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

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 $v$?

• Is the underline notation for a vector? What is the double-underline notation? What is the colon? – Ben Crowell Oct 30 '14 at 1:48
• @BenCrowell underline is a vector, double-underline is a tensor (Cauchy stress tensor and strain tensor here). Colon is the double-dot product of tensors. – anderstood Oct 30 '14 at 1:55
• double-dot product of tensors i.e., contracting on both indices? – Ben Crowell Oct 30 '14 at 1:59
• @BenCrowell Yes, so that it gives a scalar, but that does not really matter here :) – anderstood Oct 30 '14 at 2:07
• Can someone provide to me an interpretation of the integrand $\vec s \cdot \mathrm{d}\vec F$? Or at the very least $\mathrm{d}\vec F$? My mind fails.. – BMS Oct 30 '14 at 5:05

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.

• I upvoted now that you added the dependencies in $s$. That seems to be (part of) what I'm looking for because is shows that F and s do not play a symmetric role. TY – anderstood Oct 30 '14 at 2:24
• @anderstood: I've also added an alternative look at the work when trying to keep $\mathbf v$ there instead of keeping $\mathbf F$ there. – Kyle Kanos Oct 30 '14 at 2:25

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.

• Sure, but why is it that $W$ has a physical meaning and $W'$ doesn't really? Of course I could see it the other way round "we define W by ..." and that's all. – anderstood Oct 30 '14 at 1:58
• Another way to put this is that in integration by parts, you don't just have $\int u d v=\int v d u$; there is also a $uv$ term. Yet another way to see that this doesn't make sense is that $F$ is a function of $s$, but $s$ isn't a function of $F$. And yet another way: $d s$ means an inifinitesimal displacement, which makes sense, but $d F$ would be an infinitesimal force, which doesn't make sense. – Ben Crowell Oct 30 '14 at 1:58
• @BenCrowell Thank you for these enlightening explanations. I think I am trying to find a symmetric relationship between $F$ and $s$ (or $v$) while it is not possible. This idea that they are symmetric comes from my understanding of symplectic mechanics (see my edit). I must have something misunderstood. – anderstood Oct 30 '14 at 2:22
• Well, I really can't help much with the more sophisticated aspects of it, but I hope I helped you understand why it doesn't make practical sense with few words. Also, note that $dF$ does make sense sometimes: remember that, for a control volume in a reversible, steady state process, $\delta W = -v\ \text{d} p$. – André Chalella Oct 30 '14 at 11:00

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.

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.

• Thank you for making things clearer. So if I get it right, a force can be seen as a covector $F_i$. It defines a 1-form $\mathbb{F}$, i.e. a smooth section of the cotangent bundle ($M\longrightarrow T^\star M$). Also, $x\in M$ and $v\in T_x M$. By definition, the work is the integral of the 1-form $\mathbb{F}$. This 1-form is more naturally described in terms of the basis of $T^\star M$: $\mathbf{d}x^i$, naturally in the sense that it does not depend on a set of local coordinates. Agreed? – anderstood Oct 30 '14 at 18:48
• Additionally, in the particular case of linear elasticity, $F_i=kx^i$ (I'm not sure about the $x^i$ instead of $x_i$, there should be a metric tensor involved...?) so $\mathrm{d}F_i=k\mathrm{d}x^i$ and $\mathbb{F}=F_i\mathrm{d}x^i=x^i\mathrm{d}F_i$, so $x$ and $F$ can be inverted in the definition of $W$. For the same reason, $\int \sigma:\mathrm{d}\varepsilon=\int \varepsilon:\mathrm{d}\sigma$ in linear elasticity (and small strains). Does it seem correct? – anderstood Oct 30 '14 at 19:29