Heat, Mass, and Momentum Transfer

Vector and Index Notation

Motivation

All Engineering tasks are three-dimensional problems.
Often we try to simplify the problem by neglecting some component of the system, or exploiting a symmetry to reduce its dimensionality.
For example, for the flow of a fluid between two plates (top-right), we ignore the direction into the plane (the problem is now 2D), then we assume the flow doesn't change in the $x$-direction (now a 1D problem).
For problems with an axis of symmetry, such as annular flow (bottom-right) we can often ignore the rotational component ($\theta$) and axial component ($z$).

Motivation

However, we must start from a 3D description, as some problems are fully three-dimensional and need to be treated using a general approach.
In this course, all systems reduce to one dimension, but the basics of later 3D problems are being taught which can be solved using CFD (see later courses).
We need to be fluent in the language/notation of three dimensional problems.
The natural mathematical language for describing three dimensional problems is that of Matrices and Vectors.

Representations

You should be familiar by now with vector notation.
A position vector, $\boldsymbol{r}$, is the direction and distance to a point, relative to some origin .
A velocity vector, $\boldsymbol{v}$, is the direction and speed an object is travelling.
The common definition of all vectors is that it is an encoding of a direction and a magnitude!
You also need an origin for vectors to make sense, even for velocity.
Q: How fast are you moving sitting in your seat?
A: 0 mph relative to the seat, 1000 mph relative to the centre of the earth, 67 000 mph relative to the sun, 490 000 mph relative to our galaxy centre, $10^9$ mph relative to the cosmic background radiation.

Representations: Rectangular

There are many ways a vector can be represented…
For a position vector, we can write coordinates: \begin{align*} \boldsymbol{r} = \left[x,y,z\right] \end{align*}
This is the most trivial Cartesian or rectangular coordinate representation, but we could pick any number of other coordinate systems.
For example, on a sphere like the earth, we might use latitude, longitude, and altitude. \begin{align*} \boldsymbol{r}=\left[\text{lat},\,\text{lon},\,\text{alt}\right] \end{align*}
We don't usually use these as they're a bit archaic, but we do use radians and meters.

Representations: Spherical

This angular coordinate system is called spherical or polar coordinates and is useful for systems where there is a point symmetry . \begin{align*} \boldsymbol{r} = \left[r,\theta,\phi\right] \end{align*}
For example, this coordinate system is useful for understanding a burning spherical(ish) piece of coal, or a spherical(ish) solid falling in a liquid.
More about symmetry later in the course, but there is one more coordinate system to introduce…

Representations: Cylindrical

In cylindrical coordinates, we have a rotation $\theta$ about the axis $\boldsymbol{z}$: \begin{align*} \boldsymbol{r} &= \left[r, \theta, z\right] \end{align*}
This coordinate representation is the most useful for systems where we have a rotational symmetry, such as a pipe.
But regardless of the representation we use (Cartesian, cylindrical, spherical/polar), the same information of direction and magnitude is always encoded.
Spherical and cylindrical coordinates are collectively grouped together as curvelinear coordinate systems, and defining operations in these coordinate systems is problematic as they change depending on the coordinate system.

Representations: Cylindrical

If we don't want the basic operations to change¹, the position vector may be defined as follows: \begin{align} \boldsymbol{r}=r_r\,\hat{\boldsymbol{r}}+ r_\theta\,\hat{\boldsymbol{\theta}} + r_z\,\hat{\boldsymbol{z}} \end{align}
You should note that $r_\theta\neq\theta$, but instead is distance of the vector $\boldsymbol{r}$ projected along the unit vector $\hat{\boldsymbol{\theta}}$ illustrated in the diagram.
It is not particularly convenient to work with this coordinate frame as $\hat{\boldsymbol{\theta}}$ and $\hat{\boldsymbol{r}}$ ALSO depend on your coordinates, so we derive everything in rectangular coordinates (which are easy) and use look-up tables for other coordinate systems.

¹ Dimensional Analysis: To prove that basic operations MUST change, consider the units of each term of $\boldsymbol{r}\cdot\boldsymbol{r}=r_r^2+r_\theta^2+r_z^2$. Units of terms that are added together must have the same units but they are inconsistent if ${r}_r=r$ (length), ${r}_\theta=\theta$ (angle), and $r_z=z$ (length), thus the usual definition of the dot product does not apply when $\boldsymbol{r}$ is represented as $\left[r,\,\theta,\,z\right]$.

