# Do excited states contribute to temperature?

Consider a gas of molecules, each of which is in their ground state. Each moves with velocity vi, and their average kinetic energy is the temperature of the gas.

Now imagine each molecule absorbs a photon and is put into the same excited state, but each molecule still has the same velocity vi. Has the temperature of the gas changed?

• It's not in thermodynamic equilibrium yet since in equilibrium each degree of freedom should have same average energy. But sometimes it is possible to define individual temperatures for separate degrees of freedom, e.g., in magnetized plasma it can be different temperatures for motion parallel to the magnetic field and perpendicular to it. May 23 '18 at 1:20

Now imagine each molecule absorbs a photon and is put into the same excited state,

Temperature is a thermodynamic variable, and in a gas is connected to the kinetic energy distribution of the individual molecules:$${KE}_{\text{avg}}=\left<{\frac{1}{2}mv^2}\right>=\frac{3}{2}kT \,.$$

but each molecule still has the same velocity vi. Has the temperature of the gas changed?

You forget conservation of momentum, each photon also carries momentum, the velocity of the molecules will be modified,increased or decreased depending on the direction of the photon and molecule vector, so statistically the kinetic temperature may go up, particularly if the photons are in a beam.

BUT:

Molecules are not stable in a higher excited state, and cannot maintain a thermodynamic equilibrium for measuring the temperature, as they will start decaying to the lower state according to the lifetime of the state.

So the answer to the title is: no, the thermodynamic equilibrium cannot be reached so as to measure changes in temperature.