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As a geometric quantity, the value of an "angle" can be determined and expressed in a coordinate-free way: Given three pairwise space-like events, "$A$", "$B$" and "$C$", and given the positive real numbers $\frac{s^2[ A C ]}{s^2[ A B ]}$, $\frac{s^2[ A C ]}{s^2[ B C ]}$ and $\frac{s^2[ B C ]}{s^2[ A B ]} = \frac{s^2[ A C ]}{s^2[ A B ]} / \frac{s^2[ A C ...


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There are both physical and formal reasons to introduce the spin connection. Physically, we know that there are spin 1/2 particles. A spin 1/2 field cannot be described by anything built from 4-vector fields. You can realize this for example by that 4-vector fields (and so anything built from them) returns to their original value after a $2\pi$ rotation ...


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We will work with the metric supplied, $$\mathrm{d}s^2 = A(r)^2\mathrm{d}t^2 -B(r)^2 \mathrm{d}r^2 -r^2 \mathrm{d}\theta^2 - r^2 \sin^2\theta \, \mathrm{d}\phi^2$$ I will assume that omitting the fourth spatial coordinate (which I have added) is simply a typo in the original post. I have also redefined the arbitrary functions for convenience. We choose a ...


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If it is a photon, then you know that $x^{a}x_{a}$ is always zero. You have measured the spatial direction of the photon with your mirror apparatus, so you know the values of the $x^{i}(0)$ at some time${}^{1}$, which we will call zero. Furthermore, we can infer the value of $x^{0}$ at this time from the fact that $x^{a}x_{a} = 0$. Then, all we need is to ...


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Your suspicions are correct: It is wrong! At least as it is written presently. Let us start form the embedding of manifold $$\imath_t : F_t \ni p \mapsto p \in M\:.$$ It induces and embedding of corresponding tangent bundles: $$T\imath_t : TF_t \ni (p,v) \mapsto (p, d\imath_t (v)) \in TM$$ The latter can only preserve the vectors tangent to $F_t$ seen ...


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I'm guessing that when you say: What is now the spatial distance between the two particles? You mean the proper distance. The coordinate distance is of course just $L$. The proper distance is the distance you would measure if you sat at radius $R+L$ and let out a tape measure until it reached radius $R$. To calculate the proper distance start with the ...


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I want to add my personal understanding of the concept of reference frame. In the articles: Marmo, G., Preziosi, B. (2006). The Structure Of Space-Time: Relativity Groups. International Journal of Geometric Methods in Modern Physics, 03(03), 591-603. Marmo, G., Preziosi, B. (2005) Objective existence and relativity groups. Symmetries in Science XI, ...


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The question is somewhat ill-defined. This answer is based on what I gather the OP is asking. No. Generically, the covariant derivative of a tensor is zero only if the entire tensor is zero. Note how the covariant derivative is defined (taking example of a 1-form) $$ (\nabla_\mu A)_\nu = \partial_\mu A_\nu - \Gamma^\lambda_{\mu\nu} A_\lambda $$ Now, ...


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Re your edited question, this is just simple spherical geometry. If the initial separation is $d$ then the separation at time $t$ is $d \cos(vt/r)$, where $r$ is the radius of the sphere, $v$ is the vehicle speed and $t$ is time. The diagram shows a cross section through the poles. The vehicle is driving north at a velocity v, so the distance it drives in ...


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I don't think there's an a priori reason, but there's certainly a good a posteriori empirical one: if you make two long, straight parallel things, they neither meet nor diverge away from one another. It was presumably this empirical fact that led Euclid to introduce his parallel postulate, although he probably wouldn't have seen it as empirical. Moreover, ...


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The limit of non euclidean geometries, as the radius goes large, is euclidean. It's like relativity. Unless you have fancy equipment or fast things, the euclidean newtonian model does quite fine. If you model space on hyperbolic geometry, with a curvature the same size of the earth, the observable universe would fit inside a sphere of radius 432000 miles. ...


