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Vector calculus in higher dimensions

  1. Jun 7, 2009 #1
    in 2d, "curl"(grad(f)) = 0 "curl" is the operation that green's tehorem talks about.

    In 3D, curl(grad(f)) = 0 and div(curl(F)) = 0.

    We may consider vector calculus in 4 spatial dimensions, for vector fields F:R^4 -> R^4. what is "curl" like in 4D, since curl is actually only difined in 3D. I think there would be no curl in 4D because there's no cross product in 4D. instead there would be 2 operators related by Stokes's theorem for general manifolds.
    my conjecture is that in 4 dimensions, some other operator D2 exists such that div(D2(F)) = 0. Also there exists another operator D1 such that D2(D1(f)) = 0 and D1(gradient(f))=0. Div would still be from an operator from a vector field to a scalar field.
    the divergence theorem would be relating 4-volumes to 3-volumes(boundary of 4-volume)

    Do they exist? what is the nature of such an operator (in cartesian) and how can i visualize it?
     
    Last edited: Jun 7, 2009
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  3. Jun 7, 2009 #2

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    The generalization of vector calculus to general higher dimensional manifolds is the calculus of differential forms. Curl, div, grad all become special cases of a single operator called the 'exterior derivative' d.
     
  4. Jun 7, 2009 #3
    yes, all differential operators are special cases of exterior derivative, but in 4 or higher dimensions would they still take the general sequence of gradient:{f: R^4 -> R} -> {f:R^4 -> R^4}, ... then some other differentail operators ..., then divergence: {f:R^4 -> R^4} -> {f:R^4 -> R} ? or would some of the operators split in a different way? (an analogy for lower dimensions is how div and curl are actually the same in 2D, but they become different operators in 3D.)
     
  5. Jun 7, 2009 #4

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    d acting on a function (which gives a 1-form) is the analog of gradient, d acting on a 1-form (which gives a 2-form) is the analog of curl and d acting on a 2-form (which gives a 3-form) is the analog of divergence.

    In general, d takes a p-form to a (p+1)-form.

    div and curl are not the same in 2D.
     
    Last edited: Jun 7, 2009
  6. Jun 7, 2009 #5
    oh, i meant that both div and curl can be described by green's theorem in 2D:
    div:
    [tex]\int_C -M \,dx + L\, dy = \iint_{D} \left(\frac{\partial L}{\partial x} + \frac{\partial M}{\partial y}\right)\, dA = \iint_D\left(\nabla\cdot\mathbf{F}\right)dA[/tex]

    curl:
    [tex]\int_{C} L\, dx + M\, dy = \iint_{D} \left(\frac{\partial M}{\partial x} - \frac{\partial L}{\partial y}\right)\, dA = \iint_D |\nabla\times\mathbf{F}| dA[/tex]
     
    Last edited: Jun 7, 2009
  7. Jun 7, 2009 #6

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    Green's theorem involves the curl. I'm not sure what you mean by "both div and curl can be described by green's theorem". All the integral theorems of vector calculus are special cases of the general stokes theorem

    [tex] \int_{\Omega} d\omega = \oint_{\partial \Omega} \omega [/tex]
     
    Last edited: Jun 7, 2009
  8. Jun 7, 2009 #7
    im being a little sloppy here... you rotate L and M (L,M) -> (-M,L) so that the tangent vectors become normal to the boundary, so that (L dx+ M dy) -> (-M dx + L dy). plug -M into L, and plug L into M, in to the green's tehorem, and you will get the theorem for 2D divergence.
    we got wrapped up in (metaphorical) tangents, now could you answer my question


    a little clarification on opening post:
    in 1D there is only one integral theorem, the fundamental theorem of calculus. (special case of line integral theorem)
    in 2D there are the line integral theorem and Green's theorem
    in 3D there are the line integral theorem and the kelvin-stokes (curl) theorem and gauss's divergence theorem
    in 4D: line integral theorem (true for all dimensions) and ????

    do not tell me that they are special cases of stokes's theorem. i already know that, thank you very much. i want the specific formulas for the 4D integral theorems and the 4 dimensional vector calculus operations corresponding to them.
    template:
    "fundamental theorem of calculus(derivative): [tex]\int _a ^b f'(x) dx = f(b)-f(a)[/tex]
     
    Last edited: Jun 7, 2009
  9. Jun 16, 2009 #8
    in 4D: line integral theorem (true for all dimensions) and ????

    do not tell me that they are special cases of stokes's theorem. i already know that, thank you very much. i want the specific formulas for the 4D integral theorems and the 4 dimensional vector calculus operations corresponding to them.

    do I really have to repeat myself to be heard??

    I should have titled this "vector calculus in 4D"
     
  10. Apr 17, 2011 #9
    Hmmmm... I thought about it for a while and came up with some integral calculus formulas for arc length, surface area, hyperarea, etc
    ∭_R▒〖√(1+〖(∂t/∂x)〗^█(2@) 〖+(∂t/∂z)〗^2+(∂t/∂y)^2 ) dV〗
    That is sa in 4 dimensions
     
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