Understanding the Greenhouse (Gas) Effect

[arkie87]

And that would happen when the last atmosphere layer; atmosphere from certain height to top, which average temperature is equal to the average surface temperature it will need to emit the same radiation back to space.

I dont think that part is true. It is just a thermal resistance network. Temperature is like voltage and heat is like current. In order for the atmosphere to release the same current (heat) as it receives from the sun, it must be at a lower voltage (temperature) than the surface, since current (heat) is flowing from the surface to the atmosphere, and there is a/many (thermal) resistance(s) between the surface and the atmosphere.

Imagine this example: an isothermal blackbody radiating surface surrounded by an isothermal (i.e. really well mixed) pure greenhouse atmosphere. Imagine the atmosphere is 10x thicker than it needs to be to block all outgoing surface IR radiation. Under these conditions, there should be no greenhouse effect, since the upper most layer is emitting heat to space (and back to the planet) at the same temperature as the surface (and it doesnt cool since it is isothermal/well-mixed).

Here, I am making a simplification/assumption that the average heat flux escaping the gas is equivalent to that of a blackbody radiator at the same temperature despite the fact that the gas emits equally in all directions, not just vertically or into space. However, since layers below the edge near the edge also emit radiation in all directions, some of which end up escaping to space, it is possible that the net effect is still to emit, on average, the same heat flux a blackbody radiator would. If this is not the case, that is fine, but it would only have the effect of reducing the effective emisivity of the radiating layer to space; thus, a slight qualifier is needed: if the surface emisivity were equal to this effective emisivity, there would be no greenhosue effect.

I think this is a very good example to discuss since it really describes how the greenhouse effect works (and DOESNT work)...

We dont know what might be the co2 trigger in those times, co2 only last 30 years, maybe something trigger but before the temperature rise enoght, the amount of co2 decay.

Why do the triggers matter; if the positive feedback exists, it doesnt matter what triggers it? each individual CO2 molecule might only last 30 years before it is absorbed by the ocean or into a plant, but that doesnt matter if another one (or two!) take its place. In water cycle, residence time of each molecule might be much shorter but there is usually always another molecule to take its place, since humidity is always rising up and down.

Yeah, but we were talking about the -33 degrees that earth should have without greenhouse.

I dont know what you were talking about, but i was talking about the temperature swings from day to night only when referring to atmosphere; and only the fact that the earth is warmer than the -33C when referring to greenhouse gasses.

If you have a planet with sea and you dont have atmosphere (lets pretend the water remains liquid XD), if you place a thermometer 1 to 10 meters above the sea even in the night side, then it will have the same temperature than the sea, because it receives heat by radiation.

It would have to be facing the sea and insulated from the lack of radiation from cold space.

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Even on a cloudless day, the California coastline is (usually!) a lot cooler than the Mojave Desert, so obviously clouds can't be the primary factor. There's only one other explanation: greenhouse gases absorbing infrared on its way DOWN.

And whenever heat is absorbed by a gas before reaching the ground, where does it go? Same direction a hot-air balloon does. Up.

Ok, this is what i thought you were suggesting. Please cite a reference that shows IR intensity from sun is non-negligible... I have one here that shows it is negligible, so this cannot be the mechanism:

I suspect you might argue that the IR radiation the water vapor in the atmosphere is absorbing isnt coming from the sun, but from our own atmosphere. This is backwards, as the net effect is always to transfer heat from hot to cold. Since the upper atmosphere is colder than the lower atmosphere and surface, it is not possible that the upper atmosphere is warming the lower atmosphere i.e. net heat flow is from upper (cold) to lower (hot). If it were, then you have discovered a way to get free energy!

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Call me paranoid, but just to make sure no misunderstanding is happening: I said LACK OF water vapor is the reason deserts are hotter.

Sorry for my lazy ambiguous writing. Yes, i meant lack of.

Edited August 13, 2015 by arkie87

[Ralathon]

However, one thing that has confused me for a bit is this:

Consider a blackbody radiator in the shape of a Kerbal separated by a volume of isothermal greenhouse gas (well mixed) from a IR camera. As the pressure of the gas is increased, the image of the Kerbal will slowly disappear; however, the image will not be dark. It will change from a picture of a Kerbal to a uniform bright light, since the gas absorbs and emits the light. Thus, while the light from the Kerbal will not reach the other side of the detector, plenty of light will reach the IR camera (it will just be diffuse).

Thus, there appears to be a limit on the maximum opacity, since the outer-most layer will always radiate to space (even if the original light waves were absorbed and re-emitted).

However, i suppose what causes greenhouse effect is when the outermost layer is very cold (i.e. when the outer most layer is thermally insulated from the surface), such that the amount of radiation leaving the planet is less than would leave if the surface could radiate to space directly.

Does that sound about right?

The outflow of energy will always be the same as the inflow (if you have a stable situation). But the greenhouse effect makes it harder for the heat to escape. It's important to make a distinction between heat and temperature here.

An analogy I always like to use is a water bucket with a valve at the bottom. You pour water into the bucket at a constant rate (visible sunlight). The water fills up the bucket until the outflow via the valve (IR radiation) equals the inflow. If you now throttle the needle valve (increase the greenhouse effect) less water will escape the bucket. So the water level starts to rise until the outflow is once more equal to the inflow. The water level is equivalent to the temperature in this analogy. Even if we double the greenhouse effect we have just as much outflow as inflow. But the earth stores more joulles before radiating them and thus the surface temperature will be higher.

[arkie87]

The outflow of energy will always be the same as the inflow (if you have a stable situation). But the greenhouse effect makes it harder for the heat to escape. It's important to make a distinction between heat and temperature here.

An analogy I always like to use is a water bucket with a valve at the bottom. You pour water into the bucket at a constant rate (visible sunlight). The water fills up the bucket until the outflow via the valve (IR radiation) equals the inflow. If you now throttle the needle valve (increase the greenhouse effect) less water will escape the bucket. So the water level starts to rise until the outflow is once more equal to the inflow. The water level is equivalent to the temperature in this analogy. Even if we double the greenhouse effect we have just as much outflow as inflow. But the earth stores more joulles before radiating them and thus the surface temperature will be higher.

This does not at all address the points and questions i have raised in the quoted text.

I asked what is the derivative of f(x)=x^2*cos(exp(x)) with respect to x, and you answer that the derivative of f(x) with respect to y is 0... true, but it is a lot simpler to calculate and doesnt answer my question...

Edited August 13, 2015 by arkie87

[Ralathon]

This does not at all address the points and questions i have raised in the quoted text.

I asked what is the derivative of f(x)=x^2*cos(exp(x)) with respect to x, and you answer that the derivative of f(x) with respect to y is 0... true, but it is a lot simpler to calculate and doesnt answer my question...

Then state your question more clearly. The way I read your post is that you're confused how the greenhouse effect works since an opaque sphere around a kerbal would radiate just as much heat as a naked kerbal. You then go off on some hypothesis about the temperature of the outer layer and that this causes less heat radiation.

So my post explains how the same energy inflow and outflow can nevertheless result in higher surface temperatures.

Is my reading comprehension completely off here?

[AngelLestat]

I dont think that part is true. It is just a thermal resistance network. Temperature is like voltage and heat is like current. In order for the atmosphere to release the same current (heat) as it receives from the sun, it must be at a higher voltage (temperature) than the surface, since there is a/many (thermal) resistance(s) between the surface and the atmosphere.

Not.. because the surface is behind a big greenhouse effect, if venus atmosphere would be releasing so much heat to space as its surface temperature, then it will have an input of 650 w/m2 and a output of 16000 w/m2.

Something that cant happens.

Yeah you can make an anology between heat conductive transfer and electrical circuits, the math is the same, but your hypothesis is wrong.

You just need to radiate the same energy you receive.

Imagine this example: an isothermal blackbody radiating surface surrounded by an isothermal (i.e. really well mixed) pure greenhouse atmosphere. Imagine the atmosphere is 10x thicker than it needs to be to block all outgoing surface IR radiation. Under these conditions, there should be no greenhouse effect, since the upper most layer is emitting heat to space (and back to the planet) at the same temperature as the surface (and it doesnt cool since it is isothermal/well-mixed).

Not sure if I follow you, there is not possible case (except for this particular case) where you will have higher temperature at higher height than the surface.

Here, I am making a simplification/assumption that the average heat flux escaping the gas is equivalent to that of a blackbody radiator at the same temperature despite the fact that the gas emits equally in all directions, not just vertically or into space. However, since layers below the edge near the edge also emit radiation in all directions, some of which end up escaping to space, it is possible that the net effect is still to emit, on average, the same heat flux a blackbody radiator would. If this is not the case, that is fine, but it would only have the effect of reducing the effective emisivity of the radiating layer to space; thus, a slight qualifier is needed: if the surface emisivity were equal to this effective emisivity, there would be no greenhosue effect.

You can ignore the radiation than goes down because it will bounce up again.

Lets make an example: your incomming radiation is 300w/m2, so it should out 300w/m2 to find equilibrium.

Using:

Your average temperature should be 270 k (-6 celcius degree)

So is the average atmosphere height from top to certain height, where your thermal mass sum is 270k.

