Wikipedia:Reference desk/Archives/Mathematics/2014 August 14
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August 14
[edit]Parallel curves in geometric algebra?
[edit]Is there a way to represent the problem of parallel curves in geometric algebra? JMP EAX (talk) 22:59, 14 August 2014 (UTC)
- I haven't gone through the exercise, but intuitively seems straightforward: start with a curve in a two-dimensional space described as a position vector function of some scalar parameter. Finding the tangent vectors of the curve requires differentiation of the vector function. Aside from that, normalization, rotation and scaling are straightforward operations in geometric algebra, and these can be used to find a parallel curve. Maybe not quite what you asked. It would be interesting to determine a curve is always parallel to its parallel – not answered by the article as far as I can see. —Quondum 01:59, 16 August 2014 (UTC)
- Yeah, the differentiation (tangent) is probably the non-straightforward part. There's a (relatively recent) book that probably covers that, but I haven't read it: John Snygg (2011). A New Approach to Differential Geometry using Clifford's Geometric Algebra. Springer. ISBN 978-0-8176-8282-8. JMP EAX (talk) 20:22, 17 August 2014 (UTC)
- The differentiation is straightforward: it is the derivative with respect to the single parameter. No partial derivatives needed; no different from normal vector calculus. Snygg seems to cover the concept in §7.1 as in a sense a velocity along the curve. If the position vector is expressed in two components, the components of the tangent vector can be found as the derivative of each component with respect to the parameter. —Quondum 05:52, 18 August 2014 (UTC)
- Well, it's not that trivial in general (greater than 2 dimensions), but the Frenet–Serret formulas can be expressed in Clifford/geometric algebra. JMP EAX (talk) 12:04, 26 August 2014 (UTC)
- The differentiation is straightforward: it is the derivative with respect to the single parameter. No partial derivatives needed; no different from normal vector calculus. Snygg seems to cover the concept in §7.1 as in a sense a velocity along the curve. If the position vector is expressed in two components, the components of the tangent vector can be found as the derivative of each component with respect to the parameter. —Quondum 05:52, 18 August 2014 (UTC)
- Yeah, the differentiation (tangent) is probably the non-straightforward part. There's a (relatively recent) book that probably covers that, but I haven't read it: John Snygg (2011). A New Approach to Differential Geometry using Clifford's Geometric Algebra. Springer. ISBN 978-0-8176-8282-8. JMP EAX (talk) 20:22, 17 August 2014 (UTC)