2

In the absence of evidence to the contrary we tend to assume the simplest possible description for physical systems. Suppose the spatial scalar curvature had the non-zero value $S$. Immediately we have the question: why is it $S$ and not $S + 0.001$ or $S - 0.001$ or any other value. There must be some mechanism for making the curvature exactly $S$ and that ...


3

Maxwell's equations in curved spacetime are written in the form $$\begin{split}\nabla_a F^{ab} &= - 4\pi J^b,\\ \nabla_{[a} F_{bc]} &= 0,\end{split}$$ with $F$ the Faraday two-form, $J^a$ the current four-vector, $\nabla$ the covariant derivative and $[]$ denotes antisymmetrization of the indices. In terms of exterior calculus they become: $$ ...


1

So on in three-dimensional Euclidean space we have an isomorphism between vectors and 1-forms, the usual way $$\eta_\mu = g_{\mu\nu} \eta^\mu.$$ We also have an isomorphism between 1-forms and 2-forms, given by $\star : dz\mapsto dx\wedge dy$ and cyclically. This isomorphism has a fancy name, the Hodge dual, if you want to know about it in general. Then if ...


1

Without specifying any particular scenario, and ignoring any proportionality constants, simply consider some general differential form $\omega$, and let this represent the electric flux through a closed surface which bounds some volume V. In classical electromagnetism, the Gauss law tells us that the flux through a closed surface is proportional to the ...


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An isometry of ${\cal{M}}$ is a transformation $\phi:{\cal{M}}\rightarrow{\cal{M}}$ such that the following condition hold: \begin{equation} g_{p}(v,v)=g_{\phi(p)}(\phi_{*}v,\phi_{*}v) \end{equation} where $v\in T_{p}{\cal{M}}$ and $\phi_{*}v\in T_{\phi(p)}{\cal{M}}$ and $\phi_{*}$ is the pushforward. Which is basically the statement that the the metric is ...


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When you write $\omega(X,Y)$ this gives the impression that $\omega$ is a 2-form! It is $d\omega$ that is a 2-form, if $\omega$ is a 1-form. The action of the 1-form $\omega =\omega_\mu dx^\mu$ on the vector field $X = X^\mu \partial_\mu$ is simply $$\omega(X) = \omega_\mu X^\mu.$$ (Since you use index notation, you probably know about relativity and this ...


1

In gauge theory, the connection (the potential in the electromagnetic case) is defined over a fiber bundle with fibers in some suitable Lie Group. Yang Mills theories are theories with Lie Group $SU(N)$, while in electromagnetism the group is $U(1)$. So for all these theories, the fiber in the fiber bundle is a Lie Group. The case where I know there is ...


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Without getting complicated, Einstein predicted and proved that electro-magnetic wave/photons describe a "curvature" of space-time known as a "geodesic relative to a body of mass. Nothing escapes geodesics that we know so far.


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You're on the right track. Hints: Recall that under a diffeomorphism $f$, the tensor transformation law tells us that the metric transforms as $g\to g_f$ where \begin{align} (g_f)_{\mu\nu}(f(x)) = g_{\alpha\beta}(x)\partial_\mu (f^{-1})^\alpha(f(x))\partial_\nu (f^{-1})^\beta(f(x)) \end{align} which, sending $x\to f^{-1}(x)$ can be re-written as ...


-1

Write $$g^\mathfrak{ab} = x^\mathfrak{a}_\mu x^\mathfrak{b}_\nu g^{\mu\nu}.$$ Consider everything as a function of ${x'} = x + \xi$ and Taylor expand to first order.


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The dynamics of a large class of mechanical systems can be described as a geodesic motion in some ambient space. This is the essence of the Kaluza-Klein theory. The basic and most elementary example is the case of a charged particle in $3D$ coupled to a magnetic field which can be described as a neutral particle geodesically moving in a background metric ...


