Is there any derivation of Newton's second law?

[dubey suraj]

is there any derivation of Newton second law?

 

[Orodruin]

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.

 

[HallsofIvy]

Who decided which law was "first" and which "second"?

 

[H Smith 94]

I'm assuming you mean Newton's second law in the form
$$F = ma.$$ 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 $p$:
$$\boxed{\mathbf{F} \propto \frac{d\mathbf{p}}{dt}},$$ (call this equation 1), where $\mathbf{p} = m\,\frac{d\mathbf{r}}{dt}$ for constant mass. So
$$\mathbf{F} \propto m\, \frac{d^2\mathbf{r}}{dt^2},$$ where $\frac{d^2\mathbf{r}}{dt^2} = \mathbf{a}$ which means that
$$\mathbf{F} \propto m \mathbf{a}.$$ Now, the equation for this proportionality must have a constant $k$, which means
$$\mathbf{F} = k\,m\mathbf{a},$$ where Newton defined $k \equiv 1$, conveniently. Therefore
$$\mathbf{F} = m\mathbf{a}.$$ 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
$$\mathbf{p}(t) = m(t)\,\dot{\mathbf{r}}(t).$$ Note that the dot represents the rate of change of the quantity. This makes the rate of change of momentum
$$\frac{d}{dt}\mathbf{p}(t) = \frac{d}{dt}\left(m(t)\,\dot{\mathbf{r}}(t)\right).$$ 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,
$$\frac{d}{dt}\mathbf{p}(t) = \dot{\mathbf{r}}(t)\,\frac{d}{dt}m(t) + m(t)\frac{d}{dt}\dot{\mathbf{r}}(t).$$ Now, since $\frac{d}{dt}\dot{\mathbf{r}}(t) = \mathbf{a}(t)$ and $\dot{\mathbf{r}}(t) = \mathbf{v}(t)$, this becomes
$$\frac{d}{dt}\mathbf{p}(t) = \mathbf{v}(t)\,\frac{d}{dt}m(t) + m(t)\mathbf{a}(t).$$ We know $\mathbf{F}=k\frac{d}{dt}\mathbf{p}(t) = \frac{d}{dt}\mathbf{p}(t)$ (from above), so
$$\mathbf{F}(t) = \mathbf{v}(t)\,\frac{dm(t)}{dt} + m(t)\mathbf{a}(t),$$
$$\mathbf{F} = \mathbf{v}\,\frac{dm}{dt} + m\mathbf{a}.$$ This is Newton's second law in its most general and explicit form.

 

[Andy Resnick]

is there any derivation of Newton second law?

No- Newton's second law is a definition.

 

[dx]

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

$$E = P^2/2m + V(X)$$

So dE/dt = 0 implies

$$\frac{P}{m} \frac{dP}{dt} + \frac{dX}{dt} \frac{dV}{dX} = 0$$

Or

$$\frac{dP}{dt} = F$$

More generally,

$$\frac{d}{dt} E(X, P) = \frac{\partial E}{\partial X} \frac{dX}{dt} + \frac{\partial E}{\partial P} \frac{dP}{dt} = 0$$

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

$$\frac{dP}{dt} = - \frac{\partial E}{\partial X}$$

$$\frac{dX}{dt} = \frac{\partial E}{\partial P}$$

The relativistic generalization of dE/dt = 0 is

$$D_m T^{m n} = 0$$

which are the equations of mechanics in their most complete form, much more comprehensive than F = ma.

 

[Chestermiller]

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.

