Tag Info

Hot answers tagged

14

This is a general property of waves. If you have waves reflecting off a clamped point (like waves running on a string that you pinch hard at one point), the waves get phase inverted. The reason is the principle of superposition and the condition that the amplitude at the clamped point is zero. The sum of the reflected and transmitted wave must be the ...


11

I) The substitution $f=r\psi$ is the standard substitution to get a radial 3D problem to resemble a 1D problem, see e.g. Ref. 1. II) From the perspective of the normalization of the wavefunction $\psi(r)$, a $1/r$ singularity of $\psi(r)$ at $r=0$ is fine because $|\psi(r)|^2$ is suppressed by a Jacobian factor $r^2$ coming from the measure in 3D spherical ...


9

In general, boundary conditons must be adapted to the real situation. Zero boundary conditions are just for the sake of simplicity. But they are realistic only when the field is really zero for some definite reason. If the boundary is at infinity, zero boundary conditions means that everything of interest happens in a finite domain and cannot be noticed ...


6

It is a very delicate matter to decide when solutions are unphysical or not. A classic example is Dirac's discovery of anti-particles: he found them as negative energy eigenstates for a relativistic Hamiltonian. A less insightful theorist might have discarded the negative energy solutions as unphysical, although we now know that those solutions mean a great ...


6

I) Initial value problems and boundary value problems are two different classes of questions that we can ask about Nature. Example: To be concrete: an initial value problem could be to ask about the classical trajectory of a particle if the initial position $q_i$ and the initial velocity $v_i$ are given, while a boundary value problem could be to ask ...


6

Different boundary conditions represent different models of cooling. The first one states that you have a constant temperature at the boundary.This can be considered as a model of an ideal cooler in a good contact having infinitely large thermal conductivity The second one states that we have a constant heat flux at the boundary. If the flux is equal zero, ...


6

The two solutions are different because they have different boundary conditions. In the first case, the equation is indeed $$ \frac{\partial^2u}{\partial t^2} = c^2 \nabla^2 u = c^2 \left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) u. $$ Here though we specify $u(t,x=x_0)$ to be some value ...


5

Topological degeneracy is only defined in the thermodynamic limit on a closed manifold. The ground state degeneracy of a finite-sized system or on an open manifold is not "topological", and can not be called topological degeneracy. Considering your examples. (1) The ground state degeneracy is ill-defined with open boundary condition. Because there might be ...


5

One may view both (i) the infinite wall $$\tag{1} V(x)~=~\left\{\begin{array}{ccc}\infty & \text{for} & x>0, \\ 0 & \text{for} & x\leq 0, \end{array} \right. $$ and (ii) the delta function potential $$\tag{2} V(x)~=~A\delta(x),$$ as an appropriate limit of a finite barrier wall $\tag{3} V(x) ~=~ V_0 ...


5

In the Lorentzian case: I am not aware of anyone studying it, and don't know explicit counterexamples off-hand. But I have doubts on the uniqueness. With the Lorentzian case, the nature candidate to draw comparisons with is the wave equation. And we see that on something as simple as the unit square $[0,1]\times [0,1]$, the wave equation with vanishing ...


5

There is a standard book which contains everything about electrostatics, the Laplace/Poisson equation and boundary conditions: Classical Electrodynamics by J. D. Jackson. Get the book from the library of your choice, read all chapters labeled "Electrostatics", and you will find the answers to all your questions (if you are simulating this, you need to know ...


4

The Fourier transform of $y\left(x,t\right)$ from the time to the frequency domain is given by $Y\left(x,\omega\right)=\int_{-\infty}^{\infty}y\left(x,t\right)e^{i\omega t}dt$ and satisfies the differential equation: $$ EI\frac{\partial^{4}Y\left(x,\omega\right)}{\partial ...


4

The second condition is saying that there is no discontinuity in the slope of the rope at the junction. In other words, there is no "kink" in the rope. Imagine if this assumption were to fail in the following way: $$ \frac{\partial D_1}{\partial x}(0,t) = -1, \qquad \frac{\partial D_2}{\partial x}(0,t) = 1 $$ Then near the origin, the rope would look ...


4

The most difficult part is to actually get a set of consistent boundary conditions in the first place - this requires a combination of educated guessing, physical insights, prior experience with related problems, detailed calculations and trial-and-error. In short, it is a bit of an art. However, once you have a set of boundary conditions (as in your case ...


4

A naive guess would be that there isn't any real difference. The theoretical logic behind my guess is that at low-energies FQH states are described by 2+1D Chern-Simons theories, which are topological gauge theories. Although the bulk does not have any local degrees of freedom, the boundary does. This is because in the presence of a boundary $\partial M$ ...


4

To explicitly verify this, one solves the problem for a box of finite depth $V_0$. If you additionally assume the wavefunction and its first derivative to be continuous across the potential step, the solution becomes a matter of Solve the Schrödinger equation in the distinct regions in- and outside of the box. Match $\phi$ and $\phi'$ at the potential ...


