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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

That's right, there is the following equation: $$X^{\mu}=Z^{\mu}(\sigma)) (dX^{\mu}=\partial_{a}Z^{\mu}(\sigma)d\sigma^{a})$$ for 2d surface embedded into, let say, a flat finite-dimensional target space inside the Polyakov and Nambu-Goto action. The main/only "reason" why people (mostly QFTs theorists who would like to call themselves and be called by ...


0

Nakahara's Geometry, Topology and Physics has two chapters covering Fiber Bundles up to Connections on Fiber Bundles with a few applications in Gauge theories. If it is your first time learning Fiber Bundles i would recommend this books, it's rigorous and has a lot of physics-motivated examples.


0

The Lagrangian must be a gauge invariant and Lorentz invariant object that can be integrated over the entire spacetime $\Sigma$. So, we must first obtain an $n$-form (for $n$ the dimension of spacetime), and all that we have for that is the gauge field $A$, which itself transforms in an ugly way under gauge transformation. The only object we can build out ...


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} - ...


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 ...


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 ...


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

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 ...


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 ...


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, ...


0

Here is one way to derive the geodesic equations from the Euler-Lagrange equations. First consider a natural Lagrangian system $(M,L)$, where $L\in C^\infty(TM)$. Let $g$ be a Riemannian metric. Suppose in our mechanical system the net force is zero. That is, the Lagrangian is just equal to the kinetic energy, $$L(p,V_p)=\frac{1}{2}mg_p(V_p,V_p)$$In ...


0

I think the answer to this is basically yes. You do have to be careful, because in general, you can't interpret inner products in relativity as measures of whether something is "orthogonal" to something else in the Euclidean sense. E.g., a lightlike vector has a zero inner product with itself. This is because the metric isn't the Euclidean metric. However, ...


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 $$ ...


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 ...


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 ...


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

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 ...


0

In 3 dimensions, all the data in the Riemman tensor is contained in the Ricci tensor as the Weyl tensor vanishes. So it suffices to compute the Ricci tensor and then using the decomposition given here.


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 ...


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 ...


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 ...


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 ...


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})\\ ...


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 ...


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 ...


0

I think you're misinterpreting what the no-hair theorems say. There's a no-hair theorem for stationary electrovac solutions. It applies to electrovac solutions, not to solutions containing any matter field you like; in fact, there are known counterexamples if you include certain types of matter fields. Also, it's a theorem about stationary solutions. A ...


0

I've since found out that where I'm going wrong is that I don't need to find the absolute derivative. I've been told on another physics forum that this problem is framed in terms of Riemann normal coordinates that makes it OK to assume $\frac{D^{2}\xi^{\mu}}{dt{}^{2}}=\frac{d^{2}\xi^{\mu}}{dt{}^{2}}$. Apparently, this is because the distance the cars travel ...


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 ...


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 ...


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 ...


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

Nature exists withouth mathematics. Mathematics is just a language tool (a way of visualizing things about how nature behaves). Observations of nature show us that things try to flow to especific regions of space. By definitions, this regions are described as lower potentials (the more potential, the more capacity of doing things). Gradient of the potential ...


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

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 ...


1

A full 3D model might be too complex for this case, because then you need to know everything about the wall, and the tree, and the interaction between the two in great detail. I think a simplified approach might be more suitable. I'm not sure if this is oversimplified, but let me take a swing at it: If we assume the tree is supported by both the ground and ...



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