# Tag Info

## New answers tagged differential-geometry

1

The solution that you wrote in your last (not numbered) equation is not a basis of a Hilbert space of sections because the phase factor: $(-1)^n e^{2i\pi A}$ depends on $n$. The phase factor should not depend on $n$. Please see your (correct) equation (2) defining the boundary conditions, in which the phase factor does not depend on $n$. Thus there is no ...

0

Curvature is just a form of strain, and strain produces stress. Knowing the curvature, you can compute the strain field from which you can obtain the stress field using linear elasticity theory. In the case of fracturing, the stress causes cracks to propagate which results in fracturing. The wiki article on fracture mechanics is reasonably detailed.

0

In the geometrical optics approximation light ray is represented by a null geodesic. Therefore you only need to find a null geodesic connecting points $(t_0,0,0,0)$ and $(t_1,x,0,0)$ for some $t_1$ (and this condition will determine $t_1$ uniquely). This is probably quite easy to do directly in this case, but in general for investigation of null curves in ...

2

Diffeomorphism Invariance Let $M$ be a smooth manifold. Let $\phi: M \to M$ be a diffeomorphism. A simple property of the Einstein equations is $$g \in \otimes^2 TM \text{ is solution to vacuum Einstein equation} \implies \text{ so is } \phi^*g$$ To see that this is true, simply pull back both sides of the Einstein equation by $\phi$, and use the ...

0

A general diffeomorphism does not map geodesics to geodesics. Some simple counter examples You can a build diffeomorphism on the Euclidean plane by imagining putting one finger on a tablecloth at point $x$ and dragging it. This map is clearly smooth, a smooth inverse is constructed by dragging your finger back. Any geodesic on the plane (a line) passing ...

0

Existence of smooth structures You're asking when does a (topological) manifold $M$ fail to be covered by a smooth atlas. Another way to phrase this is "when does a manifold admit a smooth structure". This is a well-studied problem. It turns out examples of manifolds which do not admit smooth structures only occur in dimensions $\ge 4$ (see e.g. ...

0

The general story Here's an attempt to formalize how physicists build spacetime manifolds. Let $N$ be one of $\mathbb R^n$ Some dimension $n$ product manifold of $S^1$ and $\mathbb R$ (corresponding to periodic solutions). Pick one such $N$. Now Take stress tensor $T$ defined on some open subset $U \subset N$ and some boundary conditions (e.g. ...

0

Non-coordinate bases are always used when coupling to fermions. You could simply look at the maths literature, but as a physicist you'll want to look at textbooks on supergravity. Take a look at words such as "vielbein", "frame" and "spin connection". As to literature, there's Freedman/van Proeyen "Supergravity", chapter 7. Alternatively Ortin's "Gravity and ...

2

You don't. Two given spacetimes can have their metrics written in the same way but may have different coordinate ranges. A simple example is just a spacetime with spatial coordinate identified , such as the cylinder spacetime : $$ds^2 = -dt^2 + d\theta$$ Identical to Minkowski space, which is its universal cover. Of course, two things to watch out for : ...

1

Prahar is correct, but here are two more things to note. If spacetime is an $n$-dimensional Lorentzian manifold $(M,g)$, let $\{E_1,\dotsc, E_n\}$ be an orthonormal frame, i.e. $E_i\in\Gamma(TM), T_pM=\mathrm{span}\,\{E_i\lvert_p\}$, and $g(E_i,E_j)=\eta_{ij}$, where $\eta=\mathrm{diag}\,(-1,1,\dotsc, 1)$ is the Minkowski matrix. Then, if $M$ is orientable ...

0

To my knowledge, there is no formal map from string states to spacetime fields. It is merely a (well-motivated) heuristic. The first "hint" of a spacetime formulation of string theory appears when we observe that the massless states of string theory neatly fall into representations of the massless little group of the (26/10) dimensional spacetime, hence are ...

3

Take a future-directed timelike curve $\gamma= \gamma(\tau)$, $\tau$ being the proper time along $\gamma$ in the spacetime $M$. Assume that $p = \gamma(0)$ is the initial point of $\gamma$. Fermi coordinates adapted to $\gamma$ are constructed this way. Consider an orthonormal basis of $T_pM$ with $e_0$ parallel to $\dot{\gamma}$. Transport the basis ...

0

The question is given as a practice problem in Poisson's book "Relativists Toolkit" (p. 159, problem #7) along with some helpful hints. The tensor $\gamma_{ab}$ (which is not a tensor at all) must be defined in the following way: Let $h_{ab}$ be the induced 3-metric on the spacelike hypersurface $\Sigma$, and let on its boundary, $\partial \Sigma$, be ...

