IPhO 2017 · Theory problem 3

Cosmic Inflation

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Due to the relative movement of galaxies observed from the earth, the wavelength of visible spectrum of a particular galaxy differs from its original wavelength, which is known as the electromagnetic Doppler effect. One expects, for a collection of galaxies, to a random distributions of wavelength shifts: some positive (red shift) and some negative (blue shift). However, observations show that all, except for a nearby group of galaxies, are red shifted. This must be true even if the observation take place on different point in the universe. As a conclusion, our universe must be expanding. On the other hand local irregularity of the universe can be neglected on scales of more than 100 Mpc, in which 1 pc = 3.26 light-years. Averaged over large scales, the clumpy distribution of galaxies becomes more and more isotropic (independent of direction) and homogeneous (independent of position). Therefore we can assume the universe as a matter having a uniform mass density \(\rho\) and is expanding.

A. Expansion of Universe

Figure 1

For a simple model of our universe, let us consider an expanding uniform-density sphere embedded in a medium of a much larger sphere with the same density. Let say at some time, the radius of the smaller sphere is \(R_s\). To express the expansion of the sphere, the time dependency of the radius \(R(t)\) can be expressed by scale factor \(a(t)\), that is \(R(t)=a(t)R_s\).

Using Newton’s law of gravity to evaluate velocity of a mass element on the sphere boundary according the model of our universe, one can obtain a set of Friedmann equations:

\[ \left(\frac{\dot a}{a}\right)^2 =A_1\rho(t)-\frac{kc^2}{R_s^2a^2(t)} \tag{1} \]

where \(k\) a dimensionless constant, and \(c\) is velocity of light.

Part Question Points
A.1 Determine the constant \(A_1\) in the equation (1) 1.3 pt.

The discussion so far is non-relativistic. But in fact, it can be extended to a relativistic system by reinterpreting \(\rho(t)c^2\) as total energy density (excluding the gravitational potential energy). In this relativistic system derives the 2nd Friedmann equation:

\[ \dot\rho+A_2\left(\rho+\left(\frac{p}{c^2}\right)\right)\frac{\dot a}{a}=0 \tag{2} \]

using the 1st law thermodynamics of an adiabatic system, where \(p\) denotes the pressure on the sphere.

Part Question Points
A.2 Determine the constants \(A_2\) in the equation (2) 0.9 pt.

To solve Eqs. (1) and (2), one should assume a relation \(p=p(\rho)\), such as \(p(t)/c^2=w\rho(t)\), where \(w\) is a constant. There is also a factor \(H=\dot a/a\) being called Hubble parameter. The present values of parameters are usually symbolized by subscript 0 such as \(t_0,\rho_0,H_0,a_0\) and so on. For simplicity, we take \(a_0=1\).

Universe is believed to start from a big explosion called Big-Bang that produces radiation of relativistic particles. During its expansion, the universe is cooling down and the particles in it become non-relativistic. However, the recent observations clarify that the present universe is dominated by cosmological constant energy density. For the case of photon, as the universe is expanding, the photon’s wavelength expands proportionally to the scale factor.

Part Question Points
A.3 For each of the following three cases determine the resulting value of \(w\): (i) a universe filled only with radiation (i.e. photon energy), (ii) a universe filled only with non-relativistic matter and (iii) a universe with constant energy density. 1.2 pt.
A.4 In the case of \(k=0\), find \(a(t)\) for each case of (i) to (iii) being mentioned in A.3. Use the initial condition \(a(t=0)=0\) for case (i) and (ii), and use the condition \(a_0=1\) for case (iii). 1.2 pt.

Constant \(k\) in Eq. (1) refers to classification of spatial geometry of the universe. Its value can be \(k=+1\) for positive-curvature universe (closed), \(k=0\) for flat universe (infinite), and \(k=-1\) for negative-curvature universe (open, infinite). Let define a density ratio \(\Omega=\rho/\rho_c\), where \(\rho_cc^2=H^2/A_1\) is critical energy density. Note that \(A_1\) is obtained from problem A.1.

