# Tag Info

41

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 ...

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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 $\mathrm{d} ... 12 No, it cannot be enough. Stokes' theorem says that the volume ($\Omega$) integral of$d\omega$, a form that is the exterior derivative of another one (of$\omega$), may be written as a surface integral. But it doesn't allow us to rewrite the volume integral of a general integrand (which isn't the exterior derivative of anything) such as the Lagrangian ... 11 In German, this property is known as the Transformationssatz, but I do not know any appropriate translation for it. This is, however, a special case of coordinate tranformations changing the measure by the determinant of their Jacobian, since obviously$\frac{\partial y_i}{\partial x_j} = A_{ij}$. That it is the determinant that plays a role in the ... 10 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 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 ... 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 ... 9 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 ... 8 There are two that I know of in the context of state estimation. The first is for estimating the mean of$Pand 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 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 $$\lim_{\epsilon\searrow 0} ... 8 a_x \Delta t = \Delta v_x = v_{xf} - v_{xi} \Delta x = v_{x,average}\Delta t = v_{xi}\Delta t + \dfrac{1}{2}a_x (\Delta t)^2 \Rightarrow v_{x,average} = v_{xi} + \dfrac{1}{2}a_x \Delta t = v_{xi} + \dfrac{1}{2}(v_{xf} - v_{xi}) = \dfrac{v_{xf}+ v_{xi}}{2} Is there a geometric interpretation or does it just work out mathematically? 7 Technically, the equation$$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}$$In other words, a quantity ... 7 It's an integral over a closed line (e.g. a circle), see line integral. In particular, it is used in complex analysis for contour integrals (i.e closed lines on a complex plane), see e.g. example pointed out by Lubos. Also, it is used in real space, e.g. in electromagnetism, in Faraday's law of induction (part of the Maxwell equations, written in an ... 6 1) As OP basically notes, an n-dimensional delta function transforms under change of variables f:\mathbb{R}^n \to \mathbb{R}^n with (the absolute value of) an inverse Jacobian$$ \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)}), $$where the sum \sum is over all zeroes x_{(0)} of f, ... 6 It depends what you want to calculate. As you rightly note, delta functions are not dimensionless, so that including one in your integral will change its dimensionality: you will be calculating something rather different! Most of the time this won't matter if you do it right, but you do need to think about what you want to calculate. The integral \int ... 6 In theory: Lebesgue integrable functions form a Banach space, whereas Riemann integrable functions do not. This causes problems in, e.g., Quantum mechanics if we try to work with Riemann integrable functions instead of Lebesgue integrable functions... (We want the functions to form a Banach space so we can use good old fashioned linear algebra to solve ... 6 This is not an equality, strictly speaking. Looks like your lecturer used spherical coordinates. If the integrand is spherically symmetric, i.e. it only depends on the magnitude of \mathbf{p}, then the integration over the angular coordinates is trivial and just gives you the solid angle subtended by a sphere, 4\pi. 6 WARNING: The function is not absolutely integrable for n>1, so the integral strongly depends on how you decide to compute it if you break the integration into iterated integrals. Use instead cylindric coordinates. k = (z, \vec{r}), where \vec{r} \in \mathbb R^{n-1} and z\in \mathbb R. You have this way, assuming that x is directed along z: ... 6 A simple reference problem Suppose we want to analyse the problem of a forced harmonic oscillator. Denote as \phi(t) the time dependent position of the oscillator. The oscillator experiences two forces, the spring force -k\phi(t) and an external force F_{\text{ext}}(t). Newton's law says$$ \begin{align} F(t) &= m a(t) \\ -k \phi(t) + ... 5 Hint: It seems the substitutionz \longrightarrow z^{\prime}=r-z$will work to prove the second equality of eq. (8). (Geometrically, this corresponds to a reflection of the$z$-axis around$z=\frac{r}{2}$.) 5 You can do even better. The ham sandwich theorem says you can simultaneously bisect any three volumes in space with a single plane. You can have a plane through the center of mass that bisects the mass and the surface area. Make one volume a tiny sphere around the center of mass, one a very thin outer skin, and the other the rest of the object. The proof ... 5 1. Since$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 ... 5 It is just a more compact notation. It is implied by the integration element$dA$that you are integrating over the surface. 5 Since the force is a function of distance, you need to integrate: $$F = kx\\ W = \int F\ dx\\ W = \int k\ x\ dx\\ W = \frac12kx^2$$ Add signs as needed... Your work considered the force to be constant - and that's not how springs work. 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 ... 5 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: $$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta v$$ $$thus$$ ... 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 ...

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