# Units of Einstein-Hilbert Action

Action is a quantity with units energy $\times$ time $=[kg \frac{m^2}{s}]$. The Einstein-Hilbert action is $$S_{EH}=\frac{c^4}{16\pi G}\int \sqrt{-g}R d^4x$$ Looking only at the units in this action: \begin{eqnarray} [S] &=& \left[ \frac{\left(\frac{m}{s}\right)^4}{\frac{m^3}{kg s^2}}\right]\left[ m^{-2} m^4\right] \\ &=& \left[ \frac{kg \times m^3}{s^2}\right] \end{eqnarray} Where the Ricci scalar has units of $m^{-2}$, and the spacetime metric has no units. Where does the extra unit of $m/s$ in my calculation come from, or did I forget to cancel something else?

• Hint: is $\mathrm dx^0=\mathrm dt$ or $\mathrm dx^0=c\mathrm dt$? – AccidentalFourierTransform Mar 27 '17 at 15:32
• So should my second term in brackets read $\left[ m^{-2} m^3s \right]$ for $[R][dxdydz][dt]$? – Bob Mar 27 '17 at 15:38
• @AccidentalFourierTransform $c=1$ tho – Ryan Unger Mar 27 '17 at 15:42
• @ocelouvsky that is true when working in natural units, but here I am explicitly working in SI units. – Bob Mar 27 '17 at 15:50
• @0celouvsky ...and $8\pi G=1$ – AccidentalFourierTransform Mar 27 '17 at 15:52

The equation you wrote - which is the same mentioned in Wikipedia, as of today - assumes that $dx^0=dt$.
It is however generally smarter to have all 4 coordinates share the same units, so most (I would say all) tensors have components sharing the same units. For example: in lorentian coordinates Riemann and Ricci tensors have units $[m^{-2}]$ and $g_{\mu\nu}$ is dimensionless.
Therefore it is customary to use $dx^0=cdt$, as you assumed. But now the action becomes
$$S_{EH}=\frac{c^3}{16\pi G}\int \sqrt{-g}R d^4x$$