# Transformation matrixes and tensors

1. May 2, 2013

### joda80

Hi All,

I have a question about transformation matrices (sorry about the typo in the title). The background is that I've spent some time learning differential geometry in the context of continuum mechanics and general relativity, but I'm unable to connect some of the concepts.

So I have this book about general relativity (by T. Fliessbach), where it says that the transformation matrices of the local diffeomorphisms considered in general relativity are no tensors (justifying writing the indices above each other, rather than 'from northwest to southeast'). So far, so good.

But in continuum mechanics, there is a quantity called the "deformation-gradient tensor", which maps e.g., a material line segment from the initial configuration to a later configuration. Interpreting this map as a passive transformation (change of basis), this tensor is, in my understanding, precisely the transformation matrix from material coordinates to spatial coordinates (material coordinates are initially Cartesian but follow along with the flow and thus become curvilinear). So if the material coordinates are denoted by $a^{\nu}$ and the fixed spatial (Cartesian) coordinates by $x^{i}$, then a material line segment transforms like this:
$$d{x}^{i} =\frac{\partial x^{i}}{\partial a^{\nu}} da^{\nu},$$
where the transformation matrix is just the deformation gradient.

The fact that in continuum mechanics, this general transformation matrix (from arbitrary curvilinear coordinates to Cartesian coordinates) is a tensor, but presumably not so in general relativity, confuses me.

In Marsden and Hughes' textbook (mathematical foundations of elasticity), it is stressed that the deformation gradient,
$$F^{i}_{\cdot \nu} = \frac{\partial \Phi^{i}}{\partial a^{\nu}}$$
does not involve the covariant derivative because $\Phi$ is not a vector but a point mapping. Could this have something to do something with it?

I hope the problem is somewhat understandable.

Johannes

Last edited: May 2, 2013
2. May 2, 2013

### micromass

You should write "/" instead of "\" in your tex-brackets.

3. May 2, 2013

### WannabeNewton

What do you mean by transformation matrix of the local diffeomorphism? If you mean the matrix representation of the pushforward $F_{*}:T_{p}(M)\rightarrow T_{F(p)}(N)$ of a local diffeomorphism $F:M\rightarrow N$ where $M,N$ are smooth manifolds then $F_{*}$ is linear sure but to be a tensor it has to map to $\mathbb{R}$. So in particular if $F$ is just a scalar valued function $f:M\rightarrow\mathbb{R}$ then $f_{*}:T_{p}(M)\rightarrow \mathbb{R}$ will just be the differential of $f$ at $p$ denoted $df(p)$ and is trivially a tensor as it is just a linear functional on $T_{p}(M)$ (when defined on an open subset of $M$ this is in fact a one-form).

Last edited: May 2, 2013
4. May 2, 2013

### joda80

@Micromass: Thanks, now it works!

@Newton: Indeed, this is what I meant. What you say makes sense (also consistent with Wald's General Relativity, p. 437), and I already suspected that the statement in Fliessbach's textbook might be wrong, but wanted to make sure.

If you're not bored yet, could you comment on the "position vector" not really being a vector? I think I can see that a point on a manifold (like an event in spacetime) does not need coordinate systems/tangent spaces in order to exist, but once we define these properties, isn't the location an honest vector? [And why is the deformation gradient formulated in terms of the partial, rather than covariant derivative? Isn't the gradient a classical example of a non-tensor, usually motivating the introduction of Christoffel symbols and the covariant derivative?]

As always: Thanks!

5. May 2, 2013

### WannabeNewton

No because the concept of a position vector requires you to be able to move vectors from one point to another (the "origin" or "reference point") freely without changing the vector at all but on curved manifolds this is not possible in general. There is also no unambiguous way to transport a vector from one tangent space to another.

I'm assuming $\Phi^{i}:\mathbb{R}^{n}\rightarrow \mathbb{R}$ in which case the covariant derivative operator $\nabla_{a}$ reduces to the regular partial derivative $\partial_{a}$ because $\mathbb{R}^{n}$ is flat.

6. May 4, 2013

### Staff: Mentor

The problem is one of terminology. The term deformation gradient tensor seems to imply that a gradient is involved. It is an unfortunate choice of terminology. There is no gradient involved in the sense you are accustomed to.

