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15

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 ...


13

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 ...


11

Indeed the problem with boundary conditions, generally speaking, is not well-posed. There are boundary conditions admitting no curves or admitting many curves, satisfying both these conditions and Euler-Lagrange equations. Examples. (1) Think of a particle constrained to stay on a smooth sphere where it can freely move. If you assign the North and the ...


10

Not necessarily. Consider this function as an example: $$\psi(x) = \frac{C\sin x^2}{\sqrt{x^2 + 1}}$$ This function is square-integrable and asymptotes to zero as $x\to\pm\infty$, but its derivative goes to $2\cos x^2$ in the same limit. In quantum mechanics, we often assume that real systems are represented by wavefunctions which have no interesting ...


10

Loosely speaking, the gradient of a scalar field (such as the electrostatic potential) points in the direction of that field's greatest change. Since no change occurs in the field when you go along the surface, the gradient shouldn't have a component in that direction. Here is another intuitive explanation: Imagine for a moment that the electric field was ...


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 ...


8

Particles do not minimize their action. Instead, they minimize their action given certain boundary conditions. We can only apply the action principle when we know the start and end points ahead of time. If we know that a particle will be at location $x_i$ at time $t_i$ and that it will be at location $x_f$ at time $t_f$, then the particle takes a path of ...


7

Actually, the extra path is not irrelevant. If you put a light bulb at A and a $4\pi$ detector (this means $4\pi$ steradian coverage, i.e. it detects incoming light in any direction) at B, the detector will see light along both paths: direct, and bounced off the mirror, which is exactly the result you got from Fermat's Principle. If you want to exclude the ...


7

This is version two of my proof. The OP discovered a sign error in my first attempt that revealed my argument to be circular. The correct proof is below. Not surprisingly, this has to do with the signature of the spacetime metric not being positive definite. Furthermore, this issue is very subtle. It should be noted that this result is quoted partially in ...


7

In the examples you've given, the boundary conditions simply say "don't have infinite energy" and "don't be non-normalizable". These don't really have a physical interpretation. Moreover, no boundary conditions ever have an interpretation like "initial position and velocity" because the time-independent Schrodinger equation describes stationary states. ...


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

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 ...


6

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 ...


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

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 ...


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 ...


6

The reason is the same as why the electric field inside a conductor is zero: if it isn't zero, the free electrons undergo a force and move (rearrange) until they don't feel a force any more. If the electrons don't feel a force, the electric field must be zero. At the surface of a conductor, the free electrons feel a force perpendicular to the surface, but ...


6

The book's sentence "we can always force a system to have discrete states by enclosing it in a sufficiently large box" is perhaps a little unclearly worded. Putting the system into a box of any size will force the system to have discrete states. The author's point is that it's always possible to choose a box size that's big enough that the box's existence ...


6

Einstein's equation is $$G_{ab} = 8 \pi T_{ab}.$$ The left-hand side of the equation, $G_{ab}$, is called the Einstein tensor. It is an expression involving second derivatives of the metric, $g_{ab}.$ The right-hand side of the equation, $T_{ab}$, is called the Stress-Energy tensor. Each stress-energy tensor is associated with a unique matter ...


6

From the Wikipedia article on sound: In physics, sound is a vibration that propagates as a typically audible mechanical wave of pressure and displacement. To fully understand how is air vibrating in an open pipe, you have to consider not only the acoustic pressure wave, $$\frac{\partial^2 p}{\partial x^2}=\frac{1}{c^2}\frac{\partial^2 p}{\partial t^...


5

The usual derivation of the Euler-Lagrange equations forces us to assume that both the initial and final conditions are fixed. However, when one actually derives the equations, he sees that there are differential equations in time so from the knowledge of the initial state, including the velocities (or whatever derivatives are needed to specify the initial ...


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 1_{[0,a]}(x)=\left\{\begin{array}...


5

Multiple classical solutions to Euler-Lagrange equations with pertinent/well-posed boundary conditions (such solutions are sometimes called instantons) are a common phenomenon in physics, cf. e.g. this related Phys.SE post and links therein. In optics, it is well-known that already e.g. two mirrors can create multiple classical paths.


5

This is, unfortunately, not a simple task in general. My experience on non-reflecting boundary conditions is for the Navier-Stokes equations, but you should be able to do a similar approach for your system. As you noted, a fixed boundary ($u=0$) will lead to one type of wave while a free boundary ($\partial u/\partial x = 0$) leads to another wave. What you ...


5

The assumption missing in all these statements is that there are boundary conditions assumed to be given. E.g. for Poisson's equation $\Delta f = \rho$, the solution is unique for Dirichlet and/or Neumann boundary conditions, see e.g. section 1.9 in Jackson's "Classical Electrodynamics".


4

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 ...


4

By mirroring $V(x)$ about $x = 0$, i.e., by setting $V(-x) = V(x)$, the wavefunction can be taken to be even or odd. The even solution satisfies the Neumann boundary condition since the derivative of an even function is odd and thus zero at $x = 0$.


4

Since this has just been asked again, let me attempt an intuitive explanation. The real explanation is of course to match $\vec{E}$ and $\vec{B}$ at the interface and the direction of the reflected wave drops out, but this isn't especially intuitive. Let's calculate the ratio $E_r/E_i$ as a function of the ratio $n_t/n_i$, and let's start with the ...


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 $\partial M$ ...



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