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Sorry, its been a while and my calculus was never good. This is really a very elementary question which I am unable to un -complicate from its shroud of notation. My difficulty is how does this parametrization thing take place,i.e how does the integral in terms of $t$ come to be, and how to convince myself that the coordinates of any curve in D dimensions can be written as single valued functions of a single parameter.

If $dl^2 = \delta_{ij} dx^i dx^j$ then why is $$\int_{P_1}^{P_2} dl = \int_{t_1}^{t_2} \sqrt{\delta_{ij} \dot{x}^i \dot{x}^j} dt\qquad \dot{x}^i\equiv \frac{dx^i}{dt}$$

Additions. $x^i$ are the rectangular coordinates in Euclidean space. So a position vector $\vec{r} = ( x^1,x^2,\cdots x^D)$. $\delta_{ij}$ is the Euclidean metric tensor defined in the same way as the Kronecker delta. An infinitesimal line element is defined as $dl^2 = \delta_{ij} dx^i dx^j$. $P_1$ and $P_2$ are two points in this space then the length of some curve $C$ between $P_1$ and $P_2$ $$\Delta L =\int_{P_1}^{P_2} dl$$

After this, the problematic bit is that "A curve in D dimensional euclidean space can be described as a subspace of the D dimensional space where the D coordinates $x^i$ are given by single valued functions of the parameter $t$. So this can be written as $P_1 = x(t_1)$ and $P_2 = x(t_2)$

$$\Delta L =\int_{t_1}^{t_2} \sqrt{\delta_{ij} \dot{x}^i \dot{x}^j} dt\qquad \dot{x}^i\equiv \frac{dx^i}{dt}$$

I can relate to what I did in two dimensions for parametrizing a circle in a plane by $x^0 = R \cos t$ and $x^1 =R\sin t$, but I just reasoned by taking the derivative as a fraction and cancelling numerators and denominators.

Note on convention: $\displaystyle g^i_j a_ib^j = \sum_{1\leq i\leq D}\quad \sum_{1\leq j\leq D}g^i_j a_i b^j$

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There is a lot missing in this question. What are $P_1$ and $P_2$, and what do they have to do with $t_1$ and $t_2$? What is $\delta_{ij}$? – Gerry Myerson Jun 16 '11 at 0:00
@Gerry $\delta_{ij}$ is the kronecker delta. – kuch nahi Jun 16 '11 at 0:26
Also there is this summation convention, right? Mathematicians would write $dl^2 = \sum_{i,j}\delta_{ij}\dot{x}^i\dot{x}^j$, and if that is the Kronecker delta, we get $dl^2 = \sum_j(\dot{x}^j)^2$. – GEdgar Jun 16 '11 at 1:54
@Gedgar Yes, the book follows what it calls tensor notation or einstein summation convention where repeated indices are summed over from $0$ to $D$ where $D$ is the dimension of the space – kuch nahi Jun 16 '11 at 2:00
I don't quite get what your question is... – ziyuang Jun 16 '11 at 2:27
up vote 5 down vote accepted

I think we have three questions here:

  1. How come a curve $\gamma$ in ${\mathbb R}^n$ has a parametrization $f:\ t\mapsto x(t)=(x_1(t),\ldots, x_n(t))$ $\ (a\leq t\leq b)\ $?

  2. Where does the length formula $L(\gamma)=\int_a^b |\dot x(t)|\ dt$ come from?

  3. How does this formula look like in a more general setting where we have a "metric tensor" $g_{ik}\ $?

Ad 1: It is the very idea of a curve in any space $X$ of being produced by a continuous map $f:\ t\mapsto x(t)$ of a certain interval $I\subset{\mathbb R}$ into $X$. Depending on circumstances there will be some comments to this definition, e.g., one considers one-one-continuous reparametrizations as defining the same curve (see below), or one insists that $f$ is essentially injective, etc.

A parametric representation of $\gamma$ is tantamount to an exact timetable. For each instant $t\in I$ we are told where the moving point $x(t)$ is. On the other hand a "curve" is a static geometrical object which is just lying there, and there is no physical motion in time involved. This means that one and the same curve has an infinite number of admissible representations, corresponding to different timetables. To account for this one calls two representations equivalent if they differ only by a homeomorphism of the time scales. Insofar as the formula for the length of $\gamma$ refers to a chosen parametrization one has to prove that the value is the same for all equivalent parametrizations.

