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**Vectors and Tensor Operations in Polar Coordinates**

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Many simple boundary value problems in solid mechanics (such as those that tend to appear in homework assignments or examinations!) are most conveniently solved using spherical or cylindrical-polar coordinate systems.

The main drawback of using a polar coordinate system is that there is no convenient way to express the various vector and tensor operations using index notation – everything has to be written out in long-hand. In this section, therefore, we completely abandon index notation – vector and tensor components are always expressed as matrices.

**Spherical-polar coordinates**

**1.1** **Specifying points in spherical-polar
coordinate****s**

To specify points in space using spherical-polar
coordinates, we first choose two convenient, mutually perpendicular reference
directions (**i** and **k** in the picture). For example, to specify position on the
Earth’s surface, we might choose **k**
to point from the center of the earth towards the North Pole, and choose **i** to point from the center of the earth
towards the intersection of the equator (which has zero degrees latitude) and
the Greenwich Meridian (which has zero degrees longitude, by definition).

Then,
each point P in space is identified by three numbers, shown in the picture above. **These
are not components of a vector**.

In words:

*R*
is the distance of P from the origin

is the angle
between the **k** direction and OP

is the angle between the **i **direction and the projection of OP onto a plane through O normal
to **k**

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**By convention, we choose , and **

**1.2** **Converting between Cartesian and
Spherical-Polar representations of points**

When
we use a Cartesian basis, we identify points in space by specifying the
components of their position vector relative to the origin (*x,y,z*), such that When we use a spherical-polar coordinate
system, we locate points by specifying their spherical-polar coordinates

The formulas below relate the two representations. They are derived using basic trigonometry

**1.3 Spherical-Polar representation of vectors**

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When
we work with vectors in spherical-polar coordinates, we abandon the {**i,j,k**} basis. Instead, we specify vectors as components in
the basis shown in the figure. For example, an arbitrary vector **a **is written as ,
where denote the components of **a**.

The basis is different for
each point *P*. In words

points along OP

is tangent to a line of constant longitude through P

is tangent to a line of constant latitude through P.

For example if polar-coordinates are used to specify points on the Earth’s surface, you can visualize the basis vectors like this. Suppose you stand at a point P on the Earths surface. Relative to you: points vertically upwards; points due South; and points due East. Notice that the basis vectors depend on where you are standing.

You
can also visualize the directions as follows.
To see the direction of ,
keep and fixed, and increase *R*. P is moving parallel to . To see the direction of ,
keep *R* and fixed, and increase .
P now moves parallel to . To see the direction of ,
keep *R* and fixed, and increase . P now moves parallel to . Mathematically, this concept can be expressed
as follows. Let **r** be the position vector of P.
Then

By definition, the `natural basis’ for a coordinate system is the derivative of the position vector with respect to the three scalar coordinates that are used to characterize position in space (see Chapter 10 for a more detailed discussion). The basis vectors for a polar coordinate system are parallel to the natural basis vectors, but are normalized to have unit length. In addition, the natural basis for a polar coordinate system happens to be orthogonal. Consequently, is an orthonormal basis (basis vectors have unit length, are mutually perpendicular and form a right handed triad)

**1.4 Converting vectors between Cartesian and Spherical-Polar bases**

Let
be a vector, with components in the spherical-polar basis . Let denote the components of **a** in the basis {**i,j,k**}.

The two sets of components are related by

while the inverse relationship is

Observe that the two 3x3 matrices involved in this transformation are transposes (and inverses) of one another. The transformation matrix is therefore orthogonal, satisfying , where denotes the 3x3 identity matrix.

**Derivation: **It is easiest to do the transformation by expressing
each basis vector as components in {**i,j,k**}, and then substituting.
To do this, recall that ,
recall also the conversion

and finally recall that by definition

Hence, substituting for *x,y,z* and differentiating

Conveniently we find that . Therefore

Similarly

while , so that

Finally, substituting

Collecting terms in **i**, **j**
and **k**, we see that

This is the result stated.

To show the inverse result, start by noting that

(where we have used ). Recall that

Substituting, we get

Proceeding in exactly the same way for the other two components gives the remaining expressions

Re-writing the last three equations in matrix form gives the result stated.

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**1.5 Spherical-Polar representation of tensors**

The
triad of vectors is an orthonormal basis (i.e. the three basis
vectors have unit length, and are mutually perpendicular). Consequently, tensors can be represented as
components in this basis in exactly the same way as for a fixed Cartesian basis
. In particular, a general second order tensor **S** can be represented as a 3x3 matrix

You can think of as being equivalent to , as , and so on. All tensor operations such as addition, multiplication by a vector, tensor products, etc can be expressed in terms of the corresponding operations on this matrix, as discussed in Section B2 of Appendix B.

