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Still don't understand why we need F=dp/dt

  1. Nov 23, 2006 #1
    I've tried reading the previous threads about Newtons second law and I've must say that I still don't understand why it's not just a definition. That is, if we want to it would be possible to just replace F with dp/dt everywhere in physics and make F=dp/dt unnecessary. For example we would instead have

    [tex] \frac{dp}{dt} = G \frac{m_1 m_2}{r^2} [/tex]

    and for electromagnetism

    [tex] \frac{d\textbf{p}}{dt} = q (\textbf{E} + \textbf{v} \times \textbf{B}) [/tex]

    In other words, the concept of "force" is intuitive and all, but still fundamentally redundant.
     
    Last edited: Nov 23, 2006
  2. jcsd
  3. Nov 23, 2006 #2
    I dont understand what the purpose of replacing F with dP/dt would solve.
     
  4. Nov 23, 2006 #3
    depends what you're dealing with, can be handy to express force with a dt if you wanna integrate the other side wrt time I guess. Other times its just superfluous to write F out like that.
     
  5. Nov 23, 2006 #4

    Hurkyl

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    Why use velocity, when you could use dx/dt instead? Why use momentum when you could use mdx/dt instead? Why use mass when you could use [itex]\int_R \rho \, dV[/itex] instead?

    Because it's convenient, and often conceptually useful.
     
  6. Nov 23, 2006 #5

    radou

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    Another major threat to the concept of classical physics. :tongue2:
     
  7. Nov 23, 2006 #6
    Yes I know it practical in the same way as having mass instead of an integral over the density. But F=dp/dt is usually claimed to be a fundamental law, why is that? I mean, this is never the case for [tex] m = \int \rho dV [/tex]
     
  8. Nov 23, 2006 #7

    russ_watters

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    What if dp and dt are both zero?
     
  9. Nov 23, 2006 #8

    vanesch

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    Feynman has a nice discussion about that in the beginning of his "Lectures". Essentially, F = dp/dt is an inspired definition. He compares it to G = d (mx)/dt, which he calls a "gorce" and which is a rather uninspired definition.

    By defining F = dp/dt, Newton linked the interaction-between-bodies part with the kinematics in such a way that the interaction-between-bodies part took on an easy and simple form.
     
  10. Nov 24, 2006 #9
    Newton's second law does not say [tex]F = \frac{dp}{dt}[/tex]. What it says is [tex]\sum \vec{F} = \frac{d\vec{p}}{dt}[/tex]. We can't just replace "F" in all the equations defining forces with [tex]\frac{dp}{dt}[/tex] 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:

    [tex]\frac{d\vec{p}}{dt} = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})[/tex]

    Here, the equations for each force separately are true when written with F's, but not when written with [tex]\frac{dp}{dt}[/tex].
     
  11. Nov 24, 2006 #10

    vanesch

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    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:

    [tex]F = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})[/tex]

    simply because that's what seems to be equal to [tex] \frac{dp}{dt}[/tex]

    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 [tex] F = \frac{dp}{dt}[/tex], and that, if you analyse the situation, that it turns out that this equals also [tex]\frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})[/tex]

    It is just an amazing property of nature, that the different terms in there ALSO correspond to momentum changes in OTHER situations. The [tex]\frac{GmM}{r^2}[/tex] term for instance, occurs in another situation, as the [tex]\frac{dp}{dt}[/tex] which looks a bit like the one we're studying, but without charge.
    The term [tex] q (\vec{E} + \vec{v} \times \vec{B})[/tex] occurs in still another situation as the [tex]\frac{dp}{dt}[/tex], 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 [tex]\frac{dp}{dt}[/tex] has a remarkable property in certain circumstances: for a given situation, it equals the vector sum of the [tex]\frac{dp}{dt}[/tex] 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: [tex]F = \frac{GmM}{r^2} \hat{r} + q (\vec{E} + \vec{v} \times \vec{B})[/tex]. We mentally have split it up into different terms, which correspond to the changes of momenta that would have occured 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 [tex]\frac{dp}{dt}[/tex]
     
  12. Nov 24, 2006 #11
    You have a stationary weight on a contracted spring. You substitute the force made from the spring with what?
     
  13. Nov 24, 2006 #12

    arildno

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    It is crucial to note that vanesch used the word "particle" here, i.e, that we are dealing with an object where we do not regard internal structure as important.