If this will be the venus case, it will be aprox from 58km height to top height. I may be wrong.. is the first time I think in this, and I may oversimplifying this problem.

Why do the triggers matter; if the positive feedback exists, it doesnt matter what triggers it? each individual CO2 molecule might only last 30 years before it is absorbed by the ocean or into a plant, but that doesnt matter if another one (or two!) take its place. In water cycle, residence time of each molecule might be much shorter but there is usually always another molecule to take its place, since humidity is always rising up and down.

Yeah you might right, but as I said.. I dont know..

I dont know what you were talking about, but i was talking about the temperature swings from day to night only when referring to atmosphere; and only the fact that the earth is warmer than the -33C when referring to greenhouse gasses.

But you should stop thinking in day time and night time.. You should always think in average..

Because all climate change graph and energy budget graph takes into account average temperatures and average power by w/m2.

THat is why when scientist said that X planet should be -33 degrees colder that it is without greenhouse effect, the fact that has or not atmosphere should not matter much.

It would have to be facing the sea and insulated from the lack of radiation from cold space.

The thermometer should have the same sea temperature if it wants to radiate the same amount than receives.

Also the thermometer is so small, than the tempearature difference between the sea side and space side does not matter.

Edited August 13, 2015 by AngelLestat

[arkie87]

Not.. because the surface is behind a big greenhouse effect, if venus atmosphere would be releasing so much heat to space as its surface temperature, then it will have an input of 650 w/m2 and a output of 16000 w/m2.

Something that cant happens.

Yeah you can make an anology between heat conductive transfer and electrical circuits, the math is the same, but your hypothesis is wrong.

You just need to radiate the same energy you receive.

WOAH! Sorry. Huge typo. Apparently compulsively re-reading my posts a few times isnt enough!! I have corrected it in the post. Please reread now...

Not sure if I follow you, there is not possible case (except for this particular case) where you will have higher temperature at higher height than the surface.

That is a very interesting graph; how is it possible that higher up it gets hotter? Solar particles? Ozone?

However, this graph has nothing to do with my question... my question is about an isothermal atmosphere and surface.

You can ignore the radiation than goes down because it will bounce up again.

Do you have a mathematical source that proves this to be the case that it perfectly balances and that an isothermal semitransparent media always emits just like a blackbody perpendicular to its containment sphere? If not, do you have a better/more-detailed logical argument as to why all the components still sum to one; i would have thought it might be between 0.5 to 1.

This is good to know though...

But you should stop thinking in day time and night time.. You should always think in average..

Because all climate change graph and energy budget graph takes into account average temperatures and average power by w/m2.

THat is why when scientist said that X planet should be -33 degrees colder that it is without greenhouse effect, the fact that has or not atmosphere should not matter much.

I understand. We are agreeing. At some points in this thread we were arguing about temperature swings (like in deserts), and were discussing thermal mass/atmosphere. At other points, we were discussing average, and were discussing greenhouse gasses and effects.

The thermometer should have the same sea temperature if it wants to radiate the same amount than receives.

The thermometer presumably has two sides. So whereas on the side facing the sea, it absorbs and emits an equal amount of radiation, yielding zero net heat flow, on the other side, it emits but does not receive, thereby making the other side cooler. So unless there is a large thermal resistance between them, the temperature sensor will measure a lower temperature.

Also the thermometer is so small, than the tempearature difference between the sea side and space side does not matter.

The smaller the thermometer, the smaller the thermal resistance between the side facing the sea and the side facing space; therefore, all things being equal, the error is magnified as the sensor gets smaller...

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Then state your question more clearly. The way I read your post is that you're confused how the greenhouse effect works since an opaque sphere around a kerbal would radiate just as much heat as a naked kerbal. You then go off on some hypothesis about the temperature of the outer layer and that this causes less heat radiation.

So my post explains how the same energy inflow and outflow can nevertheless result in higher surface temperatures.

Is my reading comprehension completely off here?

First of all, i never said ANYTHING about the Kerbal being naked

I dont know if your reading comprehension is off, but you are definitely not answering the question i asked. I understand how heat transfer works (perhaps better than most, as I have a degree in it), and I think the entire discussion up until now should indicate that. Your analogy is a good one for explaining how the greenhouse effects works, but you should carefully read my question, which i will repeat here:

Imagine this example: an isothermal blackbody radiating surface surrounded by an isothermal (i.e. really well-mixed) pure greenhouse atmosphere. Imagine the atmosphere is 10x thicker than it needs to be to block all outgoing surface IR radiation. Under these conditions, there should be no greenhouse effect, since the upper most layer is emitting the same amount of heat to space as the surface would be if there was no atmosphere to begin with. The fact that it is absorbed and re-emitted millions of times before finally escaping the planet doesnt matter, since the NET effect is radiation from the surface/atmosphere leaving the planet.

So actually, for the greenhouse effect to exist, it requires that the upper atmosphere be cooler than the surface (not that it "can nevertheless result in higher surface temperatures")

Edited August 13, 2015 by arkie87

[AngelLestat]

WOAH! Sorry. Huge typo. Apparently compulsively re-reading my posts a few times isnt enough!! I have corrected it in the post. Please reread now...

a)I dont think that part is true. It is just a thermal resistance network. Temperature is like voltage and heat is like current. In order for the atmosphere to release the same current (heat) as it receives from the sun, it must be at a lower voltage (temperature) than the surface, since current (heat) is flowing from the surface to the atmosphere, and there is a/many (thermal) resistance(s) between the surface and the atmosphere.

b)Imagine this example: an isothermal blackbody radiating surface surrounded by an isothermal (i.e. really well mixed) pure greenhouse atmosphere. Imagine the atmosphere is 10x thicker than it needs to be to block all outgoing surface IR radiation. Under these conditions, there should be no greenhouse effect, since the upper most layer is emitting heat to space (and back to the planet) at the same temperature as the surface (and it doesnt cool since it is isothermal/well-mixed).

c)Here, I am making a simplification/assumption that the average heat flux escaping the gas is equivalent to that of a blackbody radiator at the same temperature despite the fact that the gas emits equally in all directions, not just vertically or into space. However, since layers below the edge near the edge also emit radiation in all directions, some of which end up escaping to space, it is possible that the net effect is still to emit, on average, the same heat flux a blackbody radiator would. If this is not the case, that is fine, but it would only have the effect of reducing the effective emisivity of the radiating layer to space; thus, a slight qualifier is needed: if the surface emisivity were equal to this effective emisivity, there would be no greenhosue effect.

Yeah I still dont get it

Maybe someone else can help me to answer this.

There are many english words together, my brain can not assimilate all.

a) ok.. that "is true", but I guess is a bit messy the analogy, you need to take the surface as a heat source which leads to more questions.

What it means well mixed? and isothermal? How can you have the same temperature at different height?

There is not such thing as 100% greenhouse layer (perfect IR mirrow), the higher temperatures always will located on the bottom (because they can not go deeper) and everything radiates up from there. Meanwhile if you rise altitude, you receive radiation from down but lower radiation from up, because you will have always a point up where the temperature will be lower.

Maybe I dint get your question.

C) heh, super hard to follow you there. Not sure why you bring up the emisivity.

All the text block seems like a very hard way to explain something as greenhouse, lets see if someone else understand the point you want to made and answer you.

That is a very interesting graph; how is it possible that higher up it gets hotter? Solar particles? Ozone?

http://www.windows2universe.org/kids_space/temp_profile.html

Do you have a mathematical source that proves this to be the case that it perfectly balances and that an isothermal semitransparent media always emits just like a blackbody perpendicular to its containment sphere? If not, do you have a better/more-detailed logical argument as to why all the components still sum to one; i would have thought it might be between 0.5 to 1.

Not, I just use my common sense, and not sure if is right.

But what I did is this:

You always radiate heat against the background space radiations, which is 3k. So if you have a gas at just 100 kelbins, you still radiate 97⁴ , but the thermal mass (density) at that height is so low, that is almost nothing.

More down you go, tempearature of your atmosphere rise at the same time than density.

In my calculation I ignore the thermal mass and I just focus in the temperature gradient.

Lets divide the atmosphere of venus in layers of 10 km starting from 100km top (venus atmosphere reach 200 km, but we can ignore them because those extra 100km are negligible)

I will make up numbers just to visualize the mechanic: (there are not reals and there are not based on real venus data)

first layer will radiate only 1 w/m2

second layer 2w/m2

third 10 w/m2

four 47 w/m2

fifth 240 w/m2

Total= 300 w/m2

Something like that, is just an example, the idea behind will be to imagine what portion of the atmosphere emits the last radiation to the space without "bouce" in superior layers. But of course that is not complety true.

The thermometer presumably has two sides. So whereas on the side facing the sea, it absorbs and emits an equal amount of radiation, yielding zero net heat flow, on the other side, it emits but does not receive, thereby making the other side cooler. So unless there is a large thermal resistance between them, the temperature sensor will measure a lower temperature.

well yeah, depending thermometer position, I was imagine a clasic case, thermometer normal to the sea surface (mercury capsule pointing down, scale up).

It will measure a bit lower, but not half, because even if receive from one side and radiate from the other, radiation works T⁴, which a small decrease emits a lot lower than the energy is receiving.

