Chapter 1

Fundamentals of Radiative Transfer

These notes build the first radiative-transfer quantities from a simple thought experiment. The goal is to keep the discussion close to the physical picture: a bundle of light, a shutter, a collector, and the amount of energy that reaches the collector.

Scope of this part: descriptions of light, the definition of specific intensity, radiation energy density, radiation flux, radiation pressure, and the first qualitative statement about interaction with matter.

Chapter Contents

A short map of the chapter before the full derivations begin.

  1. 1Descriptions of Light
  2. 2Quantities for Radiative Transfer
  3. 2.1Specific Intensity
  4. 2.2Radiation Energy Density
  5. 2.3Radiation Flux
  6. 2.4Radiation Pressure
  7. 2.5Frequency-Integrated Intensity
  8. 3Interaction with Matter
  9. Ref.References

1. Descriptions of Light

Before defining any radiative-transfer quantity, we first decide how we are describing light. The same physical light can be discussed in more than one language.

1. Photon description

Light can be counted as photons. This is useful when atoms absorb or emit individual packets of energy.

2. Electromagnetic wave description

Light is also an electromagnetic wave. This is the language of electric and magnetic fields.

3. Macroscopic ray description

When we look at a beam or a bundle of light, we can follow rays. This is the geometric-optics limit, and this is the language used here.

2. Quantities for Radiative Transfer

The central idea is to start from one carefully measured bundle of light, define its specific intensity, and then build the usual radiation quantities from that definition.

2.1 Specific Intensity

To define specific intensity, we begin with a controlled thought experiment. Place some emitting material on the left, a small shutter in front of it, and a collector on the right. The region between the shutter and the collector is taken to be pure vacuum.

Now open the shutter only for a small time interval Δt. Let the area of the opening be A1. Also arrange the experiment so that only a narrow interval of frequency, Δν, is allowed through.

The collector has area A2 and is placed a distance r away. During this time interval, and only in this small frequency band, suppose the collector receives an amount of energy ΔE.

The collected energy is not arbitrary. It depends on the factors we deliberately controlled in the experiment:

If the shutter is open for longer, more light is collected, so ΔE is proportional to Δt.

If the allowed frequency interval is wider, more radiation is included, so ΔE is proportional to Δν.

If the shutter area is larger, a wider bundle passes through, so ΔE is proportional to A1.

If the collector area is larger, it catches more of the bundle, so ΔE is proportional to A2.

If the collector is moved farther away, the same collector occupies a smaller angular size as seen from the shutter, so the collected energy decreases as 1/r2.

Putting these dependences together gives

ΔE ∝ Δt Δν A1 A2 / r2.

The ratio A2/r2 is the solid angle subtended by the collector at the shutter, for a small nearly perpendicular collector:

ΔΩ = A2 / r2.

So the same proportionality can be written in the cleaner directional form

ΔE ∝ Δt Δν A1 ΔΩ.

The proportionality constant is the quantity we want to define. It is called the specific intensity, denoted by Iν.

Drawn shutter experiment diagram An emitting material sends a narrow beam through a small shutter of area A1. A collector of area A2 receives the beam after a vacuum distance r. Thought experiment material emits light shutter area A1 r collector area A2 ΔΩ pure vacuum: no interaction with matter
The shutter setup defines the measured bundle: area A1, time Δt, bandwidth Δν, and solid angle ΔΩ.
Solid angle decreases with distance A fixed collector covers a larger angle when nearby and a smaller angle when farther away. near farther away larger solid angle smaller solid angle source Same area, smaller angle when it is far away
A detector of fixed area subtends a smaller solid angle when it is moved farther away.

We therefore write the collected energy as

ΔE = Iν Δt Δν A1 ΔΩ.

Solving this relation for Iν gives the definition of specific intensity:

Iν = ΔE / (Δt Δν A1 ΔΩ).
Definition. Specific intensity is the amount of radiation energy collected per unit time, per unit frequency interval, per unit area perpendicular to the beam, and per unit solid angle.
Quantity Meaning
Δt How long the shutter is open.
Δν The small frequency interval that is allowed through.
A1 The area of the shutter or beam cross-section.
ΔΩ The small solid angle accepted by the collector.
Iν Specific intensity, the brightness per time, area, frequency, and solid angle.
[Iν] = energy / (time frequency area solid angle)
cgs: erg s-1 Hz-1 cm-2 sr-1

2.1.1 Constancy in Vacuum

If the bundle does not interact with matter, and if the frequency is fixed, then the specific intensity stays constant along the ray.

A distant collector may receive less total energy because it covers a smaller solid angle, but that is a geometrical effect. It is not a loss of specific intensity along the ray.

