Invariance of Inner product between 4-velocity under general coordinate transformation

Question

I know that inner product between 4-velocity is invariant under Lorentz transformation and I know that inner product between any 2 vectors under general coordinate transformation is invariant. Therefore, the inner product between two 4-velocities $u^{i}=\left(d x^{i} / d \tau\right)$ should also be invariant under arbitrary coordinate transformation. In my book it is written that $u_{i} u^{i}=1$ and here is my try to get this result : $$d s^{2}=g_{a b} d x^{a} d x^{b}=g_{a^{\prime} b^{\prime}} d x^{a^{\prime}} d x^{b^{\prime}}=d s^{\prime 2}$$ $$d s'^{2}=g_{0^{\prime} 0^{\prime}} d t'^{2}=g_{0^{\prime} 0^{\prime}} d \tau^{2}$$ $$\begin{aligned} u^{a} u_{a} &=\frac{d x^{a} d x_{a}}{(d \tau)^{2}} \\ &=g_{0^{\prime} 0^{\prime}}\frac{d x^{a} d x_{a}}{d s^{2}} \\ &=g_{0^{\prime} 0^{\prime}} \end{aligned}$$ Here I have assumed that clock is moving in some gravitational field in the frame O' and used $ dt'=d \tau$. Clearly my result is wrong as $g_{0^{\prime} 0^{\prime}}$ is not invariant but I am not getting where I have done something silly.

Answer 1

Working in the (+ - - -) convention. For simplicity, assume the metric is diagonal (this proof still holds if it's not, it's just longer):

\begin{equation} d \tau^2 = g_{00} dt^2 + g_{xx} dx^2 + g_{yy} dy^2 + g_{zz} dz^2 \\ 1 = g_{00} \left(\frac{dt}{d\tau} \right)^2 + g_{xx} \left( \frac{dx}{d\tau}\right)^2 + g_{yy} \left(\frac{dy}{d\tau}\right)^2 + g_{zz} \left( \frac{dz}{d\tau} \right)^2 \\ 1 = g_{00} \left(\frac{dx^0}{d\tau} \right)^2 + g_{xx} \left( \frac{dx^1} {d\tau}\right)^2 + g_{yy} \left(\frac{dx^1}{d\tau}\right)^2 +g_{zz} \left( \frac{dx^2}{d\tau} \right)^2 \\ 1 = g_{00} \left(u^0 \right)^2 + g_{xx} \left( u^1\right)^2 + g_{yy} \left(u^2\right)^2 +g_{zz} \left( u^3 \right)^2\\ 1 = g_{\mu \nu} u^{\mu} u^{\nu} \implies u^{\mu} u_{\mu} = 1. \end{equation}

I think your mistake is on line 2, but I don't quite understand what you are doing there so I'm not sure I can help.

Answer 2

The definition of proper time you used is wrong. To find the proper time, you can't just substitute it into the "regular" time. In special relativity, you don't notice this problem as $g_{00}$ has unitary norm, but in general relativity you have to remember that

$ d\tau^2=ds'^2 = g_{\mu\nu}dx^\mu dx^\nu \Rightarrow u^{a} u_{a} =\frac{d x^{a} d x_{a}}{(d \tau)^{2}} =\frac{d x^{a} d x_{a}}{d s^{2}} =1$

wich is invariant. See the example in this discussion or the definition on Wikipedia.