Humerous(?) interlude

A mathmatician, a physicist, and an engineer were each given the same red rubber ball and told to find the volume.

The mathmatician carefully measured the diameter and evaluated a triple integral.
The physicist filled a beaker with water, put the ball in the water, and measured the total displacement.
The engineer looked up the model and serial numbers in his red-rubber-ball table.

Cartesian/Rectangular coordinates (from the exam datasheet)
where $s$ is a scalar, $\bm{v}$ is a vector, and $\bm{\tau}$ is a tensor. \begin{align*} \nabla s = \nabla_i s &= \left[\frac{\partial\,s}{\partial x},\, \frac{\partial\,s}{\partial y},\, \frac{\partial\,s}{\partial z}\right] \\ \nabla^2 s = \nabla_i\nabla_i s &=\frac{\partial^2\,s}{\partial x^2} + \frac{\partial^2\,s}{\partial y^2}+ \frac{\partial^2\,s}{\partial z^2} \\ \nabla\cdot\bm{v} =\nabla_i v_i &= \frac{\partial\,v_x}{\partial x} + \frac{\partial\,v_y}{\partial y}+ \frac{\partial\,v_z}{\partial z} \\ \nabla \cdot \bm{\tau} &= \nabla_i\, \tau_{ij} \\ \left[\nabla \cdot \bm{\tau}\right]_x &= \frac{\partial\,\tau_{xx}}{\partial x} + \frac{\partial\, \tau_{yx}}{\partial y} + \frac{\partial\, \tau_{zx}}{\partial z} \\ \left[\nabla \cdot \bm{\tau}\right]_y &= \frac{\partial\,\tau_{xy}}{\partial x} + \frac{\partial\, \tau_{yy}}{\partial y} + \frac{\partial\, \tau_{zy}}{\partial z} \\ \left[\nabla \cdot \bm{\tau}\right]_z &= \frac{\partial\,\tau_{xz}}{\partial x} + \frac{\partial\, \tau_{yz}}{\partial y} + \frac{\partial\, \tau_{zz}}{\partial z} \\ \bm{v}\cdot \nabla \bm{v} &= v_i\,\nabla_i\,v_j \\ \left[\bm{v}\cdot \nabla \bm{v}\right]_x &= v_x\frac{\partial\,v_x}{\partial x} + v_y\frac{\partial\,v_x}{\partial y} +v_z\frac{\partial\,v_x}{\partial z} \\ \left[\bm{v}\cdot \nabla \bm{v}\right]_y &= v_x\frac{\partial\,v_y}{\partial x} + v_y\frac{\partial\,v_y}{\partial y} +v_z\frac{\partial\,v_y}{\partial z} \\ \left[\bm{v}\cdot \nabla \bm{v}\right]_z &= v_x\frac{\partial\,v_z}{\partial x} + v_y\frac{\partial\,v_z}{\partial y} +v_z\frac{\partial\,v_z}{\partial z} \end{align*}

