Recent content by LocationX

Once again, my fundamental question has not been answered. It's is obvious that I do not understand blackbody radiation. I know what it describes, but those are only words without substance. Saying a blackbody is a probability distribution for all wave lengths is the same as saying BE...

and neither is hydrogen gas... but we see discrete spectra

Ok, sorry. My question is actually related to why does hydrogen gas produce a discrete spectra and blackbodies produce a continuous spectra. I'm looking for a physical, microscopic reasoning regarding this. As in what happens inside atoms of blackbodies that allows this continuous emission of...

Who said I had a bunch of bosons? We keep a piece of steal at a temperature that makes it glow red, thus we have just confined it to emit a certain wavelength. What's wrong with that?Again, we heat to a very specific temperature T=X>>0K. Boltzmann distribution tells us the number of states...

So a Sodium atom does not have discrete energy levels? Even if NIST has measured these very energy levels to precise eV INCLUDING spin orbit coupling and hyperfine splitting? What you're saying doesn't make sense. How would you even include the coupling to vacuum in the Hamiltonian? Are you...

And the harmonic oscillators represent the atoms in a lattice? For example heating steal will cause it to glow various colors depending on the temperature. Then this glow is due to the phonon vibrations? "Each state corresponds to a different frequency of vibration, hence a different...

Why does the detector not smear the energy for hydrogen gas when it is heated? We see the emission lines very clearly thought a diffraction grating. 3. Is a bunch of hydrogen not a macroscopic body? By mandating that it has to be macroscopic, the we shouldn't see spectra lines for hydrogen gas...

Where do energy bands come from? And also what about the second part of that question? and the sun is a "solid".. yet it behaves like a blackbody. It would be nice to have some concrete answers because this flaky terminology is just confusing

You have a macroscopic piece of solid carbon. Each carbon atom has the same energy levels thus the emission lines should correspond to the same discrete energy levels. Would we still see a continuous spectra? When we say 100 atoms, does this include any atoms? What about mercury gas? Or...

Why is it that the formula for compton scattering does not include the binding energy for an electron to the nucleus? Seems like the scattered electron can have a continuous range of energies from 0 to h/mc. Why isn't this quantized?

Why do we see a continuous spectrum of radiation for blackbodies when we know that energy in atoms are quantized thus only emitting discrete energy levels?

I get it now... if g'(x_k)>0 then this integral still holds true: \int ^{g'(x_k)\epsilon} _{-g'(x_k)\epsilon } \frac{\delta(u) du}{g'(x)} if g'(x_k)<0, then this integral becomes: \int ^{-g'(x_k)\epsilon} _{g'(x_k)\epsilon } \frac{\delta(u) du}{g'(x)}=-\int ^{g'(x_k)\epsilon}...

I get: g(x) \approx g(a)+g'(a)(x-a) expand around x_k g(x_k - \epsilon ) \approx g(x_k)+g'(x_k)((x_k - \epsilon ) - x_k)^2 \int ^{g'(x_k)\epison ^2} _{g'(x_k)\epison ^2} \frac{\delta(u) du}{g'(x)} ?? Still a bit lost (sorry..)

g(x_k\pm\epsilon)\approx 0 Honestly, I'm not really sure where to go with this. I was thinking that the delta would give a -1 for g'<0 but that's not the case. I'm now thinking this: u=g(x) \rightarrow x=g^{-1}(u) \int ^{g(x_k + \epsilon)} _{g(x_k - \epsilon)} \frac{\delta(u) du}{g'(x)}...

\int ^{g(x_k + \epsilon)} _{g(x_k - \epsilon)} \frac{\delta(u) du}{g'(x)} something like that? So basically the evaluated integral is \frac{1}{g'(x_k)}