A question concerning Jacobians

["Don't panic!"]

Apologies for perhaps a very trivial question, but I'm slightly doubting my understanding of Jacobians after explaining the concept of coordinate transformations to a colleague.

Basically, as I understand it, the Jacobian (intuitively) describes how surface (or volume) elements change under a particular coordinate transformation. In essence, they characterise the difference in how area (or volume) is measured in the two different coordinate systems (I think this is right?!)

When one (informally) derives the Jacobian (aka Jacobian determinant) corresponding to a coordinate transformation the usual approach is to consider a mapping between two coordinate systems (sticking to $2D$ for simplicity), $(u,v)$ and $(x,y)$ respectively, defined by $$x=x(u,v)\, , \;\; y=y(u,v)$$ and then consider a square in the $uv$ plane whose area is given by $\Delta u\Delta v$. One then maps this area into the $xy$ plane, and to a linear approximation this corresponds to a parallelogram with sides $\left(\frac{\partial x}{\partial u}\Delta u\, ,\, \frac{\partial y}{\partial u}\Delta u\, ,\,0\right)$ and $\left(\frac{\partial x}{\partial v}\Delta v\, ,\, \frac{\partial y}{\partial v}\Delta v\, ,\,0\right)$. The corresponding area in the $xy$ plane can then be approximated as $$\Delta A\approx \bigg\lvert\left(\frac{\partial x}{\partial u}\Delta u\, ,\, \frac{\partial y}{\partial u}\Delta u\, ,\,0\right)\times\left(\frac{\partial x}{\partial v}\Delta v\, ,\, \frac{\partial y}{\partial v}\Delta v\, ,\,0\right)\bigg\rvert =\bigg\lvert\left(\frac{\partial x}{\partial u}\, ,\, \frac{\partial y}{\partial u}\, ,\,0\right)\times\left(\frac{\partial x}{\partial v}\, ,\, \frac{\partial y}{\partial v}\, ,\,0\right)\bigg\rvert\Delta u\Delta v$$ We then define the Jacobian as $J(u,v)=\left(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}\right)$ such that $$\bigg\lvert\left(\frac{\partial x}{\partial u}\, ,\, \frac{\partial y}{\partial u}\, ,\,0\right)\times\left(\frac{\partial x}{\partial v}\, ,\, \frac{\partial y}{\partial v}\, ,\,0\right)\bigg\rvert\Delta u\Delta v =\bigg\lvert\left(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial y}{\partial u}\frac{\partial x}{\partial v}\right)\bigg\rvert\Delta u\Delta v =\lvert J(u,v)\rvert\Delta u\Delta v$$ Taking the infinitesimal limit the approximation becomes exact and we have that $$dA=\lvert J(u,v)\rvert dudv$$ This procedure can then be extended to higher dimensional cases.

Now, assuming that I've explained the above correctly, my issue is that often in textbooks this relation is written as $$dxdy=\lvert J(u,v)\rvert dudv$$ Of course, in the derivation above where we considered finite changes in coordinates, the area in the $xy$ plane is not simply $\Delta x\Delta y$ (since the map transforms the square in the $uv$ plane to a parallelogram in the $xy$ plane to linear approximation), but when we take the infinitesimal limit we then express the infinitesimal area element as $dxdy$. Is this simply because, in the $xy$ plane we are integrating over the independent variables $x$ and $y$ and so we simply denote the area element as $dxdy$, and as such the infinitesimal area elements in the $uv$ and $xy$ planes are related as $$dxdy=\lvert J(u,v)\rvert dudv$$ I maybe getting myself into a fuss over nothing, but I just want to clarify things before I give out false/incorrect information.

 

[andrewkirk]

Yes, the statement $dxdy=\lvert J(u,v)\rvert dudv$ is a textual substitution rather than an equation, since, strictly speaking, neither side is a valid mathematical object.

What it means is that when you have an integral of the form $\int\int f(x,y)dxdy$ and you have an appropriate transformation to coordinates $u,v$, the integral will be equal to $\int\int f(x(u,v),y(u,v)) \lvert J(u,v)\rvert dudv$

 

["Don't panic!"]

