Bill_K said:
No, in fact energy is conserved.
This seems to disagree with the previous statement:
Bill_K said:
Dickfore, You know as well as I do that the kinetic energy of the decay products must be positive, but here's my point.
It is true that total relativistic energy is conserved, but kinetic energy is NOT. This is because part of the rest energy of the original nucleus is converted into kinetic energy of the products.
We can look at it in another way: Binding energy for a nucleus is defined as:
<br />
B = \sum{\mathrm{rest \; energies \; of \; protons \; and \; neutrons}} - \mathrm{rest \; energy \; of \; nucleus}<br />
This is a positive quantity. The bigger it is, the more stable an isotope is. Why? Because it will take that much energy to disintegrate the nucleus into its free constituents. Conversly, we would gain that much energy if we assembled the nucleus from its constituents.
Now, you can think of a reaction in the following way:
1) Disintegrate nucleus X. You need to spend B(X) amount of energy to do so;
2) Use some of the protons and neutrons to make Y. You gain B(Y) amount of energy;
3) Use the rest of the protons and neutrons to make Z. You gain B(Z) amount of energy.
All in all, you had gained:
<br />
Q = B(Y) + B(Z) - B(X)<br />
amount of energy. If Q > 0, the reaction is exothermic and it can spontaneously happen. The energy difference is distributed along the constituents, making sure linear momentum is conserved. If Q < 0, the reaction is endothermic and the particle X is stable from decaying through the channel described by:
<br />
X \rightarrow Y + Z<br />
Bill_K said:
So why does X → Y + Z happen and not the inverse, Y + Z → X? If there were just a single state in the decay, the inverse process would happen just as often, and an equilibrium would be reached. But in fact Y + Z represents a large number of states, and the odds of X → Y + Z happening outweighs the chance of Y + Z → X. Although the inverse reaction is possible, it would be unlikely for Y and Z to come together in just the right way to reform X.
I think you need to specify what exactly you mean by 'state' because:
Bill_K said:
This is exactly the same reason why a collection of air molecules will expand to fill a room rather than congregate in one corner. Both configurations are energetically the same, but one is statistically more likely.
the notion of state in this case means a point (or a small volume) in the 6N - dimensional phase space of the gas of N molecules. According to the laws of statistical physics, in the limit N \rightarrow \infty, there is a uniform probability density of occupying a particular volume in phase space as long as that part of the phase space is consistent with conservation laws (energy, momentum, linear momentum). Entropy is defined as the logarithm of this volume (measured in units of (2\pi \hbar)^{3 N}) (measured in units of k_{B} - Boltzmann constant).
Then, the law of increase of entropy means that the equilibrium
macroscopic state of a
macrocopic system is the one that can be realized by the biggest volume in phase space.
Bill_K said:
So it is increase in entropy that drives an isolated nucleus to decay, not an urge to decrease its energy. In an environment where Y and Z are plentiful, the inverse reaction would be more likely, resulting in a tendency to form the nucleus X, which has more binding energy, not less.
As I said, entropy (unlike energy) is a concept that has a meaning only for macroscopic systems. A nucleus is not such a system. It is an experimental fact that the half-life of a nucleus is an intrinsic property independent of the distribution of other nuclei around it. Therefore, your analogy is meaningless.
