Tensors, invariants and degrees of freedom

[Mark R]

I was recently doing some reading about tensors (specifically as applied to measurement of diffusion tensors) but the same could also applicable to the strain or stress tensor.

In these examples, a 2nd order tensor is a suitable approximation for many purposes. Such a tensor is represented as a 3x3 matrix, and due to its symmetry has 6 degrees of freedom. In practical cases of measurement of the diffusion tensor, 6 is regarded as the minimum number of measurements.

But, are there not, in fact, 5 degrees of freedom only, in these physically representative tensors?

In the case of the diffusion tensor

( Dxx   Dxy   Dxz )
( Dxy   Dyy  Dyz )
( Dxz   Dyz  Dzz )

The first invariant is the trace (Dxx + Dyy + Dzz), which is invariant under rotation. And therefore Dii + Djj + Dzz must sum to the trace for any 3 orthogonal axes (i,j and k).

The diffusional axes of Dxy, Dxz and Dyz, are orthogonal - and therefore must also sum to the trace. So surely, there is additional redundancy in the data - and only 5 measurements are required to acquire the whole 2nd-order tensor.

Am I missing something here? Or is it wrong to assume that Dxy + Dxz + Dyz should sum to the trace?

[For the moment ignore the fact that although the 2nd order diffusion tensor is not adequate to describe commonly measured diffusion patterns - although this would certainly be one reason why in practice measurements of Dxy + Dxz + Dyz wouldn't sum to the trace as measured by Dxx + Dyy + Dzz].

 

[CycloWizard]

I typed up a longer response, but the forum ate it. I don't think there is any reason why Dxy+Dxz+Dyz must sum to the trace. Otherwise, this sum would also be invariant. However, I suspect that this sum depends on the coordinate system, since the values will certainly differ from Cartesian to spherical in a non-additive manner. I've not dealt with anisotropic diffusion before, so you might take this with a grain of salt...

 

[eLiu]

As far as I know, the 3x3 matrix representation of the diffusion tensor (or strain or whatever) does not have any additional properties beyond that of a symmetric (positive definite) matrix. In that case, there is no reason to expect Dxy + Dxz + Dyz = trace.

From a purely linear algebra standpoint, that would indeed be a very odd property for a matrix to have.

 

[Mark R]

Lol. I am an idiot.

I had carelessly got the idea that Dxx, etc. to mean the magnitude of the tensor in that direction. Which is, of course, nonsense. The magnitude in the direction given by the unit vector v is given by:
M = v' D v

Which solves all my confusion, and all makes perfect sense.

 

[eLiu]

Lol. I am an idiot.

I had carelessly got the idea that Dxx, etc. to mean the magnitude of the tensor in that direction. Which is, of course, nonsense. The magnitude in the direction given by the unit vector v is given by:
M = v' D v

Which solves all my confusion, and all makes perfect sense.

I'm not quite sure what you mean by the 'magnitude of a tensor in that direction'. In stress terms, Dxx is the elongational component measured in the x-direction. But it is not necessarily the only x-force that will be felt, since Dxy and Dxz can show up as shear.

I would be hesitant to call M= v' D v (v = e_1 ,unit vector in x-dir) the magnitude of a tensor. In truth I'm not sure how you measure the magnitude of a tensor in a direction; my inclination would be ||Dv||_2 (2-norm of the matrix-vector product). The magnitude of the tensor itself is defined as the frobenius norm of the matrix D. It seems like whatever you choose/define, you'd like the magnitude to be invariant with coordinate rotations; calling "Dxx" a magnitude is problematic b/c it (in general) changes if you rotate your coordinate system.

 

[Mark R]

I would be hesitant to call M= v' D v (v = e_1 ,unit vector in x-dir) the magnitude of a tensor. In truth I'm not sure how you measure the magnitude of a tensor in a direction; my inclination would be ||Dv||_2 (2-norm of the matrix-vector product). The magnitude of the tensor itself is defined as the frobenius norm of the matrix D. It seems like whatever you choose/define, you'd like the magnitude to be invariant with coordinate rotations; calling "Dxx" a magnitude is problematic b/c it (in general) changes if you rotate your coordinate system.

