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Use natural units, the unreduced Planck mass $m_P=G^{-1/2}$, $V_0>0$ and $c\ne 0$. In a flat universe dominated by a canonical inflaton, $\varphi =(2m_P/c)\log (\beta t)$ and $a\propto t^p$ solve the inflaton equation of motion and Friedmann equation exactly when $p=16\pi /c^2$ and ${\beta }^2=V_0{c}^2/[2m_P^2(3p-1)]$. The positive-potential scaling solution requires $p>1/3$, while cosmic inflation requires $p>1$. Its constant equation-of-state parameter is $w=-1+2/(3p)$. In terms of the reduced Planck mass $M=m_P/\sqrt{8\pi }$, the exponential slope is $\lambda =c/\sqrt{8\pi }$ and the same exponent is $p=2/{\lambda }^2$.

Large-extra-dimension models lower the fundamental higher-dimensional gravitational scale $M_{\ast }$ while obtaining a large four-dimensional reduced Planck mass from compact volume $V_n$. Ordinary matter can be localized on a brane while gravity propagates in the bulk. The Planck mass from compactification volume relation explains weak effective four-dimensional gravity but leaves volume stabilization and the origin of a large internal volume as separate problems.

// Target: physics.bigb

For single-field adiabatic inflation, the Mukhanov-Sasaki variable rescales the comoving curvature perturbation by $z=a\sqrt{2ϵ}{M}_{\mathrm{Pl}}$, where $M_{\mathrm{Pl}}$ is the reduced Planck mass. Its quadratic bulk action is $\frac{1}{2}\int [(v^{\prime }{)}^2-(\nabla v{)}^2+(z^{\prime \prime }/z)v^2]$, after a boundary term is removed. A Fourier transform mode obeys $v_k^{\prime \prime }+(k^2-z^{\prime \prime }/z)v_k=0$ and has canonical Wronskian normalization. The mass scale and sound-speed factors depend on the underlying scalar theory; this definition assumes the canonical single-field action and positive slow-roll kinetic coefficient.

Write the reduced Planck mass as $M=(8\pi G{)}^{-1/2}=m_{\mathrm{pl}}/\sqrt{8\pi }$, and use natural units. Near the flat maximum, the quartic hilltop inflation trajectory has $V\simeq \alpha {\sigma }^4$ and $V^{\prime }\simeq -12\alpha {\varphi }^3$. The slow-roll approximation gives $H^2\simeq V/(3M^2)$ and $3H\dot{\varphi }\simeq -V^{\prime }$. Therefore the number of e-folds isThe lower field value dominates this integral. ThusThis is far below $\sigma$ for the assumed small symmetry-breaking scale. Strict slow roll actually fails before the minimum: $|M^2{V}^{\prime \prime }/V|\sim 1$ at ${\varphi }^2\sim {\sigma }^4/(36M^2)$. The last approach to the minimum is not described by slow roll. Using that earlier upper limit changes the dominant fifty-e-fold estimate by an order-one number of e-folds, not its leading scale dependence. No extrapolation of slow roll all the way to the minimum is needed.

Varying the canonical scalar action gives $\square \varphi +V_{,\varphi }=0$. In a homogeneous flat Friedmann-Lemaître-Robertson-Walker metric written in conformal time, $\sqrt{-g}=a^4$ and $g^{00}=a^{-2}$, soThe quadratic potential therefore givesEquivalently $\ddot{\varphi }+3H\dot{\varphi }+m^2\varphi =0$ in cosmic time. The intrinsic roll/oscillation time is $m^{-1}$ and the damping time is $H^{-1}$. If $m\gg H_0$, the scalar has long since begun rapid oscillations: averaging over a quadratic oscillation gives equal kinetic and potential energies, $⟨p⟩\simeq 0$, so it behaves as matter rather than accelerating dark energy. To remain nearly frozen or only begin rolling today requires the mass scale of quadratic quintessenceIf its energy is predominantly potential, equating $m^2{\varphi }_0^2/2$ to ${\rho }_{X0}=3M^2{H}_0^2{\Omega }_X$, with $M$ the reduced Planck mass, givesTaking the natural onset estimate $m\sim H_0$ and ${\Omega }_X\sim 0.7$ gives $|{\varphi }_0|\sim 2M\sim 5\times 10^{18}\mathrm{GeV}$, of order the Planck mass. More strongly frozen, negative-pressure evolution requires a smaller mass and correspondingly larger field. The density fixes the product $m|{\varphi }_0|$ at this accuracy, rather than a unique mass and field separately.

