# Tag Info

53

It is exactly because we have a factor of $\frac 1 2$ in the area formula of a triangle. To understand what I'm saying, consider what is the $v(t)$ graph of a particle under constant acceleration. Some say, a good plot is worth a million words! :)

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The result you've got would be better known as this: $$\int_0^t\biggl(\int_0^{t'} a\mathrm{d}t''\biggr)\mathrm{d}t' = \frac{1}{2}at^2$$ In other words, it's a derivation of the formula for uniformly accelerated motion. This derivation, or something like it, is one of the first things students in a good calculus-based introductory physics class learn. The ...

15

Take a look at the notes on lectures 1 and 2 of Geometric Numerical Integration found here. Quoting from Lecture 2 A numerical one-step method $y_{n+1} = \Phi_h(y_n)$ is called symplectic if, when applied to a Hamiltonian system, the discrete flow $y \mapsto \Phi_h(y)$ is a symplectic transformation for all sufficiently small step sizes. From your ...

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It is, in fact, a double integral! The first notation used $$\varPhi_E = \oint_S \vec{E} \cdot \mathrm{d}\vec{A} = \oint_S \vec{E} \cdot \hat{n} \ \mathrm{d}A$$ is simply a more compact notation. It's much easier to write $\mathrm{d} \vec{A}$ instead of, say, $r \ \mathrm{d}r \ \mathrm{d}\theta$ all the time. Furthermore, it's more general, as $\mathrm{d} ... 14 IMHO, the notation$\int_a^b\mathrm{d}x\,f(x)$is much cleaner than$\int_a^b f(x)\,\mathrm{d}x$, because the integration variable ($x$) and its associated integral range$(\int_a^b) are kept together. This is particularly important in lengthy and multi-dimensional integrals. Consider $$\Upsilon_{pq}(k)= \int_0^\infty\mathrm{d}x ... 14 This is notation from Distribution Theory in Functional Analysis. The theory of distributions is meant to make things like the Dirac Delta rigorous. In this context, just to give you one overview, a distribution is a functional on the space of test functions. We define the space of test functions over \mathbb{R} as \mathcal{D}(\mathbb{R}) being the ... 13 No, it cannot be enough. Stokes' theorem says that the volume (\Omega) integral of d\omega, a form that is the exterior derivative of another one (of \omega), may be written as a surface integral. But it doesn't allow us to rewrite the volume integral of a general integrand (which isn't the exterior derivative of anything) such as the Lagrangian ... 13 A simple reference problem Suppose we want to analyse the problem of a forced harmonic oscillator. Denote as \phi(t) the time dependent position of the oscillator. The oscillator experiences two forces, the spring force -k\phi(t) and an external force F_{\text{ext}}(t). Newton's law says$$ \begin{align} F(t) &= m a(t) \\ -k \phi(t) + ... 12 Here we will assume that OP is not questioning the fundamental physical principles/postulates/axioms of quantum mechanics, such as, e.g., the need to have a Hilbert spaceH$in the first place, etc; and that OP is only pondering the role of$L^2$-spaces (as opposed to, e.g.,$L^1$-spaces). Let us for concreteness and simplicity consider the 3-dimensional ... 12 In German, this property is known as the Transformationssatz, but I do not know any appropriate translation for it. This is, however, a special case of coordinate tranformations changing the measure by the determinant of their Jacobian, since obviously$\frac{\partial y_i}{\partial x_j} = A_{ij}$. That it is the determinant that plays a role in the ... 12 It's not just QFT literature. Physicists, especially adult research physicists, find this notation sensible and popular – even though it may be more popular among particle physicists than elsewhere. Formally,$dx\,f(x)$is a product of two factors and$\int$is a form of a sum. Because product is commutative, it doesn't hurt when the order is interchanged. ... 11 If the functional derivative $$\tag{1} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}$$ exists (wrt. to a certain choice of boundary conditions), it obeys infinitesimally $$\tag{2}\delta F ~:=~ F[\phi+\delta\phi]- F[\phi] ~=~\int_M \!dx\sum_{\alpha\in J} \frac{\delta F[\phi]}{\delta\phi^{\alpha}(x)}\delta\phi^{\alpha}(x).$$ OP's functional integral ... 10 The$\delta$function is not continuous, so it's a priori not differentiable. In fact, it's not even well-defined as an ordinary real-valued function, but can be made so in terms of distributions - linear maps on a space of test functions given by$f\mapsto\int\delta f=f(a)$. It's possible to sensibly define derivatives of distributions by looking at ... 10 An important example in quantum mechanics is e.g. the Hilbert space $$H~=~L^2(\mathbb{R}^3)$$ of Lebesgue square integrable wave functions$\psi$in the position space$\mathbb{R}^3$. The Lebesgue square integrable functions (as opposed to just the Riemann square integrable functions) are needed to complete the Hilbert space with respect to the square ... 10 This is not a peculiar physicist notation oddly enough. The notation allows one to interpret$1/x$as a distribution (which makes sense since it's being added to the delta distribution on the right hand side of the equation). For a suitable test function$\varphi$, one defines this distribution as $$\mathrm{pv}(1/x)(\varphi) = \lim_{\epsilon\to ... 