Assume that the rubber is cross-linked and isotropic, the random walk model gives R x , R y and R z are distributed according to a normal distribution. Therefore, they are equal in space, and all of them are 1/3 of the overall end-to-end distance of the chain: ⟨ ⟨ R x 0 2 ⟩ ⟩ = ⟨ ⟨ R y 0 2 ⟩ ⟩ = ⟨ ⟨ R z 0 2 ⟩ ⟩ = ⟨ ⟨ R 2 ⟩ ⟩ / 3 . Plugging in the change of free energy equation above, it is easy to get: Δ Δ F def ( R → → ) = − − k B T n s ⟨ ⟨ R 2 ⟩ ⟩ ( λ λ x 2 + λ λ y 2 + λ λ z 2 − − 3 ) 2 N b 2 = − − k B T n s ⟨ ⟨ R 2 ⟩ ⟩ ( λ λ x 2 + λ λ y 2 + λ λ z 2 − − 3 ) 2 R 0 2