Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Can someone help give me intuition with the Jacobian?

  1. Mar 17, 2010 #1
    When I learn something, especially in calculus, I like to get some intuition on where things come from and why they work, thing is I can't quite fully understand why the jacobian works...I'm not looking for an explanation on how to do it, I already know that, I just want to know why it works, if someone can explain or show me somewhere that explains it that would be great. I am not looking for a rigorous proof, something remotely simple would be nice.

    I read http://planetmath.org/encyclopedia/JacobianAndChainRule.html" [Broken] and have been reading wikipedia, and I understand it a bit better now, but not fully. I don't really get the crossing part of it.

    Can someone help clear things up?
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 17, 2010 #2
    The derivative of a map between Euclidean spaces at a given point is the linear transformation that best approximates the map in the vicinity of the point. The Jacobian matrix is the matrix of this linear transformation with respect to the standard bases of the domain and range spaces--that is all.

    The inverse function theorem provides some extra information about the Jacobian. Specifically, if f : R^n --> R^m is a C^1 map with invertible Jacobian matrix at a point p, then f has a C^1 inverse defined on some neighborhood of f(p); in fact, f will locally be an open mapping around p.

    There's a neat way to see how the hypothesis that the Jacobian matrix is invertible at p comes into play. It turns out that the inverse function theorem can be proved more or less with Newton's method! Recall that the recurrence relation in Newton's method from single-variable calculus requires that one divide by the derivative of f. In our case, this corresponds to multiplication by the inverse of the Jacobian matrix. Alternatively, one can modify this idea and use the principle of contraction mappings--a contraction of a complete metric space has a unique fixed point. This approach eliminates the need to carry out detailed estimations that justify the convergence of Newton's method.
    Last edited: Mar 17, 2010
  4. Mar 17, 2010 #3
    Hmm thanks, that offers interesting insight, I think it is a bit complicated for me though. I haven't even been tought about the inverse function, recurrence relation, or even Newton's method, although I did learn about Newton's method myself. Any simple explanations on why when doing the change of variables with multiple variables you have to do the partial derivatives and then the cross product?
  5. Mar 18, 2010 #4
    Basically can you explain why [tex]\partial (x,y)[/tex]/[tex]\partial (u,v)[/tex]=XuYv-XvYu?
  6. Mar 18, 2010 #5
    That symbol is defined to be the determinant of the Jacobian matrix. The expression on the right comes from the formula for the determinant of a two-by-two matrix.


    This is a difficult article, but it provides a very neat formulation of changing variables using pullbacks.
  7. Mar 18, 2010 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I will try to give you an informal intuitive picture. Suppose you are given a surface parametrically:

    [tex]\vec R(u,v) = \langle x(u,v), y(u,v), z(u,v)\rangle[/tex]

    The parameter curves on the surface in the "u direction" are given when v = constant, and in the "v direction" when u is constant.

    The partial derivatives R = Ru and R = Rv are tangent to these parameter curves. A change Δu and Δv in the u-v parameter space gives an approximate change on the surface RuΔu and RvΔv on the parameter curves just like differentials in one dimension. This tells you that an area ΔuΔv in the uv space transforms approximately to an area of a parallelogram on the surface whose sides are RuΔu and RvΔv. And, of course, the area of a parallelogram is given by the magnitude of the cross product of the side vectors. This leads to the definition of the surface element:

    [tex]dS = |\vec R_u \times \vec R_v|dudv[/tex]

    on the surface generated by the changes Δu and Δv in the parameter space. If you work out that cross product, it is the Jacobian. I like to think of it as the scaling factor of the transformation. The dudv element of area must be multiplied by the Jacobian to scale it correctly on the surface.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook