Fundamental Radiometric Quantities

Some important notations and defintion:

  1. Radiation flux: Radiant flux (or radiant power) \(\Phi\) is the total amount of electromagnetic energy (radiation) emitted, transmitted, or received per unit time, measured in Watts It measures the total power from a source across all wavelengths, often used for optical, UV, or IR radiation.

    Total energy passing through the element of area \(dA\) for time \(dt\) is given by: \(\Phi dA dt\).

    So, flux is a measure of the enrgy caried by all rays passing through a given area

  2. Isotropic source: A source of radition is called isotropic, if it emits energy equally in all directions, i.e. $$\Phi(r_1)(4\pi r_1^2) dt = \Phi(r)(4\pi r^2) dt \Rightarrow \Phi(r) \propto \frac{1}{r^2} \Rightarrow \text{statement of conservation of energy}.$$
  3. The specific (radiative) intensity is a quantity that describes the rate of radtion trnasfer of energy through a given area (perpendicular in a unit time and in a given frequency band): $$dE = I_\nu dA dt d\Omega d\nu $$

    4. For more details, check my notes (chapter-4) at following link.

Electromagnetic Radiation – Physical Foundation

Radiative transfer begins with classical electromagnetism. Electromagnetic radiation satisfies Maxwell’s equations. In free space: $$ \nabla^2 \boldsymbol{E} = - \frac{1}{c^2} \frac{\partial^2 \boldsymbol{E}}{\partial t^2} $$ The plane wave solution of this equation can be written as: $$\boldsymbol{E}(z,t) = \boldsymbol{E}_0 e^{i(kz-\omega t)}$$ where,
  • \(k = \frac{2\pi}{\lambda}\)
  • \(\omega = 2\pi \nu\)
  • \(c=\lambda \nu\)
We can write energy density of EM wave: $$u = \frac{1}{2} \varepsilon E^2 + \frac{1}{2\mu_0} B^2$$ For plane waves: \(u = \varepsilon_0 E^2\) and the energy flud i.e. Pointing vector can be written as: \(\boldsymbol{S} = \boldsymbol{E}\times \boldsymbol{H}\). Magneitude of the energy flux is given by \(S = c u \). This gives physical basis for radiative flux.

Radiometric Quantities

Satellite instruments do not measure field amplitudes, instead they measure radiance.

Spectal Radiance

Spectral radiance is the amount of radiant energy traveling in a specific direction, per unit area, per unit solid angle, and per unit wavelength (or frequency). $$ I_\nu = \frac{d\Phi}{dA ~\text{cos}~ \theta ~d\Omega ~d\nu} = \frac{dE}{dA ~\text{cos}~ \theta ~d\Omega ~d\nu ~dt} ~~~~ \text{W}\text{m}^{-2}\text{sr}^{-1}H z^{-1}. $$ where Radiant flux \(\Phi\) in terms of energy \(E\) is defined as: \(d\Phi = \frac{dE}{dt}\) Energy per unit area, per unit solid angle, per unit frequency, per unit time. Radiance is conserved along a ray in vacuum.

Relation Between Radiance and Flux: The flux is defined as:

$$\Phi_\nu = \int I_\nu \text{cos}\theta d\Omega$$ This distinction is important because radiative transfer equation governs radiance, not flux.

Interaction of Radiation with Matter

When radiation enters atmosphere, intensity changes due to:
  1. Absorption
  2. Scattering
  3. Emission
We define coefficients per unit length.

Absorption Coefficient: Let \(\alpha_\nu\) is the absorption coefficient, then:

$$dI_\nu = -\alpha_\nu I_\nu ds ~~~~ \text{m}^{-1}. $$ Where microscopically \(\alpha_nu = n \sigma_nu\). Here \(n\) is the number density and \(\sigma_nu\) is the absorption cross-section.

Scattering Coefficient: It is denoted by \(\beta_\nu^{\rm sca}\).

