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9

Let me take a stab at answering this (somewhat vague) question. You said you are interested in the analytic structure of QFT. But you also mentioned the RG, which is somewhat different. I will try to address the analytic structure of QFT and then emphasize that the renormalization group can be thought of as merely a trick to improve perturbation theory. ...

9

Ok so the way I see it, let's strip away the physics first. Then we're left with the math. So the question you're posing is, given only $\frac{d^nx(t)}{dt^n}|_{t=c} \forall n$, can we reconstruct the function $x(t)$ uniquely in some domain $(a,b)$ s.t. $c \in (a,b)$? Well the answer is no. Even if you impose a $C^\infty$ condition on $x(t)$, the example ...

8

Could this imply there is a formulation where that value comes naturally... This sentence implicitly assumes that analytic continuation is "unnatural". But the truth is the other way around: analytic continuation is one of the most natural mathematical procedures in physics. On the contrary, it's functions – especially functions of momenta or energy – ...

8

There are no exact solutions, only approximations and numerical solutions. Don't forget that orbiting black holes will radiate gravitational waves so any solution would have to include those and the corresponding decay of the orbit until the black holes coalesce.

8

This is a really interesting, but equally beguiling, question. Shock waves are discontinuities that develop in solutions of the wave equation. Phase transitions (of various kinds) are non-continuities in thermodynamics, but as thermodynamics is a study of aggregate quantitites, one might argue that the microscopic system is still continuous. However, the ...

6

According to general relativity, a pair of massive bodies that orbit each other emits gravitational waves - for analogous reasons to the reasons why accelerating charges in electrodynamics emit electromagnetic waves. So there can't be any static solutions resembling binary stars or binary black holes. The solutions have to be non-static and a complicated ...

5

Any analytic function is defined everywhere on its Riemannian surface just by its values in an arbitrarily small neighborhood of a point. The expression to be ''analytically continued'' therefore just specifies which function is meant, but it has ''direct and natural'' values everywhere on its Riemann surface. Except that not all of these values can be ...

5

Carefully following Feynman's procedure one actually finds: $$\langle x''|e^{-i\frac{(z''-z')}{\hbar}H}| x' \rangle$$ $$=\langle x'', z''|x', z'\rangle =\lim_{N\to \infty\: \epsilon \to 0} \left[\frac{m}{2\pi i \hbar \epsilon} \right]^{N/2}\int_{-\infty}^{+\infty}\cdots \int_{-\infty}^{+\infty} \left(\prod_{i=1}^{N-1} dx_i \right) ... 5 Let us for simplicity consider a bosonic string X:\Sigma\to M in the matter sector only (as opposed to the ghost sector). Here M is a target manifold. The worldsheet \Sigma is a Lorentzian manifold of real dimension 2. Locally in a neighborhood U\subseteq \Sigma of the worldsheet, we may work in a so-called conformal gauge, which means to choose a ... 5 1) OP is basically wondering how Weinberg on the middle of p. 112 can extend the integration region from^1$${\cal J}^{\pm}_{\beta}~=~ \int_{m_{\alpha}}^{\infty} \!dE_{\alpha}\frac{e^{-iE_{\alpha}t}g(E_{\alpha})T_{\beta\alpha}^{\pm}} {E_{\alpha}-E_{\beta}\pm i0^{+}}$$to include the negative real axis$${\cal J}^{\pm}_{\beta}~=~ \int_{-\infty}^{\infty} ...

5

The pole of Green's function is related to the spectrum of the particle which is propagating. One dimension for example $$\tilde{G}(\omega)= \frac{i}{\omega-(\epsilon+i\Gamma)}$$ If pure real, G(t) is some oscillation function which shows that the particle is stable. If pure imaginary, G(t) has some exponential decay behavior which shows that the particle is ...

5

For a very recent authoritative review of the numerical approach, see Centrella et. al. http://arxiv.org/abs/1010.5260 For the alternate parameterized post Newtonian approach, see Living Reviews of Relativity http://relativity.livingreviews.org/Articles/subject.html and look for articles number 2007-2, 2006-4 and 2003-6.

5

Every regularization scheme is somewhat arbitrary. There are three popular regularization schemes when it comes to path integrals and their associated perturbative divergent integrals: time slicing, mode regularization, and dimensional regularization. Time slicing is the usual procedure used to derive the path integral, and it is the discretization of time ...

4

Franz Pretorius has worked on this and developed animations. http://prl.aps.org/abstract/PRL/v95/i12/e121101 The field is numerical relativity. Matthew Choptuik also, I believe, has done work on this.