So if the temperature from the sea is 30c degree, the thermometor will measure 27c degree.

The smaller the thermometer, the smaller the thermal resistance between the side facing the sea and the side facing space; therefore, all things being equal, the error is magnified as the sensor gets smaller...

Insulation or resistence has nothing to do, it reach the point when it reach equilibrium between radiation in and out.

Edited August 13, 2015 by AngelLestat

[Ralathon]

I dont know if your reading comprehension is off, but you are definitely not answering the question i asked. I understand how heat transfer works (perhaps better than most, as I have a degree in it), and I think the entire discussion up until now should indicate that. Your analogy is a good one for explaining how the greenhouse effects works, but you should carefully read my question, which i will repeat here:

Okay then mr degree-in-thermodynamics. Let's go through it step by step.

Imagine this example: an isothermal blackbody radiating surface surrounded by an isothermal (i.e. really well-mixed) pure greenhouse atmosphere.

Is the temperature of the isothermal gas equal to the surface temperature of the blackbody here, or are they different?

Imagine the atmosphere is 10x thicker than it needs to be to block all outgoing surface IR radiation.

So, infinite*10? That's not how heat transfer works and you should know this if you have a degree. You can simplify it to a trivially easy model. 2 boundary conditions and an element with a certain thermal conductivity. The thermal conductivity varies with pressure, volume, temperature etc, but unless the number of gas molecules goes to infinite the thermal conductivity will never be zero.

Under these conditions, there should be no greenhouse effect, since the upper most layer is emitting the same amount of heat to space as the surface would be if there was no atmosphere to begin with.

And as I explained in my previous post, this has absolutely nothing to do with the greenhouse effect. The greenhouse effect is about decreasing the thermal conductivity of the atmosphere, it occurs even if the heatflow in equals the heatflow out. I can take a lump of molten lead and wrap it in insulation until it emits the same amount of heat as a mildly warm lump of lead. Kinda hard to argue that the lump of molten lead is only mildly warm, simply because it emits the same amount of thermal radiation.

The fact that it is absorbed and re-emitted millions of times before finally escaping the planet doesnt matter, since the NET effect is radiation from the surface/atmosphere leaving the planet.

it DOES matter because the thermal conductivity of the atmosphere is lower than it would be without the greenhouse gasses. So this means the object needs to have a higher temperature to give the same outflow of heat. You've insulated the blackbody, so it needs a higher temperature to result in the same heat output. So the object will be hotter.

So actually, for the greenhouse effect to exist, it requires that the upper atmosphere be cooler than the surface (not that it "can nevertheless result in higher surface temperatures")

Ah, I see what you're getting at. You're setting T₍₀₎=T_(end). You won't notice the greenhouse effect in that case. In the same vein, if you let the earth cool down to 2.76 kelvin and put it in an intergalactic void, you also wouldn't notice the greenhouse effect. But you're ignoring that not all incoming energy is in infrared. If you added a lamp that shone visible light onto your black body you would notice the greenhouse effect.

Basically your whole argument is: If the sun was shining in infrared instead of visible we wouldn't have a greenhouse effect! Which is true, but kinda trivial...

[arkie87]

Yeah I still dont get it

Maybe someone else can help me to answer this.

There are many english words together, my brain can not assimilate all.

I appreciate your effort to understand me

a) ok.. that "is true", but I guess is a bit messy the analogy, you need to take the surface as a heat source which leads to more questions.

It is a thought experiment to understand exactly how greenhouse effect works

What it means well mixed? and isothermal? How can you have the same temperature at different height?

"Well-mixed" means there is lots of vertical wind such that the atmosphere is isothermal (at uniform temperature). It is not important how its possible to have this in real life. It is a thought experiment.

There is not such thing as 100% greenhouse layer (perfect IR mirrow),

It is a simplified discussion for the purpose of the thought experiment.

the higher temperatures always will located on the bottom (because they can not go deeper) and everything radiates up from there. Meanwhile if you rise altitude, you receive radiation from down but lower radiation from up, because you will have always a point up where the temperature will be lower.

Maybe I dint get your question.

I am talking about a hypothetical (made up) case where temperature is uniform. The atmosphere, in my though experiment, does not have a temperature gradient. It is all at, say, 15C... surface, lower atmosphere, upper atmosphere, etc... I know this isnt realistic and/or the reality, but I am conducting a thought experiment right now...

C) heh, super hard to follow you there. Not sure why you bring up the emisivity.

All the text block seems like a very hard way to explain something as greenhouse, lets see if someone else understand the point you want to made and answer you.

Well, i appreciate your effort

http://www.windows2universe.org/kids_space/temp_profile.html

Thanks!

Not, I just use my common sense, and not sure if is right.

But what I did is this:

You always radiate heat against the background space radiations, which is 3k. So if you have a gas at just 100 kelbins, you still radiate 97⁴ , but the thermal mass (density) at that height is so low, that is almost nothing.

(100^4-3^4) =/= 97^4

(100^4-3^4) ~= 100^4

More down you go, tempearature of your atmosphere rise at the same time than density.

In my calculation I ignore the thermal mass and I just focus in the temperature gradient.

Lets divide the atmosphere of venus in layers of 10 km starting from 100km top (venus atmosphere reach 200 km, but we can ignore them because those extra 100km are negligible)

I will make up numbers just to visualize the mechanic: (there are not reals and there are not based on real venus data)

first layer will radiate only 1 w/m2

second layer 2w/m2

third 10 w/m2

four 47 w/m2

fifth 240 w/m2

Total= 300 w/m2

Something like that, is just an example, the idea behind will be to imagine what portion of the atmosphere emits the last radiation to the space without "bouce" in superior layers. But of course that is not complety true.

Are you starting from the top going down or vice versa? It's hard to follow what you are saying...

So it seems like you agree with me. A thermal gradient is necessary for the greenhouse effect to occur. Without it, there wouldnt be any...

well yeah, depending thermometer position, I was imagine a clasic case, thermometer normal to the sea surface (mercury capsule pointing down, scale up).

It will measure a bit lower, but not half, because even if receive from one side and radiate from the other, radiation works T⁴, which a small decrease emits a lot lower than the energy is receiving.

So if the temperature from the sea is 30c degree, the thermometor will measure 27c degree.

The T^4 dependence would cause a lot more error than that. Assuming water is 30C and space is 3K, the equation is:

(273.15+30)^4 + (3)^4 = 2*(273.15+T)^4

Solving for T, we get: T=-18.23C

Not 27C; huge error.

Insulation or resistence has nothing to do, it reach the point when it reach equilibrium between radiation in and out.

It has everything to do with it. If there is a large thermal resistance between measurement point of thermometer and space-facing surface, the top portion will be colder and will reach equilibrium with space, while the bottom (measurement point) of thermometer will reach equilibrium with ocean.

Edited August 13, 2015 by arkie87

[arkie87]

Okay then mr degree-in-thermodynamics. Let's go through it step by step.

Didnt mean to come off arrogant. Just meant to suggest you dont need to explain basic things to me.

Is the temperature of the isothermal gas equal to the surface temperature of the blackbody here, or are they different?

Intended both temperatures to be equal to each other.

So, infinite*10?

I dont understand what you are talking about here. There is a certain atmospheric thickness (given a certain ppm... or in more appropriate units: atm-cm i.e. distance*pressure) at which just about 100% of a given wavelength will be absorbed. It is exponential and is related to absorption coefficient, which has units of 1/length... If the absorption coefficient is 1 [1/cm], then 1 [cm] thick of gas will transmit exp(-1) percent of the radiation going through it. It will never equal zero, but will approach it... So if you have, lets say, 10x the absorption coefficient length such that you end up with exp(-10) being transmitted, which is non-zero, but very small...

After reading the rest of your post, it appears you think greenhouse effect is related to thermal conductivity of the air, which is why you think my response makes no sense... I think you need to read up on greenhouse effect.

You can simplify it to a trivially easy model. 2 boundary conditions and an element with a certain thermal conductivity. The thermal conductivity varies with pressure, volume, temperature etc, but unless the number of gas molecules goes to infinite the thermal conductivity will never be zero.

Thermal conductivity???? Are you serious?

And as I explained in my previous post, this has absolutely nothing to do with the greenhouse effect. The greenhouse effect is about decreasing the thermal conductivity of the atmosphere, it occurs even if the heatflow in equals the heatflow out. I can take a lump of molten lead and wrap it in insulation until it emits the same amount of heat as a mildly warm lump of lead. Kinda hard to argue that the lump of molten lead is only mildly warm, simply because it emits the same amount of thermal radiation.

I repeat: thermal conductivity? Are you serious?

it DOES matter because the thermal conductivity of the atmosphere is lower than it would be without the greenhouse gasses. So this means the object needs to have a higher temperature to give the same outflow of heat. You've insulated the blackbody, so it needs a higher temperature to result in the same heat output. So the object will be hotter.

I guess you are serious. Greenhouse effect has nothing to do with thermal conductivity... https://en.wikipedia.org/wiki/Thermal_conductivity

In addition, your logic itself is backwards. If increasing greenhouse gas concentration increased thermal conductivity, and the greenhouse effects act through thermal conduction (even though in reality, it actually acts through radiation), then greenhouse gasses should make the earth cooler, since less of a temperature gradient would be needed to release the same amount of heat for a higher thermal conductivity value. Obviously, it doesnt work this way, but im trying to point out the flaw in his (misinformed) logic.