Part Question Points
A.5 Express \(k\) in Eq.(1) in terms of \(\Omega\), \(H\), \(a\) and \(R_0\). 0.1 pt.
A.6 Find a range for \(\Omega\) that corresponds to each value of \(k=+1\), \(k=0\) and \(k=-1\). 0.3 pt.

B. Motivation To Introduce Inflation Phase and Its General Conditions

The observation of cosmic microwave background radiation (CMB) suggests that our present universe is approximately flat. The problem is that if this is true the present universe should start from exactly flat early universe, otherwise any deviation from the flatness will eventually grow over time and spoil the present flatness.

Part Question Points
B.1 Find \((\Omega(t)-1)\) as a function of time for the universe when it is dominated by radiation or non-relativistic matter (see problem A.3). 0.5 pt.

To solve the problem, at some early time in its history, the universe should undergo a constant energy density domination period which leads to an exponential expansion so called inflation period.

Part Question Points
B.2 For this constant energy density domination period, find \((\Omega(t)-1)\) as a function of time. Assume that \((\Omega(t)-1)\ll1\). 0.3 pt.
B.3 Show that condition for inflation implies several following conditions: negative pressure, accelerated expansion \((\ddot a>0)\), and decreasing Hubble radius \((d(aH)^{-1}/dt<0)\). 0.9 pt.
B.4 Show that the condition of decreasing Hubble radius can be expressed in terms of parameter \(\epsilon=-\dot H/H^2\) as \(\epsilon<1\). 0.2 pt.

Inflation occurs as long as \(\epsilon<1\) and then ends when \(\epsilon=1\). We can define e-folding number \(N\), such that \(dN=d\ln a=Hdt\) and \(N=0\) at the end of inflation.

C. Inflation Generated by Homogenously Distributed Matter

As an example of simple physical system that can generate period of inflation is a universe dominated by homogenously distributed matter. This kind of matter is called inflaton and can be characterized by a function \(\phi(t)\).

The dynamical equation of the matter can be expressed as

\[ \ddot\phi+3H\dot\phi=-V', \tag{3} \]

where \(V=V(\phi)\) is a potential function and \(V'=\frac{\partial V}{\partial\phi}\). The Hubble parameter satisfies

\[ H^2=\frac{1}{3M_{pl}^2}\left[\frac12\dot\phi^2+V\right]. \tag{4} \]

with \(M_{pl}\) is a constant called the reduced Planck mass. Inflation phase occurs during domination of potential energy \(V\) over kinetic energy \(\dot\phi^2/2\) for sufficient time such that \(\ddot\phi\) term in equation (3) can be neglected. This condition is called slow-roll approximation.

The quantities \(\epsilon\) and \(\eta_V=\delta+\epsilon\), where \(\delta=-\ddot\phi/(H\dot\phi)\), are called ‘slow-roll’ parameters.

Part Question Points
C.1 Estimate parameter \(\epsilon\), parameter \(\eta_V\), \(dN/d\phi\) in terms of potential \(V(\phi)\) and its first and second derivative (\(V'\) and \(V''\)). 1.7 pt.

D. Inflation with A Simple Potential

Predictions of any inflation model should be compared to observational constraints from CMB as follow \(n_s=0.968\pm0.006\) and \(r<0.12\), where \(r=16\epsilon\) and \(n_s=1+2\eta_V-6\epsilon\) are evaluated at \(\phi=\phi_{start}\) for inflation model being generated by a dominant matter. Assume that potential of matter takes a simple form \(V(\phi)=\Lambda^4\left(\frac{\phi}{M_{pl}}\right)^n\) where \(n\) is any integer and \(\Lambda\) is a constant.

