# Tag Info

8

A metric on a manifold $M$ is, by definition, a symmetric 2-tensor field $g$ with the property that $g_x$ is positive-definite for every $x\in M$ (plus some smoothness/continuity requirements if $M$ is smooth/topological). This ensures that the norm of a vector in a fibre of the tangent bundle to $M$ is a non-negative number, and that the angle between ...

4

@Phoenix87 is spot on, but I'll elaborate a bit. Definition 1 A spacetime $(M,g)$ is stationary if there exists a timelike Killing field $K$, i.e. a vector field $K$ such that $\langle K,K\rangle<0$ and $\mathcal{L}_Kg=0$. We shall show that Definition 1 implies the existence of local coordinates for which $g_{\mu\nu}$ is independent of time. ...

4

The "signed arc-length" is not used in relativity and I give reasons why. You are free to call and denote it in any way you like. $s$ and $\ell$ are interchangeably used to denote arc-length of space-like paths $g(\gamma',\gamma')>0$ in relativity and $\tau[\gamma]$ is used for "proper time" of $g(\gamma',\gamma')<0$ time-like paths but the notation ...

3

Recall that one-forms are defined as linear maps on vector fields to real numbers, so that for every one-form $\alpha$ and every vector field $X$, $\alpha(X)$ is a scalar function. Hence on a simple tensor $\alpha \otimes X$ we can define the contraction by $C : \alpha\otimes X \mapsto \alpha(X)$ and extend by linearity. On a tensor with more factors, for ...

3

The rough idea: take the local flow of the vector field and use it to get a new "time" coordinate. In general this will work locally, so you have to patch your manifold with small enough open subsets where you can then define the new set of coordinates where now the Killing vector field corresponds to $\partial_t$.

3

The dual is defined by the map $$\frac{\partial}{\partial x^\mu} \mapsto g_{\mu\nu}\mathrm{d}x^\nu$$ and hence the dual of $a \partial_t + b \partial_1$ is only $a \mathrm{d}x^t + b \mathrm{d}x^1$ iff the metric is Euclidean flat in the $t,1$-direction. Note: Do not write "$X=$" if you mean the dual of $X$ is equal to something. Duality/equivalence is ...

3

Technically, yes (for loose enough definition of "metric"), but there's very little point to it. Some attempts at unification of gravity and electromagnetism, including several attempts by Einstein and various co-authors, basically amount to some variation of trying to interpret the antisymmetric part $g_{[ab]}$ as the electromagnetic Faraday tensor ...

3

The separation vector is a Jacobi field because it obeys the Jacobi equation. Here I will derive geodesic deviation from scratch because I find MTW's derivation hard to follow. (Much like everything else in that book.) Definition 1. Consider a family of timelike geodesics, having the property that in a sufficiently small open region of the Lorentz ...

2

We are not entirely sure what OP's question (v4) is asking, but here are some comments: I) The Dirac belt trick demonstrates that the Lie group $SO(3)$ of 3D rotations is doubly connected, $$\pi_1(SO(3))~=~\mathbb{Z}_2.$$ II) As for the title question Are spinors somehow connected to spacetime? one answer could be: Yes, in the sense that the mere ...

2

This notation arises often in supergravity. Suppose one has a $d$-dimensional theory. The Hodge $\star$ operator has the usual definition, and $\star_p$ is the Hodge star operation defined on a $p$-dimensional sub-manifold. The question of which sub-manifold is often either explicitly stated or obvious from the context. For example, in the Klebanov-Strassler ...

2

Asymmetric tensors have been considered in the quest for a unified classical field theory. Einstein in particular went through a whole series of candidate theories. His last paper on the subject - co-authored by Bruria Kaufman, submitted 3 months before and published 3 months after his death - is about a field of this type; the theory actually was referred ...

2

Regarding why the Ricci tensor, not the Riemann, appears in the EFEs the answer is in the Newtonian theory.Very heuristically consider Newton's law $\ddot{r}=-\frac{GM}{r²}$. Now let's try to make this a more local statement. In order to do so I'll divide both sides by $r$ and some numerical factors in order for the volume $V=\frac{4\pi r³}{3}$ of some ...

2

The Einstein equations are equivalent to an relation on the Riemann tensor. Given $a, b$ that are linearly independent vectors, $$R(a \wedge b) = C(a \wedge b) + 4\pi [a \wedge T(b) - b \wedge T(a) - \frac{2}{3} T a \wedge b]$$ where $T(a)$ is the stress energy tensor acting on $a$ and $T$ is its trace, and $C$ is the Weyl ("conformal") tensor. This ...