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I) OP talks about minimizing curves (rather than higher dimensional objects) so let us concentrate on point mechanics (as opposed to field theory) with Lagrangian $L$ (rather than Lagrangian density ${\cal L}$). We conventionally call the curve parameter time $t$, although it doesn't have to correspond to any physical time variable. Let us for simplicity ...


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You have many comments to the effect that "topology is needed to describe continuity, calculus concepts, the notion of "looks like", homeomorphism and so forth". And these are all altogether right, but I'm getting that your question is about the global picture. Also, the following is mainly about a toplological or differentiable manifold; Joshphysics's link ...


2

Strictly speaking, if there can be defined charts covering a set (an atlas), you can give that set the topology induced by defining the charts to be bicontinous. That is, a set is open iff it's the domain of a chart in the maximal atlas. If your set already has a topology, the topology induced by the atlas will agree with that one under some conditions. (I ...


1

You don't need to assume that the path of least action is the path taken. You can show it from Newton's laws. See http://www.damtp.cam.ac.uk/user/tong/dynamics/two.pdf The path of least action is the path for which $F = ma$ holds at each point. This is the geodesic. This is the shortest path through space-time. You get this path from the Lagrangian. You ...


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You can derive the equations of motion (equations of geodesics) for a particle in curved spacetime by using the Lagrangian $$L = \frac{1}{2} \sum_{\mu,\nu} g_{\mu\nu} \frac{dx^\mu}{dt} \frac{dx^\nu}{dt},$$ so the answer is yes. You could regard the configuration manifold as the manifold, it need not be the physical spacetime. I would like to clarify that ...


2

The Newtonian vacuum field equation $\nabla^2 \phi = \rho$ where $\phi$ is the gravitational potential and $\rho$ is proportional to mass density also has non-trivial vacuum solutions, for example $\phi = -1/r$ for $r$ outside some spherical surface. The Maxwell equations also have non-trivial solutions. In electrostatics, precisely the same as classical ...


3

I'm amazed that there are all of these answers here and no one has pointed out that a spacetime containing nothing but (caustic-free) gravitational radiation will be singularity-free and will be Ricci flat but not Riemann flat, which is a cleaner example than the Schwarzschild metric.


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I noticed your question from a few months ago and I don't think you were given a very good answer to a very intelligent question. Perhaps this may help. The Schwarzschild metric is a vacuum solution even though as applied to our solar system, for instance, we regard the sun as the physical source of curvature. The non-vanishing of the stress-energy tensor ...


1

As Flanders says forms are what we integrate. Since the (surface) integral of D over a surface is the enclosed charge hence D must be a 2-form. Similarly, the work of taking a unit charge along a path is the contour integral of E, then the latter must be a 1-form. They just cannot be anything else.


0

III) A spinor is indeed an element of the vector space on which the representation of the Spin group acts. IV) I don't know V) The fact that the norm of the vector is null does not imply that has no nonvanishing entries. Remember that in physics applications one has a metric of indefinite signature, so contributions to the norm can (and most of the time ...


2

This question should probably be split up into several questions. Really quickly, I can answer the bit about the Newman-Penrose forumulation and tetrads. I think the best modern approach to the tetrad formalism is to treat the set of tetrads as a map from an internal orthonormal frame to an ordinary coordinate frame. So, the tetrad is a set of matrices ...


3

First let me refer you to Eric Weinberg's book where the instanton moduli space is described in more detail. Principal bundles over 4-dimensional Riemannian manifolds are classified by the second Chern class = Instanton number and the t' Hooft discrete Abelian magnetic fluxes. Please see the following Lecture notes by Måns Henningson. t' Hooft fluxes ...


1

There are a lot of rather different aspects being touched on in the question. I'll try to give some indications. But I notice that the relation of this question to actual physics is not very strong, instead the question seems to be more generally after getting a feeling for identity types in homotopy type theory (HoTT). I imagine there are other discussion ...


