Interval and proper time Is the definition of $$d s^2=-d \tau^2$$ assuming that $c=1$, so that we always have $$\left({ds\over d\tau}\right)^2=-1$$?
Is there a reason for this definition?
Don't we get an imaginary ${ds\over d\tau}$?
 A: It depends on what convention you're using for the metric's signature. Some people use the metric signature (-+++), which is what you have there. The interval is then:
$$ds^2=-dt^2+d \mathbf{r}^2$$
On the other hand, some people use the (+---) convention:
$$ds^2=dt^2-d \mathbf{r}^2$$
In this signature $ds^2=d \tau^2$. Which one you use is a matter of preference. Special/General Relativity textbooks tend to prefer (-+++), though a lot of Quantum Mechanics textbooks prefer (+---).
A: Yeah, you do. The reason $\mathrm{d}s$ and $\mathrm{d}\tau$ are defined this way is that one or the other will be real for any given path. For a spacelike interval $\mathrm{d}s$ will be real (indicative of the fact that the distances we measure in everyday life are spacelike intervals), and for a timelike interval $\mathrm{d}\tau$ will be real (since times we measure in everyday life are timelike intervals). They fill complementary roles, but they're really just two ways of expressing the same thing.
$\frac{\mathrm{d}s}{\mathrm{d}\tau}$ itself doesn't really have any physical meaning. It's actually always equal to $\pm ic$ because of the definition.
Note that all the above only applies if you use the $-+++$ metric. If you use $+---$, then $\mathrm{d}s^2 = \mathrm{d}\tau^2$ and it's not an issue.
