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11

Why is there no curvature outside this spherically symmetric, non-rotating, uncharged body that still has mass? I suspect you're getting confused by the fact that the Ricci tensor $R_{\mu\nu} = 0$ and therefore the scalar curvature $g^{\mu\nu}R_{\mu\nu} = 0$. This is always the case in regions of space where the stress-energy tensor is zero. The ...


10

The bible for the mathematical formulation of classical Mechanics, namely Foundations of Mechanics by Abraham and Marsden, defines a hamiltonian system as a triple $(M, \omega, X_H)$ where $(M, \omega)$ is a symplectic manifold, and $X_H$ is the Hamiltonian vector field corresponding to a hamiltonian function $H:M\to\mathbb R$. Now, are there typically any ...


9

I'm not going to provide a full answer here, because I don't know the answer, but I want to give some statements that illustrate quite nicely the kind of problems one would face when determining topology of anything: We know spacetime is a manifold. That means, locally, it looks just like $\mathbb{R}^4$. That's already a bummer. We can't do jack at one ...


8

A physicist would write your first equation $x^a = x^\mu e_\mu^a$. The notation $x^a$ is invariant in your terminology. The $a$ is an abstract index. It is ostensibly not supposed to be thought of as ranging over a set of numerical values, but is just a marker that indicates that $x$ is a vector (i.e., rank 1,0 tensor.) Similarly for each $\mu$, $e^a_\mu$ is ...


7

The phase space is a symplectic manifold, so any manifold $\mathcal{M}$ that admits a closed nondegenerate 2-form is a possible phase space. Now, what is necessary (or sufficent) for admitting such a form? First, as you mention, $\mathcal{M}$ must be even-dimensional. Second, $\mathcal{M}$ must be orientable. Why? Because orientability is equivalent to ...


6

Your intuition is correct that you should be able to use this formula in any coordinate system. Indeed, you have presumably done problems with the scalar potential where you have a similar formula that you can evaluate in any coordinate system. However you have to be more careful when defining what you mean by this integral in the vector case in general ...


6

Your second method is correct. To compare, say, the magnetic field with what you find in Jackson, you really need to realize that there's an assumption that you have unit basis vectors there, and that the cross product is actually a hodge dual (which will invoke factors of the square root of the determinant of the metric). These will make direct ...


5

Let's start at the beginning: The setting for relativity - be it special or general - is that spacetime is a manifold $\mathcal{M}$, i.e. something that is locally homeomorphic to Cartesian space $\mathbb{R}^n$ ($n = 4$ in the case of relativity), but not globally. Such manifolds possess a tangent space $T_p\mathcal{M}$ at every point, which is where the ...


4

Congratulations, you made me look into this for the last hour! And, unfortunately, I believe the answer is: Nope We are looking for a Ricci-flat Riemannian symmetric space, since your isometry group is a Lie group. I spent some time trying to construct the Ricci-flat manifold from the irreducible symmetric spaces given there, but couldn't figure out a good ...


4

The correct statement is that we can always construct a geodesic such that $x(s)=y(s)=0$ for every value of the affine parameter $s$. All that independently from our initial choice of the origin and orientation of orthogonal Cartesian coordinates $x,y,z$ in the $3$-manifolds normal to $\partial_t$ (the natural rest space of the considered spacetime). The ...


4

Your question is one about true mathematical duality, you just do not know it. What you are looking for is Hodge duality, which holds in the exterior algebra of any vector space equipped with an inner product and an orientation, and the differential forms one looks at in EM, GR and elsewhere are just elements of the exterior algebra of the tangent space (or, ...


3

The formulation you seek is gauge theory. It is not completely analogous to changing the metric of spacetime, but many similarities can be seen. In this, we take as our starting point a certain gauge group $G$ (In the case of EM, $\mathrm{U}(1)$), which will induce symmetries of our theory, just as the Lorentz group of special relativity is the symmetry of ...


3

The geodesic equation (GE) $$\tag{1} {d^2 x^\mu \over d\lambda^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\lambda} {dx^\beta \over d\lambda} ~=~ 0$$ depends on the parametrization: The GE (1) holds when the parameter $\lambda$ is affinely related to the arc length $s=a\lambda+b$ of the geodesic. This can e.g. be deduced from the fact that eq. (1) is ...


3

Going by a magic 8-ball a brief web search, the most important steps towards the geometrization of electromagnetism (ie its formulation as a classical Yang-Mills theory in terms of principal connections) should be: Maxwell's equations: James Clerk Maxwell, A dynamical theory of the electromagnetic field (1865) differential forms: Élie Cartan, Sur certaines ...


3

Comments to the question (v5): In General Relativity (GR), the notation $x^{\mu}$, $\mu=0,1,2,3,$ usually denotes some (local) coordinates of a (spacetime) manifold $M$. Note that $x^{\mu}$ does in general not transform as a $(1,0)$ (contravariant) tensor in the sense that $$ x^{\prime \nu}~=~\frac{\partial x^{\prime \nu}}{\partial x^{\mu}} x^{\mu} ...


