## You are here

# Derivative of Metric Tensor

Hi All,

How can we compute the derivative of metric tensor on one manifold with respect to metric tensor on another menifold?

Regards

Mohsen

»

- M. Jahanshahi's blog
- Log in or register to post comments
- 4992 reads

## Comments

## Derivative of Metric Tensor

Hi Mohsen,

I see you are a great fan of forth-order tensors :) The computation of the derivative of the square root is a very tough thing, isn't it?

The question is again of course too general. I suppose you have two distinct configurations of the same body and want to compute some "geometrical" stiffness tensor. Unfortunately, I don't see any possibility how to do it very shortly. Please, consult the following works:

J.E.Marsden, T.J.R.Hughes Mathematical foundations of elasticity. 1983

J.C.Simo, J.E.Marsden On the rotated stress tensor and the material version of the Doyle-Ericksen formula. 1984

M.Itskov The derivative with respect to a tensor: some theoretical aspects and applications. 2002

O.Kintzel Fourth-order tensors - tensor differentiation with application to continuum mechanics. Part II: Tensor analysis on manifolds. 2005

M.Itskov Tensor algebra and tensor analysis for engineers. 2009

Although I didn't find there exactly what you need, I think these works will be of some interest for you, especially those of Jerry Marsden and co-authors. In the works of Itskov you will find statements like "the derivative with respect to the metric tensor seems to be meaningless since in the Euclidean space

grepresents the identity tensor", but I wouldn't take them too seriously. Yes, it's true in each tangent space the metric tensor should act as the identity map, but a-priori it is not clear how the information about the global structure (the curvature of the manifold) may influence the derivative.Svyatoslav

## Derivative of the Square Root of a Tensor

Dear Svyatoslav,

Thanks for your interest in my questions. Actually the derivative of the square root of a symmetric tensor (in our case the derivative of

Uwith respect foC) is not all that hard. For example the following references can be used to do so:1. J.C. Simo, R.L. Taylor, Quasi-incompressible Finite Elasticity in Principal Stretches, Computer Methods in Applied Mechanics and Engineering, 85 (1991) 273-310.

2. C. Miehe, Computation of Isotropic Tensor Functions, Communications in Numerical Methods in Engineering, 9 (1993) 889-896.

About this question, I should say that it is closely related to my previous question. Let me be more specific about it. We have two spatial configurations. The pull back of derivatives with respect to metric tensors on these configurations leads to derivatives with respect to

C(two differentC's) in reference configuration. Now it is important to be able to compute the derivative of oneCwith respect to the other and it is in this derivation that the derivative ofFwith respect toCarises because there is a rather simple relation between theF's corresponding to differentC's.As we have already discussed I can compute the derivative of

Fwith respect toC. But I am not completely sure ofRbeing independent ofC.Mohsen

## Derivative of the Square Root of a Tensor

Dear Mohsen,

I'm involved as well into such things like multiplicative elasto-plasticity, nonlinear incompatibility etc. and this often led me to questions very similar to yours. Yes, of course, the use of the spectral decomposition is very natural, but the fact that we should deal somehow with Cardano's formulae has always stopped me. Ok, there is a lot of numerical algorithms for computing eigenvalues, but I think this is also very doubtful tool.

The question of mutual (in-)dependence of

CandRis very interesting. The one thing is clear -Cis independent ofR. Is this enough to postulate avice versatheorem? I have a feeling that it has something to with the frame indifference, but from the other side I'm quite sure that this is the fact of pure analysis on manifolds.Svyatoslav

## Cardano's Formula

Dear Svyatoslav,

Regarding the Literature, we are always able to find the derivative of eigenvalues and eigenprojections of a given second order tensor with respect to the tensor itself even in the case of multiple eigenvalues. So what makes the tool very doubtful?

Mohsen

## A relationship between R field and C field

Dear Mohsen and Svyatoslav,

One situation in which a relationship like the R field being a 'function' of the C field arises is when, say in a simply-connected open set around a point x of a reference, one considers a smooth deformation of the open set and the corresponding R and C fields of the resulting deformation gradient. In this case, the C *field* determines the R* field* up to a spatially uniform orthogonal tensor field. So, if one addtionally requires the value of the R field to be specified at a given point in the open set, then this situation results in the R field being uniquely determined by the C field (but only in this compatible case).

But even in this situation, the value of the constructed R field at a given point can only be considered as a function on the constrained space (perhaps an infinite dimensional manifold) of symmetric +ve definite tensor *fields* that satisfy the finite-strain compatibility conditions on the open set. So, at best, one can think of doing a derivative of a function on an infinite dimensional space, which would be a very complicated animal - the 'function' in this case is essentially a second order, nonlinear, differential operator.

I would be very surprised if someone could show me a function A that relates R(X) = A( C(X) ), where A is fixed over all X in some open ball and for all deformations of the open ball whose deformation gradient F pointwise yields the polar decomposition F(X) = R(X)U(X), where as usual U(X) is the tensor quare root of C(X).

May be this is of help in your discussion.

- Amit

## A relationship between R field and C field

Dear Professor Acharya,

Yes, you are completely right. The funny thing that I have already treated in principle that question a couple of times in the recent papers related to general incompatibility in shells and disclination structures. Please, see

S.V.Derezin, L.M.Zubov Disclinations in nonlinear elasticity. 2011

S.V.Derezin Gauss-Codazzi equations for thin films and nanotubes containing defects. 2011

The problem of finding

FfromCcan be clearly reduced to a certain Pfaffian system and the solvability of such a system is equal to the zero-curvature compatibility (Frobenius theorem).The solution in 3D is very complicated and usually involves path-ordered exponentials (matrizants or multiplicative integrals) . The same is true for the rotation tensor from the polar decomposition of

Fand it was already shown for example in the paper of ShieldR.T.Shield The rotation associated with large strains. 1973

Yes, I remember a long discussion here on imechanica related to the strain compatibility and the paper of Janet Blume. Now I understand which kind of multivaluedness you meant previous time. If the strain compatibility doesn't hold almost everywhere then the rotation tensor is a multi-valued function of strain.

Thanks a lot for your explanation.

Svyatoslav

## Re: A relationship between R field and C field

Dear Svyatoslav:

Could you send me your paper "S.V.Derezin Gauss-Codazzi equations for thin films and nanotubes containing defects. 2011"?

Thanks,

Arash

## Re: A relationship between R field and C field

Dear Professor Yavari,

Please, check your mailbox. I've tried to insert hyperlinks in my previous reply, but I don't see them at all. May be, it is not allowed to do this in comments, I don't know.

Svyatoslav