No interaction with matter, fixed frequency: Iν remains constant along the ray.
Specific intensity constant along a ray A ray passes through two imaginary surfaces. In vacuum, the same specific intensity labels the ray at both surfaces. surface 1 surface 2 Iν same Iν no absorption no emission or scattering
In vacuum, the ray carries its specific intensity unchanged when the frequency is fixed.

2.2 Radiation Energy Density

Energy density means the amount of radiation energy contained in a unit volume. To connect this with specific intensity, imagine the same small bundle of light crossing the area A1.

During the time interval Δt, light travels a distance cΔt. Therefore the piece of radiation that crosses the area A1 occupies a small cylindrical volume:

ΔV = A1 c Δt.

Now use the expression already obtained for the energy in the bundle:

ΔE = Iν Δt Δν A1 ΔΩ.

The energy density in that small frequency interval and small solid angle is energy divided by volume:

ΔE / ΔV = (Iν Δt Δν A1 ΔΩ) / (A1 c Δt)
ΔE / ΔV = (Iν / c) Δν ΔΩ.

This is the contribution to the energy density from one small frequency interval and one small solid angle. To get the radiation energy density per unit frequency, we add all directions:

\[ E_\nu = \int_{\Omega} {I_\nu\over c}\,d\Omega \]

Finally, to include all frequencies, integrate over frequency:

\[ E = \int_0^\infty E_\nu\,d\nu = \int_0^\infty\int_{\Omega}{I_\nu\over c}\,d\Omega\,d\nu \]

Some books write the radiation energy density as uν instead of Eν. The meaning here is the same: energy per unit volume per unit frequency.

Cylinder volume for radiation energy density A beam crosses area A1 during time Δt. Since light travels cΔt, the occupied volume is A1 cΔt. Energy density: light inside a small cylinder A1 cylinder length = cΔt ΔV = A1 cΔt light volume
During time Δt, the radiation crossing area A1 occupies a cylinder of length cΔt.

2.3 Radiation Flux

Energy density tells us how much radiation energy is present in a volume. Flux asks a different question: how much of that radiation energy actually flows through a surface, and in which direction?

For radiation travelling in the direction \(\hat{n}\), the small contribution to energy density is multiplied by c to become an energy flow rate. This is why flux has the physical meaning of an energy density being transported at the speed of light.

\[ \text{energy density contribution} = {I_\nu \over c}\,d\nu\,d\Omega \]
\[ \text{proper flux contribution} = \left({I_\nu \over c}\,d\nu\,d\Omega\right)c\hat{n} \]
\[ \text{proper flux contribution} = I_\nu \hat{n}\,d\nu\,d\Omega \]

Adding all directions and all frequencies gives the radiation flux vector:

\[ \vec{F} = \int_{\Omega}\int_{0}^{\infty} I_\nu \hat{n}\,d\nu\,d\Omega \]

If the intensity is already integrated over frequency, \(I=\int I_\nu\,d\nu\), then

\[ \vec{F} = \int_{\Omega} I\hat{n}\,d\Omega \]

At one particular frequency, the monochromatic flux is

\[ \vec{F}_\nu = \int_{\Omega} I_\nu \hat{n}\,d\Omega \]
Energy density transported into flux Radiation energy density moves with speed c along direction n-hat to produce a vector flux contribution. small surface element normal n θ energy density × c gives directional flux Only the directional flow through the surface counts as flux.
A small piece of radiation energy density becomes a flux contribution when it is transported at speed c along the ray direction.

2.3.1 Components of the Flux Vector

To calculate the vector flux, describe the ray direction \(\hat{n}\) using spherical angles. The polar angle is θ, measured from the z-axis, and the azimuthal angle is φ, measured around the z-axis.

The solid angle element is

\[ d\Omega = \sin\theta\,d\theta\,d\phi \]

The direction vector can be decomposed as

\[ \hat{n} = n_x\hat{x} + n_y\hat{y} + n_z\hat{z} \]
\[ n_x = \sin\theta\cos\phi \]
\[ n_y = \sin\theta\sin\phi \]
\[ n_z = \cos\theta \]

Therefore the flux vector can be written in components:

\[ \vec{F} = F_x\hat{x} + F_y\hat{y} + F_z\hat{z} \]
Spherical angles for ray direction A ray direction n-hat is described by polar angle theta and azimuthal angle phi in a three-dimensional coordinate system. x z y n θ φ dΩ = sinθ dθ dφ
The ray direction has components sinθ cosφ, sinθ sinφ, and cosθ along x, y, and z.

