# Tensor Analysis – Grinfeld – Chapter 4 – Jacobians This picture is a Clickable Link to the Amazon.co.uk website. Pavel Grinfeld’s “Introduction to Tensor Analysis and the Calculus of Moving Surfaces” is the latest book on the subject. You should see a few other books on Tensor Calculus there too.

### Basic Polar and Cartesian Transformations

Following on from Tensor Analysis – Grinfeld – Chapter 4 and Tensor Analysis – Grinfeld – Chapter 4 – Triple Derivative I take a look at coordinate transformations and the Jacobians.

To begin with, as an example, Pavel uses the two-dimensional Cartesian and the two-dimensional Polar coordinate systems :~ $r \left(x, y \right) = \sqrt{x^2 + y^2} \qquad \qquad (4.1)$ $\theta \left(x, y \right) = arctan \left( \frac{y}{x} \right) \qquad \qquad (4.2)$

There are two alternative versions of $\theta$ which I will use below :~ $\theta \left(x, y \right) = arccos \left( \frac{x}{\sqrt{x^2 + y^2}} \right) \qquad \qquad (4.2a)$
and $\theta \left(x, y \right) = arcsin \left( \frac{y}{\sqrt{x^2 + y^2}} \right) \qquad \qquad (4.2b)$ $x \left(r, \theta \right) = r \ cos \ \theta \qquad \qquad (4.3)$ $y \left(r, \theta \right) = r \ sin \ \theta \qquad \qquad (4.4)$ $r \left(x \left(r, \theta \right), y \left( r, \theta \right) \right) \equiv r \qquad \qquad (4.51)$

This just says that r is a function of x and y, which are themselves functions of r and $\theta$. So :~ \begin{aligned} r &= \sqrt{x^2 + y^2} \\ &= \sqrt{ \left(r \ cos \ \theta \right)^2 + \left( r \ sin \ \theta \right)^2 } \\ &= \sqrt{ r^2 cos^2 \theta + r^2 sin^2 \theta } \\ &= \sqrt{ r^2 cos^2 \theta + r^2 sin^2 \theta } \\ &= r \sqrt{cos^2 \theta + sin^2 \theta } \\ &= r \end{aligned}

Similarly with $\theta \$ :~ $\theta \left(x \left(r, \theta \right), y \left( r, \theta \right) \right) \equiv \theta \qquad \qquad (4.52)$ \begin{aligned} \theta &= arctan \left( \frac{y}{x} \right) \\ &= arctan \left( \frac{r \ sin \ \theta}{r \ cos \ \theta} \right) \\ &= arctan \left( tan \ \theta \right) \\ &= \theta \end{aligned}

I know that these are “circular arguments”, for r and $\theta$ but it proves that the formulation is correct and that the tranformation works. We can do the same for x and y :~ $x \left(r \left(x, y \right), \theta \left( x, y \right) \right) \equiv x$

and $y \left(r \left(x, y \right), \theta \left( x, y \right) \right) \equiv y$

(These are Exercise (44).) These are the same as $F \left(f \left(x, y \right), g \left( x, y \right) \right) = x \qquad \qquad (4.53)$

and $G \left(f \left(x, y \right), g \left( x, y \right) \right) = y \qquad \qquad (4.54)$

Here, Pavel is trying to get us used to the idea of using more general terms. The whole idea is that we work in Tensor Form when we do our mathematical analysis, on whatever the problem we need to work on, and only when we are happy with that, do we worry about the actual algebra of coordinates and their transformations. It’s the “Divide and Conquer” technique. You break the problem down into manageable chunks.