Cosmic inflation addresses the horizon problem by extending the causal past of the observed region, the flatness problem by damping the curvature contribution, and the cosmological monopole problem by diluting unwanted relics. A canonical inflaton can drive this expansion when its potential energy density exceeds twice its kinetic energy density.

Use the unreduced Planck mass $m_P=G^{-1/2}$, consistent with the coefficient in the Friedmann equation, and take $V_0>0$, $c\ne 0$. Seek the exponential-potential power-law inflation solution $\varphi =A\log (\beta t)$ and $a=a_{\ast }(t/t_{\ast }{)}^p$. The kinetic energy density is proportional to $t^{-2}$. The inflaton potential must have the same scaling, soSubstitution into the inflaton equation of motion givesThe Friedmann equation now reduces toConsequently the nontrivial positive-potential solution isA positive $V_0$ requires $p>1/3$. Accelerating expansion occurs exactly for $p>1$, givingThis is an exact solution, not a use of the slow-roll approximation. The first Hubble slow-roll parameter and equation-of-state parameter are constant:If the reduced Planck mass $M=m_P/\sqrt{8\pi }$ is used instead, the dimensionless exponential slope is $\lambda =c/\sqrt{8\pi }$ and the exponent becomes $p=2/{\lambda }^2$; these are the same solution.

For $p>1$, the comoving Hubble radius decreases as $(aH{)}^{-1}\propto t^{1-p}$. A region initially in causal contact can grow beyond that radius before the hot radiation-dominated universe begins. In the ideal power law extrapolated all the way to $t=0$, the particle horizon integral ${\int }_0^{t}d{t}^{\prime }/a(t^{\prime })$ diverges, so the solution has no finite past particle horizon. A finite, physically trustworthy inflationary interval instead requires a sufficiently large number of e-folds to enlarge an initially causal patch.

For a small curvature contribution, the Friedmann equation gives $|\Omega -1|=|k|/(a^2{H}^2)\propto t^{2-2p}\to 0$, solving the flatness problem. Conserved nonrelativistic magnetic monopoles have number density proportional to $a^{-3}$, while the inflaton energy density is proportional to $t^{-2}$; their energy-density ratio therefore falls as $t^{2-3p}$. This gives the required dilution, provided later reheating does not recreate them. Inflationary shear damping similarly suppresses anisotropic expansion.

The main failure is no automatic graceful exit or reheating. The first Hubble slow-roll parameter never reaches one, and the exponential inflaton potential has no finite minimum about which the field can oscillate. Additional dynamics are needed for a graceful exit from inflation and transfer to a hot radiation-dominated universe. The ideal solution also retains a curvature singularity at $t=0$; solving the horizon problem does not itself explain the beginning of the universe. For $c=0$, the constant inflaton potential instead permits constant-field de Sitter spacetime expansion; the logarithmic field solution is inapplicable.

For a homogeneous canonical inflaton, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires positive potential energy to dominate both kinetic energy and any competing material components. Then $P_{\varphi }\simeq -{\rho }_{\varphi }$ and the expansion is nearly vacuum-like. Exact scalar-driven acceleration requires ${\rho }_{\varphi }+3P_{\varphi }<0$, equivalently ${\dot{\varphi }}^2<V$; potential domination is a stronger condition.