Ah, I see what you're getting at. You're setting T₍₀₎=T_(end). You won't notice the greenhouse effect in that case. In the same vein, if you let the earth cool down to 2.76 kelvin and put it in an intergalactic void, you also wouldn't notice the greenhouse effect. But you're ignoring that not all incoming energy is in infrared. If you added a lamp that shone visible light onto your black body you would notice the greenhouse effect.

Basically your whole argument is: If the sun was shining in infrared instead of visible we wouldn't have a greenhouse effect! Which is true, but kinda trivial...

No... I'm not setting T_0 = T_end... and no, i'm not assuming anything about the sun... my argument holds true if the sun is shining in visible light (and indeed, this is what i actually intended in my example)... maybe i should make a graphic to illustrate my point, or something....

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Nevermind, i figured out the answer, everybody. Thank you those that were trying to help me, and i'm sorry we didnt communicate well...

Edited August 14, 2015 by arkie87

[Ralathon]

I dont understand what you are talking about here. There is a certain atmospheric thickness (given a certain ppm... or in more appropriate units: atm-cm i.e. distance*pressure) at which just about 100% of a given wavelength will be absorbed. It is exponential and is related to absorption coefficient, which has units of 1/length... If the absorption coefficient is 1 [1/cm], then 1 [cm] thick of gas will transmit exp(-1) percent of the radiation going through it. It will never equal zero, but will approach it... So if you have, lets say, 10x the absorption coefficient length such that you end up with exp(-10) being transmitted, which is non-zero, but very small...

Exactly, it will never be zero. Hence my surprise when you try to add an envelope 10 times as thick as needed for complete absorption.

After reading the rest of your post, it appears you think greenhouse effect is related to thermal conductivity of the air, which is why you think my response makes no sense... I think you need to read up on greenhouse effect.

Thermal conductivity???? Are you serious?

I repeat: thermal conductivity? Are you serious?

I guess you are serious. Greenhouse effect has nothing to do with thermal conductivity... https://en.wikipedia.org/wiki/Thermal_conductivity

Read the words "Simplified model". I am fully aware that the greenhouse effect has nothing to do with the thermal conductivity of the atmosphere. But for the purposes of this model that's exactly what's happening. It becomes harder for the thermal radiation to escape so you can simulate that effect by decreasing the thermal conductivity. Just link the thermal conductivity to the greenhouse effect via some function f(P,d,T) and you have a simple 1 element model. Since you have a degree in thermodynamics I reckoned this would be obvious for you, but I guess not...

In addition, your logic itself is backwards. If increasing greenhouse gas concentration increased thermal conductivity, and the greenhouse effects act through thermal conduction (even though in reality, it actually acts through radiation), then greenhouse gasses should make the earth cooler, since less of a temperature gradient would be needed to release the same amount of heat for a higher thermal conductivity value. Obviously, it doesnt work this way, but im trying to point out the flaw in his (misinformed) logic.

conductivity vs resistance. Learn the difference. As far as I can see I never messed them up in my post. Increasing the greenhouse effect is equivalent to decreasing the thermal conductivity in the model, which is the way I used it.

No... I'm not setting T_0 = T_end... and no, i'm not assuming anything about the sun... my argument holds true if the sun is shining in visible light (and indeed, this is what i actually intended in my example)... maybe i should make a graphic to illustrate my point, or something....

- - - Updated - - -

Nevermind, i figured out the answer, everybody. Thank you those that were trying to help me, and i'm sorry we didnt communicate well...

Good you figured it out on your own. Because your argument absolutely does not hold true if there is any energy transfer via visible light.

[arkie87]

Exactly, it will never be zero. Hence my surprise when you try to add an envelope 10 times as thick as needed for complete absorption.

Scientists often makes approximations. For instance, the boundary layer thickness over a wing. True, boundary layer technically extends out to infinity, but scientists define the minimum y-distance that U(y)=0.99*U_0. Making this approximation allows for useful results, such as extremely simple correlations for friction factor, heat transfer coefficient, etc...

If we assume something is essentially opaque when 99% of the light passing through it is absorbed, then i am saying we should have ten times that thickness, such that even if we discritize the volume into ten parts, each part is, by itself, still opaque.

I dont see whats the big deal.

Read the words "Simplified model". I am fully aware that the greenhouse effect has nothing to do with the thermal conductivity of the atmosphere. But for the purposes of this model that's exactly what's happening. It becomes harder for the thermal radiation to escape so you can simulate that effect by decreasing the thermal conductivity. Just link the thermal conductivity to the greenhouse effect via some function f(P,d,T) and you have a simple 1 element model.

I apologize for not understanding that you were trying to model radiation using thermal conductivity, but in my defense, you never specified that. All you would have had to say is: "if we create a simplified model to simulate radiation absorption and transmission via thermal conductivity" or something like that...

Since you have a degree in thermodynamics I reckoned this would be obvious for you, but I guess not...

No need to be rude and condescending. I already explained why i even bothered to mention that. If you are going to continue to be rude, i will kindly request you stop responding to my threads.

If you have ever submitted a paper to a science journal, you know that there is no assumption of competency. Expert reviewers will read what you say literally, and not give you the benefit of the doubt, even if it is obvious that you meant something else or made a mistake. Just correct the mistake and move on...

conductivity vs resistance. Learn the difference.

I am well aware of the difference...

As far as I can see I never messed them up in my post. Increasing the greenhouse effect is equivalent to decreasing the thermal conductivity in the model, which is the way I used it.

What does the following post mean:

it DOES matter because the thermal conductivity of the atmosphere is lower than it would be without the greenhouse gasses.

Does this not mean that thermal conductivity of a greenhouse atmosphere is higher than a non-greenhouse one?

Edit: after review, it appears there was some ambiguity; you meant "... thermal conductivity of the atmosphere with greenhouse gasses is lower ..." . I apologize for the confusion.

Good you figured it out on your own. Because your argument absolutely does not hold true if there is any energy transfer via visible light.

Yes, that was the problem. I moved too quickly from my "naked kerbal" example to a planet, without taking into account the fact that the surface is hit with a heat flux and must be hotter than an isothermal atmosphere in order to transfer net IR heat to it (even if the atmosphere itself is well mixed).

For the future, I think it is a good idea whenever trying to help someone who is conducting a thought experiment to explain the flaw in the thought experiment by directly using the assumptions and logic of the thought experiment rather than what happens in the real world.

It is equivalent to me claiming:

a=1

b=0

And then:

a+b = 0

And then you say: actually, a+b = 1 + 0 = 1 not 0, or, correct my assumption that b=0, by saying: actually b= - 1, and then a+b = 0;

This would have fostered a direct discussion of the thought experiment, and pointed out the flaws in calculations, assumptions, or logic.

What you did is ignore the thought experiment and just explain how greenhouse works in general, which is closer to something like saying:

c=1

d=-1

And then saying:

c+d = 0

Yes, it is true, but I am talking about a case where a=1, and b=0, so either claim b=/=0 or point out that a+b=/=0.

Edited August 14, 2015 by arkie87

[AngelLestat]

"Well-mixed" means there is lots of vertical wind such that the atmosphere is isothermal (at uniform temperature). It is not important how its possible to have this in real life. It is a thought experiment.

in that case there is not winds (convective) neither difference on density due isothermal and 100% IR block, so that looks more like a solid, in that case; the greenhouse effect only start when the real atmosphere starts.

I am talking about a hypothetical (made up) case where temperature is uniform. The atmosphere, in my though experiment, does not have a temperature gradient. It is all at, say, 15C... surface, lower atmosphere, upper atmosphere, etc... I know this isnt realistic and/or the reality, but I am conducting a thought experiment right now...

Yeah, but not sure how that helps you to understand more on the greenhouse effect.

(100^4-3^4) =/= 97^4

(100^4-3^4) ~= 100^4

yeah is correct.

Are you starting from the top going down or vice versa? It's hard to follow what you are saying...

So it seems like you agree with me. A thermal gradient is necessary for the greenhouse effect to occur. Without it, there wouldnt be any...

Yeah, top going down, until the sum of the energy release is equal to the outcomming energy. Rememer that the values are all make up, but they may be similar to that depending the atmosphere properties with a incomming sun energy from 300w/m2.

The T^4 dependence would cause a lot more error than that. Assuming water is 30C and space is 3K, the equation is:

(273.15+30)^4 + (3)^4 = 2*((273.15+T)^4+ 81)

Solving for T, we get: T=-18.23C

Not 27C; huge error.

You are right, I overestimate in my mind the quadratic factor.

I was visualizing the temperatures in centigrade degrees also, which 30 degress vs 18 degress is a big % difference to the quadratic, not if you take 300K vs 254K.

heh, for be lazy and not do the math, but I was also talking into account the thermometer position normal to the surface, which it would receive radiation from both sides (or almost), so that it would increase considerable the temperature measure it, plus other factors.