The component representation of a tensor can also be expressed in dyadic form as

Furthermore, the physical significance of the components can be interpreted in exactly the same way as for tensor components in a Cartesian basis. For example, the spherical-polar coordinate representation for the Cauchy stress tensor has the form

The component represents the traction component in direction acting on an internal material plane with normal , and so on. Of course, the Cauchy stress tensor is symmetric, with

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**1.6 Constitutive equations in spherical-polar
coordinates**

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The constitutive equations listed in Chapter 3 all relate some measure of stress in the solid (expressed as a tensor) to some measure of local internal deformation (deformation gradient, Eulerian strain, rate of deformation tensor, etc), also expressed as a tensor. The constitutive equations can be used without modification in spherical-polar coordinates, as long as the matrices of Cartesian components of the various tensors are replaced by their equivalent matrices in spherical-polar coordinates.

For example, the stress-strain relations for an isotropic, linear elastic material in spherical-polar coordinates read

**HEALTH WARNING: **If you are solving a problem involving *anisotropic *materials using
spherical-polar coordinates, it is important to remember that the orientation
of the basis vectors vary with position. For example, for an anisotropic, linear
elastic solid you could write the constitutive equation as

however, the elastic constants would need to be represent the material properties in the basis , and would therefore be functions of position (you would have to calculate them using the lengthy basis change formulas listed in Section 3.2.11). In practice the results are so complicated that there would be very little advantage in working with a spherical-polar coordinate system in this situation.

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**1.7 Converting tensors between Cartesian and
Spherical-Polar bases**

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Let
**S** be a tensor, with components

in
the spherical-polar basis and the Cartesian basis {**i,j,k**}, respectively. The
two sets of components are related by

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These results follow immediately from the general basis change formulas for tensors given in Appendix B.

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**1.8 Vector Calculus using Spherical-Polar Coordinates**

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Calculating
derivatives of scalar, vector and tensor functions of position in
spherical-polar coordinates is complicated by the fact that the basis vectors
are functions of position. The results
can be expressed in a compact form by defining the *gradient operator*, which, in spherical-polar coordinates, has the
representation

In addition, the derivatives of the basis vectors are

You can derive these
formulas by differentiating the expressions for the basis vectors in terms of {**i,j,k**}

and
evaluating the various derivatives. When differentiating, note that {**i,j,k**} are fixed, so their derivatives
are zero. The details are left as an
exercise.

The various derivatives of scalars, vectors and tensors can be expressed using operator notation as follows.

**Gradient of a scalar function: **Let denote a scalar function of position. The gradient of *f *is denoted by

Alternatively, in matrix form

**Gradient of a vector function **Let be a vector function of position. The gradient
of **v **is a tensor, which can be
represented as a dyadic product of the vector with the gradient operator as

The dyadic product can be expanded – but when evaluating the derivatives it is important to recall that the basis vectors are functions of the coordinates and consequently their derivatives do not vanish. For example

Verify for yourself that the matrix representing the components of the gradient of a vector is

**Divergence of a vector function **Let be a vector function of position. The
divergence of **v **is a scalar, which
can be represented as a dot product of the vector with the gradient operator as

Again, when expanding the dot product, it is important to remember to differentiate the basis vectors. Alternatively, the divergence can be expressed as , which immediately gives

**Curl of a vector function **Let be a vector function of position. The curl of **v **is a vector, which can be represented
as a cross product of the vector with the gradient operator as

The curl rarely appears in solid mechanics so the components will not be expanded in full

**Divergence of a tensor function. **Let be a tensor, with dyadic representation

The divergence of **S **is a vector, which can be represented
as

Evaluating
the components of the divergence is an extremely tedious operation, because
each of the basis vectors in the dyadic representation of **S **must be differentiated, in addition to the components
themselves. The final result (expressed
as a column vector) is

**2: Cylindrical-polar coordinates**

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**2.1 Specifying points in space using in
cylindrical-polar coordinates**

To
specify the location of a point in cylindrical-polar coordinates, we choose an
origin at some point on the axis of the cylinder, select a unit vector **k **to be parallel to the axis of the
cylinder, and choose a convenient direction for the basis vector **i**, as shown in the picture. We then use the three numbers to locate a point inside the cylinder, as
shown in the picture. **These are not components of a vector**.

In words

*r *is the radial distance of P from the axis of the cylinder

is the angle
between the **i** direction and the projection of OP onto the **i,j **plane

*z*
is the length of the projection of OP on the axis of the cylinder.

By convention *r>0* and

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**2.2 Converting between cylindrical polar and
rectangular cartesian coordinates**

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When
we use a Cartesian basis, we identify points in space by specifying the
components of their position vector relative to the origin (*x,y,z*), such that When we use a spherical-polar coordinate
system, we locate points by specifying their spherical-polar coordinates

The formulas below relate the two representations. They are derived using basic trigonometry

**2.3
Cylindrical-polar representation of vectors**

When
we work with vectors in spherical-polar coordinates, we specify vectors as
components in the basis shown in the figure. For example, an arbitrary vector **a **is written as ,
where denote the components of **a**.

