# Entropic gravity and equipartition theorem

Erik Verlinde on page 8 of https://arxiv.org/pdf/1001.0785.pdf derives Newton's law of gravity from the entropic perspective. My question is about the use of the equipartition theorem and why this is allowed. But, first let me write down what he does to serve as a reference to this question.

\begin{align} \Delta S&=2\pi k_B \frac{mc}{\hbar}\Delta x \tag{entropic position} \\ N&=\frac{4\pi R^2}{L^2} \tag{No. of bits on a sphere} \end{align}

Then he recalls the following known relations,

\begin{align} E&=\frac{1}{2}N k_B T \tag{equipartition theorem} \\ E&=Mc^2\\ T\Delta S &=F \Delta x \tag{entropic force} \end{align}

Using these equations, he obtains Newton's law as follows

\begin{align} E&=\frac{1}{2}Nk_BT \tag{equipartition theorem} \\ Mc^2&=\frac{1}{2}Nk_BT \tag{eliminating E}\\ T&=\frac{2Mc^2}{N k_B} \tag{solving for T}\\ \end{align}

He then injects the T obtained into the formula for entropic force

\begin{align} T\Delta S&=F\Delta x \tag{entropic force} \\ T2\pi k_B \frac{mc}{\hbar}\Delta x&=F\Delta x \tag{entropic position } \\ \frac{L^2}{4\pi R^2} \frac{2Mc^2}{k_B}2\pi k_B\frac{mc}{\hbar}\Delta x&=F\Delta x \tag{injecting T and N}\\ \left( \frac{L^2 c^3}{\hbar} \right) \frac{M m}{R^2} &= F \tag{clean up} \\ \frac{G M m}{R^2}=& F \tag{definition of G} \end{align}

**** My Question----

Why is the equipartition theorem applicable? Rather than using it, my reflex would have instead been to do the following, but I cannot find my mistake.

First I would obtain $dN$ from $N$ $$dN=\frac{8\pi R}{L^2}dR$$

Then I would relate $dS$ to $dN$ $$dS=k_B dN$$

Then I would derive as follows

\begin{align} dE&=TdS \\ dE&=k_BTdN \\ \int{dE}&=Tk_B \int{dN}\\ E&=Tk_BN \end{align}

And this is where it fails. I continue the derivation in a similar manner but with $E=Tk_BN$ instead of $E=\frac{1}{2}Tk_BN$. As a result, I obtain a definition for G which is divided by 2. I cannot find the mistake. Why am I not getting the equipartition theorem with the 1/2 factor?