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Let $f$ be a differentiable function on an open domain stable under positive rescaling, with $f(\lambda x)={\lambda }^{k}f(x)$ for $\lambda >0$. The chain rule differentiates this homogeneous function identity at $\lambda =1$ to giveIf $f$ is twice differentiable, differentiating once more gives ${\sum }_i{x}_i{\partial }_i{\partial }_{j}f=(k-1){\partial }_{j}f$. In degree one its Hessian matrix therefore annihilates the radial vector $x$. This is the identity used in the no-scale identity from degree-one homogeneity. For an extensive quantity, the degree-one formula also supplies the thermodynamic identity behind the Gibbs-Duhem equation.

Two related properties of an extensive quantity must be distinguished. Additivity says that a quantity $X$ for independent subsystems A and B obeysThe subsystems may be in different states. Extensivity expressed through a homogeneous function says that the equilibrium fundamental relation obeyswith composition held fixed. This is degree-one homogeneity along a replication ray; it does not say that internal energy is a linear function of arbitrary independent changes of entropy, volume and amount of substance.

Additivity gives integer replication by repeated addition. Subdividing a uniform bulk sample into $b$ equivalent parts gives rational scaling $a/b$; continuity then extends it to positive real factors. This argument requires negligible boundary contributions and a consistently prepared state. Conversely, a degree-one homogeneous function need not be additive between different rays: $f(x,y)=\sqrt{x^2+y^2}$ has $f(\lambda x,\lambda y)=\lambda f(x,y)$, but $f(1,0)+f(0,1)=2\ne \sqrt{2}=f(1,1)$. Thus a scaling identity alone cannot justify summing arbitrary states.

For finite samples, the surface correction to thermodynamic free energy can matter. Under geometric enlargement with volume multiplied by $\lambda$, surface area grows as ${\lambda }^{2/3}$, so a surface energy term is not a bulk extensive quantity. Replicating separated droplets retains every original surface, whereas merging them changes the total surface area. Likewise, appreciable mutual gravitational potential energy makes interacting subsystems fail the independent-copy premise. The thermodynamic limit explains when bulk scaling becomes accurate; it does not remove these qualifications for every small or strongly interacting system.

Homogeneity of degree $k$ means that multiplying all arguments by a common positive factor $a$ multiplies the output by $a^k$. A homogeneous function is the function possessing this property. This is different from degree-one positive homogeneity; investment values with constant relative risk aversion utility usually have degree $1-R$.

An intensive quantity is unchanged under replication at the same thermodynamic equilibrium. Temperature, pressure, chemical potential and the mass density of a uniform substance are examples. Independence from sample size does not mean independence from state: a material's mass density can change with temperature or pressure.

The relation to extensive quantities is particularly elegant for a differentiable bulk fundamental relation. Put $\mathbf{x}=(S,V,n)$ and writeTaking a partial derivative with respect to $x_i$ and cancelling $\lambda$ givesThe equilibrium differential $dU=TdS-pdV+\mu dn$ identifies these derivatives with temperature, minus pressure and molar chemical potential. They are degree-zero homogeneous functions. The same degree-one assumption yields the Euler relation in thermodynamics, but a surface energy contribution requires extra care.

An intensive quantity generally has no meaningful sum across samples: two equal temperatures do not add to a doubled temperature. Neither is the final temperature after thermal contact generally an unweighted average; heat capacities and the first law of thermodynamics determine the balance. Ratios of additive inventories lead to the useful weighted rules in normalizing extensive quantities.

For a homogeneous function, fix $x\in U$ and differentiate $f(\lambda x)={\lambda }^{c}f(x)$ at $\lambda =1$. The chain rule givesConversely, assume this derivative identity throughout $U$. Along a positive ray define $h(\lambda )={\lambda }^{-c}f(\lambda x)$. The ray stays in $U$ by hypothesis, and linearity of the derivative givesTherefore $h$ is constant on the connected interval $(0,\infty )$ and equals $h(1)=f(x)$. Hence $\boxed{f(\lambda x)={\lambda }^{c}f(x)}$ for every positive $\lambda$. This proves Euler differential identity characterizes homogeneity; no connectedness assumption on the whole domain is needed.

The divergence theorem states that for a continuously differentiable vector field on a bounded region with outward unit normal $\mathbf{n}$ and suitably regular, possibly piecewise smooth, boundary,For a differentiable homogeneous function, differentiate $f(k\mathbf{x})=k^{n}f(\mathbf{x})$ at $k=1$ to get Euler theorem for homogeneous functions:Only positive $k$ near one is needed for this derivative, so no interpretation of noninteger powers at negative $k$ is required. Applying the product rule gives $\nabla \cdot (f\mathbf{x})=3f+\mathbf{x}\cdot \nabla f=(n+3)f$. The volume integral of a homogeneous function is thereforeThe condition $n\ne -3$ is needed for division; at that degree the divergence theorem instead gives only zero net boundary flux. The field $f\mathbf{x}$ must also be regular on the region, or singularities must be treated by excision and limiting boundary terms. Neither issue arises for the polynomial in the final calculation.

