19 Anisotropic Extension Analysis

This chapter establishes the physical correctness of the anisotropic exchange factor transformation by proving two fundamental properties: non-negativity of all radiative quantities and conservation of energy. These proofs demonstrate that the anisotropic extension preserves the mathematical rigor and physical validity of the original isotropic formulation.

19.1 Non-Negativity of the Anisotropic Interaction Matrix

The first requirement for physical correctness is that the interaction-reflection-scattering matrix \mathbf{K} must be non-negative, as negative probabilities are not physically meaningful.

Theorem 19.1 For the anisotropic radiative transfer formulation with anisotropy matrix \mathbf{\Gamma} satisfying \mathbf{\Gamma}_{ij} \in [0, 1], the interaction-reflection-scattering matrix \mathbf{K} constructed via row-stochastic normalization is guaranteed to be non-negative: \mathbf{K}_{ij} \geq 0 for all i, j.

Proof. The proof proceeds by analyzing the inequality that must hold for \mathbf{K}_{ij} \geq 0.

Step 1: Express \mathbf{K}_{ij} in terms of construction parameters.

From (Eq. 16.10):

\mathbf{K}_{ij} = \frac{\mathbf{K}_{\mathrm{init},ij}}{\mathbf{s}_i} - \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \left(1 - \frac{1}{\mathbf{s}_i}\right)

Substituting \mathbf{K}_{\mathrm{init},ij} = \mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj}):

\mathbf{K}_{ij} = \frac{\mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj})}{\mathbf{s}_i} - \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \left(1 - \frac{1}{\mathbf{s}_i}\right)

Step 2: Multiply through by \mathbf{s}_i > 0 to establish the inequality.

For \mathbf{K}_{ij} \geq 0, we require:

\mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj}) \geq \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \cdot (\mathbf{s}_i - 1)

Step 3: Analyze the term (\mathbf{s}_i - 1).

From equation (Eq. 16.6):

\mathbf{s}_i - 1 = \sum_k \mathbf{F}_{ik} \cdot (\mathbf{\Gamma}_{ik} - 1) \cdot (1-\mathbf{P}_{kk})

Step 4: Apply the constraint \mathbf{\Gamma}_{ik} \leq 1.

Since \mathbf{\Gamma}_{ik} \leq 1 for all i, k:

\mathbf{\Gamma}_{ik} - 1 \leq 0

Since \mathbf{F}_{ik} \geq 0 and (1-\mathbf{P}_{kk}) \geq 0:

\mathbf{F}_{ik} \cdot (\mathbf{\Gamma}_{ik} - 1) \cdot (1-\mathbf{P}_{kk}) \leq 0 \quad \forall k

Therefore:

\mathbf{s}_i - 1 = \sum_k \mathbf{F}_{ik} \cdot (\mathbf{\Gamma}_{ik} - 1) \cdot (1-\mathbf{P}_{kk}) \leq 0

Step 5: Conclude non-negativity.

From Step 2, we need:

\mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj}) \geq \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \cdot (\mathbf{s}_i - 1)

Since we proved \mathbf{s}_i - 1 \leq 0 in Step 4:

\mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \cdot (\mathbf{s}_i - 1) \leq 0

And since \mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj}) \geq 0 (all factors are non-negative), we have:

\mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj}) \geq 0 \geq \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \cdot (\mathbf{s}_i - 1)

Therefore: \mathbf{K}_{ij} \geq 0 ✓

Physical Interpretation

The non-negativity proof reveals a fundamental constraint: the upper bound \mathbf{\Gamma} \leq 1 ensures \mathbf{s}_i \leq 1, which makes the normalization correction term non-positive, thereby guaranteeing \mathbf{K}_{ij} \geq 0.

Importantly, no lower bound on Γ is required for non-negativity. The full range \mathbf{\Gamma} \in [0, 1] is permissible, providing maximum flexibility for representing anisotropic reflection-scattering distributions while maintaining physical validity.

Let \mathbf{F} be a row-stochastic exchange factor matrix, and let \mathbf{b} be a vector of reflection-scattering coefficients with b_j \in [0,1) for all j. Let \mathbf{P} be the diagonal emissivity matrix with P_j = 1 - b_j.

Define the anisotropic reflection-scattering matrix with columnwise constraints:

\mathbf{B}_{\mathrm{init},ij} = b_j + r_{ij} \cdot (1 - b_j) \tag{19.1}

where r_{ij} \in [0, 1] for all i, j.

Following the normalization procedure:

Form the unnormalized interaction matrix: \mathbf{K}_{\mathrm{init},ij} = \mathbf{F}_{ij} \cdot \mathbf{B}_{\mathrm{init},ij}
Account for both reflection-scattering and absorption: \mathbf{U}_{ij} = \mathbf{K}_{\mathrm{init},ij} + \mathbf{F}_{ij} \cdot P_j
Compute row sums: s_i = \sum_k \mathbf{U}_{i,k}
Normalize to row-stochastic: \mathbf{T}_{ij} = \mathbf{U}_{ij} / s_i
Extract the reflection-scattering component: \mathbf{K}_{ij} = \mathbf{T}_{ij} - \mathbf{F}_{ij} \cdot P_j

Then \mathbf{K}_{ij} \leq \mathbf{F}_{ij} for all i, j.

Proof. We will prove that the columnwise construction guarantees \mathbf{K}_{ij} \leq \mathbf{F}_{ij} by showing that the inequality holds in the worst case, and that all other cases are less restrictive.

Part 1: Derive an expression for \mathbf{K}_{ij}

Starting from step 2 of the normalization procedure, we can write \mathbf{U}_{ij} as:

\mathbf{U}_{ij} = \mathbf{F}_{ij} \cdot \mathbf{B}_{\mathrm{init},ij} + \mathbf{F}_{ij} \cdot P_j

Factoring:

\mathbf{U}_{ij} = \mathbf{F}_{ij} \cdot (\mathbf{B}_{\mathrm{init},ij} + P_j)

From step 4, the normalized matrix is:

\mathbf{T}_{ij} = \frac{\mathbf{U}_{ij}}{s_i} = \frac{\mathbf{F}_{ij} \cdot (\mathbf{B}_{\mathrm{init},ij} + P_j)}{s_i}

From step 5, the interaction-reflection-scattering matrix is:

\begin{aligned} \mathbf{K}_{ij} &= \mathbf{T}_{ij} - \mathbf{F}_{ij} \cdot P_j \\ &= \frac{\mathbf{F}_{ij} \cdot (\mathbf{B}_{\mathrm{init},ij} + P_j)}{s_i} - \mathbf{F}_{ij} \cdot P_j \end{aligned}

Factoring out \mathbf{F}_{ij}:

\mathbf{K}_{ij} = \mathbf{F}_{ij} \cdot \left[\frac{\mathbf{B}_{\mathrm{init},ij} + P_j}{s_i} - P_j\right]

To combine the terms, we write:

\mathbf{K}_{ij} = \mathbf{F}_{ij} \cdot \left[\frac{\mathbf{B}_{\mathrm{init},ij} + P_j - s_i \cdot P_j}{s_i}\right]

Simplifying the numerator:

\mathbf{K}_{ij} = \mathbf{F}_{ij} \cdot \frac{\mathbf{B}_{\mathrm{init},ij} + P_j \cdot (1 - s_i)}{s_i} \tag{19.2}

Part 2: Establish the condition for \mathbf{K}_{ij} \leq \mathbf{F}_{ij}

We want to prove that \mathbf{K}_{ij} \leq \mathbf{F}_{ij} for all i, j.