Ad 2: Given a space $X$ with some sort of distance (infinitesimally or globally) and a curve $\gamma\subset X$ defined by an injective map $f: \ [a,b] \to X$ one may ask about the length of $\gamma$. If $X={\mathbb R}^d$ with its euclidean metric a natural procedure is the following: Approximate $\gamma$ by a sequence of segments. The sum of the lengths of these segments is then a raw value for $L(\gamma)$. This means we take a partition $P: \ a=t_0<t_1<\ldots <t_N=b$ of $[a,b]$ and compute $$L_P:=\sum_{k=1}^N |x(t_k)-x(t_{k-1})| \ \dot= \ \sum_{k=1}^N |\dot x(t_k)| (t_k-t_{k-1})$$ where $|\dot x(t)|=\bigl(\dot x_1^2(t)+\ldots+\dot x_n^2(t)\bigr)^{1/2}$. If the function $f$ is continuously differentiable then one can prove that $L_P$ converges to the integral $\int_a^b |\dot x(t)|\ dt$ when $\max_k(t_k-t_{k-1})$ goes to zero. Furthermore the value of the integral does not change under a reparametrization of $\gamma$.

Ad3: In a so-called Riemannian manifold $M$ there is no a priori global distance defined, nor do we have obvious "segments". But there is given a "Riemannian metric" which attributes to tangent vectors $X=(X_1,\ldots, X_n)$ attached to the points $p\in M$ a length $|X|=\sum_{i,k} g_{ik} X_i X_k$. You have a similar formula in euclidean $n$-space when you use nonorthogonal coordinates, but with standard coordinates you just have $g_{ik}=\delta_{ik}$ (the Kronecker-Delta). Now in Riemannian geometry it is shown that the natural formula for the length of a curve $\gamma:\ [a,b]\to M, \ t\mapsto x(t)$ is $$L(\gamma)=\int_a^b \sqrt{\sum_{i,k} g_{ik}(x(t)) \dot x_i \dot x_k}\ dt\ .$$

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So $L(\gamma) = \lim_{N\to\infty}L_P$ if the curve is continuously differentiable? And does "the value of the integral does not change under a reparametrization of $\gamma$" mean that parametrization is not unique? – kuch nahi Jun 16 '11 at 16:15
@yayu: $L(\gamma)=\sup_P \ L_P$ by definition and $L(\gamma)=\lim_{{\rm mesh}(P)\to0} \ L_P$ when $f\in C^1$. - For the other question see my edit. – Christian Blatter Jun 16 '11 at 20:48
@Christian Blatter, Nice, +1. Sorry for the necropost: is this the first fundamental form? – user99680 Feb 13 '14 at 0:56
@user99680: Given a surface $S\subset{\mathbb R}^3$ the first fundamental form is the pullback of the euclidean metric tensor $(\delta_{ik})$ in ${\mathbb R}^3$ to the $(u,v)$-parameter plane used to represent $S$. – Christian Blatter Feb 13 '14 at 8:53

An example in three dimensions would be the spiral defined by the curve $$ f(t) = (x_1(t), x_2(t), x_3(t)) = (cos(t), sin(t), t) \in \mathbb{R}^3 $$ Note that we treat $\mathbb{R}^3$ as a manifold with the canonical smooth structure and the global map $id_{\mathbb{R}^3}$.

The parameterization defines a "velocity vector $\vec{v}$" in $\mathbb{R}^3$ at every point of the curve via $$ \vec{v} := (\dot x_1(t), \dot x_2(t), \dot x_3(t)) $$

The definition of the length of the curve $len_{a, b}$ for $t \in [a, b]$ is defined to be the integral of the euclidean norm of the velocity vector $\vec{v}$ $$ len_{a, b} = \int_a^b \| \vec{v} (t) \| \; d t $$ This is just a reformulation of the general definition for curves in $\mathbb{R}^n$. The euclidean norm of $\vec{v}$ is $$ \| \vec{v} (t) \| = (\dot x_1(t)^2 + \dot x_2(t)^2 + \dot x_3(t)^2)^{\frac{1}{2}} = ((-sin(t))^2 + (cos(t))^2 + 1^2 )^{\frac{1}{2}} = \sqrt{2} $$ So the length is $$ len_{a, b} = \int_a^b \sqrt{2} \; d t $$ HTH

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