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There is a large rigorous mathematical literature on quantum field theory. This literature is not complete, in that it does not cover the QFTs of greatest interest to particle physicists, the 4d Yang-Mills theories. But it does cover a great many other models, both toy examples and statistical models used in condensed matter theory, and it also covers the ...


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This is a very good question and one that is quite difficult to answer. In physics, the use of a group is usually to describe a symmetry of the physical system. What we know of the monster is that it has a non-trivial action in 196883 dimensions. In other words, the Monster group is the symmetry of a 196883 dimensional object. The hope of Monster moonshine ...


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Yes, OP is basically correct, at least locally. For a 1st-order dynamical system on a $(2n+1)$-dimensional manifold $(M,\alpha)$ equipped with a globally defined 1-form $\alpha$, a dynamical/evolutionary vector field $X$ satisfies by definition $i_X\mathrm{d}\alpha=0$. An action functional is given by $S[\gamma]=\int\! \gamma^{\ast}\alpha$. Deviation from ...


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The last equation is just inhomogeneous differential equation for the simple harmonic motion. So use, $$x(t)=x_{inhm}(t)+Asin(\omega t)+Bcos(\omega t)$$ for further calculations.


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The coordinates are $x^\alpha$. There is no coordinate $x_\alpha$ as coordinates are not vectors and cannot have their indices raised a lowered by the metric. Instead $$ \partial_\alpha \equiv \frac{\partial}{\partial x^\alpha}, \quad \partial^\alpha\equiv g^{\alpha\beta}\partial_\beta. $$ If you want an invariant you must contract an upper index with a ...


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The result is true at least locally. I do not think that it is valid globally. I assume that $\Sigma$ is an immersed (at least) submanifold. Take $p\in \Sigma$, then there is a local coordinate system $(u,x,y,z)$ in $M$ with domain an open neighborhood of $p$ such that a neighborhood $S\subset \Sigma$ of $p$ is represented by $u=0$. Since $\Sigma$ is ...


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The Nieh-yan three-form is $N=d({\bf e}^{*a}\wedge T_a)$ where ${\bf e}^{*a}\wedge T_a$ is globally defined --- i.e on the overlap of coordinate patches it is the same scalar-valued 3-form in both patches. Consquently on a closed 4 manifold $M$ we have $$ \int_M N= \int_{\partial N}{\bf e}^{*a}\wedge T_a=0 $$ because $\partial M=\emptyset$. On the other ...


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So, if I do convolution of velocity and time, then the resultant should have units of metre x time. Am I correct? Yes. If you integrate over time, dimensionally you get the unit [length][time]. $$ \left[\int \text{velocity}\times\text{time}\times dt\right] = [v][t][dt] = [\text{length}][\text{time}] $$ But if the capacitance becomes time varying, then ...


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The curvature form on a $G$-principal bundle $P\to M$ is a $\mathfrak{g}$-valued 2-form that is equivariant and horizontal, and hence descends to a well-defined global $\mathfrak{g}$-valued 2-form on $M$. Since it is a 2-form, it has 2 ordinary Lorentz indices. To see this, pick any trivializations $\phi_i : U_i\to M$ and identity sections $s_i : U_i \to ...


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The time average of a periodic quantity $Q(t)$ is, by definition, $$\langle Q \rangle=\frac{1}{T}\int_0^TQ(t)\,dt$$ where $T$ is the period. For Keplerian orbits, it is usually easiest to change the integration variable to the angular coordinate $\psi$ and express all quantities being integrated in terms of $\psi$. The exercise thus consists of doing a ...


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Here is the problem: It is possible to show that such a vector field is identified by a pair $(f,Y)$ where … vector field $X$ is not uniquely identified by this pair $(f,Y)$, only its action on the boundary data, while the vector field itself is defined not only near the boundary but everywhere “inside” the manifold. In other words, this pair $(f,Y)$ ...


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The fibre in a tangent bundle is an entire vector space. In other words if ${\bf e}$ is a unit vector tangent to the circle at some point, then the fibre is the infinite set $\{x{\bf e}| x\in {\mathbb R}\}$ equiped with the usual vector addition and scalar multiplication operations.


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These smearing functions are closely related to the single particle wave functions. Fix a decomposition of spacetime into space times time. Fourier transform the space coordinate. In the special case $f(t,p) = \delta_0(t) \psi(p)$, which can be reached by taking a family of gaussian approximations, the operator $\phi^\dagger(f)$ creates a particle with ...


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If a self-adjoint operator has pure point spectrum, then (by one definition of pure point spectrum) its eigenfunctions form a complete basis for the Hilbert space. However, an operator may also have continuous spectrum, in which case to get a decomposition of the identity one must use the spectral measure of the self-adjoint operator. The usual thing ...


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