# Tag Info

0

Your first equation looks a bit like GR with a dilaton. IIRC, the analogous equations in supergravity will naturally have spinors coupling to the dilaton.

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First of all we need the equation: $$\require{Amsmath} 2\nabla_{[a} \nabla_{b]} K_{cd} = R_{abc}{}^e K_{ed} + R_{abd}{}^e K_{ce} = 2 R_{ab(c} K_{d)e}\tag{1}.\label{eq:KT}$$ We have: $$R_{d(ba}{}^e K_{c)e} = \frac{1}{3}(R_{db(a}{}^e K_{c)e} + R_{da(b}{}^e K_{c)e} + R_{dc(b}{}^e K_{a)e}) = \frac{1}{3} (\nabla_{[d} \nabla_{b]} K_{ac} ... 3 Indeed, f is a symmetric form, since \omega and \omega ' are Grassmann-even:$$(\text dx \wedge \text d y)\wedge (\text d z \wedge \text d t)=(\text d z \wedge \text d t)\wedge(\text dx \wedge \text d y)$$etc.. Now, to calculate the signature, you should find a basis which diagonalizes \omega, the dimension of the space is 6. A basis is given ... 0 Easy way Let me first state the straight-forward way to do this computation.$$ \langle \nabla_a \nabla_b V, \partial_c\rangle = \partial_a \langle \nabla_b V, \partial_c \rangle - \langle \nabla_aV, \nabla_a \partial_c\rangle = \partial_a (\nabla_bV)_c - (\nabla_bV)_d \Gamma_{ac}^d $$First equality follows from compatibility, second equality uses ... 2 I will answer this with a simple example. Let us consider the metric for weak gravity,$$ ds^2 = \left(1 - \frac{2GM}{rc^2}\right)c^2dt^2 - dr^2 -r^2d\Omega^2. $$The g_{tt} metric element is largest by a factor of c^2 and we have$$ \Gamma^r_{tt} = \frac{1}{2}g^{rr}\partial_r g_{tt} = \frac{GM}{r^2}. $$Now let us work with the geodesic equation that is ... 0 The claims aren't true in complete generality. For example, R^{2n} is a Kähler manifold and any vector k is a Killing vector. But the Lie derivatives of J and \omega don't vanish for a general k. However, this example was special because it was insufficiently curved. For generic and curved enough Kähler manifold, the objects J,\omega may be ... 0 If the connection you are using is torsion-free (as the Levi-Civita one), you can systematically replace the coordinate derivative for the covariant one. So (please, pay attention to signs and positions of indexes, since I could use a convention different from yours)$${\cal L}_\xi \nabla_a K^b = \xi^c\nabla_c \nabla_a K^b - (\nabla_c\xi^b) \nabla_a K^c + ...

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It might be useful to know the formula for Lie derivative of metric itself. It is $\mathcal{L}_Xg_{ab}=\nabla_a X_b+\nabla_b X_a$. Then you might notice that covariant derivative can be split into symmetric and atisymmetric part: $$\nabla _a K_b = \nabla _{(a} K_{b)}+\nabla _{[a} K_{b]}$$ The first part is half of lie derivative of metric with respect to ...

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A slightly different and, perhaps, more simple proof follows. $\def\lie{\mathit{£}}$ For $K^a$ a Killing vector, we have (Kostant formula), $$\nabla_a \nabla_b K^c = -R_{bca}{}^d K^d.$$ Using this, we may prove that the covariant and the Lie derivatives commute: $$\lie_K \nabla_a X^b = \nabla_a \lie_K X^b,$$ for arbitrary $X^a$. We have: ...

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The spatial part of a metric is just a 3-sphere: $$d\Omega_3^2 = d\chi^2 + \sin^2\chi d\Omega_2^2 \, ,$$ and in fact the metric on a $n$-dimensional sphere can always be written in a resursive way using the metric on an $(n-1)$-dimensional sphere in the same manner: $$d\Omega_n^2 = d\chi^2 + \sin^2\chi d\Omega_{n-1}^2 \, .$$ In these coordinates the ...

2

Torsion is not frame dragging. Torsion is having an anti-symmetric spacetime connection. As you do parallel transport in general relativity (GR) you drag frames the frames roll as they move. With torsion they would twist. The connection is GR is the Christopher symbols, symmetric in the two bottom indices. The torsion is an anti-symmetric tensor. It will ...

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Schaum series Differential Geometry will solve part of your problem. Search "problem book in riemannian geometry" on google and it should bring out something useful. Also see V.I. Arnold's books.

2

The acceleration should be $$a = \frac{G\cdot M}{r^2 \cdot \sqrt{1-r_s/r}}$$ with $r$ as the height above the center of mass and the Schwarzschildradius $$r_s = \frac{2\cdot G\cdot M}{c^2}$$ The force to hold the ball at rest is $$F=m\cdot a$$ As one can see it now takes an infinite force and energy to keep a body at a fixed height when $r=r_s$.

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The conditions about (i) differentiability of the functions and (ii) the maximal rank of the corresponding rectangular Jacobian matrix are regularization conditions imposed to simplify the mathematical analysis of the physical problem, in particular to legitimate the possible future use of the inverse function theorem. In the affirmative case, the ...

2

The commutator of two vector fields $n^a$ and $X^b$ is $[n,X]^a = n^b \nabla_b X^a - X^b \nabla_b n^a$. Since this vanishes, it follows that $n^b \nabla_b X^a = X^b \nabla_b n^a$, and the second step follows from there. I don't have my copy of Wald in front of me, but I'm 99% sure that the commutator of two vector fields is defined in terms of the ...

