The Jacobian of the inverse of a transformation

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

.

.

2018.08.09 Thursday (c) All rights reserved by ACHK

Chain Rule of Differentiation

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

or

\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

.

.

2018.07.15 Sunday (c) All rights reserved by ACHK

多項選擇題 6

Multiple Choices 6

這段改編自 2010 年 8 月 24 日的對話。

.

有時,一題物理 MC(多項選擇題)會,同時有數學做法和物理做法。

那時,你就先用物理方法做一次,再用數學方法做多一次,以作驗算。

(問:哪有那麼多的時間?)

之前講過,那些做法,必須透過考試前,平日多加收集和練習而來;並不是在考試中途,才花額外時間發明。

— Me@2018-05-22 06:02:40 PM

.

.

2018.05.22 Tuesday (c) All rights reserved by ACHK

Van der Waals equation 1.2

Whether X_{\text{measured}} is bigger or smaller than X_{\text{ideal}} ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

.

In the ideal gas equation derivation, the volume used in the equation refers to the volume that the gas molecules can move within. So

V_{\text{ideal}} = V_{\text{available for a real gas' molecules to move within}}

Then, when deriving the pressure, it is assumed that there are no intermolecular forces among gas molecules. So

P_{\text{ideal}} = P_{\text{assuming no intermolecular forces}}

.

These are the reasons that

V_{\text{ideal}} < V_{\text{measured}}

P_{\text{ideal}} > P_{\text{measured}}

P_{\text{ideal}} V_{\text{ideal}} = nRT

\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) \left(V_\text{measured}-nb\right) = nRT

— Me@2018-05-16 07:12:51 PM

~~~

… the thing to keep in mind is that the “pressure we use in the ideal gas law” is not the pressure of the gas itself. The pressure of the gas itself is too low: to relate that pressure to “pressure for the ideal gas law” we have to add a number. While the volume occupied by the real gas is too large – the “ideal volume” is less than that. – Floris Sep 30 ’16 at 17:34

— Physics Stackexchange

.

.

2018.05.16 Wednesday (c) All rights reserved by ACHK

Van der Waals equation 1.1

Why do we add, and not subtract, the correction term for pressure in [Van der Waals] equation?

Since the pressure of real gases is lesser than the pressure exerted by (imaginary) ideal gases, shouldn’t we subtract some correction term to account for the decrease in pressure?

I mean, that’s what we have done for the volume correction: Subtracted a correction term from the volume of the container V since the total volume available for movement is reduced.

asked Sep 30 ’16 at 15:20
Ram Bharadwaj

— Physics Stackexchange

.

Ideal gas law:

P_{\text{ideal}} V_{\text{ideal}} = nRT

However, since in a real gas, there are attractions between molecules, so the measured value of pressure P is smaller than that in an ideal gas:

P_{\text{measured}} = P_{\text{real}}

P_{\text{measured}} < P_{\text{ideal gas}}

Also, since the gas molecules themselves occupy some space, the measured value of the volume V is bigger that the real gas really has:

V_{\text{measured}} > V_{\text{real}}

P_{\text{ideal}} V_{\text{ideal}} = nRT

If we substitute P_{\text{measured}} onto the LHS, since P_{\text{measured}} < P_{\text{ideal}}, the LHS will be smaller than the RHS:

P_{\text{measured}} V_{\text{ideal}} < nRT

So in order to maintain the equality, a correction term to the pressure must be added:

\left(P_\text{measured} + a\left(\frac{n}{V}\right)^2\right) V_{\text{ideal}} = nRT

P_{\text{ideal}} V_{\text{ideal}} = nRT

If we substitute V_{\text{measured}} onto the LHS, since that volume is bigger that actual volume available for the gas molecules to move, the LHS will be bigger than the RHS:

P_{\text{ideal}} V_{\text{measured}} > nRT

So in order to maintain the equality, a correction term to the pressure must be subtracted:

P_{\text{ideal}} \left(V_\text{measured}-nb\right) = nRT

.

In other words,

V_{\text{measured}} > V_{\text{real}}

V_{\text{ideal}} = V_{\text{real}}

V_{\text{measured}} > V_{\text{ideal}}

— Me@2018-05-13 03:37:18 PM

.

Why? I still do not understand.

.

How come

P_{\text{measured}} = P_{\text{real}}

but

V_{\text{measured}} \ne V_{\text{real}}?

.

How come

V_{\text{real}} = V_{\text{ideal}}

but

P_{\text{real}} \ne P_{\text{ideal}}?

— Me@2018-05-13 03:22:54 PM

.

The above is wrong.

The “real volume” V_{\text{real}} has 2 possible different meanings.

One is “the volume occupied by a real gas”. In other words, it is the volume of the gas container.

Another is “the volume available for a real gas’ molecules to move”.

.

To avoid confusion, we should define

V_{\text{real}} \equiv V_{\text{measured}}

P_{\text{real}} \equiv P_{\text{measured}}

.

Or even better, avoid the terms P_{\text{real}} and V_{\text{real}} altogether. Instead, just consider the relationship between (P_{\text{ideal}}, P_{\text{measured}}) and that between (V_{\text{ideal}}, V_{\text{measured}}).

Whether X_{\text{measured}} is bigger or smaller than X_{\text{ideal}} ultimately depends on the assumptions and definitions used in the derivation of the ideal gas equation itself.

— Me@2018-05-13 04:15:34 PM

.

.

2018.05.13 Sunday (c) All rights reserved by ACHK