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?

  • 2
    $\begingroup$ Hint: is $\mathrm dx^0=\mathrm dt$ or $\mathrm dx^0=c\mathrm dt$? $\endgroup$ Commented Mar 27, 2017 at 15:32
  • $\begingroup$ So should my second term in brackets read $\left[ m^{-2} m^3s \right]$ for $[R][dxdydz][dt]$? $\endgroup$
    – Bob
    Commented Mar 27, 2017 at 15:38
  • $\begingroup$ @AccidentalFourierTransform $c=1$ tho $\endgroup$
    – Ryan Unger
    Commented Mar 27, 2017 at 15:42
  • $\begingroup$ @ocelouvsky that is true when working in natural units, but here I am explicitly working in SI units. $\endgroup$
    – Bob
    Commented Mar 27, 2017 at 15:50
  • $\begingroup$ @0celouvsky ...and $8\pi G=1$ $\endgroup$ Commented Mar 27, 2017 at 15:52

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.