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

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Try to use that $dx^i = \dot x^i dt$... –  Fabian Jun 20 '11 at 18:14
    
@Fabian so is it a general statement that the D coordinates can be paramterized by single valued function $f:\mathbb{R}^D\rightarrow \mathbb{R}$ is this paramterization unique? –  yayu Jun 20 '11 at 18:20
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@yayu: a (one-dimensional) curve in any D-dimensional space can be parametrized by a single-valued function $f:\mathbb{R}^D\to\mathbb{R}$. The parametrization is not unique; it can be rescaled or "translated" or subjected to any one-to-one function $g:\mathbb{R}\to\mathbb{R}$ and will result in a new parametrization. –  David Z Jun 20 '11 at 18:31
    
I think einstein should be written Einstein even when it is used in "Einstein summation convention". –  MBN Jun 20 '11 at 18:39
    
@MBN for small typos, etc.. this server allows anyone to edit posts :) –  yayu Jun 20 '11 at 18:40

3 Answers 3

up vote 1 down vote accepted

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.

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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

$$\int_0^{\pi/2} \sqrt{(-\sin t)^2 + (\cos t)^2}dt = \pi/2$$

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The 2D parametrization of a circular curve was the first thing I thought too. But this just shows that it works. Not much more –  yayu Jun 20 '11 at 18:17
    
@yayu As I said, this is the definition of length. What you do mean by "show that it works"? –  Mark Eichenlaub Jun 20 '11 at 20:01
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No it is not. A metric is defined for the locally flat space. For a one dimensional smooth curve we get the line element. Assuming a single parameter can define the curve, my question is to get the length from this parametrization. By "show that it works" I meant that you plugged and chugged for a circular segment on $D=2$ and verified it with familiar knowledge that a quadrant is a fourth of the perimeter. –  yayu Jun 20 '11 at 21:20

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.)

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That is true, but I have derived one gravely incorrect result in the past treating derivatives as fractions. I will think about this, thanks for answering. –  yayu Jun 20 '11 at 18:38
    
@yayu: yes, that happens sometimes, which I why I added that last paragraph about singular parametrizations. Basically this works as long as you can think of $\mathrm{d}l$ and $\mathrm{d}t$ as positive lengths and take the limit as they go to zero. –  David Z Jun 20 '11 at 18:41
    
@yayu: you never derive incorrect results when you use the key property of differentials (en.wikipedia.org/wiki/Differential_%28infinitesimal%29) instead of treating derivatives of fractions (as I indicated in my comment to your question). In fact the subject is called differential geometry exactly because one should think about $dx$ as being a differential (rather than a limit $\Delta x$ going to 0). –  Fabian Jun 21 '11 at 5:42

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