Yes, the statement dxdy=∣J(u,v)∣dudvdxdy=\lvert J(u,v)\rvert dudv is a textual substitution rather than an equation, since, strictly speaking, neither side is a valid mathematical object.

Yes, the mathematically rigorous method would be to use differential forms in which the Jacobian appears naturally when one maps between two coordinate systems, but I find the non-rigorous informal derivation more intuitive geometrically, especially when trying to explain the concept to someone who has never been exposed to differential geometry.
My issue is that the linear approximation of the area (mapped into the $xy$ coordinate system) is a parallelogram whose area is not, in general, given by $\Delta x\Delta y$, so why in this heuristic, informal derivation, do textbooks write $\Delta A\rightarrow dxdy$ in the "infinitesimal limit"? Wouldn't it be better to write $$\Delta A\rightarrow dA=\lvert J(u,v)\rvert dudv$$ Or is the point that any additional factor that relates $dA$ to $dxdy$ can be absorbed into the integrand, such that one can effectively write $$\Delta A\rightarrow dxdy=\lvert J(u,v)\rvert dudv$$ such that $$\int\int_{R}f(x,y)dxdy=\int\int_{D}f(x(u,v),y(u,v))dudv$$ holds? (Where $R$ is the domain of integration in the $xy$ coordinate system, and $D$ in the $uv$ coordinate system).

 

[andrewkirk]

I can't really answer the question without seeing the particular textbook you are looking at, how it uses the formulas and what the context is.

Often (usually?) it is the case with coordinate transformations that one of the two coordinate systems is orthonormal. In that case the incremental area will be rectangular in the limit and we will have $$\lim_{\Delta x,\Delta y\to 0}\frac{\Delta A}{\Delta x\Delta y}=1$$ if the orthonormal system is the $x,y$ one.

 

["Don't panic!"]

I can't really answer the question without seeing the particular textbook you are looking at, how it uses the formulas and what the context is.

This set of notes (for example): www.google.co.uk/url?sa=t&source=web&rct=j&url=http://booksite.elsevier.com/andrilli/elementary/content/jacobian.pdf&ved=0CBsQFjABOApqFQoTCNyZrMP3_MgCFUbRFAodzvQNPg&usg=AFQjCNELcdk9L9Zt34P32DbCu4OA7QbvvA&sig2=QQzeWP-RkLhZ64WGIGXBQA

 

[andrewkirk]

This set of notes (for example)

In that set of notes the $x,y$ coordinate system used on pages 2-3 is orthonormal, so the limit equation I wrote above will hold and it is valid to substitute $dydx$ for $dA$ in the area integrals.

Are you getting thrown off by the fact that the large area being integrated in Figure 1 is a parallelogram? That doesn't matter because it's only the infinitesimal patches of area that we are interested in in the integration, and they are rectangles of dimension $\Delta x\times \Delta y$.

 

["Don't panic!"]

Are you getting thrown off by the fact that the large area being integrated in Figure 1 is a parallelogram? That doesn't matter because it's only the infinitesimal patches of area that we are interested in in the integration, and they are rectangles of dimension Δx×Δy\Delta x\times \Delta y.

Yeah, I think it's because in the figure the $uv$ coordinate plane looks orthonormal and so I thought that it was in this system that the patches of area are $\Delta u\Delta v$?! Is the point though that when mapped into the $xy$ coordinate plane the $uv$ coordinate system is curvilinear (i.e. the $uv$ coordinate lines are no longer orthonormal) and so with respect to the $uv$ coordinate lines in the $xy$ coordinate plane the area is no longer given by $\Delta u\Delta v$, but (approximately) $\lvert J\rvert\Delta u\Delta v$, and as the patches of area in the $xy$ coordinate system are given by $\Delta x\Delta y$, then $\Delta x\Delta y\approx\lvert J\rvert\Delta u\Delta v$. In the infinitesimal limit we have that $dxdy=\lvert J\rvert dudv$. Is something like this correct at all or am I completely missing the point?!

