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

Question

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 $dW=x \cdot dF$?

Answer 1

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).

Answer 2

There are already several good answers. In this answer, we will just highlight a geometric argument.

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.

Answer 3

The answer to your question depends on how we define work.

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.

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

Answer 4

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 (infinitesimally) tiny.

Answer 5

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.

Answer 6

$F(x,t)$ 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.

Answer 7

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.

Answer 8

The two are related by integration by parts; in vector notation we could say that $$W = \mathbf F_1\cdot\mathbf r_1 - \mathbf F_0 \cdot \mathbf r_0 - \int_{\mathbf F_0}^{\mathbf F_1} \mathrm d\mathbf F \cdot \mathbf r.$$ where $\mathbf r$ is the position vector.

As other answers noted, if you're dealing with a constant force then it does not change from $\mathbf F_0$ to $\mathbf F_1$ and the latter integral is over zero domain, so that $W=\mathbf F\cdot(\mathbf r_1 - \mathbf r_0).$ In this respect it is not very well connected "microscopically" to what is happening "macroscopically", as that expression could be written $F~\Delta x$. In addition it can often be a bit difficult to phrase $\mathbf r$ as a function of $\mathbf F$ such that this integral can be done; so for example you might imagine a spring force $F = -kx$, you have to rewrite this as $x=-F/k$ to do this integral and find that $$ W = -\frac{F^2}{k} + 0 - \int_0^{F} \mathrm df~\frac{-f}{k} = -\frac{F^2}{k} + \frac{F^2}{2k} = -\frac{F^2}{2k}. $$ Compare with the same calculation for $F~\mathrm dx$ where it's just $$W = \int_0^x\mathrm dx'~(-kx') = -\frac12 k x^2.$$

It's also worth pointing out that the leading sign on $\int x~\mathrm dF$ is negative, so that $x\cdot\mathrm dF$ represents how much the linear approximation overcounts the work that you did. So if I said "well the force is 0 at $x=0$, the force is $-kx$ at $x=x$, let's just call the work $-kx^2$," this expression comes in to tell us that we overcounted by a factor of $-\frac12 kx^2$ and we need to subtract that off. In other words this is a mathematics where we have "pretended that all forces are constant" and then we need to introduce this $x~\mathrm df$ term to measure how far from constancy things were: the more $f$ changes, the more $\mathrm df$ gets involved.

Answer 9

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).