In this case, the three-acceleration vector is perpendicular to the three-velocity vector, u ⋅ ⋅ a = 0. and the square of proper acceleration, expressed as a scalar invariant, the same in all reference frames, α α 2 = γ γ 4 a 2 + γ γ 6 ( u ⋅ ⋅ a ) 2 , becomes the expression for circular motion, α α 2 = γ γ 4 a 2 . or, taking the positive square root and using the three-acceleration, we arrive at the proper acceleration for circular motion: α α = γ γ 2 v 2 r .