1

First recall that we define a co-vector field $\eta$ from a vector field $v$ via the flat map $ \eta~=~v^{\flat}.$ In components we have $\eta_k = g_{ki} v^i.$ Equivalently, we can reconstruct the vector field $v$ via the sharp map $v ~=~\eta^{\sharp}.$ In components we have $v^i = (g^{-1})^{ik} \eta_k.$ Now let us return to OP's question. As Danu writes ...


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In the language of differential forms in spacetime, the field strength $2$-form $F = E\wedge\mathrm{d}\sigma + B$ gives Gauss's law for magnetism and Faraday induction: $$\mathrm{d}F = 0\text{.}$$ Meanwhile, the electromagnetic excitation $2$-form $H = -\mathcal{H}\wedge\mathrm{d}\sigma + \mathcal{D}$ provides a natural formulation of Gauss's law and ...


2

The point is that, with your second equation, you are dealing with in a local coordinate patch, say an open set $U\subset M$ equipped with coordinates $x^\mu \equiv x^1,x^2,x^3,x^4$. Therefore, if $p\in U$, you can handle two bases of the tangent space $T_pM$. One is made of (pseudo)orthonormal vectors $e_a$, $a=1,2,3,4$ and the other is the one associated ...


0

They are your tetrad. If we take, say, Schwarzschild spacetime in Schwarzschild coordinates, one choice for the $e_{\mu}{}^{I}$ is: $e_{t}{}^{t} = \sqrt{1-\frac{2M}{r}}$ $e_{r}{}^{r} = \frac{1}{\sqrt{1-\frac{2M}{r}}}$ $e_{\theta}{}^{\theta} = r$ $e_{\phi}{}^{\phi} = r\sin\theta$ with all others zero. The key point to notice here is that $g_{ab} = ...


1

Kyle Kanos mentioned Geometric Algebra for Physicists. While geometric algebra is somewhat different in notation from differential forms, the basic concepts are all there, and in many ways, geometric algebra avoids some cumbersome things that differential forms does (I'm thinking of Hodge duality in particular). I think the notation is easier to relate to ...


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If you'd like to quickly obtain an understanding of the basics of differential forms, including their relation to connections, tangent bundles etc. I recommend the first 4 online lectures of the Perimeter Institute from the Gravitational Physics course (13/14, R. Gregory).


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I do think Jerry Schirmer answered the question in the comments, but I'll try to expand just to make clear how he explained everything. Let us consider given that special relativity is correctly described by physics in Minkowski spacetime. Then we can ask ourselves how to include gravity without violating causality, which is mandatory by the finite velocity ...


1

I'm not sure if your main interest lies on the question on the title of this thread or in the question you pose near the end of your text. I'll try to answer both, in spite of not being an expert neither on GR nor on the education on physics. Why is it, if it is true, that GR would have taken many more years to discover, had Einstein not discovered it? I ...


3

It is on page 377 in my book. May be there are different editions. You only need connectedness of the manifold. That the field is uniquely determined by its value and the value of its derivative at a point is equivalent (by linearity) to the statement that if its value and the value of its derivative are zero at a point, the field is identically zero. To ...


0

You're already there. The method would just be to solve the geodesic equation. To me it seems you have a completely flat manifold. The affine connection vanishes on a flat manifold, reducing the geodesic equation to $$ \frac{d^{2}x^{\mu}} {du^{2}} = 0 $$ where $ \mu \; \epsilon \; 1,2,3 $ and $x^{1} = x$, $x^{2}=y, $ and $ x^{3}=z$ Trivial result, but the ...


1

I've figured out the answer to the question. I write the answer here to help others who might be interested. Also, there is an assumption here that I'm not totally comfortable with, so it might be interesting to get some insight there. Let us start with the equation $$ \partial_z B_{\bar z} - \partial_{\bar z} B_z = 0 $$ The most general solution to ...



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