A Brief Introduction to Tensors and their properties
1. BASIC PROPERTIES OF TENSORS
1.1 Examples of Tensors
The gradient of a vector field is a good example of a second-order tensor. Visualize a vector field: at every point in space, the field has a vector value . Let represent the gradient of u. By definition, G enables you to calculate the change in u when you move from a point x in space to a nearby point at :
G is a second order tensor. From this example, we see that when you multiply a vector by a tensor, the result is another vector.
This is a general property of all second order tensors. A tensor is a linear mapping of a vector onto another vector. Two examples, together with the vectors they operate on, are:
The stress tensor
where n is a unit vector normal to a surface, is the stress tensor and t is the traction vector acting on the surface.
The deformation gradient tensor
where dx is an infinitesimal line element in an undeformed solid, and dw is the vector representing the deformed line element.
1.2 Matrix representation of a tensor
To evaluate and manipulate tensors, we express them as components in a basis, just as for vectors. We can use the displacement gradient to illustrate how this is done. Let be a vector field, and let represent the gradient of u. Recall the definition of G
Now, let be a Cartesian basis, and express both du and dx as components. Then, calculate the components of du in terms of dx using the usual rules of calculus
We could represent this as a matrix product
Alternatively, using index notation
From this example we see that G can be represented as a matrix. The elements of the matrix are known as the components of G in the basis . All second order tensors can be represented in this form. For example, a general second order tensor S could be written as
You have probably already seen the matrix representation of stress and strain components in introductory courses.
Since S can be represented as a matrix, all operations that can be performed on a matrix can also be performed on S. Examples include sums and products, the transpose, inverse, and determinant. One can also compute eigenvalues and eigenvectors for tensors, and thus define the log of a tensor, the square root of a tensor, etc. These tensor operations are summarized below.
Note that the numbers , , … depend on the basis , just as the components of a vector depend on the basis used to represent the vector. However, just as the magnitude and direction of a vector are independent of the basis, so the properties of a tensor are independent of the basis. That is to say, if S is a tensor and u is a vector, then the vector
has the same magnitude and direction, irrespective of the basis used to represent u, v, and S.
1.3 The difference between a matrix and a tensor
If a tensor is a matrix, why is a matrix not the same thing as a tensor? Well, although you can multiply the three components of a vector u by any matrix,
the resulting three numbers may or may not represent the components of a vector. If they are the components of a vector, then the matrix represents the components of a tensor A, if not, then the matrix is just an ordinary old matrix.
To check whether are the components of a vector, you need to check how change due to a change of basis. That is to say, choose a new basis, calculate the new components of u in this basis, and calculate the new matrix in this basis (the new elements of the matrix will depend on how the matrix was defined. The elements may or may not change if they don’t, then the matrix cannot be the components of a tensor). Then, evaluate the matrix product to find a new left hand side, say . If are related to by the same transformation that was used to calculate the new components of u, then are the components of a vector, and, therefore, the matrix represents the components of a tensor.
1.4 Formal definition
Tensors are rather more general objects than the preceding discussion suggests. There are various ways to define a tensor formally. One way is the following:
A tensor is a linear vector valued function defined on the set of all vectors
More specifically, let denote a tensor operating on a vector. Linearity then requires that, for all vectors and scalars
Alternatively, one can define tensors as sets of numbers that transform in a particular way under a change of coordinate system. In this case we suppose that n dimensional space can be parameterized by a set of n real numbers . We could change coordinate system by introducing a second set of real numbers which are invertible functions of . Tensors can then be defined as sets of real numbers that transform in a particular way under this change in coordinate system. For example
· A tensor of zeroth rank is a scalar that is independent of the coordinate system.
· A covariant tensor of rank 1 is a vector that transforms as
· A contravariant tensor of rank 1 is a vector that transforms as
· A covariant tensor of rank 2 transforms as
· A contravariant tensor of rank 2 transforms as
· A mixed tensor of rank 2 transforms as
Higher rank tensors can be defined in similar ways. In solid and fluid mechanics we nearly always use Cartesian tensors, (i.e. we work with the components of tensors in a Cartesian coordinate system) and this level of generality is not needed (and is rather mysterious). We might occasionally use a curvilinear coordinate system, in which we do express tensors in terms of covariant or contravariant components this gives some sense of what these quantities mean. But since solid and fluid mechanics live in Euclidean space we don’t see some of the subtleties that arise, e.g. in the theory of general relativity.
