2
$\begingroup$

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 :)

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$

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.