For entries where \mathbf{F}_{ij} = 0, we have:

\mathbf{K}_{ij} = 0 \cdot \frac{\mathbf{B}_{\mathrm{init},ij} + P_j \cdot (1 - s_i)}{s_i} = 0 = \mathbf{F}_{ij}

So the inequality holds trivially when \mathbf{F}_{ij} = 0.

For entries where \mathbf{F}_{ij} > 0, we can divide both sides by \mathbf{F}_{ij}:

\frac{\mathbf{B}_{\mathrm{init},ij} + P_j \cdot (1 - s_i)}{s_i} \leq 1

Multiplying both sides by s_i > 0:

\mathbf{B}_{\mathrm{init},ij} + P_j \cdot (1 - s_i) \leq s_i

Rearranging:

\mathbf{B}_{\mathrm{init},ij} \leq s_i - P_j \cdot (1 - s_i)

Expanding the right side:

\mathbf{B}_{\mathrm{init},ij} \leq s_i - P_j + P_j \cdot s_i

Factoring:

\mathbf{B}_{\mathrm{init},ij} \leq s_i \cdot (1 + P_j) - P_j \tag{19.3}

Since P_j = 1 - b_j, we have 1 + P_j = 1 + (1 - b_j) = 2 - b_j, giving:

\mathbf{B}_{\mathrm{init},ij} \leq s_i \cdot (2 - b_j) - (1 - b_j) \tag{19.4}

This is the condition we need to verify.

Part 3: Substitute the columnwise construction

From the construction formula (Eq. 19.1):

\mathbf{B}_{\mathrm{init},ij} = b_j + r_{ij} \cdot (1 - b_j)

Substituting into condition (Eq. 19.4):

b_j + r_{ij} \cdot (1 - b_j) \leq s_i \cdot (2 - b_j) - (1 - b_j)

Adding (1 - b_j) to both sides:

b_j + r_{ij} \cdot (1 - b_j) + (1 - b_j) \leq s_i \cdot (2 - b_j)

Since b_j + (1 - b_j) = 1:

1 + r_{ij} \cdot (1 - b_j) \leq s_i \cdot (2 - b_j) \tag{19.5}

This is the inequality we need to prove holds for all i, j.

Part 4: Determine the bounds on s_i

From step 3 of the normalization procedure:

s_i = \sum_k \mathbf{U}_{ik} = \sum_k \mathbf{F}_{ik} \cdot (\mathbf{B}_{\mathrm{init},ik} + P_k)

Substituting the columnwise construction:

s_i = \sum_k \mathbf{F}_{ik} \cdot (b_k + r_{ik} \cdot (1 - b_k) + P_k)

Since P_k = 1 - b_k:

s_i = \sum_k \mathbf{F}_{ik} \cdot (b_k + r_{ik} \cdot (1 - b_k) + 1 - b_k)

Simplifying the terms inside:

b_k + r_{ik} \cdot (1 - b_k) + 1 - b_k = 1 + r_{ik} \cdot (1 - b_k)

Therefore:

s_i = \sum_k \mathbf{F}_{ik} \cdot (1 + r_{ik} \cdot (1 - b_k))

Since \mathbf{F} is row-stochastic with \sum_k \mathbf{F}_{ik} = 1:

s_i = \sum_k \mathbf{F}_{ik} + \sum_k \mathbf{F}_{ik} \cdot r_{ik} \cdot (1 - b_k)

s_i = 1 + \sum_k \mathbf{F}_{ik} \cdot r_{ik} \cdot (1 - b_k) \tag{19.6}

Now we establish bounds on s_i. Since r_{ik} \in [0, 1], \mathbf{F}_{ik} \geq 0, and (1 - b_k) > 0:

The minimum value occurs when all r_{ik} = 0:

s_i \geq 1 + \sum_k \mathbf{F}_{ik} \cdot 0 \cdot (1 - b_k) = 1

The maximum value occurs when all r_{ik} = 1:

\begin{aligned} s_i &\leq 1 + \sum_k \mathbf{F}_{ik} \cdot 1 \cdot (1 - b_k) \\ &= 1 + \sum_k \mathbf{F}_{ik} \cdot (1 - b_k) \\ &= 1 + \sum_k \mathbf{F}_{ik} - \sum_k \mathbf{F}_{ik} \cdot b_k \\ &= 1 + 1 - \sum_k \mathbf{F}_{ik} \cdot b_k \\ &= 2 - \sum_k \mathbf{F}_{ik} \cdot b_k \end{aligned}

Since b_k \geq 0:

s_i \leq 2 - 0 = 2

Therefore:

1 \leq s_i \leq 2 \tag{19.7}

Part 5: Verify the inequality in the worst case

The inequality (Eq. 19.5) we need to prove is:

1 + r_{ij} \cdot (1 - b_j) \leq s_i \cdot (2 - b_j)

The left-hand side (LHS) is maximized when r_{ij} is largest: - Maximum LHS occurs when r_{ij} = 1 - Maximum LHS = 1 + 1 \cdot (1 - b_j) = 1 + 1 - b_j = 2 - b_j

The right-hand side (RHS) is minimized when s_i is smallest: - Minimum RHS occurs when s_i = 1 - Minimum RHS = 1 \cdot (2 - b_j) = 2 - b_j

In the worst case (maximum LHS and minimum RHS):

\text{LHS}_{\max} = 2 - b_j = 2 - b_j = \text{RHS}_{\min}

Therefore, the inequality holds with equality in the worst case.

Part 6: Verify the inequality holds generally

We have shown that in the worst case, LHS = RHS. Now we show that in all other cases, LHS < RHS (strict inequality).

Case 1: s_i > 1

When s_i > 1, the RHS increases:

\text{RHS} = s_i \cdot (2 - b_j) > 1 \cdot (2 - b_j) = 2 - b_j

Since the maximum possible LHS is 2 - b_j:

\text{LHS} \leq 2 - b_j < s_i \cdot (2 - b_j) = \text{RHS}

Case 2: r_{ij} < 1

When r_{ij} < 1, the LHS decreases:

\text{LHS} = 1 + r_{ij} \cdot (1 - b_j) < 1 + 1 \cdot (1 - b_j) = 2 - b_j

Since the minimum possible RHS is 2 - b_j:

\text{LHS} < 2 - b_j \leq s_i \cdot (2 - b_j) = \text{RHS}

Case 3: Both s_i > 1 and r_{ij} < 1

In this case, both effects combine: - LHS is reduced from its maximum - RHS is increased from its minimum

Therefore: \text{LHS} < \text{LHS}_{\max} = \text{RHS}_{\min} < \text{RHS}

This gives a strict inequality \mathbf{K}_{ij} < \mathbf{F}_{ij}.

