# Is the average force calculated from $F(x)$ the same as that calculated from $F(t)$?

Say a force is doing work on an object in one dimension. I could calculate the average force over the distance with

$$\frac{1}{\Delta{x}}\int_{x_1}^{x_2} F(x) \text dx$$

If I also formulated force as a function of $$t$$, I could calculate the average force over the total time period with

$$\frac{1}{\Delta{t}} \left| \int_{t_1}^{t_2} F(t) \text dt \right|$$

Obviously, the integrals themselves are not equal. One represents work, and the other represents impulse; $$\Delta{p} \neq \Delta{K}$$.

However, if $$x_1$$ corresponds to $$t_1$$ and $$x_2$$ corresponds to $$t_2$$, could I set the above expressions equal to one another? Does averaging over the interval make them equal? Put another way, I've never seen texts differentiate between the $$F_{ave}$$ in $$W = F_{ave} \cdot \Delta{x}$$ and the $$F_{ave}$$ in $$\Delta{p} = F_{ave} \Delta{t}$$. Is there a difference? If not, I'm thinking they can relate work to $$\Delta{t}$$.

Before we start, let's assign a "name" to each integral (just for convenience):

$$I_1=\frac{1}{\Delta{x}}\int_{x_1}^{x_2} F(x) \mathrm dx \qquad \text{and} \qquad I_2=\frac{1}{\Delta{t}} \left| \int_{t_1}^{t_2} F(t) \mathrm dt \right|$$

Also, throughout the answer, I will consider the force $$F$$ to be the net force on the object. I will also be assuming that the mass of the body doesn't change throughout the motion. This way we lose a bit of generality but it becomes way simpler.

## Intuition

As you already observed that the average $$I_1$$ represents the change in kinetic energy divided by the displacement ($$\Delta K/\Delta x$$). This quantity gives us the answer of the question: What constant force should I apply over the distance $$\Delta x$$ such that the body gains a kinetic energy $$\Delta K$$? In other words, it is the average force which would yield the same kinetic energy change (as the original) when applied during the same displacement.

Similarly, the second average, $$I_2$$, gives us the answer to the question: What constant force should I apply over a time $$\Delta t$$ so that the body's momenum changes by $$\Delta p$$ (or equivalently, the body's velocity chnges by $$\Delta p/m$$)?

Now, we can observe a few differences between both the averages. First, $$I_1$$ is a displacement average whereas $$I_2$$ is a time average. Secondly, the change in velocity (or momentum) cannot be independently used to find the change in kinetic energy, which implies both the averges are independent and can independently take different values. Thus there is no reason to believe that both of them are, in any way, equal. However, if the force is a constant, then both the averages will be equal to the constant force. Thus, in uniformly accelerated motion, both the averages will turn out to be equal.

## Mathematical Analysis

Let's try to analyse both the averages. It is clear that:

$$I_1=\frac{\Delta K}{\Delta x}=\frac m 2 \frac{(v_{t_2}^2-v_{t_1}^2)}{\Delta x}$$

and

$$I_2=\frac{\Delta p}{\Delta t}=\frac{m(v_{t_2}-v_{t_1})}{\Delta t}$$

Now let's find the ratio of both these averages ($$I_1/I_2$$):

$$\frac{I_1}{I_2}=\frac{\Delta t}{\Delta x} \frac{(v_{t_2}+v_{t_1})}{2}\tag{1}$$

But $$\Delta x/\Delta t$$ is the time average of velocity over the time interval $$\Delta t$$, thus equation $$(1)$$ can be rewritten as

$$\frac{I_1}{I_2}=\frac{1}{v_{\text{average}}} \frac{(v_{t_2}+v_{t_1})}{2}\tag{2}$$

Thus as you can see, this ratio need not always equal one, which implies that $$I_1=I_2$$ is not necessarily true. However, in the special case of constant acceleration (where $$F(t)$$ is a constant, which implies that $$a(t)$$ is a constant), the time average of velocity comes out to be the average of initial and final velocities, i.e.

$$v_{\text{average}}=\frac{(v_{t_1}+v_{t_2})}{2}$$

Substituting this in equation $$(2)$$, we get

$$\frac{I_1}{I_2}=1\Longrightarrow I_1=I_2$$

Note: The above analysis, in no way implies that $$I_1=I_2$$ only when the motion is a uniformly accelerated one. There can be many other scenarios where $$v_{\text{average}}=(v_{t_1}+v_{t_2})/2$$ even if the acceleration isn't a constant.

In fact, when we divided $$I_1$$ by $$I_2$$, we implicitly assumed that $$v_{t_1}\neq v_{t_2}$$. If $$v_{t_1}=v_{t_2}$$, we would get a $$0/0$$ indeterminate form. Although, we can still compare the values of $$I_1$$ and $$I_2$$ under this condition, and these simply turn out to be $$I_1=I_2=0$$. Thus no matter how you change your velocity, if the initial and the final velocities are equal, then both the will become equal to $$0$$. This is intuitive as well, since in this case, both, the work done and the momentum change are $$0$$, thus yielding $$I_1=I_2=0$$.

## Summary

Thus, both the averages, in general, aren't equal, but in the case of uniformly accelerated motion, both the averages turn out to be equal.

