Riemannian manifold

  • Thread starter whattttt
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Can someone please explain to me how this formula
integral from 0 to T of
sqrt(g_ij c'(t) c'(t))
I have seen it on wikipedia but don't know how to actually implement the formula.
 
1,005
65
(g_ij) is the metric tensor, c'(t) is, for each t, a tangent vector to the curve c(t).
With respect to a specific choice of coordinates, (g_ij) is a matrix that you premultiply and postmultiply by c'(t) to get a scalar. For example, the standard Euclidean metric in R2 with Cartesian coordinates is
[tex]\left(\begin{array}{cc}1 & 0\\0 & 1\end{array}\right)[/tex]
A typical curve in 2-dimensional Euclidean space with Cartesian coordinates has tangent vector
[tex]c'(t) = \left(\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\end{array}\right)[/tex]
Premultiplying and postmultiplying gets us the expression:
[tex]\left(\begin{array}{cc}\frac{dx}{dt} & \frac{dy}{dt}\end{array}\right) \left(\begin{array}{cc}1 & 0\\0 & 1\end{array}\right) \left(\begin{array}{c}\frac{dx}{dt}\\ \frac{dy}{dt}\end{array}\right)[/tex]
[tex]= \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2[/tex]
Different metrics will give different notions of length.
 

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