 

[andrewkirk]

Hmmm. I must say I don't think those notes are very clear. The coordinate system $u,v$ in Example 3 is not curvilinear, but I can see why you thought it might be by looking at Figure 3. Unfortunately, Figure 3 is not a depiction of Example 3 but of the more general case, in which curvature may occur. The only hint at this is where it says 'This will work in general for all change of variable transformations'.

When integrating, you can ignore curvature, because it disappears as one takes the limit. What does not disappear is the ratio of areas, which is given by the Jacobian (determinant). In the example given, because all coordinates and transformations are linear, you can skip the integration and calculate the area of the rectangle P directly as the area of the unit square in the $u,v$ system multiplied by the Jacobian det. But in the more general case, the integration must be performed, and the Jacobian multiplication is applied to each value of the integrand, giving the formula as written in post 2 above.

 

["Don't panic!"]

Hmmm. I must say I don't think those notes are very clear.

It's frustrating as several sets of notes that I've read seem to follow this approach. Are there any in particular that you can recommend that explain the concept well?!

What I find confusing is that in the subsequent example in which they look at the polar coordinate case, both coordinate systems are again orthonormal, but the Jacobian does vary as the coordinates vary, since $\lvert J\rvert =r$. Also, as we know the coordinate lines (at least the $\theta$ coordinate lines) appear curved in the Cartesian $xy$ coordinate system, i.e. the mapping between the two systems is non-linear.

 

["Don't panic!"]

Is the point that even though the two different coordinate systems are orthonormal, the mapping between them will be, in general, non-linear and so although the $uv$ coordinate lines with be rectangular in the $uv$ system, they will be mapped to curved lines in the $xy$ coordinate system and so the infinitesimal area element $dudv$ (when mapped into the $xy$ coordinate system) will no longer be equal to $dudv$ (the area will be distorted by the mapping between the two coordinate systems). The Jacobian then relates the infinitesimal area element $dxdy$ in the $xy$ coordinate system to the infinitesimal area element $dudv$ in the $uv$ coordinate system by describing how the area element $dudv$ is distorted by the mapping into the $xy$ coordinate system. This is of course given by $\lvert\frac{\partial (x,y)}{\partial (u,v)}\rvert dudv=\lvert J(u,v)\rvert dudv$.

Would this be a correct summary at all?

 

[andrewkirk]

both coordinate systems are again orthonormal,

The $u,v$ system is not orthonormal. The notes just make it look orthonormal by drawing it a particular way. A coordinate system is orthonormal iff, at every point $P$ in the region $M$ covered by the coordinate system, for all pairs of basis vectors $\vec{e}_i,\vec{e}_j$ in the basis of the tangent space $T_PM$, we have $\vec{e}_i\cdot\vec{e}_j=\delta_{ij}$. where $\delta_{ij}$ is the Kronecker Delta. That is not the case for the $u,v$ system as the inner product of its basis vectors is $( 2,1)\cdot(-1,1)=-1$ and their lengths are $\sqrt{5},\ \sqrt{2}$. Those inner products are calculated in the $x,y$ coordinate system but inner products are invariant between coordinate systems, so the same values are obtained in the $u,v$ system.

Polar coordinates are not orthonormal either, although they are orthogonal. The circumferential basis vector has non-unit length, proportional to its distance from the origin.

I think the notes confuse this issue of orthonormality by referring to $u,v$ as if they were coordinates for the original $x,y$ plane. We can define them to be that, but if we do then they are not orthonormal coordinates.

It is clearer, and more correct to describe the mapping as a mapping between two different spaces, one of which is a number plane with axes labelled $x,y$ and the other of which is a number plane with axes labelled $u,v$. We do not call $u,v$ coordinates for the first plane. They are coordinates for the second plane and in that space they are orthonormal. There is a mapping, call it $\phi$ from the $u,v$ space to the $x,y$ space, and the differential of that map is the Jacobian. With that perspective, we can think of the calculation of the area of $P$ in the $x,y$ space as a calculation of the area of the shape $\phi^{-1}(P)$ in the $u,v$ space, multiplied by the Jacobian det in order to adjust for the distortions introduced by the mapping. Where the mapping is non-linear, we need to do the calculation as an integral, rather than just a simple macro area calculation in the $u,v$ space multiplied by the Jacobian det.

