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Deriving average surface temperature of a blackbody sphere?

  1. Nov 21, 2012 #1
    I have been reading about climate change recently and I have run into a problem understanding the earth's energy budget. I will try to summarise my question but am happy to try to expand if needed. Bear in mind I am just an interested average Joe with little science background and probably a very rudimentary graps of physics.

    The basic point I am grappling with is how the temperature is calculated. I am not sure if I understand just what is being stated as various sites explain the calculation in different ways. What I am uncertain of is how an effective temperature for the earth is calculated. The argument seems to be that if the earth were a perfect blackbody with no atmosphere, the surface temperature would be around 5C. However due to earth's albedo, we adjust this figure to give us an effective temperature of about -18C. As the earth's measured surface temperature is 15C there is therefore a difference of 33C which is due to the Greenhouse Effect of the atmosphere.

    This seems to derive from a calculation that uses the average solar radiative flux of 1360 w/m2. This number is divided by 4 to give an average for the earth's surface due to the fact that the theoretical blackbody is a rotating sphere. That is, the energy intercepted is equivalent to that of a disk with the same diameter. However, the rotating sphere has an area equal to 4 times that of the disk. On average therefore we have approximately 340 w/m2 for the surface of our theoretical blackbody. This is the number used for the TOA energy flux shown in many energy budget diagrams.

    I follow that, but this seems to be just a matter of geometry. I don't see why a square metre becomes the effective area of heat transfer when a w/m2 is merely a unit of measure.

    To try to explain. As I understand it, a perfect blackbody absorbs all incident radiation and has no albedo. The critical point to me is whether this absorption occurs at the molecular level, or at a smaller or larger scale. Imagine the surface of our blackbody sphere. At a point on the equator with light shining from the sun directly at that point, how much radiation can an individual totally absorbent molecule absorb? How much is available for it to absorb?

    My guess is that it absorbs as much of the available energy as it needs to ‘heat’ up and reach thermal equilibrium. Presuming that my guess is right, let us now consider another molecule at a point very close to the pole. What will be the effect on that molecule, providing that the full disk of the sun is visible to the molecule? Bear in mind that this molecule is part of a blackbody with no atmosphere, so there is no atmospheric absorption or scattering and the body has no albedo. Does our molecule absorb more or less radiation than the one at the equator, and does it heat more or less?

    The question is, does it reach thermal equilibrium more quickly, more slowly, or at the same rate? Put another way, is there more or less radiation available to our pole based molecule when compared to our equator based molecule?

    The answer to this seems to me to determine how much energy is available to our blackbody and how hot it will be at equilibrium. If for some reason radiative flux is only effective in units of square metres then the calculation I noted above is correct. But if the flux is effective at the individual molecule level, then surely it is not?

    Another thought experiment illuminates my question another way. The GHE requires backradiation from greenhouse gas molecules. So, individual molecules can clearly 'heat' and give off energy. And an individual CO2 molecule for example at 1mm above the pole of our theoretical blackbody must logically, if exposed to the full disk of the sun, heat as much and as quickly as a molecule 1mm above the equator.

    Why would a molecule on the surface of a blackbody not behave the same way?

    Clearly, there must be something about radiation and heat transfer that I don't understand. Can anyone explain simply?
  2. jcsd
  3. Nov 21, 2012 #2


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    The meter is not special in any way here. You could do the same calculation with W/km^2, with W/(earth surface), W/squarefoot or whatever. In the same way, watt is nothing special, you could use any other unit of power.

    It does not matter.

    The number of photons per molecule and typical timescales is small.
    Consider a crystal with ~100pm^2 surface per atom and 1000W/m^2 of 500nm-photons: This corresponds to 26 photons per second and atom or 40ms per photon - extremely long compared to atomic timescales (~picoseconds).

    Depends on details of the surface. In a perfect sphere, it absorbs less photons corresponding to cos(alpha) where alpha is the angle between sun and the vector orthogonal to the surface.

    Depends on the details of the sphere. On a rotating sphere, there is no equilibrium for individual regions anyway.


    No. The surface below that molecule is colder at the pole. And keep in mind that "heat" is not defined for individual particles.

    I don't think it is useful to consider individual atoms.
  4. Nov 21, 2012 #3
    Thanks mfb but that didn't really help me. I *did* say I had no science background, so a lot of what you wrote went over my head. I think too I phrased my question poorly. However you did touch on what I am finding to be a stumbling block. In a general sense I understand the way that the GHE works. It's that number of 340 w/m2 at TOA used in energy budget diagrams.