Cylindrical coordinates (from the exam datasheet)
where $s$ is a scalar, $\bm{v}$ is a vector, and $\bm{\tau}$ is a tensor. All expressions involving $\bm{\tau}$ are for symmetrical $\bm{\tau}$ only. \begin{align*} \nabla s &= \left[\frac{\partial\,s}{\partial r},\, \frac{1}{r}\frac{\partial\,s}{\partial \theta},\, \frac{\partial\,s}{\partial z}\right] \\ \nabla^2 s &=\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial s}{\partial r}\right) + \frac{1}{r^2}\frac{\partial^2\,s}{\partial \theta^2}+ \frac{\partial^2\,s}{\partial z^2} \\ \nabla\cdot\bm{v} &= \frac{1}{r}\frac{\partial}{\partial r}\left(r\, v_r\right) + \frac{1}{r}\frac{\partial\,v_\theta}{\partial \theta}+ \frac{\partial\,v_z}{\partial z} \\ \left[\nabla \cdot \bm{\tau}\right]_r &= \frac{1}{r}\frac{\partial}{\partial r}\left(r\,\tau_{rr}\right) + \frac{1}{r}\frac{\partial\, \tau_{r\theta}}{\partial \theta} - \frac{1}{r} \tau_{\theta\theta} + \frac{\partial\, \tau_{rz}}{\partial z} \\ \left[\nabla \cdot \bm{\tau}\right]_\theta &= \frac{1}{r}\frac{\partial \tau_{\theta\theta}}{\partial \theta} + \frac{\partial\, \tau_{r\theta}}{\partial r} + \frac{2}{r} \tau_{r\theta} + \frac{\partial\, \tau_{\theta z}}{\partial z} \\ \left[\nabla \cdot \bm{\tau}\right]_z &= \frac{1}{r}\frac{\partial }{\partial r}\left(r\,\tau_{rz}\right) + \frac{1}{r}\frac{\partial \tau_{\theta z}}{\partial\theta} + \frac{\partial\, \tau_{z z}}{\partial z} \\ \left[\bm{v}\cdot \nabla \bm{v}\right]_r &= v_r \frac{\partial v_r}{\partial r} + \frac{v_\theta}{r}\frac{\partial v_r}{\partial \theta} - \frac{v_\theta^2}{r}+v_z\frac{\partial v_r}{\partial z} \\ \left[\bm{v}\cdot \nabla \bm{v}\right]_\theta &= v_r \frac{\partial v_\theta}{\partial r} + \frac{v_\theta}{r}\frac{\partial v_\theta}{\partial \theta} + \frac{v_r\,v_\theta}{r} + v_z \frac{\partial v_\theta}{\partial z} \\ \left[\bm{v}\cdot \nabla \bm{v}\right]_z &= v_r \frac{\partial v_z}{\partial r} + \frac{v_\theta}{r}\frac{\partial v_z}{\partial \theta}+v_z \frac{\partial v_z}{\partial z} \end{align*}

Check out the back of the questions booklet for the exam data sheet.

Operations

Hopefully the past two slides scared you enough that you're motivated to learn about vector operations and what they mean.
So lets define some vector operations/mathematics for them and discuss how they change in different coordinates.

We write the magnitude (or length) of a vector, $\boldsymbol{r}$, as $\left|\boldsymbol{r}\right|$.
In a rectangular coordinate system, we use a 3D version of Pythagoras' theorem¹: \begin{align*} \boldsymbol{r} &= \left[x,y,z\right] & \left|\boldsymbol{r}\right| &= \sqrt{x^2+y^2+z^2} \end{align*}
In a cylindrical coordinate system Pythagoras' theorem gives: \begin{align*} \boldsymbol{r} &= \left[r, \theta, z\right] & \left|\boldsymbol{r}\right| &= \sqrt{r^2 + z^2} \end{align*}
In a spherical coordinate system, only one component has units of length and we have: \begin{align*} \boldsymbol{r} &= \left[r, \theta, \phi\right] & \left|\boldsymbol{r}\right| &= r \end{align*}
I don't want you to derive these, I'm just driving home that you need to be careful about your coordinate systems! We will rely on look-up tables to keep us straight².

¹ The proof is very elegant, just consider only two dimensions at first, then rotate so that the 2D diagonal is one of your triangle sides along with another, as-yet unconsidered, dimension. Repeated application of the 2D pythagoras theorem gives you the N-dimensional generalisation.

² If you are not satisfied and want to understand this properly or simply enjoy mathematics, take a look at the math appendix to Transport Phenomena by Bird, Stewart, and Lightfoot.

Unit Vectors

Once we know the magnitude of a vector, we can generate a vector in the same direction, but with a unit length: \begin{align*} \hat{\boldsymbol{r}} = \frac{\boldsymbol{r}}{\left|\boldsymbol{r}\right| } \end{align*}
$\hat{\boldsymbol{r}}$ is a unit vector (a vector of length 1) pointing in the same direction as $\boldsymbol{r}$.
Unit vectors are really handy when you only want to encode direction (ignoring magnitude).
I've used them on the diagram on the right to indicate the directions the vector $\boldsymbol{r}$ would move in each of the coordinates (both Cartesian and cylindrical).

