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1

The existance of Killing fields even just for a small region is a special property of the metric. For general metrics you cannot expect to find Killing vectors. Notice that the Killing equation which should be written using covariant derivatives as $X_{\mu;\nu} + X_{\nu;\mu} = 0$ is 10 independent partial differential equations for only 4 field components so ...


0

Actually, evolution equations are even more than just second order in time : they don't depend naively on first order derivative, that is, on "velocity". This can be easily understood as the fact that there exists no privileged inertial frames. The change (that is, what is absolute) is given by acceleration and not velocity. If it depended naively on some ...


2

The Lorentz group $O(3,1)$ has spinor representations (actually $SL(2,\mathbb C)$, that is the universal cover of $O(3,1)$), as well known. The problem is that now, in general relativity, we want to deal with generic transformations. So we are working with G$L(4)$. Roughly speaking, the associated Lie Algebra $\mathfrak{gl}(4)$ doesn't admit spinor ...


3

A general vector is written as $v = v^\mu \partial_\mu$. Its norm is defined as $$ g(v,v) = g_{\mu\nu}(\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu)(v^\mu\partial_\mu,v^\nu\partial_\nu) = g_{\mu\nu}v^\mu v^\nu (\mathrm{d}x^\mu \otimes \mathrm{d}x^\nu)(\partial_\mu,\partial_\nu) = g_{\mu\nu}v^\mu v^\nu$$ where we have used linearity of the duals and ...


1

The Lagrangian for GR is $$ L \propto \int R \sqrt{-g} \, d^4 x $$ where $R$ is the Ricci scalar $$ R = R^\mu_\mu = R^{\mu \nu}_{\mu \nu} $$ So, this is a scalar which is related linearly to all the components of the Riemann tensor, and is a second-order differential of the metric $g$ of the form $$ R \sim g \partial^2 g + (\partial g)^2 $$ This is ...


0

They are not analogous. $R_{abcd}$ is just Riemann Tensor and $R_{abcd}R^{abcd}$ is Riemann Tensor squared. Mathematically they must be squared since having a Single term Riemann Tensor / Ricci Tensor in gravitational action doesn't make sense. Physically speaking they are modification of Einstein Hilbert action. They are curvature not field one shall ...


1

Mathematically, any topological (metric, or any kind of) space is a sub model of a larger space. That is, you can always find a space of larger dimension that includes the original one. Now, what is the implication of this for physics? that you can always have two theories in which in one, the smaller space is the only thing there is. But there will be also ...


1

In the Schwarzschild geometry, the Schwarzschild radius breaks naive dilation symmetry. In the simple case of a radial dilation $r \to \lambda r$, the geometry is only preserved by $R_S \to \lambda R_S$. So, it naively seems like it would be difficult to find a working dilation, even just a radial dilation. I went to some effort (as an exercise for myself) ...


1

There is no problem in treat Cristoffel symbols as tensors, because in some definitions they actually are tensors. If one defines abstractly a covariant derivative as an operator over tensors with the following properties: Linearity: $$ \nabla_c \left( \alpha A^{a_1,\dots,a_k}_{b_1,\dots,b_l} + \beta B^{a_1,\dots,a_k}_{b_1,\dots,b_l} \right)= \alpha ...


3

It doesn't. The covariant derivative is a map from $(k,l)$ tensors to $(k,l+1)$ tensors that satisfies certain basic properties. As such it cannot act on anything except tensors. The collection of components $\Gamma^a_{bc}$ does not constitute a tensor. If you got to this expression via something like $$ \nabla_d(\nabla_b A^a) = \nabla_d(\partial_b A^a) + ...


4

You basically have the right idea: the existence of a covariantly constant vector field is a big restriction on the metric. You already discovered that $A^a$ has constant norm. The next thing we find is that $A^a$ is a Killing vector, because obviously $$\nabla_{(a}A_{b)}=0.$$ Furthermore, $A^a$ is geodesic, $A^a\nabla_a A^b=0$, and hypersurface ...


1

As mentioned by Qmechanic, the answer to your questions is no. However, assuming space-time is oriented, we have the following: For any pseudo-Riemannian metric $g$, there exists a normalized time-like one-form $h^0$ and a Riemannian metric $g^R$ so that $$ g = 2h^0\otimes h^0 - g^R $$ This yields a locally Euclidean topology compatible with the manifold ...


2

I) Consider a Riemannian 3-manifold $(M,g)$ with a non-vanishing vector field $V\in \Gamma(TM)$. Define a 2-dimensional distribution as $$\tag{1} \Delta ~:=~ {\rm ran} (V)^{\perp}~\subseteq~ TM.$$ This definition (1) and the integrability condition $$\tag{2} V \cdot \nabla\times V ~=~ 0$$ for $\Delta$ relies on a choice of dot product/metric tensor $g$. ...


1

Affine spaces use vectors to model displacements between points. This fails in a curved space because displacements no longer add according to the parallelogram law, so it makes no sense to model them as vectors. It's not so much that any particular one of those axioms fails, as that the whole definition is no longer reasonable. In a curved space, ...


0

Hyperplanes can be defined by (1) a direction normal to them and (2) one point within them, so a train of parallel hyperplanes certainly encodes direction information. A good concrete model for an one-form is a wavevector $k$ that defines a linear plane wave solution of D'Alembert's equation $(\nabla^2 - c^{-2})\,\psi=0$, when the wavevector is thought of ...


2

Edit: Now that I've read the original post more closely, I'm not sure that this answer actually addresses your question. Let me know if it doesn't and I can perhaps try again. Remember: curl is only defined for vector fields and not for lone, individual vectors. With this in mind, curl is sometimes said to quantify the "vorticity" of a vector field (i.e. ...


