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Jacobian determinant and volume change

  1. Aug 3, 2010 #1
    Hello everyone,

    So, I read somewhere that the Jacobian determinant of a transformation determines the local volume change.

    Say I am in 3D space and I have the following relationship:

    F(x', y', z') = F(x, y, z) + T(x, y, z)

    The LHS gives the new position and the RHS is the old position + the displacement vector.

    So, the gradient is:

    Identity + Jacobian.

    Now, my question is: what is this Jacobian of the displacement field telling me? From what I read, it sort of represents a scaling of the local volume. However, I am struggling to see that.

    Also, is the identity matrix in the gradient formulation important to calculate this scaling factor or just looking at the Jacobian determinant of the displacement field suffice?? From what I have been reading on the net so far, the displacement field and its jacobian tells us everything about the volume change...

    I have been struggling for a while to visualize and understand this Jacobian stuff but with no luck...

    I would be really grateful for your help.

    Many thanks,

    Last edited: Aug 3, 2010
  2. jcsd
  3. Aug 4, 2010 #2


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    It's like this. First, you gotta know (show) that if you take 3 vectors v1,v2,v3 in R^3, then the volume of the parallelepiped that they span (more precisely, the volume of the convex hull of the set of points {0,v1,v2,v3}) is given, up to a ± sign, by the determinant of the matrix whose columns. I think I read in "history capsule" in a linear algebra textbook once that this is actually the property of the determinant that initially arose interest in the beast.

    Alright, so, let's take the simplest case first and consider a linear map L:R^3-->R^3 and let (e1,e2,e3) be the standard basis of R^3. Consider the parallelepiped (cube) C spanned by e1,e2,e3. How does L transforms this "unit volume element"? Well, using linearity of L, you can show that L(C) is just the parallelepiped spanned by L(e1),L(e2),L(e3). And you can compute its volume by taking the (absolute value of) the determinant of the matrix whose columns are L(e1),L(e2),L(e3).

    But since the derivative of a linear map is the linear map itself, the matrix whose columns are L(e1),L(e2),L(e3) is actually the jacobian matrix of L evaluated at the origin, DL_0. And so,

    [tex]\mathrm{Vol}(L(C))=|\det DL_0|=|\mathrm{Jac}(L)_0|[/tex]

    More generally, if C' is any parallelepiped based at 0 spanned by v1,v2,v3, then Vol(L(C'))=|det(matrix whose columns are L(v1),L(v2),L(v3))|. Then using some elementary linear algebra, you can show that Vol(L(C'))=Vol(C')Vol(L(C)).

    The moral of the story here is that all the information relative to how the volume of a parallelepiped changes is contained in the information of how the volume of the standard unit cube C changes. Namely, if you want to know the volume of L(C') for C' a parallelepiped spanned by v1,v2,v3, then just compute the volume of C' and multiply by the volume of L(C).

    Slight generalization: Consider an affine map A:R^3-->R^3 of the form A(x)=p+L(x-p), where p is some point of R^3 and L is a linear map. Then a parallelepiped based at p is of the form p+C' for C' a parallelepiped based at 0, and A(p+C')=p+L(C'), so Vol(A(p+C'))=Vol(L(C'))=Vol(C')Vol(L(C)).

    Now, consider an arbitrary smooth map F:R^3-->R^3. Near a point p of R^3 (i.e., locally near p), F is very well approximated by the affine map [itex]A_p:\mathbb{R}^3\rightarrow\mathbb{R}^3:x\mapsto F(p)+DF_p(x-p)[/itex]. This statement is made precise by the very definition of the derivative of F at p as the (only) linear map DF_p verifying

    [tex]0=\lim_{x\rightarrow p}\frac{|F(x)-F(p)-DF_p(x-p)|}{|x-p|}=\lim_{x\rightarrow p}\frac{|F(x)-A_p(x)|}{|x-p|}[/tex]

    So, if you ask how the volume of a little cube p+C' based at p changes under the transformation F, the answer is: well, approximately as the volume changes under the affine map A_p. (And the smaller the cube, the better the approximation). So, according to the above,

    [tex]\mathrm{Vol}(F(p+C'))\approx \mathrm{Vol}(A_p(p+C'))=\mathrm{Vol}(C')\mathrm{Vol}(DF_p(C))=\mathrm{Vol}(C')|\mathrm{Jac}(F)_p|[/tex]

    And that is what the jacobian determinant has to do with "local volume change".
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