The basis vectors are selected as follows

is a unit vector normal to the cylinder at P

is a unit vector circumferential to the cylinder at P, chosen to make a right handed triad

is parallel to the **k** vector.

You will see that the position vector of point P would be expressed as

Note also that the basis vectors are intentionally chosen to satisfy

The basis vectors have unit length, are mutually perpendicular, and form a right handed triad and therefore is an orthonormal basis. The basis vectors are parallel to (but not equivalent to) the natural basis vectors for a cylindrical polar coordinate system (see Chapter 10 for a more detailed discussion).

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**2.4 Converting vectors between Cylindrical and Cartesian
bases**

Let
be a vector, with components in the spherical-polar basis . Let denote the components of **a** in the basis {**i,j,k**}.

The two sets of components are related by

Observe that the two 3x3 matrices involved in this transformation are transposes (and inverses) of one another. The transformation matrix is therefore orthogonal, satisfying , where denotes the 3x3 identity matrix.

##### The derivation of these results follows the procedure outlined in E.1.4 exactly, and is left as an exercise.

**2.5 Cylindrical-Polar representation of tensors**

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The
triad of vectors is an orthonormal basis (i.e. the three basis
vectors have unit length, and are mutually perpendicular). Consequently, tensors can be represented as
components in this basis in exactly the same way as for a fixed Cartesian basis
. In particular, a general second order tensor **S** can be represented as a 3x3 matrix

You can think of as being equivalent to , as , and so on. All tensor operations such as addition, multiplication by a vector, tensor products, etc can be expressed in terms of the corresponding operations on this matrix, as discussed in Section B2 of Appendix B.

The component representation of a tensor can also be expressed in dyadic form as

The remarks in Section E.1.5 regarding the physical significance of tensor components also applies to tensor components in cylindrical-polar coordinates.

**2.6 Constitutive equations in cylindrical-polar
coordinates**

** **

The constitutive equations listed in Chapter 3 all relate some measure of stress in the solid (expressed as a tensor) to some measure of local internal deformation (deformation gradient, Eulerian strain, rate of deformation tensor, etc), also expressed as a tensor. The constitutive equations can be used without modification in cylindrical-polar coordinates, as long as the matrices of Cartesian components of the various tensors are replaced by their equivalent matrices in spherical-polar coordinates.

For example, the stress-strain relations for an isotropic, linear elastic material in cylindrical-polar coordinates read

The cautionary remarks regarding anisotropic materials in E.1.6 also applies to cylindrical-polar coordinate systems.

**2.7 Converting tensors between Cartesian and Spherical-Polar
bases**

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Let **S** be a tensor,
with components

in
the cylindrical-polar basis and the Cartesian basis {**i,j,k**}, respectively. The
two sets of components are related by

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**2.8 Vector Calculus using Cylindrical-Polar
Coordinates**

** **

Calculating
derivatives of scalar, vector and tensor functions of position in
cylindrical-polar coordinates is complicated by the fact that the basis vectors
are functions of position. The results
can be expressed in a compact form by defining the *gradient operator*, which, in spherical-polar coordinates, has the
representation

In addition, the nonzero derivatives of the basis vectors are

The various derivatives of scalars, vectors and tensors can be expressed using operator notation as follows.

**Gradient of a scalar function: **Let denote a scalar function of position. The gradient of *f *is denoted by

Alternatively, in matrix form

**Gradient of a vector function **Let be a vector function of position. The gradient
of **v **is a tensor, which can be
represented as a dyadic product of the vector with the gradient operator as

The dyadic product can be expanded – but when evaluating the derivatives it is important to recall that the basis vectors are functions of the coordinate and consequently their derivatives may not vanish. For example

Verify for yourself that the matrix representing the components of the gradient of a vector is

**Divergence of a vector function **Let be a vector function of position. The
divergence of **v **is a scalar, which
can be represented as a dot product of the vector with the gradient operator as

Again, when expanding the dot product, it is important to remember to differentiate the basis vectors. Alternatively, the divergence can be expressed as , which immediately gives

**Curl of a vector function **Let be a vector function of position. The curl of **v **is a vector, which can be represented
as a cross product of the vector with the gradient operator as

The curl rarely appears in solid mechanics so the components will not be expanded in full

**Divergence of a tensor function. **Let
be a tensor, with dyadic representation

The divergence of **S **is a vector, which can be represented
as

Evaluating
the components of the divergence is an extremely tedious operation, because
each of the basis vectors in the dyadic representation of **S **must be differentiated, in addition to the components
themselves. The final result (expressed
as a column vector) is