Does Optic Focusing Contradict the Second Law of Thermodynamics?

[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?

 

[the4thamigo_uk]

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

 

[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.

 

[Meir Achuz]

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.

 

[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? "

 

[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.

 

[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

 

[Andy Resnick]

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

 

[mgb_phys]

"Is it, that the black body can only absorb the exact same profile of light as it emits at a given temperature

No - that is the mistake in the original question

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? "

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.

 

[fluidistic]

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

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?

 

[mgb_phys]

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?

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.

 

[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.

 

[fluidistic]

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

 

[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 :

https://www.physicsforums.com/showthread.php?t=365018

 

[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..

 

[the4thamigo_uk]

As I understand it from following the previous discussion, the profile of the incident light isn't 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 can't 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?

 

[mgb_phys]

As I understand it from following the previous discussion, the profile of the incident light isn't the thing that determines the equilibrium temperature of the black body, it is the total power (the integrated spectrum) of the incident light.

Correct

The black body will come to the temperature at which it is emitting black body radiation which has the same total integrated power.

Yes

then by equating the total energy transferred between the two surfaces we then can have radiative equilibrium at different temperatures right?

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.

 

[Andy Resnick]

"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.

<snip>

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.

 

[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.

 

[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.

Please ignore my previous message.

 

[the4thamigo_uk]

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.

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?

 

[sylas]

"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.

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 T_b 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:

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?

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

 

[the4thamigo_uk]

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.

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?

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.

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 isn't 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 can't 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

Not a bad prediction really.

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?

 

[sylas]

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?

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.

 

[the4thamigo_uk]

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.

What is the reasoning though? I don't see 'in principle' why not? Practically of course you can't 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 doesn't 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.

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.

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 doesn't 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 doesn't have to be so tiny), if it is focussed somehow.

 

[Roger44]

"This was a great thought experiment"...it is indeed but I have to go to work. Back later.

 

[Andy Resnick]

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?

It's a thermodynamics/photometry argument. Rather than go through it all over again, please find relevant threads I've already posted to.

 

[mgb_phys]

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?

Yes that's correct

But what you said earlier (or what I understood you to say) was that the power emitted by two bodies in equilibrium was the same which isn't true - imagine the sun and a small rock.

 

[the4thamigo_uk]

It's a thermodynamics/photometry argument. Rather than go through it all over again, please find relevant threads I've already posted to.

Ok, I am just thinking about pure thermodynamics here...

It was stated earlier in the thread that it was impossible for a black body to raise a second black body to a higher temperature than itself. The enclosed black body scenario suggests it can if the source is maintained at a constant temperature.

For a more sensible example, and to avoid too many complicated optical arguments. We could have a source black body of any shape not necessarily a sphere... such as flat surfaces...

We could have a theoretical flat black body sat on a hot plate at temperature T, we could use four flat mirrors to angle the radiation from this surface onto a second flat black body of quarter the area. It receives 'roughly' the same power although on 1/4 of the surface area. It therefore must heat up to a higher steady state temperature. It doesn't matter that some radiation is lost in the optics, in fact all we really require is that there is a higher power per unit area hitting the second black body than emitted by the first. We quadrupled the power with the four mirrors but even if we only increased the flux falling on the second black body by 1 Watt/m2, it would still achieve a higher steady state temperature than the source. That the two surfaces are not in thermodynamic equilibrium is fine. They are not meant to be.

 

[the4thamigo_uk]

Yes that's correct

But what you said earlier (or what I understood you to say) was that the power emitted by two bodies in equilibrium was the same which isn't true - imagine the sun and a small rock.

In my original post I was talking only of my enclosed black body example only, also I said I was equating the 'total energy' (per second), i.e. power. Giving rt^2 = RT^2, for my enclosed sphere example.

I admit my notion of equilibrium was hazy. My spheres were in a steady state not thermal equilibrium.

