What's the matter with Planck mass $M_P$ in Einstein-Hilbert action?

The Einstein-Hilbert (EH) action is often written as $$S_{EH}=\frac{c^4}{16\pi G}\int d^4x \sqrt{-g}R\tag{1}$$ and often as

$$S_{EH}=\int d^4x \sqrt{-g}M_P^2 R\tag{2}.$$ Comparing (1) and (2), one has $M_P=\frac{c^2}{4\sqrt{\pi G}}$. But the definition of Planck mass is given by $M_P=\sqrt{\frac{\hbar c}{G}}$.

$\bullet$ How do we reconcile these two different expressions for $M_P$? Do I misunderstand something?

$\bullet$ Both (1) and (2) are classical action, and therefore, the action should not contain $\hbar$. In that case, how is form (2) of the action justified?

• Have you checked the dimensional consistency of each equation? – Cosmas Zachos Feb 7 '18 at 20:52

You are comparing apples with oranges.

Let's first agree on the dimensions of the relevant quantities, basically Freshman physics:

$$[G] = L^3 M^{-1} T^{-2} \\ [S]= M L^2 T^{-1}\\ [\kappa]=\left [\frac{8\pi G}{c^4}\right ]= T^2 M^{-1} L^{-1} \\ [m_p]=\left [\sqrt{\frac{\hbar c}{8\pi G}}\right ] = M \\ [R]=L^{-2} \\ [d^4x]= L^3 T.$$ The last line is what Feynman uses in his lectures that passed on to the WP article you are quoting. Landau and Lifschitz, Classical theory of fields, §93, do not use his $dx^0=dt$, but $=c~ dt$ instead, so their $[d^4x]=L^4$ which most people carry in their gut, because it is really so much more sensible, but it may lead to misconceptions.

So your formula (1) jibes just fine, $$M L^2 T^{-1}= L M T^{-2} ~~ L^{-2} ~~ L^3 T.$$

With the exception of QFTheorists and quantum gravity types, nobody needs use the reduced Planck mass defined above. Nevertheless, since power counting is second nature to them, and they are practical people, they bypass the mess by simply rewriting, instead of your (2), for a start (only!), $$S_{EH}=\frac{ 1}{ 2\kappa}\int d^4x \sqrt{-g}R = \frac{c^3 m_p^2}{2\hbar}\int d^4x \sqrt{-g}R .$$ You appreciate $\hbar$ is not really there, as it was only inserted to neutralize/eliminate itself in the Planck mass.

However, you never see this, since QFTheorists use HEP natural units, $c=\hbar=1$: their world is steeped in relativity and QM, so they rely on them as conversion units. In this world, then, everything becomes simple and wholesome, in no time, $$S_{EH}= \frac{ m_p^2}{2 }\int d^4x \sqrt{-g}R ~.$$ So wholesome, in fact, that one can grow oblivious of the underlying structure--and the Feynman-L&L notation disparity, now invisible.

The Planck mass is about the weight of a hundredth of a grain of sugar, so, then, minuscule when it comes to classical gravitational effects, but huge, O(1018 GeV), when considering energies of ordinary particles-- a particle that heavy would have its Compton wavelength recede inside its own Schwarzschild radius; the perfect background unit, then.