5
$\begingroup$

I can understand this through various examples found in the internet but I can't quite intuitively understand why the determinant of the derivative(in its most general form)-the Jacobian-gives the change of volume factor that arises when we change variables in, say, an integral.

I mean, why does the determinant of the matrix consisting of the derivatives of the original variables wrt the new variables give a number that corresponds to how much the infinitesimal volume has changed?

How can we geometrically connect the derivatives that are the components of the Jacobian with the aforementioned change of volume?

$\endgroup$
3
$\begingroup$

The intuition has (at least) two pieces:

  1. Let $f : X \to Y$ be a diffeomorphism between open domains in $R^n$. Then for $x \in X$, The Jacobian at $x$ locally approximates f by a linear change of variables. (It is the linear transformation that is arbitrarily indistinguishable from your actual diffeomorphism as you zoom into a x, i.e. take smaller and smaller neighborhoods. This is the definition of the derivative, which is what the Jacobian is. One can prove that the definition of all those partial derivitves gives this derivative - I think it is best to think of the Jacobian as THE derivative, and then think of the computation with partials as a theorem about how to compute it.)
  2. The determinant of a linear transformation measures how volume changes. EDIT: I think it is helpful to feel convinced that volume scaling is multiplicative (forgetting the formula for the determinant for a moment), and that for elementary matrices the volume scaling agrees with the determinant (where it is essentially a bunch of verifications in $\mathbb{R}^2$). Now use that every matrix can be written as a product of elementary matrices (which is intuitive, since you can think of moving the basis to a new basis by adjusting via scaling and swaps and skews two vectors at a time.) Now the only question that remains is why the formula for the determinant is multiplicative, and this is best seen for example by using the exterior powers and functoriality.

Putting these things together, one sees that if we cared about how a little area patch changed nearby a point, the determinant of the Jacobian gives a very good approximation of it. Taking it to the limit, so to speak, and integrating, we get a complete description of area distortion. The rest is (hard) calculus and analysis, and the proof is quite involved, if memory serves.

$\endgroup$
  • $\begingroup$ But, why is the second point true? I mean, this is the true essence of my question, right? $\endgroup$ – TheQuantumMan Sep 12 '16 at 6:18
  • $\begingroup$ @TheQuantumMan I updated my answer. I hope it helps. $\endgroup$ – Lorenzo Najt Sep 12 '16 at 14:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.