# Thermodynamics and gravity

A gas cloud shrinks due to gravitational forces. The gas gets hotter because the velocity (hence kinetic energy) of the gas particles gets larger. At the same time the potential energy gets smaller because the gas particles are closer to each other.

Once you get past the fact that the decrease in potential energy is greater than the increase in kinetic energy (hence the gas gets hotter even though the total energy decreases), the next question is this: what happens to the lost energy?

Can this question be answered without resorting to GR?

I don´t understand your position clearly. But I think there´s thousands of ways for the ´lost energy´ to go: photons, internal energy for gas particles,etc. Otherwise, the increase of kinetic energy should be equal to the decrease of potential energy.

russ_watters
Mentor
ddesai said:
the decrease in potential energy is greater than the increase in kinetic energy (hence the gas gets hotter even though the total energy decreases),
Where do you get that from?

See: http://math.ucr.edu/home/baez/entropy.html for the source of the question. Baez show that the the decrease in potential energy is greater than the increase in kinetic energy: overall the internal energy decreases.

This is an ideal* gas in which particles are pointlike. We don't have em radiation since we are assuming the particles don't really have internal state (or we are neglecting this effect) and there are no significant photon sources to interact with.

The only thing left is gravity. In GR you have energy radiated away as gravity waves.

But in newtonian physics? There is no such thing as gravity waves. So how do we explain the decrease in total energy?

*Not really an ideal gas, since the particles do interact with each other via gravity.

Gravity and the 1st law of thermodynamics

I have been reading the responses to the possibility of non-linear energy conversion and I too am skeptical of many, if not all of the claims out there. However, I have recently published a web page that you might find interesting http://www.geocities.com/joeiii63/
My angle is simple and straight forward. If the value of the inertial mass (read apparent mass) in an isolated system is varied during the system's cycle, then the effect is exactly the same as if the gravitational field affecting it were fluctuated between weak and strong. This can be used to cause a weight to behave as though it were lighter on its way up and heavier on its way down.
This invitation has been extended to researchers, university physics depts. (including MIT and Cambridge) and skeptics worldwide. As of yet, no one has indicated why this system wouldn't or can't work (invoking the 1st Law of Thermodynamics in this case is no different than saying that since humans can't fly, airplanes must be impossible because they would allow humans to fly).
So far the responses have been mostly polite, but similar; they all want more time to review the idea. If you understand physics, its just not that complicated.

What is meant by reversing the system is this: The weight doesn't fall up, it is lifted by the kinetic energy stored in the flywheel and the upward momentum of the weight. At the start, the weight falls in a normal flywheel configuration and its potential energy is divided between the flywheel and the weight. At the end of this half cycle, the motion of the weight is reversed, possibly by a spring, and it re-enters the system on the other side of the flywheel axle (the upward side). Yes, some velocity (and thus kinetic energy) will be lost, but not enough to overcome the large advantage provided by the fact that as the weight is raised, its deceleration due to the pull of gravity diminishes as it gets higher. Because the weight is being lifted by the variable inertia configuration, its deceleration drops from 0.5g at the bottom to 0.2g at the top. It's exactly as though the gravitational field acting on the weight is weakening.
Remember, when the weight fell, it was connected to the flywheel in the non-variable (normal) configuration, and it had a constant acceleration of 0.5g.
This means that using the variable inertia configuration, a weight can be lifted to a higher elevation than it falls from, using nothing more than the kinetic energy produced by its own fall through a non-varying inertia configuration.

If anyone should insist on knowing where the extra energy comes from when the system is reversed, they need only to figure out where the missing energy goes when the system is operated in the direction described on the web page. Therein lies the answer. It doesn't go anywhere. It just disappears. And when the system is reversed, energy simply appears. It was never anywhere.

ddesai said:
See: http://math.ucr.edu/home/baez/entropy.html for the source of the question. Baez show that the the decrease in potential energy is greater than the increase in kinetic energy: overall the internal energy decreases.

This is an ideal* gas in which particles are pointlike. We don't have em radiation since we are assuming the particles don't really have internal state (or we are neglecting this effect) and there are no significant photon sources to interact with.
Baez is not claiming that the energy of an isolated system can change. What that page means is that if the overall energy of the gas cloud decreases (which it will do by radiation) then the temperature will increase.

Yes, never assumed it was an isolated system. But you are right.

He uses the Virial Theorem to show that, under certain conditions, the total energy is equal to minus the kinetic energy or equal to the one-half the potential energy. This means the rate of change of kinetic energy with respect to total energy is -1 and the rate of change of potential energy with respect to energy is 2.

This means the potential energy will decrease faster than the kinetic energy will increase as the clound looses energy (and hence volume).

Baez relaxes the "energy is constant" constraint assuming that there is some process by which energy is taken away from the system.

At the end of ther article he asks "where does the lost energy go?" This was the question I was trying to answer.

I think I have a tentative answer. In newtonian physics, if the gas is ideal, then the energy must be constant: there is no energy loss, because there is nothing to carry the energy away.

In quantum mechanics, where atoms have internal states, you could have energy loss via inelastic collisions and the emission of photons.

But if we ignore QM and treat this as an ideal gas as Baez does, then what does he mean by "where does the energy go?" The only thing I could think of was gravity waves. Where else can the energy go in this model?

And it is surprising that you can conclude, amoung other things, that as potential energy decreases faster than the kinetic energy increases just by resorting to the Virial Theorem and not specifying how the energy is lost.

Also, the Virial theorem here assumes the existence of an inverse square law but does it still hold if there are other processes (inelastic collisions, gravity waves, or whatever) at work? Have to think about this.