We know the intensity distribution from a black-body at a fixed temperature using Planck's equation. During evaluating density of states we use a factor of 2 (for two polarization states: under-graduate text level explanation). Now suppose we have a hypothetical black-body emitting massive bosons (W+, W-, Z0, etc.) then what will be the factor? Whether it will remain same or change? If it is changed than what will happen to the intensity distribution shape?
I don't know David. I am also wondering. My understanding on this matter is only at best at undergrad level. I was recently asked this question by some other at a presentation meeting? I don't have an idea about the authenticity of the question itself. But thought better to get information from experts in this field.
But the fact of spin multiplicity of photon is correct. i checked.
Black bodies do radiate mass-energy. They lose mass as the energy radiates away. I expect that is what the (fairly) recent discussion about radiation of black holes was about.
thanks to all of you for participating in the discussion. interesting views. Will look forward to receive many more views.
BTW any reading material on this topic. Suggestions please.
How blackbody can radiate rest mass? As far I am concern it doesn't radiate rest mass we can discuss rest mass energy but not just the mass .
Basit you are absolutely correct. It will emit rest mass energy. But then there will be a difference in rest mass energy of photon and massive bosons. Also there will be a difference in density of states. I was wondering whether the end intensity vs. wavelength graph remains same or shows some sort of different nature with respect to the photon case?
Basit and Barid. When it radiates, energy leaves, and the mass remaining is smaller. It doesn't radiate mass, but mass is converted to energy which is radiated, so the mass goes away. Mass and energy are descriptions of the same thing measured in different ways.
- Badges
- Science topic