Extinction does not mean energy disappears. It means energy is removed from the specific direction of propagation. $$\text{Extinction} = \text{Absorption} + \text{Scattering}$$ In radiative transfer, the total extinction coefficient is the sum of: $$ \beta_\nu^{\rm ext} \equiv \beta_\nu= \alpha_\nu + \beta_\nu^{\rm sca} $$ Here, \(\alpha_\nu\) is the total absorption coefficient and \(\beta_\nu\) i.e \(\beta_\nu^{\rm ext}\) is the total scattering coefficient. So this is process based decompositon.

Emission Coefficient: Energy emitted per unit volume per unit solid angle per unit frequency is denoted by \(j_\nu\).

Total extinction coefficient: $$\beta_\nu = \beta_\nu^{\rm gas} + \beta_\nu^{\rm aer} + \beta_\nu^{\rm cloud}$$ This is a species-based decomposition. It separates extinction by physical contributor:
  • Gas molecules (CO₂, O₂, H₂O, Rayleigh)
  • Aerosols
  • Clouds
Each of these contributors themselves contains absorption and scattering. So more explicitly: $$ \begin{eqnarray} \beta_\nu^{\rm gas} & = & \alpha_\nu^{\rm gas} + \beta_\nu^{\rm sca,sca} \\ \beta_\nu^{\rm aer} & = & \alpha_\nu^{\rm aer} + \beta_\nu^{\rm sca,aer} \\ \beta_\nu^{\rm cloud} & = & \alpha_\nu^{\rm cloud} + \beta_\nu^{\rm sca,cloud} \end{eqnarray} $$

If we combine both decompositions:

$$\beta_\nu^{\rm ext} = \sum_i \left(\alpha_{\nu,i} + \beta_{\nu,i}^{\rm sca}\right)$$

where \(i=\) gas, aerosol, cloud etc.

For CO₂ Retrieval:

  • In clear sky: \(\beta_\nu \approx \beta_\nu^{\rm gas} \) and hence: $$\beta_\nu^{\rm gas} = \alpha_\nu^{{\rm CO}_2} +\alpha_\nu^{{\rm H}_2{\rm O}} + \beta_\nu^{{\rm Rayleigh}}$$
  • In aerosol conditions: \(\beta_\nu^{\rm aer} \) adds significant scattering, increasing photon path length.
  • In cloudy scenes: \(\beta_\nu^{\rm cloud} \) dominates and retrieval often fails.

Derivation of the Radiative Transfer Equation (RTE)

Consider a beam traveling through medium. Energy balance over infinitesimal path \(ds\): $$ \text{Change in radiance} =\text{Loss due to extinction} + \text{Gain due to emission}. $$ Mathematically: $$d I_\nu = -\beta_\nu I_\nu ds + j_\nu ds ~~~~ \Rightarrow ~~~~ \frac{dI_\nu}{ds} = -\beta_\nu I_\nu + j_\nu$$

Special Case: Reflected Solar Radiation (SWIR):

In CO₂ retrieval missions:

  • Thermal emission negligible
  • Dominant process: absorption + scattering
For pure absorption case: $$\frac{dI_\nu}{ds} = -\beta_\nu I_\nu \Rightarrow \frac{dI_\nu}{I_\nu} = -\beta_\nu ds $$ Now integrating: $$\int_{I_0}^{I} \frac{dI'}{I'} = -\int_0^L \beta_\nu ds \Rightarrow \ln\left(\frac{I}{I_0}\right) = -\tau_\nu$$ where $$\tau_\nu = \int_0^L \beta_\nu ds$$ Therefore: $$ I_\nu = I_{\nu,0} e^{-\tau_\nu} $$ This is Beer–Lambert law.
This is the scalar Radiative Transfer Equation. This equation be re-written in terms of optical depth \(\tau_\nu\) as: $$\frac{dI_\nu}{d\tau_\nu} = - I_\nu +S_\nu, $$ where \(d\tau_\nu = \beta_\nu ds\) and \(S_\nu = \frac{j_\nu}{\beta_\nu}\). Therfore, $$\tau_\nu = \int_0^L \beta_\nu ds. $$

Full Equation with Explicit Terms:

In general above RTE doesnot explain actual scenario. For example, in the case when radiation coming from other directions into the line of sight, we need to add few terms on the right side. $$ \frac{dI_\nu}{ds} = -\alpha_\nu I_\nu -\beta_\nu^{\rm sca} I_\nu + j_\nu + \int_{4\pi} \beta_\nu^{\rm sca} P(\Omega' \Omega) I_\nu(\Omega') d\Omega' $$ where \(P\) is the phase function, and is integral over all directions. This function describe, how radiation coming from direction \(\Omega'\) is redistributed into direction \(\Omega\) after scattering. In simple words: It tells you where the photon goes after it scatters.

