# Newton's second law of motion in terms of momentum

I am reading a document and in answer to the question State Newton’s second law of motion the candidate answers that The force acting on an object equals the rate of change of momentum of the object. While this is not a complete answer, the examiner picks on on a word equal and says that it is proportional instead.

Now I understand that $F=\frac{dp}{dt}$ where both $F$ and $p$ are vectors — what is "proportional" about it?

I checked my Good Old Ohanian (2nd edition) and it says explicitly "The rate of change of momentum equals force" in section 5.5 The Momentum of a Particle.

What is this examiner talking about?

-
The examiner is stupid, as usual. They are equal. –  Ron Maimon Sep 29 '11 at 17:07
This is not psychology.stackexchange but the examiner almost certainly was confusing momentum with velocity, and replacing force equals change of momentum with change in velocity in his head. Why must some person's incompetence be a question on this site? –  Ron Maimon Sep 29 '11 at 17:43
Well, I just found that in this translation of Newton's Principia, "the alteration of motion [momentum] is ever proportional to the motive force impressed" although I fail to see how it is relevant today. It seems to be more of an issue with Newton's writing style which is probably why 'proportional' had been replaced with 'equal' in (most) modern textbooks. –  The Gruffalo Sep 29 '11 at 18:06
404 link now dead. –  Qmechanic Apr 21 '13 at 10:47

## 5 Answers

This actually comes down to a question about the interpretation of units, and it's kind of a tricky issue.

On one hand, you can view any equation in physics as nothing more than a mathematical relationship between some numbers. This was the view taken in the early days of quantitative physics,* back in the 17th and 18th centuries, when the concepts of force and momentum were just being quantified. Since there was very little in the way of collaboration, the idea of a standardized unit system hadn't really taken off, so if you were doing an experiment to establish the relationship between force, momentum, and time, the units you used would have been determined by your equipment. In other words, all you would know is that you have a force meter (scale) which will give you a number proportional to the force, a "momentum meter" which will give you a number proportional to the momentum change, and a clock of some sort (perhaps a pendulum) which will give you a number proportional to the time.

Let's say it's been established that the relationship is linear. You would probably run an experiment in which you apply a certain quantity of force for a given number of ticks of the clock, and change in momentum off your measuring device. Then you would plug those numbers - it's important to notice that you're only dealing with numbers, since there are really no meaningful units to speak of - into the discrete approximation of Newton's second law:

$$F^{(N)} = K_F^{(N)}\frac{\Delta p^{(N)}}{\Delta t^{(N)}}$$

Here I've used the superscript $^{(N)}$ to indicate pure numerical value, i.e. the number you read off the scale/clock/meter. Plugging in these numbers will allow you to determine the value of the constant $K_F^{(N)}$. Hopefully it's obvious that the value of this constant will depend on how your equipment is calibrated, or in other words, which unit system you're using. If hypothetical-experimenter you decided to label your unit of force the pound $\mathrm{lb_F}$, your unit of momentum $\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}$, and your unit of time the second, you would find that

$$K_F^{(N)} = 32.174$$

Back in those days, before people started really thinking about units, all multiplicative equations in physics were thought of as simple relationships between numbers. Accordingly, they included constants of proportionality which were customized to each lab's equipment.

Eventually, as more people started doing physics, there arose a need for standardized unit system so you could compare data from different labs. Then you could write the above equation as

$$\frac{F}{\mathrm{lb_F}} = \frac{K_F}{\mathrm{lb_F}/(\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2})}\frac{\bigl(\frac{\Delta p}{\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}}\bigr)}{\bigl(\frac{\Delta t}{\mathrm{s}}\bigr)}$$

because the numerical value of a measurement is just the measurement divided by its unit. You can algebraically rearrange this to

$$F = K_F\frac{\Delta p}{\Delta t}$$

where

$$K_F = K_F^{(N)}\frac{\mathrm{lb_F}}{\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2}}$$

So Newton's second law has shifted from being a simple relationship between numbers, to being a relationship between physical quantities which are expressed as multiples of a reference value. However, you still have a conversion constant in the equation. Symbolically, it's independent of units, but you still do have to plug in a different number depending on which combination of units you want to work with. This is the sense in which $F$ is only proportional to $\frac{\mathrm{d}p}{\mathrm{d}t}$.

