# Why do the formulas for planck's units use h instead of hbar?

On the Answers.com page on Planck length, I see two almost-same formulas for the Planck length that differ only by the use of h and hbar. However, the constants are the same, and my calculator gives the correct answer for hbar instead of h, so the first use of h was probably intended to mean hbar. Why doesn't the Oxford Dictionary (and my textbook!) use hbar instead?

UPDATE: the equation (from Oxford dictionary?) I was talking about that uses h:

and the equation from Wikipedia that uses hbar, but gives the same constant for the Planck length:

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Planck Units are order of magnitude things, anyway. Since we don't have a theory of quantum gravity, we don't know its exact energy scale, so our knowledge of such things is only accurate to the scales we can get by dimensional analysis. Multiplication by pure numbers isn't going to change this. Using $\hbar$ instead of $h$ is just as 'right' either way. Of course, almost all quantum mechanics uses $\hbar$, so it would make more 'sense' to use the latter. –  Jerry Schirmer Jan 16 '11 at 21:36

the answers.com page you mentioned uses the following formula: $$L_{Planck} = \sqrt{\frac{Gh}{2\pi c^3}}$$ Note that there is the $2\pi$ factor in the denominator - so $h/2\pi$ may be simplified as the usual $\hbar$. They probably weren't able to type this character, or wanted to avoid terminology and symbols that are only known to physicists. But there is no numerical error on the answers.com page. At any rate, the definition above is equivalent to $$L_{Planck} = \sqrt{\frac{G\hbar}{c^3}}$$ which is the usual "unreduced" Planck length. See Wikipedia for the same formula:

http://en.wikipedia.org/wiki/Planck_length

Numerically, it's $1.6 \times 10^{-35}$ meters. (Update: The Oxford Dictionary of English has a wrong formula - they omitted $2\pi$ and forgot to cross the $h$, too. But they clearly mean the same Planck length.) Sometimes, people also use the "reduced" Planck length which is more fancy and "professional" in a sense: $$L_{Planck,reduced} = \sqrt{\frac{8\pi G\hbar}{c^3}}$$ Note that the $8\pi$ in the numerator may also be merged with the $\hbar$ to get $4h$ back - so the reduced Planck length is twice (because of the square root) the wrong Planck length that you would get by using $h$ instead of $\hbar$. But what is the real reason why $8\pi$ was added there?

The reason why $8\pi G$ appears instead of $G$ is because in some sense, $8\pi G$ is more natural a constant than $G$: this discussion is analogous to the treatment of $4\pi$ in electrodynamics. The constant $8\pi G$ is natural because the Einstein-Hilbert action is $$S_{EH} = \int d^D x \frac{1}{16\pi G} R\sqrt{-g}$$ The most natural coefficient would be $1/2$ instead of $1/16\pi G$ which makes it natural to set $8\pi G=1$. The reduced Planck length is somewhat longer (five times or so) - less extremely tiny. Even more often, particle physicists talk about the Planck energy and the reduced Planck energy which are close to $10^{19}$ and $10^{18}$ GeV, respectively.

The convention for the constant $G$ was originally chosen by Newton who wanted to write the gravitational force as $GMm/r^2$. Well, it would be more natural to have the factor of $4\pi$ or $8\pi$ in the denominator, $\Gamma Mm/8\pi r^2$. You can see that $\Gamma$ is simply $\Gamma=8\pi G$, and it would be natural to set $\Gamma$ equal to one.

I hope that I don't have to explain why $\hbar$ is more natural than $h$ for adult physicists. The "laymen" versions of the formulae may be simpler with $h$ - but they deal with wavelength etc. Adult physicists know that the wavelength of the sine is proportional to $2\pi$. And the most fundamental equations, such as the Schrödinger's equation or the commutators of $[x,p]$, take a simpler form in terms of $\hbar$ than $h$, of course.

Back to $G$: people had to choose the convention how to normalize $G$ in higher dimensions. The usual convention, as implicitly used above, is that the Einstein-Hilbert action always has the coefficient $1/16\pi G$. That implies that in $D$ spacetime dimensions, the force won't be $GMm/r^{D-2}$ but it will have some $D$-dependent numerical coefficients in it.

Best wishes Lubos

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Thank you very much Lubos! I understand that there should be the reduced Planck's constant in there one way or another (with hbar or with h over 2 pi). However, I do see a discrepancy between Wikipedia's equation and Oxford dict's equation, as I've updated the question to display. –  wrongusername Jan 16 '11 at 21:37
Thanks for the update, wrongusername. The Oxford Dictionary has an error - they forgot to slash the $h$, either because of insufficient fonts or incompetent writers haha. –  Luboš Motl Jan 16 '11 at 21:40