Parlyne said:
Newton's second law does not say F = \frac{dp}{dt}. What it says is \sum \vec{F} = \frac{d\vec{p}}{dt}. We can't just replace "F" in all the equations defining forces with \frac{dp}{dt} because that would only be true in the case that the force under consideration is the only force acting. If for example, electromagnetic and gravitational forces were acting, Newton's second law comes down to:
\frac{d\vec{p}}{dt} = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})
Here, the equations for each force separately are true when written with F's, but not when written with \frac{dp}{dt}.
Yes, but you could as well say that in a situation that you describe (namely where there are E and B fields as well as masses around), that:
F = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})
simply because that's what seems to be equal to \frac{dp}{dt}
You are in fact implicitly using the superposition principle for forces when "several interactions" take place. But that was exactly the usefulness of introducing the concept of force: for a given situation, we can
mentally split up the entire interaction into different 1-1 interactions (between "masses", and between "charges and field" etc...), pretend first that they act only alone (ie, considering the interaction in simplified situations, for instance, when only masses are present, and no charges, etc...), and then add them all together to find the genuine interaction.
But it is "in our minds" that this decomposition occurs.
You can just as well say that for a given setup, well, the force on a particle is given by F = \frac{dp}{dt}, and that, if you analyse the situation, that it turns out that this equals also \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})
It is just an amazing property of nature, that the different terms in there ALSO correspond to momentum changes in OTHER situations. The \frac{GmM}{r^2} term for instance, occurs in another situation, as the \frac{dp}{dt} which looks a bit like the one we're studying, but without charge.
The term q (\vec{E} + \vec{v} \times \vec{B}) occurs in still another situation as the \frac{dp}{dt}, when there are charges, but no masses present.
So in fact, what we discover here, is a specific property of nature, which says that the quantity \frac{dp}{dt} has a remarkable property in certain circumstances: for a given situation, it equals the vector sum of the \frac{dp}{dt} that occur in OTHER situations, which each seem to correspond to a certain aspect of the initial situation.
So we mentally say that in the initial situation, TWO forces work on the particle, but that's only a mental picture. Only one force works on it, which is the entire expression: F = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B}). We mentally have split it up into different terms, which correspond to the changes of momenta that would have occurred in different, simplifying situations, using this remarkable property of nature.
But there is no way to distinguish "a particle on which two equal and opposite forces act" from a "particle on which no force acts".
The two "equal and opposite" forces are just mental constructions because we think of "simplifying situations" (for instance, with only a mass left to it, and then only with a mass right to it). In each of the cases, the total force corresponds to \frac{dp}{dt}