The slow-roll approximation additionally neglects $\ddot{\varphi }$ compared with $3H\dot{\varphi }$ in the inflaton equation of motion. Define the unreduced Planck mass $m_{\mathrm{pl}}=G^{-1/2}$; it differs from the reduced Planck mass by $\sqrt{8\pi }$. Sufficient potential-flatness conditions aretogether with approach to the slow-roll attractor. The reduced equations are $H^2\simeq 8\pi V/(3m_{\mathrm{pl}}^2)$ and $3H\dot{\varphi }\simeq -V^{\prime }$.

For $V=m^2{\varphi }^2/2$, take $m>0$ and the positive-field rolling branch. ThenThe field velocity is constant to this order. Choosing the start at $t=0$ gives the quadratic-potential slow-roll solutionThe PDF puts $\pi$ inside this square root; the converted TeX incorrectly moves it outside. Next integrate $d\log a/d\varphi =H/\dot{\varphi }=-4\pi \varphi /m_{\mathrm{pl}}^2$:Thus the number of e-folds is the logarithmic expansion, not the expansion factor itself:Here ${ϵ}_V={\eta }_V=m_{\mathrm{pl}}^2/(4\pi {\varphi }^2)$. The usual quadratic-inflation endpoint estimate sets ${ϵ}_V\simeq 1$, givingThis is an endpoint estimate, not an exact result from the full equations. Exact acceleration ends at the Hubble slow-roll parameter ${ϵ}_H=1$, equivalently ${\dot{\varphi }}^2=V$. Inserting the slow-roll velocity into that last condition instead gives $\varphi \simeq m_{\mathrm{pl}}/\sqrt{6\pi }$; the different order-one coefficient illustrates the failing approximation at the endpoint. A precise value requires the full scalar and Friedmann evolution. The large initial-field contribution to $N$ is unaffected by that order-one uncertainty.

Finally $V({\varphi }_i)\sim m_{\mathrm{pl}}^4$ gives ${\varphi }_i^2\sim 2m_{\mathrm{pl}}^4/m^2$. The quadratic-inflation expansion from Planck density is thereforeFor $m\ll m_{\mathrm{pl}}$ this is enormous and the initial field is in the slow-roll regime. No unique numerical factor follows without specifying $m/m_{\mathrm{pl}}$. Extrapolating the classical model from Planck density is only a formal estimate; it does not establish control of quantum-gravity corrections at that starting scale.

Use natural units and the unreduced Planck mass $M_P=G^{-1/2}$; it is $\sqrt{8\pi }$ times the reduced Planck mass. A homogeneous canonical inflaton has ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. Vacuum domination by a scalar field requires ${\dot{\varphi }}_i^2/2\ll V({\varphi }_i)$, with other matter and the curvature term negligible in the Friedmann equation. The equation-of-state parameter is then near $-1$, and the Friedmann acceleration equation gives accelerating expansion.

For sustained slow-roll inflation, Hubble friction must dominate the inflaton acceleration: $|\ddot{\varphi }|\ll 3H|\dot{\varphi }|$, with the scalar potential changing only slightly in one Hubble time. In terms of the potential slow-roll parameter and second potential slow-roll parameter, this requiresFor the quartic scalar potential,Thus $|{\varphi }_i|\gg M_P$ supplies a broad slow-roll range when $\lambda \ll 1$. At an initial density of order the Planck density, potential domination givesso both potential slow-roll parameters are small for small $\lambda$. Small initial kinetic energy is a separate condition; the height of the scalar potential alone does not enforce it. These classical equations describe the post-Planck evolution under the model assumptions, rather than resolving quantum gravity at the initial endpoint.

Take ${\varphi }_i>0$; the negative branch follows by the symmetry $\varphi \mapsto -\varphi$. Dropping kinetic energy from the Friedmann equation and acceleration from the inflaton equation of motion givesWriting $b=M_P\sqrt{\lambda /(6\pi )}$, the quartic-potential slow-roll solution isIndeed, $\dot{\varphi }\simeq -b\varphi$, and integrating $H=\dot{a}/a$ gives $\log (a/a_i)=\pi ({\varphi }_i^2-{\varphi }^2)/M_P^2$. These expressions apply during slow-roll inflation; their formal late-time saturation must not be used beyond the end of that approximation.

