So I been reading "A first course in Rational Continuum Mechanics" by C. Truesdell and got the following confusion: torque is defined as the integral $$F_{x_0} = \int_B (x - x_0) \wedge df_{B^e},$$ here $B$ is a body, $B^e$ is it's exterior, $x = x(\bullet)$ is position function (motion), $x_0$ is a point of respect and $f_{B^e} = f(\bullet, B^e)$ is force applied to body from $B^e$.

I guess he never defined integral with respect to wedge product.

Truesdell also defined the wedge product $a \wedge b$ as alternating the tensor product $$a \wedge b = \frac 1 2(a \otimes b - b \otimes a) \in \bigwedge\nolimits^2V \subseteq V \otimes V \cong Hom(V,V), \quad V \cong \mathbb{R}^3$$ and defined other integrals as following (chap. 1.5): given $w: B \to V$, integral $\int_B w \otimes df_{B^e} \in V \otimes V$ is unique tensor such: $$\left(\int_B w \otimes df_{B^e}\right)^T a = \int_B (a \cdot w)df_{B^e}, \quad \forall a \in V,$$ and it's trace written as follows $$\int_B w \cdot df_{B^e} := \mathbf{tr}\left(\int_B w \otimes df_{B^e}\right).$$

As far as I understand integral of scalar function $\phi: B \to \mathbb{R}$ is just coordinate-wise integral $$\int \phi df = \left(\int \phi df_1, \int \phi df_2, \int \phi df_3\right), \quad f = (f_1, f_2, f_3): \Omega \to V,$$ here $\Omega$ is universe of bodies.

So my main problem is the mathematical notation (not the physics really), but since I never saw anything like that in math books I ask this question here. Also it would be good if someone may define these using Hilbert space $V$ not three-dimensional but rather infinite dimensional (just for math purposes).

P.S.: I'm a math student, and don't know much about physics and notations, so I'm really sorry for stupid questions like this. It's just to much confusion for me.

P.P.S.: I'm very bad at English so huge sorry for bad English here :)


1 Answer 1


It might be helpful to look at https://en.wikipedia.org/wiki/Differential_form.

$df_{B^e}$ is really the external force density expressed as a (vector-valued) 3-form, which can be integrated on a 3-(sub)manifold of the space, such as $B$. That integral will actually give you the total force, you can get it done for each component of the force.

The integral involving $(x-x_0)\wedge df_{B^e}$ is using $\wedge$ on the force part but not on the density part. It's just the idea of torque being the cross-product of position vector and force on an infinitesimal volume.

For this particular integral, it probably does not need a generalization from 3D to Hilbert space.


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