# Quantum Conservation versus classical conservation

If energy is conserved in all quantum mechanical interactions, how are there classical interactions in which energy is not conserved, given that classical interactions are a macroscopic approximation to the quantum mechanical interactions?

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I think your first sentence should read "If all q.m. interactions conserve energy, how are there classical non-energy-conserving interactions....". – twistor59 Dec 13 '12 at 17:54
Yes, I think I was being sloppy. – mnky9800n Dec 13 '12 at 18:08
OK I edited it for you. – twistor59 Dec 13 '12 at 19:01
Can you give an example of a "classical interaction in which energy is not conserved"? – Mark Eichenlaub Dec 13 '12 at 19:08
A block sliding on a plane at a constant velocity produces heat (from friction) yet classically the work done is zero. – mnky9800n Dec 13 '12 at 19:10
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Macroscopically, a differentiation between usable and unusable energy is sensible.

Microscopically, the energy of an atom moving with a given speed $\vec v$ is $\frac{1}{2} m v^2$, and if you have $n$ such atoms, the total kinetic energy is simply $\frac{n}{2} m v^2$. This is independent of whether all the $\vec v$ point in the same direction (if a liquid flows through a pipe) or they are directed randomly (think of a perfect gas).

Macroscopically, the first case describes a liquid whose energy is usable: You can put a fan into the pipe and ‘generate’ electricity. The kinetic energy of the gas, however, cannot be recovered (that easily): if you put a fan into the gas, there is an equal chance that an atom will hit it in the one direction as in the other, causing no effective movement at all.

To sum up: Energy is also conserved classically/macroscopically, however, some of that energy is in a state which makes it unusable to us, which is often described as heat.

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Energy is always conserved in any classical interaction. This is an universal law named the law of conservation of energy or first law of thermodynamics.

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 Could the smart guy who voted this down explain why? Thanks – juanrga Dec 14 '12 at 16:42