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Pg 163 of "Tensors, Relativity and Cosmology"

The action integral of a given matter distribution can be written in the form $$I_K=-c\int_\Omega\frac{\rho}{\sqrt{-g}}\frac{ds}{dt}\sqrt{-g} d\Omega \tag{19.48} $$ where $\frac{ds}{dt}=\sqrt{g_{kn}\frac{dx^k}{dt}\frac{dx^n}{dt}}$ and $\rho$ is the mass density of the matter distribution.

Since $$I_K=\frac{1}{c}\int_\Omega\mathcal{L}_{MK}\sqrt{-g}d\Omega \tag{1}$$

We may write $$\sqrt{-g}\mathcal{L}_{MK}=-\rho c\frac{ds}{dt} \tag{19.56}$$

Using the definition of the energy-momentum tensor $$\frac{1}{2}\sqrt{-g}T_{kn}=\partial_j{\frac{\partial{\sqrt{-g}\mathcal{L}_M}}{\partial{\partial_jg^{kn}}}}-\frac{\partial \sqrt{-g}\mathcal{L}_M}{\partial g^{kn}} \tag{2}$$

It can be shown that

$$T^{kn}=-\frac{2}{\sqrt{-g}}\frac{\partial\sqrt{-g}\mathcal{L}_{MK}}{\partial g_{kn}}=\frac{\rho c}{\sqrt{-g}} \frac{ds}{dt}u^ku^n \tag{19.57}$$

(The first term on the RHS of (2) vanishes since (19.48) does not depend on the second order derivatives of the metric tensor)

where $u^n=\frac{dx^n}{ds}$ (the author's version of the four-velocity vector)

In the pseudo-Euclidean case, the energy density: $$T^{00}=\frac{\rho c^2}{\sqrt{1-\frac{v^2}{c^2}}} \tag{19.58}$$

and the momentum density $$ T^{0 \alpha}=\frac{\rho v^\alpha}{\sqrt{1-\frac{v^2}{c^2}}} \tag{19.59}$$

*1.Shouldn't (19.56) be $-\rho c^2 \frac{ds}{dt}$ instead since there is a factor $\frac{1}{c}$ in the action integral (1)?

(the author continued the chapter by using $ -\rho c\frac{ds}{dt}$ later on so I don't know whether or not I've missed something out...)

*2. Isn't the numerator of (19.59) missing a constant c (i.e. $\rho c v^\alpha$) since $u^\alpha=\frac{v^\alpha}{c\sqrt{1-\frac{v^2}{c^2}}}$?

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  • $\begingroup$ And this is why people use natural units. $\endgroup$ – Bob Knighton Jul 2 '19 at 7:39

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