CO2 v. CH4

Subtitle: Carbon dioxide or methane? Who will win the smackdown of the greenhouse gas molecules?

I was born in a crossfire hurricane
And I howled at the morning drivin’ rain

But it’s all right now, in fact it’s a gas
But it’s all right, I’m jumpin’ jack flash
It’s a gas, gas, gas

  • The Rolling Stones

You’ve probably read that methane traps heat in the Earth’s atmosphere much more effectively than carbon dioxide. An article about these two greenhouse gases will say that over some number of years, methane traps multiple times the heat more than carbon dioxide. But, the number of years and the multiplier seems to change from one story to the next. What’s the right answer?

To understand what’s going on we need to do a bit of atmospheric physics, some chemistry, and a little math.

Atmospheric lifetime

Both molecules trap heat, and methane is a much more powerful greenhouse gas. Methane doesn’t last as long in the atmosphere compared to CO2 because it reacts with oxygen to form carbon dioxide,

CH4 + 2O2 → CO2 + 2H20.

But, both molecules last decades or even centuries in the atmosphere.

If you could somehow keep track of an individual molecule, you might be surprised to find that it disappears within days. This is because molecules are constantly exchanged between the atmosphere, the land, and the oceans according to Dalton’s Law of partial pressure.

As soon as one molecule leaves the atmosphere and enters the ocean, there’s more room for another molecule to pop back out. So there’s a constant exchange of molecules, but the total number doesn’t change very much relative to the number of molecules in the atmosphere.

Some people have misinterpreted this constant flow of gases to mean that the lifetime is much shorter than it actually is, but because there is a nearly even exchange we will ignore partial pressure and only look at the longer effective lifetimes.

How long does CO2 last in the atmosphere? Consulting my Magic 8 Ball, I get


The EPA isn’t much better. The table of lifetime averages and global warming potentials says that carbon dioxide has a 100-year global warming potential of 1, and an average lifetime of - see below * - where they explain that there are many mechanisms for removal from the atmosphere, so the answer is REPLY HAZY. TRY AGAIN LATER. Well, alright, what the footnote says is,

* Carbon dioxide’s lifetime cannot be represented with a single value because the gas is not destroyed over time, but instead moves among different parts of the ocean–atmosphere–land system. Some of the excess carbon dioxide is absorbed quickly (for example, by the ocean surface), but some will remain in the atmosphere for thousands of years, due in part to the very slow process by which carbon is transferred to ocean sediments.

The EPA page, Understanding Global Warming Potentials says, “CO2, by definition, has a GWP of 1 regardless of the time period used, because it is the gas being used as the reference.”

For methane, the average lifetime is 11.8 years, and the global warming potential (GWP) is 27.0-29.8. The GWP for CH4 comes with a double asterisk because it includes methane from both fossil and non-fossil sources.

So, the GWP for CO2 is exactly 1, but we don’t know its lifetime, and the lifetime of CH4 seems to be well known, but the GWP is the ratio of how much heating CH4 causes during its lifetime compared to CO2 which has an unknown lifetime.


CO2, CH4 and O2 in the atmosphere

Since the beginning of the industrial revolution, people have been burning a lot of stuff, mostly oil, gas, and coal. The molecules in fossil fuels are mainly composed of carbon and hydrogen atoms. When we burn these fuels, the chemical reaction is very much like the reaction of methane and oxygen forming CO2 and water, using the balanced equation shown above.

The number of carbon and hydrogen atoms varies in fossil fuels because the molecules aren’t as simple as methane. The atoms in coal or gasoline are chained together to form very long molecules, so the ratio of carbon to hydrogen varies quite a bit.

Fuel Carbon-Hydrogen Ratio

Gasoline has about two hydrogen atoms for every carbon atom, giving the reaction

2CH2 + 3O2 → 2CO2 + 2H20.

The atomic masses for each of these elements are C=12C=12, H=1H=1, and O=16O=16, so the ratio of the mass of CO2 to CH2 is 44 to 14, or about 3.14 to 1. A gallon of gasoline weighs about 6 pounds, which means it generates about 19 pounds of CO2 when burned.

The concentration of carbon dioxide, methane, and nitrous oxide in the atmosphere has been steadily increasing as we continue to burn fossil fuels. Oxygen levels have been correspondingly decreasing as a result of the reactions.


There’s still plenty to breathe, but the drop is quite noticeable in the plot.


Earth’s energy budget

To understand GWP we need to know how greenhouse gases trap heat in the atmosphere, and that means we need to look at the Earth’s energy budget. Imagine there’s a giant disk with a radius exactly that of Earth, and it’s parked just above the top of the atmosphere, blocking every bit of sunlight. There’s nothing between the Sun and the disk, so everywhere on the disk, the solar radiation is the same, ~1365 W/m2.

