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Hi, I am a string theorist and a publicist.


3h
comment Inconsistency between Helmholtz and Gibbs Free Energies
Hi Tom, well, take a mixture of water and ice at the melting point. It has a well-defined pressure, temperature, and number of molecules, but the fraction the ice may be anything - you have to add one full intensive variable (the percentage of ice) to describe the system. So G can have any value in an interval for those values. This mixture of ice and water is really an optimized replacement of some "forbidden part" of the phase diagram.
19h
answered what physical quantity do real scalar field operators create/destroy?
1d
awarded  Nice Answer
1d
awarded  Nice Answer
2d
awarded  Enlightened
2d
awarded  Nice Answer
2d
answered Is voltage always proportional to its derivative?
Sep
13
awarded  Good Answer
Sep
13
comment If water evaporates at below 100 degrees then why not stone?
Phonon, nice, but is the rate of sublimation stictly zero for a rock at room temperature, or just approximately so? Is the rate the exponential of a huge negative number? What is it and why is it large?
Sep
13
revised Inconsistency between Helmholtz and Gibbs Free Energies
added 268 characters in body
Sep
13
answered Inconsistency between Helmholtz and Gibbs Free Energies
Sep
10
awarded  Nice Answer
Sep
10
comment Showing a fourth rank tensor in $\epsilon$'s reduces to one in the metric $g$
Yes, it is never equal (by equivalent, did you mean equal?). Just substitute $\mu\nu\rho\sigma=1123$ and $x=(0,0,1,1)$, for example. Only $g_{\mu\nu}$ is nonzero among all the metric coefficients, so the first expression is zero while the second is not.
Sep
7
awarded  history
Sep
6
comment Why black body radiation is all over the frequency range
OK, I understood you well but what you wrote is wrong, agreed? You can't write "distribution" if you mean "energy of a particular particle". The distributions are fixed by the temperature and saying that "anything goes" is exactly as wrong as saying that "everything must have one frequency", although these two claims may err on the opposite sides in some sense.
Sep
6
comment Why black body radiation is all over the frequency range
The fact that the temperature is an average doesn't mean that the distributions are not unambiguously determined at thermal equilibrium. They are. Have you heard of Maxwell-Boltzmann distribution for an ideal gas, for example? You may calculate the average even for other distributions but if the distribution doesn't agree with the Maxwell-Boltzmann distribution (e.g. for a gas), then there won't be any equilibrium and you shouldn't say that the average energy you calculate determines a well-defined temperature! Out of equilibrium, the temperature isn't well-defined.
Sep
6
comment Why black body radiation is all over the frequency range
It's not true at all, Anna. Thermal equilibrium dictates all statistical distributions of energy and every other quantity, given a well-defined physical system. All the distributions are ultimately derived from the Boltzmann one. This distribution is nonzero everywhere - not peaked at particular frequencies.
Sep
6
comment Why black body radiation is all over the frequency range
What radiates are not free electrons but bound objects like atoms or - which is better to imagine - harmonic oscillators from springs around the lattice sites. They have different frequencies but each of them has $kT/2$ average energy per degree of freedom. Free electrons, unless they collide with something, are moving by a constant speed and they don't radiate. The speed of the overall motion of a particle (like a free atom or electron) is also dictated by having $kT/2$ per degree of freedom in average but it isn't "exactly" this much for each particle. There is a calculable distribution.
Sep
6
comment How is weight distributed when legs are astride?
Because I don't want to calculate these things with the numbers.
Sep
6
comment How is weight distributed when legs are astride?
You write down the equation that the total forces on all parts of the feet from the floor are equal to the weight of the person; and the total torque $\vec r_i \times \vec F_i$ relatively to the person's center of mass is zero.