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36

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! :)

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 ...

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 ...

9

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 ...

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

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 ...

7

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 $$\lim_{\epsilon\searrow 0} ... 7 Here we will assume that OP is not questioning the fundamental physical principles/postulates/axioms of quantum mechanics, such as, e.g., the need to have a Hilbert space H in the first place, etc; and that OP is only pondering the role of L^2-spaces (as opposed to, e.g., L^1-spaces). Let us for concreteness and simplicity consider the 3-dimensional ... 7 If the functional derivative$$\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). $$OP's functional integral ... 6 Technically, the equation$$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}$$In other words, a quantity ... 6 It depends what you want to calculate. As you rightly note, delta functions are not dimensionless, so that including one in your integral will change its dimensionality: you will be calculating something rather different! Most of the time this won't matter if you do it right, but you do need to think about what you want to calculate. The integral \int ... 6 This is not an equality, strictly speaking. Looks like your lecturer used spherical coordinates. If the integrand is spherically symmetric, i.e. it only depends on the magnitude of \mathbf{p}, then the integration over the angular coordinates is trivial and just gives you the solid angle subtended by a sphere, 4\pi. 5 It's an integral over a closed contour (which is topologically a circle). An example from Wikipedia:$$ \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} . $$5 Sorry, a solid angle is something different than an ordinary angle, see http://en.wikipedia.org/wiki/Solid_angle so it is not measured "with respect to anything". Solid angle \Omega measures the size of a set of directions in the 3-dimensional space via the formula$$ \Omega = \frac{A}{R^2} $$where A is the area of the intersection of all these ... 5 There's no paradox. We are on physics stackexchange, not mathematics stack exchange. Non-measurable sets are purely mathematical concepts that cannot be physically instantiated. Any medium in our universe is either made out of particles that are discrete or fields which, as far as we know, can be modeled as being continuous in our 4 dimensional space-time. ... 5 1) As OP basically notes, an n-dimensional delta function transforms under change of variables f:\mathbb{R}^n \to \mathbb{R}^n with (the absolute value of) an inverse Jacobian$$ \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)}), $$where the sum \sum is over all zeroes x_{(0)} of f, ... 5 That's equivalent simply to c\int dx/x. Switch to the Euclidean spacetime, k_0=ik_4 where (k_1,\dots k_4) is k_E; i.e. analytically continue in k_0 (Wick rotation). The integral is$$\int \frac{i\cdot d^4 k_E}{(2\pi)^4} \frac{1}{(k_E^2)^2} \exp(ik\cdot \epsilon)$$So it's proportional to the Fourier transform of 1/k_E^4. The original function is ... 4 Note that the right spelling is "principal value". The formulae aren't identical but the results are the same whenever both definitions yield a well-defined expression. What matters is that we remove the leading logarithmic divergence on both sides from x=0 and we do so in a symmetric way with respect to x\to -x. If you denote the second ... 4 If the integral I:=\int d\theta on the algebra {\cal A}  of superfunctions f(\theta)=\theta a + b should be 1) a (graded) linear operation, 2) translation invariant, i.e., \int d\theta ~f(\theta+\theta') =\int d\theta~f(\theta), 3) and if the output \int d\theta~ f(\theta) should not depend on the integration variable \theta, then it is ... 4 The Dirac delta function is often defined as the following distribution:$$\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

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 ...

4

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 ...

4

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 ...

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: $$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta v$$ $$thus$$ ...

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 ...

4

In your problem, you need integrals of kind : $I_{2n} = \int x^{2n} e^{- \large \frac{x^2}{a}} ~ dx$ Note first that $I_0 = (\pi)^\frac{1}{2} (\frac{1}{a})^ {-\frac{1}{2}}$ Now, it is easy to see that there is a reccurence relation between the integrals : $$I_{2n+2} = - \frac{\partial I_{2n}}{\partial (\frac{1}{a}) }$$ For instance, I_2 = - ...

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