Energy Flux Units – Climate Puzzles

Climate skeptics are sometimes confused by the way that the intensity of sunlight is reported. The intensity of sunlight in space at Earth’s orbit is reported as about 1361 W/m², while the average intensity of sunlight above Earth’s atmosphere is reported as about 340 W/m². The latter is 1/4 of the former. Why the difference?

It’s essential to understand how an average energy fluxes is defined. An average energy flux is the total rate of energy flow (in Joules/second, i.e., Watts) through a particular reference surface divided by the total area of that surface (in m²).

So, the numerical value of an average flux depends, in general, on area of the reference surface.

There are two distinct reference surfaces relevant to the measurement of sunlight and other climate energy fluxes. These two reference surfaces are illustrated in the figure below.

*Figure 1. Reference surfaces used in defining energy fluxes*

The two types of fluxes are defined relative to different reference surfaces. The reference surface is either

a flat disk perpendicular to the direction of sunlight, or
a spherical (or oblate spheroid) surface parallel to the Earth’s nominal surface.

Both the disk and the sphere have the same radius R, equal to the radius of the Earth. The disk has an area A_d = 𝜋R² = 1.3×10¹⁴ m², while the sphere has an area A_s = 4𝜋R². (It turns out that if you use an oblate speroid to model the shape of the Earth more accurately, the globe has an area A_g = 4.0034⋅𝜋R² = 5.1×10¹⁴ m².)

Conventionally, a flux is expressed in units of W/m², regardless of what reference surface is involved. However, this can lead to unnecessary confusion. So, in this post, I will use W/m²[disk] or W/m²[globe] as the units for fluxes, to distinguish fluxes measured relative to the disk or the globe, respectively.

The total rate at which sunlight arrives at Earth can be expressed as

P = TSI⋅A_d = MSI⋅A_g= 1.7×10¹⁷ W

where TSI ≈ 1361 W/m²[disk] is the “total solar irradiance” measured relative to a plane (or disk) perpendicular to the Sun’s rays, and MSI ≈ 340 W/m²[globe] is the “mean solar irradiance” measured relative to a reference globe surrounding the Earth.¹

Both values, 1361 W/m²[disk] and 340 W/m²[globe], correspond to the same total solar power being incident on the Earth (1.7×10¹⁷ W). They are just measured relative to different reference surfaces.

It’s a bit like measuring volume in liters vs. gallons. The two fluxes are simply different ways of expressing exactly the same information. It’s just two different units, which require a conversion factor to convert one set of units to the other. The conversion factor is

4.0034 W/m²[disk] = 1 W/m²[globe]

Although units of W/m²[disk] are used to express TSI, all other energy fluxes used in climate work are typically expressed in units of W/m²[globe]. Why is that?

Units of W/m²[disk] are only relevant to types of energy that flow in a single direction; in other words, these units are only relevant to sunlight. All other energy fluxes flow to and from Earth’s surface in a broad variety of directions, and do so day and night. Such energy flows often do not flow through the disk used in defining W/m²[disk]. So, such energy flows (e.g., the flow of longwave thermal radiation) cannot be measured in units of W/m²[disk]; they can only be measured in units of W/m²[globe].

That is why units of W/m²[globe] are generally used for reporting energy fluxes in climate studies. It’s a single uniform way of reporting and comparing energy flows.

An Earth’s Energy Budget diagram must be expressed in terms of W/m²[globe] if all entries are to be expressed in consistent units.

Climate skeptics are sometimes upset by the use of units of W/m²[globe] when reporting the average intensity of sunlight. They seem to mistakenly believe this means scientists are assuming sunlight to be uniformly distributed over the surface of the Earth, or assuming that the Sun shines equally day and night. Yet, that is not at all what is happening.

When scientists talk about an average energy flux, that is simply a coded way of talking about a total energy flow. When scientists talk about “radiative balance”, what they really mean is that the total energy arriving at and leaving a planet must be equal, if the planet is to have a stable temperature.