  • Normalization Condition: The phase function is normalized such that: $$\frac{1}{4\pi}\int_{4\pi} P(\Omega' \Omega) d\Omega = 1 $$ This ensures energy conservation during scattering.
  • Common Types of Phase Functions:
    • Isotropic Scattering: \(P= 1 \Rightarrow~~~~\) Scattering equally in all directions. Simple but rarely realistic.
    • Rayleigh Scattering (Molecules): Small particles relative to wavelength. $$P(\theta) = \frac{3}{4}(1+\text{cos}^2\theta)$$ where \(\theta\) is the scattering angle. Symmetric forward/backward Important for molecular scattering. Relevant in clear-sky CO₂ retrieval
    • Henyey–Greenstein Phase Function (Aerosols): Very commonly used: $$P(\theta) =\frac{1-g^2}{(1+g^2-2g\text{cos}\theta)^{3/2}}$$ where \(g=\langle \text{cos}\theta \rangle\) is the asymmetry parameter, and
      • \( g=0 \rightarrow \) isotropic
      • \( g>1 \rightarrow \) forward scattering
      • \( g<1 \rightarrow \) backward scattering

      For aerosols: \(g\approx 0.6 - 0.8\)

      For clouds: \(g\approx 0.85 - 0.95\)

Why Phase Function Matters for CO₂ Retrieval:

In shortwave CO₂ bands:
  • Surface-reflected sunlight dominates
  • Aerosol scattering changes photon path length
  • Forward scattering increases effective path length
  • That biases retrieved XCO₂

If the phase function is wrong:

  • → wrong angular redistribution
  • → wrong radiance simulation
  • → retrieval bias

Above equation can be written in more compact form as:

$$\frac{dI_\nu}{ds} = -\beta_\nu^{\rm ext} I_\nu + \beta_\nu^{\rm ext} S_\nu$$ where the source function is: $$S_\nu = \frac{\alpha_\nu B_\nu +\beta_\nu J_\nu}{ \beta_\nu^{\rm ext}}$$ and $$J_\nu = \frac{1}{4\pi} \int I_\nu d\Omega$$

Special Case: Pure Absorption (Important for CO₂ Retrieval): If scattering is negligible

$$\beta_\nu^{\rm ext} \approx \alpha_\nu$$ Then RTE simplifies to: $$\frac{dI_\nu}{ds} = -\alpha_\nu I_\nu + \alpha_\nu B_\nu$$ This leads directly to the classical solution: $$I_\nu(\tau) = I_\nu(0)e^{-\tau} + \int_0^\tau B_\nu e^{-(\tau - t)} dt.$$ where \(d\tau_\nu = \alpha_\nu ds\) and \(B_\nu\) is the Planck function. It represents the spectral radiance emitted by a perfect blackbody at temperature \(T\) and it is defined as: $$B_\nu(T) = \frac{2h\nu^3}{c^2}\frac{1}{e^{\frac{h\nu}{kT}}-1}$$ In above equation, \(h\) is the Planck constant, \(\nu\) = frequency, \(c\) = speed of light, \(k\) = Boltzmann constant and \(T\) = temperature. Unit of \(B_\nu\) is Wm\(^{-2}\)sr\(^{-1}\)Hz\(^{-1}\).

Why This Is Critical for CO₂ Retrieval?

In near-infrared CO₂ bands:
  • Extinction mostly from absorption
  • Optical depth depends on CO₂ concentration
  • Small errors in extinction → errors in retrieved XCO₂

Scattering (aerosols, thin clouds) changes photon path length, which biases retrievals if not modeled correctly.

