- #31
Count Iblis
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ffleming7 said:
Yes, absolute zero can be reached, although not using thermodynamical means. And no, temperature is not kinetic energy, despite what most high school physics textbooks will tell you.
Temperature is a statistical concept that arises when you describe a system consisting of many particles using a few macroscopically measurable variables like volume, pressure etc. Clearly, to be able to do statistical computations, you need to know the probabilities of the system being in some arbitrary state. It turns out that for an isolated system the laws of physics suggest that all accessible states are a priori equally likely. So, it's a bit like throwing dice, all the possible throws are equally likely. And just like dice throwing problems can be solved by counting in how many ways a certain outcomes can be realized, in physics we just neet to know how many states there are for each macrostate. I.e. given the volume, total internal energy, how many energy levels are there compatible with these macroscopic variables? All these energy levels the system can be in are then equally probable.
A minor complication is the following. If we specify the internal energy with infinite accuracy, then there usually can be only one state the system can be in (ignoring possible exact degeneracies). If the system consists of a large number of particles, then the spacing between energy levels will decrease (simply because the more particles you have, the more different states between two given total energies you can make). So, what we do is we specify a small energy interval \Delta E and when we say that the total energy is E we mean that the system must be in some energy level with an energy between E and E+\Delta E. It turns out that the choice of \Delta E does not affect the outcome of calculations when the limit of system size to inifinity is taken (the so-called thermodynamic limit).
What we need to know to do statistics is, given some energy E, how many states (energy levels) are there between energy E and E+\Delta E? It is customary to denote this quantity (so-called multiplicity fiunction) as \Omega\left(E\right).
To see how all this leads to something like "temperature", consider two systems with multiplicity functions \Omega_{1}\left(E_{1}\right) and \Omega_{2}\left(E_{2}\right). The number of states available for the combined system must be the product:
\Omega_{1}\left(E_{1}\right)\times \Omega_{2}\left(E_{2}\right)
Suppose that we bring the two systems into contact, so that energy can flow from one system to the other. If all microstates are equally likely, then in thermal equilibrium, you will get that outcome for which there are the largest number of microstates. If the total energy is E, then we can put E_{2}=E - E_{1} and compute for which value for E1 the above expression is maximal. It is more convenient to maximize the logarithm:
\frac{d \log\left[\Omega_1\left(E_1\right)\right]}{dE_1} = - \frac{d \log\left[\Omega_2\left(E-E_1\right)\right]}{dE_1}
Or, rewriting the r.h.s. in terms of E_2:
\frac{d \log\left[\Omega_1\left(E_1\right)\right]}{dE_1} = \frac{d \log\left[\Omega_2\left(E_2\right)\right]}{dE_2}
So, what we see is that the derivatives of the logarithms of the multiplicity functions w.r.t. energy become equal in thermal equilibrium. One defines the temperature as being inversely proportional to this quantity.
Now, with a little more work you can show that at zero temperature a system will be in the ground state. So, the question is really, can we put a system in its ground state? The answer is, of course we can! In principle, even for a system containing 10^{30} particles, you can put the entire system in its ground state simply by manipulating the individual particles. Quantum mechanics will not prevent you from doing that.
However, you cannot do it using thermodynamic means only (i.e. by manipulating the few macroscopic variables or by bringing the system into thermal contact with another system). The reason is that thermal contact does not work because heat flows only from hot to cold, so you would need something colder in the first place. The only other option left is by letting the system perform work. But this will leave the multiplicity function \Omega the same at best. Going to lower \Omega is statistically ruled out because all states being equally likely, lower \Omega has less probability. Now, a little less proabability would be no problem, however, at absolute zero the system has \Omega=1 because we then know it to be in the (unique) ground state. The system at even very low but nonzero temperatures will have an astronomically large value for \Omega.
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