# Proper time experienced by an observer moving along a NON-geodesic

For an observer moving along a time-like geodesic $$x^{\mu}(\lambda)$$ (parametrized by $$\lambda$$) the geodesic equations are satisfied $$\ddot{x}^{\mu}(\lambda) + \Gamma^{\mu}_{\ \; \nu\rho} \; \dot{x}^{\nu}(\lambda) \dot{x}^{\rho}(\lambda) \ = \ 0 \ .$$ (As described on page 109 of Carroll's Spacetime and Geometry, I am aware that you can always re-parametrize $$\lambda \to a \lambda + b$$ such that the RHS more generally has the form $$f(\lambda) \dot{x}^{\mu}(\lambda)$$)

As I understand it, the proper-time experienced by the observer on the geodesic maximizes the functional $$\tau \ = \ \int d\lambda\ \sqrt{ - g_{\mu\nu}(\lambda) \dot{x}^{\mu}(\lambda) \dot{x}^{\nu}(\lambda) } \ .$$

However, through use of some `external force' the observer can be put on a non-geodetic trajectory $$x_{\mathrm{non}}^{\mu}(\lambda)$$ where $$\lambda$$ parametrizes some path with $$\ddot{x}_{\mathrm{non}}^{\mu}(\lambda) + \Gamma^{\mu}_{\ \; \nu\rho} \; \dot{x}_{\mathrm{non}}^{\nu}(\lambda) \dot{x}_{\mathrm{non}}^{\rho}(\lambda) \ \neq \ 0 \ .$$ (or more generally $$\neq f(\lambda) \dot{x}^{\mu}(\lambda)$$).

My question is how does one calculate the proper time experienced by the observer moving along $$x_{\mathrm{non}}^{\mu}(\lambda)$$? Is it still through an integration of $$\int d\lambda\ \sqrt{ - g_{\mu\nu}(\lambda) \dot{x}_{\mathrm{non}}^{\mu}(\lambda) \dot{x}_{\mathrm{non}}^{\nu}(\lambda) } \ \ ?$$