How to derive Black's formula for the valuation of an option on a future?

I've got a question about 1976 Black Model and Bachelier model.

I know that a geometric brownian motion in the P measure $dS_{t}=\mu S_{t}dt+\sigma S_{t} dW_{t}^{P}$ for a stock price $S_{t}$ leads (after a change of measure) to the Black-Scholes formula for a Call:

$$C= S_{0} N(d_{1}) − Ke^{−rT} N(d_{2})$$ .

Where $d_{1} = \frac{ln(\frac{S_{0}}{K})+(r+\frac{1}{2}\sigma^{2})T}{\sigma\sqrt{T}}$ and $d_{2}=d_{1}-\sigma \sqrt{T}$

I actually don't know how's possible to get the famous black formula on a forward contract:

$$C= e^{−rT}(F N(d_{1}) − KN(d_{2}))$$ .

where now $d_{1} = \frac{ln(\frac{F}{K})+\frac{1}{2}\sigma^{2}T}{\sigma\sqrt{T}}$ and $d_{2}=d_{1}-\sigma \sqrt{T}$

Should I simply insert $F(0,T)=S_{0}e^{rT}$ in the first BS formula to get the second one?

I'm asking this because I've tried to derive the BS formula using an arithmetic brownian motion like $dS_{t}=\mu dt+\sigma dW_{t}^{P}$, and I get:

$$C= S_{0} N(d) + e^{−rT}[v n(d)-K N(d)]$$ .

where $d=\frac{S_{0}e^{rT}-K}{v}$ and $v=e^{rT}\sigma\sqrt{\frac{1-e^{−2rT}}{2r}}$ and remembering that $N(d)$ and $n(d)$ are the CDF and PDF.

but the previous substitution $F(0,T)=S_{0}e^{rT}$ doesn't seems to lead to the known result $C= e^{−rT}[(F-K)N(d)-\sigma\sqrt{T}n(d)]$

where now $d=\frac{F-K}{\sigma\sqrt{T}}$

I think I could reach the equations on forward both in the geometric brownian motion and arithmetic brownian motion using the equations

$dF=F\sigma dW_{t}^{Q}$ and $dF=\sigma dW_{t}^{Q}$ but I don't know how justify the use of them.