Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Blackbody math help

  1. Jan 7, 2004 #1

    Have a look at this site, specifically the calculation of Planck's formula. I'm sure you've seen it many times before.

    What I need help with is calculating a similar formula based on my own idea instead of the oven model. I'll try to explain what I'm thinking by refering to parts of the above site.

    Making it very short... in the oven model there are a bunch of oscillators in the walls of the oven, all oscillating at different frequencies, ie those that "fit" ([itex]\lambda = 2a, a, 2a / 3 ..[/itex]). Planck's "fix" was to quantize the energylevels and assume higher energies where less likely to be emitted.

    First, I just want to say that I think the oven model is very lacking when it comes to reflecting reality. Though perfect blackbodies don't exist, the "almosts" - ie stars - are best represented by a simple chaotic (huge) quantity of oscillators, all oscillating in random directions. There are no walls and no "fits". Unlike the oven model, a perfect blackbody would produce a perfectly continuous spectrum.

    The model we're gonna focus on is the most realistic one, ie the gas analogy.
    According to the theorem of equipartition of energy the oscillators should have an average energy

    K_\textrm{av} = \frac{mv^2}{2} = \frac{3k_{b}T}{2}

    and thereby average velocity

    v_\textrm{av} = \sqrt{\frac{3k_{b}T}{m}}

    Similar to the oven model approach we must first figure out which wavelengths the oscillators are emitting. The main frequency [tex]f_0[/tex] is emitted perpendicular to the direction of oscillation and is thereby not doppler affected. [itex]f_\textrm{max}[/itex] and [itex]f_\textrm{min}[/itex] are emitted along the axis of oscillation since this is where most dopplereffect occurs.


    f_\textrm{max} = \left(f_0 \frac{c+v_\textrm{av}}{c-v_\textrm{av}}\right)
    f_\textrm{min} = \left(f_0 \frac{c-v_\textrm{av}}{c+v_\textrm{av}}\right)

    What I'm saying is that every single oscillator in the blackbody emitts all frequencies part of the blackbody's spectrum. And unlike the oven model, the mechanism controlling the energy- and frequency-distribution is the oscillation itself. Let me explain..

    In the case with the oven the only radiation that matters is the one radiated perpendicular to the oscillators' direction of oscillation (due to the "fit"). So for an oscillator in the oven wall

    f_0 = f_\textrm{max} = f_\textrm{min}

    So at any given time, a single oscillator only radiates a single frequency.

    However, this is not the case in our gas analogy. Every single oscillator radiates in all directions at all time, and so all angles must be accounted for. But why should this mean oscillators radiate more than one frequency ? This is where dopplereffect comes in to play. Let me make an analogy with atoms and photons:

    The analogy is very simple. While the atoms in a gas bump around and vibrate, they radiate photons. Depending on where the observer is, the atom radiating a specific photon might be moving towards or away from him. This causes the photon to shift it's frequency accordingly. If the atom is moving towards the observer while emitting the photon, the photon will be blueshifted. Now, if we would have a gas with an infinite number of atoms, there would be an infinite amount of relative angles to the observer, which of course would produce a continuous spectrum. This spectrum would stretch from it's most redshifted frequency, [itex]f_\textrm{min}[/itex], to it's most blueshifted one, [itex]f_\textrm{max}[/itex].

    To this point the analogy with our blackbody gas model is excellent, but continuing will reveal some crucial differences.

    The first difference is quantum related. A quantum atom will emit photons sporadically, but a classical oscillating charge will radiate continuously in every direction. This means that it would take an infinite amount of time for a quantum atom to emit the whole continuous spectrum, while it would only take a brief moment for the classical charge to do the same.

    The most important difference from a blackbody perspective is the fact that photon intensity doesn't change with the relative angle. The observer will see as many unshifted photons as blueshifted and redshifted. If this was the case in our blackbody, the intensity wouldn't vary as a function of wavelength (As the famous Planck curve). But on the other hand, if the atoms were oscillating charges and the photons were electromagnetic waves, the intensity would vary very much indeed. In fact, Maxwell's equations tells us that the amplitude of an electromagnetic wave is a function of the angle between the oscillation and the radiated wave. This means that the intensity is highest perpendicular to the oscillation, and lowest (zero) in the direction of oscillation. This is why the Planck curve slopes down again at higher frequencies ! There are two relationships controlling the intensity curve - dopplershift and Maxwell's equations. These two are related as follows:

    More dopplershift = less intensity // No dopplershift = highest intensity

    [tex]\theta = \angle[/tex] radiation / oscillation

    [tex]D_\textrm{amount}[/tex] = The amount of dopplereffect

    [tex]D_\textrm{amount} \propto \cos{\theta}[/tex]

    Amplitude of EM field [tex]A \propto \sin{\theta}[/tex] (no radiation in direction of oscillation)

    Anybody feel like helping me formulate this properly ? (mathematically)

    Comments are appreciated...
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you help with the solution or looking for help too?
Draft saved Draft deleted

Similar Discussions: Blackbody math help
  1. What are blackbodies ? (Replies: 3)

  2. Earth as a Blackbody (Replies: 6)

  3. Blackbody radiation (Replies: 6)