My question is what does a Jacobian have to do with the change of coordinates (coordinate transformation). Why do we care about this notion to start with? Also, why should it be non-singular?
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2Have you read the Wolfram page on the Jacobian? Or the Wikipedia entry? – Kyle Kanos Jan 13 '16 at 11:21
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Yes, but I want to know why does it enter in the context above as you might have read in my question. What is the relation between that non singular Jacobian and general coordinate transformation. I also have asked about its non-singularity in this case. @KyleKanos – Beyond-formulas Jan 13 '16 at 11:34
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1Related question by OP: http://physics.stackexchange.com/q/185267/2451 Related: http://physics.stackexchange.com/q/1324/2451 and links therein. – Qmechanic Jan 13 '16 at 12:02
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- Follows from the chain rule given 1.
– Ryan Unger Jan 13 '16 at 13:25 -
@0celo7 can you please elaborate? – Beyond-formulas Jan 13 '16 at 13:36
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@Beyond-formulas The coordinate transition functions $f$ are diffeomorphisms, so $f\circ f^{-1}(x)=x$. Applying the chain rule gives $Df\cdot Df^{-1}=\mathrm{id}$. Since $Df$ and $Df^{-1}$ exist by definition of $f$, we infer $Df^{-1}=(Df)^{-1}$ exists and therefore $\det Df\ne0$. – Ryan Unger Jan 13 '16 at 13:47
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Yes, I understand this. I thought you were pointing out something else. What I was asking for is for example how is a Jacobian related to coordinate transformation in anyway? @0celo7 – Beyond-formulas Jan 13 '16 at 13:49
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You need $\det Df\ne0$ when proving theorems about e.g. orientation. – Ryan Unger Jan 13 '16 at 16:10
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Expanding on 0celo7's comment, the Jacobian is basically that $Df$ and so for an inverse function to exist, $Df$ cannot be equal to zero, since if $Df=0$, then $Df^{-1}$ is undefined and does not exist. For your next question, the Jacobian relates to coordinate transformation as it keeps the equations invariant under the transformations. – Horus Jan 14 '16 at 12:27
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The functions are "smooth" because we want to be able to speak about derivatives on our manifold $M$, and for that, it is convenient to have a smooth structure on $M$ (one could settle for a $C^k$-structure with $k$ as needed, but physicists rarely care for such details). And of course we want to be able to take derivative because we might be interested in the divergence of a vector field, for instance, or the Laplacian of some function, etc... Physics needs the notion of (spatial) change, and that's what the derivative gives. If nothing else is said, all manifolds that appear in physics are silently assumed to be smooth manifolds.
That the $x'(x)$, or rather, the $\phi_j\circ\phi_i^{-1}$, are smooth follows from the definition of a smooth atlas. That the Jacobian is non-singular follows from the $\phi_i,\phi_j$ being diffeomorphisms, and so $\phi_j\circ\phi_i^{-1}$ is also a diffeomorphism, and diffeomorphisms have non-singular Jacobians.
ACuriousMind
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Yes, it is understandable that $M$ should be a differentiable manifold and should be smooth, my question was why the mapping should be smooth (set of functions and their inverses a.k.a the $\phi_j \circ \phi_i^{-1}$). – Beyond-formulas Jan 14 '16 at 00:41
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@Beyond-formulas: That's what it means for a manifold to be smooth. A smooth manifold is one with a smooth atlas, and the functions $\phi_i$ are that atlas. – ACuriousMind Jan 14 '16 at 00:43
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Ah, also, concerning the Jacobian, so yes, I understand what you meant, but in general why would we care for it? Is it the one for example in trasnformation of a vector $V'^{\mu}(x')=\frac{\partial x'^{\mu}}{\partial x^{\nu}}V^{\nu}(x)$? – Beyond-formulas Jan 14 '16 at 00:48
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1) If the transformation itself is smooth, then the Jacobian will be smooth. This is a desirable property because...
2) The coordinate transformation tells us how the coordinates change, but the Jacobian tells us directly how the coordinate basis vectors change. For example, a transformation from Cartesian to polar coordinates would use the Jacobian to relate the basis vectors $\partial_x, \partial_y$ to the basis vectors $\partial_r, \partial_\theta$.
Ensuring that the Jacobian is invertible (non-singular) as well means that that non-degenerate volumes in one coordinate system will still be non-degenerate (i.e. nonzero, actual volumes instead of planes) in another, and that a nonzero vector field in one coordinate system will not be mapped to a zero vector field in another.
Muphrid
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