Curl of Tensor


by touqra
Tags: curl, tensor
touqra
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#1
Sep25-06, 11:12 AM
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Can we curl a stress tensor? What physically meaning will it be?
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chroot
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#2
Sep25-06, 03:41 PM
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The term "curl" usually applies to vector fields. If there is an equivalent definition of curl for tensor fields, I am not familair with it.

- Warren
robphy
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#3
Sep25-06, 03:44 PM
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Are you referring to an operation like
[tex]\nabla_{[a}T_{b]c[/tex]?

touqra
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#4
Sep25-06, 06:47 PM
P: 284

Curl of Tensor


I was thinking of something like Helmholtz's theorem, where if you specify the div and curl of a vector field, you then know everything there is to know about the field.
Maybe there's something similar for rank 2 tensor, like the stress tensor, or higher tensors.
Thrice
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#5
Oct3-06, 05:41 PM
P: 232
Quote Quote by chroot
The term "curl" usually applies to vector fields. If there is an equivalent definition of curl for tensor fields, I am not familair with it.

- Warren
Incidentally, the defn of curl resembles the antisymmetrized derivative (F in electromagnetism & the curvature tensor in GR). That's not accidental, is it?
nike^^
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#6
Nov17-06, 12:09 PM
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Hi

i think that the rotor (curl) of a bilinear tensor T can be defined as follows:

let [T] be the matrix associated with T :

t11 t12 t13
[T] = t21 t22 t23
t31 t32 t33

interpreting the raws of [T] as vectors

T1=t11*e1+t12*e2+t13*e3 => [T1]'=(t11,t12,t13)
T2=t21*e1+t22*e2+t23*e3 => [T2]'=(t21,t22,t23)
T3=t31*e3+t32*e2+t33*e3 => [T3]'=(t31,t32,t33)

we can write [T] as

[T1]'
[T] = ( [T2]' )
[T3]'

then, the rotor (curl) of T is simply :


[rotT] = ( [rotT1] , [rotT2] , [rotT3] )

where [rotT1] , [rotT2] , [rotT3] are the column matrix of the rotor (curl) of vectors T1, T2 and T3
Norman Albers
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#7
May8-07, 08:14 AM
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Quote Quote by Thrice View Post
Incidentally, the defn of curl resembles the antisymmetrized derivative (F in electromagnetism & the curvature tensor in GR). That's not accidental, is it?
Indeed, Thrice, this is not accidental. I am learning much of the nature of axial vectors, curl, and the Minkowski tensor. I need to understand the forms expressed in spherical terms and fields for magnetism.
Chris Hillman
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#8
May17-07, 08:02 PM
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Quote Quote by touqra View Post
Can we curl a [second rank] tensor?
There are various things one could mean by this, but yes, there are various ways of generalizing the "curl" from vector calculus. In the context of gtr, one particularly useful formalism involves "curl" and "div" operations on hyperslices using the induced connection in the slice.

Quote Quote by nike^^ View Post
i think that the rotor (curl) of a bilinear tensor T can be defined as follows:
You should ask yourself if your proposed operation yields a tensor.

Quote Quote by touqra View Post
I was thinking of something like Helmholtz's theorem, where if you specify the div and curl of a vector field, you then know everything there is to know about the field.
Maybe there's something similar for rank 2 tensor, like the stress tensor, or higher tensors.
Yes indeed, the Hodge decomposition, which applies to p-forms, generalizes the Helmholtz decomposition. This can be stated in various ways: one statement is that any exterior form on a compact boundaryless Riemannian manifold can be uniquely decomposed (as an orthogonal direct sum) as the sum of an exact form, a coexact form, and a harmonic form: [itex]\beta = d\alpha + \delta \gamma + \eta[/itex], where [itex]\alpha[/itex] is a coclosed (p-1)-form, [itex]\gamma[/itex] is a closed (p+1)-form, and [itex]\eta[/itex] is a harmonic p-form (thus, both closed and coclosed).

Undergraduate courses which mention the Helmholtz decomposition typically omit mention of the harmonic term by adding the additional assumption that the form to be decomposed asymptotically vanishes far from the origin (of R^3), whence by Liouville's theorem we expect the harmonic form appearing in the decomposition to vanish, which it does. See for example Frankel, Geometry of Physics.

Exterior forms are anti-symmetric tensors, so this won't apply directly to a symmetric tensor. Also, Hodge theory works best in Riemannian manifolds, not Lorentzian manifolds.


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