General Solution of RTE

Linear first-order ODE, after ignoring the term represenitng contribution from other direction into the line of sight; $$\frac{dI}{d\tau} + I = S$$ Multiply by integrating factor: \(e^\tau\) $$\frac{dI e^\tau}{d\tau} = Se^\tau$$ and integrate: $$I(\tau) = I(0) e^{-\tau} +\int_0^\tau S(t) e^{-(\tau - t)} dt$$ This is the general solution.

Folloiwng cases can be considered from this equation:

  1. Caste-1: If \(S_\nu = \text{constant}\) $$ I_\nu(\tau_\nu) = I_\nu(0) {\rm e}^{-\tau_\nu} +S_\nu\int_0^{\tau_\nu} {\rm e}^{-(\tau_\nu-\tau_\nu')} d\tau_\nu' \\ I_\nu(\tau_\nu) = I_\nu(0) {\rm e}^{-\tau_\nu} +S_\nu \left(1-{\rm e}^{-\tau_\nu}\right) \\ = I_\nu(0) {\rm e}^{-\tau_\nu} +S_\nu-S_\nu {\rm e}^{-\tau_\nu} \\ I_\nu(\tau_\nu) = S_\nu +\left(I_\nu(0)-S_\nu\right) {\rm e}^{-\tau_\nu}. $$
  2. Caste-2: For optically opaque medium, \(\tau_\nu\rightarrow \infty\), $$I_\nu (\tau_\nu\rightarrow \infty) =S_\nu$$
  3. Caste-3: For optically transparent medium \(\tau_\nu \rightarrow 0\) $$I_\nu(0) S_\nu +\left(I_\nu(0)-S_\nu\right) \Rightarrow 0= 0.$$
  4. Caste-4: For \(\tau_\nu\ll 1\): $$ I_\nu(\tau_\nu) = S_\nu +\left(I_\nu(0)-S_\nu\right)(1-\tau_\nu) \\ I_\nu(\tau_\nu) -I_\nu(0) = \left(S_\nu-I_\nu(0)\right)\tau_\nu $$
    • If \(I_\nu(\tau_\nu) < I_\nu(0) \Rightarrow\) Absorption,
    • If \(I_\nu(\tau_\nu) > I_\nu(0) \Rightarrow\) Emission.
  5. Case-5: If:
    • When \(I_\nu>S_\nu \Rightarrow \frac{dI_\nu}{d\tau_\nu} < 0 \Rightarrow\) \(I_\nu\) tends to decrease along the ray.
    • When \(I_\nu < S_\nu \Rightarrow \frac{dI_\nu}{d\tau_\nu} > 0 \Rightarrow\) \(I_\nu\) tends to increase along the ray.

    6. Thus the source function is the quantity that the specific intensity tries to approach. In this respect the transfer equation describes a "relaxation" process.

Optical Depth Decomposition

Thus: $$\tau_\nu = \tau_\nu^{\rm gas} + \tau_\nu^{\rm aer} + \tau_\nu^{\rm cloud}$$ We can derive the formula for the gas optical depth. Probability of photon absorption per unit path: $$dP = n(z) \sigma_\nu dz$$ Now integrating: $$\tau_\nu^{\rm gas} = \int_0^\infty \sigma_\nu (T,p) n(z) dz.$$ Using ideal gas law: $$n(z) = \frac{p(z)}{k_B T(z)}$$ Therefore: $$\tau_\nu^{\rm gas} = \int_0^\infty \sigma_\nu (T,p) \frac{p(z)}{k_B T(z)} dz.$$

Path Geometry in Satellite Observation

For slant path: $$\tau_\nu = \int \sigma_\nu n(s)ds$$ If zenith angle \(\theta\): $$ds = \frac{dz}{\text{cos}\theta}$$ so: $$\tau_\nu = \frac{1}{\text{cos}\theta} \int \sigma_\nu n(z) dz.$$ Thus solar zenith angle increases optical depth.

Final Fundamental Equation for SWIR CO₂ Retrieval

Measured radiance: $$I_\nu = I_{\nu,0} \text{exp}\left(-\int \sigma_\nu^{{\rm CO}_2} n_{{\rm CO}_2} ds - \tau_\nu^{\rm aer} - \tau_\nu^{\rm cloud}\right)$$ This is the complete physical foundation.

Reference

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