# Is proper time equal to the Invariant Interval or the time elapsed in the Rest Frame?

This is not a contradiction, but simply a constraint on the allowable coordinate systems that would qualify as a particle’s rest frame. You have discovered that along the worldline of the particle the metric in the particle’s rest frame must have $$g_{00}=1$$. It is not mandatory to use such coordinates, but only such coordinates will be called the particle’s rest frame.

As an example, consider the standard Schwarzschild metric for a manifold with (-+++) signature and in units where $$c=1$$ $$g=\left( \begin{array}{cccc} -\left(1-\frac{R}{r}\right) & 0 & 0 & 0 \\ 0 & \left(1-\frac{R}{r}\right)^{-1} & 0 & 0 \\ 0 & 0 & r^2 & 0 \\ 0 & 0 & 0 & r^2 \sin (\theta ) \\ \end{array} \right)$$ which is the line element $$ds^2 = -d\tau^2 = -\left(1-\frac{R}{r}\right) dt^2 + \left(1-\frac{R}{r}\right)^{-1} dr^2 + r^2 d\theta^2 + r^2 \sin (\theta ) d\phi^2$$

Now, for a particle at rest we have $$dr=d\theta=d\phi=0$$ so $$ds^2 = -d\tau^2 = -\left(1-\frac{R}{r}\right) dt^2 = g_{00} dt^2$$ Notice that for $$r=\infty$$ we have $$g_{00}=-1$$ so the proper time $$d\tau$$ is equal to the coordinate time $$dt$$. So these coordinates are a valid rest frame for an observer at rest at $$r=\infty$$. However, for $$r=10R$$ we have $$g_{00} = -0.9$$ so $$d\tau^2 = 0.9 dt^2$$ thus the coordinate time is not equal to the proper time. These same coordinates are not a valid rest frame for an observer at rest at any finite $$r$$.