    Put another way, why is it 340 w/m2 rather than 680 w/m2? Lets remove the matter of rotation. Why is the flux at any point on the lit hemisphere of a perfect blackbody sphere at the same size and location as earth stated to be 680 m/w2 rather than 1360 w/m2?

    You have said
  5. Nov 21, 2012 #4


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    The surface of earth is (approximately) ##4 \pi r^2## where r is the radius of earth. That surface emits blackbody radiation into space.
    As seen from the sun, the earth is like a disk with an area of ##\pi r^2## - all light emitted towards that disk will hit earth.
    This has an intensity of 1360W/m^2, therefore the total incoming solar power is ##1360 \frac{W}{m^2} \cdot \pi r^2##.
    In equilibrium and with the assumption of a single, constant global temperature, the same power is emitted as blackbody radiation with an unknown power PBB and an area of ##4 \pi r^2##:
    $$P_{BB} \cdot 4 \pi r^2 = 1360 \frac{W}{m^2} \cdot \pi r^2$$
    As you can see, ##P_{BB}=\frac{1}{4}1360 \frac{W}{m^2}=340\frac{W}{m^2}##.

    More atoms on the surface per incoming light. Simple geometry.
  6. Nov 21, 2012 #5
    Thank you again mfb, and my apologies if this is somewhat tedious. i guess i just don't know how to phrase the question.

    Your latest response merely restates my question. I understand that the calculation makes use of geometry. My question is more to do with why that should be so. And I think it relates more to my ignorance about how light/radiation is propogated. or even what radiation actually is.

    Is it strictly required that a surface must be perpendicular to the radiation to absorb all of that radiation? Let us take a theoretical square metre of perfectly smooth, perfectly absorbent to radiation, exhibiting no albedo material.

    If this is placed at the location of the earth, perpendicular to the suns rays, it will receive 1360 w/m2 (my apologies if my terminology is inaccurate but hopefully it's clear what I mean).

    If I then tilt my square to approximately 30 degrees away from the sun, by your statements there is less energy available to each point on the surface of that square because now there is less of it 'visible' to the sun's rays (or another way to say that is that it presents a smaller target).

    I am asking why that would be so for a perfectly absorbent non reflective surface. Not the geometrical explanation for that, but why if that surface is still exposed to the radiation from the sun, an individual point will receive less than it did before.

    If that individual point is a molecule, and that molecule were a free floating isolated molecule, it would heat up and give off heat. It should do the same if it is part of the square or if it free floating. Yet you argue that it will do so more slowly if part of the square than if it were isolated.

    What I am asking is that if my tilted square still receives the radiation from the full disk of the sun onto its surface, why should any point receive less radiation than it did previously?
  7. Nov 21, 2012 #6


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    Right. Less power for the same area.

    If you give 20 bananas to 10 monkeys (fair), each monkey gets 2 bananas.
    But if you just have 10 bananas, each monkey gets just 1 banana. It is the same situation with light. Less light for the same area => less light per area. You can divide the area into many small parts, corresponding to molecules, and nothing changes.

    A "point" in the classical sense has no surface, and will receive a power of 0.
  8. Nov 21, 2012 #7
    Less power for same area. Now we are getting closer. So it's something about light that I don't understand. Forgive me for being rather slow here. And for testing your patience. Can we dig a bit further?

    You speak of light in this context as though it's a volume. But I thought light propogates as either a wave or a particle?

    In this context, I am not seeing why light shining on the tilted surface is diminished because of the tilting. The apparent reduction in surface area is a visual effect, there is after all no change in the actual surface area. Now if light shone on that surface as a single physical stream I can understand the point - that is, the fixed volume of the stream is spread over a broader area.

    But is that how light (or radiation) is really manifested? Surely if we are talking heating, then it is each individual molecule that is excited by the radiation from the sun. Those molecules would then heat other molecules and so on until the whole object is heated and radiating at equilibrium. So isn't it the case that on the surface of the tilted square, it is the individual molecules that are being excited?

    At that level then, why is there less radiation available to a molecule on the tilted square than on the perpendicular square? Wouldn't the amount of available energy in any field of radiation from a uniformly radiating sphere be the same at every place of equal distance from that source?

    Or are you saying that in this case, radiation is actually a physical stream of particles with a finite number of particles?

    If that is so, then imagine if you will the surface of the sun. It is emitting these streams of particles from every available part of its surface, and presumably emitting these in all directions. if I take a single square metre of its surface, is it emitting particles (photons?) only perpendicular to its surface, or in all directions from horizon to horizon?

