New answers tagged potential
0
Ok, let's consider $\vec{r} \vec{dr}$, it is equal to $|\vec{r}||\vec{dr}|_\vec{r}$ where $|\vec{dr}|_\vec{r}$ is projection of $\vec{dr}$ on $\vec{r}$. If you draw $\vec{r}$ and a small (remember, you need infinitesimal!) $\vec{dr}$ you will notice that this projection is actual equal to $|\vec{dr}|_\vec{r} = d|\vec{r}|$, so $\vec{r} \vec{dr} = ...
3
Within the superposition of the ground and the first excited state, the wavefuncion oscillates between "hump at left" and "hump at right". Maybe you are asked to find the half-period of these oscillations?
2
Yes, I believe you have to think of it as if it were a semiclassical problem; you evaluate with QM the mean square velocity $\left< v^2 \right>$ of the particle, then calculate its square root; this should give you an estimate of the typical velocity of the particle. Once you have it, you divide the length of the well by it and find the time it takes ...
0
In general, what you're trying to do is called the "spectral method" for solving PDEs. Wikipedia has a little on it, including some useful references, and a solution of the Poisson equation. http://en.wikipedia.org/wiki/Spectral_method
As Peter Kravchuck says, the kernel will always be $k^{-2}$ for the Poisson equation.
In the linked PDF, ...
0
In Fourier space the Poisson equation is $k^2\phi=\rho$ (up to a convention-dependent constant factor). So in every dimension the kernel is $1/k^2$. As for the real space, it is, up to a constant $|r|,\,\log|r|,\,1/r$ in 1D,2D,3D respectively.
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 = ...
0
The amount of work done in bringing a unit positive charge fron infinity to a given point in electric field is called electric pontial at that point mathematically,
$$\phi=\frac{W}{Q}$$
Work W and charge Q.
The work done in moving a unit positive charge from the point of lower potential to point of higher potential at the point is called electric potential ...
0
I think I've understood it now .
$ds=dr$ .
but $dr<0$ and $|dr|=-dr$
Because dr is a small position vector and position vector is directed along field .
Now why I can't use ds directly is because the limits in the integral , (the upper and lower limit in integral notation) are in terms of position vector and not the displacement .
Had they been in ...
2
New version
The problem in your demonstration is when you write down $\vec{A}\cdot\vec{B} = ||\vec{A}||\,||\vec{B}||\,\cos\theta$. More exactly, in your case $||d\vec{r}||\neq dr$ because $dr<0$ when you go from $\infty$ to $r$ and a norm is positive by definition. So the sign error is introduced from 3rd to 4th line.
Old version
The demonstration on ...
1
Just to be clear, the potential energy of a particle of charge $q_2$ at a distance $r$ from a source of potential (supposidely at zero) of charge $q_1$ is the work that an external operator has to provide to bring the particle from infinity to $r$ at constant velocity.
This reads then:
$\int_{\infty}^r \vec{F}_{op}\cdot \vec{ds}$
As people have said, the ...
2
When you calculate work, you do so along a given path. Here, that path has tangent vector $d\mathbf s$. This is a vector with direction; the minus sign will ultimately come from choosing the path's orientation--inward or outward.
Edit: Aha, I think I've found the unintuitive part. The key is in the use of the coordinate $r$ to parameterize the path, in ...
-1
$$\mbox{d}\vec s = \mbox{d}r$$
Therefore,
$$\vec F\cdot \mbox{d}\vec s= F\mbox{ d}r\mbox{ }\cos\theta=F\mbox{ d}r\mbox{ }\cos\pi=-F\mbox{ d}r$$
Edit: sorry for the error where I forgot to put the magnitude sign. I did mean the magnitude sign.
$$\left|\left|\mbox{d}\vec s\right|\right| = \mbox{d}r$$
1
Yes. If you define $f=-\partial_\mu A^\mu$ then you can write the equation in the form $$ \partial_\mu\partial^\mu\psi = f$$
This is the Klein-Gordon equation with a nonzero source ($f$) and can be solved via Green's function methods. Once you have the Klein-Gordon propagator* $G(x)$ (this is derived in any e.g. quantum field theory textbook) appropriate to ...
1
Potential is negative of work done per unit charge by electrostatic force
1
$\mathbf{r}$ is a position vector and $\mathbf{s}$ is a displacement vector between two points, let say A and B. In general case, they are not equal, but they can be if we properly choose the origin of the coordinate system: A={0,0,0} or B={0,0,0} The sign depends on at which point A or B the origin is placed.
2
The electric field is a conservative vector field which implies that there exists a function $V$ for which
$$
\mathbf E = -\nabla V
$$
We call this function $V$ the electric potential. There is no mathematical need to first define potential energy. One can then physically interpret $V$ in terms of a "potential landscape" to get intuition for what it ...
0
In a conductor at equilibrium the electric potential is equal everywhere, for, if it were not, then the electrons would experience a force proportional to the gradient of the potential, causing them to redistribute themselves until the potential became homogeneous.
Top 50 recent answers are included
