> Please read up on the physics behind "black body radiation" before you try to
> argue this issue any further.
> There is a simple equation that relates to two bodies transfering heat via
> radiation.
> A perfect black body has a emissivity of 1. Everything else has less.
> A perfect black body is not only a perfect absorber of radiated energy
> but a perfect emitter.
> The confusion here is that you are relating a black hole to a perfect black
> body. However, that is a completely
> different case because it takes in photons due to a high concentration of
> gravity.
>
> 73
> Bill wa4lav
>
>
> _
"Planck's formula predicts that a black body will radiate energy at all
frequencies, but its intensity rapidly tends to zero at high frequencies
(low wavelengths).# For example, a black body at room temperature (300
kelvins) with one square meter of surface area will emit a photon in the
visible range once every minute or so, meaning that for most practical
purposes a black body at room temperature does not emit in the visible
range. Significance of this fact for the derivation of Planck's law from
experimental data, and for the substantiation of the law by the data, is
discussed in^[18]
<http://en.wikipedia.org/wiki/Planck%27s_law_of_black-body_radiation#cite_note-Ribaric-17>
Ultimately, Planck's assumption of energy quantization and Einstein's
photon hypothesis became the fundamental basis for the development
ofquantum mechanics <http://en.wikipedia.org/wiki/Quantum_mechanics>."
from;
http://en.wikipedia.org/wiki/Planck's_law_of_black-body_radiation
<http://en.wikipedia.org/wiki/Planck%27s_law_of_black-body_radiation>
Also from http://en.wikipedia.org/wiki/Black_body;
"Inphysics <http://en.wikipedia.org/wiki/Physics>, a*black body*is an
idealizedobject <http://en.wikipedia.org/wiki/Physical_body>that absorbs
allelectromagnetic radiation
<http://en.wikipedia.org/wiki/Electromagnetic_radiation>falling on it.
Black bodies absorb and incandescently
<http://en.wikipedia.org/wiki/Incandescence>re-emit radiation in a
characteristic, continuous spectrum. Because no light (visible
electromagnetic radiation) is reflected or transmitted, the object
appears black when it is cold. * However, a black body emits a
temperature-dependent spectrum
<http://en.wikipedia.org/wiki/Electromagnetic_spectrum>of light.
Thisthermal radiation
<http://en.wikipedia.org/wiki/Thermal_radiation>from a black body is
termed*blackbody radiation*. In the blackbody spectrum, the shorter the
wavelength, the higher the frequency, and the higher frequency is
related to the higher temperature. Thus, the color of a hotter object is
closer to the blue end of the spectrum and the color of a cooler object
is closer to the red."
See *. So your argument carries no weight, as we are talking about black
material not black body, and as described above thermal radiation is
negligible # from a black surface.
Adrian ... vk4tux
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