Re: In energy transformation even on subatomic levels is thermal enengy lost

Bruce,

Good question. The quick answer is "sometimes". As it happens, I recently answered another question concerning the appplication of entropy and the second law of thermodynamics to the energy of a single particle. What I explained in my answer is that thermodynamics applies to systems of particles, which have a natural tendency to evolve to less organized, more "random" states. I included a simple numerical example explaining entropy, which is one of the most mis-understood concepts of physics. I suggest you take a look at it before reading the rest of my answer.

Let me invoke another popular analogy for entropy, involving a pool table and a videocamera. Imagine that you've just racked up 15 balls at one end of the table and placed the cue ball on the other end. Everything is in a nice, neat pattern. Now shoot your cue ball at the other 15, and watch the balls scatter all over the place, running into one another and the bumpers on the sides. Even starting from the same initial conditions, no two breaks will be exactly alike. So the "broken" solution can be said to have more states available to it, and therefore higher entropy.

Now take your camera and zoom in on just a small area of the table where two balls are colliding elastically. You could take that collision and run it in reverse, and it would still look perfectly normal. Two particles on their own do not create additional "disorder" by colliding elastically. It is only when you zoom out over the whole table that you see which way time is really going, because you would never expect the balls to somehow coalesce from a random distribution back to the initial "racked up" state.

So, in a handwaving way I am offering you an informal test for determining whether a process increases entropy (and therefore loses thermal energy, as per your question). If you can "run the tape backwards" and the process looks equally probable, then the system does not increase entropy.

So what about subatomic processes? Well, many of them evolve elastic processes where an equal number of particles enter and leave the interaction point:

e -  +   gamma --> e-   +   gamma       (electron photon scattering)

e+   +   e-   -->  gamma + gamma      (electron-positron annihilation)

e   +   u  -->  neutrino + d       (weak electron-quark scattering)

and so forth. The thing about all these examples is that if you turn the arrow the other way, the reverse process is also perfectly valid, so you wouldn't know one direction from the other. So I would not consider these processes to increase entropy (and thus lose thermal energy).

On the other hand, there are also inelastic processes, such as this electron-proton "Deep Inelastic Scattering" process:

e- + proton  --->  e-  + many hadrons from scattered quark and proton remnant

In a process like this, a couple of particles collide and create a huge shower of outgoing particles that themselves decay into even more particles. So this is an event that wouldn't look at all likely if you ran it in reverse. Therefore I would say that these subatomic collisions do increase entropy, and therefore lose thermal energy.

So to summarize, subatomic collisions can be elastic or inelastic. Collisions that are inelastic, or which have more particles going out of the interaction point than come in, will have a greater entropy, and therefore lose thermal energy according to the classical definition. Otherwise it is possible for an elastic process to be reversible, and therefore not increase entropy.

I hope this helps. Please contact the Mad Scientist Network again if you need more clarification.

Regards,

Sam Silverstein