 

[the4thamigo_uk]

I was thinking again about the enclosed black body thought experiment. It seems for a 'steady state' when the outer body is maintained at constant temperature, we can equate the net energy flow for the inner body and make the following equation :

total power emitted from outer body = total radiative power lost by inner body.

However, if neither body is maintained at a constant temperature we could argue that we could still make this identity. Niaevely we will get the same result, that the temperatures of the two bodies in thermal equilibrium are different. Which is a contradiction in terms.

It would seem that the crucial difference is that there is a finite distance between the inner surface of the outer body and the outer surface of the inner body. Hence the radiation must travel a distance (R-r) and therefore take a finite time (R-r)/c.

So we have a subtle set of differential equations :

\frac{dQ_{i}(t)}{dt} = k_{i}\frac{dT_{i}(t)}{dt} = 4\pi R_{o}^{2} \sigma T_{o}^{4} (t - \Delta t) - 4\pi R_{i}^{2} \sigma T_{i}^{4} (t)

\frac{dQ_{o}(t)}{dt} = k_{o}\frac{dT_{o}(t)}{dt} = 4\pi R_{i}^{2} \sigma T_{i}^{4} (t - \Delta t) - 4\pi R_{o}^{2} \sigma T_{o}^{4} (t)

\Delta t = \frac{R_{o} - R_{i}}{c}

Q - net heat flow into body
T - temperature of body
t- time (brackets denote function arguments)
k - specific heat of body
R - radius of body
subscripts denote _{i}nner and _{o}uter black body

It seems to me that this must be the crucial difference between the two systems. Problem is I have no idea how to go about solving this set of equations? What techniques are required when there is a shift \Delta T to the variables? I wonder if there is potential for a harmonic type damped oscillation to thermal equilibrum or would it be a nice smooth decline?

 

[Dale]

We could have a theoretical flat black body sat on a hot plate at temperature T, we could use four flat mirrors to angle the radiation from this surface onto a second flat black body of quarter the area. It receives 'roughly' the same power although on 1/4 of the surface area. It therefore must heat up to a higher steady state temperature. It doesn't matter that some radiation is lost in the optics, in fact all we really require is that there is a higher power per unit area hitting the second black body than emitted by the first. We quadrupled the power with the four mirrors but even if we only increased the flux falling on the second black body by 1 Watt, it would still achieve a higher steady state temperature than the source. That the two surfaces are not in thermodynamic equilibrium is fine. They are not meant to be.

This is a great idea for a simplification. I would recommend that you actually go through the geometry rather than using the "roughly" argument. Since you are violating the 2nd law of thermo I think it merits more than a hand waving argument.

My guess is that when you do the actual geometry you find that the 2nd law wins out after all, but I could be wrong.

 

[the4thamigo_uk]

The flat plate idea is just to give a more real world example than the enclosed BB idea. They are both equally valid test cases. Geometrical arguments shouldn't be used to prove or disprove thermodynamics anyway as they are not general enough. As I said before I also don't believe there are any violations of the laws of thermodynamics as there is an external heat supply (namely the hot plate).

 

[the4thamigo_uk]

So we have a subtle set of differential equations :

\frac{dQ_{i}(t)}{dt} = k_{i}\frac{dT_{i}(t)}{dt} = 4\pi R_{o}^{2} \sigma T_{o}^{4} (t - \Delta t) - 4\pi R_{i}^{2} \sigma T_{i}^{4} (t)

\frac{dQ_{o}(t)}{dt} = k_{o}\frac{dT_{o}(t)}{dt} = 4\pi R_{i}^{2} \sigma T_{i}^{4} (t - \Delta t) - 4\pi R_{o}^{2} \sigma T_{o}^{4} (t)

\Delta t = \frac{R_{o} - R_{i}}{c}

Q - net heat flow into body
T - temperature of body
t- time (brackets denote function arguments)
k - specific heat of body
R - radius of body
subscripts denote _{i}nner and _{o}uter black body