In such calculations involving the balance of total energy, it doesn’t matter if the amount of energy arriving or leaving is different at different locations. There is no assumption that sunlight shines uniformly, or that energy leaves uniformly. Average energy fluxes are just a coded way of talking about the total energy arriving and leaving. And, for such purposes, the location at which energy arrives or leaves doesn’t matter.

So, talking about average energy fluxes, and doing so in units of W/m²[globe], is proper, and involves no assumptions about how uniformly energy is distributed.

I hope this explanation has been at least a little bit helpful in making sense of how the intensity of sunlight can be reported as both 1361 W/m²[disk] and 340 W/m²[globe].

Addendum

One reader wrote to ask “Why is a flat disk needed, when sunlight actually falls on the lit hemisphere of the planet, not on a flat disk?”

The answer is: there is no need for a flat disk―it’s simply a convenient but optional way of simplifying the calculations.

The first point to understand is the Total Solar Irradiance (TSI) is rigorously defined as the power per unit area experienced by an illuminated surface perpendicular to the direction of sunlight. It’s how much energy per unit area that falls on the ground at the equator at noon. Most of the lit hemisphere receives sunlight much less intense than what is suggested by TSI.²

In particular, if $\theta$ is the angle between the direction of sunlight and the local vertical, then the local intensity of sunlight incident on a horizontal surface will be $\mathrm{TSI} \cdot \cos\theta$ . This means the intensity has a peak value of TSI where the sun is directly overhead ( $\theta=0$ ), but goes to zero at the transition between day and night ( $\theta=\pi/2$ radians or 90º).

One can use calculus to calculate the total power $P$ of the sunlight which falls on the lit hemisphere:

$P = \mathrm{TSI} \cdot R^2 \cdot \oint \cos\theta \, \dd\Omega = \mathrm{TSI} \cdot R^2 \cdot 2\pi \int_0^{\pi/2} \cos\theta \sin\theta \, \dd\theta = \mathrm{TSI} \cdot \pi R^2 = \mathrm{TSI} \cdot A_d \,.$

In the above, $R$ is the radius of the planet, $R^2 \cdot \oint X(\theta)\,\dd\Omega = R^2 \cdot 2\pi\int_0^{\pi/2}X(\theta)\sin(\theta)\,\dd\theta$ is the integral of $X$ over the lit hemisphere, and $A_d = \pi R^2$ .

Many people in the general public don’t understand calculus sufficiently to do an integral over the lit hemisphere. Fortunately, there is a simpler way to do the calculation.

The simpler approach involves noticing that the amount of sunlight which lands on the lit hemisphere of the planet is precisely the same as the amount of sunlight which passes through a flat disk of the same radius. Given that fact, it easily follows that $P = \mathrm{TSI} \cdot A_d$ . This is the same answer that was calculated above using calculus. Yet, using the flat disk there was no need to do calculus or any sort of complex calculation in order to find the formula for the total power of sunlight, P.

That’s what I mean when I say that the “flat disk” is simply a convenient but optional way of simplifying the calculation.

1
TSI is technically defined relative to any flat surface perpendicular to the direction of sunlight. A flat disk with the same radius as the planet is simply an example of a surface perpendicular to the direction of sunlight; this example is convenient because it provides a means of calculating the total sunlight reaching the planet.
2
The illumination of the lit hemisphere has an average value of TSI/2, and the illumination of the dark hemisphere is 0, resulting in a global average illumination of MSI = TSI/4. This assume the planet is a perfect sphere; taking a more realistic planetary shape into account yields MSI = TSI/4.0034.

1
TSI is technically defined relative to any flat surface perpendicular to the direction of sunlight. A flat disk with the same radius as the planet is simply an example of a surface perpendicular to the direction of sunlight; this example is convenient because it provides a means of calculating the total sunlight reaching the planet.
2
The illumination of the lit hemisphere has an average value of TSI/2, and the illumination of the dark hemisphere is 0, resulting in a global average illumination of MSI = TSI/4. This assume the planet is a perfect sphere; taking a more realistic planetary shape into account yields MSI = TSI/4.0034.