A Brief Introduction to Tensors and their properties
1. BASIC PROPERTIES OF TENSORS
1.1 Examples of Tensors
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
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
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
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
We could represent this as a
Alternatively, using index
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
have probably already seen the matrix representation of stress and strain
components in introductory courses.
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.
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
the same magnitude and direction, irrespective of the basis used to represent u, v,
1.3 The difference between a matrix and a tensor
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,
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
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
specifically, let denote a tensor operating on a vector. Linearity then requires that, for all vectors
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.
tensor of rank 1 is a vector that transforms as
tensor of rank 1 is a vector that transforms as
tensor of rank 2 transforms as
tensor of rank 2 transforms as
A mixed tensor of
rank 2 transforms as
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
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
all vectors u. (Clearly, this maps u onto a vector parallel to a
with magnitude )
The components of in a basis are
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
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.
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
notation we would write
of a tensor and a vector
Let u be a vector and S a
second order tensor. Then
is a vector.
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
is also a vector. In component form
Observe that (unless S is symmetric).
Let T and S be two second
order tensors. Then is also a tensor.
Denote the components of U, S
and T by matrices.
Alternatively, using index notation
Note that tensor products, like matrix products, are not
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
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
More formally, let be any Cartesian basis. Then
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.
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
determinant of a tensor is defined as the determinant of the matrix of its
components in a basis. For a second
In index notation this
Note that if S and T are two tensors, then
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.
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.
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 ).
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
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.
practice, the most commonly used invariants are:
Eigenvalues and Eigenvectors (Principal values and
Let S be a second order tensor.
The scalars and unit vectors m which satisfy
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.
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.
eigenvalues of a second order tensor are computed using the condition . This yields a cubic equation, which can be
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
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
that T= . Substituting in the preceding equation
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 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:
eigenvectors of a symmetric tensor with distinct eigenvalues are orthogonal. To
see this, let be two eigenvectors, with corresponding
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
eigenvectors can then be found by back-substitution into .
To do this, note that the matrix equation can be written as
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
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
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
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
An orthogonal tensor must
a tensor with is known as a proper orthogonal tensor. Orthogonal
tensors also have some interesting and useful properties:
tensors map a vector onto another vector with the same length. To see this, let u be an arbitrary vector.
Then, note that
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
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
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
given n and ,
R can be computed from
W is the skew tensor that has n as its dual vector, i.e. . In index notation, this formula is
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
We can then set and define . U is
clearly symmetric, and also .
To see that R is orthogonal note
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.