Part Question Points
D.1 Calculate \(\phi_{end}\) at the end of inflation. 0.5 pt.
D.2 Express \(r\) and \(n_s\) in terms of e-folding number \(N\) and integer \(n\). Estimate the value of \(n\) that is closest to observational values \(r\) and \(n_s\). Take \(N=60\) in your calculation. 0.9 pt.

Official English text of the problem.

Answers

The results below use the dimensionally consistent critical density \(\rho_c=H^2/A_1\). Write \(\Delta=\Omega-1\). For D.2, let \(\mathcal N=\ln(a_{\mathrm{end}}/a_{\mathrm{start}})=-N_{\mathrm{start}}>0\); the requested 60 e-folds means \(\mathcal N=60\).

  • A.1: \(\displaystyle \boxed{A_1=\frac{8\pi G}{3}}\).

  • A.2: \(\boxed{A_2=3}\).

  • A.3: \(\boxed{w_{\mathrm{radiation}}=\tfrac13,\quad w_{\mathrm{matter}}=0,\quad w_{\mathrm{constant\ density}}=-1}\).

  • A.4: On the expanding branch, \[ \boxed{ \begin{aligned} \text{(i)}\quad a(t)&=(2H_0t)^{1/2}=(t/t_0)^{1/2},\\ \text{(ii)}\quad a(t)&=\left(\tfrac32H_0t\right)^{2/3}=(t/t_0)^{2/3},\\ \text{(iii)}\quad a(t)&=e^{H_0(t-t_0)}. \end{aligned}} \] Here \(H_0^2=8\pi G\rho_0/3\), with \(H_0t_0=1/2\) in (i) and \(2/3\) in (ii).

  • A.5: \(\displaystyle \boxed{k=(\Omega-1)\left(\frac{aHR_0}{c}\right)^2}\).

  • A.6: \(\boxed{k=+1:\Omega>1;\quad k=0:\Omega=1;\quad k=-1:\Omega<1}\). For nonnegative density, the last range is \(0\leq\Omega<1\).

  • B.1: To leading order near flatness, relative to a reference time \(t_*\), \[ \boxed{ \Delta(t)\simeq \begin{cases} \Delta_*(t/t_*),&\text{radiation},\\ \Delta_*(t/t_*)^{2/3},&\text{non-relativistic matter}. \end{cases}} \]

  • B.2: With \(H_\Lambda=\sqrt{8\pi G\rho_\Lambda/3}\), \[ \boxed{\Delta(t)\simeq\Delta_i e^{-2H_\Lambda(t-t_i)}}. \]

  • B.3: Constant density gives \[ \boxed{p=-\rho c^2<0,\qquad \frac{\ddot a}{a}=\frac{8\pi G\rho}{3}>0,\qquad \frac{d}{dt}(aH)^{-1}<0.} \] More generally, accelerated expansion requires \(p<-\rho c^2/3\).

  • B.4: \[ \boxed{\frac{d}{dt}(aH)^{-1}=\frac{\epsilon-1}{a}<0 \quad\Longleftrightarrow\quad \epsilon<1.} \]

  • C.1: In the slow-roll approximation, \[ \boxed{ \epsilon\simeq\frac{M_{pl}^2}{2}\left(\frac{V'}{V}\right)^2,\qquad \eta_V\simeq M_{pl}^2\frac{V''}{V},\qquad \frac{dN}{d\phi}\simeq-\frac{V}{M_{pl}^2V'}.} \]

  • D.1: For \(n>0\), on the positive-field branch, \[ \boxed{\phi_{\mathrm{end}}\simeq\frac{n}{\sqrt2}M_{pl}}. \] The formal threshold is \(\lvert\phi\rvert\simeq |n|M_{pl}/\sqrt2\). Nonpositive \(n\) does not give the assumed end of slow-roll inflation in this model.