I think the solutions must oscillate. As \frac{dQ_{i}(t)}{dt} tends to zero we end up with the following tending to equality :

4\pi R_{o}^{2} \sigma T_{o}^{4} (t - \Delta t) = 4\pi R_{i}^{2} \sigma T_{i}^{4} (t)
4\pi R_{i}^{2} \sigma T_{i}^{4} (t - \Delta t) = 4\pi R_{o}^{2} \sigma T_{o}^{4} (t)

which is essentially of the form:

X(t - \Delta t) = aY(t)
Y(t - \Delta t) = bX(t)

So X will eventually become proportional to what Y was a short time ago and Y will eventually become what X was a short time ago. Obviously it wouldn't work if X and Y both became flat. I would hope that the oscillations are damped though, to allow them to come to the same temperature.

Id like to be able to do this experiment to find out what really happens... I suppose the initial oscillations might be detectable although over time I would expect the oscillations to tend to a very high frequency 1 / \Delta T (as it is light speed that is mediating the heat transfer) for a lab sized experiment...

 

[Dale]

Geometrical arguments shouldn't be used to prove or disprove thermodynamics anyway as they are not general enough.

In the context of the conversation I don't understand this comment at all. In post 29 you specifically posited that you could get thermal energy transfer from a large cold blackbody to a small hot blackbody by using mirrors. That is inherently a geometrical argument.

As I said before I also don't believe there are any violations of the laws of thermodynamics as there is an external heat supply (namely the hot plate).

If you have passive radiative heat transfer from a cold blackbody to a hot blackbody then you are violating the 2nd law of thermodynamics by definition.

 

[the4thamigo_uk]

In the context of the conversation I don't understand this comment at all. In post 29 you specifically posited that you could get thermal energy transfer from a large cold blackbody to a small hot blackbody by using mirrors. That is inherently a geometrical argument. If you have passive radiative heat transfer from a cold blackbody to a hot blackbody then you are violating the 2nd law of thermodynamics by definition.

No, I wonder, does anyone here know what I am talking about? /

Im simply trying to show that it is theoretically possible to direct the radiative power from one source black body at fixed temperature T onto a second smaller black body, in order that the smaller black body achieves a 'steady state' temperature greater than T. The source black body is NOT passive, it is supplied heat from a hot plate. It all seems fairly obvious to me now, but some people disagree it seems?

Avoiding geometry, from only a thermodynamic argument, I 'could' simply state, that these two black bodies are the only things that exist in the world and that all the radiative power that each body emits is directed onto the surface of the other body. This is sufficient for the argument. There really is no need for geometry/optics.

Does anyone disagree with this on a 'purely thermodynamic' basis? This is what I want to find out. If its wrong, then its wrong, but exactly why is it wrong?

However, some people seem have countered that this is impossible, by invoking geometrical arguments about mirrors/lenses and such. Hence, to challenge this geometical argument, and 'only' in order to challenge this geometrical argument, we can use the enclosed black body example, which does not rely on mirrors/lenses. I think it is a fairly robust example.

Unfortunately again, some people may still think that this is an unrealistic counter-example even though it is a perfectly good enough example as far as I can see. We could easily build this experiment I believe.

In fact what the enclosed BB example shows is a perfect example, in that 'All' the radiative power from each body is directed to the other body. But, in fact you can relax this condition anyway. You 'dont even' need to capture all the radiation from the source body. All you need to show is that there is a way to increase the flux emitted from the surface of the source BB, such that it illuminates the entire surface of a smaller black body. This is what the mirrors example shows. You 'dont even' have to double/quadruple the flux, all you need to do is increase it by some amount. The flat BB is a sufficient example I think.

So, for a large BB maintained at T by a source of energy, it *is* possible to heat a smaller BB such that it achieves a higher temperature in a steady state. The two bodies do not have to be in thermal equilibrium as there is heat supplied.

