Demonstration of the possible reduction from an accelerated frame of
reference to a Galilean frame of reference over a very short duration or interval.

We pose:
ΔTo²=ΔTr²+ΔEt²
ΔTo²=ΔTr²+Δx²/c²
ΔTo²=ΔTr²+(1/4)(a²/c²)(Tr2²-Tr1²)² ΔTo²=ΔTr²+(1/4)(a²/c²)[(Tr2+Tr1)(Tr2-Tr1)]² ΔTo²=ΔTr²+(1/4)(a²/c²)(Tr2+Tr1)².ΔTr² ΔTo²=ΔTr²[1+(1/4)(a²/c²)(Tr2+Tr1)²] ΔTo²=ΔTr²[1+(1/4)(a²/c²)(2Trm)²]
ΔTo²=ΔTr²[1+(a²/c²)(Trm)²]
ΔTo²=ΔTr²(1+Vri²/c²)

Let ΔTo=ΔTr.sqrt(1+Vr²/c²)

and Voi=Vri/sqrt(1+Vri²/c²)

As in the Galilean frames of reference.

Note that the instantaneous speed given for frames accelerated by
physicists is not correct, and that the correct formula is simply Voi/c=(1+c²/2ax)^(-1/2)

It's a lot less than what they say.

R.H.

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