Lecture 10: Metric Manifolds (International Winter School on Gravity and Light 2015)

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Recommends ctsmd · 2019-06 · Physics Stack Exchange · on “GR visualization”

I'm watching some GR lectures by Schuller (more or less rushing through them so bear with my ignorance here please) in Lecture 10: Metric Manifolds [this video] . He's talking about geodesics in a manifold with a metric and that you can basically require the geodesics to be straight lines for a person living in this manifold. So from a metric you derive a covariant derivative. Let's talk about some $2$ -dimensional person living in a $2$ -Sphere (FIG. $1$ ) [LINK] I suppose this person couldn't see the mountain, so he would just see something like a reparameterization. I've indicated two possible geodesics as arrows. (FIG. $2$ ) Now when we require the geodesics to be the straight lines does the

Cites WillG · 2018-07 · Physics Stack Exchange · on “Clean proof that the inverse metric tensor really is the inverse”

In Frederic Schuller's lectures on GR, he defines [this video] the metric tensor as a $(0,2)$-tensor field satisfying: $g(X,Y)=g(Y,X) \,\,\,\forall X,Y$ $\flat (X)$ is a $C^\infty$-isomorphism, where $\flat(X):=g(X,\cdot)$ He then defines [this video] the inverse metric tensor as the $(2,0)$-tensor field satisfying: $g^{-1}(\omega,\sigma)=\omega(\flat^{-1}(\sigma))$ He then claims without proof that $(g^{-1})^{am}g_{mb}=\delta^a_b$. Can someone offer a clean proof of this statement, starting from Schuller's definitions? I began with this: $$\begin{align} (g^{-1})^{am}g_{mb} &= g^{-1}(e^a,e^m)\cdot g(e_m,e_b) \\ &= e^a(\flat^{-1}(e^m))\cdot \flat(e_m)(e_b) \end{align}$$ If I could just massage this to

Cites Antoni Parellada · 2019-09 · Physics Stack Exchange · on “Basic question about units of velocity and speed of a curve on a smooth manifold”

Frederic Schuller says that velocity has units in Hertz in The WE-Heraeus International Winter School on Gravity and Light [this video] . He says: \begin{align} [v^a]&=\frac 1 T \\ [g_{ab}]&=L^2 \\ \Big[\sqrt{g_{ab \left(v^av^b\right)}}\Big]&=\frac L T \end{align} with $g_{ab}$ corresponding to the metric tensor on a Riemannian manifold. Velocity of a parametrized curve on a point of a manifold is just the differential operators at that point forming the tangent space, and evaluated by a test function on a chart to Euclidean space. The test function doesn't need to have any connections to time; hence, can we conclude that the only way the velocity components will have units of 1/time is if the cu

Cites amilton moreira · 2021-04 · Physics Stack Exchange · on “Second derivative of a function in a manifold”

Suppose we have a function curve $\gamma(t)$ on a manifold $M$ . Define the function $$f(t)=\frac{1}{2}g_{ab}\dot\gamma(t)^a\dot\gamma(t)^b$$ Introducing coordinates $x^i$ the first derivative of the function along the curve is $$\frac{df}{dt}=\frac{\partial f}{\partial x^i}\dot\gamma(t)^i$$ I am confused in how we calculate the second derivative . In this lecture Lecture 10: Metric Manifolds [this video] to express the Euler-Lagrange equation it is calculated like this $$\frac{d^2f}{dt^2}=\frac{\partial^2 f}{\partial x^i\partial x^j}\dot\gamma(t)^i\dot\gamma(t)^j +\frac{\partial f}{\partial x^i}\ddot\gamma(t)^i$$ but in this book Relativity on Curved Manifolds [LINK] page 278, in the taylor expa