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I'm studying rigid body mechanics and I've seen several proofs of properties related to total angular momentum, kinetic energy, etc. that all regard discrete set of points. For example, to show that in an inertial frame $\frac{\text d \mathbf {L}}{\text d t}=\mathbf{\Gamma}^{\text {ext}}$, one writes down the sum $$\mathbf{L}=\sum \mathbf{r} \times \mathbf{p},$$ and does the required derivatives.

How can such arguments be extended with rigour to continuous rigid bodies? To give an example, in "Rudin, Principle of Mathematical Analysis", in the chapter of Riemann Stieltjes integral, there's an example regarding the moment of inertia of a "straight line" body, that can be defined uniquely with the Riemann-Stieltjes integral $$\int _0 ^{\ell} x^2\, \text dm(x). $$ Altough it's restricted to a "straight-line" body, the result is very elegant.

Back to the example of angular momentum, I think that one could define the total angular momentum as $$\mathbf{L} =\int _{V} \mathbf {r} \times \mathbf {v(\mathbf{r})} \,\text d m(\mathbf{r}),$$where the integral is extended over the volume of the body (actually I've never seen a definition of total angular momentum for continuous rigid bodies, since it always appears as $$\mathbf{L} = \mathbf{I}\mathbf{\omega},$$ and, of course, this is the useful identity).

So, how can the discrete arguments be extended to continuous arguments?

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You recognize that $\vec{L} = I \vec{\omega}$ is a definition for $I$ and not for $\vec{L}$. – ja72 May 30 '13 at 19:21
Yes. In fact my point was that, when talking of continuous rigid bodies, I've never come across to an expression such as $\mathbf{L}=\int \mathbf{r}\times \mathbf{v} dm$, but I've always seen it expressed as $L=I\omega$ that, ofcourse, is a theorem and not a definition. – pppqqq May 30 '13 at 19:29
It is right here on section 1.4. – ja72 May 30 '13 at 20:21
and here on equation (24) – ja72 May 30 '13 at 20:27
and my favorite from Harvard at – ja72 May 30 '13 at 20:28

Let's look at the case of angular momentum.

In physics, "continuous" mass distributions are usually specified by a mass density. The mass density is defined as the function $\rho:\mathbb R\times\mathbb R^3\to\mathbb R$ that specifies the mass density of the system at any time $t$ and any point $\mathbf x$ in space. This means that it satisfies the following two properties

  1. At every time $t\in\mathbb R$, $\rho(t,\mathbf x) = 0$ for all $\mathbf x$ at which there is no mass.

  2. Given any volume $V$, the total mass $m(t,V)$ contained in that region at time $t$ is given by the following integral

$$ m(t,V) = \int_{\mathbb R^3} d^3 x\, \rho(t,\mathbf x) $$ In addition to the density function, we assume that there is a velocity function $\mathbf v(t,\mathbf x)$ that at any time $t$ gives the velocity of the point in the body passing through position $\mathbf x$ in space. Equipped with such functions, we can define the total angular momentum of the system at time $t$ as follows $$ \mathbf L(t) = \int_{\mathbb R^3}d^3 x\,\mathbf x\times \mathbf v(t,\mathbf x)\rho(t,\mathbf x) $$ Given this expression, one can then prove any "continuous" analog of a statement about angular momentum that one pleases.

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Thank you for your answer. Yes, this was basicly my same idea (what does $d^3 x$ mean?), but now I wouldn't know how to act on the last integral. Could you give, as an example, a complete proof that $ \frac{\text d \mathbf{L} }{\text d t}=\mathbf{\Gamma}$? – pppqqq May 30 '13 at 18:51
@Kazz8 $d^3x = dx dy dz$ is the volume element in three, Euclidean dimensions. I'll try to include that proof as soon as I can (I haven't actually tried it yet.) – joshphysics May 30 '13 at 20:41

Consider an infinitesimal mass element of mass $dm\left(\vec{r}\right)$ at position $\vec{r}$ and with velocity $\vec{v}$ . Its angular momentum is: $d\vec{L}=\vec{r}\times\vec{v}dm\left(\vec{r}\right)$

If there are multiple such masses, then we can sum over them to find the total angular momentum of the system. $$\vec{L}_{total}=\sum_{i=1}^{N}\vec{r}_{i}\times\vec{v}_{i}dm\left(\vec{r}_{i}\right)$$

Now, this summation can be written as an integral over the distribution of such masses. For a continuous mass distribution, the sum approaches the integral when $N\rightarrow\infty$ and the “mesh” of the mass distribution volume approaches zero. This statement is as rigorous as you want it to be and is the essence of the argument that is used to define Riemann integrals from sums of finite elements.

