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Note that $\mathbf r(t)$ is the trajectory (a priori unknown) of a charged particle in an external electric field. Now consider the ansatz $\mathbf r(t) = \mathbf r_0(t) - \mathbf a(t) \cos \Omega t$, which is motivated by the solution for a homogeneous electric field $\mathbf E(t) = \frac{m\Omega^2}{q} \mathbf a \cos \Omega t$. Here $\mathbf r_0(t)$ is a ...
$\Phi_c(t)$ is the directional derivative of $L(c(t),\dot{},t)$ along $W(c(t))$, so by the chain rule $$\Phi_c(t) = \sum_i\frac{\partial L}{\partial \dot{q}^i}W^i = \sum_i\frac{\partial L}{\partial \dot{q}^i}\frac{\partial \phi^i}{\partial s}$$ There is then an inexplicable notational shift from $\frac{\partial\phi^i}{\partial s}$ to $\frac{\partial ... 0 Have you ever seen Feynman's QED: The strange theory of light and matter lectures? For integer$n$the Bessel functions can be derived as$$J_n(x) = \frac1{2\pi} \int_{-\pi}^\pi d\theta ~e^{i(n\theta -x\sin \theta)}$$ For$x \gg n$we can immediately start to see that this complex number spirals a lot in the complex plane; the sum is therefore usually zero ... 0 Hint. ( I give absolutely no guarantee about not making calculational mistakes!) \begin{eqnarray*} L(\tau ) &=&\frac{1}{\tau }\int_{-\tau /2}^{+\tau /2}dy\left( 1-\frac{1}{ \sqrt{\pi }\sigma }\int_{-\tau /2}^{+\tau /2}dx\exp [-\frac{(y-x)^{2}}{ \sigma ^{2}}]\right) e^{-iPy}\rho e^{+iPy}=L_{1}(\tau )-L_{2}(\tau ) \\ L_{1}(\tau ) ... 0 Conformal maps tend to be the exception rather than the rule. In general, if a transformation$T:S\to S$on some$n$-dimensional space$S$is of interest in physics, it will not be conformal. (Indeed, there's generally no guarantee of a useful notion of angle in that space, but even if there is such a guarantee then$T$is still not likely to be conformal.) ... 0 My query is this: From the following two methods, which is correct method to find error/uncertainty in DOP? If you know the individual errors for each p and q, then I would use (b) and propagate the errors since the relationship is not linear. Be careful with averages though. It might be better to use the median, for example, if you have one or two ... 1 Here is a purely geometrical way to think about this Edward says it is possible to cut a wedge out of a flat spacetime and glue the edges together. So in my mind this looks like a paper cone. A cone is flat precisely because it can be created by rolling up a flat sheet. Rolling up preserves the metric on the interior of the sheet (not on the boundaries ... 0 The most intuitive way to express this, as far as I know, is to start by taking the limit of the ratio of the circumference of a circle about the singularity to its radius, with the radius tending to zero. This ratio must be$2\pi$for a manifold, and the conical deficit indicates that the space-time is singular at the centre of the circle. You may find more ... 1 There are attempts to use nonstandard analysis (e.g., Albeverio) or Colombeau algebras, but these haven't been developed very far. I haven't seen anything in terms of surreal numbers, but they may probably substitute for the infinitesimals in nonstandard analysis. 2 I think you make it sound much more mysterious than it is. The relativistic distribution function is $$f_p = \frac{1}{(2\pi)^3}\exp(-(\mu-u\cdot p)/T)\,$$ where$u_\alpha$is the 4-velocity of the fluid,$p_\alpha$is the 4-momentum of the particle,$T$is temperature, and$\mu$is the chemical potential. This is sometimes called the Juttner ... 0 Here is a fun paper on using Möbius inverse formula. Nan-xian Chen, Modified Möbius inverse formula and its applications in physics, Phys. Rev. Lett. 64, 1193 (1990). 1 What came to be called "discrete torsion" is simply the data that makes the B-field gerbe be equivariant over the orbifold. This was clarified by Eric Sharpe, see the references here: Eric Sharpe, Discrete Torsion and Gerbes I (arXiv:hep-th/9909108) Discrete Torsion and Gerbes II (arXiv:hep-th/9909120) Discrete Torsion, Quotient Stacks, and String ... 3 (1) Yes, take${\cal H} = L^2(\mathbb R, dx)\oplus L^2(\mathbb R, dx)$and thereon$\left(X (\psi, \phi)\right)(x,y) := (x\psi(x),y\phi(y))$. We have$\sigma(X)=\sigma_c(X)$and the degeneracy is just$2$. (2) Yes, use the example (1) with a countably infinite copies of$L^2(\mathbb R, dx)$and use the Hilbertian direct sum of Hilbert spaces. (There are ... 1 I can only answer the mathematical part of your question (or make a stab at it). We could say that by describing a space as an orbifold, the singularities are taken care of by somehow declaring them to be under control. Where a manifold is a topological space that may be very complicated, but locally looks very nice, namely like$\mathbb R^n\$, an orbifold ...