spatial variable r ∗ = r ∗ ( r ) defined through then we have with u = u ( t, r ∗ ), v = v ( t, r ∗ ) and F ( r ∗ ) := g$_{00}$ ( r ) /f 2 where r is a function of r $^{∗}$. We want to show that u and v satisfy a wave equation. If we consider the combinations (14)+(15) ± 2(16) we arrive at the following equations While taking the square roots of the above equations only those choices of the sign are allowed for which the determinant of the Jacobian of the transformation ˜ x := ( v, u, ϑ, ϕ ) −→ x = ( t, r, ϑ, ϕ ) does not vanish identically. Hence, we require Clearly, there are four possible choices of the sign. If we consider for example the case we would obtain ∂$_{t}$u = ∂$_{t}$v, ∂$_{r}$$_{∗}$u = ∂$_{r}$$_{∗}$v which implies det( J ) = 0. The same happens if we consider On the other hand the choice leads to the equations 6 ∂$_{t}$u = ∂$_{r}$$_{∗}$v, ∂$_{r}$$_{∗}$u = ∂$_{t}$v. (20) ∂$_{t}$u + ∂$_{r}$$_{∗}$u = ∂$_{t}$v + ∂$_{r}$$_{∗}$v, ∂$_{t}$u − ∂$_{r}$$_{∗}$u = − ∂$_{t}$v + ∂$_{r}$$_{∗}$ v, ∂$_{t}$u + ∂$_{r}$$_{∗}$u = − ∂$_{t}$v − ∂$_{r}$$_{∗}$ v, ∂$_{t}$u − ∂$_{r}$$_{∗}$ u = − ∂$_{t}$v + ∂$_{r}$$_{∗}$v. ∂$_{t}$u + ∂$_{r}$$_{∗}$ u = ∂$_{t}$v + ∂$_{r}$$_{∗}$v, ∂$_{t}$u − ∂$_{r}$$_{∗}$u = ∂$_{t}$v − ∂$_{r}$$_{∗}$ v, det( J ) = det   ∂$_{t}$v ∂$_{r}$$_{∗}$ v ∂$_{t}$u ∂$_{r}$$_{∗}$u   = ∂$_{t}$v ∂$_{r}$$_{∗}$u − ∂$_{r}$$_{∗}$ v ∂$_{t}$u = 0 . (19) ( ∂$_{t}$v − ∂$_{r}$$_{∗}$ v ) 2 = ( ∂$_{t}$u − ∂$_{r}$$_{∗}$ u ) $^{2}$. (18) ( ∂$_{t}$v + ∂$_{r}$$_{∗}$v ) 2 = ( ∂$_{t}$u + ∂$_{r}$$_{∗}$ u ) $^{2}$, (17) ( ∂$_{t}$v ) 2 − ( ∂$_{t}$u ) 2 = F ( r ∗ ) , (14) ( ∂$_{r}$$_{∗}$v ) 2 − ( ∂$_{r}$$_{∗}$u ) 2 = − F ( r ∗ ) , (15) ∂$_{t}$v ∂$_{r}$$_{∗}$v − ∂$_{t}$u ∂$_{r}$$_{∗}$ u = 0 (16) dr ∗ dr = 1 g$_{00}$ ( r ) (13)