# Deriving average surface temperature of a blackbody sphere?

by Graeme M
Tags: average, blackbody, deriving, sphere, surface, temperature
Mentor
P: 9,632
 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.
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.

 The critical point to me is whether this absorption occurs at the molecular level, or at a smaller or larger scale.
It does not matter.

 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?
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).

 Does our molecule absorb more or less radiation than the one at the equator, and does it heat more or less?
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.

 The question is, does it reach thermal equilibrium more quickly, more slowly, or at the same rate?
Depends on the details of the sphere. On a rotating sphere, there is no equilibrium for individual regions anyway.

 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.
No.

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

 Clearly, there must be something about radiation and heat transfer that I don't understand.
I don't think it is useful to consider individual atoms.
P: 48
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
 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.
Why?

Mentor
P: 9,632

## Deriving average surface temperature of a blackbody sphere?

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}##.

 Why?
More atoms on the surface per incoming light. Simple geometry.
Mentor
P: 9,632
 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).
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.

 an individual point will receive less than it did before.
A "point" in the classical sense has no surface, and will receive a power of 0.
HW Helper
Thanks
P: 7,915
 Quote by Graeme M I am not seeing why light shining on the tilted surface is diminished because of the tilting.
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.
 P: 48 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. Correct?
HW Helper
Thanks
P: 7,915
 Quote by Graeme M 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.
No. If you move the frame closer to the source then the frame gets a bigger share of the total output of the source.
 Mentor P: 9,632 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 Thumbnails
 P: 48 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?
 Mentor P: 9,632 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.
 P: 48 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!! :)
PF Patron
P: 2,140
 Quote by Graeme M 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?
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)