    I'd have assumed in all directions horizon to horizon. If radiation is emitted in all directions, and it travels in a straight line forever (disregarding effects like gravity etc), then light emitted from the top of the sun's disk (or visible hemisphere really) will in effect shine 'down' on our tilted square.

    That is, two rays of light emitted one from the top and one from the bottom of the sun exactly parallel to the plane of the tilted square/sun relationship would presumably still be at the same distance apart when they reach our tilted square. So equally, a ray emitted from the top of the disk that is angled 'down' relative to our plane would tend to strike our square fully perpendicularly?

    What is wrong with my mental construction here?
  9. Nov 21, 2012 #8


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    Imagine a square wire frame placed in the path of the light, at right angles to the direction of the light. The total power in the light passing through the frame cannot be affected by how a surface beyond the frame is tilted. So that same power must be landing on, and spread across, the area bounded by the shadow of the frame on the surface. But as the surface is tilted, that area bounded by the shadow increases, so the power per unit area must decrease.
  10. Nov 22, 2012 #9
    That does help admittedly. But it is still just restating my question in a different way.

    Let me change slightly your reply:

    Imagine a square wire frame placed in the path of the light, at right angles to the direction of the light. The total power in the light passing through the frame cannot be affected by its relationship to a surface beyond the frame. So that same power must be landing on, and spread across, the area bounded by the shadow of the frame on the surface. But as the frame is moved closer to the source of the light, that area bounded by the shadow increases, so the power per unit area must decrease.

  11. Nov 22, 2012 #10


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    No. If you move the frame closer to the source then the frame gets a bigger share of the total output of the source.
  12. Nov 23, 2012 #11
    Yes, I see that. You are quite right. Nonetheless it is still missing the point of my question.

    I think my issue is this. I do understand the geometrical aspect. That is very simple. And all of the answers above have simply restated in different ways that aspect. But I have no problem with that. Rather, when I sat and thought about it, it seemed to me that the question is how does radiation propagate and 'heat' an object.

    I didn't see in my mind that it had to occur at a macro level like a square metre. The fact that a disk and sphere intercept the same amount of light and cast the same shadow seems a quite separate matter and more of an optical one.

    I had assumed that matter is heated by interactions at the atomic/molecular level rather than at some broader scale. If so, then what is different about how radiation from the sun might heat a molecule on the equator of our theoretical body compared to a molecule say halfway to the pole.

    The argument advanced above is that of a geometrical explanation of a static quantity. For example, a rubber sheet covered with X dots per square metre will have fewer dots per square metre if stretched over a hemisphere. Quite obvious.

    But why should radiation act like that? Until it strikes a molecule, it is of the same power at every point equidistant from the sun.

    Let me try to illustrate my thinking. A quick caveat of sorts - I know that the power falls off in a strict relationship with distance, but I doubt it is a significant amount over the few thousand km represented by the distance from the earths equator to the pole. So I am assuming largely the same power of 1360 w/m2 in the following scenario.

    Imagine a location at the equator of an imaginary blackbody sphere, and another location much closer to the pole, perhaps 2/3 of the way there. Now, remove the sphere. If we had one or more molecules at each of those locations, I would think they will each be affected by the sun's radiation in the same way, that is there is no special condition associated with the location of any of those molecules (or think of it another way - with no molecules present, how much power is 'available' in the radiation extant at those points?)

    There is not 1/2 or some other fraction available to the one/s at the more poleward location surely? It is the same amount for all (remember we are disregarding the small difference due to distance).

    Now let us return our imagined sphere to our mental model - why should the molecules at the same location which are now part of the sphere behave differently simply by being part of the sphere? Why is there now less radiation (power) available to those molecules?

    Remember our sphere is completely absorbent, non-reflective, with no atmosphere. Does heating really occur at some macro scale when we have bodies larger than a few free-floating molecules? Or is it that radiation and heating are far more complex than I know about? That is certainly the case, yet the argument advanced is a simple one - the power of radiation and it's ability to heat, diminishes on a spherical surface as one moves further toward the 'pole'. And it is argued this is a geometrical effect.

    I cannot see why that is so. Note I am not arguing against the fact, simply observing that when I think about it, it is not immediately clear to me why it is so.
    Last edited: Nov 23, 2012
  13. Nov 24, 2012 #12


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    Does that sketch help?
    The number of photons which hit each atom per time is smaller in the second case.

    If that does not help, I really cannot understand your problem.

    Attached Files:

  14. Nov 24, 2012 #13
    Thanks mfb. Your diagram makes sense of course and it has helped me crystallize what it is that I have in mind. What your diagram illustrates is what would happen with a single stream of light in which all 'rays' are parallel. Then we would indeed have a discrete quantity of photons per unit area and as we tilt the target and in effect make the target area smaller it receives a smaller number of photons in total.