 

["Don't panic!"]

It is clearer, and more correct to describe the mapping as a mapping between two different spaces, one of which is a number plane with axes labelled x,yx,y and the other of which is a number plane with axes labelled u,vu,v. We do not call u,vu,v coordinates for the first plane. They are coordinates for the second plane and in that space they are orthonormal. There is a mapping, call it ϕ\phi from the u,vu,v space to the x,yx,y space, and the differential of that map is the Jacobian. With that perspective, we can think of the calculation of the area of PP in the x,yx,y space as a calculation of the area of the shape ϕ−1(P)\phi^{-1}(P) in the u,vu,v space, multiplied by the Jacobian det in order to adjust for the distortions introduced by the mapping. Where the mapping is non-linear, we need to do the calculation as an integral, rather than just a simple macro area calculation in the u,vu,v space multiplied by the Jacobian det.

Thanks for the clear explanation, that's kind of what I was thinking, but didn't express it very well (sorry, I meant in my explanation that in the two respective planes the $xy$ and $uv$ coordinates are orthonormal). Is it correct to say that the coordinate lines of the $uv$ coordinate system will, in general, be mapped to curved lines in the $xy$ plane (for example, in polar coordinates, the $\theta$ coordinate line is mapped to a closed curve, i.e. a circle, in the $xy$ plane) and in doing so distorting the infinitesimal area element $dudv$ as it is mapped into the $xy$ plane (this is then related to the infinitesimal area element $dxdy$ via the Jacobian as $\lvert J\rvert dudv$)?

 

[andrewkirk]

Yes that sounds like a reasonable characterisation.

 

["Don't panic!"]

Yes that sounds like a reasonable characterisation.

OK great. Thanks for your help.

One more thing. Is it fair to say that the reason why the derivation is done this way, i.e. considering the $uv$ coordinate system to be orthogonal (when in general it won't be), is to aid intuition and the derivation, as it allows one to consider the area element in the $uv$ system, $\Delta u\Delta v$, to be rectangular, which is then distorted as it is mapped into the $xy$ plane (we map into the $xy$ plane in the derivation so that we can determine how the area element $\Delta u\Delta v$ is distorted when mapped into the $xy$ plane and directly see how this relates to an area element of the $xy$ plane $\Delta x\Delta y$).
Mapping the area element $\Delta u\Delta v$ into the $xy$ plane then allows one to directly relate an element of area, measured with respect to the $uv$ coordinate axes, with an element of area measured with respect to the $xy$ axes.

 

[zinq]

There are two viewpoints that can each be used to imagine what is going on, and one of these is distinctly more useful.

a) One is undergoing a "coordinate change" — changing the pair of numbers (coordinates) used to refer to any point — from one coordinate system to another.

b) The other, much more useful one, is that of a mapping or transformation: Simply carrying each point to some (usually other) point.

If we use the viewpoint of a mapping, then to say that "the Jacobian (intuitively) describes how surface (or volume) elements change under a particular coordinate transformation. In essence, they characterise the difference in how area (or volume) is measured in the two different coordinate systems"* can be rather confusing.

There *is* no difference between how area or volume is measured in two coordinate systems: They are simply two ways of referring to the points in some space. That doesn't give them the right to differ in how they measure area or volume or length or angle, for that matter.

On the other hand, a mapping will take the area near some point P and multiply it by some constant. Assuming the mapping is differentiable, the ratio of the area near P to the area of its image by the mapping will approach a constant in the limit ( as the area gets arbitrarily close to P).

This ratio will be the Jacobian determinant. Depending on the mapping, the value of the Jacobian determinant can be even 0 or negative (since area turned "upside down" is counted negatively by the integral.

_________________________________________
* What is a "coordinate transformation"? Is this conceptually a change of coordinates, or is a mapping? Whether a) or b) is used above, it is immensely helpful to avoid mixing the two concepts.

 

["Don't panic!"]

