Length of a curve in D dimensional euclidean space In a book I am reading on special relativity, the infinitesimal line element is defined as $dl^2=\delta_{ij}dx^idx^j$ (Einstein summation convention) where $\delta_{ij}$ is the euclidean metric. Next, if we have some curve C between two points $P_1$ and $P_2$ in this space then the length of the curve is given as $\Delta L = \int_{P_1}^{P_2}dl$
I am having trouble deriving the next statement, which I quote:

A curve in D-dimensional Euclidean
  space can be described as a subspace
  of the D-dimensional spce where the D
  co-ordinates $x^i$ are given by single
  valued functions of some parameter
  $t$, in which case the length of the
  curve from $P_1=x(t_1)$ to
  $P_2=x(t_2)$ can be written
  $$\Delta L = \int_{t_1}^{t_2}\sqrt{\delta_{ij}
\dot{x}^i \dot{x}^j} dt \qquad
 \mbox{where}\; \dot{x}^i\equiv
 \frac{dx^i}{dt}$$

 A: You can derive the correct results when you use the key property of differentials $$dx_i=\dot{x}_i dt.$$ Note that
$\Delta L$ is invariant under reparameterization $t'=f(t)$ as you can check easily (this is in fact the reason why you can write it as $\int d\ell$ without any reference to a parametrization). However, to calculate the length $\Delta L$ it is advisable to introduce some (arbitrary) parameterization. If you are interested in unique parameterizations: there exists also a unique parameteriziation with respect to arclength which has some nice feature.
A: I think you should take that as the definition of the word "length".  I wouldn't try to derive it at all.
It is basically saying that if, for example, you want to know the length of the unit circle in the first quadrant, set
$$x^1 = \cos t$$
$$x^2 = \sin t$$
$$\dot{x}^1 = -\sin t$$
$$\dot{x}^2 = \cos t$$
and do
$$\int_0^{\pi/2} \sqrt{(-\sin t)^2 + (\cos t)^2}dt = \pi/2$$
A: Well, $\mathrm{d}l$ represents an infinitesimal length along the curve. $\mathrm{d}t$ also represents an infinitesimal length along the curve, although if the parametrizations $t$ and $l$ are different, the two infinitesimal lengths are not going to be the same. You can write the identity $\mathrm{d}l = \frac{\mathrm{d}l}{\mathrm{d}t}\mathrm{d}t$ and substitute in the definition of $\mathrm{d}l$:
$$\frac{\mathrm{d}l}{\mathrm{d}t}\mathrm{d}t = \frac{\sqrt{\delta_{ij}\mathrm{d}x^i\mathrm{d}x^j}}{\mathrm{d}t}\mathrm{d}t$$
Now, since $\mathrm{d}x^{i(j)}$ and $\mathrm{d}t$ are positive lengths, you can do a little algebraic manipulation:
$$\frac{\mathrm{d}l}{\mathrm{d}t}\mathrm{d}t = \sqrt{\frac{\delta_{ij}\mathrm{d}x^i\mathrm{d}x^j}{\mathrm{d}t^2}}\mathrm{d}t = \sqrt{\delta_{ij}\dot x^i\dot x^j}\mathrm{d}t$$
This is not 100% mathematically rigorous, but in physics we think of derivatives and differentials as the limit of finite lengths and ratios, so it makes physical sense at least. And as long as your parametrization isn't singular, the math should hold up. (If you do have a singular parametrization, then I believe the result you're asking about still holds, although you need to resort to more precise math to prove it.)