It has everything to do with it. If there is a large thermal resistance between measurement point of thermometer and space-facing surface, the top portion will be colder and will reach equilibrium with space, while the bottom (measurement point) of thermometer will reach equilibrium with ocean.

Yeah, but we are talking about a thermometer.. that thin piece of glass that we all know.

Or a thermocouple, which it has less resistence.

If you are talking about something alive, then the blood will manage to transfer heat.

Well, not sure what else we can discuss about the greenhouse effect, all points are pretty well clarify.

Edited August 14, 2015 by AngelLestat

[GeneralVeers]

Ok, this is what i thought you were suggesting. Please cite a reference that shows IR intensity from sun is non-negligible...

Wikipedia https://en.wikipedia.org/wiki/Sun says the following:

Sunlight at the top of Earth's atmosphere is composed (by total energy) of about 50% infrared light, 40% visible light, and 10% ultraviolet light.

Note that the above is sunlight before it enters the Earth's atmosphere. Infrared is the number one component. Any greenhouse gas that absorbs infrared will absorb some of that before it hits the ground.

That's not how greenhouse gases work. The a greenhouse gas is something that absorbs strongly in the IR range, but is a weak absorber in the visible. If it blocks as much incoming radiation as outgoing, it ain't a greenhouse gas, by definition.

Not the case at all. A greenhouse gas (or a greenhouse itself!) only needs to retain heat. A greenhouse isn't bright--it's warm. In fact, that would make the perfect example. An actual greenhouse doesn't work by trapping infrared radiation. It works by retaining warm air and preventing that air from circulating to the cooler outside air. Some greenhouse gases do follow your definition, but they don't have to do that in order to meet the job criteria.

A greenhouse gas only needs to retain enough heat to keep the Earth warm and habitable. If it loses more than it blocks, fine and dandy.

The reason deserts have higher temperatures isn't because of water vapour in the air, it's because of a lack of liquid water on the ground. Liquid water is an excellent heat sink, it has a specific heat capacity of 4180 J/kgK, whereas sand, rock or soil are all of the order of 1000 J/kgK, often less:

Then how do you explain the fact that your sidewalk isn't as hot as a desert? Remember a few posts ago when I made the comment about putting your hand on the sidewalk or on the street pavement? Outdoor sidewalks and pavements are definitely warmer than your front lawn, yet they're still a lot cooler than a desert. Why? Liquid water in the ground doesn't explain it. Rather the opposite, in fact. Concrete and asphalt are specifically formulated to have as little water content as possible (to prevent cracking and pothole-ing).

Desert and concrete: same water content (i.e. nearly none at all) but vastly different result.

As Arkie correctly pointed out in the previous thread, the phase change of water vapour does have some moderating effect, but let's take a case study of three cities with similar average temperatures, but varying humidity and proximity to the coast:

Why should all cities everywhere be the same?

Clearly water vapour in the atmosphere isn't the sole driving factor behind temperature or the rate at which it swings, as you're trying to make out.

I disagree. Well, in the strict technical sense water vapor isn't the only factor, but it outmuscles all the others so much that there's no problem with treating it like it is. The Earth's atmosphere contains twenty to forty times as much CO2 as water vapor, yet water vapor is responsible for something like 50% of the ENTIRE GREENHOUSE EFFECT WORLDWIDE. In fact, some estimates range as high as SEVENTY percent (the lowest I've seen is thirty-five percent). My humorous line about water vapor being the Chuck Norris of greenhouse gases was NOT a joke. File it under "funny but true".

Edited August 14, 2015 by WedgeAntilles

[KSK]

I disagree. Well, in the strict technical sense water vapor isn't the only factor, but it outmuscles all the others so much that there's no problem with treating it like it is. The Earth's atmosphere contains twenty to forty times as much CO2 as water vapor, yet water vapor is responsible for something like 50% of the ENTIRE GREENHOUSE EFFECT WORLDWIDE. In fact, some estimates range as high as SEVENTY percent (the lowest I've seen is thirty-five percent). My humorous line about water vapor being the Chuck Norris of greenhouse gases was NOT a joke. File it under "funny but true".

Source please. From the sources I can find, atmospheric CO2 is currently running at just over 400ppm, so about 0.04%. On the other hand average water vapour content of the atmosphere is around 2-3%.

[peadar1987]

Not the case at all. A greenhouse gas (or a greenhouse itself!) only needs to retain heat. A greenhouse isn't bright--it's warm. In fact, that would make the perfect example. An actual greenhouse doesn't work by trapping infrared radiation. It works by retaining warm air and preventing that air from circulating to the cooler outside air. Some greenhouse gases do follow your definition, but they don't have to do that in order to meet the job criteria.

https://en.wikipedia.org/wiki/Greenhouse_gas , first line: "A greenhouse gas (sometimes abbreviated GHG) is a gas in an atmosphere that absorbs and emits radiation within the thermal infrared range". Greenhouse gases don't work via the exact same mechanism as actual greenhouses.

Then how do you explain the fact that your sidewalk isn't as hot as a desert?

Because I live in Glasgow?

Remember a few posts ago when I made the comment about putting your hand on the sidewalk or on the street pavement? Outdoor sidewalks and pavements are definitely warmer than your front lawn, yet they're still a lot cooler than a desert. Why? Liquid water in the ground doesn't explain it. Rather the opposite, in fact. Concrete and asphalt are specifically formulated to have as little water content as possible (to prevent cracking and pothole-ing).

Desert and concrete: same water content (i.e. nearly none at all) but vastly different result.

Glibness aside, the main reason is convection. Large body of water nearby, reduces the air temperature, air absorbs heat from the ground.

Let's take another case study. Summit of Kilimanjaro. The air is extremely cold, and therefore holds very little water vapour. It is close to the equator, and at high altitude, which means less atmosphere to block solar radiation.

Why do you think it's not blisteringly hot up there?

Why should all cities everywhere be the same?

They shouldn't be. Some are close to the coast, some are at altitude, some are humid, the point I was trying to make is that, contrary to what you have said, humidity does not appear to be the main driving factor behind the temperature, or the temperature range, of a city.

You're the one claiming that humid cities should be cooler, and drier cities should be hotter, and that we can ignore any other factors apart from the water vapour content of the atmosphere.

I disagree. Well, in the strict technical sense water vapor isn't the only factor, but it outmuscles all the others so much that there's no problem with treating it like it is. The Earth's atmosphere contains twenty to forty times as much CO2 as water vapor, yet water vapor is responsible for something like 50% of the ENTIRE GREENHOUSE EFFECT WORLDWIDE. In fact, some estimates range as high as SEVENTY percent (the lowest I've seen is thirty-five percent). My humorous line about water vapor being the Chuck Norris of greenhouse gases was NOT a joke. File it under "funny but true".

Water vapour causes a lot of warming if it gets into the atmosphere, but that has very little to do with what you have been saying. Simply put:

-There's a lot of water around, the water cycle is pretty much saturated at current global temperatures. Any more water we emit rains out within a few days.

-The best way of increasing the amount of water the atmosphere can hold is to warm it up.

-The CO2 cycle is not saturated, any CO2 we admit stays in the atmosphere for several decades. It causes a small amount of warming.

-This warming allows more water vapour to be contained in the atmosphere, exacerbating the warming caused by the CO2. A feedback loop.

Contrary to what you have said, water vapour does not strongly absorb visible light from the sun. It is pretty much transparent in the visible range. It does, however, absorb the infra-red light radiated by the earth. This is the mechanism by which greenhouse gases such as water vapour work.

Clouds are a different matter, as they are made of droplets of liquid water, not water vapour

[arkie87]

in that case there is not winds (convective) neither difference on density due isothermal and 100% IR block, so that looks more like a solid, in that case; the greenhouse effect only start when the real atmosphere starts.

No, surface must still be warmer than atmosphere (assuming non-infinte heat transfer coefficient from surface to lower atmosphere) since it is receiving heat flux from sun, and wouldnt be able to reject any heat to lower atmosphere if it was at the same temperature as it.

Yeah, but not sure how that helps you to understand more on the greenhouse effect.

If my logic wasnt flawed, it would have pointed out something very important...

Yeah, but we are talking about a thermometer.. that thin piece of glass that we all know.

Or a thermocouple, which it has less resistence.

Well, i was imagining a thermocouple (since thats what we use in our lab). Since the thermocouple is so thin, heat will easily conduct between the two sides, so if one side is facing 3K and one side is facing 300K, the measured temperature will not be accurate at all. You need to radiatively insulate the thermocouple from the cold side

Well, not sure what else we can discuss about the greenhouse effect, all points are pretty well clarify.

Actually, i figured out how to prove (not so rigorously) that outgoing radiative flux from an opaque volume of gas is equal to a blackbody. Imagine an infinite plane of isothermal gas separated by a distance (so no convection or conduction) from a infinite planar surface at the same temperature of the gas (with a heater to maintain the temperature if needed). The system is in equilibrium, so no temperature changes would take place. Since the surface is emitting like a blackbody, and all the radiation is being absorbed by the gas, the amount radiated by the gas to surface must be equal. Therefore, gas must behave like a blackbody as well, despite the fact that radiation in gasses is volumetric and not a surface phenomena.

