Hot answers tagged potential
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:
\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 ...
8
This is fundamentally no more difficult than understanding how quantum mechanics describes particle motion using plane waves. If you have a delocalized wavefunction $\exp(ipx)$ it describes a particle moving to the right with velocity p/m. But such a particle is already everywhere at once, and only superpositions of such states are actually moving in time.
...
8
The Harmonic oscillator has the same spectrum as a weaker harmonic oscillator with a hard wall at x=0.
LATER EDIT: I see that I have to be more explicit--- the potentials
$V(x)= 2x^2 - 2$
$(x>0)$ $V(x)= x^2 - 3$ and $(x<0)$ $V(x)= \infty$
have the exact same spectrum.
7
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 ...
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
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 ...
5
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. ...
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
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 ...
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
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
While Jonathan's answer is correct, it is very general. I would like to explain the reason for the terminology in this specific problem.
When dealing with second order equations, like Newton's laws, the notion of potential is designed to give a conserved energy. This definition is natural in the inertial domain--- in the case that reversible inertial motion ...
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
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 ...
4
If you are wondering about causality, then I think that voltage difference $\Delta V$ is fundamental as it is the cause, and the current $I$ is the consequence.
If you want to have current, you need movement of the charges. The most obvious way to move charges is to act upon them with electric field, and each electric field is accopmained with voltage ...
4
Assuming you mean a macroscopic potential difference, the largest I know about was in the Nuclear Structure Facility accelerator at the Daresbury laboratory in the UK, and this was 30MV. The Wikipedia article on electrostatic particle accelerators claims this is about the highest possible in such devices.
4
We are allowed to use gauge invariance in quantum mechanics – even quantum mechanical theories with the electromagnetic 4-potential are gauge-invariant theories. However, it's not quite true that all gauge invariant quantities are functions or functionals of $F_{\mu\nu}$.
Instead, we may consider the phase
$$ \exp\left(i\oint d\vec x\cdot \vec A\right) $$
...
4
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 ...
4
No, most force fields refuse to be conservative. The path-independence is a nontrivial constraint in an arbitrarily small region of space, an arbitrary neighborhood.
If the force field is conservative, it must be
$$\nabla\times \vec F = 0$$
because $F = -\nabla\Phi$. It's clear that the curl of $\vec F$ may be nonzero even if you look at a small ...
4
I'd like to add a bit of mathematical detail the (correct) statements by DJBunk. Let a scalar function $f$ be given (let's not restrict ourselves to the electric potential). For any unit vector $\mathbf n$, we can define the directional derivative $D_\mathbf{n}$ of the function $f$ in the direction $\mathbf n$ as follows:
$$
D_\mathbf{n}f(\mathbf x) = ...
3
Yes, for infinite potential wells in any number of dimensions, the same technique can be applied. Each confinement direction gives an extra quantum number, and the wavefunction is just the product of the 1D wavefunction for each direction.
More problematic is the finite potential well, because one can't use a simple expression for the potential in more than ...
3
Hint: Use Ohm's law $I=\frac{V}{R_p}$ and the formula $\frac{1}{R_p}=\frac{1}{R}+\frac{1}{R_0}$ for parallel resistors to derive a straight line in a $I$-$\frac{1}{R}$ diagram
$$ I~=~V \left(\frac{1}{R}+\frac{1}{R_0}\right). $$
Here $R$ and $R_0$ are the variable and the fixed resistor, respectively. To answer OP's two questions:
The slope is $V=1{\rm ...
3
The charge will redistribute as follows:
$$\frac{+Q}{2}\begin{vmatrix}
\frac{+Q}{2} & \frac{-Q}{2} \\
\text{charged plate} & \text{free plate}
\end{vmatrix}\frac{+Q}{2} $$
Note that the charges on the inner and outer sides of each plate are distinct. In a capacitor (even one not in a circuit), it is mandatory that the opposing faces have ...
3
I think your confusion may be addressed by an answer to a related question that I posted a while back. (And Lubos' answer too, probably) Basically, the reason is that potential is only defined relative to a reference point. So in a sense, the real, most general formula for potential is
$$V = -\frac{kq}{r} + \frac{kq}{r_\text{ref}}$$
using your notation.
...
3
Keep in mind that the definition of potential is dependent on context. There is nothing deep going on here. If he wishes to define the potential of the system $\dot{x}=f(x)$ to be a function $V$ such that $-\frac{dV}{dx}=f(x)$, he is free to do so. Now of course, because the system itself is different, you can't expect this potential to satisfy all of the ...
3
The potential energy between two oppositely charged particles goes like $-1/r$. First of all, the potential energy doesn't scale as $1/r^2$; you must have confused it with the force which is the gradient (spatial derivative) of the potential energy. Second of all, the potential energy of a bound state – a pair or collection of particles/objects that ...
Only top voted, non community-wiki answers of a minimum length are eligible
