I am a bit confused regarding how an asset returns in a risk-neutral measure (say $\mathbb{Q}$), and subsequently its Black-Scholes-esque PDE.
In class, we learned about the approach to take when considering a geometric BM, and thus we get $\frac{\mathrm{d} S_t}{S_t}=r \mathrm{~d} t + ...$ as our SDE.
which I understand since we want the expected value of $dS/S$ to be $r \ dt$
Now, if we work on an arithmetic GM $\mathrm{d} S_t=\mu \mathrm{~d} t + \sigma \mathrm{~d} B$, would we be considering the same definition for risk-neutral measure, $\mathbb{E}_t^Q\left[\frac{\mathrm{d} S_t}{S_t}\right]=r \mathrm{~d} t$? And if so, would we get a SDE along the lines of $\frac{\mathrm{d} S_t}{S_t}=r \mathrm{~d} t + \sigma \mathrm{~d} B^Q$ (i.e. the arithmetic GM becomes a geometric GM under risk-neutral measure $\mathbb{Q}$)?
Let $D_t$ be the discount factor, a process satisfying $dD_t = -rD_t dt$. We wish to find a measure under which the discounted process $X_t = D_t S_t$ is a martingale. By Itô's lemma, observe that: $$dX_t = D_t dS_t + S_t dD_t = D_t (\mu dt + \sigma dB_t) - rS_tD_tdt = \sigma D_t \left[\frac{(\mu - r S_t)}{\sigma}dt + dB_t\right] = \sigma D_t d\widetilde{W}_t$$
where we have performed the change of measure to a probability measure $\mathbb{Q}$ which renders the process
$$d\widetilde{W}_t = \frac{(\mu - r S_t)}{\sigma}dt + dB_t$$
witnessed in Girsanov's theorem, into a $\mathbb{Q}$-Brownian motion. To get the risk-neutral dynamics of $S$ we work backwards to get:
$$dS_t = \mu dt + \sigma dB_t= \mu dt + \sigma d\widetilde{W}_t - (\mu -rS_t)dt = rS_t dt + \sigma d\widetilde{W}_t$$
In this case you do indeed get the expectation you suggested (at least when $S_t$ is away from zero).