In the modern scientific community, on the other hand, I think most people would agree that units are a human invention, and that physical quantities should exist in some sense independently of the units we choose to use for them. Taking that view, there should be some "natural" way to express the equations of physics that doesn't incorporate any "unit system artifacts" like these proportionality constants.

The way we do this is to define the units as abstract objects and develop a set of rules for manipulating them (kind of like unit vectors). We can then incorporate the conversion constants into those rules. For example, let's again consider the discrete approximation of Newton's second law, but this time without the conversion constant written into it:

$$F = \frac{\Delta p}{\Delta t}$$

You can still use seconds for time and $\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}$ for momentum in this equation. When you read the numbers off your measuring equipment and plug them into the formula, you'll do it like this:

$$F = \frac{\Delta p^{(N)}\ \mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}}{\Delta t^{(N)}\ \mathrm{s}} = \frac{\Delta p^{(N)}}{\Delta t^{(N)}}\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2}$$

Suppose you want your answer in pounds of force. You would look up the multiplication rule for $\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2} \to \mathrm{lb_F}$, which in this case can be found on Wikipedia:

$$\mathrm{lb_F} = 32.174\ \mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2}$$

(in general you might have to chain a few rules together to get the right conversion). So you wind up with

$$F = \frac{1}{32.174}\frac{\Delta p^{(N)}}{\Delta t^{(N)}}\mathrm{lb_F}$$

It works out to the same thing as before, but this time the conversion constant $K_F$ is part of the unit system, not the equation. This means that if you're not plugging actual values into the equation, you don't have to think about units or proportionality constants at all. And if you look at it this way, $F$ is actually equal to $\frac{\mathrm{d}p}{\mathrm{d}t}$.

So what's the verdict? Unfortunately, there's no unassailable answer to this question of whether Newton's second law is a proportionality or an equality. Depending on how you think about it, either answer could be valid. But I would say the "equality" answer, which corresponds to the modern view of units, is conceptually cleaner. It's accepted by all competent modern physicists, as far as I know (for mechanics, at least; electromagnetism is a whole different story), and it's certainly the interpretation we try (however unsuccessfully) to instill into introductory physics students' minds. I'd definitely agree that the examiner was being unreasonably picky (though to be fair, he did give credit for it).

*I don't have an explicit source, so I'm not entirely sure this is the way things were really developed; I'm basing my description on some fuzzy memories. That being said, the backstory does help clarify the various ways in which we treat units, so consider it historical fiction if you must.

-
Thank you David. I think now I understand the context of this question much much better. –  The Gruffalo Sep 30 '11 at 7:54

From Principia Mathematica - The Mathematical Principles of Natural Philosophy by Isaac Newton translated by John Machin - volume 1

Definition II

The quantity of motion is the measure of the same, arising from the velocity and quantitiy of matter conjunctly.

The motion of the whole is the sum of the motion of the sum of the parts; and therefore in a body double in quantity, with equal velocity, the motion is double; with twice the velocity, it is quadruple.

So this is where he defines momentum as being proportional to the produce of mass and velocity, but doesn't give his second law in the form of a similar definition. The Stanford Encylopedia of Philosophy states

The modern F=ma form of Newton's second law nowhere occurs in any edition of the Principia even though he had seen his second law formulated in this way in print during the interval between the second and third editions in Jacob Hermann's Phoronomia of 1716. Instead, it has the following formulation in all three editions: A change in motion is proportional to the motive force impressed and takes place along the straight line in which that force is impressed. In the body of the Principia this law is applied both to discrete cases, in which an instantaneous impulse such as from impact is effecting the change in motion, and to continuously acting cases, such as the change in motion in the continuous deceleration of a body moving in a resisting medium. Newton thus appears to have intended his second law to be neutral between discrete forces (that is, what we now call impulses) and continuous forces. (His stating the law in terms of proportions rather than equality bypasses what seems to us an inconsistency of units in treating the law as neutral between these two.)

