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4 dimensional curl.

  1. May 7, 2012 #1
    I would just like to know what the 4 dimensional curl is?

    I believe it is a matrix but I am not completely sure.

    Thanks for any help!!
  2. jcsd
  3. May 10, 2012 #2


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    There is nothing specifically called a "4 dimensional curl" but you might be interested in a google search on "exterior algebra" or "Grassmann algebra". This notation provides the proper mathematical generalization of various types of vector products in any number of dimensions. Beware that the situation becomes more complicated and there are many more types of products as you increase the number of dimensions of the space you're working in.

    From these algebraic structures you can also construct various types of derivatives in n-space. You can also construct differential forms which allow various types of integration that generalize the ideas of line and surface integrals from ordinary 3 dimensional vector calculus.

    The first chapter of "Differential Forms and Connections" by R.W.R. Darling covers this topic. The text is quite mathematical though and may not be suitable if you don't have some background in basic linear algebra and real analysis and feel comfortable with higher levels of mathematical abstraction. I enjoyed the text but found myself scratching my head quite a bit and it took a lot of work to get through it and feel satisfied.

    Perhaps others have recommendations.
    Last edited: May 10, 2012
  4. May 13, 2012 #3


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    Also want to point out any generalization of curl to 4 dimensions leads to a 6 dimensional value. But I'll first explain 3 dimensional curl.

    Three dimensional curl can be geometrically interpreted as a rate of rotation induced by a vector field. It can be described using a 3 by 3 matrix representing the "induced rotation" in the plane generated by each pair of axis (x, y and z). However, this matrix must be anti-symmetric. A pair of like axis doesn't generate a plane and thus the diagonal elements (x-x, y-y, and z-z) aren't used. Also any rotation in the the x-y plane is just the negative of the same rotation in the y-x plane (same goes for y-z vs z-y and z-x vs x-z). Therefore there are really only three planes needed to completely describe a rotation in 3-space, namely x-y, y-z, and z-x.

    It turns out in three dimensions each plane has exactly one normal vector, so each plane rotation can be described as a rotation about a given axis using the right hand rule. However this doesn't hold in 4 dimensions because there is no longer a single normal vector for each plane. Instead there is an entire plane of vectors perpendicular to any given plane. Therefore the common notion of "rotation about an axis" no longer makes any sense. Vector fields in dimensions other than 3 do not have a vector form of curl.

    Instead the rotation must be described from an anti-symmetric 4 by 4 matrix, which turns out to have 6 linearly independent elements. In the notation of differential forms the curl of a 4 dimensional vector field is not a vector but a bi-vector. The 6 differential rotation components are 2 dimensional objects (dx^dy, dy^dz, dz^dw, dw^dx, dx^dz, and dy^dw).

    So in general, the n-dimensional "curl" will be bi-vector with n(n-1)/2 linearly independent components. It just so happens that when n is 3 we get 3*(3-1)/2 = 3 so there is an isomorphism between the bi-vector form and the vector form. Vector fields in [itex]\mathbb{R}^n[/itex] with n something other than 3 do not have a vector form curl, but all have a bi-vector form curl.
    Last edited: May 13, 2012
  5. May 13, 2012 #4


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    In the context of physics (relativity theory), the 4D curl of a 4vector A is a second rank antisymmetric tensor F whose double covariant components (6 of them being independent) in a coordinate basis are

    [tex]F_{\mu\nu} = \nabla_{\mu}A_{\nu} - \nabla_{\nu}A_{\mu} [/tex]

    Standard stuff.
  6. May 13, 2012 #5


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    in differential form notation, I believe "curl" is a differentiation operator on (co)vector fields, i.e. it is d applied to a one form,

    specifically it is d(pdx + qdy + rdz) = dp^dx + dq^dy + dr^dz

    = (∂p/∂x dx + ∂q/∂y dy + ∂r/∂z dz)^dx +.....obvious stuff, and since dx^dx = 0, we get

    = ∂q/∂y dy^dx + ∂r/∂z dz^dx + similar stuff.....

    this operator can be applied to any degree forms in any dimension in the exact same way. you could call any of them higher dimensional curls. you only have to know dx^dx = dy^dy = dz^dz = dw^dw = 0...

    and dx^dy = -dy^dx,.....
  7. May 13, 2012 #6


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    I'm not familiar with that notation. Is it the same as the anti-symmetric tensor...

    [itex]\dfrac{\partial A_i}{\partial x_j} - \dfrac{\partial A_j}{\partial x_i}[/itex]

    This is stuff I'm interested in learning at some point but haven't had the time.
  8. May 14, 2012 #7


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    For General Relativity (no torsion and coordinate basis), yes

    [tex] \nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu} = \partial_{\mu}A_{\nu} - \partial_{\nu}A_{\mu} [/tex]

    , so it doesn't matter is the curvature is present or not.
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