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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 ${ds\over d\tau}$?

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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 (+---).

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Thanks, elfmostat! I use (-+++). May I ask: how does an imaginary ${ds\over d\tau}$ make physical sense? –  Harrold Apr 16 '13 at 23:40
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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.

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