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Question Number 218265 by SdC355 last updated on 04/Apr/25

$$\mathrm{can}\mathrm{interpret}\mathrm{the}\mathrm{metric}\mathrm{Tensor}{\mathbf{g}}_{\mu \nu }\mathrm{is}$$$$\mathrm{kinda}\mathrm{distance}\mathrm{function}\mathrm{at}\mathrm{curved}\mathrm{Surface}??$$$$\mathrm{ex}.\mathrm{Euclidean}\mathrm{space}{\mathbf{g}}_{\mu \nu }=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$$$$\mathrm{Sphere}{\mathbf{g}}_{\mu \nu }=\left(\begin{array}{ccc}1& 0& 0\\ 0& r^2& 0\\ 0& 0& r^2{\sin }^2\left(\theta \right)\end{array}\right)$$

Answered by MrGaster last updated on 05/Apr/25

$${\mathrm{g}}_{\mu \nu }\triangleq {ds}^2=g_{\mu \nu }{dx}^{\mu }◻{dx}^{\nu }$$$${\delta }_{\mu \nu }{dx}^{\mu }{dx}^{\nu }={dx}^2+{dy}^2+{dz}^2$$$${\int }_C\sqrt{{\delta }_{\mu \nu }\frac{{dx}^{\mu }}{dt}\frac{{dx}^{\nu }}{dt}}dt=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2}$$$${\mathbb{S}}^2\left(r\right):{\mathrm{g}}_{\mu \nu }=\left(\begin{array}{ccc}1& 0& 0\\ 0& r^2& 0\\ 0& 0& r^2{\sin }^2\theta \end{array}\right)\Rightarrow {ds}^2={dr}^2+r^{2}d{\theta }^2+r^2{\sin }^2\theta d{\varphi }^2$$$$\mathrm{Constraint}r=\mathrm{const}\Rightarrow {ds}^2=r^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$$$${\int }_{{\theta }_1}^{{\theta }_2}\sqrt{g_{\theta \theta }{\left(\frac{d\theta }{dt}\right)}^2}dt=r({\theta }_2-{\theta }_1)$$$${\int }_0^{2\pi }\sqrt{g_{\varphi \varphi }{\left(\frac{d\varphi }{dt}\right)}^2}dt=2\pi r\sin \theta$$$${\mathrm{\Gamma }}_{\mu \nu }^{\lambda }=\frac{1}{2}{g}^{\lambda \sigma }({\partial }_{\mu }{g}_{\sigma \nu }+{\partial }_{\nu }{g}_{\sigma \mu }-{\partial }_{\sigma }{g}_{\mu \nu })\Rightarrow {▽}_{\mu }{g}_{\nu \rho }=0$$$$\mathrm{Riem}\left(g\right)={\partial }_{\rho }{\mathrm{\Gamma }}_{\mu \nu \sigma }-{\partial }_{\sigma }{\mathrm{\Gamma }}_{\mu \nu \rho }+{\mathrm{\Gamma }}_{\lambda \rho \sigma }{\mathrm{\Gamma }}_{\mu \nu }^{\lambda }-{\mathrm{\Gamma }}_{\lambda \sigma }{\mathrm{\Gamma }}_{\mu \nu }^{\lambda }$$$${\mathcal{L}}_{\mathrm{geodesic}}=\sqrt{g_{\mu \nu }\stackrel{.}{x}{}^{\mu }\stackrel{.}{x}{}^{\nu }}\Rightarrow \frac{D}{dt}\stackrel{.}{x}{}^{\mu }+{\mathrm{\Gamma }}_{\nu \rho }^{\mu }\stackrel{.}{x}{}^{\nu }\stackrel{.}{x}{}^{\rho }=0$$

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