Jacobian matrix of a variable transformation

[AxiomOfChoice]

Suppose I am changing variables from (x,y) to (s,t), where

<br /> \begin{align*}<br /> s &amp; = \frac 12 (x+y),\\<br /> t &amp; = y - x<br /> \end{align*}<br />

According to Wikipedia, if I want to see how the measure dx dy changes, I need to compute the Jacobian matrix J associated with this variable transformation and take its determinant. It will then follow that dx dy = \det J ds dt. The Jacobian matrix takes the form

<br /> \begin{bmatrix}<br /> \partial x / \partial s &amp; \partial x / \partial t \\<br /> \partial y / \partial s &amp; \partial y / \partial t<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> 1 &amp; -1/2 \\<br /> 1 &amp; 1/2<br /> \end{bmatrix}<br />

Is it just coincidence that the matrix J is identical to the matrix of the transformation; i.e., the matrix that shows up in the identity

<br /> \begin{bmatrix}<br /> x \\ y<br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> 1 &amp; -1/2 \\<br /> 1 &amp; 1/2<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> s \\ t<br /> \end{bmatrix}<br />

 

[chiro]

Suppose I am changing variables from (x,y) to (s,t), where

<br /> \begin{align*}<br /> s &amp; = \frac 12 (x+y),\\<br /> t &amp; = y - x<br /> \end{align*}<br />

According to Wikipedia, if I want to see how the measure dx dy changes, I need to compute the Jacobian matrix J associated with this variable transformation and take its determinant. It will then follow that dx dy = \det J ds dt. The Jacobian matrix takes the form

<br /> \begin{bmatrix}<br /> \partial x / \partial s &amp; \partial x / \partial t \\<br /> \partial y / \partial s &amp; \partial y / \partial t<br /> \end{bmatrix}<br /> =<br /> \begin{bmatrix}<br /> 1 &amp; -1/2 \\<br /> 1 &amp; 1/2<br /> \end{bmatrix}<br />

Is it just coincidence that the matrix J is identical to the matrix of the transformation; i.e., the matrix that shows up in the identity

<br /> \begin{bmatrix}<br /> x \\ y<br /> \end{bmatrix} = <br /> \begin{bmatrix}<br /> 1 &amp; -1/2 \\<br /> 1 &amp; 1/2<br /> \end{bmatrix}<br /> \begin{bmatrix}<br /> s \\ t<br /> \end{bmatrix}<br />

Do you understand what the physical interpretation of the Jacobian is?

 

[HallsofIvy]

No, it is not coincidence. Since this is a linear transformation, its coefficients measure how much the measurement is stretched or contracted along the axes. And that determines how you will measure area.

 

[Trifis]

So it is the Jacobian always identical to the transformation matrix (rotations, reflections, scalling, shearing, projections etc) between two arbitrary vector spaces? If yes up to what extent does this hold?

 

[HallsofIvy]

So it is the Jacobian always identical to the transformation matrix (rotations, reflections, scalling, shearing, projections etc) between two arbitrary vector spaces? If yes up to what extent does this hold?

As long as the transformation is linear (which it must be to be written as a matrix) that is true. The "Jacobian" is most often used with non-linear transformations. In that case it is similar to the derivative in Calculus, which you can use to find the "tangent approximation" to a function at a given point, since you can use the Jacobian, at a point, to find the linear transformation that best approximates the non-linear transformation around that point.

 

[Trifis]

As long as the transformation is linear (which it must be to be written as a matrix) that is true. The "Jacobian" is most often used with non-linear transformations. In that case it is similar to the derivative in Calculus, which you can use to find the "tangent approximation" to a function at a given point, since you can use the Jacobian, at a point, to find the linear transformation that best approximates the non-linear transformation around that point.

So you are suggesting that linearity implies the global nature of the transformation. Otherwise the Jacobian yields a local transformation (and if the derivative is not zero, local reversibility too).

Why don't we use the Jacobian as the standard transformation matrix for every transformation then? I think it has to do with the fact that those transformations, which do not map cartesian coordinate systems to curvilinear and are rather like geometric alterations in the same vector space or at least at the same "type" of coordinates, do not need the extra information about local properties provided by the Jacobian...

But then again if the Jacobian encloses all kinds of linear transformations I do not understand why such an important property is not usually mentioned right after its definition.

 

[Trifis]

Moreover I would like to know what role plays the Jacobian in transformation between affine spaces...

 

[HallsofIvy]

The Jacobian is, by its nature, linear, so can only represent general transformations locally, just as the slope of a straight line can give information about the corresponding linear function globally while the derivative of a non-linear function only gives local information.

 

[Trifis]

But in our case the Jacobian does give us global information since it is identical with the transformation matrix, as you stated above...

 

[Muphrid]

While what HallsofIvy said is correct, I think you're getting confused. The Jacobian descries a local, linear approximation of a general (potentially nonlinear) transformation. When the Jacobian is the same everywhere, then the transformation itself is linear.

I think Jacobians aren't talked about very much because a lot of the transformations we consider are already linear--for example, the rigid rotation of a coordinate system. Nonlinear transformations carry a little more overhead, but the Jacobian is a very powerful tool and essential to understanding the general theory of transformations.

 

[Trifis]

To me then the Jacobian is also the generalization of every transformation matrix of linear algebra. That is what I wanted to hear.

I have two last questions. Firstly does it really matter if our transformation is passive or active? E.g. does the Jacobian remain the same between a rotation and a change of coordinates from cartesian to polar (consider the 2D case since in 3D the Euler angles complicate the matter) ? And secondly (I've already stated that question) does the Jacobian work with homogenous coordinates and by extension to affine spaces as the general transformation matrix (the way it does by linear transformations) ?

 

[Muphrid]

Passive and active transformations ought to be related through transpose, I believe.

You ask if the Jacobian remains the same under a change of coordinates and a rotation; I would say no, because each of those has its own Jacobian matrix and so the matrices will multiply one another for a net Jacobian.

Does it work with homogeneous coordinates? I would say yes. One could model the action of the Jacobian on the added dimension as the identity and then have it do what it should to other dimensions.

 

[Trifis]

@Muphrid is the Jacobian matrix of the Lorentz transformation identical with the transformation matrix we are familiar with?
Despite the fact that this is the maths thread, I ask you this because I believe that you are well-versed with relativity...

 

[Muphrid]

Yes, it naturally is. Usually, Lorentz transformations are global--the same everywhere--but you can instead define a local Lorentz transformation that boosts vectors and tensors by different amounts or directions at every point in spacetime. If this local boost is differentiable, then the Jacobian at every point in space is Lorentz transformation, but the Jacobian still has functional dependence on spacetime position and evaluates to a different matrix at each point in space.