[Alchemist]

Here are some more of considerations on gas absorption and thermal radiation.

How is the thermal radiation spectrum for a solid surface calculated? For each wavelength it's blackbody radiation intensity at this wavelength multiplied by what fraction of falling radiation of this wavelength the surface absorbs (as opposed to reflecting).

What about a layer of gas? It's very similar: For each wavelength it's blackbody radiation intensity at this wavelength multiplied by what fraction of assing radiation of this wavelength the gas layer absorbs (as opposed to letting through).

interesting conclusion - a backbody surrounded by a completely isothermal atmosphere of the same temperature will emit the same spectrum as without the atmosphere (no matter of the absorption spectrum of the gas), because the atmosphere will emit to the space exactly the same spectrum it absorbs of the radiation coming from the ground.

Wait, what to put for the temperature in nonisothermal case (to calcuate the thermal radiation of the real atmosphere)? Well, it's wavelength-dependent as well.

In first approximation (ignoring higher volume radiation intensity of warmer lower layers... which doesn't do too much against exponential dimming) if you take the data on what fraction of this wavelength radiation leaves the atmosphere and find effective mean value, you get the line fraction of atmosphere mass about which is inversely proportion to the absorption coefficient at this wavelength (sufficient to say, for high absorbance region of the spectrum, the actual altitudes wouldn't differ too much despite significant changes in this coefficient, just because atmospheric density falls exponentially as well). For further simplification of the model you can even get effective mean value for this altitude and use it (and the temperature at this altitude) in the thermal calculations.

Can we already build a simplified thermal exchange model? Actually, yes.

There are two paths of losing surface heat to space. One is the direct thermal radiation at wavelengths not absorbed by the atmosphere (rather significant for Earth, but most likely completely blocked on Venus), then there's path that leads through the atmosphere, which you can reduce to 3 components:

1) surface-low atmosphere heat exchange. Combines all the non-radiative forms as well as thermal radiation of highly absorbed wavelengths (there's an effective minimal altitude for radiative heat exchange, but for highly absorbed wavelengths it's rather low, so let's not bother too much about variation in heat conductance mechanisms)

2) heat conductance in the atmosphere (which includes both molecule collision and thermal radiation reabsorbtion) - can be boiled down to some heat insulator effect

3) effective upper edge from where you get most wavelengths (of those that atmosphere absorbs) emitted directly (that is with low chance of further absorption) to space.

Now if you think of the system when the upper edge is mostly losing heat (although some of the incoming radiation is absorbed at that level as well...) and there's insulator (atmosphere) between it and the hot surface, there must be a temperature gradient in this heat-conducting part of the atmosphere. Now look at the the atmospheric temperature graph and that's exactly what you see in the troposphere! With the effective upper edge from which most emission happens for wavelengths absorbed by greenhouse gases somewhere about the upper edge of the troposphere (above that there's not enough of these gases to trap heat energy, but the few molecules that there are get in thermal equilibrium with the passing IR radiation resulting in temperature stabilization). Of course, there's also abrupt drop in humidity at that level that helps with releasing most of the trapped heat, but it's both source (large drop in greenhouse gas concentration) and consequence (moisture turning into high-altutude clouds due to very low temperature) of this edge of the troposphere.

Why does the temperature rises again in the stratosphere? It's upper edge is the zone where O2 and O3 absorb the most incoming UV.

And here's the answer to what would be if not for greenhouse effect - there wouldn't be much of the troposphere (I wouldn't claim that there would be none, since greenhouse gas IR reabsorption isn't the only heat transfer mechanism in the lower atmosphere, but the layer efficiently exchanging heat with the surface would be much thinner). Probably, there would be thick tropopause region.

[GeneralVeers]

Source please. From the sources I can find, atmospheric CO2 is currently running at just over 400ppm, so about 0.04%. On the other hand average water vapour content of the atmosphere is around 2-3%.

The source was Wikipedia--but I think I misread the page or something. CO2, around .04%, water vapor anywhere from near-zero to five percent.......

Hmm.....after staring at those last nine words for a minute, I think comparing by mass or by volume is a bad idea. Water vapor varies too widely for the worldwide average to be useful. Be that as it may, the line about water vapor being responsible for half of the entire planet's greenhouse effect remains unchanged (I double-checked to make sure I didn't misread THAT part of Wikipedia.....)

Greenhouse gases don't work via the exact same mechanism as actual greenhouses.

Then perhaps they should rename "greenhouse gases" to something else......(that's not really a "perhaps" actually)

Because I live in Glasgow?

Heheh. That would do it!

Try my old hometown as an example then. I used to live in the Los Angeles area of southern California, which is within driving distance of the Mojave Desert. Los Angeles area sidewalk: uncomfortably hot to the touch. Pavement: VERY hot to the touch.

Mojave desert sand and rocks? OWWWWWW!!!

My current hometown, which is very far (i.e. hundreds of miles) from any ocean? Sidewalk: uncomfortably warm. Pavement: very hot. But, surprisingly, no different from southern California. Can you explain that? I can: I theorize water vapor produced by human beings piping water in from elsewhere to use for their farms and sinks and bathtubs and watering their lawns. Probably farms being the major source.

Glibness aside, the main reason is convection. Large body of water nearby, reduces the air temperature, air absorbs heat from the ground.

Better explanation:

Large body of water nearby. Water evaporates into the air. Air absorbs much more heat from the ground because it has water vapor in it.

Even better explanation than the first one:

Large body of water nearby. Water evaporates into the air. Because water vapor is lighter than air, it goes

up.

Where it can't convect heat from the ground because it's not touching the ground. Instead it absorbs some infrared radiation

before

it reaches the ground. Its low density continues to carry it up, where it radiates the heat into space.

Let's take another case study. Summit of Kilimanjaro. The air is extremely cold, and therefore holds very little water vapour. It is close to the equator, and at high altitude, which means less atmosphere to block solar radiation.

Why do you think it's not blisteringly hot up there?

Letsee......higher albedo (reflectivity) due to ice and snow? Cooling of exposed rock via conduction to colder rock that's covered by ice and snow? Evaporation of ice and snow off the peaks? (thinner air equals faster evaporation, and said evaporation produces strong cooling, just as evaporation off your skin does when you sweat) There ya go, dude: three alternative explanations. Most likely all three of these are active up there to some degree. Certainly circulation is a factor, because it's an exposed peak that probably experiences frequent high winds. But I don't see any reason for that to be Number One.

Or......what if the ground itself on Kilimanjaro (when exposed to the Sun instead of covered by ice or snow) IS blisteringly hot during the day? I've done some mountain hiking myself, so I know that at high altitudes the Sun can be uncomfortably hot.

Got any sources listing measurements of surface temperatures on Kilimanjaro? I looked--couldn't find any. I think the numbers would be interesting--not to settle this argument (that ain't likely to happen! ) but simply for the science.

They shouldn't be. Some are close to the coast, some are at altitude, some are humid, the point I was trying to make is that, contrary to what you have said, humidity does not appear to be the main driving factor behind the temperature, or the temperature range, of a city.

Because you're comparing cities in different regions. Indeed, in different countries. One of your examples was Delhi; bad choice. Almost all of India is kept warmer than other regions at the same latitude because the Himalayas block cold Asian winds.

Did you note how I compared my old hometown to the nearby Mojave Desert? Instead of comparing Delhi and Jeddah and Bangkok, how about comparing Delhi to, say, the nearby Thar Desert? Jeddah to the (somewhat) nearby Sahara Desert? From what I've read so far, the Thar and the Sahara both follow a similar pattern: much hotter maximum temperatures than both Delhi and Jeddah, and much, MUCH colder minimums! Probably something worth noting, is that both of these deserts apparently can have very warm nights during the summer. But when they do freeze, they freeze like a BOSS.

Water vapour causes a lot of warming if it gets into the atmosphere

And cooling. Don't forget that part: cooler daytime and warmer nighttime is my central theme here.

-There's a lot of water around, the water cycle is pretty much saturated at current global temperatures. Any more water we emit rains out within a few days.

-The CO2 cycle is not saturated, any CO2 we admit stays in the atmosphere for several decades.

The amount of time a greenhouse gas stays in the atmosphere? Meaningless. The fancy term for that is a "semi-attached figure".

The relevant figures are these two: how much of each gas is going into the atmosphere per unit time, and how much is being taken out per unit time. The really important number is how much the total amount of a greenhouse gas is changing, and the lifetime of an individual molecule tells you nothing about this process. In fact, if human beings suddenly stopped emitting CO2 entirely? The average "lifetime" of a CO2 molecule would be exactly the same.....(the mention of the long "lifetime" of a CO2 molecule is merely a scare tactic used by environmental activists)

-This warming allows more water vapour to be contained in the atmosphere, exacerbating the warming caused by the CO2. A feedback loop.

And also exacerbating the cooling caused by water vapor. Anti-feedback loop. Remember, water vapor has a much higher specific heat than carbon dioxide; it takes a lot more heat to warm one cubic centimeter of water vapor one degree, than to warm a cubic centimeter of carbon dioxide by one degree. When carbon dioxide absorbs infrared, it warms very quickly--and that heat gets absorbed by anything nearby that happens to be cooler.....

