{5/3}. \]

Thus

\[ a_{\mathrm{grav}} =\frac{3Gm_N^2}{5R_0} =\frac35\frac{\hbar c}{R_0} \left(\frac{m_Nc^2}{M_Pc^2}\right)^2. \]

Using the supplied constants,

\[ a_{\mathrm{grav}} =\frac35\frac{197\ \mathrm{MeV\,fm}}{1.1\ \mathrm{fm}} \left(\frac{939}{1.22\times10^{22}}\right)^2 =6.3656\times10^{-37}\ \mathrm{MeV}. \]

To one significant digit,

\[ \boxed{a_{\mathrm{grav}}\approx6\times10^{-37}\ \mathrm{MeV}}. \]

For \(Z=0\) and \(N=A\), the Coulomb term vanishes and the symmetry term is \(-a_{\mathrm{sym}}A\). Ignoring the surface term as instructed,

\[ B_{\mathrm{tot}}(A) =(a_V-a_{\mathrm{sym}})A+a_{\mathrm{grav}}A^{5/3}, \]

or

\[ \frac{B_{\mathrm{tot}}}{A} =-7.9\ \mathrm{MeV}+a_{\mathrm{grav}}A^{2/3}. \]

Taking the finite threshold to mean the onset of binding relative to \(A\) free neutrons, the condition is

\[ B_{\mathrm{tot}}>0. \]

Consequently,

\[ A>A_c,\qquad A_c= \left(\frac{a_{\mathrm{sym}}-a_V}{a_{\mathrm{grav}}}\right)^{3/2}. \]

Retaining the unrounded value of \(a_{\mathrm{grav}}\) during this calculation,

\[ A_c= \left(\frac{7.9}{6.3656\times10^{-37}}\right)^{3/2} =4.3721\times10^{55}. \]

Thus

\[ \boxed{A_c\approx4\times10^{55}}. \]

The distinction between this binding threshold and a strict two-drop fission comparison is essential here: for \(Z=0\) and with the surface term ignored, \(B_{\mathrm{tot}}(A)-2B_{\mathrm{tot}}(A/2)\) equals \((1-2^{-2/3})\,a_{\mathrm{grav}}A^{5/3}\), which is positive for every \(A\), so the two-drop comparison gives no finite threshold.

C.1

During free fall through height \(\Delta h\), Newtonian energy conservation gives, to the required accuracy,

\[ \frac12v^2=g\Delta h=\Delta\phi, \]

so

\[ v^2=2\Delta\phi. \]

In the instantaneous inertial frame of the freely falling observer, clock-II moves with speed \(v\). Special-relativistic time dilation gives

\[ \Delta\tau_{\mathrm{II}} =\Delta\tau_F\sqrt{1-\frac{v^2}{c^2}}. \]

Using the clock comparison prescribed in the problem, \(\Delta\tau_F=\Delta\tau_{\mathrm{I}}\), and expanding to first order,

\[ \begin{aligned} \Delta\tau_{\mathrm{II}} &=\Delta\tau_{\mathrm{I}} \sqrt{1-\frac{2\Delta\phi}{c^2}}\\ &\approx\Delta\tau_{\mathrm{I}} \left(1-\frac{\Delta\phi}{c^2}\right). \end{aligned} \]

Therefore,

\[ \boxed{ \Delta\tau_{\mathrm{II}} \approx\Delta\tau_{\mathrm{I}} \left(1-\frac{\Delta\phi}{c^2}\right) }. \]

The lower clock runs more slowly.

C.2

At position \(x\), the distance from the white dwarf is

\[ r(x)=\sqrt{x^2+d^2}, \]

and its gravitational potential is

\[ \phi(x)=-\frac{GM_{\mathrm{WD}}}{\sqrt{x^2+d^2}}. \]

The given weak-field expression implies

\[ \frac{1}{c_{\mathrm{eff}}} \approx\frac1c\left(1-\frac{2\phi}{c^2}\right). \]

The travel time without the white dwarf is \((x_E-x_N)/c\). Subtracting it from the actual travel time along the same straight path gives

\[ \begin{aligned} \Delta t &=\int_{x_N}^{x_E} \left(\frac{1}{c_{\mathrm{eff}}}-\frac1c\right)dx\\ &\approx-\frac{2}{c^3}\int_{x_N}^{x_E}\phi(x)\,dx\\ &=\frac{2GM_{\mathrm{WD}}}{c^3} \int_{x_N}^{x_E}\frac{dx}{\sqrt{x^2+d^2}}. \end{aligned} \]

An equivalent form of the supplied antiderivative is \(\ln(x+\sqrt{x^2+d^2})\). Thus, at first order in the gravitational potential,

\[ \Delta t\approx\frac{2GM_{\mathrm{WD}}}{c^3} \ln\left( \frac{x_E+\sqrt{x_E^2+d^2}} {x_N+\sqrt{x_N^2+d^2}} \right). \]

For \(x_E\gg d\),

\[ x_E+\sqrt{x_E^2+d^2}\approx2x_E. \]

For \(x_N<0\) and \(|x_N|\gg d\), rationalizing avoids cancellation:

\[ x_N+\sqrt{x_N^2+d^2} =\frac{d^2}{\sqrt{x_N^2+d^2}+|x_N|} \approx\frac{d^2}{2|x_N|}. \]

Therefore,

\[ \boxed{ \Delta t\approx \frac{2GM_{\mathrm{WD}}}{c^3} \ln\left(\frac{4x_E|x_N|}{d^2}\right) }. \]

This is a positive gravitational propagation delay. The logarithm is dimensionless, and \(GM_{\mathrm{WD}}/c^3\) has units of time.

