# Tag Info

## Hot answers tagged differential-geometry

6

I'll show you how to do this for the 2-plane in polar coordinates. Once you work this out, it should be doable to work it out in your case. You start with the metric $$ds^{2} = dr^{2} + r^{2}d\theta^{2}$$ Since the geodesics of this metric (i.e., straight lines) minimizes distance, we know that the geodesics are an extremum of: $$I = \frac{1}{2}\int ds ... 6 The strategy is to recall the geodesic equation,$$ \frac{d^2x^\lambda}{dt^2}+\Gamma^\lambda_{\,\mu\nu}\frac{dx^\mu}{dt}\frac{dx^\nu}{dt}=0\tag{1} From your Lagrangian, you'll end up with equations of the form \begin{align} \ddot{\psi}&=f(\psi,\,\theta,\,\phi,\,\dot{\psi},\,\dot{\theta}\,\dot{\phi})\\ ... 4 Rising bubbles of air in a liquid oftentimes are anything but spherical. These bubbles have haphazard shapes because they are rising and because they are interacting with other nearby bubbles. The combination of drag, turbulence, and mutual interactions prevents those bubbles from taking on a nice, simple spherical shape. Here's a rather non-spherical ... 4 It's Stokes's theorem. Consider a field F = dA + A \wedge A such that A is pure gauge at infinity, that is, \lim_{x\to\infty} A(x) = \omega\, d \omega^{-1} for some \omega : S^3 \to SU(2) \sim S^3 where \omega is a function on the 3-sphere because the limit can depend on the direction out to infinity. In differential forms the first expression is ... 4 They're not zero in general. For example, take flat Euclidean space in two dimensions, in polar coordinates: ds^2=dr^2+r^2d\phi^2 $$for which the nonzero Christoffel symbols are$$ \Gamma^r_{\phantom\phi\phi\phi}=-r,\quad \text{and} \quad \Gamma^\phi_{\phantom\phi \phi r}=\Gamma^\phi_{\phantom\phi r \phi}=\frac1r $$Then \Gamma^\lambda_{\phantom\phi ... 4 Please let me first refer you to the following review by I. L. Shapiro, which contains a lot of theoretical and phenomenological information on spacetime torsion. The answer will be mainly based on this review. In the basic Einstein-Cartan theory, in which the antisymmetric part of the connection is taken as independent additional degrees of freedom, the ... 3 The first reason is that your "distance" between geodesics is measured by a parallely propagated direction \partial/\partial \phi. If you take a look at the sphere, the difference \Delta \phi does not correspond to the distance between the points on the geodesics. The distance between them would be measured by arc-lengths of great circles. But you are ... 3 In general, the statement that \nabla_\mu V^\mu transforms as a scalar does not quite fix the transformation properties of V^\mu. Rather, the most general such transformation would be$$V^\mu \mapsto V'^\mu + C^\mu,$$where V'^\mu = \frac{\partial x'^\mu}{\partial x^\nu} V^\nu is the ordinary vector transformation law, and C^\mu is any quantity ... 3 I think I good book for that may be C. J. Isham's Modern Differential Geometry for Physicists. I haven't gotten to the chapter of fiber bundles, but what I've read seems to be quite rigorous. And as it is written for physicists, I think it could please your needs. 3 The metric tensor is the only additional thing you need to define a Riemannian manifold, beyond what you'd need in order to define the manifold as a differentiable manifold anyway even if it wasn't Riemannian. In particular, topology is important. For example, just specifying that a 2-D manifold has a metric that's Euclidean everywhere is insufficient to ... 3 Think of an infinitesimal Diff as of a translation where the the shift is space dependent, x^\mu\rightarrow x^\mu+\epsilon^\mu(x). Now, you get that the generators are L_\epsilon=\epsilon^\nu(x)\partial_\nu since L_\epsilon x^\mu=\epsilon^\mu(x). They form an infinite space since \epsilon^\mu(x) is a function that can be expanded in infinitely many ... 2 Here's a way to find the Riemann tensor of the 3-sphere with a lot of intelligence but no calculations. At any point p on a sphere, all directions look the same. Therefore there can be no privileged vector at a point p. Now consider the eigenvalue problem for the Ricci tensor,$$R^\alpha{}_\beta x^\beta = \lambda x^\alpha.$$Since no vector is better ... 2 Surely there must be a better approach than this? Certainly; it's the Cartan formalism which employs differential forms. Consider your case of a sphere, with a metric tensor,$$ds^2 = r^2dr^2+r^2\sin^2 \theta \, d\theta^2$$We can choose an orthonormal basis e^a such that ds^2 = \eta_{ab}e^{a}e^{b}, and e^a = e^a_\mu dx^\mu. For our case: ... 2 Yes, you're exactly right, \omega_b {C^b}_{ac} = {C^b}_{ac} \omega_b. In general, the order of factors doesn't matter in a tensor expression like this. This is Wald, so technically you're supposed to think of those expressions as using abstract index notation instead of involving components in any particular basis, but it's also valid to think of them as ... 2 For the sake of the explanation I will assume you mean a gas bubble in a liquid*. David Hammen names a few conditions for a bubble to be spherical, in fact you could summarize these all as: for a bubble to be spherical the surface tension has to dominate over other forces (per unit length). If surface tension is indeed dominant than the pressure in the ... 2 As Wikipedia says, a great circle is a circle formed by the intersection of a sphere and a plane that passes through the center of the sphere. The great circles parametrized by \theta =\tau and \phi =\text{const} are not all the great circles. They are only the `vertical' great circles, that is, the great circles formed by intersection with a plane ... 2 Incompleteness of a coordinate system is not a canonical definition as that, for instance, of geodesical (in)completeness. It simply means that the domain of the coordinate system does not cover the whole manifold (and perhaps there are several inequivalent extensions of the initial manifold represented by the given domain of the coordinate system). If a ... 2 In the context of general relativity it is often stated that one of the main purposes of tensors is that of making equations frame-independent. Question: why is this true? Actually this isn't quite true. General relativity doesn't have frames of reference (except locally, which is trivially true because GR is the same as SR locally). A better way of ... 2 I) In Palatini f(R) gravity, the Lagrangian density is$$ {\cal L}~=~ \sqrt{-g} f(R), $$with$$R~:=~ g^{\mu\nu} R_{\mu\nu}(\Gamma),$$and where \Gamma^{\lambda}_{\mu\nu}=\Gamma^{\lambda}_{\nu\mu} is an arbitrary torsionfree^1 connection. II) As OP mentions, the word Palatini refers to that the metric g_{\mu\nu} and the connection ... 1 Start with the lower expression:$$ (c/2)\eta^{bc}\eta^{ae}\partial_{a}\left(g_{be,c} + g_{ce,b} - g_{bc,e}\right) - (c/2)\eta^{ae}\eta^{bc}\partial_{c}\left(g_{be,a} + g_{ae,b} - g_{ba,e}\right).\\ = \frac{c}{2}\eta^{bc}\eta^{ae}\left(g_{be,ca} + g_{ce,ba} - g_{bc,ea}-g_{be,ac} - g_{ae,bc} + g_{ba,ec}\right)\\ =\frac{c}{2}\eta^{bc}\eta^{ae}\left(g_{ce,ba} - ...

