# Tag Info

22

In general, the answer is no. This type of inverse problem is sometimes referred to as: "Can one hear the shape of a drum". The following extensive exposition by Beals and Greiner discusses various problems of this type. Despite the fact that one can get a lot of geometrical and topological information from the spectrum or even its asymptotic behavior, this ...

11

The "simplest" classical explanation I know is the van der Waals interaction described by Keesom between two permanent dipoles. Let us consider two permanent dipoles $\vec{p}_1$ (located at $O_1$) and $\vec{p}_2$ located at $O_2$. Their potential energy of interaction is: U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...

10

Since the energy spectrum does not depend on the absolute position $\vec{r}=\vec{a}$ of the delta potential, we may assume that $\vec{a}=\vec{0}$. Therefore, in its current formulation (v1), OP is effectively saying that The attractive 1D delta potential $V(x) = -A\delta(x)$, $A>0$, has exactly one bound state. The same is true for the 3D delta ...

10

Wavefunctions are found by solving the time-independent Schrödinger equation, which is simply an eigenvalue problem for a well-behaved operator: $$\hat{H} \psi = E \psi.$$ As such, we expect the solutions to be determined only up to scaling. Clearly if $\psi_n$ is a solution with eigenvalue $E_n$, then $$\hat{H} (A \psi_n) = A \hat{H} \psi_n = A E_n ... 8 All bound states can typically be chosen to have real-valued wavefunctions. The reason for this is that their wavefunction obeys a real differential equation,$$ -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf r)+V(\mathbf r)\psi(\mathbf r)=E\psi(\mathbf r)and therefore for any solution you can construct a second solution by taking the complex conjugate ... 7 1) OP wrote (v1): [...] and thus this leads me to believe that {\bf A} should be somehow connected to momentum, [...]. Yes, in fact the magnetic vector potential {\bf A} (times the electric charge) is the difference between the canonical and the kinetic momentum, cf. e.g. this Phys.SE answer. 2) Another argument is that the scalar electric ... 7 If you want to generalize a potential to a class that's broader than the simple \frac12 k_2 x^2, it is tempting as a first step to include a small perturbation of the form \frac13k_3x^3. Unfortunately, this drastically changes the structure of the potential, because it becomes unbounded from below. Thus, you might get a slightly perturbed behaviour ... 6 Everything you've probably learned about capacitors, especially including the statement that opposite plates of the capacitor carry opposite charges, applies only to a capacitor in a circuit. If your capacitor is floating, so that the plates are not connected to anything, the charge on the plates is not going to change. If you hook up only one plate to a ... 6 Electrons will flow against the electric field lines because their charge is negative, and the electric field thus exerts a force \mathbf{F}=q\mathbf{E} on them which is in the opposite direction. Thus electric field lines inside the wire go from the positive to the negative terminal and the electron flow goes from the negative to the positive terminal. ... 6 Both sides of the equation given are vectors and so represent 3 equations which are, on a Cartesian basis: E_x + \frac{\partial A_x}{\partial t} = -\frac{\partial V}{\partial x}  E_y + \frac{\partial A_y}{\partial t} = -\frac{\partial V}{\partial y}  E_z + \frac{\partial A_z}{\partial t} = -\frac{\partial V}{\partial z}  6 The full mathematical statement is as follows: Theorem If two particles exert a mutual conservative force \mathbf{F}_{12} and \mathbf{F}_{21} which is independent of any other degree of freedom of any bigger system they're part of, and obeys Newton's third law as \mathbf{F}_{12}+\mathbf{F}_{21}=\mathbf{0}, with the forces collinear to the ... 5 You are right that the result you see is due to the chain rule. The author uses either spherical or cylindrical coordinates, so $$r = \sqrt{x^2 + y^2 + z^2}$$ or $$r = \sqrt{x^2 + y^2}$$ which you can differentiate to obtain \frac{\partial{r}}{\partial{x}} = \frac{x}{r} ... 5 Yes, u is indeed the potential energy. And yes, you can calculate the force acting on a particle by calculating the gradient of the potential energy field at the position the particle is in. Computationally you will want to calculate the force on particle 1, by taking the gradient at the position particle 1 is in, of the potential energy field created by ... 5 When there is no resistance, as is the case with an ideal wire, any value of current satisfies Ohm's Law: V = I R since both V=0 and R=0. UPDATE: But isn't V is like what causes the current? Perhaps a mechanical analogy of the resistor will help. Consider the dashpot where the velocity of the arm is analogous to current while the force acting ... 5 You missed a term in expanding the upper-indexed metric. The full version is below: \begin{align} \tilde{\Gamma}^\lambda_{\mu\nu} & = \frac{1}{2} \tilde{g}^{\lambda X} \left(\partial_\mu \tilde{g}_{\nu X} + \partial_\nu \tilde{g}_{\mu X} - \partial_X \tilde{g}_{\mu\nu}\right) \\ & =\frac{1}{2} \tilde{g}^{\lambda\sigma} \left(\partial_\mu ... 5 Let me first comment that the statement electric fields cancel while the electric potentials just add up algebraically is not actually correct. Electric fields add due to the principle of superposition (see the section on superposition in the wikipedia article). However, when two electric field vectors are of the same magnitude but point in ... 5 Every system likes to decrease its electrostatic energy. The charges on the plates are almost in stable equilibrium. The charges on the opposite plates attract them, and the charges on the same plate repel them with almost the same force. However, a capacitor has fringe fields: These may be negligible when calculating the field inside a capacitor, but ... 5 This is your circuit: The current that comes from the source, when reaches the point that must choose it's way, sees no difference between the two paths (symmetry) , so half of it flows through one way and the other part flows in the second way. It means that, I_1=I_2 , So the potential difference across yellow resistors is the same. It means that the ... 4 EDIT: Put simply, potential difference is the work done by electrostatic force on a unit charge, while EMF is the work done by anything other than electrostatic force on a unit charge. I don't like the term "voltage". It seems to mean anything measured in volts. I'd rather say electric potential and electromotive force. And the two are fundamentally ... 4 That schematic is quite confusing; draw it with higher voltages at the top, lower voltages at the bottom, and signals flowing from left to right per standard procedure. It's also helpful to label components and nodes so that you can solve the equation symbolically. I haven't labeled some of the nodes here because they can be represented by the names of ... 4 OP wrote (v1): I hardly feel that one out of two is a sufficient justification for the introduction of a potential. I) Here we would like to point out that there exists a velocity-dependent generalized potentialU~=~q(\phi - {\bf v}\cdot {\bf A}) $$for the Lorentz force$$ {\bf F}~=~ q({\bf E} + {\bf v}\times {\bf B}) . $$Here \phi is ... 4 Your Question all but includes the right search term for an Answer from Wikipedia, "Conservative Forces", which gets you to http://en.wikipedia.org/wiki/Conservative_Forces. There's even what you ask for, a proof. There's also another link to http://en.wikipedia.org/wiki/Conservative_vector_field, which gives some quite good visualizations that will probably ... 4 To be concrete, let us here assume that the dissipative force is a friction force$$\tag{1} {\bf F}~=~-k {\bf v} $$proportional to the velocity {\bf v}=\dot{\bf r} of the point particle. Recall that a velocity dependent potential U=U({\bf r},{\bf v},t) of a force {\bf F} by definition satisfies$$\tag{2} {\bf F}~=~\frac{d}{dt} \frac{\partial ...

