# Physics of a cold and hot top

Imagine two tops made up of exactly one thousand atoms. One is kept at 4 degrees Kelvin, the other at room temperature.
1. Would they weigh the same given an arbitrarily precise scale in the Earth's gravitational field?
2. If both were spun at exactly 100 rpm, would they have the same moment of inertia?

Here is my own initial idea. Gravitational forces bind to energy density. As such, the hotter top would weigh more on a scale because heat contributes to the energy density of the top. The rest mass of a thousand atoms is the same no matter what its temperature. As such, the cold and hot top would have the same moment of inertia.

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4 degrees what? And I guess I agree with your statement about "gravitational forces bind to energy density", but under the principle of equivalence, I think the '$m$' in $I=\sum mr^2$ is the same as the '$m$' in $mg$. –  levitopher May 13 '13 at 5:32
However, the $r$ in $I = mr^2$ will be slightly larger in the room temperature top due to thermal expansion –  Lagerbaer May 17 '13 at 2:54

The rest mass of a thousand atoms is the same no matter what its temperature

Nope. Here's an equation that you probably have heard of:

$$E=m_0c^2$$

As you supply heat to a body at rest, the rest mass of the bulk body increases. One way of looking at it is that the individual molecules are no longer at rest and have kinetic energy (which contributes to relativistic mass).

Thus, for the purposes of both inertia and gravity, a heated body has more mass in a relativistic framework. So moment of inertia will increase (if we neglect thermal expansion).

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I think my mini-struggle was where to put the heat energy. The equation I always think about is (pseudo-latex): m^2 c^4 = E^2 - P^2 c^2. The heat can be thought of as adding a bit to E term. As such the invariant mass remains invariant to all observers, and all observers will agree that the cold atoms have a lower mass than the hot atoms. –  sweetser May 19 '13 at 4:01
@sweetser: Yep. (btw latex works in comments, use dollar symbols). That equation is usually the best way of looking at such things. $E=mc^2$ gets confusing if the particle is moving too much :) –  Manishearth May 19 '13 at 6:19