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**A Brief Introduction to Tensors and their properties**

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**1. BASIC PROPERTIES OF TENSORS**

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**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**

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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**

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where *d***x** is an infinitesimal line element in
an undeformed solid, and *d***w** is the vector representing the
deformed line element.

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**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 *d***u**
and *d***x **as components. Then,
calculate the components of *d***u** in terms of *d***x **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**.

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**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**

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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*

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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**

**Tensor components**.

Let
be a Cartesian basis, and let **S **be a second order tensor. The components of **S** in may be represented as a matrix

where

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.

#### Addition

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

The product

is also a vector. In component form

#### or

#### Observe that (unless S is symmetric).

#### Product of two tensors

Let **T **and **S **be two second
order tensors. Then ** **is also a tensor.

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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.

#### Transpose

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

**Trace**

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**.

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In index notation, the trace is written

**Contraction**.

**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

Observe that

**Determinant**

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

**Inverse**

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

Recall that

so multiplying both sides by [Q] shows that

so, comparing with the first of equation (1)

as stated.

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**

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*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.

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In practice, the most commonly used invariants are:

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**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.

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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 .

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**The Cayley-Hamilton Theorem**

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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

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**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:

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*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

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** 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 d**x). **

· 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.