1

The expression for the length of a path $\gamma : I = [a,b] \to \mathcal{M}$ $$ L[\gamma] = \int_\gamma \sqrt{g_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu}$$ is a mnemonic expression (by this I mean that it tells you how you would calculate it in all cases, but that it is not the "rigorous" meaning) for taking the length of the tangent vector at every point ...


4

You seek a 1-form $A$ on $\mathbb{R} - \{0\}$ such that $\mathrm{d}A = B$. On all of $\mathbb{R} - \{0\}$, $\mathrm{d}B = 0$, so this could exist. But, since you have magnetic flux, you require that the integral of $B$ over any 2-sphere around the origin should be $g$. Therefore, by Stokes' theorem, $$ g = \int_{S^2}B = \int_{S^2} \mathrm{d}A = ...


3

The magnetic flux through any closed surface enclosing the origin is just $g$ (the magnetic charge enclosed). If the magnetic field comes from a vector potential $\vec{B}=\nabla\times\vec{A}$, this surface integral by Stokes' theorem is an integral of $\vec{A}$ around the boundary of the surface. But the surface is closed, so has no boundary, so the answer ...


0

user23873 answered my question in the comments. I quote: " Try reading the book 'Geometry, Topology and Gauge Fields: Foundations', the author (Naber) have this discussion right on the introductory chapter and points how the impossibility to define a proper vector potential on ℝ3−0 is linked with it's topology (the second homotopy group is non-trivial), and ...


0

There is an answer to your question that seems overlooked and worth mentioning: Your intuition is exactly CORRECT. The electric field intensity E is a 1-form and magnetic flux density B is a 2-form giving you $\nabla\times E=-\dfrac{\partial B}{\partial t}$ and $\nabla \cdot B=0$ The excitation fields,displacement field D and magnetic field intensity H, ...


2

Advanced Classical Field Theory (2009) by Giachetta, Mangiarotti, Sardanashvily remarks on p. 248: A non-compact world manifold admits a Dirac spinor structure if and only if it is parallelizable. For a compact world manifold $X$, its Euler characteristic and the second Stiefel-Whitney class $w_2$ must be zero, and its first Pontryagin number ...


2

Comments to the question (v4): The question formulation seems to talk about affine parametrizations before applying the principle of stationary action. In the context of Riemannian$^1$ geometry, an affine parametrization of a (not necessarily geodesic) curve means by definition that the arc-length $s$ and the curve parameter $\lambda$ are affinely related ...


0

After thinking about this a bit I realize the question was due to a rather silly misconception,the converse was not mentioned in the usual references because it's trivial from the proof of the usual direction. For future reference I will include a quick version here. $$F = L^2/2$$ $$ EL[F] \equiv \frac{d}{ds}(\frac{\partial}{\partial \dot{x^\mu}}F)- ...


3

Orientation. Let a lagrangian $L$ that is a local function of position and velocity be given. For a parameterized path $x$ on $M$, define a corresponding action $S$ as follows: \begin{align} S[x] = \int_a^b ds\,L(x(s), \dot x(s)). \end{align} Let $\delta x$ be a fixed-endpoint variation, then a standard computation shows that the corresponding variation ...


1

The property you are referring to is called geodesic completeness. It is an important concept in the study of singularities in general relativity. There are somewhat trivial examples of geodesic incompleteness where you are just "missing" part of the spacetime, i.e. you could have Minkowski space with a point removed. In these cases you usually consider ...


1

It is not true. Sometimes the spacetime itself and therefore the geodesics can be extended. Consider for example the manifold described by (t,x,y,z) with $x,y,z> 0$ and Minkowski metric. This is nothing more than Minkowski spacetime truncated to a smaller region, but it is a perfectly valid manifold for all purposes. In the diagram we see schematically ...


2

It's the chain rule: $$ \partial_x f(y(x)) = \partial_y f(y(x)) \cdot \partial_x y(x)$$ Your vector field $A^\mu$ depends on the worldsheet coordinates only through the worldsheet coordinates $x^\mu$. Thus, when how $A^\mu$ behaves under an infinitesimal shift on the world-sheet you need to take into account how $A^\mu$ depends on $z$ - that's $\partial_z ...


3

This is not true. There are trivial counter-examples. For example, take $\mathbb{R}$ with the trivial metric. Then, $\gamma :(0,1)\rightarrow \mathbb{R}$ defined by $\gamma (t):=t$ is a geodesic, but it is also clearly extendable. Indeed, it is extended by $\tilde{\gamma}:\mathbb{R}\rightarrow \mathbb{R}$ defined by $\tilde{\gamma}(t):=t$.


5

Let's look at your last line: Why not simply postulate a global coordinate system and let the coordinate mapping from $M$ to $\mathbb{R}^n$ and the resulting metric tensor describe the geometry? P.S. By Rn I mean the set of all n-tuples with the usual definitions of sum and scalar multiple, not Euclidean n-space. $\mathbb{R}^n$ has a particular, simple ...


2

First of all, the metric tensor is one additional piece of structure one inserts on a smooth manifold to measure lenghts and angles. The metric is indeed not present in all applications of Differential Geometry to Physics (see e.g. Lagrangian Mechanics). In that case, it is important to know also how to deal with manifolds without metric tensors. Now, about ...


1

All curves can be parameterised by an affine parameter (commonly written as $\lambda$ in the GR books I have). However only for timelike curves does the parameter $\lambda$ have a physical meaning i.e. it's the elapsed time $\tau$ shown on a clock by the observer following the curve. So there's no mathematical difference between parameterising a timelike ...


1

Yes, the proper time along a timelike curve in relativity is very much analogous to the length of a curve. Just as the length of a curve is invariant under rotations, the proper time along a curve is invariant under Lorentz transformations. One difference with conventional length is that although a straight line is the shortest length between two points, a ...



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