Conclusion

We have shown that: 1. When \mathbf{F}_{ij} = 0: \mathbf{K}_{ij} = 0 = \mathbf{F}_{ij} ✓ 2. When \mathbf{F}_{ij} > 0 and worst case (s_i = 1, r_{ij} = 1): \mathbf{K}_{ij} = \mathbf{F}_{ij} ✓ 3. When \mathbf{F}_{ij} > 0 and not worst case: \mathbf{K}_{ij} < \mathbf{F}_{ij} ✓

Therefore, the columnwise construction \mathbf{B}_{\mathrm{init},ij} = b_j + r_{ij} \cdot (1 - b_j) with r_{ij} \in [0, 1] guarantees that \mathbf{K}_{ij} \leq \mathbf{F}_{ij} for all i, j. \square

The columnwise construction (Eq. 19.1) can be expressed in terms of an anisotropy matrix \mathbf{\Gamma} as:

\mathbf{B}_{\mathrm{init},ij} = b_j + \mathbf{\Gamma}_{ij} \cdot (1 - b_j) \tag{19.8}

where \mathbf{\Gamma}_{ij} \in [0, 1] for all i, j ensures \mathbf{K} \leq \mathbf{F}.

When \mathbf{\Gamma}_{ij} = 0, element j exhibits maximum anisotropy (minimum reflection-scattering) for radiation incident from element i.

When \mathbf{\Gamma}_{ij} = 1, element j exhibits isotropic behavior (maximum reflection-scattering) for radiation incident from element i.

19.2 Physical Interpretation

The columnwise construction has several important physical properties:

Locality of constraints: Each element j’s anisotropic behavior is governed only by its own reflection-scattering coefficient b_j, not by other elements in the system. This respects the physical independence of material properties across the domain.

High reflectivity limit: As b_j \to 1 (highly reflective element), we have:

\mathbf{B}_{\mathrm{init},ij} = b_j + \mathbf{\Gamma}_{ij} \cdot (1 - b_j) \to 1

regardless of \mathbf{\Gamma}_{ij}. Highly reflective elements are forced toward isotropic behavior, which is physically reasonable since a perfect reflector cannot exhibit directional selectivity.

Low reflectivity limit: As b_j \to 0 (weakly reflective element), we have:

\mathbf{B}_{\mathrm{init},ij} = b_j + \mathbf{\Gamma}_{ij} \cdot (1 - b_j) \to \mathbf{\Gamma}_{ij}

which spans the full range [0, 1] as \mathbf{\Gamma}_{ij} varies. Low-reflectivity elements have complete anisotropic freedom.

Effective reflection-scattering coefficient: For a given element j with reflection coefficient b_j, the effective reflection-scattering coefficient for radiation incident from element i is \mathbf{B}_{\mathrm{init},ij}, which ranges from: - Minimum: b_j (when \mathbf{\Gamma}_{ij} = 0) - Maximum: 1 (when \mathbf{\Gamma}_{ij} = 1)

The anisotropy factor \mathbf{\Gamma}_{ij} controls a smooth interpolation between these extremes, providing intuitive control over directional reflection-scattering effects.

19.3 Energy Conservation for the Anisotropic Formulation

The second requirement for physical correctness is conservation of energy at steady state. At equilibrium, the sum of all source fluxes must equal zero: \mathbf{1}^T\mathbf{C}\mathbf{j} = 0.

19.3.1 Key Identity: The Foundation of Energy Conservation

The proof of energy conservation relies on establishing that the fundamental eigenvalue relationship from the isotropic case is preserved in the anisotropic formulation.

Lemma 19.1 For the anisotropic formulation with row-stochastic normalization, the key identity holds:

\mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1}

where \mathbf{b} is the vector of reflection-scattering coefficients.

Proof. Step 1: Analyze the row sums of \mathbf{K}.

From the row-stochastic property of T:

\sum_j \mathbf{T}_{ij} = 1

Therefore:

\sum_j \mathbf{K}_{ij} = \sum_j \mathbf{T}_{ij} - \sum_j \mathbf{F}_{ij} \cdot \mathbf{P}_{jj}

\sum_j \mathbf{K}_{ij} = 1 - \sum_j \mathbf{F}_{ij} \cdot \mathbf{P}_{jj}

Step 2: Express in terms of reflection-scattering coefficients.

Since \mathbf{P}_{jj} = 1 - \mathbf{b}_j:

\sum_j \mathbf{K}_{ij} = 1 - \sum_j \mathbf{F}_{ij} \cdot (1 - \mathbf{b}_j)

Step 3: Compute [(\mathbf{I} - \mathbf{K})\mathbf{1}]_i.

[(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = 1 - \sum_j \mathbf{K}_{ij}

Substituting from Step 2:

[(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = 1 - \left(1 - \sum_j \mathbf{F}_{ij} \cdot (1 - \mathbf{b}_j)\right)

[(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = \sum_j \mathbf{F}_{ij} \cdot (1 - \mathbf{b}_j)

Step 4: Recognize the right-hand side.

[(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = [\mathbf{F}(\mathbf{1} - \mathbf{b})]_i

Therefore: \mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1} ✓

Critical Insight

This theorem is the cornerstone of energy conservation for the anisotropic formulation. The row-stochastic normalization ensures that despite the full matrix structure of \mathbf{K} (rather than the column-constant structure in the isotropic case), the fundamental relationship between \mathbf{F}, \mathbf{K}, and the reflection-scattering coefficients is preserved.

This means that all subsequent eigenvalue properties of the isotropic case carry over directly to the anisotropic case.

19.3.2 Energy Conservation Theorem

Theorem 19.2 Let \mathbf{F} be the row-stochastic exchange factor matrix, let \mathbf{\Gamma} be the anisotropy matrix with \mathbf{\Gamma}_{ij} \in [0, 1], and let \mathbf{b} = b\mathbf{1} be the uniform vector of reflection-scattering coefficients for some scalar b \in [0,1). Define the system matrices via the anisotropic exchange factor transformation.

Then for any mixed system \mathbf{M}\mathbf{j} = \mathbf{h} where \mathbf{M} is assembled from rows of \mathbf{C} and \mathbf{D}, the sum of the source fluxes equals zero:

\mathbf{1}^T \mathbf{C} \mathbf{j} = 0

regardless of the choice of rows or non-negative boundary conditions in the mixed system.

Proof. The proof follows directly from Lemma 19.1.

Step 1: Establish the eigenvalue relationship for \mathbf{A} + \mathbf{R}.

From Lemma 19.1: \mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1}

Left-multiply by (\mathbf{I} - \mathbf{K})^{-1}:

(\mathbf{I} - \mathbf{K})^{-1}\mathbf{F}(\mathbf{1} - \mathbf{b}) = \mathbf{1}

Left-multiply by \mathbf{P} (where \mathbf{P} = (1-b)\mathbf{I} for uniform b):

\mathbf{P}(\mathbf{I} - \mathbf{K})^{-1}\mathbf{F}(\mathbf{1} - \mathbf{b}) = \mathbf{P}\mathbf{1}

Step 2: Simplify using matrix definitions.

The left side equals \mathbf{A} + \mathbf{R}:

From the definitions:

\begin{aligned} \mathbf{A} &= \mathbf{P}(\mathbf{I}-\mathbf{K})^{-1}(\mathbf{F}-\mathbf{K}) \\ \mathbf{R} &= \mathbf{P}(\mathbf{I}-\mathbf{K})^{-1}\mathbf{K} \end{aligned} \tag{19.9}

it is obtained that:

\mathbf{A} + \mathbf{R} = \mathbf{P}(\mathbf{I}-\mathbf{K})^{-1}\mathbf{F} \tag{19.10}

Therefore:

(\mathbf{A} + \mathbf{R})(\mathbf{1} - \mathbf{b}) = \mathbf{P}\mathbf{1}

The right side simplifies:

\mathbf{P}\mathbf{1} = (1-b)\mathbf{1} = \mathbf{1} - \mathbf{b}

Step 3: Establish the eigenvalue property.

