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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 $v$?

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    $\begingroup$ Is the underline notation for a vector? What is the double-underline notation? What is the colon? $\endgroup$ – Ben Crowell Oct 30 '14 at 1:48
  • $\begingroup$ @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. $\endgroup$ – anderstood Oct 30 '14 at 1:55
  • $\begingroup$ double-dot product of tensors i.e., contracting on both indices? $\endgroup$ – Ben Crowell Oct 30 '14 at 1:59
  • $\begingroup$ @BenCrowell Yes, so that it gives a scalar, but that does not really matter here :) $\endgroup$ – anderstood Oct 30 '14 at 2:07
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    $\begingroup$ 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.. $\endgroup$ – BMS Oct 30 '14 at 5:05
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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.

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  • $\begingroup$ 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 $\endgroup$ – anderstood Oct 30 '14 at 2:24
  • $\begingroup$ @anderstood: I've also added an alternative look at the work when trying to keep $\mathbf v$ there instead of keeping $\mathbf F$ there. $\endgroup$ – Kyle Kanos Oct 30 '14 at 2:25
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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.

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  • $\begingroup$ 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. $\endgroup$ – anderstood Oct 30 '14 at 1:58
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    $\begingroup$ 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. $\endgroup$ – Ben Crowell Oct 30 '14 at 1:58
  • $\begingroup$ @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. $\endgroup$ – anderstood Oct 30 '14 at 2:22
  • $\begingroup$ 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, $\delta W = -v\ \text{d} p$. $\endgroup$ – André Chalella Oct 30 '14 at 11:00
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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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – anderstood Oct 30 '14 at 18:48
  • $\begingroup$ 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? $\endgroup$ – anderstood Oct 30 '14 at 19:29

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