Re: confusing page on Wien's displacement; also question between matter-energy and radiation
- From: "Jim Black" <ghytrfvbnmju7654@xxxxxxxx>
- Date: 7 Nov 2005 11:18:47 -0800
Autymn D. C. wrote:
> http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/wien3.html#c1 says
> that the /peak/ /wavelength/ against blackbody intensity /changes/ when
> different measurement scales are used, such as for linear wavelength,
> linear frequency, frequency squared, logarithmic (?), and median (?).
> That makes absolutely no sense. It's intensity, which is countable and
> invariant with scale!
It's not the intensity (energy per time per area) that is maximized at
lambda_peak, but intensity per wavelength (or, as the page points out,
it could be intensity per frequency, etc.).
If you filtered a beam of sunlight so that only the light at exactly
500 nm could get through, nothing would get through at all. If you let
light between 499 and 501 nm through, the intensity of the light that
got through would be small but nonzero. If you let light between 498
and 502 nm through, the intensity would be about twice what it was in
the 499-501 nm case.
So it makes no sense to talk about the intensity of the light at 500
nm. But we can consider the intensity of the light between 500 nm -
d(lambda) / 2 and 500 nm + d(lambda) / 2. The limit of
intensity between 500 nm - d(lambda) / 2 and 500 nm + d(lambda) / 2
-------------------------------------------------------------------
d(lambda)
as d(lambda) approaches zero is the intensity per unit wavelength of
the light at 500 nm. This has units not of energy/time/area, but of
(energy/time/area)/length.
You can do the same thing with frequency, but you get different
results; thus, the web page's statement.
> And the proof has bad form by equating a product
> of length and temperature with a dimension of length.
Judging from the fact that they give the same expression units of
meters * Kelvin above, I'd assume they just forgot the "* K" on the
later ones. If you're concerned about this, you could e-mail the
authors of the page -- I'm sure it would be trivial for them to fix it.
> My off asking is, when I compare the cinetic energy of a molecule in
> terms of temperature with the blackbody radiation peak energy in terms
> of temperature, I find that one particle would radiate with a peak
> energy that's greater than its own energy! I realise that an
> explanation is that radiation from a bunch of particles is taken from
> parts of energies from all of them,
The energy (3/2)kT is merely the average kinetic energy of a molecule.
Some molecules have more kinetic energy, and some molecules have less.
> but what would the radiation curve
> look like with only a few particles to one?
It would depend a great deal on what the particles were. The reason
the blackbody spectrum only depends on the temperature is that light
has interacted with the matter in the blackbody many times before it
reaches the viewer. Each time, some of the light is absorbed and some
more light is emitted. The light eventually comes into thermal
equilibrium with the matter.
In a hot, low-density gas, the light has very little opportunity to be
re-absorbed before it reaches the viewer, so it does not emit a
blackbody spectrum. You would probably see a line spectrum instead.
> example: 1 eV -> 7736 K ~> 3.3 eV
>
> -Aut
.
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