Your notion and the equality in the book are consistent, since $$ \delta (a^2 t^2 - |x|^2 ) = \frac{1}{2|at|} (\delta (at-|x|) + \delta (at+|x|) ) $$ (I suppose an assumption is being made that $at\geq 0$).
Michael Engelhardt
- 6.7k
- 2
- 18
- 39