# What are the units of the Bekenstein bound?

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?

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## 1 Answer

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.

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In that case, the units seem to be measured in meters: Top: 2*pi*R is area, which is m^2. And E is J. Bottom: h-bar is J*s. And c is m/s. (m^2 * J) / (J*s * m/s) = m. –  Luke Burns Feb 18 '13 at 23:19
Ah, nevermind. I'm mixing up the Area in the Bekenstein-Hawking entropy formula with the 2pi*r in the Bekenstein bound. –  Luke Burns Feb 18 '13 at 23:51
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