# Tag Info

3

Equation (13) expresses the metric on an embedded hypersurface given by the relations $y^k = y^k(x^a)$. However, the equation for the inverse metric (4-th equation) is in general not correct. Take for example a hypersurface defined by: $y^1 = x^1$, $y^2 = x^2$, $y^3 = x^2$. In our case, the partial derivative of $x^2$ with respect to $y^2$ or $y^3$ ...

4

Let there be given a manifold $(M,\nabla)$ equipped with a (not necessarily torsionfree) tangent bundle connection $\nabla$. OP is asking: Is it possible that the local coordinate expression for the covariant derivative of a co-vector/one-form $\lambda =\lambda_a \mathrm{d}x^a$ (in some local coordinate system $x^a$) could be on the form $$\tag{1} ... 4 For p=1, CTC's do not exist in Minkowski spacetime. In other 1+3 spacetimes, in principle they are admitted in the absence of further requirements (like globally hyperbolicity) on the causal structure of the spacetime. They must be present if the spacetime is compact, for instance. For p\geq 2, the answer is obviously YES. Consider a manifold M with ... 2 There is a really nice derivation of this identity using differential forms, and it completely avoids all the messiness of the Christoffel symbols. The nice thing about differential forms is that the exterior derivative can be computed using any derivative operator, so it allows us to compare the expressions we get using the covariant derivative to the ... 2 This is based on the observation that, given some vector V^\mu,$$\nabla_\mu V^\mu=\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}V^\mu)$$We can show explicitly that this is true:$$\nabla_\mu V^\mu=\partial_\mu V^\mu +\Gamma^\mu_{\mu\lambda}V^\lambda$$Let's examine the last term:$$\Gamma^\mu_{\mu\lambda}=\frac{1}{2}g^{\mu\rho}(\partial_\mu ...

2

Given a pseudo-Riemannian manifold $(M,g)$, the Laplace-Beltrami operator acts on scalar functions. The formula for the Laplace-Beltrami operator follows from the formula $$\Gamma^{\nu}_{\mu\nu}=\partial_{\mu}\ln\sqrt{|g|}$$ for the Levi-Civita connection.

0

For a start, you might have a look at the paper "Differential forms for scientists" by J.B. Perot, Journal of Computational Physics, 2014, Vol. 257 Part B, I found it useful.

2

This is true in general, and there is a very nice geometrical reason why. First use that the Lie derivative satisfies the Leibniz rule, $$£_N(q_{ab} p^{ab})=(£_Nq_{ab})p^{ab}+q_{ab}£_Np^{ab}$$ to rewrite the integral as $$\int d^3x (£_N q_{ab})p^{ab}= \int d^3x\,£_N(q_{ab}p^{ab}) - \int d^3x\,(£_N p^{ab})q_{ab}$$ Now note that the first integrand on ...

2

I've always liked the interpretation you get from the Raychaudhuri equation. It shows you that the Ricci tensor tends to cause geodesics to focus together. If you begin with a family of geodesics with tangent vector $u^a$, you can define the expansion $\theta\equiv \nabla_a u^a$ which measures the rate at which geodesics are spreading out or converging ...

0

It sounds like they are talking about perturbing around a cosmological solution to Einstein's equations. The starting point in cosmology is an assumption that the universe is spatially homogeneous and isotropic, so the metric can depend only on an overall, time dependent scale factor. Then to make things interesting and match what realistically happens, ...

7

Two general methods come to mind: Prove that the Riemann tensor takes the form of equation 3.191, i.e. $$R_{abcd} = \frac{R}{d(d-1)}(g_{ac}g_{bd}-g_{ad}g_{bc})$$ If you are handed a metric, this should in principle be a straightforward calculation. If the metric is actually maximally symmetric, the calculation of the Riemann tensor usually turns out to ...

11

No, general relativity is based on something called "intrinsic curvature", which is related to how much parallel lines deviate towards or away from each other. It doesn't require embedding space-time in a higher dimensional structure to work. You're thinking of something called "extrinsic curvature". In fact, many examples of extrinsic curvature - including ...

3

Nope, spacetime curvature says nothing about the dimensionality. Your intuition here is probably wrong because human imagination needs 'some dimension to bend into' in order for something to be curved (i.e. an embedding in a higher-dimensional space). This is just our lack of imagination showing, though.

0

Background Energy means Non-Relativistic Mass-Energy Tensor (Or, simply Non-relativistic Matter). I don't know which book has written it this stupid way, but such thing (background energy density as function of time) is considered to prove conservation of Non-relativistic Matter (when talking about inflation). The term "Background" here is conventional from ...

4

As a general rule, compactifaction on a Calabi-Yau $n$-fold results in a preservation of $2^{(1-n)}$ parts of the original supersymmetry. If you start with $\mathcal{N}=4$ and compactify on a 2-fold, you preserve $2^{1-2}=1/2$ of the original supersymmetry, i.e. $\mathcal{N}=2$. One now has to realize what a Calabi-Yau $n$-fold is: it is a Kähler manifold ...

4

Here is a belated reply. (I come across this question only now, by chance. This was posted right when our daughter was born, which was kind of distracting for me...) The quick answer to the question is the following somewhat remarkable statement Identity types in the new foundations of mathematics in homotopy type theory correspond in physics to spaces of ...

1

Regardless of the form of whichever (holonomic) constraints you may have, non-autonomous systems are most naturally understood from a field theoretical viewpoint. More precisely, one should understand Lagrangian mechanics as Lagrangian field theory in $0+1$ dimensions (that is, the space-time manifold is just the real time line). There, coordinates over each ...

4

I'll write this as an answer so that the math is more clear. So given an (p,q)-tensor $T^{\mu_1\cdots\mu_p}{}_{\nu_1\cdots\nu_q}$, this one transforms as: T'^{\mu'_1\cdots\mu'_p}{}_{\nu'_1\cdots\nu'_q}=\frac{\partial x^{\mu'_1}}{\partial x^{\mu_1}}\cdots \frac{\partial x^{\mu'_p}}{\partial x^{\mu_p}}\frac{\partial x^{\nu_1}}{\partial ...

3

Yes, you can have such "isolated" isometries. Consider the real line $\mathbb R$ and the inversion mapping $x\to -x$. This isometry does not arise from a killing vector because it's not "continuously connected to the identity."

5

The group of isometries of a given connected smooth (semi) Riemannian manifold is always a Lie group. However, a Lie group can include subgroups of discrete isometries that, barring the identity, cannot be represented by continuous isometries and thus they have no Killing vectors associated with them. (Actually, only some elements of the connected component ...

Top 50 recent answers are included