# Kinetic energy, heat, and reference frame

1. Jul 21, 2010

### FDGSa

I'm having some trouble reconciling the following facts: 1) that kinetic energy depends on v _squared_, 2) at the same time energy is conserved in all reference frames, and 3) reference frames transform linearly in v in classical mechanics (galileo transform). I've basically been able to boil down my confusion to the following thought experiment:

A "train" weighing 1 kg moving at 10 m/s slows down to 1m/s by braking. In the frame of the railroad tracks it loses 10^2 - 1^2 = 99 joules of kinetic energy which presumably is all transferred into the railroad tracks in the form of heat.

On the other hand, in the frame of a person walking 1m/s along the side of the train, it slowed down from 9m/s to 0 m/s, losing 9^2 - 0 = 81 joules of energy into heat in the tracks.

This can't both be right - all observers should be able to agree on how much thermal energy the track has!!

The only thing I can think of so far is that perhaps the idea of the "thermal energy" of an object only applies in a frame where that object is stationary. However I do not find this explanation satisfactory since an object should conduct heat to it's surroundings at a rate proportional to the temperature difference regardless of whether it is stationary or moving.

2. Jul 21, 2010

### hikaru1221

First, the train's mass should be 2kg. Check your calculation

Heat by friction = work by friction = work by friction on train Wt + work by friction on ground (or earth) Wg. In the first frame in which the ground is stationery, Wg=0. In the second one, Wg is not zero.

Denote m and M the masses of the train and the earth respectively (note that m<<M). Take m=2kg for convenience. Let's assume that the earth behaves like a mass point, so that we can ignore the rotation effect for simplicity; the final conclusion won't change. Consider the second frame. The train decelerates from 9m/s to 0m/s. The initial speed of the earth is 1m/s.
_ Due to the momentum conservation, we have: m*9 + M*1 = m*0 + M*v.
_ Therefore, the final speed of the earth is: v = 1 + 9m/M.
_ The work by friction done on the earth:
Wg = 0.5*M*(v^2-1^2) = 0.5*M*(18m/M + 81*(m/M)^2) = 0.5*M* 18m/M (approx. as m<<M) = 9m = 18 (J).
Now you see that Wg=18 J, Wt=81 J, the total heat = Wg+Wt = 99 J as expected.

We can also prove that since friction is internal force between train and earth, work rate (power) done by friction on the system train-earth only depends on the relative velocity between earth and train, which means it is independent from the reference frame chosen.

3. Jul 21, 2010

### FDGSa

Thank you for the response. So then if the heat capacity of the rails is 1 K/J, the rails would in fact heat up by only 81 K, and the other 18 J would go into mechanically changing the momentum of the rails/earth.

I somehow had convinced myself that the work done to change the velocity of the earth could be neglected by a scaling argument when it cannot.

In fact, it seems this is a circuitous proof that kinetic energy must depend on v squared (otherwise the math wouldn't work out such that everyone agrees on the change in temperature of the rails).

4. Jul 21, 2010

### hikaru1221

Uhm, no. Friction, which is the cause for momentum change, transfers kinetic energy into heat. The amount of heat in this case is 99 J, not 81 J.

If you say there is only 81 J that goes to heat in the rails, then in the 1st reference frame, there will be 99 J of heat going to the rails! You see the problem?

All 99 J of kinetic energy goes to heat in the train and the rail. In fact, we cannot conclude anything about how much heat each receives without further information (what kind of information? I don't know; I'm no expert). All we know is the total heat = 99 J.

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