Purcell says that taking the surface integral of the magnetic field ##\textbf{B}## over the surfaces ##S_{1}, S_{2}, S_{3},...## below is a good way of finding the average of the volume integral of ##\textbf{B}## in the neighborhood of these surfaces.
More specifically, he says in page...
@Stephen Tashi
No where in the notes is it indicated that the graph is the plot of ##P(E)## against##E/U##. Had it been the case, the x-axis would not have been labeled ##E##. Besides, it is clearly indicated under the graph that it is the plot of##P(E)## against the energy...
@vanhees71
Correct me if I'm wrong, but do you mean by this that the standard deviation can be approximated by the mean when we have a very small standard deviation?
The question was not really about why ##\frac{\Delta E_{RMS}}{\langle E\rangle}## can be approximated by ##\frac{1}{\sqrt{N}}##. But, rather, about why the author says the width of the distribution is, approximately, the ratio of the standard deviation to the mean ##\frac{\Delta...
In these lecture notes about statistical mechanics, page ##10##, we can see the graph below.
It represents the distribution (probability density function) of the kinetic energy ##E## (a random variable) of all the gas particles (i.e., ##E=\sum_{i}^{N} E_{i}##, where ##E_{i}## (also a random...
@Orodruin
Small side question, does this mean the EM wave has an infinite number of photons? If this is the case, and we assume that EM wave's energy is the sum of the energy of the photons, that would mean the EM wave has infinite energy, wouldn't it?
So the relation ##E=pc## applies to...
@Orodruin I wouldn't qualify the book as "not good", it is even still used today at MIT.
@PeroK
The total energy ##E_{radiant}## is the energy of the electromagnetic wave (light pulse for example). We know for example the energy of a photon depends only on the frequency, so varying the power...
In A.P. French's Special relativity the author said,
The mass and length of the box are irrelevant here.
He said the momentum of the radiation is ##E_{radiation}/c##. We know that the momentum of a single photon with energy ##E_{photon}## is ##p_{photon}=E_{photon}/c##.
So is...
@Orodruin So if I understood correctly, ##m## in ##p=mc## is the relativistic mass ##\gamma m_{0}##?
But then this is in contradiction with the way the author describes ##m## in ##E=mc^{2}##. He said,
In A.P. French's Special Relativity, the author said the following,
As I understand, photons are massless, so I don't think the last equation above applies to photons, but then, when deriving it, he used an equation proper to photons (##E=pc##).
So in which context is ##m=p/c## valid?
##\gamma## is the proportionality factor between the friction force ##\textbf{f}_{r}## and the velocity ##\dot{\textbf{x}}(t)##, i.e., ##\textbf{f}_{r}=-\gamma \dot{\textbf{x}}(t)##. The friction force is a viscous one.
@BvU For example, for a damped, driven harmonic oscillator with a natural frequency ##\omega_{0}##, the cutoff frequencies are at ##\sqrt{\omega_{0}^2+\frac{\gamma ^2}{4}}\pm \frac{\gamma}{2}##, where ##\gamma## is the damping coefficient. The equation of motion of the oscillator is given by...
The ##Q## factor of an oscillating system is defined as ##\omega_{r}/\Delta \omega##, where ##\omega_{r}## is the resonant frequency, and ##\Delta \omega## the resonance width. As I understand, ##Q## measures how sharp the resonance curve is.
Why is it that the resonance curve gets...