3

Although the radiation-dominated (RD) era is long in comparison to the matter-dominated (MD) and $\Lambda$-dominated ($\Lambda$D) eras, it is nice to have an answer which can be adapted easily for any cosmological era. If we assume that the Universe is permeated by a perfect fluid we may use the equation of state w = \frac{P}{\rho}, ...

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To find points where local flatness breaks down, a general strategy is to calculate the curvature tensor $R_{\mu\nu\rho\sigma}$, and find the locus of various singularities (points where $R_{\mu\nu\rho\sigma}$ becomes unbounded). Except in rare cases (i.e. unusual cancellation), singular behavior of $R_{\mu\nu\rho\sigma}$ is apparent in the Ricci scalar ...

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We can write down the Lagrangian for this problem with general $F$: (restricted to the equatorial plane) $$L = -F(r)\dot{t}^2 + F(r)^{-1}\dot{r}^2 + r^2 \dot{\phi}^2$$ Now using the conserved quantities (energy and angular momentum), and using the convenient substitution $u = \frac{1}{r}$ and rewriting the problem as a differential equation for $u = ... 0 Calculating the geodesic equations in something like De Sitter is not hard, but it's already been done so I'll just link them. For instance, I found a thesis by someone name Chris Ripkin: http://www.ru.nl/publish/pages/760966/thesis_chris_ripken.pdf Go to chapter 3, "Geodesics". Since De Sitter is maximally symmetric, the geodesics will have constant ... 2 A standard reference packed to the brim with other references to everything you ever wanted to know about anomalies is "Anomalies in Quantum Field Theory" by Bertlmann. This particular topic is what comprises part of chapter 11 there. I'll highlight the main points, but this is a technical topic for which you'll have to go to the references and follow all of ... 0 Just a note, the map$z\mapsto -\overline{z} $is also an isometry of the upper half plane model, so the above statement about the group of isometries of$\mathbb{H}^2$being$PSL_2(\mathbb{R})$is not quite correct - these are precisely the orientation preserving isometries (they are indeed the only conformal/holomorphic ones). One needs to look at the ... 2 Therefore you need to calculate the future light cone $$LC_{proper}=\int_{a(t_0)}^{a(t_1)} \frac{c\cdot a(t_1)}{\alpha^2\cdot H(\alpha )} \, \text{d}\alpha$$ In comoving coordinates you divide that by the scale factor of the time at absorption $$LC_{comoving}=\frac{LC_{proper}}{a(t_1)}$$ with H as the Hubble parameter $$H(a)=H_0\cdot ... 2 The solution that you wrote in your last (not numbered) equation is not a basis of a Hilbert space of sections because the phase factor: (-1)^n e^{2i\pi A} depends on n. The phase factor should not depend on n. Please see your (correct) equation (2) defining the boundary conditions, in which the phase factor does not depend on n. Thus there is no ... 0 Curvature is just a form of strain, and strain produces stress. Knowing the curvature, you can compute the strain field from which you can obtain the stress field using linear elasticity theory. In the case of fracturing, the stress causes cracks to propagate which results in fracturing. The wiki article on fracture mechanics is reasonably detailed. 2 In the geometrical optics approximation light ray is represented by a null geodesic. Therefore you only need to find a null geodesic connecting points (t_0,0,0,0) and (t_1,x,0,0) for some t_1 (and this condition will determine t_1 uniquely). This is probably quite easy to do directly in this case, but in general for investigation of null curves in ... 3 Diffeomorphism Invariance Let M be a smooth manifold. Let \phi: M \to M be a diffeomorphism. A simple property of the Einstein equations is$$ g \in \otimes^2 TM \text{ is solution to vacuum Einstein equation} \implies \text{ so is } \phi^*g $$To see that this is true, simply pull back both sides of the Einstein equation by \phi, and use the ... 2 A general diffeomorphism does not map geodesics to geodesics. Some simple counter examples You can a build diffeomorphism on the Euclidean plane by imagining putting one finger on a tablecloth at point x and dragging it. This map is clearly smooth, a smooth inverse is constructed by dragging your finger back. Any geodesic on the plane (a line) passing ... 0 Existence of smooth structures You're asking when does a (topological) manifold M fail to be covered by a smooth atlas. Another way to phrase this is "when does a manifold admit a smooth structure". This is a well-studied problem. It turns out examples of manifolds which do not admit smooth structures only occur in dimensions \ge 4 (see e.g. ... 0 The general story Here's an attempt to formalize how physicists build spacetime manifolds. Let N be one of \mathbb R^n Some dimension n product manifold of S^1 and \mathbb R (corresponding to periodic solutions). Pick one such N. Now Take stress tensor T defined on some open subset U \subset N and some boundary conditions (e.g. ... 0 Non-coordinate bases are always used when coupling to fermions. You could simply look at the maths literature, but as a physicist you'll want to look at textbooks on supergravity. Take a look at words such as "vielbein", "frame" and "spin connection". As to literature, there's Freedman/van Proeyen "Supergravity", chapter 7. Alternatively Ortin's "Gravity and ... 2 You don't. Two given spacetimes can have their metrics written in the same way but may have different coordinate ranges. A simple example is just a spacetime with spatial coordinate identified , such as the cylinder spacetime :$$ds^2 = -dt^2 + d\theta$$Identical to Minkowski space, which is its universal cover. Of course, two things to watch out for : ... 0 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 ...

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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 ...

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