2.3.2 Radiation Symmetric Around One Axis

Now consider the important case from the notes: the radiation field depends on θ but not on φ. In notation,

\[ I(\theta,\phi)=I(\theta) \]

This does not yet mean the radiation is completely isotropic. It only means that if we rotate around the z-axis, the intensity looks the same. In that situation the sideways components cancel by symmetry.

\[ F_x=0,\qquad F_y=0,\qquad F_z\neq 0\ \text{in general}. \]
\[ F_y = \int_{0}^{\pi}\int_{0}^{2\pi} I(\theta)\sin\theta\sin\phi\sin\theta\,d\theta\,d\phi \]
\[ F_y \propto \int_{0}^{2\pi}\sin\phi\,d\phi = 0 \]

The same cancellation happens for Fx. The only possible remaining component is along the symmetry axis:

\[ F_z = \int_{\Omega} I(\theta)n_z\,d\Omega \]
\[ F_z = \int_{0}^{\pi}\int_{0}^{2\pi} I(\theta)\cos\theta\sin\theta\,d\theta\,d\phi \]
\[ F_z = 2\pi\int_{0}^{\pi} I(\theta)\cos\theta\sin\theta\,d\theta \]

It is common in astrophysics to define

\[ \mu=\cos\theta,\qquad d\mu=-\sin\theta\,d\theta \]

When θ = 0, μ = 1. When θ = π, μ = -1. Changing variables gives the standard flux expression:

\[ F_z = 2\pi\int_{-1}^{1} I(\mu)\mu\,d\mu \]
Azimuthal symmetry cancels sideways flux Rays arranged symmetrically around the z-axis cancel their x and y components but can leave a z component. z same θ, different φ sideways parts cancel If I(θ,φ) = I(θ) then rotations in φ do not change the intensity. F_x = 0, F_y = 0 F_z can remain non-zero
Azimuthal symmetry means equal sideways contributions exist in opposite directions. They cancel in x and y, but the z components can add.

2.3.3 Completely Isotropic Radiation

If the intensity is the same in every direction, then

\[ I(\theta,\phi)=I \]

This is the fully isotropic case. There is as much radiation moving upward as downward, and as much moving left as right, so the net flux must vanish. The integral shows the same thing:

\[ F_z = 2\pi\int_{-1}^{1} I\mu\,d\mu \]
\[ F_z = 2\pi I\int_{-1}^{1}\mu\,d\mu = 0 \]
\[ \text{For isotropic radiation:}\qquad \vec{F}=0 \]

This is a useful intuition: a room filled equally with light from all directions can have large energy density, but zero net flux because there is no preferred direction of flow.

For isotropic radiation, the energy density becomes especially simple:

\[ E = {1\over c}\int_{\Omega}I\,d\Omega \]
\[ E = {I\over c}4\pi \]
\[ E = {4\pi I\over c} \]

2.3.4 Mean Intensity

The mean intensity is the angular average of intensity over all directions:

\[ J = {\int_{\Omega} I(\theta,\phi)\,d\Omega \over \int_{\Omega}d\Omega} \]

For isotropic radiation, the intensity is independent of angle, so

\[ J = {\int_{\Omega} I\,d\Omega \over \int_{\Omega}d\Omega} \]
\[ J = I\,{\int_{\Omega}d\Omega \over \int_{\Omega}d\Omega}=I \]

2.3.5 Example: Flux at the Surface of a Star

Now take the example from the notes. Imagine standing on the surface of a star. Radiation is leaving the star from the inside toward the outside. At the surface, assume there is no incoming radiation from outside.

For outward directions, 0 ≤ θ ≤ 90°, so 0 ≤ μ ≤ 1. Let this outgoing intensity be I+.

For inward directions, 90° < θ ≤ 180°, so -1 ≤ μ < 0. At the stellar surface we take I- = 0, because there is no incoming radiation in this simplified picture.

\[ F_z = 2\pi\int_{-1}^{1} I(\mu)\mu\,d\mu \]
\[ F_z = 2\pi\left[\int_{0}^{1} I^+\mu\,d\mu + \int_{-1}^{0} I^-\mu\,d\mu\right] \]
\[ I^- = 0 \]
\[ F_z = 2\pi\int_{0}^{1} I^+\mu\,d\mu \]

If the outgoing intensity I+ is constant over the outward hemisphere, then

\[ F_z = \pi I^+ \]
Outgoing radiation at a stellar surface At the surface of a star, outgoing rays occupy the outward hemisphere and incoming rays are assumed absent. stellar surface n_z θ I+ I- = 0 Only the outward hemisphere contributes. 0 ≤ μ ≤ 1 gives outgoing radiation.
At the stellar surface, outward rays contribute positive flux. With no incoming radiation, the inward hemisphere has I- = 0.