There *is* no difference between how area or volume is measured in two coordinate systems: They are simply two ways of referring to the points in some space. That doesn't give them the right to differ in how they measure area or volume or length or angle, for that matter.

But the area or volume element in the $uv$ coordinate system is certainly different form the area or volume element in the $xy$ coordinate system. I thought the whole point was that to determine the area of a region $R$ in the Cartesian coordinate system $(x,y)$ one can either divide the region up into rectangular area elements $dxdy$ (since the system is Cartesian) and then integrate over these elements to get the area. Or, one can define a mapping between the Cartesian system and some other coordinate system $(u,v)$ (in general, curvilinear) and in doing so the region $R$ in the $(x,y)$ system is mapped into a region $S$ in the $(u,v)$ system. The coordinates of the region $R$ in the $(x,y)$ system can then be parametrised by the "new" coordinates $(u,v)$, such that the points in $R$ are now given by $(x,y)=(x(u,v),y(u,v))$. Pictorially, we can view this as "overlaying" the $uv$ coordinate lines on the region in the $xy$ plane. In general, the mapping between the two coordinate systems will be non-linear and so the $uv$ coordinate lines in the $xy$ plane will be curves. As such, each element of area of the region $R$, relative to the $uv$ coordinate lines will no longer be a rectangle, but to best linear approximation, a parallelogram. We can then relate the two area elements by calculating the sides of such a parallelogram. I thought this was the idea behind it all?!

In summary, I thought that it was the way in which we measure the area that changed (e.g. summing up parallelograms rather than rectangles), but not the actual area (since this is independent of any coordinates we assign to it), which is why we need to account for this change in area measure to ensure that we obtain the same area in both cases.

 

[andrewkirk]

Or, one can define a mapping between the Cartesian system and some other coordinate system (u,v)(u,v) (in general, curvilinear) and in doing so the region RR in the (x,y)(x,y) system is mapped into a region SS in the (u,v)(u,v) system.

This is the sort of terminology that it's really best to avoid. The standard approach in this branch of maths is for 'coordinate systems' to refer to mappings between the space in which you are trying to measure the area and subsets of Euclidean space. In that case it is meaningless to talk about the 'area of the shape in the u,v coordinate system.' The coordinate system is a mapping, not a space, and it does not contain any areas. There is only one area to be discussed, and that's the area P in the original space.

If you want to talk about the area of something other than the shape P in the diagram, you need to talk in terms of mappings to other spaces, which is what I think zinq is referring to as (b). So we talk about the u,v space, not the u,v coordinate system. It may sound like pedantry, but it's a surefire way to avoid the sort of confusion that this set of notes has caused.

 

["Don't panic!"]

If you want to talk about the area of something other than the shape P in the diagram, you need to talk in terms of mappings to other spaces, which is what I think zinq is referring to as (b). So we talk about the u,v space, not the u,v coordinate system. It may sound like pedantry, but it's a surefire way to avoid

So in terms of Jacobians, is the idea that we are mapping between two spaces?

 

[andrewkirk]

So in terms of Jacobians, is the idea that we are mapping between two spaces?

You can approach it either way, which I think is the (a) vs (b) split to which zinq was referring. If you have not done any differential geometry yet, I think approach (b), which is as a mapping between two spaces and the relationships of two areas, one in each space, will be easier to understand.

 

["Don't panic!"]

You can approach it either way, which I think is the (a) vs (b) split to which zinq was referring. If you have not done any differential geometry yet, I think approach (b), which is as a mapping between two spaces and the relationships of two areas, one in each space, will be easier to understa

It confuses me though that in approach (b) one talks about a mapping between the two spaces, $uv$ space to $xy$ space, and then one considers the coordinates in both $uv$ space and $xy$ space to be orthogonal (for simplicity), and since the mapping will in general be non-linear, the coordinate lines $u$ and $v$ in the $uv$ space are mapped to curves of constant $u$ and constant $v$ in the $xy$ space, and then to determine how an element of area in the $uv$ space relates to an element of area in $xy$ space we consider linear approximations in changes of the $x$ and $y$ coordinates, parametrized in terms of $u$ and $v$, etc. (following the rest of the procedure described in earlier posts)... Is this the correct description or have I completely misunderstood it completely?!

