# Curl of Tensor

1. Sep 25, 2006

### touqra

Can we curl a stress tensor? What physically meaning will it be?

2. Sep 25, 2006

### chroot

Staff Emeritus
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

3. Sep 25, 2006

### robphy

Are you referring to an operation like
$$\nabla_{[a}T_{b]c$$?

4. Sep 25, 2006

### touqra

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.

5. Oct 3, 2006

### Thrice

Incidentally, the defn of curl resembles the antisymmetrized derivative (F in electromagnetism & the curvature tensor in GR). That's not accidental, is it?

6. Nov 17, 2006

### nike^^

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

7. May 8, 2007

### Norman Albers

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.

8. May 17, 2007

### Chris Hillman

Curl and all that

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.

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: $\beta = d\alpha + \delta \gamma + \eta$, where $\alpha$ is a coclosed (p-1)-form, $\gamma$ is a closed (p+1)-form, and $\eta$ 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.

Last edited: May 17, 2007