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I am working on basic physics definitions. Given a particle at position $r$ (in some coorinate reference system) upon which acts a force $F$, the $torque$ $\tau$ is defined by \begin{equation} \tau:=r\times F \end{equation} Now let's suppose that we have a continuous body bounded by a region $D$; let also $\rho(r)$ be the mass distribution and let $F(r)$ be a force field. I'd like to derive a definition of torque suitable to this situation. Any help?

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  • $\begingroup$ As strange as the body might be, it'll have an inertia tensor. Let's call it $I$. Then, $$L^\mu = I^\mu_{\,\nu} \omega^\nu ,$$ and $$\tau^\mu = d_t L^\mu .$$ $\endgroup$ – QuantumBrick Oct 29 '14 at 14:20
  • $\begingroup$ Thank you, but I was hoping to get away with it by some appropriate integral, like $\int_Dr\times\frac{d^2r}{dt^2}\rho(r)dV$ or something like that. $\endgroup$ – marco trevi Oct 29 '14 at 14:21
  • $\begingroup$ Use the vector formula of my previous answer and substitute the inertia tensor by it's full expression, taking care of also substituting $\omega = v/r$. Differentiate it with respect to time, and there's your integral. $\endgroup$ – QuantumBrick Oct 29 '14 at 14:29
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    $\begingroup$ @QuantumBrick - if you combine your two comments into an answer - with an explicit integral expression of the inertia tensor - then I think it would fully answer the question. And get an upvote from me... $\endgroup$ – Floris Oct 29 '14 at 15:28
  • $\begingroup$ Well, the moment of inertia tensor is rather simple $I_{\mu\nu}=\int_D \mathrm{d}V\rho(r)(r^2\delta_{\mu\nu}-r_{\mu}r_{\nu})$ $\endgroup$ – surajshankar Oct 29 '14 at 16:07
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I'll write my comments here as a full answer, as suggested by Floris. I won't use the moment of inertia tensor: it's simpler from pure angular momentum of each point particle.

We know that

$$\vec{L} = (\vec{r} \times \dot{\vec{r}})\,m .$$

So, for a point particle,

$$d\vec{L} = (\vec{r} \times \dot{\vec{r}})\, dm .$$

Noting that $\rho = \frac{dm}{dV}$,

$$d\vec{L} = (\vec{r} \times \dot{\vec{r}})\, \rho(\vec{r}) \, dV ,$$

which brings us to

$$\vec{L} = \int_\Sigma (\vec{r} \times \dot{\vec{r}})\, \rho(\vec{r}) \, dV ,$$

where $\Sigma$ is the region where the body's volume is defined on. Since the torque $\tau$ is the time derivative of $\vec{L}$, by assuming mass density doesn't vary in time we get

$$\vec{\tau} = \int_\Sigma \left(\dot{\vec{r}} \times \dot{\vec{r}} + \vec{r} \times \ddot{\vec{r}} \right)\, \rho(\vec{r}) \, dV \\ = \int_\Sigma (\vec{r} \times \ddot{\vec{r}} )\, \rho(\vec{r}) \, dV $$

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$ \newcommand{\r}{\mathbf{r}} \newcommand{\F}{\mathbf{F}} \newcommand{\g}{\mathbf{g}} \newcommand{\t}{\boldsymbol\tau} $Let me start by defining $\g(\r)$ to be a position dependent force per unit mass. Then the force per unit volume $d\F$ is given by $d\F(\r) = \g(\r) \rho (\r) dV$. The torque per unit volume $d\t$ is given by $d\t = \r \times d\F$. The total torque is then given by $\t=\int d\t = \int \r \times d\F = \int \r \times \rho \g dV$.

Given this definition for torque, and the independent definition for angular momentum, we can then make a meaningful state about relating the torque and angular momentum (given only central forces are at play).

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