Write $r=\sqrt{x^2+y^2}$. In the cone, $0\le z\le \alpha$ and $0\le r\le z/\alpha$. The polynomial $f=z^4+{\alpha }^4{r}^4$ has degree four. Its direct volume integral isOn the lateral cone the outward normal is proportional to $\alpha {\mathbf{e}}_r-{\mathbf{e}}_z$, so $\mathbf{x}\cdot \mathbf{n}$ is proportional to $\alpha r-z=0$. Thus the lateral contribution vanishes. The cap is $z=\alpha$, $0\le r\le 1$, with outward normal ${\mathbf{e}}_z$, soMultiplying by $1/(n+3)=1/7$ reproduces $4\pi {\alpha }^5/21$, verifying the formula. The cap edge and apex have zero surface measure, and the piecewise smooth version of the divergence theorem applies.

Write $K=k_1+k_2+k_3>0$ and $C=ϵ+1-c_s^2$. The inflationary scalar Fourier mode has real late-time value $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{c}_s{k}^3}$ and $u_k^{\prime \ast }(\tau )=A_k{c}_s^2{k}^2\tau e^{ik{c}_s\tau }$. Thus the time integrand, before the factor $-2i$, isThe Bunch-Davies vacuum prescription supplies damping in the past. Equivalently, use a positive regulator $d$ and then take the limit:The real part of $-2i$ times this integral is $4/(c_s^3{K}^3)$. Including the six Wick contractions yields the cubic time-derivative curvature bispectrumThe resulting primordial bispectrum is a homogeneous function of degree $-6$ in the wavenumbers, as expected for scale-independent background parameters.

The printed limit $c_s\ll 0$ is not a physical sound-speed limit for these modes. The normalization and positive-frequency choice assume $c_s>0$; a negative value is outside the positive sound-speed convention used by these modes. The likely intended limit is $0<c_s\ll 1$, which is analyzed explicitly here rather than silently substituted for the PDF.

To compare primordial non-Gaussianity, normalize by the power spectrum, not only by the raw three-point amplitude. HereAt an equilateral bispectrum configuration, the exact normalized amplitude isThus $c_s\to 1$ gives an amplitude of order $ϵ$, consistent with weak non-Gaussianity in single-field slow-roll inflation. For $0<c_s\ll 1$, this vertex gives an amplitude of order one, enhanced relative to the slow-roll limit but not diverging as $c_s^{-2}$ after power-spectrum normalization. At fixed $H,ϵ,M_{\mathrm{Pl}}$ the raw bispectrum does scale as $c_s^{-2}$, but $P_{\zeta }^2$ has the same scaling. The often-quoted larger small-sound-speed amplitudes of more general inflationary actions require their actual interaction coefficients and other vertices; they do not follow from the lone supplied vertex. These conclusions concern its tree contribution within the stated weak-perturbation approximation.

The bispectrum shape function isAt fixed perimeter, the arithmetic-geometric mean inequality bounds this by $1/27$, attained only at an equilateral bispectrum configuration. In a squeezed bispectrum configuration with $k_1=q\ll k_2\simeq k_3=k$, it behaves as $q/(8k)\to 0$. At a flattened bispectrum configuration $(k,k,2k)$ it is finite, $1/32$, with no boundary singularity. Thus the shape prefers equilateral triangles and suppresses squeezed triangles, with a broad finite flattened contribution. This is a shape preference, not a complete signal-to-noise calculation: actual observational weighting also depends on cosmological transfer functions, covariance and accessible triangles. If $C=0$, this vertex produces no bispectrum at any configuration; if $C<0$, the signed shape reverses while its absolute triangle preference is unchanged.

Work on an open domain with $xy>0$ and $x\ne 0$, where the displayed square root is real and differentiable. Split $u=v+h$ with $v=\varphi (xy)$ and $h=\sqrt{xy}\psi (y/x)$. For the first term, the chain rule givesso $x^2{v}_{xx}-y^2{v}_{yy}=0$.

For the second term, set $t=y/x$. Its first partial derivatives areThey show $x{h}_x+y{h}_y=h$, the degree-one case of Euler theorem for homogeneous functions. Differentiating that identity with respect to each variable yieldsThe mixed partial derivatives agree here by the chain rule. Multiplying these equations by $x$ and $y$, respectively, proves the second derivative identity for a degree-one homogeneous function:Adding the results for $v$ and $h$ verifies the partial differential equationEquivalently, $h(\lambda x,\lambda y)=\lambda h(x,y)$ for positive $\lambda$, so the same proof follows directly from its homogeneous function property. The statement is local to a domain where the given expression is defined; it does not require assigning values on $x=0$.

The scaling relation says that $z$ is a homogeneous function of degree $n$. Fix $(x,y)$ and differentiate the identity with respect to the positive parameter $\lambda$. The chain rule givesSetting $\lambda =1$ proves the Euler theorem for homogeneous functions identityDifferentiate once more with respect to $\lambda$, using equality of the mixed partial derivatives for the assumed twice continuously differentiable function:At $\lambda =1$ this is the second-order Euler identity for homogeneous functions:The repeated parameter differentiation accounts for the factor $n(n-1)$, rather than $n^2$.

