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Is there any derivation of Newton's second law?

  1. Apr 28, 2015 #1
    is there any derivation of newton second law?
     
    Last edited by a moderator: Apr 28, 2015
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  3. Apr 28, 2015 #2

    Orodruin

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    No. Physics is an experimental science and as with all physical laws, it is postulated and then evaluated based on how well it predicts what we can observe. Newton's second law is very (very) good at predicting observations.
     
  4. Apr 28, 2015 #3

    HallsofIvy

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    Who decided which law was "first" and which "second"?
     
  5. Apr 28, 2015 #4

    H Smith 94

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    I'm assuming you mean Newton's second law in the form
    [tex] F = ma. [/tex] This is not strictly Newton's second law.

    Newton's second law was originally set up to be force ##F## is proportional to the rate of change of momentum ##p##:
    [tex] \boxed{\mathbf{F} \propto \frac{d\mathbf{p}}{dt}}, [/tex] (call this equation 1), where ##\mathbf{p} = m\,\frac{d\mathbf{r}}{dt}## for constant mass. So
    [tex] \mathbf{F} \propto m\, \frac{d^2\mathbf{r}}{dt^2}, [/tex] where ##\frac{d^2\mathbf{r}}{dt^2} = \mathbf{a}## which means that
    [tex] \mathbf{F} \propto m \mathbf{a}. [/tex] Now, the equation for this proportionality must have a constant ##k##, which means
    [tex] \mathbf{F} = k\,m\mathbf{a}, [/tex] where Newton defined ##k \equiv 1##, conveniently. Therefore
    [tex] \mathbf{F} = m\mathbf{a}. [/tex] However, this only works for invariant mass.

    Time-variant mass

    Let's consider where mass depends on time (for, say, a rocket which loses mass as it burns fuel), ##m(t)##, as well as position, ##\mathbf{r}(t)##. Take Newton's second law in its simplest form (equation 1, above) and consider the momentum, which is now
    [tex] \mathbf{p}(t) = m(t)\,\dot{\mathbf{r}}(t). [/tex] Note that the dot represents the rate of change of the quantity. This makes the rate of change of momentum
    [tex] \frac{d}{dt}\mathbf{p}(t) = \frac{d}{dt}\left(m(t)\,\dot{\mathbf{r}}(t)\right). [/tex] Since both these quantities are functions of time, we must use the product rule ##\frac{d}{dx}\left(f(x)\cdot g(x)\right) = g(x)\frac{d}{dx}f(x) + f(x)\frac{d}{dx}g(x)##. So,
    [tex] \frac{d}{dt}\mathbf{p}(t) = \dot{\mathbf{r}}(t)\,\frac{d}{dt}m(t) + m(t)\frac{d}{dt}\dot{\mathbf{r}}(t). [/tex] Now, since ##\frac{d}{dt}\dot{\mathbf{r}}(t) = \mathbf{a}(t)## and ##\dot{\mathbf{r}}(t) = \mathbf{v}(t)##, this becomes
    [tex] \frac{d}{dt}\mathbf{p}(t) = \mathbf{v}(t)\,\frac{d}{dt}m(t) + m(t)\mathbf{a}(t). [/tex] We know ##\mathbf{F}=k\frac{d}{dt}\mathbf{p}(t) = \frac{d}{dt}\mathbf{p}(t)## (from above), so
    [tex] \mathbf{F}(t) = \mathbf{v}(t)\,\frac{dm(t)}{dt} + m(t)\mathbf{a}(t), [/tex]
    [tex] \mathbf{F} = \mathbf{v}\,\frac{dm}{dt} + m\mathbf{a}. [/tex] This is Newton's second law in its most general and explicit form.
     
  6. Apr 28, 2015 #5

    Andy Resnick

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    No- Newton's second law is a definition.
     