2

Same as with the symplectic form: $\omega(v) = (u_\omega,v)$ defines the isomorphism between 1-forms and vector fields. When the metric is Euclidean the dual basis to an orthonormal basis corresponds to the basis itself.

2

I have never seen that second definition before. The first definition is standard. For a Riemannian manifold the metric tensor is positive definite, that is $$g(u,u)>0\quad\forall u\ne0$$ We have the standard relation (tensor product omitted) $$\mathrm{d}s^2=g_{ij}\mathrm{d}x^i \mathrm{d}x^j$$ Let $t\in\mathbb{R}$ be a curve parameter and ...

2

General relativity can be constructed from the following principles: The Principle of Equivalence Vanishing torsion assumption ($\nabla_XY-\nabla_YX=[X,Y]$) The Poisson equation (or any other equivalent Newtonian mechanics equation) Explanations: The Equivalence Principle can be used to show that spacetime is locally Minkowskian, i.e. the laws of ...

2

I would recommend using Mathematica to calculate curvatures, unless there's a good reason to do it by hand (for example, perhaps you want to calculate the curvatures for a metric while keeping the dimension general). It's not hard to write your own code to do this, and I think it's a nice idea actually. I also have found this code to be very useful: ...

2

No, a magnetic monopole a la the Dirac string does not "violate" gauge symmetry. Rather, the statement "we have a magnetic monopole" means only that we are forced to consider the gauge theory not on the whole spacetime, but on the spacetime with the location of the magnetic monopole removed. Why? Because, at the location of the magnetic monopole, the curl of ...

1

One of the biggest surprises that General Relativity has given us is that under certain circumstances the theory predicts its own limitations. There are two physical situations where we expect for General Relativity to break down. The first is the gravitational collapse of certain massive stars when their nuclear fuel is spent. The second one is the far past ...

1

We have the frame $\{e_\mu\}_{\mu=0,\dotsc,3}$ in terms of which the velocity vector is $v=v^\mu e_\mu$. There are a few properties of the affine connection which I would like to summarize: $$\nabla_{fX}Y=f\nabla_XY$$ $$\nabla_X(fY)=f\nabla_XY+X(f)Y$$ $$\nabla_{e_\mu}e_\nu=\Gamma^\lambda{}_{\mu\nu}e_\lambda$$ Using this, let's get to work. We have ...

1

Comments to the question (v3): Given a manifold $M$, if a smooth vector field $X\in \Gamma(TM)$ does not vanish in a point $p\in M$, then one may choose a local coordinate neighborhood $U\subseteq M$ of $p$, with local coordinates $(x^1, \ldots, x^n)$, so that $X=\frac{\partial}{\partial x^1}$. This procedure is sometimes called stratification or ...

1

If you are working on a complex manifold with a Hermitian metric, then the Hodge star operator should be taken antilinear: $\star(\alpha + i\beta) = \star\alpha - i\star\beta$. If you work on a real manifold without a complex structure by itself, and you extend the scalars to the complex numbers, it may be that there is no harm in taking it to be linear.

1

The Hodge star operation acts on differential forms. Numbers, real or complex, transform as 0-forms. The Hodge dual of a 0-form will result in something proportional to the volume form of the manifold. In detail, for a $d$-dimensional manifold, $\star 1 = \text{vol}_d = \sqrt{|g|}dx^1 \wedge ... \wedge dx^d$, and the Hodge operation commutes with ...

1

In general, having an asymmetric matrix for a metric won't really help, because only its symmetric part will contribute to the norm of any given vector. Take some finite-dimensional real vector space $V$ with an inner product $\langle\cdot, \cdot\rangle$ represented by some matrix $g_{ij}$ in a given basis $\beta$. Then for any vector $v$ with components ...

1

I think the concept of charts might be leading you astray here. A simple, practical and intuitive way to define the dimension of a manifold is the number of numbers you would need to locate a point on that manifold. For the case of a sphere, you need two, commonly written as $\theta, \phi$. For $\mathbb{R}^3$, you need 3, $x,y,z$. Sure you can embed $S^2$ ...

1

If by "straight line" you mean "linearly varying the coordinates from the beginning point $(x_0,y_0)$ to the endpoint $(x_1,y_1)$," then the trajectory is just that. You can parametrize it as $(x,y) = (1-\lambda) (x_0,y_0) + \lambda (x_1,y_1)$ for $\lambda$ running from $0$ to $1$. Call this path $C$. The distance is  \int\limits_C \sqrt{ds^2} = \int_0^1 ...

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