If there is no heat supplied, then the black bodies will (eventually) achieve the same temperature, based on some system of differential equations like those given above.

Does anyone agree with this? ( or are you all bored by it? D ) ...

If you have passive radiative heat transfer from a cold blackbody to a hot blackbody then you are violating the 2nd law of thermodynamics by definition.

Second law of thermodynamics

'Heat generally cannot flow spontaneously from a material at lower temperature to a material at higher temperature.' (Clausius)

For illustration take the enclosed black body example. The condition for steady state is that the (total power emitted by inner BB) = (total power emitted by the outer BB). There is *zero net heat flow* between the two BBs.

It might be the wording of the Clausius statement that is confusing, a BB at non-zero temperature will always emit radiation, yes? There is nothing to stop it? Its charges are jiggling about so it generates radiation. Radiative energy therefore is 'spontaneously flowing' from each body, it is just that they are exactly equal in the 'steady state'

 

[sylas]

Does anyone disagree with this on a 'purely thermodynamic' basis? This is what I want to find out. If its wrong, then its wrong, but exactly why is it wrong?

You are wanting the transfer of energy to occur by spontaneous heat flow, from a cold object to a hotter one. It doesn't matter that there's an energy source to maintain the temperature of the colder object. It still violates thermodynamics.

The problem is that there's no way to focus the heat the way you want. We've looked at a couple of concrete examples, using the Sun, where the size of the Sun means that when you focus it the image also has a finite size. Whatever geometry you pick, you aren't going to get a flux of energy greater than what is coming from the surface already, so you will not be able to focus the heat as tightly as you require.

It is a bit like perpetual motion machines. People can come up with all kinds of new designs and clever tricks, and eventually with analysis we can see where the design fails, but the general rule for knowing you can't do it is not based on evaluations of all designs, but rather on the general thermodynamic principle that no design, no matter how clever, is going to violate.

In the same way, you can try picking your geometries however you like. We know in advance, from the second law, that there's going to be some way in which the design fails to heat something up to hotter that the source of your heat flow. If you propose a particular design in enough detail we might be able to explain where it fails and that may help you appreciate the power of the general principle.

But we know that we're always going to be able to find some flaw in any specific concentrating or focusing system trying to get that heat to flow into a hotter object.

Cheers -- sylas

 

[the4thamigo_uk]

In the same way, you can try picking your geometries however you like. We know in advance, from the second law, that there's going to be some way in which the design fails to heat something up to hotter that the source of your heat flow. If you propose a particular design in enough detail we might be able to explain where it fails and that may help you appreciate the power of the general principle.
Cheers -- sylas

Ive just said above at the bottom of my previous post that there *isnt* a violation of the second law as far as I can see. I don't think I am being dumb.

I think my suspicion that nobody really reads forum posts before replying to them is true...

 

[Yeti08]

The radiation exchange between blackbodies, from object i to object j is given by

q_{ij}=A_{i}F_{ij}\sigma\left(T_{i}^{4}-T_{j}^{4}\right)

where A is the surface area, F_{ij} is the view factor from i to j, and \sigma is the Stefan-Boltzmann constant. From the above equation, which can be derived from more basic principles, it is obvious that when there is no temperature difference, there is no heat exchange, thus no further change in temperature. Furthermore, the view factor are restricted by reciprocity:

F_{ij}A_{i}=F_{ji}A_{j}

So you can't have all the heat transfer from i go to j and have all the the heat transfer from j go to i. For your BB example, the view factor from the BB to an enclosure would be one as you said. However, the enclosure would be at least partially radiating to itself thus the view factor would have to be less than one.

 

[sylas]

Ive just said above at the bottom of my previous post that there *isnt* a violation of the second law as far as I can see. I don't think I am being dumb.

I think my suspicion that nobody really reads forum posts before replying to them is true...

