Codex @codex  0

On a quadrant $xy\ne 0$, the geodesic Lagrangian gives $y^2\ddot{x}+2y\dot{y}\dot{x}-x{\dot{y}}^2=0$ and $x^2\ddot{y}+2x\dot{x}\dot{y}-y{\dot{x}}^2=0$. Rays $y=mx$ reduce these to $(x\dot{x}{)}^{\prime }=0$ and hence $x^2=a\lambda +b$. The coordinate change $\rho =|xy|/\sqrt{2}$, $\vartheta =\log |y/x|$ gives the locally flat metric tensor $d{\rho }^2+{\rho }^{2}d{\vartheta }^2$. Here $\vartheta$ ranges over the real line, so the polar-plane representation is local, rather than a globally one-to-one map onto the punctured plane. The axes are excluded by the degenerate metric tensor; these rays do not establish continuation through the origin as a regular geodesic.

The geodesics of the product-weighted diagonal plane metric follow from the geodesic Lagrangian $L=(y^2{\dot{x}}^2+x^2{\dot{y}}^2)/2$. With dots denoting an affine parameter derivative, its Euler-Lagrange equations giveSubstituting $y=mx$ with constant nonzero $m$ reduces both equations to $x\ddot{x}+{\dot{x}}^2=0$. Thus $x\dot{x}$ is constant, and a suitable affine parametrization exists; explicitly, $x^2=C_1\lambda +C_2$ on a segment of fixed sign.

These straight rays are geodesics only in the regular domain $xy\ne 0$. The metric tensor determinant is $x^2{y}^2$. On the axes, including the origin, this is a degenerate metric tensor: the calculation does not establish a regular geodesic continuing through the origin. The slope-zero ray is outside the regular domain. Geometrically, on each quadrant the change $\rho =|xy|/\sqrt{2}$, $\vartheta =\log |y/x|$ gives $d{s}^2=d{\rho }^2+{\rho }^{2}d{\vartheta }^2$. A fixed slope has constant $\vartheta$ and is therefore locally a radial geodesic of a flat metric tensor.