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

can interpret the metric Tensor g_(μν) is kinda distance function at curved Surface ?? ex. Euclidean space g_(μν) = ((1,0,0),(0,1,0),(0,0,1) ) Sphere g_(μν) = ((( 1),( 0),( 0)),(( 0),( r^2 ),( 0)),(( 0),( 0),(r^2 sin^2 (θ))) )

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

Answered by MrGaster last updated on 05/Apr/25

g_(μν) ≜ds^2 =g_(μν) dx^μ □dx^ν δ_(μν) dx^μ dx^ν =dx^2 +dy^2 +dz^2 ∫_C (√(δ_(μν) (dx^μ /dt) (dx^ν /dt)))dt=(√((x_2 −x_(1 ) )^2 +(y_2 −y_1 )^2 +(z_2 −z_1 )^2 )) S^2 (r):g_(μν) = ((1,0,0),(0,r^2 ,0),(0,0,(r^2 sin^2 θ)) )⇒ds^2 =dr^2 +r^2 dθ^2 +r^2 sin^2 θ dφ^2 Constraint r=const⇒ds^2 =r^2 (dθ^2 +sin^2 θ dφ^2 ) ∫_θ_1 ^θ_2 (√(g_(θθ) ((dθ/dt))^2 ))dt=r(θ_2 −θ_1 ) ∫_0 ^(2π) (√(g_(φφ) ((dφ/dt))^2 ))dt=2πr sin θ Γ_(μν) ^λ =(1/2)g^(λσ) (∂_μ g_(σν) +∂_(ν ) g_(σμ) −∂_σ g_(μν) )⇒▽_μ g_(νρ) =0 Riem(g)=∂_ρ Γ_(μνσ) −∂_σ Γ_(μνρ) +Γ_(λρσ) Γ_(μν) ^λ −Γ_(λσ) Γ_(μν) ^λ L_(geodesic) =(√(g_(μν) x^. ^μ x^. ^ν ))⇒(D/dt)x^. ^μ +Γ_(νρ) ^μ x^. ^ν x^. ^ρ =0

$$\mathrm{g}_{\mu\nu} \triangleq{ds}^{\mathrm{2}} ={g}_{\mu\nu} {dx}^{\mu} \square{dx}^{\nu} \\ $$$$\delta_{\mu\nu} {dx}^{\mu} {dx}^{\nu} ={dx}^{\mathrm{2}} +{dy}^{\mathrm{2}} +{dz}^{\mathrm{2}} \\ $$$$\int_{{C}} \sqrt{\delta_{\mu\nu} \frac{{dx}^{\mu} }{{dt}}\:\frac{{dx}^{\nu} }{{dt}}}{dt}=\sqrt{\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}\:} \right)^{\mathrm{2}} +\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)^{\mathrm{2}} } \\ $$$$\mathbb{S}^{\mathrm{2}} \left({r}\right):\mathrm{g}_{\mu\nu} =\begin{pmatrix}{\mathrm{1}}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{{r}^{\mathrm{2}} }&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta}\end{pmatrix}\Rightarrow{ds}^{\mathrm{2}} ={dr}^{\mathrm{2}} +{r}^{\mathrm{2}} {d}\theta^{\mathrm{2}} +{r}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta\:{d}\phi^{\mathrm{2}} \\ $$$$\mathrm{Constraint}\:{r}=\mathrm{const}\Rightarrow{ds}^{\mathrm{2}} ={r}^{\mathrm{2}} \left({d}\theta^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \theta\:{d}\phi^{\mathrm{2}} \right) \\ $$$$\int_{\theta_{\mathrm{1}} } ^{\theta_{\mathrm{2}} } \sqrt{{g}_{\theta\theta} \left(\frac{{d}\theta}{{dt}}\right)^{\mathrm{2}} }{dt}={r}\left(\theta_{\mathrm{2}} −\theta_{\mathrm{1}} \right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \sqrt{{g}_{\phi\phi} \left(\frac{{d}\phi}{{dt}}\right)^{\mathrm{2}} }{dt}=\mathrm{2}\pi{r}\:\mathrm{sin}\:\theta \\ $$$$\Gamma_{\mu\nu} ^{\lambda} =\frac{\mathrm{1}}{\mathrm{2}}{g}^{\lambda\sigma} \left(\partial_{\mu} {g}_{\sigma\nu} +\partial_{\nu\:} {g}_{\sigma\mu} −\partial_{\sigma} {g}_{\mu\nu} \right)\Rightarrow\bigtriangledown_{\mu} {g}_{\nu\rho} =\mathrm{0} \\ $$$$\mathrm{Riem}\left({g}\right)=\partial_{\rho} \Gamma_{\mu\nu\sigma} −\partial_{\sigma} \Gamma_{\mu\nu\rho} +\Gamma_{\lambda\rho\sigma} \Gamma_{\mu\nu} ^{\lambda} −\Gamma_{\lambda\sigma} \Gamma_{\mu\nu} ^{\lambda} \\ $$$$\mathcal{L}_{\mathrm{geodesic}} =\sqrt{{g}_{\mu\nu} \overset{.} {{x}}\:^{\mu} \overset{.} {{x}}\:^{\nu} }\Rightarrow\frac{{D}}{{dt}}\overset{.} {{x}}\:^{\mu} +\Gamma_{\nu\rho} ^{\mu} \overset{.} {{x}}\:^{\nu} \overset{.} {{x}}\:^{\rho} =\mathrm{0} \\ $$

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