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5

Notation $\renewcommand{braket}[2]{\langle #1 | #2 \rangle}$ First let us define the term "1-form": A 1-form is a linear function which takes a single vector as it's argument. That is it. It is no more confusing or complicated than that single statement. Consider a vector $v$. Given any other vector $w$ we can form the inner product $\braket{v}{w}$. ...


4

The only orthonormal coordinate basis is the Cartesian coordinate basis. The basis vectors for the, e.g., polar coordinate basis are orthogonal but not normalized. That doesn't mean that one can't normalize the polar basic vectors to get the polar unit basis but such a basis isn't a coordinate basis. For the Cartesian coordinate basis, the basis vectors ...


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The spin connection has a geometric meaning - it is a connection associated to a particular non-coordinate frame, which diagonalizes the metric. Here's how: Let $M$ be our spacetime. The tangent bundle $TM$ may be thought of as the associated bundle to an $\mathrm{SO}(n)$-principal bundle, where the $\mathrm{SO}(n)$ matrices represent ordered orthonormal ...


3

The Gauge Theory of Gravity (GTG) by Lasenby, Doran and Gull has a background spacetime with fields on it. It is basically derived from the same physical principles but as a background theory. It ends up not being the same theory, for instance it doesn't have the same isotropic solutions, and I think it does not allow time travel and such (unlike General ...


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A point to emphasize here is that one cannot separate the rest mass of the nucleons and electrons from that of field: rest masses aren't additive in this way and one can only state the rest mass of a system as a whole. The mass-energy of the fields is already built into the rest mass of the system. The rest mass of an electron includes the energy of the ...


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First, terminology: Symmetry groups are not "defined on domains". Symmetry groups exist in the abstract, and they are then represented on certain spaces. If we have a spacetime manifold $\mathcal{M}$, then the fields are functions $$ f : \mathcal{M} \to V$$ where $V$ is some vector space upon which a representation $\rho : \mathrm{SO}(1,3)\to ...


3

Gravity is very, very weak compared to the electromagnetic force. Whatever curvature of space-time is induced inside an atom would be far too weak to hold mass-energy together at that scale. The electromagnetic force is 1/137 the strength of the strong nuclear force, while Gravity is 6 * 10^-39 the strength of the strong nuclear force. Here is a link that ...


3

First some terminology: A non-degenerate 2-form $\omega$ is called an almost symplectic structure. A closed 2-form $\omega$ is often called a pre-symplectic structure. If the 2-form $\omega$ is both non-degenerate and closed, it becomes a symplectic structure. In the non-degenerate case, the closedness condition $$\mathrm{d}\omega~=~0\tag{C}$$ is ...


2

First, we will compute $\text{div}\,F$. The partial derivatives are given by$${{\partial F}\over{\partial x_i}} = q {\partial\over{\partial x_i}}\left({{x_i}\over{r^3}}\right) = q\left({1\over{r^3}} - {{3r^2 {{x_i}\over{r}} x_i}\over{r^6}}\right) = 0.$$Thus, $\text{div}\,F = 0$ away from the origin. Consider now a ball $B$ of radius $r$ centered at the ...


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MS Mohamed et al begin with a "standard vector calculus formulation of the NS equations", $$ \frac{∂ \bf u}{∂ t} - \mu ∆ {\bf u} + ({\bf u} \cdot \nabla) {\bf u} + \nabla p = 0 $$ $$ \nabla \cdot {\bf u} = 0 $$ and use the rotational form, $$ \frac{∂ \bf{u}}{∂ t} + \mu \nabla \times \nabla \times {\bf u} - {\bf u}\times(\nabla \times {\bf u}) + \nabla p^d ...


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If $m$ < $n$, then your $F$ is a function from $m$-dimensional flat space to $n$ dimensional flat space, which maps $\mathbb{R}^m$ onto a $m$ dimensional surface in $\mathbb{R}^n$. If the motion of a particle in $\mathbb{R}^n$ is restricted to this surface, then obviously a force is required to keep it on that surface. Without such a force, it follows a ...


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The area of an event horizon is an invariant, so it doesn't matter what coordinate system you use to calculate it. The area is of physical interest because it's proportional to the entropy of the black hole, and in a second law kind of way that means the area cannot decrease.


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I don't believe there is a general relationship for an arbitrary Lorentzian manifold, such that $R_{abcd}R^{abcd} \propto R$ The only relationship that I can think of is the Weyl tensor invariant. Let us denote the Weyl tensor $C_{abcd}$. The quantity you have mentioned above is the Kretschmann scalar, which is denoted: $K = R_{abcd}R^{abcd}$. One can then ...


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I don't think there is any universal "intuition" to tap into, aside from that which comes from practice. You perhaps need to explore different physics texts in the electromagnetic department. I for example loathed Jackson as a learning text: it is comprehensive and useful as a reference for refreshing knowledge, but not good at conveying it. Volume 2 of the ...


1

Comments to the question (v3): Note that the gamma matrices are covariantly conserved$^1$ $$\nabla_{\mu}\gamma^c~=~\omega_{\mu}{}^c{}_b\gamma^b+\frac{1}{4}\omega_{\mu}{}^{ab}[\gamma_{ab},\gamma^c] ~=~0,\qquad \nabla_{\mu}\gamma^{\nu}~=~0,\tag{1}$$ cf. e.g. Ref. 1. Consider the vector current $$J^{\mu}~:=~\bar{\psi}\gamma^{\mu}\psi,\tag{2}$$ where $\psi$ ...


1

Suppose you have a skewed coordinate system, with vectors that have components $v^i$ defined relative to some basis vectors $\hat e_i$ such that $\vec v = v^i \hat e_i$ in the Einstein summation convention (we sum over any index which is repeated once above, once below). The meaning of a skewed coordinate system is that $\hat e_i \cdot \hat e_j \ne ...


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The name "1-form" is only used because there are 2-forms and 3-forms and so on. An equivalent but probably better name for 1-forms in this case is that of "dual vector." Any vector space will have associated with it the dual space consisting of all linear functions that map vectors into scalars. This is all a dual vector is: a thing that linearly maps ...


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If you have a coordinate system you could move along a coordinate, which indicates some vectors you could use for a basis. These vectors might be orthogonal, that depends on your coordinates (think, does the metric look diagonal in those coordinates)? But even if your coordinates are orthogonal then you still have to pick a magnitude for these vectors. ...


1

Technically speaking, manifolds are by definition topological spaces, which resemble locally an inner-product space. Since there are vector spaces (with a dot product) of infinite dimension, then there shall be infinately-dimensional manifolds as well. The infinity of the dimension is not a problem for the tensors as well - each multi-linear function over a ...


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GR can be recast into an equivalent but conceptually quite different form, using teleparallel gravity. This approach introduces the Weitzenboeck connection, which has no curvature, but has torsion. The presence of torsion indicates that gravity is not geometrized. Recall that in GR, we can always choose a locally inertial coordinate system such that the ...



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