 

[andrewkirk]

to determine how an element of area in the

uv space relates to an element of area in

xy space we consider linear approximations in changes of the

x and

y coordinates, parametrized in terms of

u and

Yes. When the mapping is non-linear and the shape P is not simple we can think of P as broken up into lots of tiny tiles that are the images, under the mapping function F from the uv space to the xy space, of tiny rectangles $[u+\delta u]\times[v,v+\delta v]$ in the uv space. The area contribution to P from each tile is $\delta u\delta v|J_F(u,v)|$. That area is, as you say, a linear approximation to the true area of $F([u+\delta u]\times[v,v+\delta v])$, which approaches exactness as the tile size approaches zero.

 

["Don't panic!"]

Yes. When the mapping is non-linear and the shape P is not simple we can think of P as broken up into lots of tiny tiles that are the images, under the mapping function F from the uv space to the xy space, of tiny rectangles [u+δu]×[v,v+δv][u+\delta u]\times[v,v+\delta v] in the uv space. The area contribution to P from each tile is δuδv|JF(u,v)|\delta u\delta v|J_F(u,v)|. That area is, as you say, a linear approximation to the true area of F([u+δu]×[v,v+δv])F([u+\delta u]\times[v,v+\delta v]), which approaches exactness as the tile size approaches zero.

OK, so I'm understanding the idea correctly then in that case (apart from incorrect syntax), right?

Is it correct then to say that we split up the area P in two different ways, first into tiny rectangles $\delta x\delta y$ and secondly into tiny tiles (which are the images under F from the $uv$ space to the $xy$ space), $\lvert J(u,v)\rvert\delta u\delta v$ . In the limit as these tiles and rectangles become infinitesimal, integrating over each of them should give the same result as they are covering the same area, hence $\int\int_{R}dxdy =\int\int_{S}\lvert J(u,v)\rvert dudv$.

 

["Don't panic!"]

To add to this. conceptually, is it correct to say that we can either directly calculate the "actual" area $P$ in the $xy$ space by integrating over infinitesimal rectangles $dxdy$ over the region $R$, or we can consider a map between this space and another space, so-called $uv$ space. Both coordinates are described by a set of orthonormal coordinates, $(x,y)$ and $(u,v)$, respectively, such that the infinitesimal area elements in each respective space are given by $dxdy$ and $dudv$. The area $P$ corresponds to some area $P'$ in the $uv$ space.
We wish to calculate the area $P$ in the $xy$ space however, and so to do this with respect to the coordinates $(u,v)$ defined in the $uv$ plane, we must map the $(u,v)$ coordinate lines into the $xy$ space and calculate the corresponding area element with respect to the $(x,y)$ coordinates (as functions of the $(u,v)$ coordinates). This will give us the area element corresponding to changes along the $u$ and $v$ coordinate lines in the $xy$ space.
In essence, is the reason that we write $$dxdy =\rvert J\lvert dudv$$ to symbolically denote that $\rvert J\lvert dudv$ is the area element relative to the $uv$ coordinate lines, but in the $xy$ space (a sort of mapping of the area element, $dudv$, in the $uv$ space, into the $xy$ space, i.e. $"dudv\mapsto\lvert J\rvert dxdy"$)?!

Sorry, I feel like I've got myself into a right tangle here. I thought I understood the concept before, but now I'm really not sure?!

 

[andrewkirk]

@"Don't panic!" Yes post 22 is correct. Post 23 looks alright too but I would add that the images under the mapping F of the coordinate lines in the uv space give a curvilinear grid in the xy space that defines the tiles that are used to integrate the shape P in the xy space using the Jacobian. eg say you make a grid in the uv space with a mesh width of 0.001, then the lines that are the images in xy-space of those gridlines under the map F make a (not necessarily orthogonal or even linear) grid in xy-space, made up of the integration tiles. For a (non-orthogonal, potentially curved) tile T in xy-space, the ratio of its area to the area of its (square) preimage in uv-space is $|J|$, to first order.

