# Is this a coordinate transfomation?

Consider the redefinition $$d\zeta_1=ad\theta,~d\zeta_2=a\sin\theta d\phi$$ so that the distance between two points on the surface of a sphere of radius $$a$$ becomes $$dl^2=a^2(d\theta^2+\sin^2\theta d\phi^2)=d\zeta_1^2+d\zeta_2^2=\delta_{ij}d\zeta_id\zeta_j$$ Can the first equation be called a coordinate transformation? Have I really transformed the line element on the surface of a sphere to Euclidean metric? But I'm told that transforming the metric tensor to $$\delta_{ij}$$ via a coordinate transformation is possible only for a flat surface.

• If you really did find a coordinate transformation from $θ,φ$ to $ζ_1, ζ_2$, why are you failing to display it? – Cosmas Zachos Mar 22 '19 at 21:25
• @CosmasZachos In the first equation I displayed it. Why is this not a coordinate transformation is my question. Because they are not finite transformations? – mithusengupta123 Mar 23 '19 at 0:34
• A coordinate transformation is a map of variables, right? – Cosmas Zachos Mar 23 '19 at 1:54

Assume you did have a coordinate transformation. Look at your conjectural $$\zeta_2(\theta,\phi)$$, and work out its $$d\zeta_2= \frac{\partial\zeta_2}{\partial\theta}d\theta + \frac{\partial\zeta_2}{\partial\phi}d\phi ~.$$ Compare with your peculiar ad hoc identification, $$\frac{\partial\zeta_2}{\partial\theta}=0, \qquad \frac{\partial\zeta_2}{\partial\phi}=a \sin \theta.$$ The first equation is solved by all functions of only $$\phi$$ , but the second by $$a\phi \sin \theta + b(\theta)$$, so there is no such $$\zeta_2$$.