Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Line integrals of differential forms

  1. Oct 29, 2016 #1
    Consider a curve ##C:{\bf{x}}={\bf{F}}(t)##, for ##a\leq t \leq b##, in ##\mathbb{R}^{3}## (with ##x## any coordinates). oriented so that ##\displaystyle{\frac{d}{dt}}## defines the positive orientation in ##U=\mathbb{R}^{1}##. If ##\alpha^{1}=a_{1}dx^{1}+a_{2}dx^{2}+a_{3}dx^{3}## is a ##1##-form on ##\mathbb{R}^{3}##, then its integral or line integral over ##C## becomes

    ##\displaystyle{\int_{C}\ \alpha^{1} = \int_{C}\ \sum\limits_{i}a_{i}(x)dx^{i}}##

    ##\displaystyle{=\int_{a}^{b} F^{*}\bigg[\sum\limits_{i}a_{i}(x)dx^{i}\bigg]}##

    ##\displaystyle{=\int_{a}^{b} F^{*}\bigg[\sum\limits_{i}a_{i}(x(t))\frac{dx^{i}}{dt}\bigg]dt}##

    Thus ##\displaystyle{\int_{(U,o;F)}\alpha^{p}:=\int_{(U,o)}F^{*}\alpha^{p}}## is the usual rule for evaluating a line integral over an oriented parameterized curve! We may write this as

    ##\displaystyle{\int_{C}\alpha^{1}=\int_{a}^{b}\alpha^{1}\left(\frac{d{\bf{x}}}{dt}\right)dt}##

    and so the integral of a ##1##-form over an oriented parameterized curve ##C## is simply the ordinary integral of the function that assigns to the parameter ##t## the value of the ##1##-form on the velocity vector at ##{\bf{x}}(t)##.





    What does it mean for ##\displaystyle{\frac{d}{dt}}## to define the positive orientation in ##U=\mathbb{R}^{1}##
     
  2. jcsd
  3. Oct 29, 2016 #2

    Orodruin

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    ##d/dt## is a vector on the real line.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Line integrals of differential forms
  1. Differential form (Replies: 3)

  2. Differential forms (Replies: 8)

Loading...