# Tag Info

## New answers tagged potential

0

Radio frequency Paul traps confine charged particles without applying any magnetic fields (As in Penning traps), but since confining charged particles only using electrostatic forces is impossible according the the Earnshaw's theorem, a quasi-static approach is taken, and the particles are trapped dynamically. Radio frequency ion traps do this by forming a ...

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If the electric field is conservative, you can integrate along any path between the specified points to obtain the potential. All paths will yield the same answer. This makes it easier, because you can pick any path that is convenient for you - or easiest to integrate. Recall that a conservative field is defined by: $$\nabla \times E = 0$$ So one ...

1

The main point that although a pointwise convergent Fourier series of cosine modes is an even function $\psi(-x)=\psi(x)$, it does not have to be differentiable at $x=0$. A pointwise convergent infinite sum of differentiable functions is not necessarily a differentiable function. More generally, as OP already mentions, the wave function $\psi(x)$ is not ...

1

In the 1d particle in the box the energy of the particle should be completely determined by the momentum of the particle that you observe correct? The Hamiltonian in the position basis is $$\hat H = \begin{cases}-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}, & 0\lt x \lt a\\\infty, & \text{otherwise} \end{cases}$$ and the energy ...

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The momentum operator $P$ in the infinite well can be defined as a self-adjoint operator by infinitely many ways with respect to the boundary conditions by: $$P_\theta=-i\hbar\frac{d}{dx}\\ \mathcal{D}(P_\theta)=\left\{\psi\in \mathcal{H}^1[0,a]:\psi(a)=e^{i\theta}\psi(0)\right\},$$ where $\mathcal{H}^1[0,a]$ is the Sobolev space, on the interval $[0,a]$. In ...

1

There are several (equivalent) ways to look at this. One is to say that for any conservative force $\mathbf{F}$, one can define the potential energy Ep as an associated potential field such as $\mathbf{F}=-\frac{\partial Ep}{\partial r}$, or maybe more formally $\mathbf{F}=-\nabla(Ep)$. That's no more than a definition of the potential energy. ...

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electric field strength is $$E=\frac Fq=\frac Vd$$ with $V$=voltage, $d$=distance between charged plates \begin{align} \frac Fq&=\frac Vd \\ Fd&=qV \end{align} but $Fd$=energy $$\therefore {\rm energy}=qV$$

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I'll give this a shot though I'm uncertain if this will clarify or cloud the issue. electrons are people, OK current is the number of people, No, that's not the correct analogy. Current is a flow and a people (electron) current is a flow of people (electrons) and the amount of current is the number (Coulombs) of people (electrons) passing a ...

1

Since your question is kind of pictorial, i will try to stick with this and avoid any deeper mathematical / physical explanations. electrons are people ok with that current is the number of people no, that's a mistake. Actually current is is this case the number of people crossing a line on the street per time. Like Steeven already mentioned in ...

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But what is potential in this example? In your people-in-the-street analogy, the electric potential would be a shop with sale in one end of the street, or maybe an accident or something else interesting. People tend to move towards the interesting end (less potential) and away from the other and less interesting end (higher potential). Just like a stone ...

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The choice of the constants $A$ and $B$ depends on the form of the solution. You could have denote one pair of constanst by $A$ and $B$, and the other by $C$ and $D$. A possible solution is $\psi (x)=A\sin(kx)$. In complex form, the sine is: $$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$ If you've never proved this formula, try it using $e^{ix}=\cos x+i\sin x$. ...

1

I also don't quite understand what you mean by "imaginary $\psi$", but let me give you some general, more mathematically correct view. In general, what we measure is the spectrum of a self-adjoint operator. For the energy, this operator is of course the Hamiltonian. Now, since it is self-adjoint, its spectrum will lie on the real line. To have stability, we ...

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I ended up spending an hour or so wading through the papers, so here's my main conculsion from that: No, I don't think the critique of the papers is wrong; Nor do I think that the basic algebraic arguments in the three papers addressed by the critique are wrong either. The authors describe a particular "decomposition" of the electromagnetic fields into a ...

1

Refer to: http://www.encyclopediaofmath.org/index.php/Hulthen_potential According to this it's a model of potential which behaves similar to Coulomb potential. I don't know how it's derrived, but it was derrived by a German dude. This paper compares both the Coulomb Potential and Hulthen potential: http://adsabs.harvard.edu/full/1954AuJPh...7..365M ...

0

I think your question is really: Why is Gibbs-Duhem equation not valid for inhomogeneous and/or small systems? This is related to why Thermodynamics does not apply to these systems. Examples, discussion and a Thermodynamical theory for these systems can be found in Terrell Hill's works link. But in short: both inhomogeneous and few-body systems have more ...

2

That's probably for charged solid sphere, not a cylinder. In any case, setting the potential at infinity as zero, we have for $r>R$: $$V(r)-V(\infty)=-\int_\infty^rE(r')dr'\implies V(r)=\frac{Q}{4\pi \epsilon_0}\frac{1}{r}$$ For $r<R$, we got: $$V(r)-V(\infty)=-\int_\infty^rE(r')dr'=-\int_\infty^RE(r')dr'-\int_R^rE(r')dr' \implies \\ ... 3 I too was confused by this difference between gravity and electromagnetism. Hopefully the following clears things up. The gravitational potential a distance r from a mass M is$$ \phi_g=-\frac{GM}{r}, $$the gravitational field is$$ {\bf g} = - \nabla \phi_g, $$and the gravitational potential energy (of two masses M and m separated by a distance ... 1 Well, the electric field \vec E is different from the force field \vec F a test charge will feel. That difference is exactly the charge of the test particle. That force field is given by the gradient of a function, too$$ q \vec E = \vec F = - \frac{\mathrm d}{\mathrm d r} W where I use the letter $W$ in order not to have confusing notation. The ...

4

I think you are reading a lot into what is a minor distinction. Strictly speaking I suppose the gravitational potential is the energy per unit mass, i.e. $m=1$ in your first equation, while the gravitational potential energy is the potential times the mass. In practice no-one I know has ever bothered to make the distinction because it's usually obvious what ...

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 ...

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