This energy on the disk is called the solar constant, but it can vary a bit depending on how many sunspots there are. Again, HAZY. The variation is only about 0.1%, so for a simple model, we could use 1365 W/m2.

Let’s call the radius of the Earth ReR_e, which is also the radius of the imaginary disk. The total area of the disk is πRe2\pi R_e^2. The surface area of a sphere is 4πr24 \pi r^2, so the Earth’s surface area is 4πRe24 \pi R_e^2, which is exactly four times the surface area of the disk. On average, the surface of the Earth absorbs 1/41/4 the energy arriving on the disk or 341.25 W/m2. It’s like living on a rotisserie. Every 24 hours we get baked on one side.

Brian Rose teaches climate science at the University of Albany and has published an online textbook, The Climate Laboratory which includes Python Jupyterlab notebooks on GitHub. In the first chapter, there’s a figure showing the Earth’s energy budget.


Before getting into a lot of detail about all of these energy flows, notice at the bottom where it says that the net energy absorbed is 0.9 W/m2. That’s our problem right there. We’ve got more energy coming in than going out, so the Earth is heating up.

In Economics and the Stefan-Boltzmann Law, we showed how to calculate temperature (in degrees Kelvin) from energy,

T=(jσ)(1/4)T = \left( \frac{j^*}{\sigma} \right)^{(1/4)}

where jj^* is energy and σ=5.670374419e8Wm2K4\sigma = 5.670374419e-8 \frac{W}{m^2 \cdot K^4} is the radiation constant. Physicists have been able to measure σ\sigma very precisely, but getting exact values for the Earth’s energy budget is much more difficult which is why you’ll see slightly different values in the figures.

The outgoing longwave radiation (OLR) is 238.5 W/m2 (the upward pointing arrow on the right side of the figure). From this, we can calculate a the emission temperature, which is the average temperature of the Earth measured from a point in space outside the atmosphere,

TOLR=(238.55.670374419e8)1/4=254.66°K=18.49°C.T_{OLR} = \left( \frac{238.5}{5.670374419e-8} \right)^{1/4} = 254.66\degree K = -18.49\degree C.

This would be true, except that the atmosphere blocks some outgoing radiation, so the equation needs to be modified to include a transmissibility factor, τ\tau,

j=τσT4.j* = \tau \sigma T^4.

The actual measured temperature is T=288°KT = 288\degree K.

The single-layer model of the atmosphere

A very basic model of the atmosphere can help explain the Earth’s energy balance. This model treats the atmosphere as if each square meter from the ground to the top of the atmosphere acts independently, and there’s no circulation between adjacent columns and no convection within the column.

Sunlight enters the top of the column, some is reflected away, and the remainder reaches the ground. The ground reradiates energy in the form of infrared, or longwave, light. The atmosphere is treated as a single layer that absorbs the infrared (IR) and reradiates it in all directions. Half of the IR returns to the Earth, and the other half escapes to space.

For the model to make sense, the total energy in must equal the total energy out.


Using the solar constant of 1365 W/m2 divided by 4 gives S=341.25S = 341.25 W/m2. The Earth’s albedo α=0.3\alpha = 0.3, meaning that about 30% of the incoming sunlight is reflected away by clouds, ice, the ground, and oceans.

To balance the energy equation, the total energy arriving at the surface must be equal to the total outgoing infrared,

(1α)S+σTa4=σTe4.(1-\alpha)S + \sigma T_a^4 = \sigma T_e^4.

The infrared energy must also balance, and since half of the IR leaving the surface is reflected back to the surface by the atmosphere,

σTa4=12σTe4.\sigma T_a^4 = \frac{1}{2} \sigma T_e^4.

Factoring out σ\sigma from both sides and taking 4th roots, Te=1.19TaT_e = 1.19 T_a.

There’s one other factor that needs to be included in this simple model, the emissivity ϵ\epsilon of the atmosphere. The emissivity of a material is a measure of how well the material emits heat by radiation. A perfect emitter is called a blackbody, so emissivity is defined as the ratio of the energy radiated from the material to the energy radiated from a blackbody at the same temperature.

Including emissivity in the equations adds a second outgoing pathway from the top of the atmosphere because the fraction of the outgoing IR from the Earth’s surface that isn’t absorbed passes directly through.


Including the emissivity term, the energy balance equation for the Earth’s surface is

(1α)S=(1ϵ)σTe4+ϵσTa4(1-\alpha)S = (1-\epsilon) \sigma T_e^4 + \epsilon \sigma T_a^4

and the balance for the atmosphere becomes

ϵσTa4=12ϵσTe4.\epsilon \sigma T_a^4 = \frac{1}{2} \epsilon \sigma T_e^4.

