9

Structurally, it looks like this: Start by rescaling one or both of the position/momenta pairs such that the kinetic-energy term has both masses equal (while also retaining $[x_i,p_i]=i\hbar$), and then find a rotation in the rescaled $x_1,x_2$ plane that will eliminate the coupling terms. That rigid rotation will be mirrored in the momentum plane, but ...


7

A mode in physics is, generally speaking, the spatial part of a waveform. A Fourier mode, more specifically, is a wave that oscillates sinusoidally in space. Thus, when we write the Fourier transform of a wave function $f(\mathbf r,t)$ as $$ f(\mathbf r,t) = \int\tilde f(\mathbf k,t)e^{i\mathbf k\cdot \mathbf r} \: \mathrm d\mathbf k, $$ the quote you give, ...


6

On the top of Figure we have $\:n+1\:$ ideal springs and $\:n\:$ particles in equilibrium. The constants of the springs are $\:k_{\rho}\: (\rho=1,2,\cdots, n+1) \:$ with equilibrium lengths $\:\ell_{\rho}\:(\rho=1,2,\cdots, n+1 )\:$ and the particle masses $\:m_{\rho}\:(\rho=1,2,\cdots, n)$. Disturbing the system from this equilibrium, the equation ...


6

There's a sign error in your equations of motion. The Lagrangian of the system will be $$L=T-U= \frac{m}{2} \left( \dot{x_1}^2 + \dot{x_2}^2 \right)-\frac{k}{2} \left( L + x_1 - x_2 \right)^2$$ So the equation of motion for $x_1$ is: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1}-\frac{\partial L}{\partial x_1}=0 \\ m\ddot{x}_1+k( L + x_1 - x_2 )=0 \\ m\...


5

First of all, to have standing waves you must be talking about a wave carrying system with spatial extent: something like a guitar string. Such a system has a set of possible vibrational modes. The first two modes are shown in the figure. We can describe each mode shape with a function $\phi_n(x)$ where $n$ is a label that indexes the modes, ie. $n$ is an ...


5

As usual, whenever you have a complex-valued amplitude $\tilde{\mathbf{E}}(\mathbf r)$, it's a stand-in for a physical field that's obtained as its real part, i.e. $$ \mathbf E(\mathbf r,t) = \mathrm{Re}\left( \tilde{\mathbf{E}}(\mathbf r)e^{-i\omega t}\right). $$ The presence of nontrivial imaginary parts of $\tilde{\mathbf{E}}(\mathbf r)$ signifies ...


5

The Hamiltonian is a quadratic form both in the momenta and in the coordinates. You can decouple the problem by using the principal axis theorem on the coordinates to obtain generalized coordinates. For this, first a scaling of the coordinates to equalize the coefficients of the kinetic energy is necessary. The whole approach corresponds to "diagonalizing" ...


4

The best thing to do is to look at the eigenvectors. This will tell you all you need to know. Anyway, $0$ corresponds to all the masses rotating in unison. Another mode corresponds to neighbouring masses moving closer and farther away from each other, symmetrically. Here you can choose a generalized coordinate $x$ to be the distance between one pair of ...


4

The equation that dictates the behavior of mechanical systems such as oscillators is Newton's Second Law. It is a second order differential equation, and in the case of oscillators it is linear and has the general form $$\ddot{\vec{x}} = A\vec{x} .$$ This means that for each degree of freedom you have, another coordinate will be assigned to it and the ...


4

My first instinct as a physicist is always to ask whether symmetry impacts you at all. From my understanding, you had three ions spaced in a quadratic potential (and, I assume, interacting with each other). It seems clear to me that you must have started in a configuration like so: | o --- o --- o | That is, the ions would have been arranged ...


3

"Normal" in the context of oscillators simply means "periodic" – periodic solutions and the frequencies and other aspects associated with them. It's like in "he breathes normally" – the breathing seems to be periodic. "Quasinormal ones" are those whose time dependence is $\exp(-\Gamma t) \sin (\omega t)$, i.e. they have some exponential decrease aside from ...


3

It's true that a hollow conductor has a minimum cutoff frequency. However, a hollow conductor is not a black body. A black body has perfect absorption of radiation at all frequencies, while a perfect conductor perfectly reflects all radiation. A black body emits radiation according to the Planck law. Since the black body absorbs all incoming radiation, but ...


3

It's not that the final solution looks like that. Rather, you are looking for all the solutions of that form (normal modes) for two reasons: They are easy to find You can afterwards decompose any motion into a sum of normal modes. This comes from writing your equations of motion in the normal basis*. So first you work out the normal modes by assuming a ...


3

I guess this question might spawn a very large spectrum of answers, I'll try to take on the (simple) mathematical side of dealing with a system of first order ODE instead of an higher order one. In this sense (as in many others!) the Hamiltonian formulation of a dynamical problem, like the one you have posed as an example, gives in an easier way plenty of ...


3

You can use the orthogonality of the sinusoids to do the inversion. First, multiply both sides by $\sin (l a n)$ where $l$ is an integer: $$y_n(t) \sin(l a n) = \sum_k A_k(t) \sin(k a n) \sin(l a n).$$ Now sum both sides over $n$ $$\sum_n y_n(t) \sin(l a n) = \sum_{k,n} A_k(t) \sin(k a n) \sin(l a n).$$ On the right hand side, the sum over $n$ gives $\...