The exact end of accelerating expansion is ${ϵ}_H=-\dot{H}/H^2=1$, equivalently ${\dot{\varphi }}^2=V$ for the scalar-dominated flat model. The quartic-inflation endpoint estimate uses the leading potential condition ${ϵ}_V\simeq 1$, givingThe numerical coefficient is an endpoint estimate: the slow-roll approximation breaks down there. Imposing $|{\eta }_V|=1$ estimates when slow roll loses accuracy and gives a different order-one coefficient, not a different exact acceleration condition.

The slow-roll e-fold count isThe quartic-inflation expansion from Planck density combines this initial estimate with the conventional endpoint to giveFor small $\lambda$ the dominant term is enormous; the order-one endpoint uncertainty does not affect that conclusion.

To estimate the reheating temperature, assume prompt, efficient reheating converts the end-of-inflation energy into a thermal radiation bath. With the same endpoint convention, $V_e\simeq \lambda M_P^4/(4{\pi }^2)$, and equating this to the radiation energy density givesThe quartic-inflation prompt-reheating estimate fixes the scaling; the endpoint kinetic energy changes the coefficient by order one. Without an inflaton coupling and its thermalization history, $V$ alone cannot specify the actual reheating temperature.

Finally, during cosmic inflation the comoving Hubble radius decreases:An initially causally connected patch is enlarged enough that the present observable region can lie within it. Its portions may subsequently be far outside one another's Hubble radii while retaining their common earlier thermal history. This solves the horizon problem if the number of e-folds is sufficiently large, conventionally of order sixty for the relevant thermal history; $N\sim 2\pi /\sqrt{\lambda }$ can readily exceed that. The necessary count depends on reheating and later expansion. The causal mechanism and the conversion back to a hot universe are discussed in David Tong's account of inflation.

Kaluza-Klein theory explains apparently four-dimensional physics as the low-energy limit of a higher-dimensional theory with compact spatial directions. In units $\hslash =c=1$, take a circle coordinate $y\sim y+2\pi R$. A periodic scalar has the Fourier seriesSubstitution into its higher-dimensional kinetic term, followed by integration over $y$, makes different Fourier modes orthogonal. The internal derivative contributes $n^2/R^2$ to the lower-dimensional squared mass, soThe quantized internal momentum produces the Kaluza-Klein tower. A real higher-dimensional field satisfies ${\varphi }_{-n}={\varphi }_n^{\ast }$; the tower does not introduce independent complex modes for both signs. More general compact spaces replace $n^2/R^2$ by Laplacian eigenvalues, with boundary conditions determining the spectrum.

At energies $E\ll R^{-1}$, nonzero Kaluza-Klein modes cannot be produced and may be integrated out, leaving a four-dimensional effective field theory. Dimensional reduction retains the zero modes at leading order; integrating out the massive modes also generates interactions suppressed by the Kaluza-Klein compactification scale. The higher-dimensional metric decomposes into a four-dimensional metric, vectors from mixed components, and scalar fields from internal components. For a circle, these include an Abelian gauge field and a radion. Internal coordinate shifts induce the vector's gauge symmetry, while the radion describes the radius. Pure circle reduction alone does not automatically give the chiral matter spectrum of the Standard Model.

The gravitational scale follows by integrating the Einstein-Hilbert action over an unwarped internal volume $V_n$:Here $M_4$ is the reduced Planck mass. For $n$ equal circle radii, $V_n=(2\pi R{)}^n$. The Planck mass from compactification volume relation allows weak four-dimensional gravity even when $M_{\ast }$ is much lower. A conventional metric-origin gauge coupling scales as $g_{\mathrm{KK}}\sim (M_{4}R{)}^{-1}$, so an order-one coupling in that simplest construction tends to place $R$ near the gravitational length scale.