 

["Don't panic!"]

For a (non-orthogonal, potentially curved) tile T in xy-space, the ratio of its area to the area of its (square) preimage in uv-space is |J||J|, to first order.

So is am I right in saying that the area element $\delta x\delta y$ in this case corresponds to a linear approximation of the area of a tile T in $xy$ space measured relative to the $u$ and $v$ coordinate curves in $xy$ space. To linear approximation this is then related to the corresponding (rectangular) area element $\delta u\delta v$ in $uv$ space by $\delta x\delta y\approx\lvert J\rvert\delta u\delta v$. Then, as you say $$\lim_{(\delta x,\delta y),(\delta u,\delta v)\rightarrow 0}\frac{\delta x\delta y}{\delta u\delta v}=\lvert J\rvert$$ i.e. In the infinitesimal limit, the ratio of the image (relative to the image $u$ and $v$ coordinate curves in $xy$ space) to the area of its preimage in $uv$ space is $J$ (to first order).

 

[andrewkirk]

So is am I right in saying that the area element $\delta x\delta y$ in this case corresponds to a linear approximation of the area of a tile T in $xy$ space measured relative to the $u$ and $v$ coordinate curves in $xy$ space.

Sort of. It's important to note that they are different tiles. The expression $\delta x\delta y$ refers to a small square tile in xy space and the expression $|J|\delta u\delta v$ refers to a tile that will be approximately a parallelogram, that will in general not be rectangular and will even have slightly curved edges. The two tiles have the same area and one can interpret the equality in such a way that they overlap and share a vertex, but that is all.

My view is that the expression $dxdy=|J|dudv$ is misleading, ambiguous and should be avoided. What it is intended to convey is that in certain circumstances one can substitute one side of the 'equality' for the other in a formula without changing the result. It is not a genuine equality. It can be given a meaning as an approximation, as per the previous paragraph, but it's not a proper equation.

The genuine equation that covers such situations is

$$\int_{-\infty}^\infty\int_{-\infty}^\infty f(x,y) dx dy=\int_{-\infty}^\infty\int_{-\infty}^\infty f(x(u,v),y(u,v))|J(u,v)| du dv$$

where $x(u,v),y(u,v)$ are $x,y$ written as functions of $u,v$. This equation says that the integral of the function $f$ in the xy-plane is the same regardless of whether we calculate it relative to a tiling of tiny squares of the form $[x,x+\delta x]\times[y,y+\delta y]$ or relative to a tiling of tiny almost-parallelograms of the form $F([u,u+\delta u]\times[v,v+\delta v])$ where $F$ is continuous bijection from the uv plane to the xy plane.

 

["Don't panic!"]

Sort of. It's important to note that they are different tiles. The expression $\delta x\delta y$ refers to a small square tile in xy space and the expression $|J|\delta u\delta v$ refers to a tile that will be approximately a parallelogram, that will in general not be rectangular and will even have slightly curved edges. The two tiles have the same area and one can interpret the equality in such a way that they overlap and share a vertex, but that is all.

My view is that the expression $dxdy=|J|dudv$ is misleading, ambiguous and should be avoided. What it is intended to convey is that in certain circumstances one can substitute one side of the 'equality' for the other in a formula without changing the result. It is not a genuine equality. It can be given a meaning as an approximation, as per the previous paragraph, but it's not a proper equation.

The genuine equation that covers such situations is

$$\int_{-\infty}^\infty\int_{-\infty}^\infty f(x,y) dx dy=\int_{-\infty}^\infty\int_{-\infty}^\infty f(x(u,v),y(u,v))|J(u,v)| du dv$$

where $x(u,v),y(u,v)$ are $x,y$ written as functions of $u,v$. This equation says that the integral of the function $f$ in the xy-plane is the same regardless of whether we calculate it relative to a tiling of tiny squares of the form $[x,x+\delta x]\times[y,y+\delta y]$ or relative to a tiling of tiny almost-parallelograms of the form $F([u,u+\delta u]\times[v,v+\delta v])$ where $F$ is continuous bijection from the uv plane to the xy plane.