C.3

At the two conjunctions, projection of the separation \(L\) onto the line of sight gives

\[ x_N=-L\cos\varepsilon \quad\text{at maximum delay}, \]

and

\[ x_N=+L\cos\varepsilon \quad\text{at minimum delay}. \]

At either conjunction,

\[ d=L\sin\varepsilon,\qquad \sqrt{x_N^2+d^2}=L. \]

Since Earth is far from the binary, we may use the same \(x_E\) at both configurations to the retained accuracy. The integral result from C.2, before assuming a negative \(x_N\), gives

\[ \Delta t\approx \frac{2GM_{\mathrm{WD}}}{c^3} \ln\left(\frac{2x_E}{L+x_N}\right). \]

Hence

\[ \Delta t_{\max}\approx \frac{2GM_{\mathrm{WD}}}{c^3} \ln\left(\frac{2x_E}{L(1-\cos\varepsilon)}\right), \]

whereas

\[ \Delta t_{\min}\approx \frac{2GM_{\mathrm{WD}}}{c^3} \ln\left(\frac{2x_E}{L(1+\cos\varepsilon)}\right). \]

Subtracting eliminates both \(x_E\) and \(L\):

\[ \boxed{ \Delta t_{\max}-\Delta t_{\min} \approx\frac{2GM_{\mathrm{WD}}}{c^3} \ln\left(\frac{1+\cos\varepsilon}{1-\cos\varepsilon}\right) }. \]

For \(\varepsilon\ll1\),

\[ 1-\cos\varepsilon\approx\frac{\varepsilon^2}{2}, \qquad 1+\cos\varepsilon\approx2. \]

Thus the requested small-angle form is

\[ \boxed{ \Delta t_{\max}-\Delta t_{\min} \approx\frac{2GM_{\mathrm{WD}}}{c^3} \ln\left(\frac{4}{\varepsilon^2}\right) =\frac{4GM_{\mathrm{WD}}}{c^3} \ln\left(\frac{2}{\varepsilon}\right) }. \]

Here \(\varepsilon\) is in radians. The logarithmic growth at small inclination corresponds to the light passing increasingly close to the white dwarf, within the stated approximations.

C.4

Reading the graph gives approximately

\[ \Delta t_{\max}\approx+22\ \mu\mathrm{s}, \qquad \Delta t_{\min}\approx-23\ \mu\mathrm{s}. \]

The measurable range is therefore approximately

\[ \Delta t_{\max}-\Delta t_{\min}\approx45\ \mu\mathrm{s}. \]

An arbitrary constant offset in the plotted delays does not affect this difference.

Using the supplied inclination,

\[ \ln\left(\frac{1+\cos\varepsilon}{1-\cos\varepsilon}\right) = \ln\left(\frac{1.99989}{0.00011}\right) \approx9.808. \]

C.3 then gives

\[ 45\ \mu\mathrm{s} \approx 2\left(5\ \mu\mathrm{s}\right) \frac{M_{\mathrm{WD}}}{M_\odot}(9.808). \]

Therefore,

\[ \frac{M_{\mathrm{WD}}}{M_\odot} \approx\frac{45}{98.08} \approx0.46. \]

To one significant digit,

\[ \boxed{\frac{M_{\mathrm{WD}}}{M_\odot}\approx0.5}, \qquad \boxed{M_{\mathrm{WD}}\approx0.5M_\odot}. \]

The supplied separation \(L\) cancels from the maximum-minus-minimum delay.

C.5

Let the stellar masses be \(M_1,M_2\), and let their separation be \(r\). Newtonian circular two-body motion obeys

\[ \omega^2r^3=G(M_1+M_2). \]

For fixed masses, either star’s orbital radius \(R\) is a constant multiple of \(r\). Consequently,

\[ \omega\propto r^{-3/2}\propto R^{-3/2}. \]

Comparing with \(\omega=\chi R^p\),

\[ \boxed{p=-\frac32}. \]

C.6

As the binary loses orbital energy, its orbital radius decreases. From C.5,

\[ \omega\propto R^{-3/2}, \]

so the orbital frequency, and hence the gravitational-wave frequency, increases.

The given amplitude dependence yields

\[ \text{amplitude}\propto R^2\omega^2 \propto R^2R^{-3} =R^{-1}. \]

Thus the amplitude also increases as \(R\) decreases. The required waveform has an expanding envelope and progressively closer oscillations:

\[ \boxed{\text{Profile (b)}}. \]

The solution was checked against the official solution.