  • D.2: \[ \boxed{ r\simeq\frac{16n}{4\mathcal N+n},\qquad n_s\simeq1-\frac{2(n+2)}{4\mathcal N+n}.} \] Equivalently, the denominators are \(n-4N_{\mathrm{start}}\) in the statement’s signed convention. At \(\mathcal N=60\), \[ \begin{array}{c|cc} n&r&n_s\\ \hline 1&0.0664&0.9751\\ 2&0.1322&0.9669 \end{array} \] If \(r<0.12\) is enforced, \(n=1\) is the closest positive-integer choice. No positive integer satisfies both quoted observational bounds.

Solution

A.1

Consider a mass element on the boundary of a comoving sphere of radius \(R=aR_s\). By Newton’s shell theorem, its acceleration depends only on the enclosed mass, \[ M=\frac{4\pi}{3}\rho R^3. \] In the non-relativistic model, this enclosed mass is constant. Newton’s equation is \[ \ddot R=-\frac{GM}{R^2}. \] Multiplying by \(\dot R\) and integrating gives the conserved specific mechanical energy \[ \frac12\dot R^2-\frac{GM}{R}=\mathcal E. \] Therefore \[ \left(\frac{\dot R}{R}\right)^2 =\frac{2GM}{R^3}+\frac{2\mathcal E}{R^2} =\frac{8\pi G}{3}\rho+\frac{2\mathcal E}{a^2R_s^2}. \] Since \(\dot R/R=\dot a/a\), comparison with (1), identifying \(2\mathcal E=-kc^2\), yields \[ \boxed{A_1=\frac{8\pi G}{3}}. \]

A.2

Let the physical volume of the sphere be \[ \mathcal V=\frac{4\pi}{3}(aR_s)^3. \] Its total internal energy is \(U=\rho c^2\mathcal V\). The first law for an adiabatic system gives \[ dU=-p\,d\mathcal V. \] Thus \[ c^2\mathcal V\,d\rho+(\rho c^2+p)\,d\mathcal V=0. \] Using \(\dot{\mathcal V}/\mathcal V=3\dot a/a\), \[ \dot\rho+3\left(\rho+\frac{p}{c^2}\right)\frac{\dot a}{a}=0. \] Hence \[ \boxed{A_2=3}. \]

A.3

For constant \(w\), the continuity equation becomes \[ \frac{\dot\rho}{\rho}=-3(1+w)\frac{\dot a}{a}, \] so, using \(a_0=1\), \[ \boxed{\rho=\rho_0a^{-3(1+w)}}. \]

For radiation, the photon number density decreases as \(a^{-3}\). Each photon has energy \(hc/\lambda\), and the given wavelength scaling \(\lambda\propto a\) makes its energy proportional to \(a^{-1}\). Therefore \[ \rho c^2\propto a^{-4}, \] which gives \[ 3(1+w)=4,\qquad \boxed{w=\frac13}. \]

For non-relativistic matter, the energy is dominated by approximately constant rest energy per particle. The number density decreases as \(a^{-3}\), giving \[ \rho\propto a^{-3},\qquad \boxed{w=0}. \] This is the pressureless-matter approximation.

For constant energy density, \(\dot\rho=0\). In an expanding universe, the continuity equation requires \[ \rho+\frac{p}{c^2}=0, \] and hence \[ \boxed{w=-1}. \]

A.4

For \(k=0\), \[ H^2=\frac{8\pi G}{3}\rho =H_0^2a^{-3(1+w)}. \] Choosing the expanding branch, \[ \dot a=H_0a^{1-\frac32(1+w)}. \]

For \(w\neq-1\), integration with \(a(0)=0\) gives \[ a^{\frac32(1+w)} =\frac32(1+w)H_0t. \]

Consequently, for radiation, \[ \boxed{a(t)=(2H_0t)^{1/2}}. \] The normalization \(a(t_0)=1\) implies \(2H_0t_0=1\), so \[ a(t)=\left(\frac{t}{t_0}\right)^{1/2}. \]