4

OP's question is essentially pondering (in the context of the holomorphic/coherent state path integral) if a pair of variables is a complex conjugate pair or$^1$ truly independent variables. Notation in this answer: In this answer, let $z,z^{\ast}\in \mathbb{C}$ denote two independent complex numbers. Let $\overline{z}$ denote the complex conjugate of $z$. ...

3

Suppose that for all $z$ in some open set $Z$ of complex numbers containing $z_0$, the Hamiltonian $H(z)$ is a compact perturbation of the self-adjoint $H(z_0)$ depending analytically on $z$. Then, for every simple eigenvalue $E_0$ of $H(z_0)$ and associated normalized eigenstate $\psi_0$, there exist a complex neighborhood $N$ of $z_0$ and unique functions ...

3

Some rules for the cross product are: $$A \times (\beta B + \gamma C) = \beta (A \times B) + \gamma (A \times C)$$ $$(e_x \times e_y) \cdot e_z = 1$$ The equation $$(B \times C)\cdot A = d$$ is antisymmetric with respect to transpositions, meaning that if you take any two of those vectors and switch them, you multiply $d$ by $-1$. These rules ...

3

A way of physically thinking about this is that a two body problem in general relativity does not generally have closed orbits. If one of the bodies is very large and the other a small satellite the problem is integrable. The periapsis (perihelion) advance of the small satellite is repeated with each orbit, which makes the problem integrable. If the two ...

3

What quantities supposed to be $C^\infty$? I don't know if it answers you question, but AFAIK smooth functions are nice and useful tool to describe many aspects of the physical world. However, I don't see why they should be considered as fundamental in any sense. When it comes to QFT, even there you often encounter Dirac delta (and you can't get rid of it ...

3

In dimensional regularization, $d$ is a complex number, not a true dimension. The $d$-dimensional integrals of a rational function are defined for any complex $d$ with sufficiently negative real part (the threshold depending on the integrand), and therefore can be analytically continued to a (provably meromorphic) function for all $d$. For a concise, ...

2

To perform the analytic continuation of the step function, start with the second equation stated in the question. Since the step function's derivative is the Dirac delta function, substitute the integration of the Dirac delta for the step function. Then perform the Fourier transformation on the integral. After simplifying this equation, it is equal to zero. ...

2

One way to look at it is simply a mathematical trick that encodes the boundary conditions of the Schroedinger equation. An alternative and only slightly more intuitive view is the following. In order to obtain only outgoing solutions it is essential to assume that the scattering potential is slowly switched on adiabatically. Formally one can achieve this by ...

2

Let $\mathscr{H}$ be a separable Hilbert space. Suppose that the Hamiltonian $H$ is a densely defined self-adjoint operator with domain $D(H)\subset \mathscr{H}$. Then for any $\phi\in D(H)$, $i\partial_t e^{-it H}\phi=He^{-itH}\phi$, where $e^{-itH}\phi$ is the unique solution of the Schrödinger equation. Now, $e^{-itH}$ is a unitary operator for any ...

2

Comment to the question (v2): Yes, the authors of Ref. 1 are cheating. They are not using Wick's theorem (although one in principle could do so). They know that the modified Bessel function $I_0$ of first kind has an asymptotic series expansion in terms of a generalized hypergeometric function$^1$ $$e^{\frac{1}{4g}} ... 2 Use fig 13.2 of [2] as reference. Taking the example Qmechanic uses, the idea is that I(\omega) = \int_{\zeta_a}^{\zeta_b}\frac{Z(\omega)}{\zeta - \xi(\omega)} d\zeta I(\omega) needs to be analytically continued from \omega_1 \rightarrow \omega_2. The pole of the integrand travels from \zeta = \xi(\omega_1) \rightarrow \zeta = \xi(\omega_2) in ... 2 Complex analysis is very useful in potential theory, the study of harmonic functions, which (by definition) satisfy Laplace's equation. One way to see this connection is to note that any harmonic function of two variables can be taken to be the real part of a complex analytic function, to which a conjugate harmonic function representing the imaginary part of ... 1 I) Let there be given a meromorphic function \zeta \mapsto F_{w}(\zeta) in the \zeta-plane with a single (not necessarily simple) pole at the position \zeta=\xi(w), where \xi is a holomorphic function, and w\in \mathbb{C}  is an external parameter. Ref. 1 is considering the contour integral$$\tag{A}I_{\Gamma,w} ~=~ \int_{\Gamma} \! d\zeta ...

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