    If we are dealing with an "extended object" (essentially something composed of many "particles"), then we may distinguish between an object upon two equal, opposite forces acts, and one upon which no force acts.

    In the first case, the object will be in a state of stress, but in the other case, that need not be the case.
     
  14. Nov 24, 2006 #13

    Gokul43201

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    Even in light of Aharonov-Bohm?
     
  15. Nov 24, 2006 #14
    Well with dp/dt of course, why wouldn't that work?
     
  16. Nov 24, 2006 #15

    vanesch

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    This is only in the case when the points of action of the two forces are different. But if that is the case, the two forces can also be identified as the dp/dt in the two different situations where the extended body has been cut into two pieces (which is actually even the definition of the stress tensor: the change of momentum that one piece would undergo if cut away from the other part of the extended body).

    So also in that case, a force is always equal to a dp/dt in a different situation (here, the body cut into two pieces).

    The situation I was more thinking off, was:

    Take a big mass M1, at distance R1 to the left, and a big mass M2, at distance R2 to the right.
    We say that the force on a point particle, or even to a small (way smaller than R1 and R2) but extended body, equals "the gravitational force" due to M1, plus "the gravitational force" due to M2, and write:
    dp/dt = F1 + F2

    However, F1 is simply dp/dt of the same body, with only M1 present, and F2 is simply dp/dt of the same body, with only M2 present. It is a feat of nature that the third, and different, situation, with M1 and M2 present, has:

    (dp/dt)_3 = (dp/dt)_1 + (dp/dt)_2

    with 3: the situation where M1 and M2 are present
    2: the situation with only M2 present
    1: the situation with only M1 present.

    It is rather amazing that these 3 different setups have any relationship, at first sight. Given this aspect, it becomes interesting, to give a special name to (dp/dt)_1 and (dp/dt)_2.
    We call it "force".

    But note that in situation 3, individually, F1 and F2 don't have any physical meaning as such, and only their sum counts. There is no internal stress in the extended body which suffers from F1 - F2.
    (there will be a small tidal effect, but it is not F1 - F2).
     
  17. Nov 24, 2006 #16

    vanesch

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    I'm only talking in the frame of classical mechanics here :cool:

    The only point I wanted to make, is that, strictly speaking, we don't need the concept of force in classical mechanics, but that it is a damn useful definition, and that its usefulness comes about by a peculiar property of nature (in classical mechanics), which relates momentum changes between different situations, but that we could limit ourselves, if we wanted to, to talk purely about momentum changes and their properties.

    However, this would be a clumsy way of doing things, because we would then not make explicit the specific property of nature which gives us the vectorial composition of forces due to 1-1 interactions (which is nothing else but this composition property of momentum changes).

    I maintain however, that for *every* use of a force, we can find a situation where it corresponds to a dp/dt. Maybe I'm wrong, but I don't think so (yet).
     
    Last edited: Nov 24, 2006
  18. Nov 24, 2006 #17

    vanesch

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    You consider the situation where the spring is cut, and the force is the dp/dt of the weight.
     
  19. Nov 24, 2006 #18
    I fail to see how that changes anything. Fundamentally, what vanesch stated is still true. In the situation of an extended object composed of many "particles", the stress is just a consequence of the complexitivity of the constructed object. This is not in any way a fundamental principle. Take friction for example, this is not a fundamental property either, just a phenomenon that occurs for many-particle systems.
     
  20. Nov 24, 2006 #19

    This is precisely my opinion as well, and so far I've never seen a convincing argument against it.

    The main conseqence of this is that if we were to try to properly axiomatize classical physics, Newtons second law together with the concept of force would NOT be included as a _fundamental_ axiom/law, instead it would be a definition used in the formulation building on top of the _real_ axioms/laws.
     
  21. Nov 24, 2006 #20

    vanesch

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    Well, yes, some books do it that way (Alonso and Finn for instance, define force as dp/dt). That said, it is nevertheless a very inspired definition which captures something essential about (classical) nature. I find Feynman extremely illuminating in that respect, where he introduces his "gorce" (=d (mx) / dt), which leads him nowhere.

    The definition of F = dp/dt captures the essential simplicity of the composition of the different dp/dt for different "elementary" situations, which turns F into such a useful concept.

    In Landau and Lif., "force" is an entirely auxillary concept, because they start from the beginning from a variational principle as the fundamental axiom of mechanics.
     
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