PHY-101: Vector Differentiation

Vector Differentiation

What is Differentiation?

Simply put, Differentiation is the rate of change of a function with respect to a variable. For example, velocity tells us how fast and in which direction an object is moving, which is the rate of change of the object in space with respect to time. Mathematically, $$v=\frac{dx}{dt}$$where x is the space coordinate in 1D and t is time.

In a way, mathematicians like to define is, the slope of the line tangent to the graph of \(f(x)\)

Fig1: The blue line shows the function f(x) and the red line is the first derivative of the function

To understand how to calculate a derivative of a function I'll give you a very basic formula widely known as the First principle $$f '(x)=\lim_{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$where \(f '(x)\) is the first derivate, \(f(x)\) is the function and \(\Delta x\) is the change we are calculating the function over.

You might encounter a few notations while dealing with derivatives of a function.

Let's say it is natural to write \(y=f(x)\) and,

\(\Delta y\)=\(\Delta f\) = \(f(x)-f(x_{0})\)=\(f(x_{0}+\Delta x)-f(x_{0})\)

We would read this as "Delta y" or "Delta f" or the "change in y".

If then we divide both sides by \(\Delta x\)=\(x-x_{0}\), we get two expressions, $$\frac{\Delta y}{\Delta x}=\frac{\Delta f}{\Delta x}$$ Taking the limit as \(\Delta x \rightarrow 0\), we get $$\frac{\Delta y}{\Delta x}\rightarrow \frac{dy}{dx}(Leibniz' \hspace{1mm}notation)$$or$$\frac{\Delta f}{\Delta x}\rightarrow f '(x_{0}) \hspace{2mm} (Newton's \hspace{1mm}notation)$$

Other equally valid notation are : \(\frac{df}{dx},f ', and \hspace{2mm} Df\)

We call the above class of differentiation finding the Ordinary Derivates of a function. In terms of a vector when we calculate the derivative we just repeat the process across all components where the component does depend on the variable against which the derivative is sought after.

Let us understand it using the velocity example again,

In a space R(t), there is a position vector r(t)=\(x(t) \hat i +y(t) \hat j +z(t) \hat k \) a point to be noted is all the components and the position vector all depend on the variable t.

the velocity then, is the rate of change of the position vector, thus

\(v=\frac{dr}{dt}=\frac{dx}{dt}\hat i+\frac{dy}{dt}\hat j+\frac{dz}{dt}\hat k \)

and the acceleration will be \(a=\frac{dv}{dt}=\frac{d^2r}{dt^2}\)

While mathematicians will be interested in finding even higher-order differentials we physicists have no need to go higher than the second order. They do appear but are a rare sight in physics as we can almost find all the important information about an object using second-order Differential equations.

Now that you know the essence of differentiation here are some formulas you can use to simply the questions:

$$\begin{aligned}&1. \frac{d}{du}(A+B)=\frac{dA}{du}+\frac{dB}{du}\\ \\ &2. \frac{d}{du}(A\cdot B)=A \cdot \frac{dB}{du}+\frac{dA}{du}\cdot B \\ \\ &3. \frac{d}{du}(A \times B)=A \times \frac{dB}{du}+\frac{dA}{du}\times B\\ \\ &4. \frac{d}{du}(\phi A)=\phi \frac{dA}{du}+\frac{d\phi}{du}A\\ \\ &5. \frac{d}{du}(A\cdot B  \times C)=A\cdot B\frac{dC}{du}+A\cdot \frac{dB}{du}\times C+\frac{dA}{du}\cdot B\times C\\ \\ &6. \frac{d}{du}(A\times (B\times C))=A\times (B\times \frac{dC}{du})+A\times (\frac{dB}{du}\times C)+\frac{dA}{du}\times(B\times C) \end{aligned}$$

Partial Derivatives

If A is a vector depending on more variables on more than one variable, say x, y, z for example, we then write A=A(x, y, z).

The partial derivative of A with respect  to x is defined as $$\frac{\partial A}{\partial x}=\lim_{\Delta x\to 0}\frac{A(x+\Delta x, y, z)-A(x, y, z)}{\Delta x}$$if this limits exist then similarly, $$\frac{\partial A}{\partial y}=\lim_{\Delta y\to 0}\frac{A(x, y+\Delta y, z)-A(x, y, z)}{\Delta y}$$$$\frac{\partial A}{\partial z}=\lim_{\Delta z\to 0}\frac{A(x, y, z+\Delta z)-A(x, y, z)}{\Delta z}$$

Partial derivative only focuses on one variable, so if you derivate partially for x; y and z terms will not be affected, similarly, we can say for y and z terms too.

While it may seem vague looking just at these mathematical expressions, when you start solving problems I promise it will seem like child's play.

Now that we understand what Partial Derivative is and how we can do Vector Differentiation, it is time for us to go into the application of this Math in Physics.

Del Operator (\(\nabla\))

It is probably one of the most used operators in physics. What this means is$$\nabla=\hat i \frac{\partial}{\partial x}+\hat j \frac{\partial}{\partial y}+\hat k \frac{\partial}{\partial z}$$It is a partial derivative operator which calculates the derivative of each component of a vector in that corresponding axis. It is useful in defining three quantities which arise in practical applications: gradient, divergence and curl. It is also known as nabla.