You are correct to say that the deformation gradient tensor is a mapping, which maps a differential position vector joining two neighboring material points at time zero into the differential position vector joining these same two material points at time t. It is not necessary to express either of the two differential position vectors in terms of cartesian coordinates. Any coordinate system (or systems) is adequate. But, one of the problems is that the differential position vector at time t is not situated at the same location in space as the differential position vector at time zero. This is OK if you are working exclusively with cartesian coordinates, but, if you are either using two different coordinate systems or a curvilinear coordinate system, the coordinate basis vectors at the location at time zero are not the same as the coordinate basis vectors at the location at time t. So you are stuck simultaneously working with two sets of coordinate basis vectors.

There is a simple mathematical way of dealing with all this if you are interested in my continuing with this discussion. But, in terms of your original question, I think that that is now adequately addressed.

Chet

7. May 7, 2013

### joda80

Hi Chet,

Thanks for your comments. I certainly am interested in more discussion. Were you hinting at two-point tensors?

It would seem that whenever we map a tangent vector from one manifold to another, we would have the problem that we need to express the vector on each manifold w.r.t. different bases. So is it true that in general those "push-forwards" are two-point tensors?

I'm still not sure I understand why this tensor isn't defined in terms of the covariant derivative. If we were to change the basis, wouldn't we need the covariant derivative to ensure the object transforms like a tensor? Or is this where the two-point nature comes in (since there are two different bases, maybe we don't need the covariant derivative? Perhaps this makes no sense... clearly, I have to wrap my brain around this issue some more...).

Thanks,

Johannes

8. May 8, 2013

### Staff: Mentor

Hi Johannes.

I never heard the term two-point tensor before, but the term seems to capture the essence of what we are talking about. When using the deformation gradient tensor, it is customary to express this tensor with respect to two different coordinate bases, one for the undeformed configuration of the body, and the other for the deformed configuration of the body. I'll try to be more precise below.

In analyzing the mechanical constitutive behavior of solids and fluids experiencing large deformations, the first step is to mathematically describe the kinematics of the deformation. This is accomplished using the deformation gradient tensor. It is typical to use an embedded material coordinate system in mathematically describing the kinematics of the deformation. The embedded material coordinate system serves at a label for each and every particle of the material. It's analogous to inscribing a grid onto a sheet of rubber that you are about to deform (stretch in some arbitrary way). The material coordinates usually correspond to the spatial coordinates of the various material particles of the body at time zero (prior to the deformation), as measured using some convenient spatial coordinate system, such as cartesian coordinates, cylindrical coordinates, spherical coordinates, etc. It is not necessary to use the same spatial coordinate system to describe the locations of the various material points of the body during the deformation, provided that the spatial coordinates used for the deformed configuration of the body can be expressed in terms of the material coordinates at time zero.

In what I'm going to discuss, I'm going to use upper case letters to represent parameters for the material coordinate system, and lower case letters to represent parameters for the spactial coordinate system describing the locations of the material particles for the deforming body:

Xi=material coordinate for a material particle (at time zero)
$\vec{A_i}$=coordinate basis vector at material particle location at time zero
$\vec{A^i}$=duel basis vector at material particle location at time zero
$$\vec{A_i}\centerdot\vec{A^j}=\delta_i^j$$

xi=spatial coordinate of material particle location at time t
$\vec{a_i}$=coordinate basis vector for spatial coordinate system at material particle location at time t
$\vec{a^i}$=duel basis vector for spatial coordinate system at material particle location at time t
$$\vec{a_i}\centerdot\vec{a^j}=\delta_i^j$$

The spatial coordinates of a material point within the body at time t are uniquely related to the material coordinates of the same material point at time zero by:
$$x^i=x^i(X^1,X^2,...., t)$$
Alternately, the material coordinates of a material point within the body at time zero can be uniquely related to the spatial coordinates of the same material point at time t by:
$$X^i=X^i(x^1,x^2,...., t)$$
If there are two neighboring material points at Xi and Xi+dXi in the initial configuration of the body, these and these same two material points are located at xi and xi+dxi in the deformed configuration of the body, the differential differences in the coordinates in the two configurations of the body are related by the chain rule:
$$dx^i=\frac{\partial x^i}{\partial X^j}dX^j$$

A more general way of representing this and capturing the kinematic essence of the deformation is to let $\vec{dS}$ represent an arbitrary differential position vector between two material points in the initial undeformed configuration of the body, and $\vec{ds}$ represent the differential position vector between the same two material points in the deformed configuration of the body (at time t). The general mapping between $\vec{dS}$ and $\vec{ds}$ is provided via the deformation gradient tensor $\vec{F}$:
$$\vec{ds}=\vec{F}\centerdot\vec{dS}$$

