Tag Info

Hot answers tagged

5

You see, when you have a pendulum with friction you account it by including a force $\vec{F}_r=-b\vec{v}$. Then your differential equation for the pendulum is $$ml\ddot{\theta}=-mg\theta-bl\dot{\theta}\iff\ddot{\theta}+\frac{b}{m}\dot{\theta}+\frac{g}{l}\theta=0$$The solution of this differential equation depends on the values of $b$, $m$ and $l$ but since ...


3

your result is correct $$ [a_k, a_q] = -2 a_k a_q $$ which is consistent with $$ [a_k, a_k ]= - 2 a_k a_k = 0 $$ because $$ a_k a_k = \frac{1}{2}\{a_k, a_k \} = 0 $$ And in general you can use $$ [A,B] = 2AB - \{A,B\}$$ which would also give $$[a_k^\dagger, a_q] = 2a_k^\dagger a_q - \delta_{kq}$$


3

Repeatedly applying this relation to the ground state is exactly what you need to do. There's nothing more to it.


3

I don't think you need quantum mechanics to understand what's going on in dipole-induced dipole interaction. The basic mechanism is quite simple and just the details of the calculations change by switching to a quantum description. Polarizable molecule in an external field So first things first. Let us consider a simple model of polarizable molecule as ...


3

This formula is actually Euler's formula for planar graphs, and holds for all Feynman diagrams regardless of what theory we are in. The proof proceeds by induction and is easy if we first disregard the case of crossing lines: Observe that a one-loop graph has two vertices, one loop, and two internal lines, so the formula holds. Observe that a ...


3

Line (2) states that the Fourier transform of a function is zero. Then that function is also zero, at least almost everywhere.


3

.... We are told that all the subsequent collisions involving the balls and floor are elastic. We are asked to determine the maximum height to which the small sphere will rise on the rebound. The problem does not mention any radii, but if we did know the radius of each of the spheres, would it be valid to bypass conservation of linear ...


2

The integrals are difficult but not impossible, unless I've made a mistake with WolframAlpha. The result is: $$E = \frac{\sigma}{\pi \epsilon_0} \arctan\left( \frac{ab}{4r\sqrt{(a/2)^2+(b/2)^2+r^2}} \right)$$ When $a,b \to \infty$ the whole arctangent goes to $\pi/2$ and we recover $E=\frac{\sigma}{2\epsilon_0}$, which is definitely encouraging. And I ...


2

The distribution of speeds in an ideal gas is given by the Maxwell-Boltzmann distribution. There are a variety of average speeds e.g. the most probable speed, the mean speed and the root mean square speed. Which one you use will depend on the application. The two equations you give are for the RMS speed: $$ \sqrt{\langle v^2 \rangle} = \sqrt{\frac{3kT}{m}} ...


2

The OP is correct in stating that the Fourier transform $$\xi(\omega) = \int\mathrm{d}t\, \mathrm{e}^{\mathrm{i}\omega t} \xi(t), $$ vanishes upon averaging over realisations, $\langle \xi(\omega)\rangle = 0$, so long as we assume that the noise is also zero on average in the time domain, $\langle \xi(t)\rangle = 0 $. However, the noise is not only ...


2

The current is the conventional current in the opposite direction to the electrons current but they are the same thing , the current is the same in series connections. If the two lamps are identical they will give the same amount of light.


2

Since the lamps are in series, the electric current through each is identical; all of the current out of one lamp is in to the other lamp; if there is a flow through one, there is a flow through the other. It cannot be that there is a flow through one and not the other (in the context of this simple model). Thus, if the lamps are identical, their ...


2

You idea is exactly right. With that diagram, you need to figure out where the center of gravity is. The position of the center of gravity is a direct function of the number of bricks. In general, when you have $n$ bricks with equal offsets, the center of gravity is exactly midway between bricks $1$ and $n$ (by symmetry). The first brick is in contact with ...


2

You assume that the screen is perfectly absorbing when you use this field. By uses this field you are assuming that the electromagnetic field pass through the material without reflection. Furthermore, you are assuming that you don't have field in the another side of the surface. $$ \textbf{E}=E_0e^{i(kx-\omega t+\delta)} \hat{\textbf{y}}, \ \ x<0 $$ $$ ...


2

By normalization condition you get$$\int_0^{2\pi}\frac13+c^*c+\frac1{\sqrt3}c^*e^{i \phi} + \frac1{\sqrt3}ce^{-i \phi}=2\pi$$ Now we know that $e^{i\theta}=\cos{\theta}+i\sin{\theta}$ thus its integration over a period of $2\pi$ is 0. Thus our equation reduces to $$cc^*=\frac{2}{3}$$ Thus any complex number who's magnitude or modulus is $\sqrt{\frac{2}{3}}$ ...


2

By the very functioning of an open switch, there can be no current through the switch, and there is no constraint on the voltage across the terminals of the switch. This of course means that there cannot be a current through the entire branch containing R4 (and therefore the resistance R4 plays no role in this case). The current then flows through the rest ...