-1

The coordinate invariant volume element/measure on a manifold with metric $g$ is $$d^d x \sqrt{|g|}$$ By coordinate invariance, I mean that if I choose to work in a different coordinate system $x'$, then both the metric determinant changes as does the measure $d^dx$. But they change in a way so as to cancel each other out. In other words $$d^d x' ... 6 Your intuition that the Einstein equations are equations for the metric tensor, not for the manifold is mostly on the right track, but the details are wrong. That core bit of intuition is best phrased, I think, as saying that the Einstein equations are local equations for the geometry of the manifold. That is, they tell you that, whatever manifold your ... 1 The square root of the determinant of the metric can be understood as a particular function of the components of the metric g_{ab}$$\sqrt{-g} =f(g_{ab})$$By the chain rule we of course have$$\nabla_a \sqrt{-g} = \nabla_a f(g_{bc}) = \frac{df}{d g_{bc}} \nabla_a g_{bc}$$But we know that \nabla_a g_{bc}=0 so that of course \nabla_a \sqrt{-g} =0. This ... 0 I'd say none. If you are talking about a Riemannian cylinder, its metric (say, induced from its embedding in \Bbb{R}^{3}) is$$ g=dz^{2}+d\phi^{2} $$If the two were conformally equivalent, a conformal transformation (z,\phi)=(z(t,x),\phi(t,x)) would pull back the metric as$$ g'=\bigg(\frac{\partial z}{\partial t}dt+\frac{\partial z}{\partial ...

2

Here's a heuristic calculation: Let $\{E_i\}$ be an orthonormal frame ($g(E_i,E_j)=\epsilon_i\delta_{ij}, \epsilon_i=\pm 1$). Then $\mu$ is the canonical volume form $\sqrt{g}\,\mathrm{d}x^1\wedge\cdots\wedge\mathrm{d}x^n$ iff $\mu(E_1,\dotsc, E_n)=1$. Then $$(\nabla_X\mu)(E_1,\dotsc,E_n)=\nabla_X(\mu(E_1,\dotsc,E_n))-\sum \mu(E_1,\dotsc,\nabla_X ... 4$$ \partial_t\left(\phi_{t}^{*}\rho_{t}\mathrm{d}\omega\right)= \phi_{t}^{*}\left(\it\unicode{xA3}_X\left(\rho_{t}\mathrm{d}\omega\right)\right)+\phi_{t}^{*}\left(\left(\partial_{t}\rho_{t}\right)\mathrm{d}\omega\right)\tag 1$$This is quite easy, the fundamental reason for the result above is that \quad\quad the function ... 6 Comments to the post (v2): Note that \sqrt{|g|} transforms as a density rather than a scalar under general coordinate transformations. In particular, the covariant derivative of \sqrt{|g|} does not necessarily coincide with the partial derivative of \sqrt{|g|}. Here is a heuristic explanation using local coordinates. The Levi-Civita connection is ... 0 First of all, don't think about embeddings: there is no notion (in GR) that spacetime is embedded in some higher-dimensional flat space (let alone a five-dimensional space): it's just what it is. This is not to say embeddings are not interesting, but it is a red herring to worry about them too much. So, straight lines. As you say, in general there is no ... 2 I) Assuming that the variational problem for the action S=\int \! d^nx~{\cal L} is well-posed (with appropriate boundary conditions), the field-theoretic Euler-Lagrange (EL) equations read in general$$\tag{1} 0~\approx~\frac{\delta S}{\delta \phi^{\alpha}} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} -\sum_{\mu} \frac{d}{dx^{\mu}} \frac{\partial ...

0

Well, by a process of elimination, if you have your time orientation (a vector field $\tau^\mu$), and you have two equivalence classes, for $g(X,\tau) > 0$ and $g(Y, \tau) <0$, the only remaining possibility for a third class is that $g(Z,\tau) = 0$. But any vector tangent to a timelike vector will be spacelike. It can be proven thusly : if you have ...

0

Physically, a tensor is just a collection of vectors that physicists study as a single object because the vectors are related to one another in a physically "interesting" way. Tensors don't possess additional properties to the ideas of magnitude and orientation of a vector; all that can be done is to assign a vector to certain arrangements of vectors of ...

6

Parametrize the particle's worldline w.r.t. $t$: $$~x^\mu (t) = (t,1)$$ Its four-velocity is: $$u^\mu =\frac{d x^\mu}{d \tau}$$ To evaluate this we use the fact that: $$d \tau^2 = coshx~dt^2-dx^2$$ Also use $x=1$ and $dx=0$: $$d \tau^2 = cosh(1) dt^2$$ Therefore: u^\mu = \frac{d x^\mu}{d \tau} = \frac{d t}{d \tau} \frac{d x^\mu}{d t} = ...

2

You misunderstood the meaning of $H^{\bullet}_{S^1}(U_1) = \mathbb{C}[\Omega]$. This does not mean $H^i_{S^1}(U_1) = \mathbb{C}[\Omega]\forall i$, it means that the cohomology ring (with multiplication given by the cup product) is given by the polynomial algebra in one element that has degree 2. Translated back into the individual degrees $H^i$ this ...

5

Throughout the question I will use $p(T_1)$ and $p(T_2)$ to denote the 4-momentum of the baseball at times $T_1$ and $T_2$, $\mathbf{v}_1$ and $\mathbf{v}_2$ to represent the spatial component of its physical velocity, and $a(T_1)$ and $a(T_2)$ to represent the scale factor of the Universe at these times. The homogeneity and isotropy of the Universe mean ...

5

The first thing that must be said is that the question is not really specific enough: Applications to what exactly are you looking for? To me, a book on algebraic geometry and mirror symmetry, and how it relates to mirror symmetry as physicists know it, is very relevant and interesting. However, I have the feeling that this is not exactly what you're looking ...

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