Contrary to what you have said, water vapour does not strongly absorb visible light from the sun.

I never said water vapor absorbs strongly in visible light. Disagree? QUOTE ME.

I said water vapor absorbs strongly in infrared.

It is pretty much transparent in the visible range. It does, however, absorb the infra-red light radiated by the earth.

And it also absorbs the infrared light incoming from the Sun. Why are people constantly leaving that part out???

[peadar1987]

Then perhaps they should rename "greenhouse gases" to something else......(that's not really a "perhaps" actually)

Greenhouses work by two separate methods. One is by trapping warm air and stopping heat loss by convection, the other is that they transmit visible light but block infrared, greenhouse gases only work by one of these methods.

Heheh. That would do it!

Try my old hometown as an example then. I used to live in the Los Angeles area of southern California, which is within driving distance of the Mojave Desert. Los Angeles area sidewalk: uncomfortably hot to the touch. Pavement: VERY hot to the touch.

Mojave desert sand and rocks? OWWWWWW!!!

My current hometown, which is very far (i.e. hundreds of miles) from any ocean? Sidewalk: uncomfortably warm. Pavement: very hot. But, surprisingly, no different from southern California. Can you explain that? I can: I theorize water vapor produced by human beings piping water in from elsewhere to use for their farms and sinks and bathtubs and watering their lawns. Probably farms being the major source.

What altitude are you at? If you're higher up, air temperatures are going to be cooler, which keeps the ground cooler.

Better explanation:Large body of water nearby. Water evaporates into the air. Air absorbs much more heat from the ground because it has water vapor in it.

As I said in the other thread, water vapour content doesn't affect the specific heat capacity of the air in any meaningful way. At 30 degrees, a change from 0% to 100% humidity changes the specific heat capacity from 1.007 to 1.04 J/kgK. As Arkie pointed out, the main dampening effect is the phase change of water, but you need liquid water around to undergo a phase change for this to actually have an effect.

Even better explanation than the first one:

Large body of water nearby. Water evaporates into the air. Because water vapor is lighter than air, it goes

up.

Where it can't convect heat from the ground because it's not touching the ground. Instead it absorbs some infrared radiation

before

it reaches the ground. Its low density continues to carry it up, where it radiates the heat into space.

True to a certain extent, in that water vapour will rise, but not to an altitude where it can radiate significantly more heat to space:

Almost all of the water vapour will have condensed out by 5km altitude, and most a a far lower altitude, by convection, not radiation.

And of course, no matter what altitude it is at, it will still absorb more radiation in the earth's emission bands than in the sun's.

Letsee......higher albedo (reflectivity) due to ice and snow? Cooling of exposed rock via conduction to colder rock that's covered by ice and snow? Evaporation of ice and snow off the peaks? (thinner air equals faster evaporation, and said evaporation produces strong cooling, just as evaporation off your skin does when you sweat) There ya go, dude: three alternative explanations. Most likely all three of these are active up there to some degree. Certainly circulation is a factor, because it's an exposed peak that probably experiences frequent high winds. But I don't see any reason for that to be Number One.

But then you get a chicken and egg situation. Why is Kilimanjaro cold? Because it has a high albedo. Why does it have a high albedo? Because it is covered in snow. Why is it covered in snow? Because it is cold.

Or......what if the ground itself on Kilimanjaro (when exposed to the Sun instead of covered by ice or snow) IS blisteringly hot during the day? I've done some mountain hiking myself, so I know that at high altitudes the Sun can be uncomfortably hot.

Got any sources listing measurements of surface temperatures on Kilimanjaro? I looked--couldn't find any. I think the numbers would be interesting--not to settle this argument (that ain't likely to happen! ) but simply for the science.

None, unfortunately, but because the air temperature is consistently below zero, it's going to keep the surface temperature down as well, by shifting the equilibrium between convection and radiation.

Why is the air temperature lower? Because it's further from the earth's surface, and therefore less re-radiated heat is trapped at that altitude by greenhouse gases.

Because you're comparing cities in different regions. Indeed, in different countries. One of your examples was Delhi; bad choice. Almost all of India is kept warmer than other regions at the same latitude because the Himalayas block cold Asian winds.

Did you note how I compared my old hometown to the nearby Mojave Desert? Instead of comparing Delhi and Jeddah and Bangkok, how about comparing Delhi to, say, the nearby Thar Desert? Jeddah to the (somewhat) nearby Sahara Desert? From what I've read so far, the Thar and the Sahara both follow a similar pattern: much hotter maximum temperatures than both Delhi and Jeddah, and much, MUCH colder minimums! Probably something worth noting, is that both of these deserts apparently can have very warm nights during the summer. But when they do freeze, they freeze like a BOSS.

I picked based on humidity. I could easily have compared Delhi and Mumbai as well. And I'll throw in Bikaner in the Thar desert as well.

Delhi: http://en.tutiempo.net/climate/2014/ws-421820.html

Mumbai: http://en.tutiempo.net/climate/2014/ws-430030.html

Bikaner: http://en.tutiempo.net/climate/2014/ws-421650.html

Average temperatures in Mumbai are actually warmer than in Bikaner. Both Bikaner and New Delhi have comparable average temperatures in spite of Bikaner being dry, and New Delhi being extremely humid.

And cooling. Don't forget that part: cooler daytime and warmer nighttime is my central theme here.

The amount of time a greenhouse gas stays in the atmosphere? Meaningless. The fancy term for that is a "semi-attached figure".

The relevant figures are these two: how much of each gas is going into the atmosphere per unit time, and how much is being taken out per unit time. The really important number is how much the total amount of a greenhouse gas is changing, and the lifetime of an individual molecule tells you nothing about this process. In fact, if human beings suddenly stopped emitting CO2 entirely? The average "lifetime" of a CO2 molecule would be exactly the same.....(the mention of the long "lifetime" of a CO2 molecule is merely a scare tactic used by environmental activists)

Absolutely not meaningless. Residence time is one of the key factors in determining the response of a system, along with level of saturation. The water cycle in the atmosphere is saturated and water vapour has a short atmospheric lifetime, so it is hard to upset the balance by pumping water into the atmosphere, it will return to the equilibrium state, and do so quickly, therefore claims that "water vapour is a stronger greenhouse gas than CO2", while technically true, aren't particularly relevant to the AGW debate in the way that they are usually framed by deniers.

The carbon cycle is not saturated, and the atmospheric lifetime is far longer, so it is far easier to increase the amount of CO2 in the atmosphere by just pumping it out.

Put simply, rain increases quickly when you pump out lots of water vapour. Plant growth and geological carbon sequestration don't increase to such a degree when you pump out more CO2.

And also exacerbating the cooling caused by water vapor. Anti-feedback loop. Remember, water vapor has a much higher specific heat than carbon dioxide; it takes a lot more heat to warm one cubic centimeter of water vapor one degree, than to warm a cubic centimeter of carbon dioxide by one degree. When carbon dioxide absorbs infrared, it warms very quickly--and that heat gets absorbed by anything nearby that happens to be cooler.....

The thermal mass of the system does not affect the equilibrium point, just how long it takes to get there.

I never said water vapor absorbs strongly in visible light. Disagree? QUOTE ME.

I said water vapor absorbs strongly in infrared.

And it also absorbs the infrared light incoming from the Sun. Why are people constantly leaving that part out???

Posting this image again:

The majority of incident solar radiation is not in the infra-red range. The majority of the earth's reemitted thermal radiation is.

While water vapour will absorb incident solar radiation very well, and you can see the absorption bands very clearly in both graphs, the bands are far wider and stronger in the infra-red range reemitted by the earth. If you add water vapour, sure, you will marginally decrease the incident energy from the sun, but you will drastically decrease the amount of energy able to escape from the earth's surface.

[arkie87]

The amount of time a greenhouse gas stays in the atmosphere? Meaningless. The fancy term for that is a "semi-attached figure".

The relevant figures are these two: how much of each gas is going into the atmosphere per unit time, and how much is being taken out per unit time. The really important number is how much the total amount of a greenhouse gas is changing, and the lifetime of an individual molecule tells you nothing about this process. In fact, if human beings suddenly stopped emitting CO2 entirely? The average "lifetime" of a CO2 molecule would be exactly the same.....(the mention of the long "lifetime" of a CO2 molecule is merely a scare tactic used by environmental activists)

I have to agree with Wedge here. Residence time is not nearly as important as fractional changes per time. Yes, residence time is related to fractional changes, but what is directly used by modelers is fractional changes per time, not residence time.

And also exacerbating the cooling caused by water vapor. Anti-feedback loop. Remember, water vapor has a much higher specific heat than carbon dioxide; it takes a lot more heat to warm one cubic centimeter of water vapor one degree, than to warm a cubic centimeter of carbon dioxide by one degree. When carbon dioxide absorbs infrared, it warms very quickly--and that heat gets absorbed by anything nearby that happens to be cooler.....

Specific heat (and even latent heat) from water/water-vapor is not a negative feedback in the way you described. All it effects is thermal mass i.e. how quickly the system reacts to changes in forcing, but not the steady-state solution.

