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

Integrating a 2-form on a 2-cube

  1. Aug 18, 2008 #1

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I read, again in Spivak's Calculus on Manifolds, that the integral of 1-form over a 1-cube is equivalent to a line integral. And indeed, if I consider the 1-form w = Pdx + Qdy on R², and c a given 1-cube in R², I find that

    [tex]\int_c\omega = \int_0^1 F(c(t))\cdot c'(t)dt[/tex]

    where F=(P,Q), which is the integral of the vector field F along the curve c.

    Next, I read that the integral of a 2-form over a 2-cube is equivalent to a surface integral. Let [itex]\omega = Pdx\wedge dy + Qdx\wedge dz + Rdy\wedge dz[/itex] be a 2-form onR³ and c be a 2-cube in R³. I am guessing that [itex]\int_c\omega[/itex] corresponds to the flux integral of the vector field F = (P,Q,R) through c([0,1]²). Recall that this integral is given by

    [tex]\iint_{[0,1]^2}F(c(s,t))\cdot \left(\frac{\partial c}{\partial s}\times \frac{\partial c}{\partial t}\right) dsdt=\iint_{[0,1]^2}P(c(s,t))\left| \begin{array}{cc}
    \frac{\partial c^2}{\partial s} & \frac{\partial c^3}{\partial s} \\
    \frac{\partial c^2}{\partial t} & \frac{\partial c^3}{\partial t}
    \end{array} \right| - Q(c(s,t))\left| \begin{array}{cc}
    \frac{\partial c^1}{\partial s} & \frac{\partial c^3}{\partial s} \\
    \frac{\partial c^1}{\partial t} & \frac{\partial c^3}{\partial t}
    \end{array} \right| + R(c(s,t))\left| \begin{array}{cc}
    \frac{\partial c^1}{\partial s} & \frac{\partial c^2}{\partial s} \\
    \frac{\partial c^1}{\partial t} & \frac{\partial c^2}{\partial t}
    \end{array} \right| \ dsdt[/tex]

    Alright, so I get something close to this, but not quite, so I'm wondering where I'm mistaken... I say,

    [tex]\int_c\omega = \int_{[0,1]^2}(P\ \circ \ c) c^*(dx\wedge dy)+(Q\ \circ \ c) c^*(dx\wedge dz)+(R\ \circ \ c) c^*(dy\wedge dz)[/tex]

    Now, [itex]c^*(dx^i\wedge dx^j))[/itex] is a 2-form on [0,1]² so it is of the form [itex]f_{ij} \ ds\wedge dt[/itex] for some 0-form f_ij on [0,1]². In order to find f_ij, let p belong to [0,1]² and [itex]v_{p}, \ w_{p} \in \mathbb{R}^2_{p}[/itex].

    Using the definitions, I find that

    [tex]c^*(dx^i\wedge dx^j)(v_p,w_p)=dx^i\wedge dx^j(c_*(v_p),c_*(w_p))=dx^i\otimes dx^j(c_*(v_p),c_*(w_p))-dx^i\otimes dx^j(c_*(w_p),c_*(v_p))[/tex]

    where [itex]c_*(v_p)=(Dc(p)(v))_{c(p)}[/itex], and therefor

    [tex]dx^i(c_*(v_p))=\frac{\partial c^i}{\partial s}(p)v^1+\frac{\partial c^i}{\partial t}(p)v^2=\left(\frac{\partial c^i}{\partial s}(p)ds+\frac{\partial c^i}{\partial t}(p)dt\right)(v_p)[/tex]

    So I conclude that

    [tex]c^*(dx^i\wedge dx^j)(v_p,w_p)=\left(\frac{\partial c^i}{\partial s}(p)dx+\frac{\partial c^i}{\partial t}(p)dt\right)\wedge \left(\frac{\partial c^j}{\partial s}(p)dx+\frac{\partial c^j}{\partial t}(p)dt\right)=\left(\frac{\partial c^i}{\partial s}(p)\frac{\partial c^j}{\partial t}(p)-\frac{\partial c^i}{\partial t}(p)\frac{\partial c^j}{\partial s}(p)\right) ds\wedge dt(v_p,w_p)[/tex]

    and hence

    [tex]f_{ij}=\frac{\partial c^i}{\partial s}\frac{\partial c^j}{\partial t}-\frac{\partial c^i}{\partial t}\frac{\partial c^j}{\partial s}=\left| \begin{array}{cc}
    \frac{\partial c^i}{\partial s} & \frac{\partial c^j}{\partial s} \\
    \frac{\partial c^i}{\partial t} & \frac{\partial c^j}{\partial t}
    \end{array} \right|[/tex]

    Finally,

    [tex]\int_c\omega = \iint_{[0,1]^2}P(c(s,t))\left| \begin{array}{cc}
    \frac{\partial c^1}{\partial s} & \frac{\partial c^2}{\partial s} \\
    \frac{\partial c^1}{\partial t} & \frac{\partial c^2}{\partial t}
    \end{array} \right| + Q(c(s,t))\left| \begin{array}{cc}
    \frac{\partial c^1}{\partial s} & \frac{\partial c^3}{\partial s} \\
    \frac{\partial c^1}{\partial t} & \frac{\partial c^3}{\partial t}
    \end{array} \right| + R(c(s,t))\left| \begin{array}{cc}
    \frac{\partial c^2}{\partial s} & \frac{\partial c^3}{\partial s} \\
    \frac{\partial c^2}{\partial t} & \frac{\partial c^3}{\partial t}
    \end{array} \right| \ dsdt[/tex]

    So, 3 things differ: the factors of P and R and the sign of Q.

    Thanks!
     
  2. jcsd
  3. Aug 19, 2008 #2
    I'm not seeing any problem here. Modulo any sign errors that I didn't pick up, what you've shown is that the integral of a 2-form on a 2-chain in 3-space [itex]\int_c\omega[/itex] is the flux integral of the vector field F=(R,-Q,P), where P, Q and R are the coordinate components of the 2-form.

    Integrating forms on chains is a generalization of curve and surface integration, but it's not necessarily a direct correspondence. Some adjustment of the various components usually has to be performed.

    Incidentally, calculus on manifolds is not Spivak's invention. His goal with this book was to provide a short, problem-based introduction to the subject. A great companion book to his is Loomis and Sternberg's Advanced Calculus, which you can get for free here: http://www.math.harvard.edu/~shlomo/.
     
  4. Aug 19, 2008 #3

    quasar987

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Ok, sweet! And thanks for the book link.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Integrating a 2-form on a 2-cube
Loading...