How does the Jacobian relate 3D to 2D? May be it is simple, but I'm on Google for hours without finding a clue. I'm reading an article in computer vision where the optical flow equation is $\nabla I\cdot v + {dI \over dt} = 0 $ and for the 3D version it is $\nabla  I\cdot [J_\pi V]+{dI \over dt}=0$ 
I understand most of the paper but this sentence "where the associated 3D displacement of a point is related to the motion of its projection by the $2 \times 3$  Jacobian matrix $J_\pi={\partial p \over \partial P}$ "
$P$ is a 3D point and $p$ is a 2D one, $v$ is 2d vector and $V$ is a 3D one, $\pi$ is the projection matrix (I think), I don't understand this Jacobian part, thanks in advance for any help or guidance.

EDIT: Link to the original article has been added.
 A: $I$ is a function on the photograph, $R^2\times R$. It depends both on the position on the 2-D screen and time. 
The map $\pi$ in your article is a map from $R^3 \to R^2,$ not necessarily a matrix, or even a "projection" in the sense used by mathematicians.

Technical Note: There is actually a subspace $M\subset R^3$ where $M$ is a 2-dimensional subspace of $R^3$, which is possibly time-dependent. The paper assumes that this map $\pi$ is a smooth diffeomorphism when restricted to the $M$ onto a subspace of $R^2$.

$V$ is a time-dependent vector field on $R^3$. (Or $M$.) 
$v$ is the induced vector field on $R^2$ by $\pi.$ (Actually, only on the image of $M$ through $\pi$.)
A vector $V$ at the point $P$ on $M$ can be thought of as a derivative of a path based at the point $P$.
Heuristically,
$V = \left.\dfrac{d}{dt}\right|_{t=0} (P + Vt +O(t^2))$
How does this give you a vector in $R^2$? A path in $M$ turns into a path in $R^2$ by the map $\pi$. And we can take the derivative of the induced path:
$v = \left.\dfrac{d}{dt}\right|_{t=0} \pi(P + Vt +O(t^2)) = (D\pi_P V)$
The right hand side ($D\pi_P$) represents the total derivative of $\pi$ at $P$.

Technical Note 2: The whole bit about $\pi$ being one-to-one was so that you  actually get a well-defined vector field on the image. If $\pi$ isn't one to one, then a point $p$ could have two different pre-images, and there could be two different vectors at those two different pre-images. 

Because $\pi$ was a function from $R^3$ into $R^2$, $D\pi_P$ can be represented by a Jacobian matrix of dimension $2\times 3$ matrix:
$\pi(x,y,z) = ( u(x,y,z), v(x,y,z))$, where $u$ and $v$ and the horizontal and the vertical components of the function $\pi$. Then,
$J_\pi = \left[\begin{matrix} \dfrac{\partial u}{\partial x} & \dfrac{\partial u}{\partial y} & \dfrac{\partial u}{\partial z} \\ \dfrac{\partial v}{\partial x} & \dfrac{\partial v}{\partial y} & \dfrac{\partial v}{\partial z}\end{matrix}\right]$
Notice that $P = (x,y,z)$ and $p = \pi(P) = (u,v)$. Therefore, the notation $\dfrac{\partial p}{\partial P}$ makes sense.
In short, $D\pi_P = J_\pi = \dfrac{\partial p}{\partial P}$ essentially represent the same exact thing (the latter two expressions are a little bit sloppy with the base point), and
$$ v = J_\pi V$$ by definition. 
So basically, the two equations are identical. The only difference is whether you think of the equation in terms of the 2-dimensional vector field $v$ or in terms of the 3-dimensional vector field $V$. But $v$ allows you to recover $V$, so it doesn't make any difference.