1.5 Creating a tensor using a dyadic product of two vectors.
Let a and b be two vectors. The dyadic product of a and b is a second order tensor S denoted by
with the property
for all vectors u. (Clearly, this maps u onto a vector parallel to a with magnitude )
The components of in a basis are
Note that not all tensors can be constructed using a dyadic product of only two vectors (this is because always has to be parallel to a, and therefore the representation cannot map a vector onto an arbitrary vector). However, if a, b, and c are three independent vectors (i.e. no two of them are parallel) then all tensors can be constructed as a sum of scalar multiples of the nine possible dyadic products of these vectors.
2. OPERATIONS ON SECOND ORDER TENSORS
Let be a Cartesian basis, and let S be a second order tensor. The components of S in may be represented as a matrix
The representation of a tensor in terms of its components can also be expressed in dyadic form as
This representation is particularly convenient when using polar coordinates, or when using a general non-orthogonal coordinate system.
Let S and T be two tensors. Then is also a tensor.
Denote the Cartesian components of U, S and T by matrices as defined above. The components of U are then related to the components of S and T by
In index notation we would write
Product of a tensor and a vector
Let u be a vector and S a second order tensor. Then
is a vector.
Let and denote the components of vectors u and v in a Cartesian basis , and denote the Cartesian components of S as described above. Then
Alternatively, using index notation
is also a vector. In component form
Observe that (unless S is symmetric).
Product of two tensors
Let T and S be two second order tensors. Then is also a tensor.
Denote the components of U, S and T by matrices. Then,
Alternatively, using index notation
Note that tensor products, like matrix products, are not commutative; i.e.
Let S be a tensor. The transpose of S is denoted by and is defined so that
Denote the components of S by a 3x3 matrix. The components of are then
i.e. the rows and columns of the matrix are switched.
Note that, if A and B are two tensors, then
Let S be a tensor, and denote the components of S by a matrix. The trace of S is denoted by tr(S) or trace(S), and can be computed by summing the diagonals of the matrix of components
More formally, let be any Cartesian basis. Then
The trace of a tensor is an example of an invariant of the tensor you get the same value for trace(S) whatever basis you use to define the matrix of components of S.
In index notation, the trace is written
Inner Product: Let S and T be two second order tensors. The inner product of S and T is a scalar, denoted by . Represent S and T by their components in a basis. Then
In index notation
Observe that ,
and also that ,
where I is the identity tensor.
Outer product: Let S and T be two second order tensors. The outer product of S and T is a scalar, denoted by . Represent S and T by their components in a basis. Then
In index notation
The determinant of a tensor is defined as the determinant of the matrix of its components in a basis. For a second order tensor
In index notation this would read
Note that if S and T are two tensors, then
Let S be a second order tensor. The inverse of S exists if and only if , and is defined by
where denotes the inverse of S and I is the identity tensor.
The inverse of a tensor may be computed by calculating the inverse of the matrix of its components. Formally, the inverse of a second order tensor can be written in a simple form using index notation as
In practice it is usually faster to compute the inverse using methods such as Gaussian elimination.
Change of Basis.
Let S be a tensor, and let be a Cartesian basis. Suppose that the components of S in the basis are known to be
Now, suppose that we wish to compute the components of S in a second Cartesian basis, . Denote these components by
To do so, first compute the components of the transformation matrix [Q]
(this is the same matrix you would use to transform vector components from to ). Then,
or, written out in full
To prove this result, let u and v be vectors satisfying
Denote the components of u and v in the two bases by and , respectively. Recall that the vector components are related by
Now, we could express the tensor-vector product in either basis
Substitute for from above into the second of these two relations, we see that
so multiplying both sides by [Q] shows that
so, comparing with the first of equation (1)
In index notation, we would write
Another, perhaps cleaner, way to derive this result is to expand the two tensors as the appropriate dyadic products of the basis vectors
Invariants of a tensor are scalar functions of the tensor components which remain constant under a basis change. That is to say, the invariant has the same value when computed in two arbitrary bases and . A symmetric second order tensor always has three independent invariants.
Examples of invariants are
1. The three eigenvalues
2. The determinant
3. The trace
4. The inner and outer products
These are not all independent for example any of 2-4 can be calculated in terms of 1.
In practice, the most commonly used invariants are:
Eigenvalues and Eigenvectors (Principal values and direction)
Let S be a second order tensor. The scalars and unit vectors m which satisfy
are known as the eigenvalues and eigenvectors of S, or the principal values and principal directions of S. Note that may be complex. For a second order tensor in three dimensions, there are generally three values of and three unique unit vectors m which satisfy this equation. Occasionally, there may be only two or one value of . If this is the case, there are infinitely many possible vectors m that satisfy the equation. The eigenvalues of a tensor, and the components of the eigenvectors, may be computed by finding the eigenvalues and eigenvectors of the matrix of components.