• "in the case of uniformly accelerated motion, both the averages turn out to be equal." But you didn't need this mathematical analysis to see that. If you have uniform acceleration, you have a constant force. Of course any average of a constant is just the same constant. – knzhou May 13 '20 at 19:45
• @knzhou Yes :) I agree my analysis took the unnecessary and the long route, however I wanted it to be as general as possible. I didn't want to leave out any other non trivial (but simple) situation where both the averages equalised. It may be seen as my obsession with rigour. However, the point you raised is definitely worth addressing in the "Intuition" section, thanks! – user258881 May 13 '20 at 19:50
• interestingly it's not just uniform acceleration - it's just a comparison of an average with two initial values, there are whole families of functions for which it's true - we could add any odd function centred at time $t=\frac{t_1+t_2}{2}$ with period $T=t_2-t_1$ – zephyr Jun 19 '20 at 18:56
• to be more specific, any function $b(t)$ which obeys $b(-t)=-b(t)$ and $b(\frac{1}{2}T)=0$ can be added to a linear acceleration without changing $\frac{I_1}{I_2}$ - no reason why any phsyical system would do this though (as far as I know) – zephyr Jun 19 '20 at 19:04

No. They aren’t the same. To see why consider the displacement (1D) being parametrised by time in a way to give $$x(t)=\alpha t^2$$ where alpha is a constant to fix the dimensions. From Newton’s second law this gives us $$F(x)=2\alpha m$$, then,

$$\frac{1}{\Delta x}\int F(x)dx= \frac{1}{\Delta x}\int_{x(t_1)}^{x(t_2)} 2\alpha m dx=2\alpha m\\ \frac{1}{\Delta t} \int F(x(t))dt= \frac{1}{\Delta t}\int_{t_1}^{t_2}\alpha t^2dt= \frac{\alpha}{3}\left(\frac{{t_2}^3-{t_1}^3}{t_2-t_1}\right)$$

Clearly the two are unequal. What gives? It’s the measure of the integral that’s different in the two cases. Clearly $$F(x(t))=F(t)$$ as $$x$$ is parametrised by $$t$$. However in general $$dx\ne dt$$ and isn’t fixed by normalising the measure either as their rate can differ! And that makes all the difference.

• Ah, yes. Thank you for the good example. It makes sense that the average force "per unit time" is incomparable to the average force "per unit distance." Distance and time are different measures, so we're comparing apples to oranges. – iRove May 13 '20 at 19:04
• Your example doesn't obey Newton's Second Law. $\ddot{x} = 2 \alpha$ is a constant, but $F(x)$ is not constant unless $x$ is. – Michael Seifert Jun 19 '20 at 18:32
• Oops! @MichaelSeifert thanks for that. I have corrected the disobedient equation. – Superfast Jellyfish Jun 19 '20 at 20:01
• The second integral needs to be corrected as well. In fact, for uniformly accelerated motion (which is what you have here if $x \propto t^2$), the average of the force over distance and over time will be the same. It's only when the force varies over the motion that the answers will be different. – Michael Seifert Jun 19 '20 at 21:28

The answer is no, they are not the same things.

There is a tacit assumption when texts say things like "the average force," about what the average is performed over. Averaging over position is not the same as averaging over time.

You can always construct simple examples in which it just so happens that they are the same, but that is of limited utility.

Your reasoning is valid with the important caveat that you yourself realized: this works if $$x_1$$ corresponds to $$t_1$$ and $$x_2$$ corresponds to $$t_2$$, and as long as the force over time and the force over distance both remain differentiable during that span of time and space. You can't equate the two expressions in a general sense and expect that it will hold for all situations.

To draw a physical analogy, imagine a drag racer at the starting line. If we measure $$t_1$$ at the moment when the light turns green and the force of his car's rocket begins to propel him forward, then the average force will be equivalent in the two expressions, because the force vector will be applied for the entire time and for the entire distance in a way that is equivalent and differentiable.

If you measure $$t_1$$ as being 10 seconds prior to the green light, however, then the force is zero for ten seconds, and that brings down the average in your time-based expression. It does not bring down the average in the distance-based expression because the car is in the same place. You also won't be able to calculate the integral sum for your time-based expression without adding two separate integrals together: one expression for the zero-velocity force at the starting line, plus your integral above which applies to the race itself. At the point when the green light flashes, the engines ignite and the force vector goes from zero to rocket. At that moment, the force over time is not differentiable - it has undergone a transition from nothing into something. This type of non-differentiable change in your system would invalidate the equivalence between your two expressions.

• Your claim that the quantities are equivalent so long as the force is differentiable is false. See my answer to this question for a counterexample. – Michael Seifert Jun 19 '20 at 18:44

To show constructively that these quantities are not equal in general, let's consider an object undergoing simple harmonic motion over a quarter-cycle (i.e., from the equilibrium position to maximum displacement.) In this case, we have $$F(x) = - k x, \qquad x(t) = A \sin \omega t, \qquad t \in [0, \pi/(2\omega)],$$ with $$\omega = k/m$$ (though this won't matter in the end.)

The average over displacement is $$\langle F \rangle_x = \frac{1}{A} \int_0^A (-kx) \, dx = - \frac{1}{A} \left[ \frac{1}{2} kx^2\right]_0^A = - \frac{1}{2} k A,$$ while the average over time is $$\langle F \rangle_t = \frac{2 \omega}{\pi} \int_0^{\pi/2\omega} (-k A \sin \omega t) \, dt = - \frac{2 k A \omega}{\pi} \left[ - \frac{\cos \omega t}{\omega} \right]_0^{\pi/2\omega} = -\frac{2}{\pi} k A.$$ Since $$\pi \neq 4$$,[citation needed] these results are different.