Other Operations

Again, noting that the following definitions only apply for Cartesian coordinates:
Vector addition/subtraction is straightforward: \begin{align*} \boldsymbol{a}\pm\boldsymbol{b}=\left[ \begin{matrix} a_x \\ a_y \\ a_z \end{matrix}\right] \pm \left[ \begin{matrix} b_x \\ b_y \\ b_z \end{matrix}\right] = \left[ \begin{matrix} a_x \pm b_x \\ a_y \pm b_y \\ a_z \pm b_z \end{matrix}\right] \end{align*}
The dot (or scalar) product is also well known: \begin{align*} \boldsymbol{a}\cdot\boldsymbol{b}=\left[ \begin{matrix} a_x \\ a_y \\ a_z \end{matrix}\right] \cdot \left[ \begin{matrix} b_x \\ b_y \\ b_z \end{matrix}\right] = a_x b_x + a_y b_y +a_z b_z \end{align*}
It gives the product of the angle between two vectors and their lengths, but physically it is often used to measure one vector against another. E.g., $\bm{a}\cdot\hat{\bm{b}}$ is actually the length of $\vec{a}$ in the direction of $\hat{\vec{b}}$!

Other Operations

The final operation is the dyadic product, which creates a matrix out of two vectors: \begin{align*} \boldsymbol{a}\boldsymbol{b} = \left[\begin{matrix} a_x b_x, & a_x b_y, & a_x b_z\\ a_y b_x, & a_y b_y, & a_y b_z\\ a_z b_x, & a_z b_y, & a_z b_z\\ \end{matrix} \right] \end{align*} This operation appears once in the momentum balance equation, unfortunately that equation is used all the time. We'll talk about its physical meaning later when its introduced.
These are enough vector notation basics to get us started, but you may have noticed that the notation is quite clumsy.
It takes quite a lot of work to write out what the result of a single dyadic operation is, imagine if we actually tried to do some maths/calculus with it!
We would like a more compact notation to work in, one which lets us write everything above in just a couple of symbols.
The best notation would be more clear than vector notation, but still only use a few symbols, letting you easily work out complicated statements.

Index Notation

It turns out that one of Einstein's lesser known achievements was that he invented a very convenient notation for vectors, called index notation .
His idea was this: you often perform dot products and it is quite convenient to write it using a summation operator: \begin{align*} \boldsymbol{a}\cdot\boldsymbol{b} &= a_x b_x + a_y b_y +a_z b_z\\ &=\sum_{i=x,y,z} a_i b_i \end{align*}
But the true genius of Einstein was revealed when he realised he wouldn't even need to write the sum symbol, $\sum_{}$ !
He realised that every time there is a sum, there is a repeated index (the $i$ appears twice in the expression).
So he said just forget to write the sum there if the index is repeated! \begin{align*} \boldsymbol{a}\cdot\boldsymbol{b}=\sum_{i=x,y,z} a_i b_i\equiv a_i b_i \end{align*}

Index Notation

But this means whenever an index is not repeated, it is a so-called free index , or index you can pick or choose.
For example, what does a simple vector look like in index notation? \begin{align*} \boldsymbol{a} = a_i \end{align*}
The index $i$ is not repeated, so we know that we have the choice to pick its value, thus there are three separate values here, i.e., this is a vector.
What about vector addition? \begin{align*} \boldsymbol{a}+\boldsymbol{b} = a_i + b_i \end{align*}
A small extension is needed here: Indices are only summed over if they are repeated within a single multiplicative term. Thus the above result is a vector and there is no implicit sum (as expected for the sum of two vectors).
If we want the $x$ value of the (vector) result, we just set the free index to $i=x$ : \begin{align*} \left[\boldsymbol{a}+\boldsymbol{b}\right]_x = a_x + b_x \end{align*}

Index Notation

A more complicated example: lets take the following expression: \begin{align*} \left(\boldsymbol{a}\cdot\boldsymbol{b}\right)\boldsymbol{c} = \sum_{i=x,y,z} a_i b_i \left[ \begin{matrix} c_x \\ c_y \\ c_z \end{matrix}\right] \end{align*}
We can drop the sum over $i$ as it is a repeated index, but what to do about $\boldsymbol{c}$ ? We just leave it as a new free index $j$ : \begin{align*} \sum_{i=x,y,z} a_i b_i \left[ \begin{matrix} c_x \\ c_y \\ c_z \end{matrix}\right] \equiv a_i b_i c_j \end{align*}
As the $j$ index is free (only appears once in a multiplicative term), it means this result is a vector, and the components are obtained by setting $j=x,y,z$ !