Now, let's consider how the above two equations can be used to obtain the deformation gradient tensor in component form. To do this, we first express the differential position vectors $\vec{dS}$ and $\vec{ds}$ in component form in terms of the coordinate basis vectors existing at the locations of these two material particles at times zero and t, respectively:
$$\vec{dS}=\vec{A_i}dX^i$$
$$\vec{ds}=\vec{a_i}dx^i$$
From the above, it follows that:
$$\vec{ds}=\vec{a_i}dx^i=(\frac{\partial x^i}{\partial X^j}\vec{a_i}\vec{A^j})\centerdot (\vec{A_k}dX^k)=\vec{F}\centerdot\vec{dS}$$
Within this framework, the deformation gradient tensor is expressed in component form by:
$$\vec{F}=\frac{\partial x^i}{\partial X^j}\vec{a_i}\vec{A^j}$$
So, the deformation gradient tensor is represented as a two-point tensor (i.e., in terms of the coordinate basis vectors at two different points in space).

Continuing, if we take the dot product of $\vec{ds}$ with itself, we obtain

$$ds^2=(\vec{A_l}dX^l)\centerdot (\frac{\partial x^m}{\partial X^n}\vec{A^n}\vec{a_m})\centerdot (\frac{\partial x^i}{\partial X^j}\vec{a_i}\vec{A^j})\centerdot (\vec{A_k}dX^k)=\vec{dS}\centerdot \vec{G}\centerdot \vec{dS}$$
where $\vec{G}$ is the Cauchy Green tensor:

$$\vec{G}=\vec{F^T} \centerdot \vec{F}=(\frac{\partial x^m}{\partial X^n}\vec{A^n}\vec{a_m})\centerdot (\frac{\partial x^i}{\partial X^j}\vec{a_i}\vec{A^j})=g_{mi}\frac{\partial x^m}{\partial X^n}\frac{\partial x^i}{\partial X^j}\vec{A^n}\vec{A^j}$$
where $\vec{F^T}$ is the transpose of the deformation gradient tensor. Note that, in this framework, the Cauchy Green tensor is expressed in terms of the basis vectors for the material coordinate system at the initial location of the material points at time zero.

9. May 11, 2013

### joda80

Hi Chet,

Thank you for your detailed response! So my thoughts about the transformation behavior:

When changing the coordinate basis of a "conventional" vector gradient $\partial A^i / \partial x^k$, with $x^i = x^i(\xi^j)$ the tensor behavior is destroyed because not only the derivative needs to be transformed, but also the vector component:

$$\frac{\partial A^i}{\partial x^k} = \frac{\partial \xi^j}{\partial x^k} \frac{\partial}{\partial \xi^j} \left( \frac{\partial x^i}{\partial \xi^{m}}\hat{A}^m \right)$$

The transformation in the bracket makes this quantity a non-tensor.

However, if we transform a two-point tensor, we change each basis "separately", so e.g., changing the material basis:

$$\frac{\partial x^i}{\partial X^J} = \frac{\partial {\bar{X}}^K}{\partial X^J} \frac{\partial x^i}{\partial {\bar{X}}^K}$$

That is, $x^i$ does not need to be transformed, and is thus unaffected by this change of the material basis. An analogous result holds when we change the spatial basis. The result is that the deformation gradient indeed obeys the tensor transformation law, although one transformation matrix refers to the material basis, and the other one to the spatial basis.

Does that sound about right? I'm not entirely sure because in Marsden/Hughes' textbook, it says: "Notice that the coordinate expression of the deformation gradient, $\partial \Phi^i / \partial X^K$ does not involve any covariant derivatives. This is because $\Phi$ is not a vector but rather is a point mapping of B to S" (B and S are the deformed configuration and the fixed Cartesian space, respectively). I'm not sure I understand this explanation, or how this would be consistent with my reasoning above.

Thanks!

10. May 12, 2013

### Staff: Mentor

Hi Johannes,

I have several more things to say about this, but I'm going to be tied up the next couple of days with mother's day and grandchildren visiting. I'll get back with you as soon as I can. I will say now however, that what we are dealing with here is much more like a coordinate transformation than an actual gradient in the conventional sense.

Chet

11. May 16, 2013

### Staff: Mentor

Hi Johannes,

As I alluded to in my previous post, what we are dealing with here is much more akin a coordinate transformation than an actual gradient in the conventional sense. What I want to do is write out the equations for a coordinate transformation so that we can compare them with what we have for a deformation gradient.

