bigg(\frac{\Omega r_s}{c}\bigg)} =  0\\
&\implies \sinh^{2}{\bigg(\frac{\Omega r_s}{c}\bigg)}= \frac{1}{2}\\
&\implies \frac{\Omega r_s}{c} = \arcsinh{\bigg(\frac{1}{\sqrt{2}}\bigg)}\\
&\implies \frac{\Omega r_s}{c} = \ln{\bigg(\frac{1+\sqrt{3}}{\sqrt{2}}\bigg)}\bigg[ \text{ As we know that } \arcsinh{(x)}=\ln{\bigg(x+\sqrt{x^2 +1}\bigg)}\bigg]\\
&\implies\boxed{ r_s = \frac{c}{\Omega}~\ln{\bigg(\frac{1+\sqrt{3}}{\sqrt{2}}\bigg)}}
\end{align*}

Here we have taken $\Omega = 1 $ rad/s so we have $r_s=3 \times 10^{8}\times\ln{\bigg(\frac{1+\sqrt{3}}{\sqrt{2}}\bigg)}= 1.975 \times 10^{8} $ m


Now as mentioned before after a particular point the slope of the graph of $v(r_s)$ vs $r_s$ decreases, we have found that the third order derivative i.e. $\frac{d^{3}(v(r_s)}{dr_s^{3}}=0$ at $r_s = 1.975 \times 10^{8}$ m, if $\Omega = 1$ rad/s. That means after $r_s = 1.975 \times 10^{8}$ m as $r_s$ increases $v(r_s) \rightarrow c$ and $a(r_s) \rightarrow 0$.  


Now for value of $r_s$, the corresponding value of $v(r_s)$ at $\Omega = 1 $ rad/s is given by
\begin{align*}&~~~~~~~~~~~v(r_s) = c \times \tanh{\bigg(\frac{r_s}{c}\bigg)} \text{m/s}\\
&\implies v(r_s) = c \times \tanh{\bigg[\ln{\bigg(\frac{1+\sqrt{3}}{\sqrt{2}}\bigg)}\bigg]}~ \text{m/s}\\
&\implies v(r_s) = c \times \frac{1} {\coth{\bigg[\ln{\bigg(\frac{1+\sqrt{3}}{\sqrt{2}}\bigg)}\bigg]}} \text{ m/s}\\
&\implies v(r_s) = c \times \frac{1} {\coth{\bigg[\ln{\bigg(\sqrt{\frac{(1+\sqrt{3})^{2}}{2}}\bigg)}\bigg]}}~\text{m/s}\\
&\implies v(r_s) = c \times \frac{1} {\coth{\bigg[\frac{1}{2}\ln{\bigg(\frac{\sqrt{3}+1}{\sqrt{3}-1}\bigg)}\bigg]}}~\text{m/s}\\
&\implies v(r_s) = \frac{c}{\sqrt3} ~\text{m/s} = 1.732 \times 10^{8}~\text{m/s} ~~ \text{\bigg(As we know that,}~\arccoth{(x)}= \frac{1}{2} \ln{\bigg(\frac{x+1}{x-1}\bigg)}~\forall~x>1   \bigg)
\end{align*}
Hence, we can say that when the magnitude of $v(r_s)$ reaches value of $1.732 \times 10^{8}$ m/s or around $57.7\%$ of the speed of light $c$, the magnitude of tangential acceleration $a(r_s)$,  attains peak value, beyond which, it decreases to zero.
\pagebreak

\section{Lorentz Invariance of proven relativistic transformation }
The theory of Lorentz Invariance says that, the laws of physics are invariant under
a transformation between two co-ordinate frames moving at constant constant velocity w.r.t. to each other i.e. the transformation we formulated here should be invariant at both frame of references. Whenever we talk about the transformation of spherical co-ordinates from \textbf{$(ct,r,r\theta,r_s\phi)$} to \textbf{$(ct',r',r\theta',r_s\phi')$}, the value four dimensional length, let's say that the space-time interval, denoted by $S^2$ should be invariant at both frame of reference.\\
Therefore we have to prove that,
 $$ S^2 = (ct')^2-(r')^2-(r'\theta')^2-(r_s'\phi')^2 = (ct)^2-(r)^2-(r\theta)^2-(r_s\phi)^2 $$
 
