3 Vector Fields and One-Form Fields, 1.2

Posted on December 27, 2023 by rpflee

p. 21

…

$\displaystyle{(Df(x)) b(x) \ne (Df(x)) \Delta x}$

Instead, $\displaystyle{(Df(x)) b(x)}$ is a generalization of $\displaystyle{(Df(x)) \Delta x}$ .

However, how to calculate $\displaystyle{(Df(x)) b(x)}$ ?

By this:

$\displaystyle{f(x + \Delta x) \approx f(x) + (Df(x)) \Delta x}$

Then:

$\displaystyle{(Df(x)) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}}$

So:

$\displaystyle{(Df(x)) b(x) = \left( \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \right) b(x)}$

$b$ is written as subscript to capture the meaning of “with respect to $b$ “. The original directional derivative uses the same convention:

So the spatial rate of change of $\displaystyle{f}$ along the direction of the vector $\displaystyle{\mathbf{v}}$ is

$\displaystyle{\begin{aligned} D_{\mathbf{v}}(f) &= \frac{\left(\delta f\right)_{\mathbf{v}}}{|\mathbf{v}|} \\ &= \frac{1}{|\mathbf{v}|} \left( \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial x} \delta y \right) \\ &= \frac{\partial f}{\partial x} \frac{\delta x}{\sqrt{(\delta x)^2 + (\delta y)^2}} + \frac{\partial f}{\partial x} \frac{\delta y}{\sqrt{(\delta x)^2 + (\delta y)^2}} \\ &= \left(\nabla f\right) \cdot \frac{\mathbf{v}}{|\mathbf{v}|} \\ &= \left(\nabla f\right) \cdot \hat{\mathbf{v}} \\ \end{aligned}}$

$\displaystyle{D_{\mathbf{v}}(f)}$ is called directional derivative.

— Me@2016-02-06 09:49:22 PM

— Me@2023-12-27 01:37:32 PM

— Functional Differential Geometry

3 Vector Fields and One-Form Fields, 1

Posted on September 30, 2023 by rpflee

Chain Rule of Differentiation, 2

Functional Differential Geometry

p. 21

In multiple dimensions the derivative of a function is the multiplier for the best linear approximation of the function at each argument point:

$\displaystyle{f(x + \Delta x) \approx f(x) + (Df(x)) \Delta x}$

In other words:

$\displaystyle{ \begin{aligned} f(x + \Delta x) &= f(x) + a_1 (\Delta x) + a_2 (\Delta x)^2 + \cdots \\ (Df(x)) &= a_1 \\ \end{aligned} }$

The derivative $Df(x)$ is independent of $\Delta x$ .

… the product $\displaystyle{(Df(x))\Delta x}$ is invariant under change of coordinates …

$\displaystyle{(Df(x)) \Delta x}$ is the directional derivative of $\displaystyle{f}$ at $\displaystyle{x}$ with respect to $\displaystyle{\Delta x}$ .

$\displaystyle{(Df(x)) b(x) \ne (Df(x)) \Delta x}$

Instead, $\displaystyle{(Df(x)) b(x)}$ is a generalization of $\displaystyle{(Df(x)) \Delta x}$ .

— Me@2023-09-29 11:48:36 PM

The Jacobian of the inverse of a transformation

Posted on August 9, 2018 by rpflee

The Jacobian of the inverse of a transformation is the inverse of the Jacobian of that transformation

In this post, we would like to illustrate the meaning of

the Jacobian of the inverse of a transformation = the inverse of the Jacobian of that transformation

by proving a special case.

Consider a transformation $\mathscr{T}: \bar{x}^i=\bar{x}^i (x^1,x^2)$ , which is an one-to-one mapping from unbarred $x^i$ ‘s to barred $\bar{x}^i$ coordinates, where $i=1, 2$ .

By definition, the Jacobian matrix J of $\mathscr{T}$ is

$J= \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{pmatrix}$

Now we consider the the inverse of the transformation $\mathscr{T}$ :

$\mathscr{T}^{-1}: x^i=x^i(\bar{x}^1,\bar{x}^2)$

By definition, the Jacobian matrix $\bar{J}$ of this inverse transformation, $\mathscr{T}^{-1}$ , is

$\bar{J}= \begin{pmatrix} \displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^1}{\partial \bar{x}^2}} \\ \displaystyle{\frac{\partial x^2}{\partial \bar{x}^1}} & \displaystyle{\frac{\partial x^2}{\partial \bar{x}^2}} \end{pmatrix}$

On the other hand, the inverse of Jacobian $J$ of the original transformation $\mathscr{T}$ is

$J^{-1}=\displaystyle{\frac{1}{ \begin{vmatrix} \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \end{vmatrix} }} \begin{pmatrix} \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} & \displaystyle{-\frac{\partial \bar{x}^1}{\partial x^2}} \\ \displaystyle{-\frac{\partial \bar{x}^2}{\partial x^1}} & \displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}} \end{pmatrix}$

If $\bar{J} = J^{-1}$ , their (1, 1)-elementd should be equation:

$\displaystyle{\frac{\partial x^1}{\partial \bar{x}^1}}\stackrel{?}{=}\displaystyle{\frac{1}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}-\displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}\displaystyle{\frac{\partial \bar{x}^2}{\partial x^1}} }} \bigg( \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}} \bigg)$

Let’s try to prove that.

