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: $$d\tau^2 = \sum_{i,j=1}^4 g_{ij}(\mathbf{q}) dq_i dq_j$$

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:

$$\tau = \int_{u_0}^{u_1}\sqrt{g_{ij}\dot{q}_i\dot{q}_j}\ du$$