Ah ok, so the "equality" $dxdy=\lvert J\rvert dudv$ is simply to convey the fact that integrating over one is equivalent to integrating over the other (in their respective domains), i.e. $$\int\int_{R}f(x,y)dxdy=\int\int_{S}f(x(u,v),y(u,v))dudv$$ either way the integrals evaluate to the same result, right?

 

[andrewkirk]

Yes

 

["Don't panic!"]

Yes

Great, thanks very much for all your help!

 

[zinq]

"... a mapping between the two spaces, uv space to xy space ..."

No, I am talking about a mapping from a space (a set of points that form a curve, a surface, or a higher-dimensional manifold) to itself.

Whenever there is the possibility of a "coordinate change", that can only occur where we are talking about the same set of points, but using an old and a new set of coordinates to refer to points. (This is just renaming points, but the points themselves don't change.)

I probably should not have downplayed the importance of a coordinate change — my case a) above. It is actually important to understand this since such things arise often, for example in switching between cartesian and polar coordinates in the plane.

But what can be very confusing if you're not careful is getting the two ideas of

a) a coordinate change​

and

b) a transformation of a space to itself​

mixed together.

Here a) is merely a renaming of the same points. Technically speaking, we start with a (usually) open set U of a manifold of say dimension = n that has "old" coordinates defined by a map f from U into the Euclidean space of the same dimension:

f: U → Rⁿ​

where f is the map assigning to each point of U some particular choice of coordinates, an n-tuple of numbers, as a way of referring to that point. For best results, the space U should usually not be thought of as having some special set of intrinsic coordinates (unless it is originally defined as a subset of Euclidean space).

Note that, in order to make sense, the coordinate map f (sometimes called a "chart" in mathematics) must be one-to-one. I.e., it must assign any given n-tuple of numbers to at most one point. This means that f has an inverse function from its image f(U) back to U:

f^-1: f(U) → U

A "coordinate change" arises when there is a different, or "new", coordinate map for U:

g: U → Rⁿ.​

Then if you know what the old f coordinates of a point are, a "coordinate change" tells you what the new g coordinates are.

This coordinate change may be thought of specifically as the mapping

g _° f^-1: f(U) → g(U)

that takes as input the f (old) coordinates of a point and that outputs the g (new) coordinates of the same point.

Moral of this tale: It is almost always helpful to express your thoughts in terms of mappings (functions) between one set or space and another (possibly itself). Keep the concepts of a point, and its coordinates (which are just its name), separate. This usually clarifies what is going on.

P.S. At the risk of stating the obvious: Many spaces cannot be described with a single set of coordinates. The ordinary unit sphere S² in R³, for instance. (That's because the sphere has no smooth, or even continuous, one-to-one mapping into the plane.)

But every space that is a "manifold" of some dimension n — the main spaces of interest in physics — can be covered by "coordinate patches" each of which has a perfectly nice one-to-one mapping into Euclidean space Rⁿ. In the case of the sphere S² again, this can be done with as few as two patches:

U₁ = S² - {(0, 0, 1)}

and

U₂ = S² - {(0, 0, -1)}.

 

["Don't panic!"]

Keep the concepts of a point, and its coordinates (which are just its name), separate. This usually clarifies what is going on.

Thanks for your detailed answer to my follow up comments.

If we are considering a mapping between two coordinate systems describing the same set of points making up an area in some space (manifold) doesn't the Jacobian (determinant) describe how the (coordinate) unit of area changes between the two coordinate systems?

Isn't it correct that we can use the inverse of a given coordinate map to parametrise the patch on the manifold (that is the preimage of the original coordinate map), and so each given coordinate system defines an (infinitesimal) unit of (coordinate) area that can be used to tile the area that we a considering on the manifold and subsequently integrate over to determine a value of area. The Jacobian determinant describes how these tiles are distorted as we map between two different coordinate systems, but of course we must obtain the same value for the area on the manifold (as the area exists independently of any coordinate system) and so both cases lead to the same result (upon integration).

Sorry, I feel I haven't really explained what I mean very well, but hopefully you get the gist of it.