1. Dec 20, 2009

### the4thamigo_uk

Ive heard a black body described as 'a body which absorbs all the radiation that falls upon it'. This seems to me to contradict the idea that a black body emits exactly as well as it absorbs light.

I can understand that if you shine light matching a blackbody curve (at a given temperature) on a black body that it should be capable of absorbing all that light. Furthermore it would absorb the light as heat and immediately re-mit the exact same profile of light.

But imagine instead a profile of light that is greater in intensity than the blackbody curve for every wavelength. Indeed, consider a profile that does not even match the 'shape' of the blackbody curve i.e a flat distribution of light. The first definition would suggest that the black body would be able still to absorb all the light. Is this the case? If so, the temperature of the black body would rise and re-emit the light as a new blackbody curve for the higher temperature. If not, then are we saying a black body can 'reflect' some of the excess light that it is not capable of absorbing?

So, what do we mean when we say a black body absorbs as well as it emits?

Is it, that the black body can only absorb the exact same profile of light as it emits at a given temperature, or is it the weaker statement that the black body aborbs all profiles of light, but that it then re-emits a blackbody profile that has the same 'total power' (over all wavelengths) as the incident light?

Last edited: Dec 20, 2009
2. Dec 21, 2009

### the4thamigo_uk

Anyone got any ideas on this? Is my question clear?

3. Dec 21, 2009

### Andy Resnick

Not really- a blackbody is a material object which, at thermal equilibrium, emits a spectrum of light identical to blackbody radiation.

Blackbody radiation is a very specific spectrum of light that can be assigned a thermodynamic temperature.

Emission = absorption due to the requirement of equilibrium.

4. Dec 21, 2009

### clem

At a fixed temperature, any object emits energy at the same rate it absorbs energy.
Therefor, a body that absorbs the best will emit the best.

5. Dec 21, 2009

### Roger44

This is true I believe: "..... the black body aborbs all profiles of light, but that it then re-emits a blackbody profile that has the same 'total power' (over all wavelengths) as the incident light? "

6. Dec 21, 2009

### MaxwellsDemon

A blackbody is one of those physics idealizations, like a frictionless surface or a massless string, that help us to better understand a phenomenon. I'm sure you know that a hot metal rod will glow red at high temperatures and will glow white when heated to even higher temperatures. It turns out that all bodies with temperature emit electromagnetic waves. If its not at absolute zero, a body will give off some sort of light because of its temperature. Most glow at frequencies which are not visible to humans...infrared, radio, microwave, ultraviolet, etc. Physicists wanted to understand the mathematical relationship between temperature and the frequency of the light given off. Most objects also reflect some of the light that hits them...this is what allows us to see objects. A blackbody is an idealization where none of the observed light from the object is due to reflection...all of it is generated from within by its temperature. Conflict between theoretical prediction and experimental observation of the frequencies given off and the temperature of blackbodies early in the 20th century helped lead to the development of quantum physics.

7. Dec 21, 2009

### Roger44

Yes, but we know that already. The quite interesting question is ;

"Is it, that the black body can only absorb the exact same profile of light as it emits at a given temperature, or is it the weaker statement that the black body aborbs all profiles of light, but that it then re-emits a blackbody profile that has the same 'total power' (over all wavelengths) as the incident light? "

If we put a continuous red laser light generator inside a "black body", it would seem to me obvious that the body would heat up and remit at the black body profile.

If we fire the laser light at "a body which absorbs all the radiation that falls upon it", I suppose we are in the same situation, but i'm not sure

8. Dec 22, 2009

### Andy Resnick

Since laser radiation is nonthermal light, that question is not well posed.

9. Dec 22, 2009

### mgb_phys

No - that is the mistake in the original question

Yes

Imagine a steel furnace, you can heat the steel with microwaves or RF radiation at long wavelengths but the steel will glow in the visible emitting the blackbody curve of a material at it's melting point.

10. Dec 22, 2009

### fluidistic

I'm curious about the meaning of your sentence. Does that mean that I can't heat a black body (or gray body) using a laser?

11. Dec 22, 2009

### mgb_phys

Yes you can, but you migth confuse yourself mixing statements about a blackbody and a non-blackbody source.

For example, you can only heat a blackbody target using a blackbody source to the same temperature as the source. So however much you concentrate the light with lenses/mirrors you can't use the sun to heat something to a higher temperature than the surface of the sun. Because at that point the target would emit back at the sun.

But you can use a non-blackbody source, such as a laser or an RF transmitter to radiatnly heat something to a higher temperature and then that something would emit as a blackbody.
So a piece of steel at room temperature is only emitting very little infrared, you can use an infrared 10.6um CO2 laser to cut through the steel by heating it to it's melting point, where it will emit in the visible as any blackbody at 1100C would.