In a Cartesian coordinate system the leading electric dipole field givesThereforeThe force is downwards on the equator and upwards on the negative axial side. A free positive electric charge released from rest at the left-hand point initially accelerates downwards. As it moves into $x<0,y<0$, the force also acquires a positive $x$ component. Its path bends toward the right; the force continually changes as its position changes. An initial velocity was not specified, so the following more precise trajectory assumes release from rest and neglects radiation reaction.

There is a useful exact result within the point-electric dipole approximation. Put $A=k_{e}pQ>0$; the potential energy is $U=Ay/r^3$, a homogeneous function of degree $-2$. Thus $\mathbf{r}\cdot \mathbf{F}=2U$. For a particle of mass $m$, Newton's second law and conservation of energy implyAt release, both kinetic energy and potential energy vanish, so $E=0$; also $d(r^2)/dt=0$. Consequently $r=d$ throughout the motion. Parametrize the lower semicircle by $x=-d\cos \alpha$, $y=-d\sin \alpha$. ThenThe electric charge gains speed until the bottom point, where the force is inward and the speed is $\sqrt{2A/(m{d}^2)}$. It then slows, reaches $(d,0)$ at rest, and reverses. Thus, under these assumptions,Its trajectory is not an electric field line: the force determines acceleration, rather than velocity direction. For the finite two-charge electric dipole, corrections of order $a^2/d^2$ perturb the circular trajectory; an arbitrary nonzero initial velocity also changes it.

For a twice continuously differentiable homogeneous function $h(x,y)$ of degree one, Euler theorem for homogeneous functions gives $x{h}_x+y{h}_y=h$. Differentiation first with respect to $x$ and then, separately, with respect to $y$ givesMultiplying by $x$ and $y$, respectively, shows that $x^2{h}_{xx}=y^2{h}_{yy}=-xy{h}_{xy}$. This identity can verify a second-order partial differential equation without expanding a complicated homogeneous expression twice.

The virial theorem relates the average value of a function along a trajectory to the forces that bind it. For an isolated system of particles interacting through Newtonian gravitational potential energy, it gives $2⟨K⟩+⟨U⟩=0$: the average kinetic energy is half the magnitude of the average negative potential energy. An eccentric Kepler orbit makes the distinction vivid: kinetic energy and potential energy vary throughout the orbit, although their sum is constant. The relation concerns their time averages, rather than their values at every instant. It underlies the virial mass estimate and the negative heat capacity of a virialized gravitational system.

Use constant positive particle masses $m_i$, positions ${\mathbf{r}}_i$, mechanical momenta ${\mathbf{p}}_i=m_i{\dot{\mathbf{r}}}_i$, and total kinetic energy $K={\sum }_i{\mathbf{p}}_i^2/(2m_i)$. The auxiliary quantity $G(t)$ is a sum of dot products, not the gravitational constant, which we denote here by $G_N$:This finite-$T$ identity is exact wherever Newton's second law holds along a regular trajectory; ${\mathbf{F}}_i$ must include every force. The endpoint term is what a finite observation window leaves behind. For the infinite-time theorem, suppose the averages on the left exist and $(G(T)-G(0))/T\to 0$. A sufficient condition is uniformly bounded positions and velocities for finitely many particles. More generally $G(T)=o(T)$ suffices, even without those individual bounds. Bounded positions alone do not supply a bound on momenta, especially near a singular potential energy; nor does a small endpoint term prove thermodynamic equilibrium. Periodic motion with $G(T)=G(0)$ gives the identity without an endpoint term at that finite $T$. For an isolated system, use an inertial centre of mass frame so translational drift does not contribute bulk kinetic energy.^[ref]

For a differentiable potential energy that is a homogeneous function of degree $q$, choose its additive constant so that the scaling law really holds. With conservative forces, the Euler theorem for homogeneous functions givesThus a harmonic oscillator has $q=2$ and $⟨K⟩=⟨U⟩$. An isolated system with Newtonian gravitational potential energy hasEach interacting pair occurs once, and $U=0$ at infinite separation. Here the total energy $E=K+U$ is conserved. An external field or a mixture of different scaling powers requires its own force term; one cannot assign the whole potential energy a single degree $q$. The virial theorem for a homogeneous pair potential states the corresponding general energy relation.

For a sufficiently regular homogeneous function $f$ of degree $n$ on a three-dimensional region, Euler theorem for homogeneous functions gives $x\cdot \nabla f=nf$. Hence $\nabla \cdot (fx)=(n+3)f$, and the divergence theorem gives the displayed reduction when $n\ne -3$. At degree $-3$ only the zero-flux identity follows, so division by $n+3$ is unavailable. Singular functions additionally require an excision or limiting argument; the surface contribution at a removed singularity must not be silently dropped. On the sloping side of a cone with its apex at zero, $x$ is tangent to the side and $x\cdot n=0$, so only the cap contributes to this formula.