2.4 Radiation Pressure

Radiation pressure is analogous to gas pressure. In a gas, molecules hit a surface and transfer momentum to it. Radiation can do the same thing because photons also carry momentum. The difference is that here the particles are photons, moving at speed c, and their directions are described by the unit vector \(\hat{n}\).

For one photon of frequency \(\nu\), the energy and momentum magnitude are

\[ \epsilon = h\nu = {hc\over \lambda}, \qquad p = {\epsilon\over c} = {h\nu\over c}. \]

If the photon is moving in direction \(\hat{n}\), its momentum is a vector:

\[ \vec{p} = {h\nu\over c}\,\hat{n}. \]

Now define a photon number density per unit frequency per unit solid angle. Call it \(\psi_\nu\). Since \(I_\nu/c\) is the radiation energy density per unit frequency per unit solid angle, dividing by the energy of one photon gives the number density of photons:

\[ \psi_\nu = {I_\nu/c\over h\nu} = {I_\nu\over c h\nu}. \]

Therefore the total photon number density is obtained by adding all directions and all frequencies:

\[ \psi = \int_{\Omega}\int_0^\infty \psi_\nu\,d\nu\,d\Omega = \int_{\Omega}\int_0^\infty {I_\nu\over c h\nu}\,d\nu\,d\Omega. \]
Momentum flux through a surface A photon beam crosses a tilted surface. The crossing rate depends on the component along the surface normal, while the carried momentum component is measured along a chosen i direction. surface normal x_j photon ray velocity c n-hat x_i p = (hν/c) n-hat momentum component p_i = (hν/c) n_i crossing factor: c n_j θ photons in dν dΩ number density: ψν dP_ij = (number crossing) × (i-momentum carried)
Radiation pressure is momentum flux. Pij means the i-component of photon momentum crossing a surface whose normal is in the j-direction.

Pressure is force per unit area, and force is rate of change of momentum. So pressure can be read as momentum crossing unit area per unit time. For radiation this becomes a momentum flux.

Take photons in the small range \(d\nu\,d\Omega\). Their number density is \(\psi_\nu\,d\nu\,d\Omega\). Through a surface normal to the \(j\)-axis, only the velocity component \(c n_j\) carries photons across the surface.

\[ \text{number crossing unit area per unit time} = \psi_\nu c n_j\,d\nu\,d\Omega. \]

Each photon carries \(i\)-component of momentum

\[ p_i = {h\nu\over c}n_i. \]

Multiplying the number flux by the momentum component gives the \(ij\)-component of the radiation pressure tensor:

\[ dP_{ij} = \left(\psi_\nu c n_j\,d\nu\,d\Omega\right)\left({h\nu\over c}n_i\right). \]
\[ dP_{ij} = \psi_\nu h\nu\,n_i n_j\,d\nu\,d\Omega. \]
\[ P_{ij} = \int_{\Omega}\int_0^\infty \psi_\nu h\nu\,n_i n_j\,d\nu\,d\Omega. \]

Using \(\psi_\nu = I_\nu/(c h\nu)\), the photon energy cancels in a clean way:

\[ P_{ij} = \int_{\Omega}\int_0^\infty {I_\nu\over c}\,n_i n_j\,d\nu\,d\Omega. \]

The pressure is therefore not only one number in the most general case. It is a tensor, because momentum can have one component while crossing a surface with another orientation:

\[ \mathbf{P} = \begin{pmatrix} P_{xx} & P_{xy} & P_{xz}\\ P_{yx} & P_{yy} & P_{yz}\\ P_{zx} & P_{zy} & P_{zz} \end{pmatrix}. \]

If we choose the \(z\)-direction as the important normal direction, then the ordinary radiation pressure along that normal is

\[ P_{zz} \equiv P = \int_{\Omega}\int_0^\infty {I_\nu\over c}\,n_z^2\,d\nu\,d\Omega. \]

Axially symmetric radiation field

Now take the case used in the flux derivation: the radiation field depends on \(\theta\), but not on \(\phi\). Then

\[ I(\theta,\phi)=I(\theta), \qquad \mu=\cos\theta, \qquad d\Omega=\sin\theta\,d\theta\,d\phi. \]

The direction components are

\[ n_x=\sin\theta\cos\phi,\qquad n_y=\sin\theta\sin\phi,\qquad n_z=\cos\theta=\mu. \]

The \(x\) and \(y\) flux components vanished in the flux section because opposite azimuthal directions cancelled. For pressure, the squared direction factors remain, so the transverse diagonal components are not zero.