Consider equations

$\bar{x}^1 = \bar{x}^1(x^1,x^2)$

$\bar{x}^2 = \bar{x}^2(x^1,x^2)$

Differentiate both sides of each equation with respect to $\bar{x}^1$ , we have:

$A := 1=\displaystyle{\frac{\partial \bar{x}^1}{\partial \bar{x}^1}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$

$B := 0 = \displaystyle{\frac{\partial \bar{x}^2}{\partial \bar{x}^1}=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}}$

$A \times \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}:~~~~~C := \displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}=\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}+\frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^2}{\partial x^2}}$

$B \times \displaystyle{\frac{\partial \bar{x}^1}{\partial x^2}}:~~~~~D := \displaystyle{0=\frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial x^1}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}+\frac{\partial \bar{x}^2}{\partial x^2}\frac{\partial x^2}{\partial \bar{x}^1}\frac{\partial \bar{x}^1}{\partial x^2}}$

$D-C:$

$\displaystyle{ \frac{\partial \bar{x}^2}{\partial x^2}= \bigg( \frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^2}{\partial x^1}\frac{\partial \bar{x}^1}{\partial x^2}\bigg) \frac{\partial x^1}{\partial \bar{x}^1}}$ ,

results

$\displaystyle{ \frac{\partial x^1}{\partial \bar{x}^1}}=\frac{\displaystyle{\frac{\partial \bar{x}^2}{\partial x^2}}}{\displaystyle{\frac{\partial \bar{x}^1}{\partial x^1}\frac{\partial \bar{x}^2}{\partial x^2} - \frac{\partial \bar{x}^1}{\partial x^2}\frac{\partial \bar{x}^2}{\partial x^1}}}$

…

— Me@2018-08-09 09:49:51 PM

Chain Rule of Differentiation

Posted on July 15, 2018 by rpflee

Consider the curve $y = f(x)$ .

$\displaystyle{\frac{d}{dx}}$ is an operator, meaning “the slope of the tangent of”. So the expression $\displaystyle{\frac{dy}{dx}}$ , meaning $\displaystyle{\frac{d}{dx} (y)}$ , is not a fraction.

In order words, it means the slope of the tangent of the curve $y = f(x)$ at a point, such as point $A$ in the graph.

d_2018_07_15__21_31_32_PM_

The symbol $dx$ has no relation with the symbol $\displaystyle{\frac{dy}{dx}}$ . It means $\Delta x$ as shown in the graph. In other words,

$dx = \Delta x$

The symbol $dy$ also has no relation with the symbol $\displaystyle{\frac{dy}{dx}}$ . It means the vertical distance between the current point $A(x_0, y_0)$ , where $y_0 = f(x_0)$ , and the point $C$ on the tangent line $y = mx + c$ , where $m$ is the slope of the tangent line. In other words,

$dy = m~dx$

$\displaystyle{dy = \left[ \left( \frac{d}{dx} \right) y \right] dx}$

The relationship of $\Delta y$ and $dy$ is that

$\displaystyle{\Delta y = \frac{dy}{dx} \Delta x + \text{higher order terms}}$

$\displaystyle{\Delta y = \frac{dy}{dx} dx + \text{higher order terms}}$

$\Delta y = dy + \text{higher order terms}$

Similarly, for functions of 2 variables:

$\displaystyle{\Delta f(x,y) = \frac{\partial f}{\partial x} \Delta x + \frac{\partial f}{\partial y} \Delta y + \text{higher order terms}}$

$\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy}$

For functions of 3 variables:

$\displaystyle{df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz}$

$\displaystyle{\frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} + \frac{\partial f}{\partial z}\frac{dz}{dt}}$

— Me@2018-07-15 09:30:29 PM

Gradient 1.2

Posted on May 1, 2016 by rpflee

Distance vs Displacement, 2

The physical reason of “the magnitude of the gradient vector represents the spatial rate of change” of a scalar field is that $\displaystyle{\frac{\partial f}{\partial x}}$ represents the spatial rate of change of a scalar field along the $\displaystyle{x}$ direction.