The important point is that however Newton expresses his second law, he doesn't use units but rather proportions and this has been interpreted as the impressed force being proportional to the rate of change of momentum, rather than being equal to it. Nowadays we define force in terms of units so that one unit of force equals one unit of rate of change of momentum which means the constant of proportionality equals one.

Therefore we could argue that the examiner is correct for emphasising that Newton didn't give his his second law as an equation with units, but neither did he explicitly state what it was in the form of a definition.

-
The point is that the constant of proportionality, if you choose your units stupidly, is independent of anything. In these circumstances, you should drop the constant. –  Ron Maimon Sep 29 '11 at 17:19
Thank you for your answer John. I understand that we could say F ~ dp but I do not understand why we should. The clear example is Hooke's law with its F ~ x. Obviously it does require a constant of proportionality because otherwise it would fail dimensional analysis. In the Newton's second law the constant of proportionality is dimensionless number one. Saying "force is proportional to the rate of change of momentum" sounds like sophistry to me. Am I missing something really important here? –  The Gruffalo Sep 29 '11 at 17:36
Ok, even though I can't find it now, I am pretty sure Newton says: the impressed force is the change in the quantity of motion (momentum). –  Ron Maimon Sep 29 '11 at 17:38
@ron yes you're right, although he doesn't give his second law in words in Principia Mathematica. –  John McVirgo Sep 29 '11 at 21:32
@gruffalo, the examiner is just emphasising that Newton didn't give an equation with units. –  John McVirgo Sep 29 '11 at 21:33

If you put $F=\frac{dp}{dt}$ your a saying th same as the candidate.

$F=K\frac{dp}{dt}$

Actually there should be a constant of proportionality. It is like in the case of the electrostatic force. $F=\frac{K q_1 q_2}{r^2}$. $K$ is the constant. But depending of the units you choose, $K$ could be $1$ (gaussian system) or $9 \times 10^9$(international system).

-
I am saying the same as the candidate, indeed. My question is why the examiner is not happy with this. You said there should be a constant of proportionality but you did not explain why. I do not think there should be one and this is why I posted my question. Although I am not familiar with the electrostatic force yet, I am not sure how reasoning by analogy would help me here. Nevertheless, thank you for trying to help. –  The Gruffalo Sep 29 '11 at 17:20

There should be a constant of proportionality. Because the very word "proportional" means the left side quantity increases or decreases strictly inline with the right side quantity. The constant is the number by which the ratio of increment or decrement is decided.

For example: If you eat 1 KG butter, you will gain weight for sure. So a food technician might put it in a way "Human weight gain is directly proportional to the weight of the butter consumed". But it is not guaranteed that 1 kg butter will lead to 1kg weight gain. So he can do further research on the human body & digestion capabilities & may find out that it could cause 0.3 kg weight gain.

Now he can put it in an equation as (Human Weight gain by consuming butter = 0.3* 1 KG Butter). Assuming that all humans have the same digestive capabilities, 0.3 becomes a constant.

So for a statement with "Proportional" there should be a constant if you're trying to write an equation. In cases where the constant is one, then it doesn't matter.

In case of F~ma, I am not sure why no constant is mentioned anywhere. But I guess, it is written as f=ma, probably by experiments they would have found out that F is equals the m*a, making the constant value as 1. Also another expression W (weight)=mg (mass*acceleration due to gravity) also does not show any constant, probably making the constant value as 1.

--- apologies for any grammatical mistakes, as I am not a native English speaker.

-

There are two things in this context :

1. First taking k =1 is OK as long as you use F = ma to define unit of F as 1 Newton. (All proportionality constants such as the one in Coulomb law are not taken to be unity.)

2. When done as above, k can be taken as unit-less But if you opt to use another force equation such as Hookes law (F = C.x to define unit of F, then k in F = k ma will have units.)

-

## protected by Qmechanic♦Apr 21 '13 at 10:49

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?