My terminology may not be right.

Essentially, the 2nd order tensor (at least in the case of the diffusion tensor) can be visualised as an ellipsoid. The radius of the ellipsoid in direction v, gives the diffusion coefficient as measured solely in the direction v - let's call it D[size=-3]v[/size]. More precisely, D[size=-3]v[/size] = v' D v. Hence I've called it a 'magnitude' because I don't really know what else to call it.

In the case of the diagonal components of the tensor - the calculations simplifies dramatically. (I've specified v' below solely for the practicality of getting them on one line)
If v' = (1 0 0), then D[size=-3]v[/size] = D[size=-3]xx[/size].
If v' = (0 1 0), then D[size=-3]v[/size] = D[size=-3]yy[/size]
etc.

My mistake was thinking:
if v' = (1/sqrt 2, 1/sqrt 2, 0) then D[size=-3]v[/size] = D[size=-3]xy[/size].
When it isn't. Hence my confusion that you could have 2 measurements of the tensor giving different tensors simply because of a coordinate change - which would be a serious problem.
Recognising that D[size=-3]v[/size] = v' D v. The correct equality is D[size=-3]v[/size] = 1/2 Dxx + Dxy + 1/2 Dyy.

It is then trivial to prove that D[size=-3]i[/size] + D[size=-3]j[/size] + D[size=-3]k[/size] = tr (D) for any orthogonal axes i,j,k.

 

[TecHNooB]

I typed up a longer response, but the forum ate it. I don't think there is any reason why Dxy+Dxz+Dyz must sum to the trace. Otherwise, this sum would also be invariant. However, I suspect that this sum depends on the coordinate system, since the values will certainly differ from Cartesian to spherical in a non-additive manner. I've not dealt with anisotropic diffusion before, so you might take this with a grain of salt...

always tragic

 

[eLiu]

My terminology may not be right.

Essentially, the 2nd order tensor (at least in the case of the diffusion tensor) can be visualised as an ellipsoid. The radius of the ellipsoid in direction v, gives the diffusion coefficient as measured solely in the direction v - let's call it D[size=-3]v[/size]. More precisely, D[size=-3]v[/size] = v' D v. Hence I've called it a 'magnitude' because I don't really know what else to call it.

In the case of the diagonal components of the tensor - the calculations simplifies dramatically. (I've specified v' below solely for the practicality of getting them on one line)
If v' = (1 0 0), then D[size=-3]v[/size] = D[size=-3]xx[/size].
If v' = (0 1 0), then D[size=-3]v[/size] = D[size=-3]yy[/size]
etc.

My mistake was thinking:
if v' = (1/sqrt 2, 1/sqrt 2, 0) then D[size=-3]v[/size] = D[size=-3]xy[/size].
When it isn't. Hence my confusion that you could have 2 measurements of the tensor giving different tensors simply because of a coordinate change - which would be a serious problem.
Recognising that D[size=-3]v[/size] = v' D v. The correct equality is D[size=-3]v[/size] = 1/2 Dxx + Dxy + 1/2 Dyy.

It is then trivial to prove that D[size=-3]i[/size] + D[size=-3]j[/size] + D[size=-3]k[/size] = tr (D) for any orthogonal axes i,j,k.

Oooooh, ok. Yeah I think we grew up with different terminology, but I understand what you're getting at now. Cool; glad everything got cleared up.

You can visualize any symmetric matrix as an ellipse (and any general mxn matrix as a mapping from a sphere in R^n to an ellipse in R^m. The characterizing decomposition is the singular value decomposition (SVD), A = UEV^T, U and V are orthogonal matrices, called the left & right singular vectors respectively. E is a diagonal matrix of the singular values, which are defined to be positive. (For A=A^T, U=V so this is same as the eigenvalue decomposition*.) A mxn matrix maps from n-space to m-space; so you can think of V as an unit orthogonal basis (a hyper-sphere) for n-space. E provides elongation/contractions & U re-expresses this shape (now a hyper-ellipse) in m-space.

*Not quite true: if A is positive (semi) definite, then it's true. If not, then you may need to flip a few basis vectors to remove the minus signs on the eigenvalues to make it a valid SVD.