# Do solids have translational energy?

Along with having vibrational energy, do both crystalline and amorphous solids also have translational energy?

I ask because I've always understood solids to have just vibrational motion/energy. But I've heard it said that, along with having vibrational energy, solids also have translational energy (because temperature is defined as a measure of the average molecular translational energy in a system). If this is true, how can solids have translational energy if they don't have translational motion?

• If they aren't translating, there is no translational energy. But, when you through your book across the room in frustration there is translation and translational energy. Oct 4, 2015 at 19:03
• @JonCuster but if the book always remains still, how is the temperature of that object achieved if temperature is defined as a measure of the average molecular translational energy? Oct 4, 2015 at 19:08
• Because all the molecules are jiggling around happily, with no net translation of the book as a whole... Oct 4, 2015 at 19:10
• @JonCuster when you say 'jiggling around', are you referring to the molecular vibrations that occur due to atoms in a molecule being in periodic motion? Therefore, temperature due to vibrational energy? Oct 4, 2015 at 19:20
• I think your original source is causing problems by using the word "translational" in this context. I've never heard it used that way. One usually says something like "The thermal energy of a solid is the total kinetic energy of the vibrations of all of the atoms in the solid". As your intuition is telling you, "translation" is usually reserved for the motion of the center of mass. Oct 4, 2015 at 20:50

The same way any substance can have transnational energy.

It arises because the energy of motion is proportional to the square of the velocity so even if the average is zero, the average of the square is not.

If the mean velocity of the particles is $\bar v$ then the instantaneous transnational energy would be:

$$\frac12 m\sum_i (v_i-\bar v)^2$$

$$\frac12 m\sum_i (v_i^2 -2v_i\,\bar v + \bar v^2)$$

$$\frac12 m \sum_i v_i^2- m \sum_i v_i\,\bar v + \frac12 \, n \, m \, \bar v^2$$

$$\frac12 m \sum_i v_i^2- m \, n \, \bar v^2 + \frac12 \, n \, m \, \bar v^2$$

$$\frac12 m \sum_i v_i^2- \frac12 m \, n \, \bar v^2$$

So this is the kinetic energy of all the particles minus the kinetic energy of the bulk movement. Even when the bulk movement is zero there's still the vibrational velocity that contributes to the energy.