Black Body

On 08/12/2016 14:29, David Levy wrote:
Thanks for your great explanation.
Martin Brown;1329318 Wrote:

The purpose of the small hole and the large internal area is to allow
there to be a reservoir of the black body radiation at the internal
characteristic temperature of the nominal black body source which isn't

too badly perturbed by losing energy by radiation out of the small hole.

It is a practical limitation rather than anything else.

Yes, that is clear.

But it isn't at all clear to me that you understand either the meaning
of either "temperature" or of "black body".

Martin Brown;1329318 Wrote:

For what sort of temperature? Everything radiates thermal energy but
there is a world of difference between a black body at 300K and one at
3000K (it starts to get difficult to find stuff that doesn't melt!).
The temp is not critical. It can be 3000K or even 3K.
We only discuss on a principle issue.

Martin Brown;1329318 Wrote:

I don't see any reason why in principle you couldn't take a series of
solid state LEDs each with 50nm bandwidth and construct a moderately
good approximation to BB radiation over a fair range of wavelengths. It

gets harder at the UV short wave end and at unpopular IR wavelengths.

You really need to describe what you are trying to do.

O.K.
I would like to understand the following:
Let's use one cube of insulated enclosure, and set inside only one LED.
Let's assume that the ambient temp is 0K.
After operating the LED, I assume that we should get some radiation
above 0K with black body signature.

No. You will get the light emission of the LED lets say a green one at
530nm +/-50nm plus some thermal black body radiation emitted by the LED
as a result of the waste heat dissipated in its body (the LED being only
20% efficient). Assuming here that the clear plastic is a decent
approximation to a black body at its operating temperature.

Now, let's set two similar cubes (with one LED in each one) next to each
other and eliminate the shared wall between them. Hence, we should
still get an insulated enclosure which is double in its size and has two
LEDs.
So, does it mean that this insulated enclosure should also have a black
body radiation?

Everything always has thermal radiation characterised by its temperature
and emisivity as a function of wavelength.

A black body is the name we give to an idealised perfect absorber and
also a perfect emitter of radiation.

The opposite of a black body is a perfect mirror. We can realise them to
quite a good approximation in most wavebands by choosing the right
materials. Chicken wire for radio waves or aluminium for light.

Do you expect that the internal radiation in that bar will also have a
black body signature?

If there is a LED there emitting a specific wavelength then obviously
not. You would have to paint the LED black or encase it in opaque
material so that all light at a specific wavelength was thermalised.

Your question pretty much amounts to asking "if I have monochromatic
yellow light is it the same as a black body?". The answer is *NO*.

But you could contrive to build a device using a series of LEDs at
different wavelengths to approximate the curve of a black body emitter
even though the process by which their radiation is produced is
non-thermal. Much the same way RGB pixel displays can display white.

--
Regards,
Martin Brown

Do you expect that the internal radiation in that bar will also have a black body signature?

Originally Posted by 'Poutnik[_5_

Sure. There is no cavity with thermal radiation absent. If cavity opening is negligible, the radiation can be considered as BB. Note that if multiple cavities are connected, so there is option of radiative equilibrium between them, the requirement of the negligible opening is not valid any.

This bar goes to infinity.
So the question is as follow:
Do we need to close it at the end of the infinity?
If I understand it correctly:
1. Theoretically, There is no end for infinity. Hence, if the radiation is moving in the direction of the infinity – it will never bump a wall. Therefore, there is no need to close it.
2. The Energy of the radiation is decreasing as it moves. If it moves to the infinity, then at some point it should be virtually Zero. If the energy of that radiation is Zero, then it will never come back even if we set a wall. Hence, again – there is no need to close it at the infinity.
So, my conclusion is – if we look at a bar which goes to the infinity – it must have five walls which fulfill the requirements for insulated enclosure while the one which goes to the infinity can stay open without damaging the Black body signature.
Do you agree with that?

Dne 09/12/2016 v 13:04 David Levy napsal(a):


This bar goes to infinity.
So the question is as follow:
Do we need to close it at the end of the infinity?

No, if we can consider radiations incoming through the opening
and reflected back outside as negligible.

If I understand it correctly:
1. Theoretically, There is no end for infinity. Hence, if the radiation
is moving in the direction of the infinity – it will never bump a
wall. Therefore, there is no need to close it.

It is not related to infinite arrangement.
Also, you cannot ignore multidirectional aspect.

2. The Energy of the radiation is decreasing as it moves.

No, it does not.

[...] then at some point it should be virtually Zero. If the
energy of that radiation is Zero, then it will never come back even if
we set a wall. Hence, again – there is no need to close it at the
infinity.

You may forgot the infinity.

So, my conclusion is – if we look at a bar which goes to the
infinity – it must have five walls which fulfill the requirements
for insulated enclosure while the one which goes to the infinity can
stay open without damaging the Black body signature.
Do you agree with that?

The only criteria that matters is full absoption of incident radiation
no matter of the directions.


--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.

Originally Posted by 'Poutnik[_5_

2. The Energy of the radiation is decreasing as it moves.
No, it does not.

[...] then at some point it should be virtually Zero. If the
energy of that radiation is Zero, then it will never come back even if
we set a wall. Hence, again – there is no need to close it at the
infinity.
You may forgot the infinity.