Index Notation

You have to be careful that certain terms have the correct free indices, for example: \begin{align*} \boldsymbol{a}+\boldsymbol{b}\neq a_i + b_j \end{align*}
But these mistakes are easy to spot, as each multiplicative term in the expressions we work with should have exactly one of every free index.
Technically, it doesn't matter which letter you use as the free index either: \begin{align*} (\boldsymbol{a}\cdot\boldsymbol{b})\boldsymbol{c} &= a_i b_i c_j\\ &= a_j b_j c_i\\ &= a_k b_k c_j\\ &=… \end{align*}
But you must be consistent in the order of the free indices when converting, i.e. \begin{align*} \boldsymbol{A} = A_{ij} \neq A_{ji} \end{align*}

Index Notation

To recap:

An expression with no free indices is a scalar: \begin{align*} \boldsymbol{a}\cdot\boldsymbol{b} = a_i b_i \end{align*}
An expression with one free index per additive term is a vector: \begin{align*} \boldsymbol{a} + \boldsymbol{b}= a_i + b_i \end{align*}
An expression with two free indices is a matrix, e.g. \begin{align*} \boldsymbol{A} &= A_{ij} \\ \boldsymbol{a}\boldsymbol{b} &= a_i b_j = \left[\begin{matrix} a_x b_x, & a_x b_y, & a_x b_z\\ a_y b_x, & a_y b_y, & a_y b_z\\ a_z b_x, & a_z b_y, & a_z b_z\\ \end{matrix} \right] \end{align*}
In general, an expression with $X$ free indices is an $X$-order tensor.

Index Notation

There is one common vector operation which is quite awkward to describe in index notation.
The cross product, denoted using $\boldsymbol{a}\times\boldsymbol{b}$ , is an operation that is unique to 3D problems. It is defined as follows: \begin{align*} \boldsymbol{a}\times\boldsymbol{b} = \left[\begin{matrix} a_y b_z - a_z   b_y\\ a_z b_x - a_x   b_z\\ a_x b_y - a_y   b_x \end{matrix} \right] = \boldsymbol{\mathrm{\epsilon}}_{ijk} a_j b_k \end{align*}
The index notation representation relies on the use of the Levi-Civita symbol ($\boldsymbol{\mathrm{\epsilon}}_{ijk}$ ), which is a 3rd-order tensor:
Fortunately, we will not need to use this operation often in this course (1 occurrence, not on an exam). I'm just putting it here for completeness.

Index Notation

The last topic to discuss regarding vector notation is vector calculus.
This is a rich topic, which we can only skim in the time we have.
In particular, we will focus on the del (or vector differential) operator, $\nabla$ (this is also the greek symbol nabla).
This is the vector equilivent of a derivative and is defined in Cartesian coordinates as follows: \begin{align*} \nabla =\left[\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right] \end{align*}
We can also define it in index notation, but first we have to define the generic position vector: \begin{align*} \boldsymbol{r} =\left[x,y,z\right] \end{align*}
And then we can write the del operator in index notation: \begin{align*} \nabla =\frac{\partial}{\partial r_i} \end{align*}

Index Notation

To help us understand what the del operator is, we'll look at the 1,2 and 3 dimensional versions.
In 1D, if we have a function $f(x)$ , then $\nabla f=\frac{\partial f}{\partial x}$ , and it is the slope of the function.
In 2D, if we have a function $g(x,y)$ , then $\nabla g=\left[\frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}\right]$ . This is a vector which points in the direction of the steepest increasing slope, with a magnitude equal to the slope (arrows).
So if you're standing on a slope, and calculate the vector $\nabla altitude$ , it is a vector in the direction of uphill with a length proportional to the gradient.

Index Notation

In 3D, the best analogy for the del operator is to think of a bottle of perfume in a large room.
After some time of the bottle being opened, there is a concentration gradient of perfume in the air with the strongest smell at the source/bottle.
The vector $\nabla c(x,y,z)$ , where $c$ is the concentration (or smell) at your location in the room, will point in the direction of the bottle of perfume.
The magnitude, $\left|\nabla c(x,y,z)\right|$ will equal to how quickly the scent increases in strength as you approach the bottle.
We will find the del operator very useful in the following lectures as a way of simplifying our equations.

Learning Objectives

To understand the representations of vectors and that they encode directions and magnitudes.
To be aware of Cartesian, cylindrical and spherical coordinate systems.
To know that each vector operation is defined differently for different coordinate systems.
To know the magnitude, dot, dyadic and cross products for Cartesian systems.
To be able to work in Einstein's index notation!
To know the definition of the del operator in Cartesian systems and its index notation form.