(\mathbf{A} + \mathbf{R})(\mathbf{1} - \mathbf{b}) = (\mathbf{1} - \mathbf{b})

Therefore, (\mathbf{1} - \mathbf{b}) is an eigenvector of \mathbf{A} + \mathbf{R} with eigenvalue 1.

Step 4: Derive the nullspace property of \mathbf{C}.

Taking the transpose:

(\mathbf{1} - \mathbf{b})^T(\mathbf{A}^T + \mathbf{R}^T) = (\mathbf{1} - \mathbf{b})^T

Rearranging:

(\mathbf{1} - \mathbf{b})^T(\mathbf{I} - \mathbf{A}^T - \mathbf{R}^T) = \mathbf{0}^T

Therefore:

(\mathbf{1} - \mathbf{b})^T\mathbf{C} = \mathbf{0}^T

Step 5: Apply uniformity to obtain energy conservation.

For uniform \mathbf{b} = b\mathbf{1}:

(\mathbf{1} - b\mathbf{1})^T\mathbf{C} = (1-b)\mathbf{1}^T\mathbf{C} = \mathbf{0}^T

Since (1-b) > 0:

\mathbf{1}^T\mathbf{C} = \mathbf{0}^T

Therefore, for any vector \mathbf{j}:

\mathbf{1}^T\mathbf{C}\mathbf{j} = \mathbf{0}^T\mathbf{j} = 0

This establishes energy conservation. ✓

Equivalence to Isotropic Case

Notice that this proof is identical to the proof for the isotropic case (see Theorem 12.3). The only difference is that we first had to establish Lemma 19.1 for the anisotropic formulation. Once that key relationship is proven, everything else follows exactly as before.

This demonstrates the elegance of the row-stochastic normalization approach: it preserves the fundamental algebraic structure that guarantees energy conservation, regardless of the complexity of the anisotropic distribution \mathbf{\Gamma}.

19.3.3 Non-Uniform Reflection-Scattering Coefficients

For completeness, we state the energy conservation result for mixed boundary conditions with non-uniform reflection-scattering coefficients.

Theorem 19.3 Let \mathbf{F} be row-stochastic, let \mathbf{\Gamma}_{ij} \in [0, 1], and define the system matrices via the anisotropic exchange factor transformation. Consider a mixed system \mathbf{M}\mathbf{j} = \mathbf{h} where:

Rows I of \mathbf{M} are taken from \mathbf{C} with corresponding \mathbf{h}_I = \mathbf{0}
Reflection-scattering coefficients on rows I can be non-uniform: 0 \leq b_i < 1 for all i \in I
Rows J of \mathbf{M} are taken from \mathbf{D} with corresponding \mathbf{h}_J \geq \mathbf{0}
Reflection-scattering coefficients on rows J are uniform: \mathbf{b}_j = \gamma < 1 for all j \in J

Then the energy conservation property holds:

\mathbf{1}^T \mathbf{C} \mathbf{j} = 0

Proof. The proof is identical to Theorem 12.12 for the isotropic case, using the eigenvalue relationship (\mathbf{1} - \mathbf{b})^T\mathbf{C} = \mathbf{0}^T established in the proof of Theorem 19.2.

The constraint \mathbf{C}_I\mathbf{j} = \mathbf{0} eliminates dependence on non-uniform coefficients in rows I, while uniformity on rows J ensures that (1-\gamma)\mathbf{1}^T\mathbf{C}_J\mathbf{j} = 0, which implies \mathbf{1}^T\mathbf{C}_J\mathbf{j} = 0 since (1-\gamma) > 0.

Combined, these yield \mathbf{1}^T\mathbf{C}\mathbf{j} = 0. ✓

19.4 Summary of Physical Correctness

The anisotropic extension of the exchange factor transformation satisfies both fundamental requirements for physical correctness:

Non-Negativity: All elements of \mathbf{K}, \mathbf{A}, \mathbf{R}, and consequently all radiative quantities (\mathbf{j}, \mathbf{e}, \mathbf{g}_a, \mathbf{r}), are non-negative for physically valid inputs (Theorem 19.1).
Energy Conservation: The sum of source fluxes equals zero at steady state, ensuring global energy balance (Theorem 19.2, Theorem 19.3).

19.4.1 Key Insight: Preservation of Algebraic Structure

The critical insight enabling these results is that the row-stochastic normalization procedure preserves the fundamental identity:

\mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1}

This identity is the cornerstone of the isotropic formulation’s mathematical structure. By maintaining this relationship in the anisotropic case, the row-stochastic normalization ensures that:

The eigenvalue structure of \mathbf{A} + \mathbf{R} is preserved
The nullspace property of \mathbf{C} is maintained
Energy conservation follows automatically for uniform (or appropriately mixed) boundary conditions

19.4.2 Comparison with Isotropic Case

Property	Isotropic	Anisotropic	Status
\mathbf{K} structure	Column-constant	Full matrix	Generalized
\mathbf{K} \geq 0	Trivial	Requires proof	✓ Proven
Key identity	Direct	Via normalization	✓ Proven
Energy conservation	Direct	Via key identity	✓ Proven

The anisotropic extension represents a true generalization: it maintains all essential properties while accommodating arbitrary directional distributions within the physical constraint \mathbf{\Gamma} \in [0, 1].

19.5 Conclusion

The anisotropic exchange factor transformation preserves the mathematical rigor and physical validity of the isotropic formulation. Through careful construction via row-stochastic normalization, the method guarantees:

Non-negative radiative quantities for all physically valid inputs
Energy conservation at steady state
Analytical solvability through matrix inversion
Backward compatibility with the isotropic case

These properties make the anisotropic extension suitable for rigorous scientific computation in radiative transfer applications requiring directional resolution of reflection-scattering processes.