  7. Apr 28, 2015 #6

    dx

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    Today, we take conservation laws as fundamental, so Newton's second law is a consequence of the conservation laws. For a particle,

    [tex] E = P^2/2m + V(X) [/tex]

    So dE/dt = 0 implies

    [tex] \frac{P}{m} \frac{dP}{dt} + \frac{dX}{dt} \frac{dV}{dX} = 0 [/tex]

    Or

    [tex] \frac{dP}{dt} = F [/tex]

    More generally,

    [tex] \frac{d}{dt} E(X, P) = \frac{\partial E}{\partial X} \frac{dX}{dt} + \frac{\partial E}{\partial P} \frac{dP}{dt} = 0[/tex]

    With some more thought which I won't go into, you can get Hamilton's equations:

    [tex] \frac{dP}{dt} = - \frac{\partial E}{\partial X} [/tex]

    [tex] \frac{dX}{dt} = \frac{\partial E}{\partial P} [/tex]

    The relativistic generalization of dE/dt = 0 is

    [tex] D_m T^{m n} = 0 [/tex]

    which are the equations of mechanics in their most complete form, much more comprehensive than F = ma.
     
    Last edited: Apr 28, 2015
  8. Apr 28, 2015 #7
    I totally agree. Newton's 2nd law is an empirical relationship that Newton formulated based on his laboratory experiments. The mass m was the proportionality constant between force and acceleration.

    Chet
     
  9. Apr 28, 2015 #8

    Orodruin

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    It is no surprise that you can get Newtonian mechanics out of Hamiltonian or Lagrangian mechanics or vice versa. It all boils down to the fact that you have to postulate something and test it.

    I also think that this is far beyond the level of the OP and thus will not help the OP's undrstanding.
     
  10. Apr 28, 2015 #9

    dx

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    Actually, my post showed that you can get both Hamiltonian and Newtonian mechanics from conservation laws, not Newtonian mechanics from Hamiltonian mechanics.

    The OP can ignore the relativity and Hamiltonian part, but I think it is helpful for students to appreciate the logical structure of mechanics that the conservation laws (or equivalently, the symmetries) are the fundamental things. There is nothing more to mechanics than conservation laws.
     
  11. Apr 28, 2015 #10
    As we can't ask Newton anymore, we will never know. But the original wording suggests, that he started from conservation of momentum.

    If he started from an empirical based relationship between force, mass and acceleration, it would have been obvious to write, that force is proportional to the product of mass and acceleration. As mass and acceleration were familiar concepts, everybody would have understand it. Why should he turn it with an additional step into a proportionality to the change of momentum if nobody except himself ever heard about differential calculus?

    Of course the next question is: Ho did he derived conservation of momentum? And again the answer is: We will never know for certain.
     
  12. Apr 28, 2015 #11

    Orodruin

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    I am not going to argue against this, my point is that from what I read out of the OP, I do not think that he/she is at the level where such an appreciation can be reached. The next question is going to be "can you derive the conservation of the quantity X?" I think this needs to be done on a much more fundamental level regarding how experimental sciences are done.
     
  13. Apr 28, 2015 #12

    Philip Wood

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    (1) Re posts 5 and 7, the difficulty of regarding N2L as empirical is how you measure force, though I have to admit that the high school approach of accelerating a trolley using identically extended identical springs in parallel (one by itself, them 2 in parallel, then 3 in parallel) would satisfy me as the basis of showing that rate of change of momentum is proportional to force. This relies on two 'common notions' about force: that equally stretched springs exert equal forces, and that forces pulling in the same direction add together like scalars. No doubt a purist wouldn't grant these, and would have to regard N2L as a definition of force in terms of acceleration or in terms of rate of change of momentum.

    (2) Re post 4 I've never been convinced that in Galilean/Newtonian physics there is any distinction between F = ma and F = d(mv)/dt. This is because, in Newtonian physics mass is is conserved in any interaction, and the only way a body can lose or gain mass is by gaining it from, or losing it to, another body or bodies. It's my contention that we should keep track of a specific part of a body with a fixed mass (or of specific parts with fixed masses) and apply N2L to it or them, even if it or they leave the original body. Rocket problems, for example, can easily be solved this way, without ever applying N2L to a body of varying mass or differentiating mv as a product.
     