 Now we take the transformation of $\phi$ Co-ordinate,   
\begin{center}
    
$\begin{bmatrix}
ct' \\
r' \\
r'\theta'\\
r_s'\phi'
\end{bmatrix}= \begin{bmatrix}
\cosh \left(\frac{\Omega r_{s}}{c}\right) &0 &0& -\sinh \left(\frac{\Omega r_{s}}{c}\right)\\
0 & 1& 0 & 0 \\
0 & 0 & 1 & 0\\
-\sinh \left(\frac{\Omega r_{s}}{c}\right) & 0 & 0 & \cosh \left(\frac{\Omega r_{s}}{c}\right)

\end{bmatrix}\begin{bmatrix}
ct \\
r \\
r\theta\\
r_s\phi
\end{bmatrix}$
\end{center} 

Now following the transformation laws we have,
\begin{align*}& S^2 = (ct')^2-(r')^2-(r'\theta')^2-(r_s'\phi')^2 \\
&~~~ = \left[\cosh \left(\frac{\Omega r_{s}}{c}\right)ct-\sinh \left(\frac{\Omega r_{s}}{c}\right)r_s\phi \right]^{2}-(r)^2-(r\theta)^2- \left[ -\sinh \left(\frac{\Omega r_{s}}{c}\right)ct + \cosh \left(\frac{\Omega r_{s}}{c}\right)r_s\phi \right]^{2}\\
\\
&~~~ = \left[\cosh^{2} \left(\frac{\Omega r_{s}}{c}\right)c^2 t^2 + \sinh^{2} \left(\frac{\Omega r_{s}}{c}\right)r_s^{2}\phi^{2}- 2\cosh \left(\frac{\Omega r_{s}}{c}\right)\sinh \left(\frac{\Omega r_{s}}{c}\right)ctr_s\phi\right] -(r)^2-(r\theta)^2\\
&~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~-\left[ \sinh^{2} \left(\frac{\Omega r_{s}}{c}\right)c^2 t^2 +\cosh^{2} \left(\frac{\Omega r_{s}}{c}\right)r_s^{2}\phi^{2}- 2\cosh \left(\frac{\Omega r_{s}}{c}\right)\sinh \left(\frac{\Omega r_{s}}{c}\right)ctr_s\phi \right]\\
\\
&~~~  = c^2 t^2\left[\cosh^{2} \left(\frac{\Omega r_{s}}{c}\right)- \sinh^{2} \left(\frac{\Omega r_{s}}{c}\right)\right] - (r)^2-(r\theta)^2 - r_s^{2}\phi^{2}\left[\cosh^{2} \left(\frac{\Omega r_{s}}{c}\right)- \sinh^{2} \left(\frac{\Omega r_{s}}{c}\right)\right]\\
&~~~ = (ct)^2-(r)^2-(r\theta)^2-(r_s\phi)^2 ~~\text{\bigg(As we know that,}~\cosh^{2}(x)- \sinh^{2}(x) = 1 \bigg)
    \end{align*}
So, we can see that the transformation of $\phi$ Co-ordinate is Lorentz Invariant. Similarly, we can show that the transformation for $\theta$ Co-ordinate will also be Lorentz Invariant.     


 
\section{Validity of Special Theory of Relativity in Curvilinear co-ordinates}

Applicability of Special Theory of Relativity (STR) is not centralized upon accelerated motion to be considered, it is based upon how you define acceleration in a coordinate system. Since Cartesian coordinate system is rectilinear, acceleration is simply defined as rate of change of velocity along any component of x-axis, y-axis and z-axis respectively. In other words, it is also defined as the double derivative of displacement along any component with respect to time. STR is easily applicable since all is there up to when there is no velocity varying with time.

Now it is true that circular motion is accelerated and involves centrifugal force, but the real cause of acceleration is due to change in