Tag Info

Hot answers tagged

53

Here is a circuit representing the system. $R_{wire}$ is the resistance of the section of wire between the bird's legs. $R_{bird}$ is the resistance of the bird (which you can measure by sticking the two probes of the multimeter to the bird's two feet - if the cable is insulated, you will have to add the resistance of the insulation as well). When the ...


23

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 ...


23

The potential difference between two points on a wire carrying a current is given by Ohm's Law, $V = R\cdot I$. Since wires used for long-distance power transmission have, by design, a very low resistance per unit length, and the distance between the two extremities of your hands is very small (~10cm), even for large currents the potential difference is ...


11

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 ...


11

Potential is a special case of a more general construction in differential geometry. Let's start abstractly and we'll get to the potentials again at the end. Differential forms The framework of differential forms provides a basis for integration on arbitrary manifold. Differential $p$-forms are totally antisymmetric covariant $p$-tensors. What's special ...


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: \begin{equation} U(\vec{p}_1,\vec{p}_2,\vec{O_1 O_2}) = -\vec{p}_1\cdot \vec{E}_2 = ...


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 ...


9

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 ...


9

The gradient of a scalar is again vector.


9

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 ...


8

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 ...


8

Short version: In the infinite potential well, $E \geq 0$ (because $V_{min}=0$, and $E \geq V_{min}$). In your finite potential well, it sounds like you are looking for bound states, in which case $E < 0$, so you absorb the negative into the square root. Long version: When you are tackling a QM problem, first you should figure out the admissibility of ...


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 ...


7

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 ...


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

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

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

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 ...


6

With a delta function potential, the particle is free on either side of the barrier: $$ \psi(x)=\begin{cases}\psi_L(x)=A_re^{ikx}+A_le^{-ikx} \\ \psi_R(x)=B_re^{ikx}+B_le^{-ikx}\end{cases} $$ where $A_i,\,B_i$ are constants such that $A_r+A_l=B_r+B_l$ (i.e., $\psi(x)$ satisfies the continuous function condition). But at the barrier we have the issue that ...


6

I suppose you could view your situation as a circuit with two parallel resistors; those are a part of the cable and the bird. However the resistance of the cable is many orders of magnitude smaller than the bird's (the cable is effectively a short-circuit), so there won't be any appreciable current through the bird.


5

Hanging from a power line you should be as safe as a bird. The voltage difference is between the lines (e.g. in a 3-phase system) and between the line and ground. This voltage difference exists across the insulators and pole, as well as through the air to ground. These voltage differences are obviously small enough to avoid striking an arc, hence no current ...


5

You are right that the result you see is due to the chain rule. The author uses either spherical or cylindrical coordinates, so \begin{equation} r = \sqrt{x^2 + y^2 + z^2} \end{equation} or \begin{equation} r = \sqrt{x^2 + y^2} \end{equation} which you can differentiate to obtain \begin{equation} \frac{\partial{r}}{\partial{x}} = \frac{x}{r} ...


5

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 ...


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

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

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 ...



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