For non-relativistic matter, \[ \boxed{a(t)=\left(\frac32H_0t\right)^{2/3}}. \] Here \(3H_0t_0/2=1\), giving \[ a(t)=\left(\frac{t}{t_0}\right)^{2/3}. \]

For constant energy density, \(H=H_0\) is constant. Integrating \(\dot a/a=H_0\) and imposing \(a(t_0)=1\), \[ \boxed{a(t)=e^{H_0(t-t_0)}}. \]

A.5

Setting \(k=0\) in (1) determines the critical mass-equivalent density: \[ \rho_c=\frac{H^2}{A_1}=\frac{3H^2}{8\pi G}. \] Thus the printed definition \(\rho_cc^2=H^2/A_1\) is dimensionally inconsistent. The corresponding critical energy density is \[ \rho_cc^2=\frac{H^2c^2}{A_1}. \]

Using the consistent definition, \[ \Omega=\frac{A_1\rho}{H^2}. \] Divide (1) by \(H^2\): \[ 1=\Omega-\frac{kc^2}{a^2R_s^2H^2}. \] Since \(a_0=1\), the present radius is \(R_0=R_s\). Therefore \[ \boxed{k=(\Omega-1)\frac{a^2H^2R_0^2}{c^2}}. \]

A.6

The factor multiplying \(\Omega-1\) in A.5 is positive. Hence \(k\) and \(\Omega-1\) have the same sign:

Spatial curvature Density ratio
\(k=+1\), closed \(\Omega>1\)
\(k=0\), flat \(\Omega=1\)
\(k=-1\), open \(\Omega<1\)

With nonnegative energy density, the open-universe range is \(0\leq\Omega<1\).

B.1

From A.5, \[ \Delta(t)\equiv\Omega(t)-1 =\frac{kc^2}{R_0^2a^2H^2} =\frac{kc^2}{R_0^2\dot a^2}. \] This identity is exact.

To study a small departure from flatness, evaluate \(\dot a\) using the flat-universe solutions of A.4 to leading order. If \[ a\propto t^\alpha, \] then \[ \dot a^2\propto t^{2\alpha-2}, \qquad \Delta\propto t^{2-2\alpha}. \]

During radiation domination, \(\alpha=1/2\), so \[ \boxed{\Delta(t)\simeq\Delta_*\frac{t}{t_*}}. \]

During matter domination, \(\alpha=2/3\), so \[ \boxed{\Delta(t)\simeq\Delta_*\left(\frac{t}{t_*}\right)^{2/3}}. \]

Thus the magnitude of any small, nonzero departure from flatness grows during either era. These power laws apply while curvature remains a small correction to the expansion.

B.2

Let the constant density be \(\rho_\Lambda\), and define \[ H_\Lambda^2=\frac{8\pi G}{3}\rho_\Lambda. \] For \(|\Omega-1|\ll1\), the curvature contribution to (1) is small, so \[ H\simeq H_\Lambda,\qquad a(t)\simeq a_i e^{H_\Lambda(t-t_i)}. \] Substitution into \[ \Delta=\frac{kc^2}{R_0^2a^2H^2} \] gives \[ \boxed{\Delta(t)\simeq\Delta_i e^{-2H_\Lambda(t-t_i)}}. \] Inflation therefore suppresses departures from flatness exponentially.

B.3

Differentiate (1), and use the continuity equation: \[ 2H\dot H =A_1\dot\rho+\frac{2kc^2H}{R_0^2a^2}, \] so \[ \dot H =-\frac{3A_1}{2}\left(\rho+\frac{p}{c^2}\right) +\frac{kc^2}{R_0^2a^2}. \] Adding \(H^2\) from (1) eliminates the curvature term: \[ \frac{\ddot a}{a} =\dot H+H^2 =-\frac{A_1}{2}\left(\rho+\frac{3p}{c^2}\right). \] Using A.1, \[ \boxed{\frac{\ddot a}{a} =-\frac{4\pi G}{3}\left(\rho+\frac{3p}{c^2}\right)}. \]

For positive density, accelerated expansion requires \[ \boxed{p<-\frac13\rho c^2}, \] which in particular requires negative pressure.

