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18

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 ...

10

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 ...

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 ...

7

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 ...

7

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 ...

7

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 ...

6

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 space $H$ 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 ...

5

It's an integral over a closed contour (which is topologically a circle). An example from Wikipedia: \begin{align} \oint_C {1 \over z}\,dz & {} = \int_0^{2\pi} {1 \over e^{it}} \, ie^{it}\,dt = i\int_0^{2\pi} 1 \,dt \\ & {} = \Big[t\Big]_0^{2\pi} i=(2\pi-0)i = 2\pi i, \end{align} .

5

First choose a direction for u, along the z-axis. Then the integral is $$I = \int {1\over (x^2 + y^2 + A z^2 + B)^{5/2} } dx dy dz$$ Rescale z by $\sqrt{A}$ to get rid of A and restore rotational invariance. $$I = {1\over \sqrt{A}} \int {1\over (x^2 + y^2 + z^2 + B)^{2.5}} dx dy dz$$ Now you do find the B dependence immediately from rescaling x y and ...

5

Sorry, a solid angle is something different than an ordinary angle, see http://en.wikipedia.org/wiki/Solid_angle so it is not measured "with respect to anything". Solid angle $\Omega$ measures the size of a set of directions in the 3-dimensional space via the formula $$\Omega = \frac{A}{R^2}$$ where $A$ is the area of the intersection of all these ...

5

So, the properties of the derivative of the delta function can be shown relatively quickly though the following ansatz: Consider a function $\delta(x)$ such that $\delta(x) = \frac{1}{a^{2}}(x+a)$ if $-a<x<0$ and $\delta(x) = \frac{1}{a^{2}}(a-x)$ if $0<x<a$, and $\delta(x) = 0$ elsewhere. It is easy to see that $\delta(x)$ has area 1 ...

5

There's no paradox. We are on physics stackexchange, not mathematics stack exchange. Non-measurable sets are purely mathematical concepts that cannot be physically instantiated. Any medium in our universe is either made out of particles that are discrete or fields which, as far as we know, can be modeled as being continuous in our 4 dimensional space-time. ...

5

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 $f$, ...

5

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 $\int ... 5 That's equivalent simply to$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 ... 4 The Dirac delta function is often defined as the following distribution: $$\int_a^b \delta(x - x_0) F(x)\mathrm{d}x = \begin{cases}F(x_0), & a < x_0 < b \\ 0, & \text{otherwise}\end{cases}$$ where$F$is a suitable test function. Its derivative is then defined as $$\int_a^b \delta'(x - x_0) F(x)\mathrm{d}x = -\int_a^b \delta(x - x_0) ... 4 If the integral I:=\int d\theta on the algebra {\cal A} of superfunctions f(\theta)=\theta a + b should be 1) a (graded) linear operation, 2) translation invariant, i.e., \int d\theta ~f(\theta+\theta') =\int d\theta~f(\theta), 3) and if the output \int d\theta~ f(\theta) should not depend on the integration variable \theta, then it is ... 4 That looks correct to me. Consider the basic property of the delta functions$$ \int dx f(x) \delta(x-a) = f(a). $$Nothing forbids f(x) to be a composite function, for example f(x) \equiv g(x)\delta(x-b), so f(a) = g(a) \delta(a-b). Hence we get,$$ \int dx f(x) \delta(x-a) \equiv \int dx \, g(x)\delta(x-b) \delta(x-a) = g(a)\delta(a-b). $$4 So you were on the right track with integrating over r and over t. Here's how you could do it: The acceleration at any radius, r (if we assume Earth is a point mass) is:$$a=-{GM\over r^2}$$The minus sign is because the acceleration is anti-radial. Then you can do the following:$$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta vthus$$... 4 Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough$$ E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)} $$For your assumption that the motion is strictly radial and downwards you have v(t) = dr(t)/dt < 0 so you can solve for dr(t)/dt and get an ordinary first order ... 3 In theory: Lebesgue integrable functions form a Banach space, whereas Riemann integrable functions do not. This causes problems in, e.g., Quantum mechanics if we try to work with Riemann integrable functions instead of Lebesgue integrable functions... (We want the functions to form a Banach space so we can use good old fashioned linear algebra to solve ... 3 Grassman d\theta has opposite mass dimension to \theta, which is why the notation is not 100% optimal, it confuses on this issue. But if you know how to evaluate the integral, that it goes like the derivative, then you know how change of scale works, and it's the opposite of normal change of scale:$$\int d(k\theta) f(k\theta) = {1\over k} \int d\theta ... 3 The decomposition$p=\lambda \tilde\lambda$is only valid for null vectors, indeed. In loop integrals, the loop momentum may generally be off-shell but because of the delta function in these integrals, only the null values of the momentum contribute, so it's enough to deal with the null momenta and they can be factorized to the spinors in this way. The ... 3 Note that the right spelling is "principal value". The formulae aren't identical but the results are the same whenever both definitions yield a well-defined expression. What matters is that we remove the leading logarithmic divergence on both sides from$x=0$and we do so in a symmetric way with respect to$x\to -x$. If you denote the second ... 3 As far as I know, there are a lot of kinds of density - you mentioned volume density, but there is linear density (amount of mass for unit of length), for example. If your string is very-very thin, there is no sense in definition that density = mass / volume, cause very thin string doesn't have volume and you should use linear density = mass / length or ... 3 It is simply a matter of notation. The$p_1$(and hence$E_1$and$E_2$) in $$\int d\Pi_2=\int d\Omega\frac{p_1^2}{16\pi^2E_1E_2}(\frac{p_1}{E_1}+\frac{p_1}{E_2})^{-1}$$ is no longer an integration variable; it has the fixed value that satisfies the delta function$\delta(E_{cm}-E_1-E_2)$in the previous integral. The factor ... 3 As David pointed out in his answer, the circuits are called the differentiator and the integrator respectively. They use operational amplifiers (op amps) to do this. An op-amp isn't necessarily a "simple circuit" to build, since it consists of several transistors and hundreds of resistors and capacitors, but since it's extremely easily available as an IC, ... 3 Define the LHS of the equation above: $$I=\int d^d q\frac{1}{(q^2+m_1^2)((q+p_1)^2+m_2^2)((q+p_1+p_2)^2+m_3^2)}$$ The first step is to squeeze the denominators using Feynman's trick: $$I=\int_0^1 dx\,dy\,dz\,\delta(1-x-y-z)\int d^d q\frac{2}{[y(q^2+m_1^2)+z((q+p_1)^2+m_2^2)+x((q+p_1+p_2)^2+m_3^2)]^3}$$ The square in$q^2$may be completed in the ... 3 Let's first assume that "how much fluid" means "the mass of fluid." Let$\rho(t, \mathbf x)$denote the volume mass density of the fluid at time$t$and position$\mathbf x$in space and let$\mathbf v(t, \mathbf x)\$ denote the velocity vector field of the fluid. The vector field $$\mathbf j = \rho\mathbf v$$ gives the mass per unit area, per unit time ...

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