What are the units of the Bekenstein bound?

Question

Working with the Wikipedia definition of the Bekenstein bound:

$S \leq \frac{2 \pi R k_bE}{\hbar c}$

$2\pi R \ $ is $m^2$

$k_b$ is $\frac{J}{K}$

$E$ is $J$

$\hbar$ is $J*s$

$c$ is $\frac{m}{s}$

$\frac{m^2 \frac{J}{K} J}{(Js \frac{m}{s})} = m \frac{J}{K}$

Am I overlooking something?

Answer 1

In theoretical physics, entropy is typically dimensionless. For example, instead of defining $S=k_B \log W$, we would define $S=\log W$.

This is precisely what has been done in this equation: $\hbar \cdot c$ has units of $J \cdot m$, which cancels the units up top.

See also: http://www.scholarpedia.org/article/Bekenstein_bound

Answer 2

$2\pi R$ is meters,not meters squared.