In the brane-world scenario, ordinary matter and its gauge fields can instead be localized on a three-dimensional spatial brane, while gravity explores the bulk. The absence of ordinary-matter excitations along the extra dimensions then permits larger radii than a model in which all fields propagate there. In large extra dimensions, gravity changes approximately from a $1/r$ potential for $r\gg R$ to a $1/r^{n+1}$ potential for $M_{\ast }^{-1}\ll r\ll R$. For illustration, $M_{\ast }$ of order a TeV and $n=2$ give $R$ of order $0.1$ mm; one extra dimension would require an astronomically large radius.

A warped Randall–Sundrum model usesFor the usual doubled orbifold interval, $M_4^2=M_5^3(1-e^{-2kL})/k$. The relevant lengths are the bulk curvature radius $k^{-1}$, the proper separation $L$, and the inverse warped excitation scale, of order $(k{e}^{-kL}{)}^{-1}$. Thus the first mass need not equal $1/L$. Compact volume dilutes gravity; warping redshifts local masses. Both mechanisms require consistent field localization and radius stabilization; choosing a large radius or separation is not itself a dynamical explanation of it.

The hierarchy problem concerns the enormous separation between the electroweak scale and a possible fundamental ultraviolet scale such as the reduced Planck mass: $v\simeq 246\mathrm{GeV}$ whereas $M_{\mathrm{Pl}}\simeq 2.4\times 10^{18}\mathrm{GeV}$. For an elementary Higgs field, heavy states or a physical high-energy ultraviolet cutoff generically generateHere $\kappa$ represents the appropriate dimensionless couplings. Maintaining a small physical Higgs mass then requires a very accurate cancellation against its bare mass parameter. The problem is sensitivity to actual high-energy physics, rather than merely the presence of a quadratic divergence in one choice of regulator.

The cosmological constant problem concerns the much smaller gravitationally observable vacuum energy. In the Einstein equation, a constant term in the effective action contributes to the cosmological constant, with ${\Lambda }_{\mathrm{eff}}={\Lambda }_{\mathrm{bare}}+{\rho }_{\mathrm{vac}}/M_{\mathrm{Pl}}^2$ in the usual normalization. Generic zero-point contributions scale as ${\Lambda }_{\mathrm{UV}}^4$, and vacuum transitions also change ${\rho }_{\mathrm{vac}}$. The dark-energy density is only of order $10^{-47}{\mathrm{GeV}}^4\sim (\text{a few meV}{)}^4$. The puzzle is why the total remains so small despite much larger individual contributions. A constant shift of the nongravitational Hamiltonian cannot remove its gravitational effect.

Supersymmetry relates bosonic and fermionic degrees of freedom, including their interactions. Their opposite loop signs cancel the quadratic contributions to a scalar field mass in the unbroken theory. With soft supersymmetry breaking, the surviving correction is controlled by the mass splitting, schematicallyThus a sufficiently low soft-breaking scale protects a small electroweak scale against a much higher ultraviolet scale. This is the useful supersymmetric response to the hierarchy problem.

Exact global supersymmetry also cancels vacuum energy: the supersymmetry algebra expresses the Hamiltonian as a positive sum of anticommutators, so a vacuum annihilated by every supercharge has zero energy. Equivalently, with canonical kinetic terms,vanishes in a supersymmetric vacuum. However, ordinary particles and their superpartners cannot have the identical mass spectrum of unbroken supersymmetry, so a useful phenomenological theory needs supersymmetry breaking. Its vacuum energy is generically of order the fourth power of the breaking scale; even a scale of order $10^2\mathrm{GeV}$ gives about $10^8{\mathrm{GeV}}^4$, enormously above the required value. In supergravity, a negative term can cancel positive breaking contributions:but arranging a tiny residual is a new tuning. Consequently supersymmetry can protect the hierarchy after soft breaking, while its exact vacuum-energy cancellation does not by itself survive realistic breaking.