No, I read it all. I am saying you are incorrect, and there is a violation of the second law. I appreciate that you don't see that as yet, but I think you are wrong, and I AM reading your posts.

Cheers -- sylas

 

[atyy]

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?

Is it the work needed to hold the mirrors in place?

 

[sylas]

Is it the work needed to hold the mirrors in place?

No; it is the unstated assumption that the mirrors will be able to focus the light from the large ball of the Sun into a sufficiently small area that the flux will exceed what is at the surface of the Sun. The case 3 is intended to illustrate this.

Cheers -- sylas

 

[atyy]

No; it is the unstated assumption that the mirrors will be able to focus the light from the large ball of the Sun into a sufficiently small area that the flux will exceed what is at the surface of the Sun. The case 3 is intended to illustrate this.

Huh! That is cute!

 

[the4thamigo_uk]

No, I read it all. I am saying you are incorrect, and there is a violation of the second law. I appreciate that you don't see that as yet, but I think you are wrong, and I AM reading your posts.
Cheers -- sylas

You should be a politician... )

Sylas I would find it really helpful if you could examine the steps in my argument below and tell me at which point you disagree :

If you say, as you did above, that the general principles of thermodynamics are the reason my argument fails, then the argument must be proved to fail without reference to optics, since the laws of thermodynamics are independent of optics.

* If you refute this, then you are saying that it is 'optics' and not pure 'thermodynamics' that is the problem with the argument. I would be more amenable to this, but you cannot then tell me that the argument fails because of the second law of thermodynamics. It would fail because of optics alone. Furthermore, I would also hope to counter with my enclosed black body example which relies on geometry to prove that it is still theoretically possible *
If you accept this, we can assume the ideal gedanken experiment with the following conditions :

(1) two BBs in a vacuum of different surface areas
(2) The larger BB (L) is maintained at temperature T by an external heat source
(3) the smaller BB (S) is passive
(4) all the radiation from the larger one is directed onto the surface of the small one
(5) all the radiation from the smaller one is directed onto the surface of the larger one

* If you refute these assumptions then why? Bear in mind that you can't refute (4) or (5) since you have accepted that optics/geometry does not play a part in a thermodynamic argument *

At 'steady state' for body S :

(total radiative power emitted by S) = (total radiative power received from L)

i.e. 'total radiative power' meaning total integrated flux from the BB over its entire surface

* If you refute this then you are saying that there is net heat loss/gain from S so its temperature would change hence it is not in a steady state, hence you are refuting the possibility of a steady state *
and by assumptions (4) above :

(total radiative power received from L) = (total radiative power emitted by L)

Hence by Stefans Law :
(area of S) * sigma * (temp of S)^4 = (area of L) * sigma * (temp of L)^4

* If you are refuting Stefans Law then I would ask you on what grounds you are doing so *
By implication of the formula above :

(temp of S) = (temp of L) 4throot[ (area of L) / (area of S) ]

Hence if (area of L) > (area of S) then (temp of S) is greater than (temp of L)

* If you refute the conclusion then it is a logical absurdity *

And for the second law...

It states :

'Heat generally cannot flow spontaneously from a material at lower temperature to a material at higher temperature.', Clausius

* I will assume you arent going to refute this *
A black body at non-zero temperature always emits radiation

* If you refute this then you refute the idea of a black body emitter*

At steady state, S is transferring all its radiation onto the surface L
At steady state, L is transferring all its radiation onto the surface S

* If you refute this you are saying something is preventing a BB from radiating when at 'steady state' and you are now refuting (4) and (5) which have already been accepted*
At steady state by the above calculations, there is 'zero net power/heat transfer' between the bodies so the second law is satisfied

* If you refute this, then you are presumably saying that you are wanting no radiation to be transferred, in either direction, between the bodies. I am not sure what else you can disagree with. *

The heat transfer is zero but the BBs are different temperatures!

* If you refute this then it is a logical absurdity *

 

[sylas]

You should be a politician... )

Now that hurt!