For a coordinate transformation, let the two sets of coordinates be xi and yi, and the corresponding two sets of coordinate basis vectors be $\vec{a_i}$ and $\vec{b_i}$, respectively. These two sets of coordinates apply at the same location in space. The sets of coordinates are related by
$$x^i=y^i(y^1,y^2,....)$$
If $\vec{ds}$ represents a differential position vector between two neighboring points in space, then
$$\vec{ds}=\vec{a_j}dx^j=\vec{a_j}\frac{\partial x^j}{\partial y^i}dy^i=\vec{b_i}dy^i$$
from which it follows that
$$\vec{a_j}\frac{\partial x^j}{\partial y^i}=\vec{b_i}$$
The equation for $\vec{ds}$ can also be written as
$$\vec{ds}=\vec{a_j}dx^j=(\frac{\partial x^j}{\partial y^k}\vec{a_j} \vec{b^k})\centerdot (\vec{b_i}dy^i)$$
But since $\vec{ds}=\vec{b_i}dy^i$, the previous equation becomes:
$$\vec{ds}=(\frac{\partial x^j}{\partial y^k}\vec{a_j} \vec{b^k})\centerdot \vec{ds}$$
This indicates that the transformation tensor $(\frac{\partial x^j}{\partial y^k}\vec{a_j} \vec{b^k})$ must be some form of the metric (identity) tensor $\vec{I}$:
$$\vec{I}=\frac{\partial x^j}{\partial y^k}\vec{a_j} \vec{b^k}$$

This form of the transformation tensor $(\frac{\partial x^j}{\partial y^k}\vec{a_j} \vec{b^k})$ should be compared with the form of the deformation gradient tensor that I presented in my previous long posting. Note the similarity. Both relationships involve two sets of coordinate basis vectors. However, in the form of the transformation tensor presented here, the two sets of coordinate basis vectors apply to the same location in space. In the form of the deformation gradient tensor discussed in the earlier post, the two sets of basis vectors apply to two different locations in space: the location of the parcel of material at time zero, and the location of the parcel at time t. Another difference is that an equation analogous to $\vec{a_j}\frac{\partial x^j}{\partial y^i}=\vec{b_i}$ does not apply to the coordinate basis vectors involved with the deformation gradient tensor. This is because the deformation gradient tensor maps the vector $\vec{dS}$ into the deformed vector $\vec{ds}$, while the ordinary transformation tensor maps the vector $\vec{ds}$ into itself. But, my point is that the comparison between the transformation tensor and the deformation gradient tensor is very compelling. One final remark is that, if we substitute the relatinship $\vec{a_j}\frac{\partial x^j}{\partial y^i}=\vec{b_i}$ into the equation for the transformation tensor, we obtain:
$$\vec{I}=\vec{b_k} \vec{b^k}$$
This brings out much more clearly the fact that the coordinate transformation tensor is simply a form of the metric (identity) tensor.

I would like to also mention that, in practice, it is often much more convenient to work in terms of the inverse of the deformation gradient tensor $\vec{F^{-1}}$, rather than the deformation gradient tensor itself $\vec{F}$. In this case, we write:

$$\vec{dS}=\vec{F^{-1}}\centerdot \vec{ds}$$
where, as in the previous post, $\vec{dS}$ is the differential position vector between two material points at time zero, and $\vec{ds}$ represents the differential position vector between the same two material points at time t. In this equation, the inverse of the deformation gradient tensor is expressed as:
$$\vec{F^{-1}}=\frac{\partial X^i}{\partial x^j}\vec{A_i}\vec{a^j}$$

In this case, if we take the dot product of $\vec{dS}$ with itself, we obtain:
$$dS^2=\vec{ds}\centerdot (\vec{F^{-T}}\centerdot \vec{F^{-1}})\centerdot \vec{ds}$$
where
$$\vec{F^{-T}}\centerdot \vec{F^{-1}}=G_{mi}\frac{\partial X^m}{\partial x^n}\frac{\partial X^i}{\partial x^j}\vec{a^n}\vec{a^j}$$
where $G_{mi}=\vec{A_m}\centerdot \vec{A_i}$.

Note that, in this form, the tensor $\vec{F^{-T}}\centerdot \vec{F^{-1}}$ is expressed in terms of basis vectors for the parcel of material at time t, rather than in terms of basis vectors for the parcel at time zero. This is important because in deformational stress analysis, the stress tensor must be expressed in terms of basis vectors at the location of the parcel at time t.

12. May 22, 2013

### joda80

Thank you Chet, for your detailed responses. I have to re-read your exposition a few more times, but it seems you are pointing to the same direction as I was heading (regarding the transformation of the basis vectors of the deformation gradient).

Cheers,

Johannes