St. Petersburg - beautiful city of Dostoyevsky -19C today!
January 28, 2019
'El Greco Und Ich' - in hardback book and audiobook! Such a dream... out on 14th August 2018 in Germany.
August 12, 2018
All welcome - Longhorsley, Northumberland. Excited - onstage with 3 wonderful authors!
September 22, 2019
November 3, 2017
This is what can happen if you search for a word like 'molecular' to complement a phrase you have written....
Strictly speaking, there are no perfectly elastic collisions between atoms and molecules. Some part of the energy/momentum will be converted into photons. So, in reality, a real gas in a perfectly isolated volume is always two gases: one gas made of massive particles and one gas made of photons. The two components will be in thermodynamic equilibrium with each other. However, the energy density in the photon gas is usually so small, that it doesn't matter for the purposes of thermodynamics at the "room temperature" scale. – CuriousOne Oct 1 '14 at 2:36
So I just put the numbers into the formula for the photon gas energy density u=π2k415c3ℏ3T4u=π2k415c3ℏ3T4, and if I didn't make a mistake, at 300K the energy density is approx. 10−16J/m310−16J/m3. This means that around 10000∗300K10000∗300K the photon gas energy density rises to 1J/m31J/m3 and at 30 million K it is already 10,000J/m310,000J/m3. Given the energy densities in magnetically confined fusion reactors these are enormous quantities (given the fact that this photon gas also escapes at the speed of light!) and fusion reactor physics has to control the loss of energy trough photons very carefully.