0 Members and 1 Guest are viewing this topic.
There was a severe problem in classical physics with black body radiation. Any object with a higher temperature than its surroundings will lose heat by radiation. The hotter the object, the more it will radiate. A black body absorbs all frequencies, so it should radiate equally at all frequencies. But that does not happen. Black bodies emit more radiation at some wavelengths than at others. Planck decided to tackle this problem. He solved it by proposing that heat radiation is emitted only in specific amounts, which he called quanta.
Why the need for quantum phyiscs?
I suggest you take it up with Warwick university then as their website says the following:- A black body is a hypothetical object that is able to emit and absorb electromagnetic radiation of all possible wavelengths.The observed frequency spectrum (variation of intensity of radiation versus frequency) was different from that predicted by the theory.The theory predicted that the spectrum should be the same for all black bodies.
Quote from: DoctorBeaver on 17/07/2007 23:17:49I suggest you take it up with Warwick university then as their website says the following:- A black body is a hypothetical object that is able to emit and absorb electromagnetic radiation of all possible wavelengths.The observed frequency spectrum (variation of intensity of radiation versus frequency) was different from that predicted by the theory.The theory predicted that the spectrum should be the same for all black bodies.A black body is an unknown mass of unknown density. It's density's contain unknown quantities of Electrons and other Photon absorbing particles. To try and determine what it's electromagnetic responses are is an act of utter madness. Every black body can have a different response and probably does. To estimate the electromagnetic response of a black body is the same as guessing it's mass.Warwick guesses a lot...maff
You say that the energy curve of black body radiation has been observed experimentally but if a black body is an ideal radiator and ideal models don't ever exist, how did physicists observe this curve?
I am confused about this simulated black body. Is the radiation going into the cavity being absorbed by the 'black body?'
Also, if black bodies absorb all radiation how do they emit any radiation?
Isn't their namesake from the fact that they DON'T emit any radiation and are thus black?
Do any of you know of the mathematical representation of why black bodies are the most efficient emitters of radiation?
I understand the absorbtion properties and their correlation with color (color is a direct result of absorbtion properties of objects). But what I don't understand is the emission and how it relates to the color of objects (effectively how emission relates to absorbtion). Emission is due to the inherent vibration of molecules which result in temperature. Molecules consist of charges and vibrating (oscillating) charges generate electromagnetic waves. The frequency of these waves dependes on the kinetic energy of the vibrating charge
and thus the molecule. Therefore temperature (average kinetic energy) directly affects the frequency of the electromagnetic waves emitted. I understand all of this. But how do the absorbtion properties of matter affect its emission properties? And shouldn't the most efficient absorber of radiation absorb most of the waves emitted from anywhere but the surface of the substance?
Granted, something that reflects all radiation would reflect the radiation around inside the material.
But that radiation is still in existence while the ones absorbed in a black body are not. So, shouldn't the radiation emitted from inside a bright, shiny object eventually escape, thus making the object a more efficient emitter of radiation than a black body?
Look at the gradient in spatial coordinates first:x∂/∂x + y∂/∂y + z∂/∂zAll coordinates here are measuring position. You're trying to change variables so that two of your coordinates measure angles and one measures a position. In order to do so, you need to scale the angular derivatives by something that tells you how much changing that angle will change your position. For example, a change in the azimuthal angle θ by a small amount dθ means a spatial change given by dr=rdθ (see figure). Since the other spatial variable (the zenith angle) has an extra projection involved, dr=r sinθ dφ.If you'll notice these scaling factors (r, r sin θ) appear in the gradient in spherical coordinatesr∂/dr + θ(1/r) ∂/∂θ + φ 1/(r sin θ) ∂/∂φ.These scaling factors are the reason why the Laplacian looks different in different coordinates.Calculating the Laplacian gets messy. You have to dot the gradient above with another gradient, so you need to use the vector identity: Divergence[fv]=f Divergence[v]+Gradient[f]•v, and keep in mind that you can act on the unit vectors r, θ, and φ with derivatives as well. Hope that makes some sense.
Heh. I assumed since the question started off "I'm trying to solve the Schrodinger equation in 3-D for the Hydrogen atom..." that he wanted a technical answer.