2

As has been discussed in many questions around here (e.g. here), relativity tells us only about local properties and behavior of a space-time. There are some exceptions when we make global assumptions - if we have a space of globally and strictly constant positive curvature, non-trivial topology is imminent because the space has to be the 3-sphere ...


2

Descent equations for the trace anomaly in arbitrary dimensions are derived in this paper: http://arxiv.org/abs/0706.0340. Since this article is very technical, I think that in order to get a better understanding of the principle, it would be useful to go through the chapters on gravitational anomalies in Bertlmann's book "Anomalies in Quantum Field ...


2

Well, by the presentation you give, you're going to have $\frac{d^{2}x^{i}}{d\tau^{2}}\neq 0$, because you have those $\Gamma_{0i}{}^{j}$ terms. For instance ${\ddot y} + 2\frac{\dot a}{a}{\dot y}{\dot t} = 0$ (I abuse notation and mean the obvious things with dots, but obviously, $a = a(t(s))$ and $y=y(s)$) The condition you want is ...


2

You have your implications confused. A form $\sigma$ is closed if $d\, \sigma = 0$. A form $\sigma$ is exact if $\sigma = df$. The statement that $d^2 = 0$ is the same as that every exact form is closed. The opposite is not true in general, there are many cases where a closed form is not exact. The existence of forms that are closed but not exact depends on ...


2

The formula given is horribly unenlightening because it does not seem to use the fundamental fact about differential forms that they are alternating and thus adds $r$ equal terms, it also does not provide the connection to index notation that it supposedly tries to. Let us first understand the idea of the interior product. An $r$-form is something with $r$ ...


1

Notation: I will use overdot for differentiation with respect to $\tau$, overtilde for partial differentiation with respect to $x^0 = t$, and prime for partial differentiation with respect to $x^1 = r$. (Edit: removed overloading of $\lambda$, sorry.) I assumed a general $\nu = \nu(t,r)$; reading the question more carefully, they're functions of $r$ only, ...


1

I wondered about this one myself a while back. I'm not absolutely positive about this but it is definitely in the ballpark. Here's what I know for the background: I believe the first paper on exotic spheres in physics was by Witten [Commun. Math. Phys., 100, 197–229 (1985)] and centered around the idea that exotic spheres can be interpreted as ...


1

The names of these creatures are a true mess and there are mainly two independent notation schemes: the mathematical and the physical one. Let $P \to M$ be a $G$-principal bundle. Then $G$ is called the structure group by mathematicians and the gauge group by physicians The (infinite-dimensional) group of automorphism of $P$, or equivalently the group of ...


1

This is not quite a complete answer, but more of an overly large comment on terminology. The definitions from nLab don't agree with Giovanni Giachetta, Luigi Mangiarotti, Gennadi Sardanashvily: Advanced Classical Field Theory, which I'll summarize briefly: The authors call the group $G$ of a (principal) $G$-bundle the structure group. This is standard ...


1

The boundary term of the Einstein-Hilbert action is given by, $$S= \frac{1}{8\pi G}\int_{\partial M} \! \!\mathrm{d}^3x \, \sqrt{-h} \, K$$ where $h$ is the metric on the boundary of the manifold, i.e. $\partial M$, and $K$ is the trace of the extrinsic curvature. Specifically, we have, $$K=\nabla_a n^a$$ where $n^a$ is a unit normal to the surface ...


1

I admit I am a bit confused by your terminology, but here is how I learned it: Let $P$ be a $G$-principal bundle and $\Sigma$ a spacetime. gauge group: The fibers of the $G$-principal bundle over the spacetime, i.e. the group $G$. (Local) group of gauge transformations: The group of diffeomorphisms $t : P \rightarrow P$, which are fiber-preserving and ...


1

The action you're considering yields Einstein's equations in vacuum, so $R=0$ (this follows immediately from contracting Einstein's equations). Therefore the action vanishes on shell.


1

It seems to me that a good starting point is the geodesic equation: [...] This apparently refers to some particular (image of) curve $\gamma$; indeed to some particular time-like curve $\gamma$ for which $$\int_{\gamma} d \tau = \Delta \tau \mid_{\gamma} ~ \gt 0.$$ Given two (not necessarily distinct) (images of) time-like curves $\gamma$ and $\psi$ ...


1

I think it is better to argue from the curvature tensor $R_{ab}{}^\mu{}_\nu$. It is defined by $$R_{ab}{}^\mu{}_\nu x^\nu = (\nabla_a \nabla_b - \nabla_b \nabla_a)x^\mu$$ so it tells you the degree to which covariant derivatives along the $a$ and $b$ axes do not commute. You can see formally from this that curvature requires two dimensions, so it does not ...


1

In order for a manifold to be curved (intrinsic curvature), it must be of dimension $\geq 2$, which means that at least two principal curvatures must be non-zero, since Gauss's curvature is the product of them. This cannot be done with only one curved basis vector or dimension, as you put it; actually, there's no way to define intrinsic curvature in ...



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