Is there a fundamental reason not to define the work vice-versa

Question

My question arises from something which has never been really clear: in continuum mechanics, why is strain energy defined as: $$W=\int_\Omega \underline{\underline{\sigma}}:\mathrm{d}\underline{\underline{\varepsilon}}$$ rather than $$W=\int_\Omega \underline{\underline{\varepsilon}}:\mathrm{d}\underline{\underline{\sigma}}$$

I think this question is closely related to a "more general" question: that of the work of a force, defined by: $$W=\int_\mathcal{C} \underline{F}\cdot\mathrm{d}\underline{s}$$

Why do we never talk about the symmetric relation: $$W'=\int_\mathcal{C} \underline{s}\cdot\mathrm{d}\underline{F}$$

I'm not asking for explanations on the commonly used definitions but if there is a fundamental reason why their are not defined the "other way round".

Edit Additions to explain why it's unclear to me: Correct me if I am wrong: the energy can be seen as a linear form over the velocities or displacements (which live in a vector space) to give scalars called forces (which live on the dual vector space). Is it correct to say that this relation can be "symmetrized" to define a linear form over the forces to yield velocities?

Why do we write $$W=\int Fv\,\mathrm{d}t = \int F\,\mathrm{d}s\qquad\text{ rather than}\quad =\int v\,\mathrm{d}G$$ where $G$ would be a primitive of $F$, as the displacement $s$ is the primitive of $v$?

Answer 1

The reason the relationship $$ W=\int\mathbf s\cdot d\mathbf F $$ doesn't work is because Work is defined as the result of a force $\mathbf F$ on a point that moves along a distance. The point follows a curve $\mathbf s$ with a velocity $\mathbf v$. The small amount of work, $\delta W$, that occurs of the instant of time $dt$ is $$ \delta W=\mathbf F(\mathbf s)\cdot\mathbf v(\mathbf s)dt $$ Integrating both sides, $$ W=\int\mathbf F(\mathbf s)\cdot\mathbf v(\mathbf s)dt $$ since $\mathbf v=d\mathbf s/dt$, this is $$ W=\int\mathbf F(\mathbf s)\cdot\frac{d\mathbf s}{dt}dt\equiv\int\mathbf F(\mathbf s)\cdot d\mathbf s $$ Alternatively, $\mathbf F=m\mathbf a$, so this would give us $$ W=m\int \mathbf a\cdot\mathbf v\,dt $$ Since $\mathbf a=d\mathbf v/dt$, this is really $$ W=m\int d\mathbf v\cdot\mathbf v=\frac12mv^2 $$ which brings us back to the work-energy theorem. Note though that this is still not $\mathbf v\cdot d\mathbf F$, it's something entirely different.

Answer 2

Because, according to your definitions, if I strain a rubber bar with constant force until it rips apart, I haven't done one joule of work to it.

Answer 3

For starters, these are not the same thing. The integration by parts rule makes this fairly obvious:

$$\int_i^f y\,\mathrm{d}x = y_f x_f - y_i x_i - \int_i^f x\,\mathrm{d}y$$

But then you might be wondering what makes $\int \vec{F}\cdot\mathrm{d}\vec{s}$ the "right" definition for work while $\int \vec{s}\cdot\mathrm{d}\vec{F}$ is the "wrong" one. In a nutshell, the "wrong" definition depends strongly on how you define $\vec{s}$. If you just let $\vec{s}$ be the position, then you get different results for $\int\vec{s}\cdot\mathrm{d}\vec{F}$ depending on where you choose the origin of your coordinate system to be. Physics shouldn't work that way. On the other hand, $\int\vec{F}\cdot\mathrm{d}\vec{s}$ only involves the differences between coordinates, and thus is independent of where you put the origin.

Answer 4

There are already many good answers. Besides the fact that the standard definition of work directly relates to the work-energy theorem and the notion of potential energy, here is a geometric argument.

I) The force $F_i(x,v,t)$, $i\in\{1,2,3\},$ transforms as $(0,1)$ co-vector

$$\tag{1} F_i ~=~\sum_{j=1}^3F^{\prime}_j \frac{\partial x^{\prime j}}{\partial x^i} , \qquad i\in\{1,2,3\}, $$

under spatial coordinate transformations

$$\tag{2} x^i~\longrightarrow x^{\prime j}~=~f^j(x). $$

This means that

$$\tag{3} \mathbb{F}~=~\sum_{i=1}^3F_i ~\mathbb{d} x^i$$

is a one-form, which is independent of local coordinates, cf. e.g. this Phys.SE post.

II) On the other hand, both the quantity

$$\tag{4} \sum_{i=1}^3x^iF_i\quad\text{and}\quad\sum_{i=1}^3x^i \mathbb{d}F_i$$

depend on coordinate system. Therefore geometrically, it is usually not so useful to know that the one-form (3) can be written as

$$\tag{5} \mathbb{F}~=~ \mathbb{d} \sum_{i=1}^3x^iF_i - \sum_{i=1}^3x^i \mathbb{d}F_i;$$

or equivalently when integrated along a curve $\gamma:[0,T]\to \mathbb{R}^3$, that work can be written as

$$W\tag{6} ~=~\int_{\gamma}\mathbb{F}~=~ \left[\sum_{i=1}^3x^i~F_i \right]_{t=0}^{t=T}-\int_{\gamma}\sum_{i=1}^3x^i \mathbb{d}F_i , $$

which is (minus) OP's alternative formula, up to boundary terms.