The eigenvalues of a symmetric tensor are always real, and its eigenvectors are mutually perpendicular (these two results are important and are proved below). The eigenvalues of a skew tensor are always pure imaginary or zero.
The eigenvalues of a second order tensor are computed using the condition . This yields a cubic equation, which can be expressed as
There are various ways to solve the resulting cubic equation explicitly a solution for symmetric S is given below, but the results for a general tensor are too messy to be given here. The eigenvectors are then computed from the condition .
The Cayley-Hamilton Theorem
Let S be a second order tensor and let be the three invariants. Then
(i.e. a tensor satisfies its characteristic equation). There is an obscure trick to show this… Consider the tensor (where is an arbitrary scalar), and let T be the adjoint of , (the adjoint is just the inverse multiplied by the determinant) which satisfies
Assume that T= . Substituting in the preceding equation shows that
Use these to substitute for into
3 SPECIAL TENSORS
Identity tensor The identity tensor I is the tensor such that, for any tensor S or vector v
In any basis, the identity tensor has components
Symmetric Tensor A symmetric tensor S has the property
The components of a symmetric tensor have the form
so that there are only six independent components of the tensor, instead of nine. Symmetric tensors have some nice properties:
· The eigenvectors of a symmetric tensor with distinct eigenvalues are orthogonal. To see this, let be two eigenvectors, with corresponding eigenvalues . Then .
· The eigenvalues of a symmetric tensor are real. To see this, suppose that are a complex eigenvalue/eigenvector pair, and let denote their complex conjugates. Then, by definition . And hence . But note that for a symmetric tensor . Thus .
The eigenvalues of a symmetric tensor can be computed as
The eigenvectors can then be found by back-substitution into . To do this, note that the matrix equation can be written as
Since the determinant of the matrix is zero, we can discard any row in the equation system and take any column over to the right hand side. For example, if the tensor has at least one eigenvector with then the values of for this eigenvector can be found by discarding the third row, and writing
· Spectral decomposition of a symmetric tensor Let S be a symmetric second order tensor, and let be the three eigenvalues and eigenvectors of S. Then S can be expressed as
To see this, note that S can always be expanded as a sum of 9 dyadic products of an orthogonal basis. . But since are eigenvectors it follows that
Skew Tensor. A skew tensor S has the property
The components of a skew tensor have the form
Every second-order skew tensor has a dual vector w that satisfies
for all vectors u. You can see this by noting that and expanding out the tensor and cross products explicitly. In index notation, we can also write
Orthogonal Tensors An orthogonal tensor R has the property
An orthogonal tensor must have ; a tensor with is known as a proper orthogonal tensor. Orthogonal tensors also have some interesting and useful properties:
· Orthogonal tensors map a vector onto another vector with the same length. To see this, let u be an arbitrary vector. Then, note that
· The eigenvalues of an orthogonal tensor are for some value of . To see this, let be an eigenvector, with corresponding eigenvalue . By definition, . Hence, . Similarly, . Since the characteristic equation is cubic, there must be at most three eigenvalues, and at least one eigenvalue must be real.
Proper orthogonal tensors can be visualized physically as rotations. A rotation can also be represented in several other forms besides a proper orthogonal tensor. For example
· The Rodriguez representation quantifies a rotation as an angle of rotation (in radians) about some axis n (specified by a unit vector). Given R, there are various ways to compute n and . For example, one way would be find the eigenvalues and the real eigenvector. The real eigenvector (suitably normalized) must correspond to n; the complex eigenvalues give . A faster method is to note that
· Alternatively, given n and , R can be computed from
where W is the skew tensor that has n as its dual vector, i.e. . In index notation, this formula is
Another useful result is the Polar Decomposition Theorem, which states that invertible second order tensors can be expressed as a product of a symmetric tensor with an orthogonal tensor:
Moreover, the tensors are unique. To see this, note that
· is symmetric and has positive eigenvalues (to see that it’s symmetric, simply take the transpose, and to see that the eigenvalues are positive, note that for all vectors dx).
· Let and be the three eigenvalues and eigenvectors of . Since the eigenvectors are orthogonal, we can write .
· We can then set and define . U is clearly symmetric, and also . To see that R is orthogonal note that: .
· Given that U and R exist we can write so if we define then . It is easy to show that V is symmetric.
· To see that the decomposition is unique, suppose that for some other tensors . Then . But has a unique square root so . The uniqueness of R follows immediately.