I Boltzmann Distribution: Formula & Fig 2a in Document

AI Thread Summary
The discussion centers on the formula used to create the exponential Boltzmann distribution depicted in Figure 2a of a referenced document. Participants clarify that the formula is 600*exp(-βε), with β representing a parameter related to energy. The confusion arises around determining the correct values for the parameters, particularly around x=1 yielding y=200. The caption of the figure does not provide sufficient clarity, leading to the exploration of the relationship between the variables. Ultimately, the formula is simplified to exp(log(600) - βε), confirming its straightforward nature.
Physics news on Phys.org
Doesn't the caption explain it?
 
funny,
But what values to put in?
Around x=1 it seems to get y=200
e^(ln(600)-1/(1.381×10^-23*300)) = 0?
 
Since the x-axis is βε, the only free parameter is α. As you realized, it goes through ~600 at βε=0, so the formula being plotted is clearly 600*exp(-βε). Or, if you like, exp(log(600) - βε)
 
Doh! So simple, thanks!
 
Consider an extremely long and perfectly calibrated scale. A car with a mass of 1000 kg is placed on it, and the scale registers this weight accurately. Now, suppose the car begins to move, reaching very high speeds. Neglecting air resistance and rolling friction, if the car attains, for example, a velocity of 500 km/h, will the scale still indicate a weight corresponding to 1000 kg, or will the measured value decrease as a result of the motion? In a second scenario, imagine a person with a...
Scalar and vector potentials in Coulomb gauge Assume Coulomb gauge so that $$\nabla \cdot \mathbf{A}=0.\tag{1}$$ The scalar potential ##\phi## is described by Poisson's equation $$\nabla^2 \phi = -\frac{\rho}{\varepsilon_0}\tag{2}$$ which has the instantaneous general solution given by $$\phi(\mathbf{r},t)=\frac{1}{4\pi\varepsilon_0}\int \frac{\rho(\mathbf{r}',t)}{|\mathbf{r}-\mathbf{r}'|}d^3r'.\tag{3}$$ In Coulomb gauge the vector potential ##\mathbf{A}## is given by...
Thread 'Does Poisson's equation hold due to vector potential cancellation?'
Imagine that two charged particles, with charge ##+q##, start at the origin and then move apart symmetrically on the ##+y## and ##-y## axes due to their electrostatic repulsion. The ##y##-component of the retarded Liénard-Wiechert vector potential at a point along the ##x##-axis due to the two charges is $$ \begin{eqnarray*} A_y&=&\frac{q\,[\dot{y}]_{\mathrm{ret}}}{4\pi\varepsilon_0 c^2[(x^2+y^2)^{1/2}+y\dot{y}/c]_{\mathrm{ret}}}\tag{1}\\...
Back
Top