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

 

[Orodruin]

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

$$E = P^2/2m + V(X)$$

So dE/dt = 0 implies

$$\frac{P}{m} \frac{dP}{dt} + \frac{dX}{dt} \frac{dV}{dX} = 0$$

Or

$$\frac{dP}{dt} = F$$

More generally,

$$\frac{d}{dt} E(X, P) = \frac{\partial E}{\partial X} \frac{dX}{dt} + \frac{\partial E}{\partial P} \frac{dP}{dt} = 0$$

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

$$\frac{dP}{dt} = - \frac{\partial E}{\partial X}$$

$$\frac{dX}{dt} = \frac{\partial E}{\partial P}$$

The relativistic generalization of dE/dt = 0 is

$$\partial_\mu T^{\mu \nu} = 0$$

which are the equations of mechanics in their most complete form, much more comprehensive than F = ma.

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.

 

[dx]

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.

 

[DrStupid]

is there any derivation of Newton second law?

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.

 

[Orodruin]

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.

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.

 

[Philip Wood]

(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.

 

[Cruz Martinez]

Isn't the second law F = d(mv) just a very natural definition based on the law of inertia?

 

[Andy Resnick]

(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.<snip>.

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)

 

[Philip Wood]

F(t)=v(t)dm(t)dt+m(t)a(t),​

F=vdm/dt+ma.
This is Newton's second law in its most general and explicit form.

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.

 

[bhobba]

No- Newton's second law is a definition.

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

 

[bhobba]

Today, we take conservation laws as fundamental,

I thought Noethers Theorem was the explanation for conservation laws - in fact momentum and energy are in modern treatments defined using it eg momentum 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

 

[bhobba]

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.

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

Thanks
Bill

 

[dx]

I thought Noethers Theorem was the explanation for conservation laws - in fact momentum and energy are in modern treatments defined using it eg momentum 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

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

∂_μT^μν = 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)

 

[DrStupid]

But if the mass accreted is moving with velocity w before accretion, then the force equation becomes…
F= (v – w) dm/dt + ma.

Could you please provide a derivation for this equation?

 

[Philip Wood]

Here'e a simple derivation...

Let accreting body have velocity v and mass m at time t. Let force F from some external agent be acting on the system.
Let mass $\Delta m$ stick to the accreting body between time t and time [itex]t + \Delta t [/itex]. Let $\Delta m$ have velocity w at time t .

Then momentum of system at time [itex]t + \Delta t [/itex] = momentum of system at time t + F[itex]\Delta t [/itex]
So
[tex](m + \Delta m) (\mathbf{v} + \Delta \mathbf{v}) \ = m \mathbf{v} + \Delta m \mathbf{w} + \mathbf{F} \Delta t[/tex]

Multiplying out the brackets on the left, cancelling mv from both sides, and dividing through by [itex]\Delta t[/itex]...
[tex](m \frac{\Delta \mathbf{v}}{\Delta t} \+ \Delta m \frac {\mathbf{v}}{\Delta t} +\frac{(\Delta m)(\Delta \mathbf{v}}{ \Delta t} = mathbf{w} \frac{\Delta m}{\Delta t} + \mathbf{F} \Delta t[/tex]

Edit: this is unfinished. The Latex won't convert. Not even the simplest bits. Please ignore post.

 

[DrStupid]

Then momentum of system at time [itex]t + \Delta t[/itex] = momentum of system at time t + F[itex]\Delta t[/itex]

That means

$\left( {m + \Delta m} \right) \cdot \left( {v + \Delta v} \right) = m \cdot v + F \cdot \Delta t$

So
[tex](m + \Delta m) (\mathbf{v} + \Delta \mathbf{v}) \ = m \mathbf{v} + \Delta m \mathbf{w} + \mathbf{F} \Delta t[/tex]

Where does $\Delta m \cdot w$ comes from?

 

[Philip Wood]

Hello DrS
I've used Latex many times in PF, but tonight none of my equations would convert. Very frustrating. So here's my simple-minded derivation handwritten.

 

[DrStupid]

I still can't see, where $\Delta m \cdot w$ comes from.

