User:Stephen J. Brooks/greqns

In general, the proper time $d\tau$ from a point $\mathbf {q} =(q_{0},q_{1},q_{2},q_{3})$ to a (infinitesimally) nearby point $\mathbf {q} +d\mathbf {q} =(q_{0}+dq_{0},q_{1}+dq_{1},q_{2}+dq_{2},q_{3}+dq_{3})$ under a metric $g_{ij}(\mathbf {q} )$ is given by:

Swan eating grass

$d\tau ^{2}=\sum _{i,j=1}^{4}g_{ij}(\mathbf {q} )dq_{i}dq_{j}$

Generally $g_{ij}=g_{ji}$ because these metric coefficients multiply onto identical terms $dq_{i}dq_{j}=dq_{j}dq_{i}$.

$d\tau ={\sqrt {\sum _{i,j=1}^{4}g_{ij}(\mathbf {q} )dq_{i}dq_{j}}}$

Therefore if a world line is parameterised by $\mathbf {q} (u)$ from $u=u_{0}$ to $u=u_{1}$, then the total proper time experienced on that line is

$\tau =\int _{u_{0}}^{u_{1}}d\tau$ $=\int _{u_{0}}^{u_{1}}{\sqrt {\sum _{i,j=1}^{4}g_{ij}(\mathbf {q} (u))dq_{i}(u)dq_{j}(u)}}$

Now put $dq_{i}(u)={\frac {dq_{i}}{du}}(u)du$ and take the two $du$ outside the square root, giving

$\tau =\int _{u_{0}}^{u_{1}}{\sqrt {\sum _{i,j=1}^{4}g_{ij}(\mathbf {q} (u)){\frac {dq_{i}}{du}}(u){\frac {dq_{j}}{du}}(u)}}\ du$

This is just an integral involving the known functions $\mathbf {q} (u)$, $g_{ij}(\mathbf {q} )$ and the derivatives of $\mathbf {q} (u)$ (with respect to $u$).

Suppressing function dependencies, using ${\dot {}}$ for derivative with respect to the path parameter and summing repeated indices implicitly gives a more compact way of writing this:

$\tau =\int _{u_{0}}^{u_{1}}{\sqrt {g_{ij}{\dot {q}}_{i}{\dot {q}}_{j}}}\ du$