(1) two BBs in a vacuum of different surface areas
(2) The larger BB (L) is maintained at temperature T by an external heat source
(3) the smaller BB (S) is passive
(4) all the radiation from the larger one is directed onto the surface of the small one
(5) all the radiation from the smaller one is directed onto the surface of the larger one

* If you refute these assumptions then why? Bear in mind that you can't refute (4) or (5) since you have accepted that optics/geometry does not play a part in a thermodynamic argument *

I deny number 4, because it is an assumption that let's you violate the second law. As I said previously, based on the second law I am confident that if you try to be specific about the details of your optics and geometry, it will always be possible to find where your proposed design fails to achieve what you want.

You haven't actually given any optics/geometry argument. You've merely said that you can do this, but haven't provided the geometry or optics necessary.

If you ever do provide the optics and geometry, it will be possible to use the optics and geometry to show the flaw.

If you merely assert that you have some unknown design that works, then we can know, without looking at your design, that there's something wrong with it. That is because no amount of optics and geometry can violate the second law. The same applies for someone claiming to have a design for a perpetual motion machine.

Cheers -- sylas

 

[Yeti08]

The only situation that I can think of where (4) and (5) are both true is for infinite parallel plates (hence reciprocity in real, finite-size bodies), which would have equal (well, both infinite...) surface areas. It is definitely not true for your BB's and, though I can't speak for sylas, I would say that thermodynamics/heat transfer and geometry are intrinsically linked. Radiation heat transfer is very much dependent on shapes and dimensions which can be used to derive view factors (probably more of an engineering method as opposed to the basic physics).

 

[Dale]

Im simply trying to show that it is theoretically possible to direct the radiative power from one source black body at fixed temperature T onto a second smaller black body, in order that the smaller black body achieves a 'steady state' temperature greater than T.

That is exactly what you have not yet shown. You have only asserted that it is possible. Note, the word "smaller", it is inherently a geometric argument. You cannot avoid that.

we can use the enclosed black body example, which does not rely on mirrors/lenses. ... In fact what the enclosed BB example shows is a perfect example, in that 'All' the radiative power from each body is directed to the other body.

It doesn't work. If you actually sit down and draw the geometry you will find that most of the energy emitted by the enclosing object misses the small object and hits the other side of the enclosing object. This is a common effect for concave radiating surfaces.

Specifically, consider a differential element on the surface of the enclosing object. This element radiates over a half-sphere solid angle. The small object occupies a smaller solid angle which is a function of its diameter and distance. The remainder of the radiation emitted by the differential element does not hit the small object.

 

[Dale]

The radiation exchange between blackbodies, from object i to object j is given by

q_{ij}=A_{i}F_{ij}\sigma\left(T_{i}^{4}-T_{j}^{4}\right)

where A is the surface area, F_{ij} is the view factor from i to j, and \sigma is the Stefan-Boltzmann constant. From the above equation, which can be derived from more basic principles, it is obvious that when there is no temperature difference, there is no heat exchange, thus no further change in temperature. Furthermore, the view factor are restricted by reciprocity:

F_{ij}A_{i}=F_{ji}A_{j}

So you can't have all the heat transfer from i go to j and have all the the heat transfer from j go to i. For your BB example, the view factor from the BB to an enclosure would be one as you said. However, the enclosure would be at least partially radiating to itself thus the view factor would have to be less than one.

Excellent reply. That really says it all.

@ the OP: I would encourage you to study Yeti's post in detail.

