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I have a doubt on explicit calculation of proper time.

Considering that the metric is given by:

$$ds^{2} = -Adt^{2} + B^{-1}dr^{2}+Cd\Omega^{2} -2Ddtd\phi \tag{1}$$

where $d\Omega^{2}$ is the solid angle line element. Given that $r=cte$ , $a=cte$ , $\theta = cte$, then we can say that the proper time (infinitesimal) interval is given by:

$$d\tau ^{2} = g_{00}dt^{2} \equiv -Adt^{2} \tag{2}$$

I tried to calculate explicitly and I reached on:

$$\int_{0}^{\tau}d\tau = \int_{0}^{t} dt \hspace{3mm}\sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} \tag{3}$$

My doubt is:

Since $r=cte$ , $a=cte$ , $\theta = cte$ (therefore the hole square root is a fixed number $[*]$) then, can I say that the proper time is given by the follwing expression?

$$ \mathcal{T} = \int_{0}^{t} dt \hspace{3mm}\sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} =$$

$$= \hspace{3mm}\sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} \int_{0}^{t} dt = $$

$$= \sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} \hspace{3mm} t \implies $$

$$ \mathcal{T} = \sqrt{\Bigg[1+\frac{4 a cos\theta}{r}\Bigg]^{2}\Bigg[\frac{4 a^{2}+ r^{2}sin^{2}\theta }{r^{6}} -1 \Bigg]} \hspace{3mm} t $$

$$ * * * $$

$[*]$ In the software Mathematica, you would think about the $r$, $\theta$, and $a$, as quantities to insert on Manipulate structure, as just scrolling values. The real variable is therefore the time $t$.

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