1

Why can't we simply assign to every point of the Bloch sphere a phase $e^{i\phi}$? This is the idea of a section of a fibre bundle. You are considering in this case a base space $S^{2}$ with fibre $S^{1}$. Locally the fibre bundle looks like $S^{2}×S^{1}$. However you want to consider a fibration such that the global space is not the trivial product ...

1

The thing you'll notice about a sphere is that it's symmetrical. very symmetrical. No matter how you rotate it, it looks the same. the surface tension pulls the surface of the bubble into a shape that has even surface tension over the entire bubble. The shape with even surface tension is a sphere. a sphere has the smallest possible surface area for an ...

1

Formally speaking, given a (differentiable, finite dimensional) manifold $M$, then the (infinite dimensional) Lie group of (globally defined) diffeomorphisms (with composition $\circ$ as group structure) has the set $\Gamma(TM)$ of (globally defined, differentiable) vector fields as corresponding Lie algebra. This (infinite dimensional) Lie algebra ...

1

To address the first question you can consider $$\nabla_av^b := v^b ,_a$$ since $$\nabla : \Gamma(E) \rightarrow \Gamma(E\otimes T^*M)$$ where E is any section(e.g. the vector field in question) and contracting it with the tangent vector field of the curve you get $$t^av^b,_a = 0$$ this is similar to contracting a vector field with a dual vector  ...

1

When you write $V^\mu$ you mean that $V$ is a vector. Next $\nabla_\mu V^\mu$ is called divergence of a vector. Finally answering your question, a vector field with constant zero divergence is called incompressible or solenoidal – in this case, no net flow can occur across any closed surface (according to the Gauss law). If it is a constant, but not zero, ...

1

To add to Qmechanic's Answer and TwoBs's Answer and answer "....what the heck is h then? Any arbitrary function?": $h$ is pretty much arbitrary. It is wontedly taken to be at least differentiability class $C^1$ (all first derivatives continuous) so that the Lie bracket of vector fields is defined as in Qmechanic's answer. You need to assume it is of class ...

Only top voted, non community-wiki answers of a minimum length are eligible