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?
Answer
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: $$F^{(N)} = K_F^{(N)}\frac{\Delta p^{(N)}}{\Delta t^{(N)}}\tag{1}$$ 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. For example, suppose your equipment is calibrated so that:
- Your force meter measures in what we now call pounds. In other words, a reading of $3$ on the force meter corresponds to the force we would now call $3\ \mathrm{lb_F}$, and similarly for other numbers - but of course, when you take that reading, you just think of it as "$3$".
- Your momentum meter measures in what we would currently call pound-feet per second, $\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}$.
- Your time meter measures in units equivalent to modern seconds.
Then 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. The units were first thought of as reference values, and your own measurements would represent how many times the reference value you were measuring. If you were lucky enough to have equipment calibrated to match the reference value, you could just read the number right off it, but otherwise you would have to convert your measurements using an equation like $$F^{(\text{standard})} = \biggl(\frac{\text{your unit}}{\text{standard reference value}}\biggr)F^{(N)}$$ and so on for other quantities.You would find some way to determine the conversion factor $\frac{\text{your unit}}{\text{standard reference value}}$ for each of your measurement devices, and then you'd have to do this conversion each time you made a measurement.
Or, you could be smart, and put the reference values right into the equation. This is the big advance in unit systems, because your measurements are not just numbers anymore. Whereas before you would work with, say, $F^{(N)}$ or $F^{(\text{standard})}$, now you can work with the quantity $F$, where $$F = F^{(\text{standard})}(\text{standard reference value}) = F^{(N)}(\text{your unit})$$ So, again assuming that your units for force, momentum, and time correspond to the modern $\mathrm{lb_F}$, $\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}$, and $s$, respectively, you can write the above equation (1) as $$\frac{F}{\mathrm{lb_F}} = \frac{K_F}{\mathrm{lb_F}/(\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2})}\frac{\bigl(\frac{\Delta p}{\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-1}}\bigr)}{\bigl(\frac{\Delta t}{\mathrm{s}}\bigr)}$$ Then you can algebraically rearrange this to $$F = K_F\frac{\Delta p}{\Delta t} \quad\text{where}\quad K_F = K_F^{(N)}\frac{\mathrm{lb_F}}{\mathrm{lb_M}\;\mathrm{ft}\;\mathrm{s}^{-2}}$$
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.
No comments:
Post a Comment