Directional derivative has exactly the same meaning except that its direction may not be along any one of the coordinate axes.

— Me@2016-02-06 07:23:32 AM

Assume that $\displaystyle{\delta x}$ represents a displacement from point 1 to point 2 along the $\displaystyle{x}$ direction and $\displaystyle{\delta y}$ represents a displacement from point 2 to point 3 along the $\displaystyle{y}$ direction.

Denote “the value of the vector field” as “height”. Then

the height difference between point 3 and point 1

= the height difference between point 2 and point 1

+ the height difference between point 3 and point 2

That is the exact reason that the change of the $\displaystyle{f}$ due to the displacement $\displaystyle{\mathbf{v}}$ is

$\displaystyle{ \begin{aligned} \left(\delta f\right)_{\mathbf{v}} &= \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial x} \delta y \\ &= \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x}\right) \cdot (\delta x, \delta y) \\ &= \left(\nabla f\right) \cdot \mathbf{v} \\ \end{aligned}}$

The “height difference” does not care about the cause or process that introduces that height change.

— Me@2016-04-21 11:16:06 PM

Gradient

Posted on February 28, 2016 by rpflee

Assume $\displaystyle{(x, y)}$ represents the position of an object and $\displaystyle{f(x,y)}$ is a scalar field on the $\displaystyle{x}$ – $\displaystyle{y}$ plane. Then $\displaystyle{\frac{\partial f}{\partial x}}$ represents the change of $\displaystyle{f}$ per unit length along the positive $\displaystyle{x}$ direction. In other words, it is the spatial rate of change of $\displaystyle{f}$ along the $\displaystyle{x}$ direction.

Similarly, derivative $\displaystyle{\frac{\partial f}{\partial y}}$ represents the spatial rate of change of $\displaystyle{f}$ along the $\displaystyle{y}$ direction.

For an arbitrary direction, due to the nature of displacement, the change of $\displaystyle{f}$ is $\displaystyle{\delta f = \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial x} \delta y}$ when the object has finished moving $\displaystyle{\delta x}$ in $\displaystyle{x}$ direction and then $\displaystyle{\delta y}$ in $\displaystyle{y}$ direction.

Then, the spatial rate of change of $\displaystyle{f}$ is

$\displaystyle{ \begin{aligned} &\frac{\delta f}{\sqrt{(\delta x)^2 + (\delta y)^2}} \\ &= \frac{\partial f}{\partial x} \frac{\delta x}{\sqrt{(\delta x)^2 + (\delta y)^2}} + \frac{\partial f}{\partial x} \frac{\delta y}{\sqrt{(\delta x)^2 + (\delta y)^2}} \\ \end{aligned} }$

For simplicity, denote the resultant displacement as $\displaystyle{\mathbf{v}}$ :

$\displaystyle{\mathbf{v} = (\delta x, \delta y)}$

and define $\displaystyle{\nabla f(x)}$ as

$\displaystyle{\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)}$

Then, the change of the $\displaystyle{f}$ due to the displacement $\displaystyle{\mathbf{v}}$ is

$\displaystyle{\begin{aligned} \left(\delta f\right)_{\mathbf{v}} &= \frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial x} \delta y \\ &= \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial x}\right) \cdot (\delta x, \delta y) \\ &= \left(\nabla f\right) \cdot \mathbf{v} \\ \end{aligned}}$

So the spatial rate of change $\displaystyle{f}$ along the direction of the vector $\displaystyle{\mathbf{v}}$ is

$\displaystyle{\begin{aligned} D_{\mathbf{v}}(f) &= \frac{\left(\delta f\right)_{\mathbf{v}}}{|\mathbf{v}|} \\ &= \frac{\partial f}{\partial x} \frac{\delta x}{\sqrt{(\delta x)^2 + (\delta y)^2}} + \frac{\partial f}{\partial x} \frac{\delta y}{\sqrt{(\delta x)^2 + (\delta y)^2}} \\ &= \left(\nabla f\right) \cdot \frac{\mathbf{v}}{|\mathbf{v}|} \\ &= \left(\nabla f\right) \cdot \hat{\mathbf{v}} \\ \end{aligned}}$

$\displaystyle{D_{\mathbf{v}}(f)}$ is called directional derivative.

— Me@2016-02-06 09:49:22 PM

This is the reason that $\displaystyle{\nabla f}$ is in the steepest direction.

If $\displaystyle{\hat{\mathbf{v}}}$ is chosen to be parallel to $\displaystyle{\nabla f}$ , the directional derivative $\displaystyle{\left(\nabla f\right) \cdot \hat{\mathbf{v}}}$ would be maximized.

— Me@2021-08-20 05:20:02 PM

Physics Town

continues

Category Archives: differential