It is more intuitive to write $dm\left(\vec{r}\right)=m'\left(\vec{r}\right)dV=\rho\left(\vec{r}\right)dV$ and talk about Riemann integrals as opposed to their generalization, Riemann-Stieltjes. $\rho$ is identified as the volume density and in Cartesian coordinates $dV=dxdydz$ . $$\vec{L}_{total}=\sum_{i=1}^{N}\vec{r}_{i}\times\vec{v}_{i}\rho\left(\vec{r}_{i}\right)dV_{i}$$

Now we can follow the arguments that lead from a discrete sum to a Riemann integral. In the limit that $N\rightarrow\infty$ we can write:$$\vec{L}_{total}=\int_{\Omega}\left(\vec{r}\times\vec{v}\right)\rho\left(\vec{r}\right)dV$$

Where $\Omega$ is a volume encompassing the mass distribution. This is the angular momentum of the system with no further constraints on the velocities.

If the object is a rigid body rotating around an axis, then all of the infinitesimal masses from which we started will have the same angular velocity $\vec{\omega}$ whose magnitude is given by:$\omega=\frac{v}{r}$ . Without loss of generality, take the axis of rotation to be the $z$-axis. In terms of polar coordinates: $\vec{v}=v\hat{\phi}$. $$ \vec{L}_{total} = \int_{\Omega}\left(r\hat{r}\times v\hat{\phi}\right)\rho\left(\vec{r}\right)dV = \int_{\Omega}\left(r\hat{r}\times r\omega\hat{\phi}\right)\rho\left(\vec{r}\right)dV = \left(\int_{\Omega}r^{2}\omega\rho\left(\vec{r}\right)dV\right)\hat{z} = \left(\omega\int_{\Omega}r^{2}\rho\left(\vec{r}\right)dV\right)\hat{z} = \omega I\hat{z}=I\vec{\omega} $$

Note: the sum over the masses can expressed as an integral over delta-functions: $$\vec{L}_{total}=\int_{\Omega}\sum_{i=1}^{N}\vec{r}\times\vec{v}\rho\left(\vec{r}\right)\delta^{3D}\left(\vec{r}-\vec{r}_{i}\right)dV$$

where $\delta^{3D}\left(\vec{r}-\vec{r}_{i}\right)=\delta\left(x-x_{i}\right)\delta\left(y-y_{i}\right)\delta\left(z-z_{i}\right)$ is the 3-dimentional delta function.

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Hi @igphys, thank you for your answer. I think that your proof that $\mathbf{L}=I\boldsymbol{\Omega}$ is relative to a body lying in a plane, is it? Another thing: what does $\delta(x-x_i)$ mean? – pppqqq May 30 '13 at 18:58
@Kazz8, $\vec{L_{total}}=I\vec{\omega}$ as I have written, is valid for a rigid body rotating about an axis. It does not have to be a 2-dimensional body. The "Note" is extra and the arguments before it do not depend on it. $\delta\left(x-x_{i}\right)$ is the Dirac delta-function. It satisfies the relation: $\int_{x_{1}}^{x_{2}}\delta\left(x-x_{i}\right)f\left(x\right)dx=f\left(x_{i}\ri‌​ght)$ if $x_{i}\in\left[x_{1},x_{2}\right]$ and $\int_{x_{1}}^{x_{2}}\delta\left(x-x_{\star}\right)f\left(x\right)dx=0$ if $x_{i}\notin\left[x_{1},x_{2}\right]$. – igphys May 30 '13 at 21:33

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