# Anisotropic Extension Analysis {#sec-exchange_fact_anisotropic_extend} This chapter establishes the physical correctness of the anisotropic exchange factor transformation by proving two fundamental properties: non-negativity of all radiative quantities and conservation of energy. These proofs demonstrate that the anisotropic extension preserves the mathematical rigor and physical validity of the original isotropic formulation. ## Non-Negativity of the Anisotropic Interaction Matrix The first requirement for physical correctness is that the interaction-reflection-scattering matrix $\mathbf{K}$ must be non-negative, as negative probabilities are not physically meaningful. ::: {.theorem data-title="Non-Negativity of K (Anisotropic Case)" #thm-aniso-nonnegative} For the anisotropic radiative transfer formulation with anisotropy matrix $\mathbf{\Gamma}$ satisfying $\mathbf{\Gamma}_{ij} \in [0, 1]$, the interaction-reflection-scattering matrix $\mathbf{K}$ constructed via row-stochastic normalization is guaranteed to be non-negative: $\mathbf{K}_{ij} \geq 0$ for all $i, j$. ::: ::: {.proof} The proof proceeds by analyzing the inequality that must hold for $\mathbf{K}_{ij} \geq 0$. **Step 1:** Express $\mathbf{K}_{ij}$ in terms of construction parameters. From (@eq-k-simplified): $$ \mathbf{K}_{ij} = \frac{\mathbf{K}_{\mathrm{init},ij}}{\mathbf{s}_i} - \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \left(1 - \frac{1}{\mathbf{s}_i}\right) $$ Substituting $\mathbf{K}_{\mathrm{init},ij} = \mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj})$: $$ \mathbf{K}_{ij} = \frac{\mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj})}{\mathbf{s}_i} - \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \left(1 - \frac{1}{\mathbf{s}_i}\right) $$ **Step 2:** Multiply through by $\mathbf{s}_i > 0$ to establish the inequality. For $\mathbf{K}_{ij} \geq 0$, we require: $$ \mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj}) \geq \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \cdot (\mathbf{s}_i - 1) $$ **Step 3:** Analyze the term $(\mathbf{s}_i - 1)$. From equation (@eq-row-sums-expanded): $$ \mathbf{s}_i - 1 = \sum_k \mathbf{F}_{ik} \cdot (\mathbf{\Gamma}_{ik} - 1) \cdot (1-\mathbf{P}_{kk}) $$ **Step 4:** Apply the constraint $\mathbf{\Gamma}_{ik} \leq 1$. Since $\mathbf{\Gamma}_{ik} \leq 1$ for all $i, k$: $$ \mathbf{\Gamma}_{ik} - 1 \leq 0 $$ Since $\mathbf{F}_{ik} \geq 0$ and $(1-\mathbf{P}_{kk}) \geq 0$: $$ \mathbf{F}_{ik} \cdot (\mathbf{\Gamma}_{ik} - 1) \cdot (1-\mathbf{P}_{kk}) \leq 0 \quad \forall k $$ Therefore: $$ \mathbf{s}_i - 1 = \sum_k \mathbf{F}_{ik} \cdot (\mathbf{\Gamma}_{ik} - 1) \cdot (1-\mathbf{P}_{kk}) \leq 0 $$ **Step 5:** Conclude non-negativity. From Step 2, we need: $$ \mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj}) \geq \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \cdot (\mathbf{s}_i - 1) $$ Since we proved $\mathbf{s}_i - 1 \leq 0$ in Step 4: $$ \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \cdot (\mathbf{s}_i - 1) \leq 0 $$ And since $\mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj}) \geq 0$ (all factors are non-negative), we have: $$ \mathbf{F}_{ij} \cdot \mathbf{\Gamma}_{ij} \cdot (1-\mathbf{P}_{jj}) \geq 0 \geq \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} \cdot (\mathbf{s}_i - 1) $$ Therefore: $\mathbf{K}_{ij} \geq 0$ ✓ ::: ::: {.callout-note} ## Physical Interpretation The non-negativity proof reveals a fundamental constraint: the upper bound $\mathbf{\Gamma} \leq 1$ ensures $\mathbf{s}_i \leq 1$, which makes the normalization correction term non-positive, thereby guaranteeing $\mathbf{K}_{ij} \geq 0$. **Importantly, no lower bound on Γ is required for non-negativity.** The full range $\mathbf{\Gamma} \in [0, 1]$ is permissible, providing maximum flexibility for representing anisotropic reflection-scattering distributions while maintaining physical validity. ::: ::: {.theorem data-title="K ≤ F for the Columnwise Anisotropic Construction"} Let $\mathbf{F}$ be a row-stochastic exchange factor matrix, and let $\mathbf{b}$ be a vector of reflection-scattering coefficients with $b_j \in [0,1)$ for all $j$. Let $\mathbf{P}$ be the diagonal emissivity matrix with $P_j = 1 - b_j$. Define the anisotropic reflection-scattering matrix with columnwise constraints: $$ \mathbf{B}_{\mathrm{init},ij} = b_j + r_{ij} \cdot (1 - b_j) $$ {#eq-binit-columnwise} where $r_{ij} \in [0, 1]$ for all $i, j$. Following the normalization procedure: 1. Form the unnormalized interaction matrix: $\mathbf{K}_{\mathrm{init},ij} = \mathbf{F}_{ij} \cdot \mathbf{B}_{\mathrm{init},ij}$ 2. Account for both reflection-scattering and absorption: $\mathbf{U}_{ij} = \mathbf{K}_{\mathrm{init},ij} + \mathbf{F}_{ij} \cdot P_j$ 3. Compute row sums: $s_i = \sum_k \mathbf{U}_{i,k}$ 4. Normalize to row-stochastic: $\mathbf{T}_{ij} = \mathbf{U}_{ij} / s_i$ 5. Extract the reflection-scattering component: $\mathbf{K}_{ij} = \mathbf{T}_{ij} - \mathbf{F}_{ij} \cdot P_j$ Then $\mathbf{K}_{ij} \leq \mathbf{F}_{ij}$ for all $i, j$. ::: ::: {.proof} We will prove that the columnwise construction guarantees $\mathbf{K}_{ij} \leq \mathbf{F}_{ij}$ by showing that the inequality holds in the worst case, and that all other cases are less restrictive. **Part 1:** Derive an expression for $\mathbf{K}_{ij}$ Starting from step 2 of the normalization procedure, we can write $\mathbf{U}_{ij}$ as: $$ \mathbf{U}_{ij} = \mathbf{F}_{ij} \cdot \mathbf{B}_{\mathrm{init},ij} + \mathbf{F}_{ij} \cdot P_j $$ Factoring: $$ \mathbf{U}_{ij} = \mathbf{F}_{ij} \cdot (\mathbf{B}_{\mathrm{init},ij} + P_j) $$ From step 4, the normalized matrix is: $$ \mathbf{T}_{ij} = \frac{\mathbf{U}_{ij}}{s_i} = \frac{\mathbf{F}_{ij} \cdot (\mathbf{B}_{\mathrm{init},ij} + P_j)}{s_i} $$ From step 5, the interaction-reflection-scattering matrix is: $$ \begin{aligned} \mathbf{K}_{ij} &= \mathbf{T}_{ij} - \mathbf{F}_{ij} \cdot P_j \\ &= \frac{\mathbf{F}_{ij} \cdot (\mathbf{B}_{\mathrm{init},ij} + P_j)}{s_i} - \mathbf{F}_{ij} \cdot P_j \end{aligned} $$ Factoring out $\mathbf{F}_{ij}$: $$ \mathbf{K}_{ij} = \mathbf{F}_{ij} \cdot \left[\frac{\mathbf{B}_{\mathrm{init},ij} + P_j}{s_i} - P_j\right] $$ To combine the terms, we write: $$ \mathbf{K}_{ij} = \mathbf{F}_{ij} \cdot \left[\frac{\mathbf{B}_{\mathrm{init},ij} + P_j - s_i \cdot P_j}{s_i}\right] $$ Simplifying the numerator: $$ \mathbf{K}_{ij} = \mathbf{F}_{ij} \cdot \frac{\mathbf{B}_{\mathrm{init},ij} + P_j \cdot (1 - s_i)}{s_i} $$ {#eq-k-expression} **Part 2:** Establish the condition for $\mathbf{K}_{ij} \leq \mathbf{F}_{ij}$ We want to prove that $\mathbf{K}_{ij} \leq \mathbf{F}_{ij}$ for all $i, j$. For entries where $\mathbf{F}_{ij} = 0$, we have: $$ \mathbf{K}_{ij} = 0 \cdot \frac{\mathbf{B}_{\mathrm{init},ij} + P_j \cdot (1 - s_i)}{s_i} = 0 = \mathbf{F}_{ij} $$ So the inequality holds trivially when $\mathbf{F}_{ij} = 0$. For entries where $\mathbf{F}_{ij} > 0$, we can divide both sides by $\mathbf{F}_{ij}$: $$ \frac{\mathbf{B}_{\mathrm{init},ij} + P_j \cdot (1 - s_i)}{s_i} \leq 1 $$ Multiplying both sides by $s_i > 0$: $$ \mathbf{B}_{\mathrm{init},ij} + P_j \cdot (1 - s_i) \leq s_i $$ Rearranging: $$ \mathbf{B}_{\mathrm{init},ij} \leq s_i - P_j \cdot (1 - s_i) $$ Expanding the right side: $$ \mathbf{B}_{\mathrm{init},ij} \leq s_i - P_j + P_j \cdot s_i $$ Factoring: $$ \mathbf{B}_{\mathrm{init},ij} \leq s_i \cdot (1 + P_j) - P_j $$ {#eq-condition-P} Since $P_j = 1 - b_j$, we have $1 + P_j = 1 + (1 - b_j) = 2 - b_j$, giving: $$ \mathbf{B}_{\mathrm{init},ij} \leq s_i \cdot (2 - b_j) - (1 - b_j) $$ {#eq-main-condition} This is the condition we need to verify. **Part 3:** Substitute the columnwise construction From the construction formula (@eq-binit-columnwise): $$ \mathbf{B}_{\mathrm{init},ij} = b_j + r_{ij} \cdot (1 - b_j) $$ Substituting into condition (@eq-main-condition): $$ b_j + r_{ij} \cdot (1 - b_j) \leq s_i \cdot (2 - b_j) - (1 - b_j) $$ Adding $(1 - b_j)$ to both sides: $$ b_j + r_{ij} \cdot (1 - b_j) + (1 - b_j) \leq s_i \cdot (2 - b_j) $$ Since $b_j + (1 - b_j) = 1$: $$ 1 + r_{ij} \cdot (1 - b_j) \leq s_i \cdot (2 - b_j) $$ {#eq-simplified-condition} This is the inequality we need to prove holds for all $i, j$. **Part 4:** Determine the bounds on $s_i$ From step 3 of the normalization procedure: $$ s_i = \sum_k \mathbf{U}_{ik} = \sum_k \mathbf{F}_{ik} \cdot (\mathbf{B}_{\mathrm{init},ik} + P_k) $$ Substituting the columnwise construction: $$ s_i = \sum_k \mathbf{F}_{ik} \cdot (b_k + r_{ik} \cdot (1 - b_k) + P_k) $$ Since $P_k = 1 - b_k$: $$ s_i = \sum_k \mathbf{F}_{ik} \cdot (b_k + r_{ik} \cdot (1 - b_k) + 1 - b_k) $$ Simplifying the terms inside: $$ b_k + r_{ik} \cdot (1 - b_k) + 1 - b_k = 1 + r_{ik} \cdot (1 - b_k) $$ Therefore: $$ s_i = \sum_k \mathbf{F}_{ik} \cdot (1 + r_{ik} \cdot (1 - b_k)) $$ Since $\mathbf{F}$ is row-stochastic with $\sum_k \mathbf{F}_{ik} = 1$: $$ s_i = \sum_k \mathbf{F}_{ik} + \sum_k \mathbf{F}_{ik} \cdot r_{ik} \cdot (1 - b_k) $$ $$ s_i = 1 + \sum_k \mathbf{F}_{ik} \cdot r_{ik} \cdot (1 - b_k) $$ {#eq-si-formula} Now we establish bounds on $s_i$. Since $r_{ik} \in [0, 1]$, $\mathbf{F}_{ik} \geq 0$, and $(1 - b_k) > 0$: The minimum value occurs when all $r_{ik} = 0$: $$ s_i \geq 1 + \sum_k \mathbf{F}_{ik} \cdot 0 \cdot (1 - b_k) = 1 $$ The maximum value occurs when all $r_{ik} = 1$: $$ \begin{aligned} s_i &\leq 1 + \sum_k \mathbf{F}_{ik} \cdot 1 \cdot (1 - b_k) \\ &= 1 + \sum_k \mathbf{F}_{ik} \cdot (1 - b_k) \\ &= 1 + \sum_k \mathbf{F}_{ik} - \sum_k \mathbf{F}_{ik} \cdot b_k \\ &= 1 + 1 - \sum_k \mathbf{F}_{ik} \cdot b_k \\ &= 2 - \sum_k \mathbf{F}_{ik} \cdot b_k \end{aligned} $$ Since $b_k \geq 0$: $$ s_i \leq 2 - 0 = 2 $$ Therefore: $$ 1 \leq s_i \leq 2 $$ {#eq-si-bounds} **Part 5:** Verify the inequality in the worst case The inequality (@eq-simplified-condition) we need to prove is: $$ 1 + r_{ij} \cdot (1 - b_j) \leq s_i \cdot (2 - b_j) $$ The left-hand side (LHS) is maximized when $r_{ij}$ is largest: - Maximum LHS occurs when $r_{ij} = 1$ - Maximum LHS $= 1 + 1 \cdot (1 - b_j) = 1 + 1 - b_j = 2 - b_j$ The right-hand side (RHS) is minimized when $s_i$ is smallest: - Minimum RHS occurs when $s_i = 1$ - Minimum RHS $= 1 \cdot (2 - b_j) = 2 - b_j$ In the worst case (maximum LHS and minimum RHS): $$ \text{LHS}_{\max} = 2 - b_j = 2 - b_j = \text{RHS}_{\min} $$ Therefore, the inequality holds with equality in the worst case. **Part 6:** Verify the inequality holds generally We have shown that in the worst case, LHS $=$ RHS. Now we show that in all other cases, LHS $<$ RHS (strict inequality). **Case 1:** $s_i > 1$ When $s_i > 1$, the RHS increases: $$ \text{RHS} = s_i \cdot (2 - b_j) > 1 \cdot (2 - b_j) = 2 - b_j $$ Since the maximum possible LHS is $2 - b_j$: $$ \text{LHS} \leq 2 - b_j < s_i \cdot (2 - b_j) = \text{RHS} $$ **Case 2:** $r_{ij} < 1$ When $r_{ij} < 1$, the LHS decreases: $$ \text{LHS} = 1 + r_{ij} \cdot (1 - b_j) < 1 + 1 \cdot (1 - b_j) = 2 - b_j $$ Since the minimum possible RHS is $2 - b_j$: $$ \text{LHS} < 2 - b_j \leq s_i \cdot (2 - b_j) = \text{RHS} $$ **Case 3:** Both $s_i > 1$ and $r_{ij} < 1$ In this case, both effects combine: - LHS is reduced from its maximum - RHS is increased from its minimum Therefore: $\text{LHS} < \text{LHS}_{\max} = \text{RHS}_{\min} < \text{RHS}$ This gives a strict inequality $\mathbf{K}_{ij} < \mathbf{F}_{ij}$. **Conclusion** We have shown that: 1. When $\mathbf{F}_{ij} = 0$: $\mathbf{K}_{ij} = 0 = \mathbf{F}_{ij}$ ✓ 2. When $\mathbf{F}_{ij} > 0$ and worst case ($s_i = 1$, $r_{ij} = 1$): $\mathbf{K}_{ij} = \mathbf{F}_{ij}$ ✓ 3. When $\mathbf{F}_{ij} > 0$ and not worst case: $\mathbf{K}_{ij} < \mathbf{F}_{ij}$ ✓ Therefore, the columnwise construction $\mathbf{B}_{\mathrm{init},ij} = b_j + r_{ij} \cdot (1 - b_j)$ with $r_{ij} \in [0, 1]$ guarantees that $\mathbf{K}_{ij} \leq \mathbf{F}_{ij}$ for all $i, j$. $\square$ ::: ::: {.theorem data-title="Corollary: Anisotropy Matrix Constraint"} The columnwise construction (@eq-binit-columnwise) can be expressed in terms of an anisotropy matrix $\mathbf{\Gamma}$ as: $$ \mathbf{B}_{\mathrm{init},ij} = b_j + \mathbf{\Gamma}_{ij} \cdot (1 - b_j) $$ {#eq-gamma-relation} where $\mathbf{\Gamma}_{ij} \in [0, 1]$ for all $i, j$ ensures $\mathbf{K} \leq \mathbf{F}$. When $\mathbf{\Gamma}_{ij} = 0$, element $j$ exhibits maximum anisotropy (minimum reflection-scattering) for radiation incident from element $i$. When $\mathbf{\Gamma}_{ij} = 1$, element $j$ exhibits isotropic behavior (maximum reflection-scattering) for radiation incident from element $i$. ::: ## Physical Interpretation The columnwise construction has several important physical properties: **Locality of constraints:** Each element $j$'s anisotropic behavior is governed only by its own reflection-scattering coefficient $b_j$, not by other elements in the system. This respects the physical independence of material properties across the domain. **High reflectivity limit:** As $b_j \to 1$ (highly reflective element), we have: $$ \mathbf{B}_{\mathrm{init},ij} = b_j + \mathbf{\Gamma}_{ij} \cdot (1 - b_j) \to 1 $$ regardless of $\mathbf{\Gamma}_{ij}$. Highly reflective elements are forced toward isotropic behavior, which is physically reasonable since a perfect reflector cannot exhibit directional selectivity. **Low reflectivity limit:** As $b_j \to 0$ (weakly reflective element), we have: $$ \mathbf{B}_{\mathrm{init},ij} = b_j + \mathbf{\Gamma}_{ij} \cdot (1 - b_j) \to \mathbf{\Gamma}_{ij} $$ which spans the full range $[0, 1]$ as $\mathbf{\Gamma}_{ij}$ varies. Low-reflectivity elements have complete anisotropic freedom. **Effective reflection-scattering coefficient:** For a given element $j$ with reflection coefficient $b_j$, the effective reflection-scattering coefficient for radiation incident from element $i$ is $\mathbf{B}_{\mathrm{init},ij}$, which ranges from: - Minimum: $b_j$ (when $\mathbf{\Gamma}_{ij} = 0$) - Maximum: $1$ (when $\mathbf{\Gamma}_{ij} = 1$) The anisotropy factor $\mathbf{\Gamma}_{ij}$ controls a smooth interpolation between these extremes, providing intuitive control over directional reflection-scattering effects. ## Energy Conservation for the Anisotropic Formulation The second requirement for physical correctness is conservation of energy at steady state. At equilibrium, the sum of all source fluxes must equal zero: $\mathbf{1}^T\mathbf{C}\mathbf{j} = 0$. ### Key Identity: The Foundation of Energy Conservation The proof of energy conservation relies on establishing that the fundamental eigenvalue relationship from the isotropic case is preserved in the anisotropic formulation. ::: {.theorem data-title="Preservation of the Key Identity" #lem-key-identity} For the anisotropic formulation with row-stochastic normalization, the key identity holds: $$ \mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1} $$ where $\mathbf{b}$ is the vector of reflection-scattering coefficients. ::: ::: {.proof} **Step 1:** Analyze the row sums of $\mathbf{K}$. From the row-stochastic property of $T$: $$ \sum_j \mathbf{T}_{ij} = 1 $$ Therefore: $$ \sum_j \mathbf{K}_{ij} = \sum_j \mathbf{T}_{ij} - \sum_j \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} $$ $$ \sum_j \mathbf{K}_{ij} = 1 - \sum_j \mathbf{F}_{ij} \cdot \mathbf{P}_{jj} $$ **Step 2:** Express in terms of reflection-scattering coefficients. Since $\mathbf{P}_{jj} = 1 - \mathbf{b}_j$: $$ \sum_j \mathbf{K}_{ij} = 1 - \sum_j \mathbf{F}_{ij} \cdot (1 - \mathbf{b}_j) $$ **Step 3:** Compute $[(\mathbf{I} - \mathbf{K})\mathbf{1}]_i$. $$ [(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = 1 - \sum_j \mathbf{K}_{ij} $$ Substituting from Step 2: $$ [(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = 1 - \left(1 - \sum_j \mathbf{F}_{ij} \cdot (1 - \mathbf{b}_j)\right) $$ $$ [(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = \sum_j \mathbf{F}_{ij} \cdot (1 - \mathbf{b}_j) $$ **Step 4:** Recognize the right-hand side. $$ [(\mathbf{I} - \mathbf{K})\mathbf{1}]_i = [\mathbf{F}(\mathbf{1} - \mathbf{b})]_i $$ Therefore: $\mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1}$ ✓ ::: ::: {.callout-important} ## Critical Insight This theorem is the cornerstone of energy conservation for the anisotropic formulation. The row-stochastic normalization ensures that despite the full matrix structure of $\mathbf{K}$ (rather than the column-constant structure in the isotropic case), the fundamental relationship between $\mathbf{F}$, $\mathbf{K}$, and the reflection-scattering coefficients is preserved. This means that **all subsequent eigenvalue properties of the isotropic case carry over directly to the anisotropic case**. ::: ### Energy Conservation Theorem ::: {.theorem data-title="Energy Conservation (Anisotropic - Uniform Coefficients)" #thm-aniso-conservation} Let $\mathbf{F}$ be the row-stochastic exchange factor matrix, let $\mathbf{\Gamma}$ be the anisotropy matrix with $\mathbf{\Gamma}_{ij} \in [0, 1]$, and let $\mathbf{b} = b\mathbf{1}$ be the uniform vector of reflection-scattering coefficients for some scalar $b \in [0,1)$. Define the system matrices via the anisotropic exchange factor transformation. Then for any mixed system $\mathbf{M}\mathbf{j} = \mathbf{h}$ where $\mathbf{M}$ is assembled from rows of $\mathbf{C}$ and $\mathbf{D}$, the sum of the source fluxes equals zero: $$ \mathbf{1}^T \mathbf{C} \mathbf{j} = 0 $$ regardless of the choice of rows or non-negative boundary conditions in the mixed system. ::: ::: {.proof} The proof follows directly from @lem-key-identity. **Step 1:** Establish the eigenvalue relationship for $\mathbf{A} + \mathbf{R}$. From @lem-key-identity: $\mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1}$ Left-multiply by $(\mathbf{I} - \mathbf{K})^{-1}$: $$ (\mathbf{I} - \mathbf{K})^{-1}\mathbf{F}(\mathbf{1} - \mathbf{b}) = \mathbf{1} $$ Left-multiply by $\mathbf{P}$ (where $\mathbf{P} = (1-b)\mathbf{I}$ for uniform $b$): $$ \mathbf{P}(\mathbf{I} - \mathbf{K})^{-1}\mathbf{F}(\mathbf{1} - \mathbf{b}) = \mathbf{P}\mathbf{1} $$ **Step 2:** Simplify using matrix definitions. The left side equals $\mathbf{A} + \mathbf{R}$: From the definitions: $$ \begin{aligned} \mathbf{A} &= \mathbf{P}(\mathbf{I}-\mathbf{K})^{-1}(\mathbf{F}-\mathbf{K}) \\ \mathbf{R} &= \mathbf{P}(\mathbf{I}-\mathbf{K})^{-1}\mathbf{K} \end{aligned} $$ {#eq-aniso-ar-defs} it is obtained that: $$ \mathbf{A} + \mathbf{R} = \mathbf{P}(\mathbf{I}-\mathbf{K})^{-1}\mathbf{F} $$ {#eq-aniso-a-plus-r} Therefore: $$ (\mathbf{A} + \mathbf{R})(\mathbf{1} - \mathbf{b}) = \mathbf{P}\mathbf{1} $$ The right side simplifies: $$ \mathbf{P}\mathbf{1} = (1-b)\mathbf{1} = \mathbf{1} - \mathbf{b} $$ **Step 3:** Establish the eigenvalue property. $$ (\mathbf{A} + \mathbf{R})(\mathbf{1} - \mathbf{b}) = (\mathbf{1} - \mathbf{b}) $$ Therefore, $(\mathbf{1} - \mathbf{b})$ is an eigenvector of $\mathbf{A} + \mathbf{R}$ with eigenvalue 1. **Step 4:** Derive the nullspace property of $\mathbf{C}$. Taking the transpose: $$ (\mathbf{1} - \mathbf{b})^T(\mathbf{A}^T + \mathbf{R}^T) = (\mathbf{1} - \mathbf{b})^T $$ Rearranging: $$ (\mathbf{1} - \mathbf{b})^T(\mathbf{I} - \mathbf{A}^T - \mathbf{R}^T) = \mathbf{0}^T $$ Therefore: $$ (\mathbf{1} - \mathbf{b})^T\mathbf{C} = \mathbf{0}^T $$ **Step 5:** Apply uniformity to obtain energy conservation. For uniform $\mathbf{b} = b\mathbf{1}$: $$ (\mathbf{1} - b\mathbf{1})^T\mathbf{C} = (1-b)\mathbf{1}^T\mathbf{C} = \mathbf{0}^T $$ Since $(1-b) > 0$: $$ \mathbf{1}^T\mathbf{C} = \mathbf{0}^T $$ Therefore, for any vector $\mathbf{j}$: $$ \mathbf{1}^T\mathbf{C}\mathbf{j} = \mathbf{0}^T\mathbf{j} = 0 $$ This establishes energy conservation. ✓ ::: ::: {.callout-tip} ## Equivalence to Isotropic Case Notice that this proof is **identical** to the proof for the isotropic case (see @thm-spectral_radius_unity). The only difference is that we first had to establish @lem-key-identity for the anisotropic formulation. Once that key relationship is proven, everything else follows exactly as before. This demonstrates the elegance of the row-stochastic normalization approach: it preserves the fundamental algebraic structure that guarantees energy conservation, regardless of the complexity of the anisotropic distribution $\mathbf{\Gamma}$. ::: ### Non-Uniform Reflection-Scattering Coefficients For completeness, we state the energy conservation result for mixed boundary conditions with non-uniform reflection-scattering coefficients. ::: {.theorem data-title="Energy Conservation (Anisotropic - Mixed Boundary Conditions)" #thm-aniso-conservation-nonuniform} Let $\mathbf{F}$ be row-stochastic, let $\mathbf{\Gamma}_{ij} \in [0, 1]$, and define the system matrices via the anisotropic exchange factor transformation. Consider a mixed system $\mathbf{M}\mathbf{j} = \mathbf{h}$ where: - Rows $I$ of $\mathbf{M}$ are taken from $\mathbf{C}$ with corresponding $\mathbf{h}_I = \mathbf{0}$ - Reflection-scattering coefficients on rows $I$ can be non-uniform: $0 \leq b_i < 1$ for all $i \in I$ - Rows $J$ of $\mathbf{M}$ are taken from $\mathbf{D}$ with corresponding $\mathbf{h}_J \geq \mathbf{0}$ - Reflection-scattering coefficients on rows $J$ are uniform: $\mathbf{b}_j = \gamma < 1$ for all $j \in J$ Then the energy conservation property holds: $$ \mathbf{1}^T \mathbf{C} \mathbf{j} = 0 $$ ::: ::: {.proof} The proof is identical to @thm-uniform_conservation for the isotropic case, using the eigenvalue relationship $(\mathbf{1} - \mathbf{b})^T\mathbf{C} = \mathbf{0}^T$ established in the proof of @thm-aniso-conservation. The constraint $\mathbf{C}_I\mathbf{j} = \mathbf{0}$ eliminates dependence on non-uniform coefficients in rows $I$, while uniformity on rows $J$ ensures that $(1-\gamma)\mathbf{1}^T\mathbf{C}_J\mathbf{j} = 0$, which implies $\mathbf{1}^T\mathbf{C}_J\mathbf{j} = 0$ since $(1-\gamma) > 0$. Combined, these yield $\mathbf{1}^T\mathbf{C}\mathbf{j} = 0$. ✓ ::: ## Summary of Physical Correctness The anisotropic extension of the exchange factor transformation satisfies both fundamental requirements for physical correctness: 1. **Non-Negativity**: All elements of $\mathbf{K}$, $\mathbf{A}$, $\mathbf{R}$, and consequently all radiative quantities ($\mathbf{j}$, $\mathbf{e}$, $\mathbf{g}_a$, $\mathbf{r}$), are non-negative for physically valid inputs (@thm-aniso-nonnegative). 2. **Energy Conservation**: The sum of source fluxes equals zero at steady state, ensuring global energy balance (@thm-aniso-conservation, @thm-aniso-conservation-nonuniform). ### Key Insight: Preservation of Algebraic Structure The critical insight enabling these results is that the row-stochastic normalization procedure preserves the fundamental identity: $$ \mathbf{F}(\mathbf{1} - \mathbf{b}) = (\mathbf{I} - \mathbf{K})\mathbf{1} $$ This identity is the cornerstone of the isotropic formulation's mathematical structure. By maintaining this relationship in the anisotropic case, the row-stochastic normalization ensures that: - The eigenvalue structure of $\mathbf{A} + \mathbf{R}$ is preserved - The nullspace property of $\mathbf{C}$ is maintained - Energy conservation follows automatically for uniform (or appropriately mixed) boundary conditions ### Comparison with Isotropic Case | Property | Isotropic | Anisotropic | Status | |----------|-----------|-------------|--------| | $\mathbf{K}$ structure | Column-constant | Full matrix | Generalized | | $\mathbf{K} \geq 0$ | Trivial | Requires proof | ✓ Proven | | Key identity | Direct | Via normalization | ✓ Proven | | Energy conservation | Direct | Via key identity | ✓ Proven | The anisotropic extension represents a true generalization: it maintains all essential properties while accommodating arbitrary directional distributions within the physical constraint $\mathbf{\Gamma} \in [0, 1]$. ## Conclusion The anisotropic exchange factor transformation preserves the mathematical rigor and physical validity of the isotropic formulation. Through careful construction via row-stochastic normalization, the method guarantees: - **Non-negative radiative quantities** for all physically valid inputs - **Energy conservation** at steady state - **Analytical solvability** through matrix inversion - **Backward compatibility** with the isotropic case These properties make the anisotropic extension suitable for rigorous scientific computation in radiative transfer applications requiring directional resolution of reflection-scattering processes.