# Why is small work done always taken as $dW=F \cdot dx$ and not $dW=x \cdot dF$?

I was reading the first law of thermodynamics when it struck me. We haven't been taught differentiation but still, we find it in our chemistry books. Why is small work done always taken as $dW=F \cdot dx$ and not $dW=x \cdot dF$?

• I'd say Photon's answer explains it best, though some may not have enough math background to understand it well enough. The bigger picture is that, sure, you can define a quantity $x\cdot{dF}$... but it doesn't work out to be conserved in an adiabatic process that returns to its start position... which proves out to be why work is so important. (Plus $F\cdot{dx}$ applying the derivative to a fundamental dimension of the universe means it will show up in into other spots [more simply]) – JeopardyTempest Jan 31 '18 at 10:34
• I am deeply depressed by reading all the answers below. Except Steeven and Mitchell no one gave the simple, and from my point of view, the only correct answer. I just want to improve this "very nice" question. Please can someone below reply? Why small work is not defined as $dW=dF⋅dx$? – physicopath Jan 31 '18 at 13:58
• @physicopath I'm depressed by these answers, too, but evidently for a different reason. Work is the quantity of mechanical energy that crosses a system boundary. Any answer that doesn't mention energy (and most here don't) completely misses the mark. BTW $\mathrm{d}F\cdot\mathrm{d}x$ is mathematical nonsense in integral calculus. – garyp Jan 31 '18 at 14:18
• @photon does mention the energy – physicopath Jan 31 '18 at 14:20
• Actually, I think that the question is sufficiently answered. There is an intuitive answer by Mitchell (the experimentalist answer), a calculational answer by me (the theorist answer) and a very nice symmetry argument by Qmechanic (the mathematician answer). So in total, all aspects are covered. – Photon Jan 31 '18 at 16:07

Very nice question! You can see this from the second Newton's law:

$$m\ddot{\mathbf{x}} = \mathbf{F}(\mathbf{x})$$

Now I would like to integrate this equation of motion with respect to time, to arrive at the energy conservation. To do so I multiply both sides with $\dot{\mathbf{x}}$:

$$m\ddot{\mathbf{x}}\cdot\dot{\mathbf{x}} =\mathbf{F}(\mathbf{x})\cdot\dot{\mathbf{x}}$$

and finally integrate:

$$m\int dt \ddot{\mathbf{x}}\cdot\dot{\mathbf{x}} =\int dt\mathbf{F}(\mathbf{x})\cdot\dot{\mathbf{x}}$$

The l.h.s. gives me the kinetic energy. The r.h.s. gives me exactly the integral in question:

$$\frac{1}{2}m\dot{\mathbf{x}}^2 = \int d\mathbf{x}\cdot\mathbf{F}(\mathbf x)$$

So the work done by the force is the kinetic energy of the particle (up to an integration constant representing its total energy).

Definition, A force is said to do work if, when acting, there is a displacement of the point of application in the direction of the force.

In layman language, to do work you need displacement, not just force.

In the equation, $dW=x.dF$, we are considering a change in force at a constant position from a reference point (origin). As per our definition, no work is being done because there is no displacement. To understand it better, assume there is a heavy block and you are applying a variable force to it.

No matter how hard you push, you won't be able to give it some speed. The work-energy theorem says that if a force does some effective work, there is a change in kinetic energy of the body. But in our case, there is no change in kinetic energy which implies no work is done by you. This is a pretty good way to get a hang of what work is.

On the other hand, in the equation $dW=F.dx$, we are considering an infinitesimal displacement for a constant force. Work is being done here as we have a force and a displacement. If we have a variable force, we will have to break our procedure of calculating work into infinitesimal displacements for which force can be assumed constant.

$$W=\int\vec{F}.\vec{dx}=\int\vec{|F|}\vec{|dx|}\cos\theta$$

• A good answer, but I think I would reverse the emphasis and place the energy argument first, from which follows the definition that makes it work. – garyp Jan 30 '18 at 20:01

1. On one hand, work (within Newtonian mechanics) $$\mathrm{d}W=\vec{F}\cdot \mathrm{d}\vec{r}$$ is a scalar quantity, meaning that it is independent of coordinate system.

2. On the other hand, the quantity $\vec{r}\cdot \mathrm{d}\vec{F}$ depends on coordinate system. E.g. if we choose the $\vec{r}=\vec{0}$ to be the origin, the quantity vanishes.

Even if you are unswayed by physical arguments, you should think twice about introducing non-geometric quantities.

• Please please please! $\vec{F}d\vec{r}$ is scalar so is $\vec{r}d\vec{F}$! If $\vec{r}=0$ then the quantity vanish so does work vanishes if $\vec{F}=0$! – physicopath Jan 31 '18 at 13:21
• @physicopath: No. For starters, $\vec{r}$ does not transform as a $(1,0)$ tensor under coordinate transformations (even if we restrict to affine coordinate transformations). In contrast, $\vec{F}$ transforms as a $(1,0)$ tensor under general coordinate transformations. – Qmechanic Jan 31 '18 at 14:02
• @Qmechanic when I look at your answer the only thing I see is a snake eating its own tail. In addition, it surprises me that a professional like you answering such a simple question with such a simple answer, in this way. – physicopath Jan 31 '18 at 14:19
• @physicopath: In terms of transformational properties $\vec r$ is not a vector (to be more precise, it is not an element of the tangent bundle), so the scalar product involving it is not a scalar and thus not invariant under coordinate transformations. – Photon Jan 31 '18 at 16:04

The two give very different physical results. Consider a force of $1~\rm N$ applied over a distance of $1~\rm m$. The work is correctly computed as:

$$W\int^{1~\rm m}_{0~\rm m} \vec F\cdot d\vec x=\vec F\cdot\int^{1~\rm m}_{0~\rm m}d\vec x=1~\rm J$$

Trying to apply the other formula doesn't give anything sensible. The force doesn't change, so evidently $d\vec F=0$:

$$W=\int^{1~\rm N}_{1~\rm N}\vec x\cdot d\vec F=0?$$

This would imply a constant force applied over any distance always gives zero work. Clearly, this is nonsense.