One higher-dimensional proposal uses large extra dimensions. Take $n$ flat compact dimensions of common radius $R$, with gravity propagating in the bulk and Standard Model fields confined to a four-dimensional brane. Reducing the gravitational Einstein-Hilbert action givessoA fundamental gravitational scale $M_{\ast }$ near a TeV then produces apparently weak four-dimensional gravity through dilution into the large compact volume. For $n=2$ and $M_{\ast }=1\mathrm{TeV}$, $R=M_{\mathrm{Pl}}/(2\pi M_{\ast }^2)$ is about $0.075\mathrm{mm}$, using $1{\mathrm{GeV}}^{-1}\simeq 1.97\times 10^{-16}\mathrm{m}$. At distances much smaller than $R$, the gravitational potential scales as $r^{-(n+1)}$; at distances much larger than $R$ it scales as $r^{-1}$. Confining charged Standard Model fields to the brane avoids a light tower of charged Kaluza-Klein modes associated with the large radius. The primary explanatory challenge is radius stabilization: one must generate and stabilize $R{M}_{\ast }\gg 1$, rather than simply postulate a huge volume and transfer the original hierarchy into a modulus.

A second proposal is the Randall–Sundrum model. Compactify on an interval, equivalently an orbifold with branes at $y=0$ and $y=L$, and use the warped metricA negative bulk cosmological constant and suitably related brane tensions support this geometry. Integrating over the full orbifold interval $[-L,L]$ givesFor a scalar field $h$ on the brane at $L$, its induced metric gives the kinetic coefficient $e^{-2kL}$ and mass coefficient $e^{-4kL}{m}_0^2$. The canonical field is $h_c=e^{-kL}h$, henceWith $k$ and $M_5$ of the same fundamental order, an exponent $kL\simeq 35$--$37$ converts a fundamental mass into a weak-scale mass, without requiring a large proper compactification volume.

Again the separation must be selected dynamically. The original warped solution leaves a massless radion, so merely choosing $L$ does not complete the explanation. For example, the Goldberger–Wise stabilization mechanism introduces a bulk scalar field with boundary values $v_0,v_L$ and a small mass $m_{\chi }$. Its static equation and solution areFor small backreaction and large $kL$, the leading radius-dependent potential is proportional to $e^{-4kL}(v_L-v_0{e}^{-ϵkL}{)}^2$. Its leading minimum obeys $kL\simeq {ϵ}^{-1}\log (v_0/v_L)$, so a modest boundary-value ratio and a small $ϵ$ can give the required exponent. Corrections and the residual four-dimensional cosmological constant must still be controlled. Both proposals need a stable dynamical origin for the compactification geometry; their volume or warp relations alone do not supply it.

There are two distinct parts of the hierarchy problem. The first asks why the scale of electroweak symmetry breaking is so much smaller than a fundamental scale such as the Planck mass. The second, the technical hierarchy problem, asks why that separation survives quantum corrections once it is chosen. With a fundamental Higgs boson, a mass term $m_H^2{H}^{†}H$ is allowed by the ordinary gauge symmetry, so heavy thresholds can contribute additive terms of order a heavy mass squared. This is different from the chiral protection of a fermion mass and the gauge protection of a vector mass.

For example, a cutoff description of the Standard Model gives the schematic one-loop sensitivityThe regulator-dependent quadratic term is not itself an observable, but a physical heavy particle of mass $M$ coupled to the Higgs likewise produces a matching correction of order coupling squared times $M^2/(16{\pi }^2)$. Maintaining $|m_H^2|\ll M^2$ then generically requires cancellation between large unrelated terms. This is the naturalness concern; changing the regulator does not eliminate the heavy-threshold issue.

Exact supersymmetry relates bosonic and fermionic couplings and masses. Their loop contributions have opposite signs, and the relations enforce supersymmetric cancellation of quadratic divergences. Boson-fermion degeneracy in a supermultiplet also transfers fermionic mass protection to its scalar partners. It is not sufficient merely to add arbitrary scalars: the supersymmetric coupling relations are what make the cancellation persist. Perturbative superpotential non-renormalization further protects holomorphic mass parameters, although wave-function renormalization still makes canonically normalized couplings run.