We can do this circular argument thing with x and y :~ $x \left(r, \theta \right) = r \ cos \ \theta$ $r \left(x, y \right) = \sqrt{x^2 + y^2}$

and, best to use, $\theta \left(x, y \right) = arccos \left( \frac{x}{\sqrt{x^2 + y^2}} \right)$

therefore \begin{aligned} x &= \sqrt{x^2 + y^2} \cdot cos \left\{ arccos \left( \frac{x}{\sqrt{x^2 + y^2}} \right) \right\} \\ &= \sqrt{x^2 + y^2} \cdot \frac{x}{\sqrt{x^2 + y^2}} \\ & = x \end{aligned}

Similarly for y :~ $y \left(r, \theta \right) = r \ sin \ \theta$

and, best to use, $\theta \left(x, y \right) = arcsin \left( \frac{y}{\sqrt{x^2 + y^2}} \right)$

therefore \begin{aligned} y &= \sqrt{x^2 + y^2} \cdot sin \left\{ arcsin \left( \frac{y}{\sqrt{x^2 + y^2}} \right) \right\} \\ &= \sqrt{x^2 + y^2} \cdot \frac{y}{\sqrt{x^2 + y^2}} \\ & = y \end{aligned}

As I say, circular arguments, as with $r \ \textrm{and} \ \theta$. However, it is good to become fluently familiar with these coordinate transformations.

### First and Second Derivative of Transformations

Pavel differentiates Equation (4.53) and Equation (4.54) with respect to both x and y to give :~ $\frac{\partial F}{\partial X}\frac{\partial f}{\partial x} + \frac{\partial F}{\partial Y}\frac{\partial g}{\partial x} = 1 \qquad \qquad (4.55)$ $\frac{\partial F}{\partial X}\frac{\partial f}{\partial y} + \frac{\partial F}{\partial Y}\frac{\partial g}{\partial y} = 0 \qquad \qquad (4.56)$

What is being done here is that when F is differentiated w.r.t. x it must first be differentiated with respect to the sub-functions, f and g. Then this throws out a further partial derivative of f, w.r.t. x (or g, w.r.t. x ).
On the right hand side, of (4.53), when x is differentiated w.r.t. x, we get simply 1.

When F is differentiated w.r.t. y it must first be differentiated with respect to the sub-functions, f and g. Then this throws out a further partial derivative of f w.r.t. y (or g w.r.t. y ).
On the right hand side, of (4.53), when x is differentiated w.r.t. y, we get simply 0.

You might want to try to write-out the equivalent first-order derivative expressions from the results of Exercise (44), which I have given above.

Clearly equations (4.57) and (4.58) are derived in a similar way. I won’t give these here as you have to buy the book 😉 (please see the link to Amazon above).

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Exercise 45 askes us to find the the second-order partial derivatives of F and G, w.r.t. x and y.

Working with Eqn(4.55), first of all, we see that this was obtained by first differentiating F, w.r.t. X (and then w.r.t. x) and then w.r.t. Y (and then w.r.t. x). Similarly, to obtain the second derivative we need to differentiate the whole of (4.55) w.r.t. X (and then w.r.t. x) and then w.r.t. Y (and then w.r.t. x). Equation (4.55) has two terms on the LHS. I like to break things down into manageable chunks, so I will differentiate each of these two terms one at a time :~ $\frac{ \partial }{\partial x} \left[ \frac{\partial F}{\partial X}\frac{\partial f}{\partial x}\right] = \frac{\partial^2 F}{\partial X^2}\frac{\partial f}{\partial x}\frac{\partial f}{\partial x} + \frac{\partial F}{\partial X}\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 F}{\partial X \partial Y}\frac{\partial g}{\partial x}\frac{\partial f}{\partial x}$

So we are differentiating the product of a function of a function, times a function of x, the independent variable. We therefore use both the product and chain rules of differentiation.

Similarly, differentiating the second term of (4.55), w.r.t. x, gives :~ $\frac{ \partial }{\partial x} \left[ \frac{\partial F}{\partial Y}\frac{\partial g}{\partial x}\right] = \frac{\partial^2 F}{\partial Y \partial X}\frac{\partial f}{\partial x}\frac{\partial g}{\partial x} + \frac{\partial F}{\partial Y}\frac{\partial^2 g}{\partial x^2} + \frac{\partial^2 F}{\partial Y^2}\frac{\partial g}{\partial x}\frac{\partial g}{\partial x}$

If you add these two results together, you will find that you obtain Eqn(4.62). Clearly if you differentiate the RHS of (4.55) w.r.t. x you get zero, as required.
Again, you may want to go back to x and y as functions of r and $\theta$ to see how these workings pan-out, but only if you are unable to follow this here.

