I'm doing some physics for a change, and I'm trying to sort things out a bit. From the definitions of mass, torque, momentum and angular momentum I've come up with the following integrals: \begin{equation} \int_D \rho(r)dV:\qquad\text{mass}\\ \int_D r\ \rho(r)dV:\qquad\text{center of mass times total mass}\\ \int_D \|r\|^2\ \rho(r)dV:\qquad\text{moment of inertia}\\ \int_D \frac{dr}{dt}\rho(r)dV:\qquad\text{momentum}\\ \int_D r\times\frac{dr}{dt}\rho(r)dV:\qquad\text{angular momentum}\\ \int_D r\times\frac{d^2r}{dt^2}\rho(r)dV:\qquad\text{torque (not sure)}\\ \end{equation} where $dV=d\lambda^3$ is the volume form, $D$ is some bounded domain in $\mathbb{R^3}$, $r:\mathbb{R}\rightarrow\mathbb{R^3}$ is the position vector and $\rho:\mathbb{R^3}\rightarrow\mathbb{R}$ is the mass density function. They are supposed to be smooth.

All these integrals seem to originate from some common "ancestor", some kind of moment generating function. My question is: is it true? i.e. is it possible to order these integrals in some sensible way, or to derive them from a single function?


You are essentially calculating mass weighted averages of quantities defined for point particles over an extended body. As far as I know there is no convenient generating expression for mass, inertial tensor, kinetic energy, momentum, angular momentum, torque etc.

The formulas look similar because they are functions defined for point particles. The only handles you have for point particles is the path (ie $r(t)$), its derivatives in time and "external properties" like mass or charge. Most kinematic expressions are linear in the mass, so the only expressions left are things like $m r \times \dot r$ and $m \frac{1}{2}\dot r^2$ which necessarily look very similar.

In the transition to collections of point particles and extended bodies these properties are taken to be extensive (ie the total mass of two particles is the sum, and not the root of the squared sum), so the correct generalisation to many body or extended systems are expressions of the form $\int \rho r\times \dot r$ and $\int \rho \frac 12 \dot r^2$.


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.