For a realistic separation between ordinary particles and their partners, supersymmetry must be broken. Soft supersymmetry breaking introduces masses and selected positive-mass-dimension interactions without reinstating the original ultraviolet quadratic divergence. A paired propagator difference behaves at high Euclidean momentum asso a residual scalar correction is of the orderThis soft scalar-mass sensitivity controls the technical problem if the relevant soft masses and thresholds are not too far above the weak scale. Very heavy partners can still require tuning through finite and logarithmic terms. Supersymmetric cancellations alone therefore do not guarantee a naturally small weak scale for an arbitrary broken spectrum.

The origin of the hierarchy needs further dynamics. In dynamical supersymmetry breaking, an asymptotically free hidden supersymmetry-breaking sector can generate an exponentially smaller scale by dimensional transmutation,If that sector actually has no supersymmetric vacuum, its breaking can be communicated to the visible sector to produce small soft terms. The exponential creates a hierarchy without specifying a tiny dimensionless coupling at the high scale, but the existence of breaking and its mediation are additional model-building requirements. In the MSSM, running Higgs soft masses can trigger electroweak symmetry breaking. The supersymmetric mu problem remains: the allowed superpotential mass $\mu H_u{H}_d$ must also be near the soft scale, and non-renormalization protects a chosen small $\mu$ without explaining its origin.

Extra-dimensional proposals use geometry to explain why observable gravitational and weak scales differ. In large extra dimensions, assume gravity propagates in $4+n$ dimensions with fundamental scale $M_{\ast }$, while ordinary fields are localized on a brane. Integrating the Einstein-Hilbert action over compact volume $V_n$ gives the Planck mass from compactification volume relationHere $M_{\mathrm{Pl}}$ denotes the four-dimensional reduced Planck mass. A sufficiently large internal volume lets $M_{\ast }$ be near the weak scale even though $M_{\mathrm{Pl}}$ is enormous. The Higgs need then be stable only up to the lower fundamental cutoff. This explains the weakness of long-distance gravity geometrically but trades the scale question for the origin and stabilization of a large compact volume. Ordinary matter is localized to avoid an unwanted low-mass tower of its own Kaluza-Klein modes; gravitational modes, departures from four-dimensional gravity and bulk energy loss impose phenomenological requirements.

The Randall–Sundrum model instead uses a warped five-dimensional interval, conventionally the doubled $S^1/{\mathbb{Z}}_2$ orbifold with fixed branes at $0$ and $L$:For a scalar confined to the distant brane, the kinetic and mass terms carry factors $e^{-2kL}$ and $e^{-4kL}$. Canonical rescaling therefore producesA moderately large $kL$ yields an exponentially small observable mass without a comparably large compact-volume factor. The inter-brane size is the radion, so one still needs a mechanism fixing its expectation value. Consistent curvature, brane tensions and ultraviolet physics are also part of the model. Warping is a redshift mechanism; it does not provide the same cancellation of arbitrary scalar loops as supersymmetry.

Finally, protecting a Higgs mass is not the same as solving the vacuum-energy hierarchy. Exact global supersymmetry has zero-energy supersymmetric vacua, but broken supersymmetry generically leaves vacuum energy set by its breaking dynamics. In supergravity, positive auxiliary-field contributions coexist with a negative superpotential term, and obtaining a very small cosmological constant requires further structure or cancellation. Neither soft supersymmetry nor the two geometric mechanisms automatically resolves that problem. The mechanisms can also coexist in one higher-dimensional supersymmetric theory.

During radiation domination, the effective number of relativistic energy degrees of freedom, denoted $g_{\ast }$, gives the energy densityThe paper's ${\mathcal{N}}_{\mathrm{eff}}$ is this $g_{\ast }$; it is not the conventionally normalized effective number of neutrino species. With the specified reduced Planck mass, the Friedmann equation reads $H^2={\rho }_R/(3M_{\mathrm{Pl}}^2)$, whenceWrite the weak interaction conversion rate as $\Gamma =A{G}_F^2{T}^5$. The ratio $\Gamma /H$ scales as $T^3$, so it falls during cooling: conversion eventually becomes too slow to maintain neutron-proton chemical equilibrium. Equating the conversion and expansion rates givesOn suppressing the numerical coefficient, the cosmological weak freeze-out temperature is consequentlyThe one-sixth power reflects a balance between a conversion rate proportional to $T^5$ and a Hubble parameter proportional to $\sqrt{g_{\ast }}{T}^2$.

