My attempt : $$P(n) = \frac{1}{\mathcal{Z}} Exp[(n\mu -E)/\tau]$$, use $$\lambda = e^{\mu/\tau}$$, then the distribution can be written as $$P(n) = \frac{1}{\mathcal{Z}} \lambda^nExp[-E/\tau]$$
Note that the average number of particle can be written as $$<N>= \lambda \partial \lambda ( log...
How to run a numerical simulation of Laplace equation if one of the boundary condition is like this: $$V(x,y) = 0 \text{ when } x \to \infty$$
I am trying to use Python to plot the solution of this Example 3.5. in Griffins EM
If the boundary condition is not provided in the form of electric potential, how do we solve such problem?
In this case, I want to use ##V = - \int \vec{E} \cdot{d\vec{l}}##, but I don't know how to choose an appropriate reference point.
So we are actually interested in the region that ##\{(x,y,z)| z > 0, (x,y,z) \neq (0,0,d)\}##?
Or a singularity at ##z = d## do not violate ##\nabla^2 V = 0##?
I am studying the classic image problem (griffins, p. 124)
Suppose a point charge ##q## is held a distance ##d## above an infinite grounded conducting plane. Question: What is the potential in the region above the plane?
boundary conditions:
1. V = 0 when z = 0 (since the conducting plane is...
My intuition for this problem is to use divergence theorem:
## \int_V \nabla^2 u dV = \int_S \nabla u \cdot \vec{n} dS##
But note that ##\vec{n}## is perpendicular to x-y plane, and makes ##\int_S \nabla \ln s \cdot \vec{n} dS = 0##
If we take laplacian in polar coordinate directly, then...
Actually, this is chapter one, so I don't know yet. I am just beginning using this book to review the content I learned last semester. I am mainly confused about the mathematical technique he uses, ( expand the Lagrangian as a power series of ##\epsilon##) when he analyze the difference of two...
Let ##K## and ##K'## be two inertial frame, If K is moving with infinitesimal velocity relative to ##K'## , then ##v' = v + \epsilon##.
Note that ##L(v^2) - L(v'^2)## is only a total derivative of a function of coordinate and time. (I understand this part)
Because ##L' = L(v'^2) = L(v^2 +...