For completeness, I will do the remaining three second-order relationships as requested in Exercise 45, presented in a similar manner, to above, i.e. I split the two terms up :~ \begin{aligned} \frac{ \partial }{\partial y} \left[ \frac{\partial F}{\partial X}\frac{\partial f}{\partial x} + \frac{\partial F}{\partial Y}\frac{\partial g}{\partial x}\right] &= \frac{\partial^2 F}{\partial X^2}\frac{\partial f}{\partial y}\frac{\partial f}{\partial x} + \frac{\partial F}{\partial X}\frac{\partial^2 f}{\partial x \partial y} + \frac{\partial^2 F}{\partial X \partial Y}\frac{\partial g}{\partial y}\frac{\partial f}{\partial x} \\ &+ \frac{\partial^2 F}{\partial Y \partial X}\frac{\partial f}{\partial y}\frac{\partial g}{\partial x} + \frac{\partial F}{\partial Y}\frac{\partial^2 g}{\partial x \partial y} + \frac{ \partial^2 F}{\partial Y^2 }\frac{\partial g}{\partial y}\frac{\partial g}{\partial x} \end{aligned}

(Notice that you are not able to add any of the terms in the mixed-derivative, so you end up with the full six terms.)

Eqn(4.56) differentiated w.r.t. x is :~ \begin{aligned} \frac{ \partial }{\partial x} \left[ \frac{\partial F}{\partial X}\frac{\partial f}{\partial y} + \frac{\partial F}{\partial Y}\frac{\partial g}{\partial y}\right] &= \frac{\partial^2 F}{\partial X^2}\frac{\partial f}{\partial x}\frac{\partial f}{\partial y} + \frac{\partial F}{\partial X}\frac{\partial^2 f}{\partial y \partial x} + \frac{\partial^2 F}{\partial X \partial Y}\frac{\partial g}{\partial x}\frac{\partial f}{\partial y} \\ &+ \frac{\partial^2 F}{\partial Y \partial X}\frac{\partial f}{\partial x}\frac{\partial g}{\partial y} + \frac{\partial F}{\partial Y}\frac{\partial^2 g}{\partial y \partial x} + \frac{ \partial^2 F}{\partial Y^2 }\frac{\partial g}{\partial x}\frac{\partial g}{\partial y} \end{aligned}

Eqn(4.56) differentiated w.r.t. y is :~ \begin{aligned} \frac{ \partial }{\partial y} \left[ \frac{\partial F}{\partial X}\frac{\partial f}{\partial y} + \frac{\partial F}{\partial Y}\frac{\partial g}{\partial y}\right] &= \frac{\partial^2 F}{\partial X^2}\frac{\partial f}{\partial y}\frac{\partial f}{\partial y} + \frac{\partial F}{\partial X}\frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 F}{\partial X \partial Y}\frac{\partial g}{\partial y}\frac{\partial f}{\partial y} \\ &+ \frac{\partial^2 F}{\partial Y \partial X}\frac{\partial f}{\partial y}\frac{\partial g}{\partial y} + \frac{\partial F}{\partial Y}\frac{\partial^2 g}{\partial y^2} + \frac{ \partial^2 F}{\partial Y^2 }\frac{\partial g}{\partial y}\frac{\partial g}{\partial y} \end{aligned}

Eqn(4.57) differentiated w.r.t. x is :~ \begin{aligned} \frac{ \partial }{\partial x} \left[ \frac{\partial G}{\partial X}\frac{\partial f}{\partial x} + \frac{\partial G}{\partial Y}\frac{\partial g}{\partial x}\right] &= \frac{\partial^2 G}{\partial X^2}\frac{\partial f}{\partial x}\frac{\partial f}{\partial x} + \frac{\partial G}{\partial X}\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 G}{\partial X \partial Y}\frac{\partial g}{\partial x}\frac{\partial f}{\partial x} \\ &+ \frac{\partial^2 G}{\partial Y \partial X}\frac{\partial f}{\partial x}\frac{\partial g}{\partial x} + \frac{\partial G}{\partial Y}\frac{\partial^2 g}{\partial x^2} + \frac{ \partial^2 G}{\partial Y^2 }\frac{\partial g}{\partial x}\frac{\partial g}{\partial x} \end{aligned}

