What happens when you rotate yourself ? You're a frame of reference , the speed of any object is perceived by you multiplied by the radius or the distance to it, if you rotate then you cannot perceive objects farther than a critical distance d since at some distance the objects will have apparent speed equal to the speed of light. Without transformations you rotate and perceive a light source at the critical distance d , then d=ct , you rotate with speed ω then c=ωd so c=ωct t=1/ω or t=T one period so the light ray must do a full spiral before reaching you.
Suppose there's stationary light source A, rotating observer B and stationary observers C and D, the distance of A to B , A to C, B to D and C to D is the critical distance d, A shoots a signal to B and C , as soon as B and C receive the signal from A they shoot a signal to D, D measures the difference in time between the two signals. The proper time interval should be exactly one period of the rotating frame or 1/ω , when you substitute the schwarzschild metric for accelerating frames in SR by the equivalence principle or use a=v^2/r v^2=ar you get 1/γ=sqrt(1-ad/c^2) , t'=γt and d=ct , t=d/c
Δt'=γΔt 1/ω = 1/sqrt(1-ad/c^2) t 1/ω = d/sqrt(1-ad/c^2)c ω = sqrt(1-ad/c^2)c / d ω^2 d^2 = (1-ad/c^2) c^2 ω^2 d = (c^2/d -a) a' = c^2/d-a but a'=a then a=c^2/d-a a=c^2/2d The acceleration is similar for linear and circular motion.
What this means is the farthest we can observe is d=c^2/2g , inside black holes where g is large the distance is small. But the gravitational acceleration g=GM/d^2 => d=c^2/2 * d^2/GM d=d^2 c^2 /2GM d=2GM/c^2 the schwarzschild radius https://en.wikipedia.org/wiki/Schwarzschild_radius
Article On the Question of Acceleration in Special Relativity
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