12. Dec 22, 2009

### Yeti08

A blackbody will absorb all of the incident light regardless of the distribution, be it a blackbody distribution or a discrete laser output. The blackbody emission will only depend on the temperature of the object, so if it absorbs energy from incident light such that its steady state temperature is 320K the peak emitted wavelength will be at 9 micron (as per Wien's displacement law) and the distribution will be according to Planck's law.

13. Dec 22, 2009

### fluidistic

Ah thank you both, I get it. By the way I'm not the OP.

14. Dec 22, 2009

### the4thamigo_uk

Thanks for all the replies, very illuminating (please excuse the pun!)

I have followed the thread and I think it has cleared up my understanding. I was taught this stuff years ago before the internet really got going. At the time I must have thought I had understood it but since I started this thread and after reading many articles I see that I was confused on a lot of subtle issues.

I have another post in this broad area that I would welcome any comments on :

15. Dec 22, 2009

### Roger44

"you can only heat a blackbody target using a blackbody source to the same temperature as the source. So however much you concentrate the light with lenses/mirrors you can't use the sun to heat something to a higher temperature than the surface of the sun. Because at that point the target would emit back at the sun"

Are you sure of this? My first rection would be to say if you focused the light from a large 1000K black body onto a small target, then the target would rapidly rise to a temperature well above the source.

But you may well be right if we can say that whatever the power of the incident radiation, the target will re-emit more and more radiation (not nececessarily back at the source) and attain thermal equilibrium at the same temperature profile as the incident radiation. If this were true it would mean that if we doubled, and doubled again, the power falling on the target, its equilibrium temperature would remain unchanged.

I would have thought the target would have to rise to a higher temperature each time to be able to re-emit the additional incoming power, and so could finish at a temperature well above that of the source..

16. Dec 22, 2009

### the4thamigo_uk

As I understand it from following the previous discussion, the profile of the incident light isnt the thing that determines the equilibrium temperature of the black body, it is the total power (the integrated spectrum) of the incident light. The black body will come to the temperature at which it is emitting black body radiation which has the same total integrated power.

Considering focussing light from one black body onto another, consider this experiment :

I can imagine an idealised small spherical black body of radius r and temperature t, sat inside a larger hollow sphere that has its inner surface being a black body. If we also assume that no radiation comes out of the system as a whole (i.e. out of the external surface of the enclosing hollow sphere), then by equating the total energy transferred between the two surfaces we then can have radiative equilibrium at different temperatures right? implying

r * t^2 = R * T^2

This suggests that you cant heat a large black body to a temperature higher than that of a smaller black body in a closed system.

This seems strange as I am used to thermodynamic equilibrium meaning things at the same temperature? Its late maybe I am confused again?

Last edited: Dec 22, 2009
17. Dec 22, 2009

### mgb_phys

Correct
Yes

No there is no reason for an equal power transfer, they will come to the same temperature but the power will depend on the relative areas.

18. Dec 22, 2009

### Andy Resnick

This conceptual error is the root cause for the persistent belief in Archimedes' "death ray". It's fundamentally not possible due to the source being of finite size- sunlight (or your 'large 1000k blackbody') cannot be focused beyond a certain limit.

19. Dec 23, 2009

### Roger44

It's fundamentally not possible due to the source being of finite size- .....the light from a source....cannot be focused beyond a certain limit.

It can. See "Integrating Sphere" used to concentrate near 100% of the power of a light source onto a sensor placed just ouside a hole in the sphere.

20. Dec 23, 2009

### Roger44

Excuse me, my example of an integrated sphere is not relevant. All the photons produced by the source placed inside a perfect Integrating Spere have to come out of a small hole at the surface of a sphere but they come out at all angles to the tangent. So they would not, and if I'm not mistaken, could not be focussed onto a target placed some distance from the sphere.
'All angles', I don't know what the angular distibution to the tangent would be, it might be cosine, I've never asked myself this question, maybe somebody knows already, that too is an interesting question but straying from the subject.

Last edited: Dec 23, 2009
21. Dec 23, 2009

### the4thamigo_uk

I need to get this clear in my head, does this 'certain limit' come from a classical thermodynamic argument or some other physics? Could you explain why there is a limit?

If you look back at my imaginary black body inside another black body example there is no focussing going on. So in this case what is the physics?

22. Dec 23, 2009

### sylas

This was a great thought experiment. I have sorted it out in my mind taking into account some other posts in this thread. Let me try an explanation

Case 1. No focusing

Imagine a blackbody radiator with one mother of a lot of internal energy, at a temperature of 5780 K. (A bit like the Sun, for our purposes.) Let it be a sphere with surface area of 6 * 1018 m2. (That's a radius of 6.91*108 m.)