2

The equation $$\ddot{r} = -\mathbf{g},$$ is valid iff $\dfrac{h}{R_e} <<1$. The gravitational force is : $$\mathbf{F} = -m\dfrac{GM_e}{R^2}\mathbf{\hat{R}}.$$ Now one defines $r = R - R_e$ with $\dfrac{r}{R_e} <<1$. Then one has : \begin{align} \mathbf{F} &= -m\dfrac{GM_e}{R^2_e(r+R_e)^2}\mathbf{\hat{R}}\\ & = ...


2

You are specifically asking about a first order correction to the formula. Starting from $$F = \frac{GMm}{(R+h)^2}$$ for the projectile at height $h$, we can rearrange this as $$F = \frac{GMm}{R^2}\frac{1}{(1+\frac{h}{R})^2}$$ When $h\ll R$ we can use a first order Taylor expansion to write $$F = \frac{GMm}{R^2}\left(1-\frac{2h}{R}\right)$$ Finally, ...


1

Introducing: $a_ia_k^+=\delta_{ik}-a_k^+a_i$ Replacing 3 in 2 and solving: $\langle K|L\rangle=\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+|\rangle=\delta_{ik}\langle|a_ja_l^+|\rangle-\langle|a_ja_k^+a_ia_l^+|\rangle$ $\langle K|L\rangle=\langle|a_ja_ia_k^+a_l^+|\rangle$ $$ =\langle|a_j(\delta_{ik}-a_k^+a_i)a_l^+\rangle $$ $$ =\langle ...


1

Using the rule (3) you can sort the four operators in the order you prefer, by swapping terms. After each swapping, you obtain a piece with four operators, plus an additional one with two operators. And so on. In your case, by swapping the second and the third term, $$a_j a_i a_k^+ a_l^+ = - a_j a_k^+ a_i a_l^+ + \delta_{ki} a_j a_l^+ $$ and by swapping ...


1

The square bracket transformation This is just the application of chain rule. The LHS means a derivative over the primed spacial coordinates while keeping unprimed spacial and time coordinates fixed. $$\nabla'[ \rho(\mathbf{x'},t')]_{ret} = \left(\sum_i \frac{\partial }{\partial x_i'} \hat{i}\right)[\rho(x_i',x_j',x_k',t')]_{ret}\\$$ But the $\rho$ is a ...


1

I'm guessing your questions all amount to whether general relativistic effects become important at the surface of a neutron star. To answer this we can compare the flat space metric (in polar coordinates): $$ ds^2 = -c^2dt^2 + dr^2 + r^2 d\Omega^2 \tag{1} $$ with the Schwarzschild metric that describes the geometry outside a spherically symmetric mass: $$ ...


1

These problems are best solved with the Euler-Lagrange equations. Work in polar coordinates centred about the circle. Then the ring is at $(R,\theta(t))$. Its kinetic energy is, $$T = \frac{1}{2} m R^2 {\dot{\theta}}^2 $$ From the spring there is potential energy. Let $A$ be $(L,\phi)$ in polar coordinates. The potential energy is proportional to the ...


1

Let me expand on (and correct a minor error) in what I said in the alluded-to thread. Unfortunately, I do not know the mentioned article (although "The Moon's Twin", published in 1989 in "The Magazine of Fantasy and Science Fiction", discussing the Jupiter/Io system, might be it). Therefore, I can't directly address it, but I think I can say three major ...


1

When you have a lightly damped oscillator, there is a small correction to the resonant frequency. This is derived in detail on the wiki page for the harmonic oscillator. The form they give is $$\omega = \omega_0\sqrt{1 - \zeta^2}$$ Where the $Q$ (quality factor) of the oscillator is given by $Q=\frac{1}{2\zeta}$.


1

You should start by the definition of temperature $$\frac{1}{T} = \frac{\partial s}{\partial E }= k \frac{\partial \ln \Omega}{\partial E} $$ Also you have the partition function $$Z_1=\sum \exp(-bE_r) \text{ where } b=1/kT $$ Let's now assume a single magnetic dipole in a magnetic field inside a heat tank( we can assume the heat tank as the rest of the ...


1

The partition function $\mathcal Z$ is the sum over all those these exponential indexed by the state index. In German it's called "$\mathcal Z$ustandssumme", literally "state sum". You could device a scheme of indexing the possible overall energies so that you could write it as $\sum_n$, i.e. finding an injection ${\mathbb N}\to{\mathbb N}^2$, but if the ...


1

There are two important points to keep in mind when working through this problem. (1) Since the Hamiltonian for the system changes suddenly, the wavefunction just after the change is the same as the wavefunction just before the change. (2) Then energy eigenstates after the change are different from the energy eigenstates after the change. It follows that, ...


1

If the chain is folded up on itself with the initial horizontal separation then yes it can be ignored. The picture is misleading because it shows a gentle bend connecting the two sides when in idealized reality you have a discontinious sharp bend at the bottom. Remember ideally there is not flexular rigidity (no resistance to bending) and inflexibiliy (no ...



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