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seen Jan 26 '12 at 20:37

Jun
7
comment Where is spin in the Schroedinger equation of an electron in the hydrogen atom?
I dunno, Justin L. Why don't you write down the Hamiltonian (or the Schroedinger equation) you used?
Jun
7
comment Where is spin in the Schroedinger equation of an electron in the hydrogen atom?
I'll echo what dmckee said: the Schroedinger Equation does tell the complete (nonrelativistic) story, but it's up to the user to put everything in the Hamiltonian, and make sure the state vector space includes everything about the system. In this case, you approximated Hamiltonian (as explained above) to include only the Coulomb term, and you used a simplified vector space that included the spatial wavefunction but not the spinor wavefunction. That's A-OK for the Hamiltonian you used, since it has no dependence on spin.
Jun
6
answered Where is spin in the Schroedinger equation of an electron in the hydrogen atom?
Jun
2
comment Statistical physics of molecular dissociation of a diatomic gas
Thanks for modifying your answer to include the full answer. I'm always a bit uncomfortable with a "proportional to" that includes some of the functional dependence on temperature but not all of it. Your new version is much more clear.
Jun
2
comment Statistical physics of molecular dissociation of a diatomic gas
Either way, thanks for correcting your answer below. While the LaTeX names for those symbols would imply one means "similar" and the other means "proportional to", perhaps we can agree to disagree on this.
Jun
2
comment Statistical physics of molecular dissociation of a diatomic gas
@Qmechanic: Normally "is proportional to" is written as "$\propto$", no? And if you're trying to express proportionality, you need some way of communicating that the ONLY proportionality included in the equation is the $E_0$ dependence; the full dependence on $T$ is omitted, etc.
Jun
2
comment Statistical physics of molecular dissociation of a diatomic gas
I like what you've done there, but I think you need some modification of your final equation. The "approximately equals sign" isn't appropriate here because the omitted prefactor could be a number that's many orders of magnitude different than 1. The prefactor also depends on the temperature and the volume, and the functional dependance can be important. But with just a smidge more work, I think you've got the right answer.
Jun
2
comment Statistical physics of molecular dissociation of a diatomic gas
The final equation (N1^2 / N2 equals the Boltzmann factor), is incorrect. This is a big deal, since you claim the whole point is to understand that equation. To see that it's wrong, note that if you double the size of your box and double the number of particles (while keeping the temperature and density constant), your formula would predict the ratio of atoms to molecules would change. That ain't physical. The correct answer involves the densities of the two species, the Boltzmann factor, and some hbars and m's and T's.
May
27
comment What does it mean to say that the electron is a near-perfect sphere?
Chad Orzel has a delightful writeup of the experiment here: scienceblogs.com/principles/2011/05/…
May
26
comment What does it mean to say that the electron is a near-perfect sphere?
Vladimir: if the dipole moment wasn't parallel to the spin, the electron would have an additional degree of freedom. This would change the entire periodic table, since we could fit 4 electrons in the 1s orbital, etc. etc.
May
26
awarded  Editor
May
26
revised Which person can handle falls from big heights better: lighter or heavier?
added 5 characters in body
May
26
comment Which person can handle falls from big heights better: lighter or heavier?
"Better physically trained"? This is the physics stack exchange, not the physical training stack exchange :) But aside from that joke, I disagree that it's a tie even if the different weights have the same terminal velocity, due to the force of gravity at the surface (see my answer for more details).
May
26
answered Which person can handle falls from big heights better: lighter or heavier?
May
26
answered Why don't waves erase out each other when looking onto a wall?
May
26
comment Vapor pressure higher than equilibrium vapor pressure
Also, if you want strangers over the internet to diagnose why your experiment is working funny, it may be helpful to include more details (cell material, wall coating, temperatures, measurement method, etc.)
May
26
comment Vapor pressure higher than equilibrium vapor pressure
Alakali atoms can take a long time to migrate around a cell, so you may have to wait some time before the system reaches equilibrium and the pressure is dictated by the cold spot. Of course, I'm assuming that the hot spots are gravitationally elevated relative to the cold spots; if they're not, the liquid will run down and evaporate. One slightly more exotic possible cause: photodesorption.
May
26
comment What does it mean to say that the electron is a near-perfect sphere?
Jerry Schirmer is correct, but just to rephrase his answer so it answers the question as asked: "What does it mean that the electron is [almost a perfect] sphere?", it means that the electric dipole moment has been measured to be very small.
May
23
comment Is there a “Size” Cutoff to Quantum Behaviour?
Re: the baseball example, if you want to make a double-slit for a baseball, the slits have to be a few cm wide (otherwise the baseball can't get through). To see interference fringes from slits separated by a few cm, you'll need the deBroglie wavelength of to be on the order of a few cm. To get that, the baseball has to be moving incredibly slowly. So slowly that it'll take longer than the age of the universe to go through the slits. Waiting that long isn't technologically feasible. I've glossed over some math, and there are other issues as well, but I'll skip those for now.
May
23
comment Is there a “Size” Cutoff to Quantum Behaviour?
Re: the Reynolds-like cutoff: no, there isn't a known one, as described in the answer.