How to prove that time stops at the speed of light using general relativity in a flat spacetime? I know from special relativity that $t'=t/\sqrt{1-(v/c)^2}$, where t is proper time. I see from this equation that for each second of t, t' grows bigger and eventually reaches infinity as v goes to $299792458$ m/s. This means that more time passes for the observer. However, when I try to prove this in general relativity, I cannot arrive at this conclusion.
Considering only the x axis, the Minkowski metric is:
\begin{pmatrix}
-c^2 & 0\\
0 & 1
\end{pmatrix}
Plugging the variables in the equation
$$-c^2=g_{\mu\nu}v^{\mu}v^{\nu}$$ I get $$-c^2=-c^2v_t^2+v_x^2$$
Rearranging yields
$$v_t=\sqrt{(c^2+v_x^2)/c^2}$$
which means that
$$d_t/d_{\tau}=\sqrt{(1+(v/c)^2)}$$
Integrating gives
$$t=\tau \sqrt{1+(v/c)^2}$$
In this equation, while time still passes faster for the observer, it doesn't go to infinity when the object moves at the speed of light. Therefore, time doesn't stop at the speed of light.
How do you account for this difference? If I am doing something wrong, how to prove it correctly? Thanks in advance.
 A: Let's try to do the computation from the start, so you can see where you missed some things,first of all the Minkowsky metrix is:
\begin{equation}g_{\mu\nu} = \begin{pmatrix}
1 & 0\\
0 & -1
\end{pmatrix}\end{equation}
(it can also be with opposite minus sign, but I prefer this convention) and the four-velocity is defined as:
\begin{equation}
U= \frac{dX}{d\tau} = \begin{pmatrix}\ c \frac{dt}{d\tau} ,& \frac{d\vec{x}}{d\tau}\end{pmatrix} = \begin{pmatrix}\ c \gamma ,& \gamma \frac{d\vec{x}}{dt}  \end{pmatrix} = \gamma \begin{pmatrix}\ c  ,& \vec{v}  \end{pmatrix} 
\end{equation} where $X$ is the four-position vector and $\tau$ the proper time, and we have also defined $dt/d\tau \equiv \gamma$.
Now, the next axiom we are going to take into this computation is that the modulus of the four-velocity is always the speed of light, with only components in space for massless particles and with a combination of space and time components for massive particles, depending on how fast they are moving, this means:
\begin{equation}
c^2=|U|^2= g_{\mu\nu}U^\mu U^\nu =U_t^2-U_x^2
\end{equation} so, with all that we can see, that:
\begin{equation}
c^2=\gamma^2 (c^2-|\vec{v}|^2) \ \xrightarrow{} \ 1=\gamma^2 \left(1-\frac{|\vec{v}|^2}{c^2} \right)
\end{equation}
which means that:
\begin{equation}
\frac{dt}{d\tau}=\gamma= \frac{1}{\sqrt{1-\frac{|\vec{v}|^2}{c^2}}} \ \xrightarrow{} \ t = t' / \sqrt{1-\frac{|\vec{v}|^2}{c^2}}
\end{equation} as you wanted! :)
A: You have claimed that $v_x=v$, in other words that the X component of the four velocity is the classical velocity, but this is not correct and in fact you want $v=v_x/v_t.$
Once that is clarified then your derivation proceeds as $$
-c^2=(v^2-c^2)v_t^2\\
v_t = \frac1{\sqrt{1-(v/c)^2}}$$
quite directly, as desired.
Alternate ways to phrase the basic point if you are used to other notations:

*

*In terms of rapidities the four-velocity is $(c~\cosh\phi,\hat n~c~\sinh\phi)$ where $\vec v=\hat n~c\tanh\phi$


*In terms of $\gamma,\beta$ the four velocity is always $c~(\gamma,~\gamma\vec\beta)$
A: For flat spacetime and motion along the x-axis, the spacetime interval can be written
$$c^2d\tau^2 = c^2 dt^2 - dx^2$$
$$ d\tau^2 = dt^2\left( 1 - \frac{1}{c^2}\frac{dx^2}{dt^2}\right)\ , $$
$$ dt = \frac{d\tau}{\sqrt{1 - v_x^2/c^2}}\ . $$
If $v_x= dx/dt \rightarrow c$, then $dt \rightarrow \infty$ for any finite proper time interval.