Imagine it in a massive empty space filled with a bath of radiation, with a characteristic temperature of about 2.7 K. (A bit like interstellar space and the cosmic background.)

Imagine a small sphere, a super cold blackbody radiator, 1 meter in radius, situated about 1.5*1011 m from the Sun. (That's near enough to the distance of the Earth from the Sun.) Assume is it an efficient conductor of heat, so it is all one temperature.

What temperature can you expect this small sphere to reach when it comes into equilibrium with the surrounding radiation?

The radiation from the surface of the Sun will be σT4 which is 6.33*107 W/m2. This spreads out as it leaves the Sun, by the square of the distance, so by the time it gets out to the 1 meter sphere the flux is reduced to be
$$6.33 \times 10^7 \left(\frac{6.91\times 10^8}{1.5 \times 10^{11}} \right)^2 = 1343 \; W/m^2$$​
That you should recognize as the solar constant at Earth's orbit. There's negligible energy coming from space, and the sphere has to radiate all that again, but it receives energy over a cross section of pi.r^4 and radiates it over a surface area of 4.pi.r^2, so the radiated flux from the sphere is one quarter of the solar constant. From this we get the temperature Tb of the ball as follows:
$$T_b = \left( \frac{1343}{4\sigma} \right)^{0.25} = 277.4 \; K$$​
The sphere is just above the freezing point of water.

Case 2. Big mirrors for a solar furnace

Now we bring in some perfect mirrors, line them all up behind the little ball, and focus them all directly on to the ball. These mirrors have a cross section area against the Sun of that is 105 times greater than the surface area of the ball.

Without the mirrors, the ball is getting 1343/4 watts for each square meter of its own surface area. But the mirrors are getting 1343 * 105 Watts for each square meter of the ball's own surface area. So the ball has to heat up to shed 1.343 * 108 W/m2, which gives a temperature of
$$\left(\frac{1.343\times 10^8}{\sigma}\right)^{0.25} = 6976 \; K$$​

Bingo. We've heated the ball up to be hotter than the Sun. Clearly, this is wrong. It is a violation of thermodynamics. So what was the error?

Case 3. Huge mirrors for a solar furnace

To see where the above goes wrong, imagine an enormous ellipsoidal mirror enclosing the Sun and the ball, and with each one at a focus of the ellipsoid. All the energy from the Sun is now being focused on to that tiny ball.

But here's the problem. The Sun is a finite site: about 6.91*108 meters in radius. So the radiating surface is not all at the focus of the ellipsoid. The best you can possibly get has a focus that gives a pseudo-surface around the ball radiating in with a characteristic temperature of the surface of the Sun and in all directions from that virtual surface. And that, my friends, cannot be focused any more tightly to get all the radiation impinging on a ball 1m in radius.

Similarly, in case 2, the error was thinking any number of mirrors would be able to get all the light from every point on the surface of a huge ball like the Sun focused down into a tiny sphere, with an energy flux greater than the blackbody radiating surface at 5780 K.

Case 4. Huge blackbody enclosing the system.

Added in edit... This is the example proposed by the4thamigo_uk as follows:
Imagine now no focusing; just have the whole system surrounded by an enormous blackbody surface, but one which has a small heat capacity compared with the Sun. That is, to heat up the surrounding body does not require so much energy as to deplete the Sun's store of internal energy.

The Sun keeps radiating with energy having a Planck radiation spectrum of 5780 K. This will be absorbed by the surrounding surface, which will heat up comparatively quickly (because it has a low heat capacity) and begin radiating itself. When it stops heating up, it will be in equilibrium with the incoming energy, so it radiates precisely what it receives. The whole cavity will be filled with radiation at a characteristic temperature of 5780 K. The "Sun" will now be receiving back again the same energy that it is emitting, so it is no longer cooling down. Because it has a large heat capacity, its own temperature did not fall significantly in the time it took the cavity to come to equilibrium.

(Note in this example I am not thinking of any generation of new heat energy from fusion reactions; just thinking about blackbody radiators.)

A small blackbody sphere 1m in radius, anywhere in this cavity, will also come to equilibrium with the surrounding radiation bath, at the same temperature.

Cheers -- sylas

Last edited: Dec 23, 2009
23. Dec 23, 2009

### the4thamigo_uk

I need to get this clear in my head, does this 'certain limit' come from a classical thermodynamic argument or some other physics? Could you explain why there is a limit?

