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The Planck length is ${\ell }_P=\sqrt{\hslash G/c^3}$. Together with the Planck mass $m_P=\sqrt{\hslash c/G}$ it satisfies $G{m}_P/c^2={\ell }_P$.

Planck units use fundamental constants to set natural units of mass, length and time. The displayed convention sets the ordinary Planck mass to one. The related reduced Planck units convention instead normalizes the reduced Planck mass.

For $V=m^2{\varphi }^2/2$ and unreduced Planck mass $m_P=G^{-1/2}$, the potential slow-roll parameter is ${ϵ}_V=m_P^2(V^{\prime }/V{)}^2/(16\pi )=m_P^2/(4\pi {\varphi }^2)$. Its conventional value one gives the displayed endpoint estimate. Exact accelerated expansion ends at ${ϵ}_H=1$, or ${\dot{\varphi }}^2=V$ for a canonical scalar alone. Substitution of slow-roll formulas into this exact condition gives a different order-one coefficient because slow roll is failing there; a precise endpoint requires integrating the full inflaton equation of motion and Friedmann equation. The large-field e-fold estimate is insensitive to this order-one endpoint uncertainty.

For $V=m^2{\varphi }^2/2$ and unreduced Planck mass, the slow-roll e-fold count is $N=2\pi ({\varphi }_i^2-{\varphi }_f^2)/m_P^2$. A formal initial potential $V_i\sim m_P^4$ gives ${\varphi }_i^2\sim 2m_P^4/m^2$, hence the displayed large-field estimate and expansion factor $e^N$. The final-field contribution is order one in $N$. Without specifying $m/m_P$ there is no unique numerical expansion factor. The extrapolation to Planck density is a classical estimate, rather than a controlled treatment of quantum-gravity corrections.

For $V=\lambda {\varphi }^4/4$ and unreduced Planck mass $M_P$, the potential slow-roll parameter is ${ϵ}_V=M_P^2/(\pi {\varphi }^2)$. Setting it to one gives the displayed conventional endpoint. The exact end of scalar-dominated accelerating expansion is ${ϵ}_H=1$, equivalently ${\dot{\varphi }}^2=V$. Since slow-roll is failing near that point, a precise endpoint requires evolving the full inflaton equation of motion; potential and kinetic criteria evaluated with a failing approximation need not give the same order-one coefficient.

With unreduced Planck mass $M_P$ and initial potential-dominated density $\lambda {\varphi }_i^4/4\sim M_P^4$, one has ${\varphi }_i^2\sim 2M_P^2/\sqrt{\lambda }$. The slow-roll e-fold count is $N\simeq \pi ({\varphi }_i^2-{\varphi }_e^2)/M_P^2$. Thus small positive $\lambda$ produces a very large expansion factor $e^N$ in this classical quartic model. The order-one endpoint correction is negligible compared with $2\pi /\sqrt{\lambda }$; initial kinetic energy, spatial curvature and the model's validity at trans-Planckian field values are independent assumptions.

For a quartic inflaton potential, an endpoint field of order the unreduced Planck mass leaves density of order $\lambda M_P^4$. If prompt, efficient reheating converts it to equilibrium radiation, $\rho =({\pi }^2/30)g_{\ast }{T}_R^4$ gives the displayed scaling. With the potential-only endpoint estimate ${\varphi }_e=M_P/\sqrt{\pi }$, the coefficient is $[15/(2{\pi }^4){]}^{1/4}$. End-of-inflation kinetic energy changes that coefficient by order one, and delayed transfer or incomplete thermalization requires a reheating model rather than the potential alone.

For $V(\varphi )=\lambda {\varphi }^4/4$ with $\lambda >0$, use natural units and the unreduced Planck mass $M_P=G^{-1/2}$. The slow-roll equations give $H\simeq \sqrt{2\pi \lambda /3}{\varphi }^2/M_P$ and $\dot{\varphi }\simeq -M_P\sqrt{\lambda /(6\pi )}\varphi$. Integrating gives the displayed exponential field decay, valid on either sign branch, andThe potential slow-roll parameters require ${\varphi }^2\gg M_P^2$. The expression is valid only during slow-roll inflation; its formal late-time saturation does not describe the subsequent reheating or radiation era.

Here the reduced Planck mass $M_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$ is one, so the Friedmann equation for an energy density $\rho$ is $3H^2=\rho$. This differs by a factor of $\sqrt{8\pi }$ from setting the ordinary Planck mass to one; the convention must be stated in inflationary amplitudes.

For a large isolated Schwarzschild black hole with a fixed effective set of radiated species, Hawking radiation has $T\propto M^{-1}$ and luminosity $P=\alpha /M^2$ in natural units. Integrating $dM/dt=-\alpha /M^2$ gives the displayed law. A horizon-area blackbody radiation estimate gives $\alpha =1/(15360\pi )$; greybody factors and species change it. With mass-dependent $\alpha$, integrate $t={\int }_{M(t)}^{M_0}{m}^{2}dm/\alpha (m)$ instead. The semiclassical gravity approximation fails near Planck mass, so a formal zero of the cubic law does not settle the final state.

In natural units with unreduced Planck mass $M_P=G^{-1/2}$, the radiation-era Hubble parameter scales as $\sqrt{g_{\ast }}{T}^2/M_P$, while a low-energy weak interaction rate scales as $G_F^2{T}^5$. Their equality gives a decoupling scale of order one MeV. Distinct weak processes have distinct rate coefficients: neutrino decoupling and cosmological weak freeze-out of neutron-proton conversion are nearby events, not an identical precise temperature inferred from the power laws alone.