ds2=dt2−eg(x0)+f(r)[dr2+r2(dθ2+sin2θdϕ2)]{\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=ds2ds2=dt2ds2−eg(x0)+f(r)[dr2ds2+r2(dθ2ds2+sin2θdϕ2ds2)]{\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−eg(x0)+f(r)[r˙2+r2(θ˙2+sin2θϕ˙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, dds∂L∂x˙α=∂L∂xα{\displaystyle {\frac {d}{ds}}{\frac {\partial L}{\partial {\dot {x}}^{\alpha }}}={\frac {\partial L}{\partial x^{\alpha }}}}
মুক্তভাবে পতনশীল বস্তুর গতির সমীকরণ, x¨τ+Γμντx˙μx˙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 x0=t, x1=r, x2=θ, x3=ϕ{\displaystyle x^{0}=t,\ x^{1}=r,\ x^{2}=\theta ,\ x^{3}=\phi }