• Why can't you have a varying force over a constant distance? – Allure Jan 28 '18 at 7:59
• @Allure Of course you can- this answer was just one simple example showing that it gives wrong results. The second formula doesn't give the correct results for that example, either: physically that should give you zero work (as no energy is imparted if the thing doesn't move). The first formula correctly gives zero work, while the second does not. – Chris Jan 28 '18 at 10:34
• Why you can't have $$W=\int^{1~\rm N}_{0~\rm N}\vec x\cdot d\vec F$$O.o ? – user8277998 Jan 28 '18 at 11:47
• @user45914123 That would indicate that the force changed from 0 to $1~\rm N$ over time. It isn't the right expression for a constant force. – Chris Jan 28 '18 at 20:02
• @Chris Why it would represent a change with time, shouldn't it represent a change with displacement ? Why you need constant force ? Why can't a force vary with displacement sort of like electric field by a point charge - father you go lesser the force you experience ? – user8277998 Jan 28 '18 at 20:10

Because work $W$ is a force $F$ causing a change in position $\Delta x$.

Not just a force $F$ causing a position $x$. Or a change in a force $\Delta F$ causing a position $x$. Neither makes much sense. We are talking about a change in position - that is how work is defined.

And such a change $\Delta x$ is simply symbolized $dx$ when it is very, very (infinitely) tiny.

• I just do not understand why people, including @Qmechanic , trying to answer this question in these rather unintuitive ways while the answer is so simply as you put "Because work W is a force F causing a change in position Δx." not the opposite. – physicopath Jan 31 '18 at 13:09
• @physicopath Allthough I appreciate the gesture, keep in mind that the method of answering depends on how the answerer interprets the issue, how the answerer interprets the level of the OP and which approach the answerer from experience finds appropriate and most gaining. People choose differently; let's avoid calling out users by name. – Steeven Feb 1 '18 at 15:28
• "We haven't been taught differentiation but still, we find it in our chemistry books." this sentence is enough to interpret the issue. – physicopath Feb 1 '18 at 15:36

I spent some time puzzling over this because I didn't find any of the answers satisfactory. I thought for a while that $dW = dF \cdot S$ is also valid, but is rarely seen physically. However, I eventually convinced myself that indeed, $dW = F \cdot dS$, and cannot be vice versa.

The key insight is that work only happens if there is a displacement. Holding a heavy object at a constant height is not doing work (in physics). Displacement is, by definition, $dS$. If $dS$ is zero, then there cannot be work. Only $dW = F \cdot dS$ satisfies this requirement.

But what if we have a varying force acting over a constant displacement? Let's say I apply the force $F = \mathrm{sin}(x)$ over a displacement of one meter. This superficially meets the requirements: a varying force, and a constant displacement. But when we look at the obvious equation for the resulting work done

$W = \int \mathrm{sin}(x) \times 1 \mathrm{m}$

It becomes clear that even then, the formula used is actually $dW = F \cdot dS$! The equation has only one variable, which means there can only be one integration variable - $dx$ (which is equal to $dS$ in 1D). The "constant displacement" isn't constant at all. In fact that one meter is equal to $dS$

tl; dr: $dW = F \cdot dS$ is the only equation that makes physical sense. I don't know if this is the answer you're looking for, but it's the one that answered the question for me.

$F(x,t)$ can always be expressed as a function of position (and time); however, it is generally wrong to write position as a function of force (as well as, $x(F)$ will fail to describe how nature works.)

To see why: considering an object being acted by a constant force, such a function is always multi-valued, it is so ill-defined that the input is always same, while the output is always different. Hence this $x(F)$ fails to predict the motion of the object at any given moment.

Now, let's see how it is wrong with physics: suppose we manage to find the area below the $\int x(F,t)dF$. Hence we know the "work done". However, the area below an object under constant force is zero, as it is a vertical straight line, but it is plainly wrong! The object is accelerating; therefore, something must keep inputing energy to the object! The work done can't be zero!

All in all, mathematically, "$x\,dF$" is ill-defined. Physically, it describes the physics so wrong.

The answer is very simple, (small, useful) Work is defined as dW = F.dX.
There is another type of ("not useful") Work that is given by dW = X.dF.
An example of this type, is a person lifting a given weight by applying a force less than mg. As the force changes from 0 to almost mg, no (useful) work is done (the weight does not move).

• No. Work is always $dW=\vec F\cdot d\vec x$. It's not even really a definition, but a direct consequence of Newton's second law. – Chris Feb 3 '18 at 7:44

## protected by Qmechanic♦Jan 28 '18 at 11:48

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