    However the sun is very much larger than earth and is radiating from ALL of its surface in ALL directions. That should mean that we don't have a stream of parallel rays of light. Thus at say a place near the north pole of our theoretical sphere we have light rays coming from in a sense above, below and from the side.

    Do you see what I mean there?
  15. Nov 25, 2012 #14


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    It does not matter - at the north pole (and at equinox), sunlight comes from ~0° to ~0.25° over the horizon, and the whole sunlight has that tilt issue. At the equator, light comes from ~89.75° to ~90.25° over the horizon.
    You can integrate over the various angles to get the total intensity per surface area.
  16. Nov 25, 2012 #15
    Thought some more overnight, and realised something that maybe seems obvious to everyone else but didn't to me. Light from the sun will be received at any point as a 'stream' directly parallel to a line from that point to the centre of the sun. That means that an object at point A will experience light as coming directly at it, as will an object at point B whether B is 1 metre or 1 kilometre or 1000 kilometres away.

    Just why this is so rather than my earlier idea of it I cannot explain nor have I found anything in what I have read - it clearly would require me to be a) smarter and b) better educated!! :)
  17. Nov 26, 2012 #16


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    But light only travels in straight paths.
    Moreover, the sun is very far away; so from an optical point of view it is a tiny -but bright- point(almost) source. This become fairly obvious if you look at the shadows that are generated by sunlight.

    (It is also the reason why you shouldn't not take photos at mid-day, the sunlight is hitting everything from above, and unless it is overcast the shadows become very harsh)
  18. Nov 26, 2012 #17
    Well, it's the matter of sunlight travelling in a straight line that started me on this original question. Again, I have a limited understanding of how light works. But I understood that light or radiation is created by an atom or some such particle being excited and giving off a photon.

    So, if we have a large object like the sun that is emitting radiation from every point on its surface, we must have light travelling in straight lines outward from each of those points. It is not being emitted only in a straight line outward in one privileged direction surely. This is suggested by what I read about 'global warming'.

    Here the argument is that CO2 molecules are heated by absorbing long wave infrared radiation emitted by the earth's surface. They then emit radiation of their own in response in all directions, including back at the earth hence the fabled 'back radiation'.

    This implies to me that radiation is emitted in all directions from a non-directed source. Eg a torch uses a reflector to focus a beam of light, but the sun is not a torch.

    Thus if light is being emitted in all directions from the surface of the sun, it must go in a straight line for all possible directions horizon to horizon for any given point (and that by the way is not a classical point as mfb noted - I simply mean atom or whatever gives rise to radiation).

    So, the light emitted from a point at the top of the sun's disc as seen from earth must include rays that travel straight out as must a point at the bottom, and at the earth's location they must still be parallel and at equal distance apart as when they were emitted.

    But within those bounds in effect are also rays going in all other possible directions, including 'down', 'up' and sideways upon the earth.

    See my diagram below this post.

    However, from the comments above, it is clear to me that this does not happen. Somehow in fact the light from the sun when intercepted by the earth is effectively all travelling in a straight line, and this is the case for all points on the earth's surface. If there were no atmosphere, we should see a clear, sharp shadow directly parallel to that straight line, not the multi shadow or fuzzy shadow my thinking would imply.

    It then follows that the light must be travelling in a straight line relative to any point on the earth's surface. Imagine a square metre of material at distance X from the sun, if it is directly facing the sun it will receive the full power of the sun. Imagine another square placed a short distance away but still at distance X from the sun, will it be at the same angle? No. For each square, if we place a huge number of such squares at the same distance from the sun, the angle must be different one to the other. With enough squares we would create a spherical shell.

    Hence, light must always be received in a straight line relative to the point receiving it.

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  19. Nov 26, 2012 #18
    And thinking further. f95toli, you say:

    But that is literally from an optical point of view. It's what we see using a lens and a retina. If we had a very large light sensitive surface, quite flat, arranged at the point of the earth in space and perpendicular to the plane of the earth/sun, would not all points on that surface over a disc the same size as the sun receive the same amount of light ?

    In fact, I suspect that the surface would receive the same amount for the same sized disc, and then less according to some ratio as we travel further beyond that point because beyond that point the surface is effectively tilted to the rays that strike it.

    Which brings us back to my very first point that the effect of the diminution in power over the curvature of the theoretical sphere's surface is not strictly an effect of the relationship between a disc and a sphere.
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