Axially symmetric radiation pressure Spherical coordinate direction n with polar angle theta from the z axis and azimuth phi in the xy plane. z x y n projection of n on xy-plane θ φ I(θ, φ) = I(θ) independent of φ about the z-axis
Here θ is measured from the z-axis, while φ is measured in the xy-plane from the x-axis to the projection of n-hat.

For the \(z\)-component, \(n_z=\mu\). After integrating over \(\phi\),

\[ P = P_{zz} = {2\pi\over c}\int_{-1}^{1} I(\mu)\mu^2\,d\mu. \]

The corresponding energy density is

\[ E = {1\over c}\int_{\Omega} I\,d\Omega = {2\pi\over c}\int_{-1}^{1} I(\mu)\,d\mu. \]

Now derive \(P_{xx}\). Since \(n_x=\sin\theta\cos\phi\),

\[ P_{xx} = \int_0^{2\pi}\int_0^\pi {I(\theta)\over c}\,n_x n_x\,\sin\theta\,d\theta\,d\phi. \]
\[ P_{xx} = \int_0^{2\pi}\int_0^\pi {I(\theta)\over c}\,\sin^2\theta\cos^2\phi\,\sin\theta\,d\theta\,d\phi. \]
\[ \int_0^{2\pi}\cos^2\phi\,d\phi=\pi. \]
\[ P_{xx} = {\pi\over c}\int_0^\pi I(\theta)\sin^2\theta\,\sin\theta\,d\theta. \]
\[ \mu=\cos\theta,\qquad \sin^2\theta=1-\mu^2,\qquad \sin\theta\,d\theta=-d\mu. \]
\[ P_{xx} = {\pi\over c}\int_{-1}^{1} I(\mu)(1-\mu^2)\,d\mu. \]
\[ P_{xx} = {1\over 2}\left({2\pi\over c}\int_{-1}^{1} I(\mu)\,d\mu\right) - {1\over 2}\left({2\pi\over c}\int_{-1}^{1} I(\mu)\mu^2\,d\mu\right). \]
\[ P_{xx}={E-P\over 2}. \]

By the same calculation, \(P_{yy}=(E-P)/2\). The off-diagonal terms vanish because integrals such as \(\int_0^{2\pi}\sin\phi\cos\phi\,d\phi\), \(\int_0^{2\pi}\cos\phi\,d\phi\), and \(\int_0^{2\pi}\sin\phi\,d\phi\) are zero. Therefore, for an axially symmetric radiation field,

\[ \mathbf{P} = \begin{pmatrix} {E-P\over 2} & 0 & 0\\ 0 & {E-P\over 2} & 0\\ 0 & 0 & P \end{pmatrix}. \]

Isotropic radiation

For isotropic radiation, the intensity is the same in every direction. Then \(I(\mu)=I\) is constant. The energy density and pressure become

\[ E = {2\pi I\over c}\int_{-1}^{1}d\mu = {4\pi I\over c}. \]
\[ P = {2\pi I\over c}\int_{-1}^{1}\mu^2\,d\mu = {2\pi I\over c}\left({2\over 3}\right) = {4\pi I\over 3c}. \]
\[ P={E\over 3}, \qquad E=3P. \]

Thus an isotropic radiation field has the same pressure in the three coordinate directions:

\[ \mathbf{P} = \begin{pmatrix} P & 0 & 0\\ 0 & P & 0\\ 0 & 0 & P \end{pmatrix} = {E\over 3}\mathbf{1}. \]

If the radiation is strongly beamed, for example radially streaming outward, the pressure is not isotropic. The component along the beam direction can dominate, while the transverse pressure components are much smaller.

2.5 Frequency-Integrated Intensity

Specific intensity is defined at a particular frequency. If we want the total intensity over all frequencies, we integrate over frequency:

\[ I = \int_0^\infty I_\nu\,d\nu \]

3. Interaction with Matter

Up to this point the light has propagated through pure vacuum. In a medium, an atom, molecule, or other intervening material can modify the bundle.

At the qualitative level, three processes are relevant:

  1. Absorption: the atom or molecule removes energy from the beam.
  2. Emission: the material adds new light into the beam.
  3. Scattering: light is redirected, so it may leave this beam or enter it from another direction.

This provides the stopping point for the present note; the subsequent development can begin from these three processes.

Matter interacting with a bundle of light A light bundle meets material. Absorption removes energy, emission adds light, and scattering changes direction. bundle of light absorption energy removed emission new light added scattering direction changed
Absorption, emission, and scattering are the first physical mechanisms that modify a radiation bundle.

References

These references support the definitions of specific intensity, solid angle, flux, energy density, and pressure used above.