Tag Info

Hot answers tagged

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! :)


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 ...


13

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 ...


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

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 ...


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

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

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

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

$a_x \Delta t = \Delta v_x = v_{xf} - v_{xi}$ $\Delta x = v_{x,average}\Delta t = v_{xi}\Delta t + \dfrac{1}{2}a_x (\Delta t)^2$ $\Rightarrow v_{x,average} = v_{xi} + \dfrac{1}{2}a_x \Delta t = v_{xi} + \dfrac{1}{2}(v_{xf} - v_{xi}) = \dfrac{v_{xf}+ v_{xi}}{2}$ Is there a geometric interpretation or does it just work out mathematically?


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

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

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

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

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

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

Hint: It seems the substitution $z \longrightarrow z^{\prime}=r-z$ 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

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

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 ...


5

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 ...


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

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



Only top voted, non community-wiki answers of a minimum length are eligible