Eqn(4.57) differentiated w.r.t. y is :~ \begin{aligned} \frac{ \partial }{\partial y} \left[ \frac{\partial G}{\partial X}\frac{\partial f}{\partial x} + \frac{\partial G}{\partial Y}\frac{\partial g}{\partial x}\right] &= \frac{\partial^2 G}{\partial X^2}\frac{\partial f}{\partial y}\frac{\partial f}{\partial x} + \frac{\partial G}{\partial X}\frac{\partial^2 f}{\partial x \partial y} + \frac{\partial^2 G}{\partial X \partial Y}\frac{\partial g}{\partial y}\frac{\partial f}{\partial x} \\ &+ \frac{\partial^2 G}{\partial Y \partial X}\frac{\partial f}{\partial y}\frac{\partial g}{\partial x} + \frac{\partial G}{\partial Y}\frac{\partial^2 g}{\partial x \partial y} + \frac{ \partial^2 G}{\partial Y^2 }\frac{\partial g}{\partial y}\frac{\partial g}{\partial x} \end{aligned}

Eqn(4.58) differentiated w.r.t. x is :~ \begin{aligned} \frac{ \partial }{\partial x} \left[ \frac{\partial G}{\partial X}\frac{\partial f}{\partial y} + \frac{\partial G}{\partial Y}\frac{\partial g}{\partial y}\right] &= \frac{\partial^2 G}{\partial X^2}\frac{\partial f}{\partial x}\frac{\partial f}{\partial y} + \frac{\partial G}{\partial X}\frac{\partial^2 f}{\partial y \partial x} + \frac{\partial^2 G}{\partial X \partial Y}\frac{\partial g}{\partial x}\frac{\partial f}{\partial y} \\ &+ \frac{\partial^2 G}{\partial Y \partial X}\frac{\partial f}{\partial x}\frac{\partial g}{\partial y} + \frac{\partial G}{\partial Y}\frac{\partial^2 g}{\partial y \partial x} + \frac{ \partial^2 G}{\partial Y^2 }\frac{\partial g}{\partial x}\frac{\partial g}{\partial y} \end{aligned}

Eqn(4.58) differentiated w.r.t. y is :~ \begin{aligned} \frac{ \partial }{\partial y} \left[ \frac{\partial G}{\partial X}\frac{\partial f}{\partial y} + \frac{\partial G}{\partial Y}\frac{\partial g}{\partial y}\right] &= \frac{\partial^2 G}{\partial X^2}\frac{\partial f}{\partial y}\frac{\partial f}{\partial y} + \frac{\partial G}{\partial X}\frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 G}{\partial X \partial Y}\frac{\partial g}{\partial y}\frac{\partial f}{\partial y} \\ &+ \frac{\partial^2 G}{\partial Y \partial X}\frac{\partial f}{\partial y}\frac{\partial g}{\partial y} + \frac{\partial G}{\partial Y}\frac{\partial^2 g}{\partial y^2} + \frac{ \partial^2 G}{\partial Y^2 }\frac{\partial g}{\partial y}\frac{\partial g}{\partial y} \end{aligned}

Of course, all of the equations above have zero on the RHS.

It is worth noting that when Eqn(4.53) is differentiated w.r.t. x and then w.r.t. y we obtain two expressions: (4.55) and (4.56). Similarly, (4.54), differentiated w.r.t. x and then w.r.t. y gives a further two expressions: (4.57) and (4.58). So it follows that if each of (4.55), (4.56), (4.57) and (4.58) are then differentiated again, first w.r.t. x and then w.r.t. y, then we end up with eight second-order partial derivatives, as shown above.

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