 

[the4thamigo_uk]

(1) Ok, you are all denying it on the basis of optics. To prove it in all cases you would *at least* need to prove that it is false for every possible shape, size of BB, and every configuration of mirrors, lenses. Hey you then might have a new law of thermodynamics only it wouldn't be, because it isn't general enough to other modes of heat transfer (see (3) and (4) below)
(2) My argument relies on the laws of thermodynamics only and Stefans law. Not on optics. Therefore you certainly cannot deny it on the basis of the second law. If you can't follow this logic then I have no hope for you... I mean it, I really have no hope for you... )
(3) The 4 laws of thermodynamics do not concern themselves with shapes of objects, optics, geometry. They talk about heat and temperature. They don't even mention the surface area. That is Stefans law which isn't one of the 4 laws.
(4) The 4 laws of thermodynamics do not even mention conduction, convection, radiation. They are independent of the means of heat transfer.
(5) It is the 'additional laws' of heat transfer that we apply to the 4 laws of thermodynamics, we have a law of thermal conduction for example and a law of black body radiation.
(6) My argument explicitly proves that for 'purely radiative transfer' it is possible (in principle) to achieve a 'steady state' with different temperatures that satisfies the second law. You don't even really need to capture the whole energy emitted by the source object, you simply make your passive BB so much smaller to achieve the higher temperature. As said before 'steady state' is not 'thermal equilibrium' because the system has an external heat supply.
(7) For thermal convection however the situation is different. A hot body maintained at temperature T would heat up a passive body to temperature T and no higher. I think you have all got it into your head that this applies to purely radiative transport too, since purely radiative transport is an uncommon situation.
(8) The only 'geometry' in the argument is the surface area of the body. There is no necessity to define a shape at all.
(9) My argument *explicitly proves* that it *does not* violate the second law of thermodynamics as shown in my argument. I don't know how many times I can keep saying this. But yet you persist in denying it? I find this very shocking... There is no net heat transferred between the BBs in the 'steady state'
(10) The form factor laws that you quote are not the laws of thermodynamics they are laws of geometry/optics. Furthermore they do not allow for any focussing or reflection to amplify the intensity of the radiation. This can be easily achieved. I repeat there is no real necessity to capture all the emitted radiation from the source you just need to make the flux on the passive BB more intense over its whole surface.
(11) If you want a geometrical argument (it actually IS unnecessary), but since you think that everything hangs on it, I will repeat it yet again.

Geometrical argument
==============

(1) small spherical black body radius r
(2) large spherical shell radius R. Inner surface is a radiative black body.
(3) small BB is at centre of large shell (though not a requirement it makes calculations easier)
(4) outer surface of the shell is in a heat bath at constant temperature T.
(5) There is no radiation emitted to the outside world
(6) Steady state implies the same equations as above. Inner body achieves a higher steady state temperature.
(7) No net power transfer between both BBs, hence NO violation of second law.

Q.E.D. Merry christmas... I will leave you to think about it...

 

[Yeti08]

(1) No, I'm not.
(2) This has nothing to do with optics. It does have to do with geometry though. You may rely on the Stefan-Boltzmann law, but you are applying it incorrectly.
(3&4) Okay, the "4" laws make general statements.
(5) The details are worked out through additional equations. The laws by themselves are worthless unless you know how to mathematically represent the physical phenomena.
(6) No, it doesn't
(7) Are you saying the laws of thermodynamics apply differently to different modes of heat transfer?
(8) Surface area and orientation - surfaces pointing in opposite directions won't transfer heat radiatively.
(9) Yes, it does violate the law. The Clausius statement associated with the 2nd law states that "Heat generally cannot flow spontaneously from a material at lower temperature to a material at a higher temperature" (the exception is extremely small scales, though it always stands at macroscopic scales), therefore you are violating the second law no matter how you look at it, regarless the mode of heat transfer.
(10) View factors apply just as much as cross sectional area applies to conduction or orientation applies to free convection - you can't do much without them

Your geometrical argument fails for the reasons I have already stated. View factors are not optional - they are a simplification of very real effects. I just don't have the time or patience to explain their derivation through integration of solid angles between differential elements.

I will leave you to think about it...

With all due respect, I have thought about this. I've read several books and taken several classes concerning this sort of thing, and I work in a field that deals with this.