Understanding Jacobian in relation to physics

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cboyce
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I'm working with a game physics engine that uses Jacobians to resolve contact forces. It's been a few years since my physics and linear algebra classes (where we didn't get to Jacobian matrices), so what I'm reading about Jacobians is fairly overwhelming. Most of what I can find are fairly formal definitions, without any examples about what I'm specifically looking for. Can someone give me a couple simple examples about how Jacobians would apply in physics contact resolution, or point me to a resource that does? Or are they complex enough that I need to relearn all the linear algebra leading up to them to understand how to use them?
 
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Thanks for the link, I think that gives me a good idea where I need to start to understand them.
 
Don't know if this helps but if you have a simple function of a single variable
such as y = f(x) you can differentiate this to get the slope.
As there is only one direction involved, there is only one slope to chose from.
Variously we write f'(x) or dy/dx etc.

When you are dealing with a function of several variables, as you must be, there are many directions to chose from, all with different slopes available.

The Jacobian is a method of handling this, which is why the matrix contains an array of partial differential coeffiecients. If you like it is a method of resolving the slopes into as many suitable directions as are needed.
Yes linear algebra theory confirms that this is the same number as the number of independent variables.
 
I've been doing quite a bit of reading, and I've been trying to get an intuitive sense of the constraints in play. Per the example of a holonomic function at http://en.wikipedia.org/wiki/Holonomic#Examples", I can intuitively understand that x^2+x^3 - L = 0 constrains a point to someplace on a circle. But then it says given
attachment.php?attachmentid=30060&stc=1&d=1290556566.png
, L is the distance between the two positions ri and rj, but wouldn't ri -rj already give you the distance between the two positions?
 

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cboyce said:
But then it says given
attachment.php?attachmentid=30060&stc=1&d=1290556566.png
, L is the distance between the two positions ri and rj, but wouldn't ri -rj already give you the distance between the two positions?

In one dimension, yes. But note the boldface: [tex]\vec{r_i}[/tex] and [tex]\vec{r_j}[/tex] are vectors here, so [tex]\vec{r} = \vec{r_i} - \vec{r_j}[/tex] is the vector specifying the difference between the two, squaring it here is the http://en.wikipedia.org/wiki/Dot_product" of the resulting vector and itself, i.e. the x-component squared plus the y component squared and so on. So this works for any number of dimensions, by the pythagorean theorem [tex]\sqrt{x^2 + y^2}[/tex] is the distance in 2d, [tex]\sqrt{x^2 + y^2 + z^2}[/tex] in 3d and so on.
 
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