The new neutrino species undergoes thermal decoupling in cosmology when its interaction rate becomes comparable to the Hubble parameter. Equating the given rates givesWith $g_{\ast }\simeq 100$, $G_A=10^{-12}{\mathrm{GeV}}^{-2}$ and $m_{\mathrm{Pl}}\simeq 10^{19}\mathrm{GeV}$,This is an order-of-magnitude electroweak-scale estimate under the given approximation. The mass in this formula is the unreduced Planck mass, not the reduced Planck mass.

Differentiate the scalar-field Friedmann equation on the expanding branch $H>0$:The scalar equation of motion gives $V^{\prime }+\ddot{\Phi }=-3H\dot{\Phi }$. Canceling $H$ yieldsThus the Hubble parameter cannot increase for a canonical homogeneous scalar field. The mass scale here is the reduced Planck mass.

The annihilation operators kill the vacuum, and their commutator givesTherefore the scale-invariant inflationary power spectrum has the superhorizon limitThe canonical mode grows as $1/|\tau |$ outside the Hubble radius, while the physical field perturbation $f/a$ freezes. A slowly varying $H$ produces an almost scale-invariant spectrum rather than exact scale invariance.

For a slowly rolling single field, fluctuations in the inflaton clock become the comoving curvature perturbation, $\mathcal{R}\simeq -H\delta \varphi /\dot{\bar{\varphi }}$ up to sign convention. Restoring the reduced Planck mass and using $ϵ={\dot{\bar{\varphi }}}^2/(2M_{\mathrm{pl}}^2{H}^2)$ givesThis conversion presumes nonzero background roll; a strictly constant test scalar in exact de Sitter spacetime does not by itself define a finite comoving curvature perturbation through this formula. The curvature perturbations seed the later density fluctuations.

When metric fluctuations and their Einstein-gravity normalization are restored, the two graviton polarizations have the same massless mode equation. The usual total primordial tensor power spectrum is ${\Delta }_t^2=2H^2/({\pi }^2{M}_{\mathrm{pl}}^2)$, so the single-field leading ratio is $r=16ϵ$. Tensor amplitudes therefore probe the inflationary energy scale, whereas scalar amplitudes also depend on the roll rate. These tensor statements explain the relevance of the mode result; they are not derived from a scalar action with metric perturbations omitted.

The quantum harmonic oscillator has $\hat{H}=({\dot{\hat{q}}}^2+{\omega }^2{\hat{q}}^2)/2$, with $\omega >0$. Using $\hat{a}|0⟩=0$ and the canonical commutation relation givesFor $q=r{e}^{is}$, the Wronskian normalization isConsequentlyEquality requires $\dot{r}=0$ and $r^2=1/(2\omega )$. The minimum-energy normalized oscillator mode is thereforeThe constant phase $\alpha$ is irrelevant. This mode also solves the oscillator equation of motion; an arbitrary squeezed mode would have larger vacuum energy.

For inflation, introduce the Mukhanov-Sasaki variable $v=z\mathcal{R}$. Since $z$ depends only on conformal time,Integrating the cross term by parts gives the canonical bulk actionup to the boundary term $-\frac{1}{2}\int d^{3}x[(z^{\prime }/z)v^2{]}_{\mathrm{boundary}}$. The resulting Euler-Lagrange field equation and its Fourier transform areAt leading order in the slow-roll approximation, retain a small positive, nearly constant $ϵ$ in $z=a\sqrt{2ϵ}$, while approximating $a=-1/(H\tau )$ and $H$ as constant. Then $z^{\prime \prime }/z\simeq 2/{\tau }^2$, soStrict exact de Sitter spacetime would have $ϵ=0$ and would not supply the nonzero curvature kinetic coefficient assumed here; the calculation is the leading quasi-de Sitter limit, not a substitution of zero into $z$.