Orthogonal Lines and their line element

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SUMMARY

The discussion centers on the derivation of the line element on a sphere as presented in "Gravity" by Hartle. It confirms that for orthogonal lines, the line element can be expressed using a Pythagorean-like equation: (distance)^2 = (position on one line)^2 + (position on the other line)^2. The conversation elaborates on the spherical coordinates and the quadratic form that defines distances in a coordinate system, specifically detailing the calculations for dx, dy, and dz in relation to a sphere of radius R.

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  • Familiarity with quadratic forms in differential geometry
  • Knowledge of matrix representation of line elements
  • Basic concepts of orthogonality in geometry
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In one of the early chapters of Gravity by Hartle, he is developing the line element on a sphere in preparation for developing the concept of a spacetime interval. Whilst finishing up the proof Hartle sort of implicitly says that if two lines are orthogonal the line element connecting two points on the two lines can be given by a pythagorean-like equation (the distance)2=(position on one line)2+(position on the other)2.

Is the more general question of a line element being the sum of the squares of the two orthogonal lines true?

Sorry if this question is a bit messy.
 
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In general, if you have a "surface" or "space" with a coordinate system with coordinates x^1, x^2, ..., x^n, then the distance is given by a quadratic form \sum a_{ij}dx^idx^j.

Is that what you are talking about?
cgv
In particular, on the surface of a sphere of radius, R, we can identify each point by using spherical coordinates with the radial coordinate, \rho, fixed as the constant, R.

x= Rcos(\theta)sin(\phi) so that dx= -Rsin(\theta)sin(\phi)d\theta+ Rcos(\theta)cos(\phi)d\phi
y= Rsin(\theta)sin(\phi) so that dy= Rcos(\theta)sin(\phi)d\theta+ Rsin(\theta)cos(\phi)d\phi
z= R cos(\phi) so that dz= -R sin(\phi)d\phi

And now,
dx^2= R^2cos^2(\theta)sin^2(\phi)d\theta^2- 2R^2sin(\theta)cos(theta)sin(\phi)cos(\phi)d\theta d\phi+ R^2cos^2(\theta)cos^2(\phi)d\phi^2
dy^2= R^2sin^2(\theta)sin^2(\phi)d\theta^2+ 2R^2 sin(\theta)cos(\theta)sin(\phi)cos(\phi)d\theta d\phi+ R^2sin^2(\theta)cos^2(\phi)d\phi^2
so that dx^2+ dy^2= R^2sin^2(\phi)d\theta^2+ R^2cos^2(\phi)d\phi^2

And dz^2= R^2 sin^2(\phi) d\phi^2 so that dx^2+ dy^2+ dz^2= R^2sin^2(\phi)d\theta^2+ R^2d\phi^2

That can be written as \sum g_{ij}dx^idx^j or, as a matrix product, in this particular coordinate system,
\begin{pmatrix}d\theta & d\phi\end{pmatrix}\begin{pmatrix}R^2sin^2(\phi) & 0 \\ 0 & R^2\end{pmatrix}\begin{pmatrix}d\theta \\ d\phi\end{pmatrix}

There is no "d\theta d\phi" term, and the matrix is diagonal, precisely because the constant \theta and constant \phi curves are always orthogonal.
 
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I think I followed that, and I do think that is what I was talking about.

Thank you.
 

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