Why? What do you mean by -Infinity?
I had the impression that infinity means - infinity.
It means: Endlessness, infinitude, infiniteness, boundlessness, limitlessness, vastness, immensity. A number greater than any assignable quantity or countable number (symbol ∞).
How could it be that there is no reduction in the energy of radiation as it moves away from its source?
Let's use a LED.
At a distance of 1 meter it looks very bright.
At a distance of 100 meter it looks dim.
At 1000 meter it's very difficult to notice the light.
At 10,000 meter we probably won't see any light.
So, if there is a source of radiation, how could it be that at a distance of 1 KLY or even over 100 Million Billion KLY it will be the same?
Don't you think that at some point the energy of the radiation should be virtually zero?

Dne 11/12/2016 v 08:32 David Levy na2psal(a):
'Poutnik[_5_ Wrote:

-
2. The Energy of the radiation is decreasing as it moves. -
No, it does not.
-
[...] then at some point it should be virtually Zero. If the
energy of that radiation is Zero, then it will never come back even if
we set a wall. Hence, again – there is no need to close it at the
infinity.-
You may forgot the infinity.


Why? What do you mean by -Infinity?
I had the impression that infinity means - infinity.
It means: Endlessness, infinitude, infiniteness, boundlessness,
limitlessness, vastness, immensity. A number greater than any assignable
quantity or countable number (symbol ∞).

You have got mean wrong. I mean infinity size as a limit
plays no partricular role in black body scenario.


How could it be that there is no reduction in the energy of radiation as
it moves away from its source?

You may confuse the energy flow and energy.
The former decreases with distance, as the same number of photons
are passing through larger area.
The latter does not change.


Let's use a LED.
At a distance of 1 meter it looks very bright.
At a distance of 100 meter it looks dim.
At 1000 meter it's very difficult to notice the light.
At 10,000 meter we probably won't see any light.

Sure, but again ,do not confuse E and dE/dS/dt.

So, if there is a source of radiation, how could it be that at a
distance of 1 KLY or even over 100 Million Billion KLY it will be the
same?

If we consider no absorption and point like source,
the power passing through a givid solid angle
is independent on distance.

Don't you think that at some point the energy of the radiation should be
virtually zero?

No, it is just spreaded to larger area a/o volume.


--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.

Originally Posted by 'Poutnik[_5_

How could it be that there is no reduction in the energy of radiation as
it moves away from its source?

You may confuse the energy flow and energy.
The former decreases with distance, as the same number of photons
are passing through larger area.
The latter does not change.

That is clear.
However, what is the meaning of radiation which is moving for long distance?
Why it can't be represented by energy flow?

How could it be that the amplitude of the measured radiation is constant at any distance from the source?

Originally Posted by 'Poutnik[_5_

.

Don't you think that at some point the energy of the radiation should be
virtually zero?

No, it is just spreaded to larger area a/o volume.

If it is sreaded to larger area, then why the measured amplitude shouldn't decrease as we move further away from the source of energy?

Dne 12/12/2016 v 21:56 David Levy napsal(a):


That is clear.
However, what is the meaning of radiation which is moving for long
distance?

It means radiation moving for long distance.
What else should it be ?

Why it can't be represented by energy flow?

It can. But it depends on the geometry of the radiation source.

How could it be that the amplitude of the measured radiation is constant
at any distance from the source?

The more the radiation differ from radiation of omnidiractional
divergent source, the less its flow density depends on distance.
EVen if finally, as limite for distance reaching for infinity,
it converges to the reciprocal square law.

But for a photon, its energy does not depend on distance.
Neither the total energy of radiation.

-
No, it is just spreaded to larger area a/o volume.

If it is sreaded to larger area, then why the measured amplitude
shouldn't decrease as we move further away from the source of energy?

I have not said it should not.
But I do not see your point in BB context.

Each surface, BB or not BB, in equilibrium conditions
is in equilibrium with the radiation density
dE/dV = 4.sigma/c * T^4 J/m^3
where sigma is Stefan-Boltzmann constant.

What it does not absorb, but reflects, it does not emit,
so radiation net flow for the surface is zero.

--
Poutnik ( The Pilgrim, Der Wanderer )

A wise man guards words he says,
as they say about him more,
than he says about the subject.

On 09/12/2016 12:04, David Levy wrote:
-
Do you expect that the internal radiation in that bar will also have a
black body signature?-
'Poutnik[_5_ Wrote:

Sure. There is no cavity with thermal radiation absent. If cavity
opening is negligible, the radiation can be considered as BB. Note that
if multiple cavities are connected, so there is option of radiative
equilibrium between them, the requirement of the negligible opening is
not valid any.
This bar goes to infinity.
So the question is as follow:
Do we need to close it at the end of the infinity?
If I understand it correctly:
1. Theoretically, There is no end for infinity. Hence, if the radiation
is moving in the direction of the infinity – it will never bump a
wall. Therefore, there is no need to close it.
2. The Energy of the radiation is decreasing as it moves. If it moves to
the infinity, then at some point it should be virtually Zero. If the
energy of that radiation is Zero, then it will never come back even if
we set a wall. Hence, again – there is no need to close it at the
infinity.
So, my conclusion is – if we look at a bar which goes to the
infinity – it must have five walls which fulfill the requirements
for insulated enclosure while the one which goes to the infinity can
stay open without damaging the Black body signature.
Do you agree with that?

Sigh. We can add "infinity" to the long list of concepts that you
fundamentally have made no attempt to understand.


--
Regards,
Martin Brown