If you look back at my imaginary black body inside another black body example there is no focussing going on. So in this case what is the physics?

Why? I am very confused by this now...

Assuming purely radiative transfer then surely for a black body to maintain the same temperature it requires that it is absorbing the same amount of energy/second, over its who surface, as it is emitting? No? If not why not?

If this is the case, assuming a black body A maintained (i.e. on a hot plate) at temperature T and radius R and a second black body B with *no power source* at temperature t (which can change) and radius r. Also assuming we can capture all the radiation from body A and direct it to body B.

Total power emitted by A = sigma * T^4 * 4 * pi * R^2 (and this is a constant since B is being kept warm)

Net total power gained by B = (total power emitted by A) - sigma * t^4 * 4 * pi * r^2

For steady state : (Net power gained by B) = 0

Therefore :

Steady state temperature of B = t = T * sqrt(R / r )

Hence if R > r then t > T. This seems clear to me and intuitive that the temperatures must be different. Remember that the system isnt a closed system as there is heat supplied to body A. If there was no heat supplied they should *both* adjust their temperatures so they balance.

Now if body A is the sun, we have considerable practical difficulties.
(1) the earth only captures a tiny proportion of the suns energy due to its distance and the size of the earth, 1.74e17 W according to wikipedia (http://en.wikipedia.org/wiki/Sunlight#Solar_constant)
(2) The atmosphere absorbs some of the energy
(3) We cant even focus all the energy that falls on the earth (1) onto an experimental black body in the lab only a tiny fraction

What about looking at the situation as if the earth itself was our black body A? Here is my calculation :

Net power gained by earth = (solar energy received) - (radiation emitted)
= (1.74e17 W) - (sigma * T^4 * 4 * pi * R^2)

Steady state implies no net power gain.

So if R = 6.3781e6 m
sigma = 5.67e-8 J/s/m2/K4

Rearranging and plugging the numbers in we get :

T (of earth in steady state) = 4throot( 1.74e17 / (5.67e-8 * 4 * pi) ) / sqrt( R )
= 7.02978e5 / sqrt( R )
= 278K = 5 celsius

So what about if we concentrated all the solar energy received by the earth into a body with a radius 1m. We get a temperature of 70000K. Thats a pretty good death ray... but then again there would be nothing left alive to kill (

What is wrong with this argument???

24. Dec 23, 2009

### sylas

You can't focus all the energy over an area the size of the Earth into that small of a volume. See my previous post.

25. Dec 23, 2009

### the4thamigo_uk

What is the reasoning though? I dont see 'in principle' why not? Practically of course you cant but that concerns the geometry of lenses and mirrors, the absorption effects of mirrors, the size of mirrors which are external concepts to a purely thermodynamic argument I think.

Ive thought about this a lot, and I believe that the real nub of the matter is that there is no 'expectation of thermodynamic equilibrium' with the sun and a small black body on the earth. The sun has an external power source, in such a scenario my reasoning leads me to think that it is possible (in principle) to heat up a black body to a temperature hotter than a source black body.

Avoiding mirrors by taking my 'black body inside another black body' example. If the outer black body was maintained by a hot plate at temperature T, then since the condition for a 'steady state' must be that 'total power lost by inner body = total power gained by inner body', then the inner black body must achieve a higher 'steady state temperature' than the external body, as long as there is heat provided by the hot plate.

They do not achieve 'thermal equilibrium' because they are not in a closed system. If we include the hot plate and its source of energy (i.e. an electrical battery or something like), we would eventually run out of energy from the chemical reactions in the battery and hence our external black body would fall in temperature and everything would eventually end up an thermal equilibrium. Our 'steady state' would now be the same as 'thermal equilibrium' whereas previously our 'steady state' was not in 'thermal equilibrium'.

This doesnt rely on geometry or and practical considerations. I think it simply relies on the notion of when 'thermodynamical equilibrium' is not the same as the 'steady state of a system'. And this applies when there is an external energy source.

Hence my reasoning leads me to the conclusion that we can build a decent death ray if we collect enough solar energy and focus it tightly. There may be geometrical and practical limits but it seems to me that these are limits imposed by optics, geometry, engineering etc.

This is a good geometrical reason for a practical limit, but all we need to do is capture all the radiation from a body of radius R by a second body of smaller radius r. It doesnt have to be 1m2. 1mm smaller should be enough to raise the smaller body to a higher temperature slightly higher than the source.

Furthermore, I think it is not necessary to capture 'all' the radiation from the source body anyway. There is sufficient radiation hitting the earth already to supply power to a black body of a certain smaller size (which doesnt have to be so tiny), if it is focussed somehow.

Last edited: Dec 23, 2009