Gradient

Let us take \(\phi(x, y, z)\) as a scalar function which depends on the cartesian coordinates x, y and z, to be defined and differentiable at each point in a particular space. It tells us about the rate of change of one variable with respect to another. Then the gradient of phi is defined as:$$\nabla \phi=\hat i \frac{\partial \phi}{\partial x}+\hat j \frac{\partial \phi}{\partial y}+\hat k \frac{\partial \phi}{\partial z}$$

Note that \(\nabla \phi\) defines a vector field while \(\phi\) is a scalar differentiable field.

Divergence

Let \(A(x, y ,z)=A_{x}\hat i +A_{y} \hat j +A_{z} \hat k \) be a vector defined in a region of space. Then Divergence of the vector tells us about how this vector diverges in space and it is written as:$$\nabla \cdot A= \frac{\partial A_{x}}{\partial x}+\frac{\partial A_{y}}{\partial y}+ \frac{\partial A_{z}}{\partial z}$$ Note that \(\nabla \cdot A \neq A \cdot \nabla \)

We will discuss about divergence in further detail when we start Electricity and Magnetism.

Curl

Again let us take the \(\vec{A}\) defined previously. The curl of the vector will give us information on the rotation of the vector in space. We write this as:$$\nabla \times \vec{A}=(\hat i \frac{\partial}{\partial x}+\hat j \frac{\partial}{\partial y}+\hat k \frac{\partial}{\partial z}) \times (A_{x}\hat i +A_{y} \hat j +A_{z} \hat k)$$

$$\begin{aligned}&=\begin{vmatrix}\hat i & \hat j & \hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ A_{x} & A_{y} & A_{z} \end{vmatrix}\\ \\&=\begin{vmatrix} \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ A_{y} & A_{z} \end{vmatrix} \hat i - \begin{vmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial z}\\ A_{x} & A_{z} \end{vmatrix} \hat j +\begin{vmatrix} \frac{\partial}{\partial x} & \frac{\partial}{\partial y}\\ A_{x} & A_{y} \end{vmatrix} \hat k \\ \\  &=(\frac{\partial A_{z}}{\partial y}-\frac{A_{y}}{\partial z})\hat i + (\frac{A_{x}}{\partial z}-\frac{A_{z}}{\partial x})\hat j + (\frac{A_{y}}{\partial x}-\frac{A_{x}}{\partial y})\hat k \end{aligned}$$

Note that the \(A_{x}, A_{y} \hspace{1mm} and \hspace{1mm} A_{z}\) are the components of  \(\vec{A}\) and each of these components depend on \(x, y, z\).

Now some identities following the gradient, divergence and curl:

$$\begin{aligned}&1. \nabla(\phi +\psi)=\nabla \phi +\nabla \psi \\ \\&2. \nabla \cdot (\vec{A}+\vec{B})=\nabla \cdot \vec{A}+\nabla \cdot \vec{B}\\ \\&3.\nabla \times (\vec{A}+\vec{B})=\nabla \times \vec{A}+\nabla \times \vec{B}\\ \\&4.\nabla \cdot (\phi \vec{A})=(\nabla \phi)\cdot \vec{A}+\phi(\nabla \cdot \vec{A}) \\ \\&5.\nabla \times(\phi \vec{A})=(\nabla \phi)\times \vec{A}+\phi(\nabla \times \vec{A}) \\ \\&6.\nabla \cdot (\vec{A} \times \vec{B})=\vec{B} \cdot (\nabla \times \vec{A}) -\vec{A} \cdot (\nabla \times \vec{B}) \\ \\ &7.\nabla \times (\vec{A} \times \vec{B})=(\vec{B} \cdot \nabla )\vec{A}+(\vec{A} \cdot \nabla)\vec{B}-(\vec{A} \cdot \nabla)\vec{B}+A(\nabla \cdot \vec{B}) \\ \\ &8.\nabla(\vec{A} \cdot \vec{B}=(\vec{B} \cdot \nabla)\vec{A}+(\vec{A} \cdot \nabla)\vec{B}+\vec{B} \times (\nabla \times \vec{A})+\vec{A} \times (\nabla \times \vec{B}) \\ \\ &9.\nabla \cdot (\nabla \phi)=\nabla^2\phi =\frac{\partial ^\phi}{\partial x^2}+\frac{\partial ^2\phi}{\partial y^2}+\frac{\partial^2\phi}{\partial z^2}\\&\hspace{15mm}where, \nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2}\hspace{2mm} is\hspace{1mm}called\hspace{1mm}the\hspace{1mm}Laplacian\hspace{1mm}operator.\\ \\ &10.\nabla \times (\nabla \phi)=0 \\ \\&11.\nabla \cdot (\nabla \times \vec{A})= 0\\ \\&12.\nabla \times(\nabla \times \vec{A})=\nabla(\nabla \cdot \vec{A})-\nabla^2\vec{A} \end{aligned}$$

Conclusion:

Finally, I will leave you here with all the basic knowledge of vector differentiation. It might seem intimidating initially, but if you solve some practice problems, you will most definitely get it. The Identities shown above will be used constantly throughout the derivations and explanations of various entities in physics, these are merely a ploy for you to use them directly to tackle the problems as such rather than to prove them. In the next blog, we will cover Vector Integration, and a few theorems to make your life simple and then we will continue with Classical Mechanics. Until next time!