Is Entropy Reduced When Particles Combine?

1. Nov 29, 2012

referframe

Given a system of multiple free electrons. Say 2 of the electrons accidentally collide and become joined (opposite spin) by the weak force. So, the positions of those 2 electrons are now correlated.

Was the total entropy of the system reduced by those 2 electrons joining?

Thank you in advance.

2. Nov 29, 2012

Jazzdude

A system with known microstate has an entropy of 0, always.

3. Nov 29, 2012

Bill_K

Electrons do not form pairs with the weak force, so let's take a more realistic example: atoms that can form diatomic molecules. If a pair can form, a pair can also dissociate. In the equilibrium state there will be a certain ratio of atoms to molecules with a steady flux in both directions. The ratio is the one in which the chemical potentials of the two components are equal.

If your prepared initial condition is all atoms and no molecules, pairs will spontaneously form until the equilibrium ratio is attained. And anything that happens spontaneously is accompanied by an increase of entropy, not a decrease.

4. Nov 29, 2012

referframe

Interesting. You are saying that the QM amplitude and associated probability for the atoms to form molecules determines the final equilibrium ratio. That makes sense.

But I'm not sure I follow your "spontaneity = higher entropy" reasoning.

Also, if we treat our atoms and molecules as classical particles there is both an increase and a decrease in entropy. The entropy associated with the position and momentum degrees of freedom decreases because there are fewer "particles" in the end. But the entropy associated with particle type (atom or molecule) increases.

5. Nov 29, 2012

Darwin123

You can't calculate a change in entropy from a single collision. However, I think that I understand your question. Suppose you take a large number of electrons and smash them together. Your question is whether the entropy will probably increase.

The entropy will increase because the conservation laws force the interaction to spread to other particles. Other particles must interact in order to carry away the kinetic energy and the momentum of the original atoms.

I am not sure that a bound state of two electrons only can form through the weak nuclear interaction. Two electrons are negatively charged so they can not form a bound state through their electromagnetic interaction.

You conjectured a possible bound state involving the weak interaction. This has never been observed. However, the weak force is complicated. I don't know nuclear physics well enough to exclude that possibility. I conjecture that a bound state between two electrons can form through gravitational interaction. Again, I don't know general relativity well enough to discard this notion. For the sake of argument, let us assume that a bound state between two electrons is possible though any interaction at all. Let us talk about ANY two fundamental particles that can form a bound state through any sort of conservative force. I will call them electrons, but they can be any other fundamental particle.

The conservation laws have to be satisfied in this collision. Energy, linear momentum, angular momentum and charge have to be conserved. Therefore, at least one other particle has to be involved in the collision in order to conserve these properties. One can not have two electrons collide with each other and stick without a few other particles being created, destroyed, speeded up or slowed down.

You can do the calculation yourself. Use a reference frame with an origin at the center of mass of the two electrons. Determine what conditions are necessary for both electrons to stick together without violating conservation of energy and conservation of mass. Mathematically, you will find that it can't be done.

You will be changing the number of degrees of freedom by adding a particle, or especially a number of particles, to the collision. Under the conditions that allow the two electrons to stick together with high probability, the total entropy of the isolated system will increase.

Consider the special case of an electron and a positron colliding to form a bound state called a positronium. The positronium is unstable, but it can last a few microseconds. Maybe even milliseconds. The lifetime doesn't matter. The question is whether if one collides a large group of electrons at high speed together with a large number of positrons going at high speed, then will the entropy increase.

The answers would be yes. Here is why.

The electrons and the positrons each have high kinetic energy. However, the linear momentum of the electron-positron system in the center of mass reference frame is zero. If there was no third or fourth particle involved. momentum, the positronium would have zero momentum. However, that means that it has zero kinetic energy.

To be a bound state, a system of particles has to have a negative potential energy. However, the system of two electrons started with a large value of positive energy in the form of kinetic energy.

Therefore, the positive energy that was initially present in the system has to be carried away by a third particle. One way to do this is if particles form. For instance, there could be photons that form to carry the energy away. In high energy collisions, there could be additional electron-hole pairs that form during the collision. They can have kinetic energy to carry the energy away.

The additional particles will increase the entropy of the system because they represent additional degrees of freedom. Because they can be any arbitrary value of energy consistent with the conservation laws, the extra particles that form cause greater disorder.

In a high pressure system, particles could be destroyed. Believe it or not, this can cause an increase in entropy too. Under high pressure, any particle could be destroyed in the collision. Therefore, the disappearance of particles causes an increase in entropy.

Your implied example was impossible because of conservation laws. Two colliding particles can not form a bound state unless something carries away enough energy so the the two particle have a negative potential energy. The involvement of other particles necessary to remove the energy creates the entropy.

6. Nov 30, 2012

Bill_K

No, this is not the case. In nuclear alpha decay, for example, there are no other "somethings" involved. The reaction is simply N1 → N2 + α. The bound state has a positive potential energy, but the alpha particle is nevertheless held inside the nucleus by a large Coulomb barrier. It decays by penetrating the barrier. Likewise, under suitable conditions the reaction could go in reverse, in which the alpha is absorbed by N2, forming the bound state, with no other particles involved.

None of this alters the answer to the entropy argument.

7. Dec 4, 2012

Khashishi

Usually joining things together reduces the entropy. If you want to calculate the entropy of the two parts, you have to include both the internal entropy of the two parts, and the entropy allowed by the motion of the parts. A good example of this is in the Saha equation, which is used to calculate the ionization fraction of a plasma in equilibrium. Note that ionizing an atom increases the entropy, but requires energy to do so, so the reaction may or may not be spontaneous depending on the temperature.