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From the Friedmann equation $$ H^2=\frac{8\pi G}{3}\rho_m - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}. $$ The dimension of Hubble parameter $H$ is $\frac{1}{\text{time}}$ or $\frac{1}{[T]}$. Therefore, the dimension of the all terms on the right hand side of the above equation are $\frac{1}{[T]^2}$. However, for the case of scalar field or dark energy. We can write the Friedmann equation as $$ H^2 = \frac{8\pi G}{3}\left(\rho_m + \rho_{\phi} \right) = \frac{8\pi G}{3}\left(\rho_m + \frac 12 \dot\phi^2 +V(\phi) \right). $$ The dimension of the all terms are $\frac{1}{[T]^2}$. Nevertheless, I have checked the dimension of the kinetic and potential terms. I have found that the dimension of the kinetic and potential terms are not $\frac{1}{[T]^2}$.

Could you please suggest me about the dimension of the kinetic and potential terms.

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It is best understood in the Natural units. You can see for example here. In such a system of units, a scalar field has the dimensions of mass (or energy), $[\phi]=[M]=[T^{-1}]$.

Now, The LHS of your second equation is has dimensions $[M^2]$. The terms inside the parenthesis in the RHS have mass dimension $[M^4]$, as you can easily check. Also, $[G]=[M^{-2}]$. Bingo!

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  • $\begingroup$ $[G] = [M^{-2}]$ $\endgroup$ Aug 1, 2018 at 6:36

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