Action is a quantity with units energy $\times$ time $=[kg \frac{m^2}{s}]$. The Einstein-Hilbert action is \begin{equation} S_{EH}=\frac{c^4}{16\pi G}\int \sqrt{-g}R d^4x \end{equation} 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?

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

\begin{equation} S_{EH}=\frac{c^3}{16\pi G}\int \sqrt{-g}R d^4x \end{equation}

as you can easily find - for example - in the classic Landau & Lifshitz Theory of Fields.


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