# General proof of Arc Length For Parametrised Coodrdinates

1. Oct 30, 2016

### bananabandana

1. The problem statement, all variables and given/known data
Prove that, given a metric $g_{ij}$ such that $ds^{2}=g_{ij}dx^{i}dx^{j}$, where $x^{r} = x^{r}(\lambda)$ , we have the following result for the arc length:
$$L(p,q) = \int_{p}^{q} ds = \sqrt{ g_{ij} \frac{dx^{i}}{d \lambda} \frac{ dx^{j}}{d \lambda} } d \lambda$$
2. Relevant equations

3. The attempt at a solution

Honestly not sure about how to attempt this. I've had a look in Riley Hobson and Bence and Schaums Vector analysis for clues, but can't find any. Both simply state the result as self evident. Have I missed something obvious?

2. Oct 30, 2016

The $g_{ij}$ I think are equal to 1 if $i=j$ and 0 otherwise. Meanwhile, I think the statement is simply one that says the Pythagorean theorem applies for differential arc elements of the orthogonal coordinates, i.e. $(ds)^2=(dx)^2+(dy)^2+(dz)^2 +...$. Putting in a $d \lambda$ may be more evident if you let the $d \lambda=dt$. Perhaps $g_{ij}$ will take on another form for non-orthogonal coordinates, but the details of that are probably not exceedingly difficult.
It's not generally true that the off diagonal elements of the metric tensor are zero - we require that the tensor be symmetric, but that's not *necessarily* the same as it being diagonal? Even if I've got that wrong though...If you think of $S^{2}$, then the metric definitely isn't just Kronecker delta...?