d s 2 = d t 2 − e g ( x 0 ) + f ( r ) [ d r 2 + r 2 ( d θ 2 + sin 2 θ d ϕ 2 ) ] {\displaystyle ds^{2}=dt^{2}-e^{g(x^{0})+f(r)}[dr^{2}+r^{2}(d\theta ^{2}+\sin ^{2}\theta d\phi ^{2})]}
L = d s 2 d s 2 = d t 2 d s 2 − e g ( x 0 ) + f ( r ) [ d r 2 d s 2 + r 2 ( d θ 2 d s 2 + sin 2 θ d ϕ 2 d s 2 ) ] {\displaystyle L={\frac {ds^{2}}{ds^{2}}}={\frac {dt^{2}}{ds^{2}}}-e^{g(x^{0})+f(r)}\left[{\frac {dr^{2}}{ds^{2}}}+r^{2}({\frac {d\theta ^{2}}{ds^{2}}}+\sin ^{2}\theta {\frac {d\phi ^{2}}{ds^{2}}})\right]}
⇒ L = t ˙ 2 − e g ( x 0 ) + f ( r ) [ r ˙ 2 + r 2 ( θ ˙ 2 + sin 2 θ ϕ ˙ 2 ) ] {\displaystyle \Rightarrow L={\dot {t}}^{2}-e^{g(x^{0})+f(r)}[{\dot {r}}^{2}+r^{2}({\dot {\theta }}^{2}+\sin ^{2}\theta {\dot {\phi }}^{2})]}
অয়লার-লাগ্রাঞ্জ সমীকরণ, d d s ∂ L ∂ x ˙ α = ∂ L ∂ x α {\displaystyle {\frac {d}{ds}}{\frac {\partial L}{\partial {\dot {x}}^{\alpha }}}={\frac {\partial L}{\partial x^{\alpha }}}}
মুক্তভাবে পতনশীল বস্তুর গতির সমীকরণ, x ¨ τ + Γ μ ν τ x ˙ μ x ˙ ν {\displaystyle {\ddot {x}}^{\tau }+\Gamma _{\mu \nu }^{\tau }{\dot {x}}^{\mu }{\dot {x}}^{\nu }}
এখানে, α = τ = 0 , 1 , 2 , 3 {\displaystyle \alpha \ =\ \tau =0,1,2,3} এবং x 0 = t , x 1 = r , x 2 = θ , x 3 = ϕ {\displaystyle x^{0}=t,\ x^{1}=r,\ x^{2}=\theta ,\ x^{3}=\phi }