There's no difference between sound and light, in principle. Yes, in practice it is much harder to observe beats with visible light due to the high frequencies, as I already noted.
Beat frequency is a pretty basic concept in physics so you should be able to find lots of good explanations on the web -- you could start with the Wikipedia article. The beat frequency is calculated as the difference in the frequencies of the two light beams, fb = f1 - f2. So the smaller the...
This is an interesting question. It turns out that the answer is no. This is pretty straightforward to see mathematically, but for the benefit of intuition it's instructive to consider the case when the light beams have two frequencies f1 and f2 which are very close, but not quite equal. Then...
Yes, the H field inside the ferromagnetic material is much smaller because the effect of the B field is mostly canceled by the magnetization. In particular, if we forget about non-linearity, hysteresis, etc., and just assume the material is characterized by a large magnetic susceptibility...
Yes, measuring along different axes represents a different process, so obviously it needs a different circuit diagram to represent it. The "dial" symbol represents measurement along the z-axis -- you can get measurement along a different axis by combining with other gates, but you could also...
Could you provide details? I've been looking at Section IX of Siano's http://dx.doi.org/10.1016/0016-0032(85)90032-8" , and it turns out he does give a partial justification of why his method works, by assuming that physical laws are tensor equations (as they must be from rotational invariance)...
However, you could consider a neutral pion, which is its own antiparticle (and therefore can decay without interacting with any other particle), yet still has electric potential energy since it consists of charged quarks.
Really? The electric field of a point charge is
E \propto \frac{1}{r^2}
So the electrostatic energy density is
u \propto E^2 \propto \frac{1}{r^4}
So the total energy is
U = \int_0^\infty 4\pi r^2 u(r) dr
\propto \int_0^\infty \frac{1}{r^2} dr
This integral diverges at...
Well, no. Could you elaborate on why you would think this?
Sort of. It depends on the direction of the magnetic field in the Stern-Gerlach apparatus. If the magnetic field points along the z-direction, then the Stern-Gerlach apparatus is measuring the spin along the z-axis, which is...
Could I repeat my question from before: why are you so sure that Siano's extension is valid? Does it follow from rotational invariance somehow? I would certainly not be comfortable making use of some ad-hoc set of rules unless I understand how they come about.
Contrary to what IsometricPion said, it is true that the field energy of the electron is already included in its mass. This is the point of Einstein's E = mc2; any energy which the electron has in its rest frame translates into inertia (i.e. mass). (Actually, the problem in classical physics is...
No, this is not true except in special cases (despite what my not-quite-up-to-speed EM lecturer once tried to tell me!). If you have David Griffith's electromagnetism textbook, then he addresses this point in Section 6.3.2, "A Deceptive Parallel". The point he makes is that a vector field like H...
These pictures are known as "Feynman diagrams". The rules for drawing them are quite simple (e.g. see http://www2.slac.stanford.edu/vvc/theory/feynman.html" which I found on Google). Understanding how they arise from quantum field theory is pretty advanced physics though.
EDIT: I should add...
The way I like to think about the dimensionality of a quantity is that it is characterizes how the numerical value of the quantity changes when we make use of our freedom to arbitrarily scale units.
Since all the SI units like metre, second, and so forth are defined completely arbitrarily [e.g...
It sounds as though you're thinking of a band only as a range of energy values, which is not accurate. Each band basically consists an electron wavefunction for each value of the wave vector k. If the valence band and conduction band overlap, then there is a valence band wavefunction and a...