Vinogradov's equidistribution theorem

Question

Is it true that

$(\alpha p_k)$ is equidistributed on $[0,1)$ mod 1 (Vinogradov)

$\Leftrightarrow$

$(p_k)$ is equidistributed on $[0,2\pi) $mod $2\pi$ ?

$p_k$ is the kth prime and $\alpha$ is an irrational number.

Answer 1

I would think the 1st statement implies the second by taking $\alpha=1/2\pi$. The 2nd statement implies the first in the trivial sense that the first is a theorem (I take it that's what you mean when you attribute it to Vinogradov) and everything implies a theorem. But perhaps I misunderstand.