4

1) Postponing for a moment the issue of magnetic monopoles, one conventional answer is, that the gauge potential $A_{\mu}$ (as opposed to, e.g., the electric and magnetic $\vec{E}$ and $\vec{B}$ fields) constitute the true fundamental variables and (the photon field) of QED. At the classically level, by saying that $A_{\mu}$ are fundamental variables, we ...

4

1) The infinitely long wire has an infinite charge $Q=\lambda \int_{-\infty}^{\infty} \! dz = \infty$, and EM has an infinite range, so one shouldn't be surprised to learn that the result $$\phi(r)~=~ \frac{\lambda}{4 \pi \epsilon_0} \int_{-\infty}^{\infty} \frac{dz}{\sqrt{z^2+r^2}} ~=~ \frac{\lambda}{4 \pi \epsilon_0} \left[ {\rm arsinh} ... 4 Frenkel and Smit definitely make a mistake. Eq. (12.1.3) page 294 is:$$-\nabla^2 \phi(\mathbf{r}) = 4\pi \rho(\mathbf{r}) $$then immediately afterwards, Eq. (12.1.4) is "the solution of this equation" "for a single charge z at the origin":$$\phi(\mathbf{r}) = \frac{z}{4\pi |\mathbf{r}|} This is a mistake: Eq. (12.1.4) is definitely not "the solution" ...

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