4

The domain of the problem is the entire real line, not $[0,L]$. Otherwise the potential would not be specified for $x>L, x<0$. Thus, calculate the total energy of any wavefunction that does not vanish at $x = 0,L$, you will find it to be infinite. Therefore for all eigenstates that do not have infinite energy, i.e. the entire spectrum, the wavefunction ...


4

The charge is distributed uniformly on a spherical surface, but that is a function of the high degree of symmetry on the situation. In general the charge tends to accumulate most strongly near pointy bits and most weakly in depressions in the surface. There are several way to understand this. My favorite is not necessarily the most helpful for a beginner, ...


4

The integral you wrote integrates $\rho$ over the whole space. This is impossible to calculate if $\rho$ is not known in the whole space. For example, when the charge $\rho$ is known only inside some finite region enclosed by a metallic shell, the shell is known to have constant potential $\phi$ on its inner surface. This information is useless in ...


3

Look more closely at what you write. I'm not going to write it with all the partial derivative and Lagrangians, because the confusions are not dependent on that: The requirement for spatial infinity arises because $\int_\mathcal{M} \mathrm{d}\omega \neq \omega$, but $\int_\mathcal{M} \mathrm{d}\omega = \int_{\partial\mathcal{M}}\omega $ (Stokes' theorem). ...


3

This is a typical case of a problem which is clear enough physically speaking, but mathematically messy. Where rigorous results are folkloristically employed to achieve some result which, actually, would need much more care in deriving it... But presumably, mathematical details would not change the physical picture. Here the difference between theoretical ...


3

If the cross product of two vectors is zero, that means that the two vectors are parallel. That is, $\hat{n}$ is parallel to $\vec{E_1} - \vec{E_2}.$ Since $\hat{n}$ is normal to the surface, that means that $\vec{E_1} - \vec{E_2}$ is normal to the surface. That is, the only difference between the two fields lies normal to the surface. Therefore there is no ...


3

If the cross product is zero, the two vectors are parallel (here $\hat n$ and $\vec E_1 - \vec E_2$). If the difference of the two fields has only a normal component the tangential component is zero or: $\vec E_{1,t}-\vec E_{2,t} = 0$ or $\vec E_1 = \vec E_2$. A more detailed way to look at it, is by splitting the vectors into normal and tangential parts: ...


3

Suppose we impose a current density $\newcommand{\j}{\mathbf{J}}\j$, then the resulting electric field $\newcommand{\e}{\mathbf{E}}\e$ is given by $\e = \rho \j$, where $\rho$ is the resistivity. In a perfect conductor, $\rho=0$. So in a perfect conductor with some fixed current $\j$, the electric field satisfies $\e = \rho \j = 0 \j = \mathbf{0}$. I don't ...


3

A $j^\mu_a$ current is conserved if it satisfies the equations $$ \partial_\mu j^\mu_a = 0 \implies \partial_t j^0_a + \partial_i j^i_a = 0 ~~~~~~~~~~~...... (1) $$ The first equation in your question is simply a definition. We can define an object called "charge" as $$ Q_a = \int d^{d-1} x j_a^0 $$ Conservation of the current then implies that the charge ...


3

You should not imagine a virtual photon as an individual object wandering from one charged particle to another. This picture is simply inapplicable. Unfortunately, Feynman diagrams mislead people to imagine such things. Actually, Feynman diagrams are good for calculation and bad for imagination. Feynman diagrams have been introduced to help physicists to ...


3

Consider the case of a free scalar field, governed by the usual Lagrangian, $$\mathcal{L} = \frac{1}{2}\partial_\mu \phi \partial^\mu \phi - \frac{1}{2}m^2 \phi^2$$ The propagator, or equivalently Green's function for the theory is a function which can be though of as a response when we use a delta function as an input in the equations of motion, i.e. ...


3

You may think this way: take a perfect infinite crystal where Bloch theorem perfectly work and add potential which makes real crystal finite. Next question you may ask how this potential is "seen" by quasiparticles which have been obtained from infinite crystal consideration. This procedure is perfectly self-consistent and is applicable in all cases. Also, ...


3

You cannot have a total vorticity with periodic boundary conditions, since if you take a path around all of your vortices, it will have a non-zero circulation. But you have periodic bc, so you can continuously deform that path to a point, and a point has zero circulation. Mirror images are not quite the same as in electrostatics. We want periodic boundary ...


3

Dear James, there is no reason why $\psi$ should be periodic. First, if you have a problem, imagine that $\psi$ are just auxiliary variables but the true ones are the bilinears $\psi_i \psi_j$ which are still periodic. Only things such as the world sheet stress-energy tensor $T_{++}$ and $T_{--}$ have to be periodic and they are because they're bilinear in ...



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