3

Another method to solve the problems of this kind is to consider operators of the form $$ \hat{a} = c x_1 + d x_2 + e P_1 + f P_2. $$ Here $c$, $d$, $e$ and $f$ are numerical coefficients. If you find solutions of the equation $$ \left[ \hat{a}, H \right] = \varepsilon \hat{a}, $$ you will get creation and annihilation operators and energies of quanta.


3

It's been a long time since I dealt with normal modes (and it was on a completely different system), so I hope I won't get this completely wrong :-) First of all, the two equations you write at the end of your question is essentially the same equation written differently. By solving Eq.(2) you get eigenvalues (which are linked to the frequencies) and ...


3

The "$n$" in the $J_n(\kappa r)$ refers to the number of nodes in the angular direction. A complete set of eigenfunctions of $-\nabla^2$ in ${\mathbb R}^2$ are $$ \psi_{n,\kappa}(r,\theta)= e^{in\theta}J_n(\kappa r), \quad \kappa\in [0,\infty) $$ with eigenvalue $\kappa^2$.


2

This is the classical treatment to model vibrations in solids, using the analogy with vibrations of a one-or-two dimensional monatomic or diatomic chains. Which basically boils down to writing Newton's equation of motion to find out the force on each mass when the whole system constitutes of masses attached by Hookean springs, i.e. for our purposes the ...


2

The masses moving in a circle have to have a net force pulling towards the center. The force will be from the vector sum of the two springs that each mass is attached to. The force components tangential to the circular path will cancel but the components in the radial direction will add and point inwards. This will require that the springs stretch a little ...


2

The exponential dependence comes about because we are breaking the oscillatory motion into sines and cosines. Why do we choose sines and cosines? The reason is that normal modes occur about some equilibrium, and an equilibrium is a local minimum in the energy. The Hamiltonian or Lagrangian can then be Taylor expanded about this equilibrium point, and the ...


2

I'm not sure where you get: $$p(r) = A\frac{\cos kr}{r} + B\frac{\sin kr}{r}$$ ... from. The wave equation in cylindrical coordinates is: $$\frac{1}{c^2}u_{tt}=u_{rr}+\frac{1}{r}u_r+\frac{1}{r^2}u_{\theta \theta}+u_{zz}$$ With Ansatz: $$u(r,\theta,z,t)=R(r)\Theta(\theta)Z(z)T(t)$$ Separation: $$\frac{1}{c^2}R\Theta ZT''=\Theta Z T R''+\frac{1}{r}\Theta ...


2

There is a well-known isomorphism between the linear space ${\mathcal M}_{m, n}$ of $m\times n$ matrices and typical (vectorial) linear spaces ${\mathcal L}_{m\times n}$ of dimension $\text{dim}({\mathcal L}_{m\times n}) = m\times n$. Everything that is valid in ${\mathcal L}_{m\times n}$ has an equivalent in ${\mathcal M}_{m, n}$ and conversely. For this ...


2

I think the explanation is right there in the text, but it might take a bit of translation. First, notice that the time dependence of a mode field is $e^{-i\omega t}$. A mode has a single frequency. Next notice that the "mode function" obeys the eigenvalue equation and subsidiary conditions which I take to mean boundary conditions. A mode function is a ...


2

There are no restoring forces in the springs: this is easy to see, ...... I do not think that this statement is correct. The rotational mode has stretched springs but the masses do not vibrate whereas the breathing and two clapping modes do have the masses vibrating about the centre of mass. It is those stretched springs which provide the centripetal ...


2

You ask, for water containers across a variety of volumes, "...the range of notes appears to be identical regardless of vessel shape. Why? And can you describe exactly what is making the sound?" The short answer is they are definitely not identical. However, the container-dependent effects on the sound are small in comparison to the source-dependent ...


2

Flat torus is just the mathematical term for a flat space with periodic boundary conditions. A space-like $3$-torus is therefore three-dimensional ordinary space ("ordinary" meaning "without a time-like direction" here) with periodic boundary conditions, just like the book says :) You probably think "donut" when you hear the word torus. Imagine cutting such ...


2

You are correct. Normal modes are eigenmodes of the system, and thus are decoupled from one another. Assume the following differential equation $$\ddot{\boldsymbol{x}}=A\boldsymbol{x}$$ where $A$ is a symmetric matrix. It is always possible to define new coordinates $\boldsymbol{x}^{\prime}$ where the matrix is diagonal and thus the above equation is $$\...


2

The different modes of electromagnetic waves in a cavity correspond to different eigenfunctions of the solutions of the electromagnetic wave equation given the boundary conditions of the walls of the cavity. The eigenvalues correspond to the frequencies of the modes. For example, the wave equation for the electric field given as a phasor with a given ...


2

Imagine with that we have a standing electromagnetic wave inside a cavity, as on page 11 in the book by Gerry & Knight. This cavity supports electromagnetic field modes of many different frequencies, which satisfy the given boundary conditions. Now suppose that we look at a specific frequency $\omega$ i.e. a specific standing wave which is called a ...


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