Two-point Tensors

There are two separate issues here: two-point tensor and Lagrange desciption.  While two-point tensor is not a must, a Lagrange system is usually required to properly describe a solid.  A Lagrange system, in contrast to an Euler system, traces material points by its original (reference) position rather than its current position (so that a cooridnate system moves during deformation).  The former is often adopted in solid mechanics while the later in fluid mechanics, because of the physics of the two types of materials.

Solids (especially elastic solids) differ from fluids in that they have "memory": a solid knows its original state to some extent, while a fluid only cares about its current state (with some exceptions).  Therefore, to fully describe a solid, we need to specify a reference state, and measure the difference between its current state and the reference.  A two-point tensor, the deformation gradient, is naturally involved to bridge the two states.  Two point tensors can be avoided by introducing the right Cauchy-Green deformation tensor F'F (or Green strain), and the second PK stress, with both "legs" in the reference state.  Actually, these tensors will remove the rigid-body rotations, which are included in F.

However, it is imposible to describe the deformation state of an elastic solid with Eulerian tensors only, defined in the current state.  You need to tell the material where it was.

Dear Wei,

I have a question regarding to what you said.

If we consider a body just did rigid body rotation,  then the deformation gradient F=R, R is the rotation. If we suppose there is only one frame (Euleran frame), then R is not a two-point tensor, R=RijEiEj, both Ei and Ej are in the same fame

however, if we still assume there exist a Lagrangian and a Eulerian frame, then F=R=Rij*ei*Ej is a two-point tensor?

Is this correct based on what you said?

I have described these ideas with some care in my class notes on finite deformation.  Here is a list of points:

Deformation gradient 

• To understand these ideas, all you need is a homogeneous deformation of a body, in a three-dimensional space, from a reference state to a current state.
• Consider a set of material particles that forms a segment of a straight line in the space when the body is in the reference state.  Because the deformation is homegeneous, the same set of material particles forms a segment of another straight line in the space when the body is in the current state.  
• In the reference state, the segment is an element Y in a vector space U.  That is, a linear combination of two segments in U is yet another segment in U, as defined by the usual addition of vectors.  
• In the current state, the segement is an element y in a vector space V.  Note that U and V are two distinct vector spaces:  we will not form linear combination of an element in U and an element in V.  
• The deformation gradeint F is a linear map that maps an element in U to an element in V, y = FY.  The two elements represent the same set of material particles, respectively, in the reference state and in the current state.  
• The two vector spaces U and V are distinct.  We can choose different bases for the two vector spaces.  By definition, the deformation grdient is a two-point tensor.  (We define a two-point tensor as a linear map that maps an element in one vector space to an element in a different vector space.)

Geometric representation of the deformation gradient

• Consider another set of material particles that forms a unit cube in the reference state.  Because the deformation is homogeneous, the same set of material particles forms a parallelepiped in the current state.
• We choose the three edges of the unit cube as the basis for both vector spaces.
• Relative to this basis, we represent each straight segement Y in the reference state by a column, and represent the same set of material particles y in the current state by another column.  We represent the deformation gradient F by a matrix.
• Each column of the deformation gradeint F reprentsents a vector, corresponding to an edge of the parallelepiped.

Green deformation tensor

• The state of matter does not change if the body undergoes a rigid-body rotation in the current state.
• Thus, to characterize deformation without ridid-body rotation, we need to describe the size and the shape of the parallelepiped, but not the orientation of the parallelepiped.
• The size and the shape of the parallelepiped are fully determined by six quantities:  the lengths of the three edges and the angles between the three edges.  The six quantities do not form a tensor. 
• The inner products of the edges of the parallelepiped contain the same information of the lengths and angles.  Furthermore, the six inner products do form a tensor if we write these inner products as FTF.  This tensor is the Green deformation tensor.

My class notes on finite deformation develop these ideas, and also contain the development of the nominal stress tensor.

Also see a thread on linear algebra, tensors, and linear maps between linear spaces .