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General proof of Arc Length For Parametrised Coodrdinates

  1. Oct 30, 2016 #1
    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. jcsd
  3. Oct 30, 2016 #2

    Charles Link

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    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.
     
  4. Oct 30, 2016 #3
    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...?
     
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