  14. Apr 28, 2015 #13
    Isn't the second law F = d(mv) just a very natural definition based on the law of inertia?
     
  15. Apr 29, 2015 #14

    Andy Resnick

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    And that is a very nice explanation of why N2L is so difficult for many students: the law itself (F = dp/dt) is nothing more and nothing less than a precise definition of 'Force'. The complication is that N2L is also a statement connecting kinematics (measurable properties such as mass, velocity, acceleration) with dynamics (unmeasurable things: 'force laws' are very much empirical). Unfortunately, this subtle distinction is often glossed over in introductory courses, leading to conceptual problems such as:

    1) objects can 'possess' force, and force is a quantity of motion that can be transferred from one object to another.
    2) if a = 0, then F = 0.
    3) free fall and projectile motion are somehow different (because gravity is apparently present in one but not the other)
     
  16. May 1, 2015 #15

    Philip Wood

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    This is the end of post 4 with the latex stripped out, as it wouldn't perform.

    I'd contend that it's not the most general form, because it's true in Galilean/Newtonian Physics for a body accreting mass only if the extra mass is stationary (in the frame of reference in which the equation is written) before it is accreted on to the accreting body. An example would be a hailstone accreting moisture from mist through which it is moving. The equation holds for this case. But if the mass accreted is moving with velocity w before accretion, then the force equation becomes…
    F= (v – w) dm/dt + ma.
    This brings out clearly that the term in dm/dt is not due to the mass of the accreting body changing per se, but represents the rate of change of momentum of the accreted mass due to its change of velocity on accretion.

    Putting it another way, we can't write a force equation for a body of changing mass without taking account of how the extra mass was moving before it became part of the ‘body of changing mass’. In Galiean/Newtonian Physics it's very easy to show that a system's momentum is conserved only if its total mass is conserved. A body can't gain mass without gaining it from somewhere.
     
    Last edited: May 1, 2015
  17. May 1, 2015 #16

    bhobba

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    Exactly. And the first follows from the second. The real content is in the third law. So what's the physics - surely it can't be definitions. Really Newtons laws are a prescription for solving problems that says - get thee to the forces. It also contains an unstated assumption inertial frames exist where free particles move at constant velocity. What's a free particle - one not acted on by forces. So the circularity continues.

    If you want to see a presentation of Classical Mechanics without definitions masquerading as laws, exactly what an inertial frame is, what a free particle is, and why they move at constant velocity, take a look at Landau - Mechanics. It's real basis is the Principle Of Least Action (PLA) and symmetry.

    What's the basis of the PLA - its Quantum Mechanics. At rock bottom classical mechanics is really QM.

    Thanks
    Bill
     
  18. May 1, 2015 #17

    bhobba

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    I thought Noethers Theorem was the explanation for conservation laws - in fact momentum and energy are in modern treatments defined using it eg momemtum is the conserved Noether charge related to spatial translational symmetry. The reason Noether can be applied is the Principle of Least Action is true. That would seem the fundamental principle.

    Thanks
    Bill
     
  19. May 1, 2015 #18

    bhobba

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    I tend to view the symmetries as fundamental and the conservation laws following from Noether.

    Thanks
    Bill
     
  20. May 1, 2015 #19

    dx

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    Yes, the action is ultimately the most basic quantity, because the definition of energy and momentum is tied to the action. One you have the action, the symmetries and conservation laws are equivalent. At the level of formulating the dynamical equations though, I would say that the conservation law

    μTμν = 0

    is the ultimate equation of mechanics. You might find the following paper by Jacob Bekenstein interesting:

    http://arxiv.org/abs/1411.2424

    (I haven't read it fully, but its possibly interesting in the context of this topic)
     
  21. May 1, 2015 #20
    Could you please provide a derivation for this equation?
     
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