9 The short answer is that the two principal value definitions agree on sufficiently well-behaved functions, but may disagree on sufficiently singular functions. For instance, on one hand$$\lim_{\epsilon\searrow 0} \int_{\mathbb{R}\backslash[-\epsilon,\epsilon]} \frac{\mathrm{d}x}{x^3}~=~0$$is zero, while on the other hand$$\lim_{\epsilon\searrow 0} ... 9 When the velocity is not constant you have: $$x(t)=\overline{v(t)} t + x_i$$ where$\overline{v(t)}$is the average velocity from$0$to$t$. When you have constant acceleration the average velocity is $$\overline{v(t)}=\frac{v(0)+v(t)}{2}=\frac{at}{2} + v_i$$ which will give the correct result. If the acceleration is non constant you will have to do the ... 8 It's an integral over a closed line (e.g. a circle), see line integral. In particular, it is used in complex analysis for contour integrals (i.e closed lines on a complex plane), see e.g. example pointed out by Lubos. Also, it is used in real space, e.g. in electromagnetism, in Faraday's law of induction (part of the Maxwell equations, written in an ... 8 There are two that I know of in the context of state estimation. The first is for estimating the mean of$P$and is a Metropolis-Hasting MCMC algorithm here: Optimal, reliable estimation of quantum states. The second is also mainly for computing the mean (but can do other functions -- including the characteristic function of the region you are interested ... 8$a_x \Delta t = \Delta v_x = v_{xf} - v_{xi}\Delta x = v_{x,average}\Delta t = v_{xi}\Delta t + \dfrac{1}{2}a_x (\Delta t)^2\Rightarrow v_{x,average} = v_{xi} + \dfrac{1}{2}a_x \Delta t = v_{xi} + \dfrac{1}{2}(v_{xf} - v_{xi}) = \dfrac{v_{xf}+ v_{xi}}{2}$Is there a geometric interpretation or does it just work out mathematically? 8 1. Since$x\gg p$, we see that$\sin(px)$is highly oscillatory. In fact, the integral becomes $$\int_0 ^\infty \mathrm{d}p\ p \sin px \ e^{-it\sqrt{p^2 +m^2}}\sim \int_{-\infty} ^\infty \mathrm{d}p\ p\ e^{ipx-it\sqrt{p^2 +m^2}}$$ modulo some factor of$\pm2/i$. Observe now this integral resembles$\int f(p)\exp(g(p))\,\mathrm{d}p$. We find the point ... 8 In order to express the position as a function of the velocity you have to integrate with respect to time. When the velocity is constant this integral is simple, namely$vt+C$. However once the velocity becomes a function of time this integral will change and will in general not be equal to$v(t)t+C$. You actually have to integrate$v(t)$with respect to$t$... 8 Besides the reasons listed in Lubos Motl's answer, here is another reason for the$\int \!dx ~f(x)$notation: By writing the integral sign$\int_a^b$and$dx$next to each other in multiple nested integrations, it becomes more easy to trace which limits belong to which integration. This becomes particularly handy when changing the orders of integration. ... 8 The delta function$\delta(x)$has unit area, but the function$\delta(2x)$is "half as wide" and thus has half as much area; thus you can pick up extra factors from 'how fast' you cross the peak of the delta function. The general identity is $$\delta(f(x)) = \sum \frac{\delta(x-x_i)}{\big| df/dx|_{x=x_i} \big|}$$ where the$x_i$are the roots of$f$. In ... 7 Technically, the equation $$d = \frac{\mathrm{d}x}{\mathrm{d}t}t + \frac{\mathrm{d}^2x}{\mathrm{d}t^2}\frac{t^2}{2}$$ is not right. Instead, for constant acceleration, you need $$d = \left(\left.\frac{\mathrm{d}x}{\mathrm{d}t}\right|_0\right) t + \left(\left.\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\right|_0\right) \frac{t^2}{2}$$ In other words, a quantity ... 7 You know that you can pull a multiplicative constant out in front of an integral, right? $$\int cf(t)\mathrm{d}t = c\int f(t)\mathrm{d}t$$ where$f(t)$is any function of$t$, like$t^2$or$t(2\text{ s} - t)$(and$c$does not depend on$t$). Units can be part of that constant factor too. In this case, the constant factor is$4\mathrm{\ kg\ m/s^4}$. The ... 6 In theory: Lebesgue integrable functions form a Banach space, whereas Riemann integrable functions do not. This causes problems in, e.g., Quantum mechanics if we try to work with Riemann integrable functions instead of Lebesgue integrable functions... (We want the functions to form a Banach space so we can use good old fashioned linear algebra to solve ... 6 1) As OP basically notes, an$n$-dimensional delta function transforms under change of variables$f:\mathbb{R}^n \to \mathbb{R}^n$with (the absolute value of) an inverse Jacobian $$\tag{1} \delta^n(f(x))~=~ \sum_{x_{(0)},f(x_{(0)})=0 }\frac{1}{|\det(\partial f(x_{(0)}))|} \delta^n(x-x_{(0)}),$$ where the sum$\sum$is over all zeroes$x_{(0)}$of$f$, ... 6 It depends what you want to calculate. As you rightly note, delta functions are not dimensionless, so that including one in your integral will change its dimensionality: you will be calculating something rather different! Most of the time this won't matter if you do it right, but you do need to think about what you want to calculate. The integral$\int ...

6

That's equivalent simply to $c\int dx/x$. Switch to the Euclidean spacetime, $k_0=ik_4$ where $(k_1,\dots k_4)$ is $k_E$; i.e. analytically continue in $k_0$ (Wick rotation). The integral is $$\int \frac{i\cdot d^4 k_E}{(2\pi)^4} \frac{1}{(k_E^2)^2} \exp(ik\cdot \epsilon)$$ So it's proportional to the Fourier transform of $1/k_E^4$. The original function is ...

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