For the constant-density period considered here, A.3 gives \(p=-\rho c^2\). Hence \[ \frac{\ddot a}{a}=\frac{8\pi G}{3}\rho>0. \]

Finally, \(aH=\dot a\), so \[ \frac{d}{dt}(aH)^{-1} =\frac{d}{dt}\frac1{\dot a} =-\frac{\ddot a}{\dot a^2}<0. \] This proves all three requested properties.

The radius appearing in this condition is the comoving Hubble radius, \(c/(aH)\); the constant \(c\) does not affect the derivative’s sign.

B.4

Direct differentiation gives \[ \frac{d}{dt}(aH)^{-1} =-\frac{\dot aH+a\dot H}{a^2H^2} =-\frac1a\left(1+\frac{\dot H}{H^2}\right). \] Since \(\epsilon=-\dot H/H^2\), \[ \boxed{\frac{d}{dt}(aH)^{-1}=\frac{\epsilon-1}{a}}. \] Because \(a>0\), the radius decreases exactly when \[ \boxed{\epsilon<1}. \] Equivalently, \[ \frac{\ddot a}{a}=H^2(1-\epsilon). \]

C.1

First differentiate (4): \[ 2H\dot H =\frac{1}{3M_{pl}^2} \left(\dot\phi\,\ddot\phi+V'\dot\phi\right). \] Equation (3) gives \(\ddot\phi+V'=-3H\dot\phi\), so \[ 2H\dot H=-\frac{H\dot\phi^2}{M_{pl}^2}. \] Thus \[ \boxed{\dot H=-\frac{\dot\phi^2}{2M_{pl}^2}}, \qquad \epsilon=\frac{\dot\phi^2}{2M_{pl}^2H^2}. \]

The slow-roll assumptions give \[ H^2\simeq\frac{V}{3M_{pl}^2}, \qquad 3H\dot\phi\simeq-V'. \] Substituting these into \(\epsilon\), \[ \epsilon \simeq\frac{V'^2}{18M_{pl}^2H^4} =\boxed{\frac{M_{pl}^2}{2}\left(\frac{V'}{V}\right)^2}. \]

To obtain \(\eta_V\), differentiate the slow-roll field equation: \[ 3\dot H\dot\phi+3H\ddot\phi\simeq-V''\dot\phi. \] Dividing by \(3H^2\dot\phi\), \[ \frac{\dot H}{H^2}+\frac{\ddot\phi}{H\dot\phi} \simeq-\frac{V''}{3H^2}. \] The left side is \(-\epsilon-\delta\). Therefore \[ \eta_V=\delta+\epsilon \simeq\frac{V''}{3H^2} =\boxed{M_{pl}^2\frac{V''}{V}}. \]

Finally, using the statement’s definition \(dN=Hdt\), \[ \frac{dN}{d\phi} =\frac{H}{\dot\phi} \simeq-\frac{3H^2}{V'} =\boxed{-\frac{V}{M_{pl}^2V'}}. \] This expression applies along a rolling, monotonic field trajectory.

D.1

For \[ V(\phi)=\Lambda^4\left(\frac{\phi}{M_{pl}}\right)^n, \] the derivative ratios are \[ \frac{V'}{V}=\frac n\phi, \qquad \frac{V''}{V}=\frac{n(n-1)}{\phi^2}. \] Consequently, \[ \epsilon\simeq\frac{n^2M_{pl}^2}{2\phi^2}. \] Extrapolating this slow-roll expression to the end condition \(\epsilon=1\) gives \[ |\phi_{\mathrm{end}}|\simeq\frac{|n|}{\sqrt2}M_{pl}. \]

For \(n>0\), take the positive-field branch. The field rolls toward smaller \(\phi\), so \(\epsilon\) increases and reaches unity at \[ \boxed{\phi_{\mathrm{end}}\simeq\frac n{\sqrt2}M_{pl}}. \] The field has the same units as \(M_{pl}\).