However, water vapor in air can cause negative feedback by forming clouds, which reflect sunlight back into space; it is my understanding that clouds are responsible for the majority of the earth's non-zero albedo (correct me if i am wrong; and yes, ice caps reflect sunlight too, but the solar flux in the poles is much weaker and the fraction of land covered in ice/snow is only a small fraction of the total, whereas clouds can form all over the planet)

And it also absorbs the infrared light incoming from the Sun. Why are people constantly leaving that part out???

Because it should be negligible, since the fraction of energy from the Sun in IR is much less than in visible...

- - - Updated - - -

Greenhouses work by two separate methods. One is by trapping warm air and stopping heat loss by convection, the other is that they transmit visible light but block infrared, greenhouse gases only work by one of these methods.

Not sure why he is arguing this point. Greenhouse gas effect and the way greenhouses work are obviously slightly different. Greenhouse effect is just an analogy. It's like arguing that "color" in quantum mechanics is misleading, since it has nothing to do with color....

As I said in the other thread, water vapour content doesn't affect the specific heat capacity of the air in any meaningful way. At 30 degrees, a change from 0% to 100% humidity changes the specific heat capacity from 1.007 to 1.04 J/kgK. As Arkie pointed out, the main dampening effect is the phase change of water, but you need liquid water around to undergo a phase change for this to actually have an effect.

In his defense, there is liquid water on the ground, so the heat capacity of the air might not change by adding water vapor into it (directly), but water itself can absorb energy (and the oceans have much mroe thermal mass than the atmospehre itself).

Furthermore, while you need liquid water to dampen warming via evaporation, you need water vapor to damping cooling via dew point and condensation.

But then you get a chicken and egg situation. Why is Kilimanjaro cold? Because it has a high albedo. Why does it have a high albedo? Because it is covered in snow. Why is it covered in snow? Because it is cold.

Agree with Peadar here. Arguing that there is ice, so albedo is lower does not explain why there is ice to begin with...

Absolutely not meaningless. Residence time is one of the key factors in determining the response of a system, along with level of saturation. The water cycle in the atmosphere is saturated and water vapour has a short atmospheric lifetime, so it is hard to upset the balance by pumping water into the atmosphere, it will return to the equilibrium state, and do so quickly, therefore claims that "water vapour is a stronger greenhouse gas than CO2", while technically true, aren't particularly relevant to the AGW debate in the way that they are usually framed by deniers.

Well, it is my understanding (from a previous thread) that global warming via CO2 actually requires a positive feedback loop with water vapor to produce a scary amount of warming i.e. more CO2 raises temperature, which raises solubility level of water vapor levels in atmosphere (so the water vapor wont necessarily be fully removed by natural water cycles and negative feedbacks), which causes CO2 to become insoluable in ocean and enter atmosphere, which warms atmosphere, etc...

[peadar1987]

I have to agree with Wedge here. Residence time is not nearly as important as fractional changes per time. Yes, residence time is related to fractional changes, but what is directly used by modelers is fractional changes per time, not residence time.

Perhaps I've been phrasing it badly. The important thing is that the system is saturated. The short residence time of water is a consequence of that, not the cause. Shorter residence times do tend to mean you get back to an equilibrium state more quickly though.

In his defense, there is liquid water on the ground, so the heat capacity of the air might not change by adding water vapor into it (directly), but water itself can absorb energy (and the oceans have much mroe thermal mass than the atmospehre itself).

Furthermore, while you need liquid water to dampen warming via evaporation, you need water vapor to damping cooling via dew point and condensation.

That's pretty much what I've been saying. I'm just trying to spell it out, because two major components of Wedge's argument are:

-The specific heat capacity of humid air is far greater than the specific heat capacity of dry air.

-Specific heat capacity determines the steady-state air temperature.

Neither of these things are strictly true. The presence of liquid water, both as a heat sink in itself, and as a source of a phase change for temperature dampening, are far more important.

Well, it is my understanding (from a previous thread) that global warming via CO2 actually requires a positive feedback loop with water vapor to produce a scary amount of warming i.e. more CO2 raises temperature, which raises solubility level of water vapor levels in atmosphere (so the water vapor wont necessarily be fully removed by natural water cycles and negative feedbacks), which causes CO2 to become insoluable in ocean and enter atmosphere, which warms atmosphere, etc...

That's my understanding as well.

The point I was trying to make was that this isn't how deniers usually talk about the effect of water vapour. What we should be saying is:

"CO2 warming on its own could cause problems, but this warming will cause more greenhouse gases such as water vapour to be carried in the atmosphere, increasing the effect by a factor of two or three."

Whereas deniers usually frame it as:

"Water vapour is a more powerful greenhouse gas than CO2, and nobody's scared of water vapour, so we should just continue to pump out CO2 as much as we want."

[arkie87]

That's pretty much what I've been saying. I'm just trying to spell it out, because two major components of Wedge's argument are:

-The specific heat capacity of humid air is far greater than the specific heat capacity of dry air.

-Specific heat capacity determines the steady-state air temperature.

Neither of these things are strictly true. The presence of liquid water, both as a heat sink in itself, and as a source of a phase change for temperature dampening, are far more important.

Neither of those are true at all! i.e. you dont need the word "strictly"

That's my understanding as well.

The point I was trying to make was that this isn't how deniers usually talk about the effect of water vapour. What we should be saying is:

"CO2 warming on its own could cause problems, but this warming will cause more greenhouse gases such as water vapour to be carried in the atmosphere, increasing the effect by a factor of two or three."

Whereas deniers usually frame it as:

"Water vapour is a more powerful greenhouse gas than CO2, and nobody's scared of water vapour, so we should just continue to pump out CO2 as much as we want."

Well, in their defense, if CO2 is responsible for only 25-33% of the warming, and the water vapor feedback loop doesnt actually exist or is severely damped (due to the presence of negative feedback loops), then the greenhouse effect becomes a lot less scary, at least, in immediate terms (i.e. our lifetime).

Anyway, thanks for being respectful throughout this thread!

[GeneralVeers]

Well, in their defense, if CO2 is responsible for only 25-33% of the warming, and the water vapor feedback loop doesnt actually exist or is severely damped (due to the presence of negative feedback loops), then the greenhouse effect becomes a lot less scary, at least, in immediate terms (i.e. our lifetime).

Oh, it gets better! And not in a sarcasm kind of way, either. The following is genuinely good stuff.

Global warming itself should be a lot less scary to you, and keep in mind, the following proof is not some bit of quackery in a scientific paper reviewed by a gaggle of corrupt self-serving peers. The following is something the Earth has already done:

During the late Paleocene epoch, the Earth was much warmer than today. Around ten degrees warmer--Celsius. For all practical purposes (with one possible exception I have not been able to verify), none of the global-warming-alarmist doomsday scenarios panned out. The whole planet was green (literally) with life--even the poles. The north and south poles were temperate zones; the United States was tropical. Biodiversity, both plant and animal, was through the roof. Deserts and tundra were virtually nonexistent.

My sources: pretty much every single web site that comes up when you bring up the IXquick search engine and search "paleocene". Just don't make the mistake I once did--which was to get "paleocene" and "pleiocene" mixed up.

You have here an example of what actually happens on a warmer Earth. A larger percentage of the Earth's surface is farmable (because deserts and tundra shrink); wherever plants do grow, they grow healthier. What does that mean? First and foremost, more food for poor third-world countries without having to rely on imports from the world's breadbaskets (the foremost being the United States).

That happy sigh is Arkie's worries melting away.

That's all for now. For the time being, I'm going to skip replying to everybody else's posts--which is a good thing, because among other issues, I really need to give one person in here (no, Arkie--not you) a stern talking-to for committing a pretty hideous debate foul.

[Ralathon]

Oh, it gets better! And not in a sarcasm kind of way, either. The following is genuinely good stuff.

Global warming itself should be a lot less scary to you, and keep in mind, the following proof is not some bit of quackery in a scientific paper reviewed by a gaggle of corrupt self-serving peers. The following is something the Earth has already done:

During the late Paleocene epoch, the Earth was much warmer than today. Around ten degrees warmer--Celsius. For all practical purposes (with one possible exception I have not been able to verify), none of the global-warming-alarmist doomsday scenarios panned out. The whole planet was green (literally) with life--even the poles. The north and south poles were temperate zones; the United States was tropical. Biodiversity, both plant and animal, was through the roof. Deserts and tundra were virtually nonexistent.

My sources: pretty much every single web site that comes up when you bring up the IXquick search engine and search "paleocene". Just don't make the mistake I once did--which was to get "paleocene" and "pleiocene" mixed up.

You have here an example of what actually happens on a warmer Earth. A larger percentage of the Earth's surface is farmable (because deserts and tundra shrink); wherever plants do grow, they grow healthier. What does that mean? First and foremost, more food for poor third-world countries without having to rely on imports from the world's breadbaskets (the foremost being the United States).

No, global warming won't turn the earth into some venusian hellhole. But comparing the current increase in the greenhouse effect with the one during the paleocene is disingenuous. The current increase is way faster than any natural effect and thus the biospheres of the earth don't have time to respond. If you wait a couple of millenia the planet would be a lush haven. But on the short term (short term being 50-100 years) it means desertification, flooding, more extreme weather, a minor mass extinction and all around bad stuff.