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seen Dec 11 at 6:34

Feb
1
answered Is there symmetry in 2d stress tensor in linear elastic fracture mechanics?
Jan
29
answered One dimensional Schrödinger equation equation with initial condition, finding the probability of the particle's future position
Jan
22
comment What are conditions for the existence of a critical value (for a phase transition)?
The Hamiltonian has dimensions of energy. It must contain some interaction constant with dimensions of energy (in the Ising model - the interaction $J$. In liquid-gas transition - the Lenard-Jones energy scale, etc..)
Jan
22
comment How do electrical devices suck electricity?
Tiny correction: the question refers, most probably, to an AC case, where a factor of $\sqrt{2}$ has to be added.
Jan
22
comment What are conditions for the existence of a critical value (for a phase transition)?
I cannot imagine such a case. If the phase transition occurs as a function of $T$, this means that you're working with a Hamiltonian, and average expressions of the form $e^{-\beta H}$. And if you have a Hamiltonian, you have an energy scale.
Jan
22
comment What are conditions for the existence of a critical value (for a phase transition)?
If you set $k_b=1$, and there's no reason not to do so, then temperature is measured energy units. So clearly, if you multiply ALL energy scales in your system by 2, then $T_c$ will also be multiplied by 2. This is what I meant. I can't quite understand your question beyond that.
Jan
22
answered What are conditions for the existence of a critical value (for a phase transition)?
Jan
19
comment how to represent the effect of linking rigid-bodies together?
Why don't you just treat them as a single, large rigid body?
Jan
18
comment How to “read” the temperature of an abstract system?
When you couple your system to a bath, you actually allow it to exchange heat with it. The total system (your system and the bath) will be in the most probable state - the one with the highest entropy. Since the total entropy is the sum of the bath's and the system's entropy, the total entropy is maximal when $$\frac{\partial S_{bath}}{\partial E}=\frac{\partial S_{system}}{\partial E}\ .$$ This quantity is defined to be $1/T$, and the probability to find your system in an energy $E$ goes like $e^{-E/kT}$, as can be seen by taking a first order Taylor series around the maximum.
Jan
17
comment How to “read” the temperature of an abstract system?
forget what I wrote about f. It is confusing and wrong. In this formulation the free energy is given by $f=-k_bT\log Z=k_bT\log \sum_i e^{-\beta E_i}$, and the mean energy is calculated by averaging $\langle E\rangle=Z^{-1}\sum E_i e^{-\beta E_i}=-\partial_\beta \log(Z)$
Jan
17
revised How to “read” the temperature of an abstract system?
erased wrong statement
Jan
17
comment How to “read” the temperature of an abstract system?
The point is showing that coupling your system to an external bath results in a probability that goes like $e^{-\beta E}$, for some $\beta$. This is te definition of $T$, and is crux of the matter. Until you've shown this you don't know that "T will just take its value" from the bath, because $T$ is not even defined.
Jan
17
revised How to “read” the temperature of an abstract system?
book reference
Jan
17
comment How to “read” the temperature of an abstract system?
I posted a real, elaborate answer. Hope that gets the job done.
Jan
17
answered How to “read” the temperature of an abstract system?
Jan
17
revised Do amorphous metals undergo conchoidal fracture?
added 100 characters in body
Jan
17
comment How to “read” the temperature of an abstract system?
Systems flow towards probable states, that is - states that have a large number of micro-states. The (log of the) number of micro states is the entropy. So systems "want" to be in (="go to") states with high entropy, but that usually means states with high energy. The trade-off between entropy and energy is exactly the temperature. In evaporating liquid, for example, the vapor state has a higher entropy, but also a higher energy. Therefore, at low $T$ the system is liquid, but when $T$ is high enough the system prefers to be at a higher energetic state (="pay") because its entropy is higher.
Jan
16
answered Do amorphous metals undergo conchoidal fracture?
Jan
16
comment How to “read” the temperature of an abstract system?
You wrote the answer. This is the definition of $T$. In the Ising model. for example, there is no sense in talking about "average kinetic energy of the spins" or anything of that sort. I'd say that a good way to think about the temperature in this case is "how far above the ground state can I go", which is roughly equivalent to "how much energy can I pay in order to buy some entropy"?
Jan
16
answered Hydrostatic equilibrium of a star derivation