The statement permits any integer \(n\), but the assumed exit restricts the relevant cases:

  • For \(n=0\), the potential is constant and \(\epsilon\simeq0\); this model provides no slow-roll exit.
  • For \(n<0\) and \(\phi>0\), \(V'<0\), so \(\dot\phi\simeq-V'/(3H)>0\). Thus \(\epsilon\propto\phi^{-2}\) decreases. The formal \(\epsilon=1\) threshold is an entry into inflation.

The end-field result is an estimate because slow roll ceases to be a controlled approximation as \(\epsilon\) approaches unity.

D.2

For the monomial potential, C.1 gives \[ \frac{dN}{d\phi}\simeq-\frac{\phi}{nM_{pl}^2}. \] Integrating from the end of inflation, where \(N=0\), \[ N(\phi)\simeq-\frac{\phi^2-\phi_{\mathrm{end}}^2}{2nM_{pl}^2}. \]

For \(n>0\), the initial field is larger than the final field, so the statement’s \(N_{\mathrm{start}}\) is negative. Define the positive duration \[ \mathcal N\equiv-N_{\mathrm{start}} =\ln\frac{a_{\mathrm{end}}}{a_{\mathrm{start}}}. \] Then \[ \phi_{\mathrm{start}}^2 \simeq\phi_{\mathrm{end}}^2+2nM_{pl}^2\mathcal N =\frac n2(4\mathcal N+n)M_{pl}^2. \]

The slow-roll parameters at the start are therefore \[ \epsilon_{\mathrm{start}} \simeq\frac{n}{4\mathcal N+n}, \qquad \eta_{V,\mathrm{start}} \simeq\frac{2(n-1)}{4\mathcal N+n}. \] Thus \[ \boxed{r\simeq\frac{16n}{4\mathcal N+n}} \] and \[ \boxed{ n_s\simeq 1+\frac{4(n-1)-6n}{4\mathcal N+n} =1-\frac{2(n+2)}{4\mathcal N+n}}. \]

In the statement’s signed convention these become \[ r\simeq\frac{16n}{n-4N_{\mathrm{start}}}, \qquad n_s\simeq1-\frac{2(n+2)}{n-4N_{\mathrm{start}}}. \] For \(\mathcal N\gg n\), the familiar leading forms are \[ r\simeq\frac{4n}{\mathcal N}, \qquad n_s\simeq1-\frac{n+2}{2\mathcal N}. \]

Interpreting the requested \(N=60\) as a positive duration \(\mathcal N=60\), the tensor bound gives \[ \frac{16n}{240+n}<0.12 \quad\Longrightarrow\quad n<\frac{28.8}{15.88}\simeq1.814. \] Therefore the only positive integer satisfying this bound is \(n=1\).

The numerical predictions, all dimensionless, are:

\(n\) \(r\) \(n_s\) Comparison with the stated constraints
1 0.0664 0.9751 Satisfies \(r<0.12\); \(n_s\) is slightly above \(0.974\)
2 0.1322 0.9669 Satisfies the quoted \(n_s\) interval; exceeds the \(r\) bound
3 0.1975 0.9588 Outside both quoted bounds

Hence \[ \boxed{n=1\quad\text{if the tensor upper bound is enforced}.} \] Its predicted spectral index is about \(0.0011\) above the upper edge of the quoted interval. The integer \(n=2\) is closer to the central spectral index, but violates the tensor bound. No positive integer satisfies both quoted bounds simultaneously.

The solution was checked against the official solution. Where this solution differs from the official solution, the official solution contains an error or a misprint.