Substituting Ta4=2Te4T_a^4 = 2 T_e^4 into the first equation gives

(1α)S=(1ϵ)σTe4+12ϵσTe4=(1ϵ2)σTe4.\begin{aligned} (1-\alpha)S &= (1-\epsilon) \sigma T_e^4 + \frac{1}{2}\epsilon \sigma T_e^4 \\ &= \left( 1 - \frac{\epsilon}{2} \right) \sigma T_e^4. \end{aligned}

Solving for the surface temperature TeT_e,

Te=((1α)S(1ϵ2)σ)1/4.T_e = \left( \frac{(1-\alpha)S}{(1-\frac{\epsilon}{2}) \sigma} \right)^{1/4}.

The atmospheric emissivity is approximately, ϵ=0.78\epsilon = 0.78.

The single-layer model in SMath Studio

SMath Studio is a free program where you can write equations almost as you would on a piece of paper. The program keeps track of, and knows, many physical constants and standard units. By associating variables with their correct units, you can be sure that your final answer is in the correct units as well.

In the equation for TeT_e, the albedo α\alpha and atmospheric emissivity ϵ\epsilon are dimensionless, meaning they don’t have any units, but the solar constant SS is in Wm2\frac{W}{m^2}, and the radiation constant σ\sigma has units Wm2K4\frac{W}{m^2K^4}. If you set up the equations correctly in SMath Studio, TeT_e will show a temperature in degrees Kelvin, K°.K\degree.

You can plot the Earth’s temperature as a function of atmospheric emissivity and see that a small change in emissivity results in a large temperature change. Adding more greenhouse gases to the atmosphere does exactly that.


In this worksheet, I defined the temperature function T(ϵ,α,σ,S)T(\epsilon,\alpha,\sigma,S) using the equation derived above, and then created a new function f(ϵ)f(\epsilon) that only depends on the atmospheric emissivity.

I did that to plot ϵ\epsilon against T,T, and notice that I subtracted 288.2752 so that the temperature crosses the xx-axis when ϵ=0.78\epsilon =0.78. If you want to try it out yourself, you can download the worksheet from GitHub.

For help with SMath Studio, look at the wiki, the forum, or “Of Sailing Ships, Velociraptors, and Walking on the Moon”.

A search for answers

Someone must have done the research to estimate the global warming potential of methane. It seems that if we knew the GWP of a single molecule of methane compared to a single molecule of carbon dioxide, and how long each lasts in the atmosphere, we could calculate the GWP for any time, and by integrating over a period, we’d get the relative impact.

To find the answer, using some of the new AI-powered search engines might help us find relevant papers. I’ll ask the same question, “What is the global warming potential of methane as a function of atmospheric lifetime and density?” to each search engine. If you’ve installed the bookmarks described in The Academic Browser you might want to add some of these to a new folder called AI Search.

Here are links to each along with the answers returned:

From this, Elicit provided direct links to papers and gave good summaries, and Consensus gave a reasonable response along with links to journal articles. More traditional sources such as the Wikipedia page on Atmospheric Methane and a simple Google search are also useful.

In the end, I put the question into Google image search where I found this:


in the paper Methane emissions: choosing the right climate metric and time horizon.

GWP, GTP, and all that

The paper says that global warming potential has been criticized for several reasons,

Other researchers have proposed alternatives, some of which include impacts on policies or technologies. Global temperature potential (GTP) seems to be the best alternative metric that avoids the problems with GWP. GTP is the change in surface temperature at any time after a release of methane, relative to the effect of an equivalent release of CO2.

GTP is an end-point measurement, meaning that it gives the expected change in temperature at a given time after the methane has been released into the atmosphere. GWP measures the change in the radiation balance, which while useful, isn’t quite as important as the change in Earth’s temperature. But, to estimate the temperature change, GTP needs to take into consideration the exchange of heat between the atmosphere and the oceans. This introduces larger uncertainties in the model.

This plot of the relative forcing of a pulse of methane and the resulting change in temperature shows the lasting impacts of methane in the atmosphere. Notice the lag between the time of release and the peak temperature change. By the time the radiative forcing has fallen to 50% of the peak, the change in temperature forcing is delayed by 20 years because heat is exchanged between the atmosphere and the oceans. The methane in the atmosphere now will continue to increase global warming for decades.


Hopefully, this clears up a little bit of the confusion surrounding the impacts of methane in the atmosphere. Of course, most of the trapped heat goes into the oceans. To learn more about ocean heat listen to Alex Smith’s interview with Kevin Trenberth on his Radio Ecoshock show in an episode called, “How Ocean Heat Ends The Human Experiment”.