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Let $I(x)$ be an inertia tensor in matrix notation of a body in a coordinate system $x\in R^n$. Under a coordinate change $x=\phi(y)$, does the tensor transform as $Dx^TI(\phi(y))Dx$, where $Dx=\frac{\partial x}{\partial y}$ ? In other words, how can i express the inertia tensor in the $y$ coordinates alone?

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I'm not sure I understand what you mean. What's "a coordinate system $x\in R^n$"? Usually that notation would be used to denote a point or its coordinates, not a coordinate system. That raises the question what you mean by $I(x)$ -- is that supposed to indicate that the inertia tensor is position-dependent, or just that this is its matrix in a particular coordinate system? That would be a somewhat unusual notation; a subscript $x$ might be more appropriate. Lastly, $x=\phi(y)$ looks like a general coordinate transformation -- do you mean that, or just a basis change? –  joriki Jan 7 '12 at 1:46
Ok, i was a little bit hasty. The notation $I(x)$ means that the inertia tensor is dependent on coordinates $x_i$. For example, in Lagrangian mechanincs, I is a matrix whose elements are functions of the generalized coordinates $q% (except when the coordinate system is the body attached system). The map $x=\phi(y)$ is a general nonlineat (curvilinear) coordinate change. –  Jorge Jan 7 '12 at 2:19
I still don't understand the question. Could you be a bit more explicit about the setting you're presupposing? Is this a rigid body? A point mass? A system of point masses? Is this the usual inertial tensor with respect to Cartesian axes, or are you trying to find a concept of inertial tensor where not only $x$ but also $I$ is expressed with respect to generalized coordinates? If so, how would such an $I$ be defined (considering that the usual definition is an integral over quadratic polynomials in the Cartesian coordinates)? –  joriki Jan 7 '12 at 11:54
Let A be a Cartesian coordinate system (inertial frame). Consider a rigid body with coordinates $x_i$ and a coord system attached to this body (body frame). The inertia tensor w.r.t the body frame is a constant matrix. With respect to the inertial frame, the tensor depends on the coordinates $x_i$, thus $I=I(x)$. Now change coordinates $x=\phi(y)$. The question is how to express the inertia tensor w.r.t $y$. –  Jorge Jan 7 '12 at 12:17
Unfortunately you still haven't explained the most enigmatic part of the question -- what do you mean by "expressing the inertia tensor w.r.t. $y$"? Somewhat trivially, the inertia tensor as it's usually defined will "transform" as $I(x)=I(\phi(y))$, since $x=\phi(y)$ and the usual definition with respect to the inertial Cartesian frame nowhere refers to generalized coordinates. Unless you mean this trivial relationship, you'll need to explain what else you mean by "expressing the inertia tensor w.r.t. $y$". –  joriki Jan 7 '12 at 12:33

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