# Why $\sigma_{rr}-\sigma_{\theta\theta}=O(r^2)$ Goal: show the tensor-parity refinement used in [[polymeric-stress-regularity-slender-jet|slender-jet regularity arguments]]. ## 1) Scalar/velocity parity reminder For smooth axisymmetric fields near the axis (also see [[continuity-slender-jet|Continuity equation in slender jet]]), $ v_z(r,z,t),\;p(r,z,t)\;\text{are even in }r, \qquad v_r(r,z,t)\;\text{is odd in }r. $ So $ v_z=v_0+v_2 r^2+\cdots, \qquad p=p_0+p_2 r^2+\cdots, \qquad v_r=u_1 r+u_3 r^3+\cdots. $ ## 2) Axis regularity for transverse stress At $r=0$, there is no preferred direction in the $(x,y)$ plane. For a smooth rank-2 tensor this implies transverse isotropy at the axis: $ \sigma_{xx}(0)=\sigma_{yy}(0), \qquad \sigma_{xy}(0)=0. $ In cylindrical components, $ \sigma_{rr}(0)=\sigma_{\theta\theta}(0). $ ## 3) Cartesian-cylindrical relation that enforces $r^2$ For axisymmetric stress with no $r\theta$ component in the local $(\mathbf e_r,\mathbf e_\theta)$ basis, $ \boldsymbol\sigma =\sigma_{rr}\,\mathbf e_r\otimes\mathbf e_r +\sigma_{\theta\theta}\,\mathbf e_\theta\otimes\mathbf e_\theta. $ Using $ \mathbf e_r=(\cos\theta,\sin\theta), \qquad \mathbf e_\theta=(-\sin\theta,\cos\theta), $ one gets $ \sigma_{xy} =(\sigma_{rr}-\sigma_{\theta\theta})\sin\theta\cos\theta =(\sigma_{rr}-\sigma_{\theta\theta})\frac{xy}{x^2+y^2}. $ Now require $\sigma_{xy}(x,y)$ to be smooth at $(0,0)$. Since $ \frac{xy}{x^2+y^2}=O(1) $ but depends on direction, smoothness at the origin forces the prefactor to vanish at least as $x^2+y^2=r^2$: $ \sigma_{rr}-\sigma_{\theta\theta}=O(r^2). $ So the difference cannot start at $O(1)$ or $O(r)$; it starts at even order $r^2$. ## 4) Equivalent expansion form Write $ \sigma_{rr}=\Sigma_0+\Sigma_2 r^2+\Sigma_4 r^4+\cdots, \qquad \sigma_{\theta\theta}=\Sigma_0+\Theta_2 r^2+\Theta_4 r^4+\cdots, $ then $ \sigma_{rr}-\sigma_{\theta\theta}=(\Sigma_2-\Theta_2)r^2+O(r^4)=O(r^2). $