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A question concerning Jacobians

  1. Nov 6, 2015 #1
    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.
     
    Last edited: Nov 6, 2015
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  3. Nov 6, 2015 #2

    andrewkirk

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    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##
     
  4. Nov 6, 2015 #3
    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).
     
  5. Nov 6, 2015 #4

    andrewkirk

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    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.
     
  6. Nov 6, 2015 #5
  7. Nov 6, 2015 #6

    andrewkirk

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    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##.
     
  8. Nov 7, 2015 #7
    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?!
     
    Last edited: Nov 7, 2015
  9. Nov 7, 2015 #8

    andrewkirk

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    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.
     
  10. Nov 7, 2015 #9
    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.
     
    Last edited: Nov 7, 2015
  11. Nov 7, 2015 #10
    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?
     
    Last edited: Nov 7, 2015
  12. Nov 7, 2015 #11

    andrewkirk

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    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.
     
  13. Nov 8, 2015 #12
    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##)?
     
  14. Nov 8, 2015 #13

    andrewkirk

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    Yes that sounds like a reasonable characterisation.
     
  15. Nov 8, 2015 #14
    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.
     
    Last edited: Nov 8, 2015
  16. Nov 9, 2015 #15
    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.
     
    Last edited: Nov 9, 2015
  17. Nov 9, 2015 #16
    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.
     
    Last edited: Nov 9, 2015
  18. Nov 9, 2015 #17

    andrewkirk

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    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.
     
  19. Nov 9, 2015 #18
    So in terms of Jacobians, is the idea that we are mapping between two spaces?
     
  20. Nov 9, 2015 #19

    andrewkirk

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    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.
     
  21. Nov 9, 2015 #20
    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?!
     
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