To my understanding the situation is as follows:
You have two open systems A and B (the body and the external agent), interacting which each other (including mass transfer). According to the second law the forces between the systems are

$F_A = \dot p_A = m_A \cdot \dot v_A + v_A \cdot \dot m_A$
$F_B = \dot p_B = m_B \cdot \dot v_B + v_B \cdot \dot m_B$

With the third law

$F_A + F_B = 0$

and conservation of mass

$\dot m_A + \dot m_B = 0$

this results in

$m_A \cdot \dot v_A = \left( {v_B - v_A } \right) \cdot \dot m_A - m_B \cdot \dot v_B$

With Euler's F=m·a this would turn into your result, but this is not allowed for open systems, because it violates the second and the third law. If you read Newton's term "body" as "closed system" the third law would not be violated but it would still not apply. That would make forces almost useless. Therefore I can't see a justification for your equation.

 

[Philip Wood]

Leaving out, for the time being, the F delta t term in my handwritten derivation, I'm simply applying the principle of conservation of momentum to the system consisting of delta m and m. The initial momentum of delta m is (delta m) w.

Could you please tell me if you're happy with that? Sorry to want to take things a step at a time.

 

[DrStupid]

Leaving out, for the time being, the F delta t term in my handwritten derivation, I'm simply applying the principle of conservation of momentum to the system consisting of delta m and m.

Could you please tell me if you're happy with that?

I am not happy with the derivation because
- Δv is not the change of velocity of m, but the difference between the the initial velocity of m and the final velocity of the total system consisting of m and Δm and
- a and dm/dt are not the acceleration and the change of mass of the total system, but of the body only.
But if Δm goes to zero and a and dm/dt are assigned correctly, the result would be acceptable.

 

[Philip Wood]

- Δv is not the change of velocity of m, but the difference between the the initial velocity of m and the final velocity of the total system consisting of m and Δm

But Delta m sticks to m, so both m and (m + Delta m) have the same final velocity, namely (v + Delta v). So Delta v is the change in velocity of m.

a and dm/dt are not the acceleration and the change of mass of the total system, but of the body only

What I'm taking as the total system is the accreting body and the stuff it accretes in time Delta t. So the mass of the total system is constant. dm/dt is indeed the rate of change of mass of the accreting body. That's what I intend it to be. Likewise, I'm intending a to be the acceleration of just the accreting body.

 

[DrStupid]

But Delta m sticks to m, so both m and (m + Delta m) have the same final velocity, namely (v + Delta v). So Delta v is the change in velocity of m.

OK, yes that's correct.

What I'm taking as the total system is the accreting body and the stuff it accretes in time Delta t. So the mass of the total system is constant. dm/dt is indeed the rate of change of mass of the accreting body. That's what I intend it to be. Likewise, I'm intending a to be the acceleration of just the accreting body.

Yes, that's what I mean above. The equations apply to the conservation of momentum of the total system but a and dm/dt do not refer to the total system but just to a part of it.

Thus I agree with the derivation for F=0 and Δm→0.

 

[Philip Wood]

Good. Hope that my treatment of a non-zero F (from outside the system) acting on the accreting body is also acceptable. I argue that in time delta t it delivers momentum F delta t.

 

[DrStupid]

Good. Hope that my treatment of a non-zero F (from outside the system) acting on the accreting body is also acceptable. I argue that in time delta t it delivers momentum F delta t.

Currently you just described the conservation of momentum for the inelastic collision of m and Δm within the isolated system. Including an external force and considering the transfer of Δm into the system the situation changes completely. In this case the momentum of the system is no longer conserved. It changes according to

$\Delta p = p\left( {t + \Delta t} \right) - p\left( t \right) = \left( {m + \Delta m} \right) \cdot \left( {v + \Delta v} \right) - m \cdot v = F \cdot \Delta t$

I still do not see how you get Δm·w into your equation.

 

[Philip Wood]

considering the transfer of Δm into the system

As I said above, my system is the accreting body and the stuff it accretes in time delta t. So mass delta m isn't transferred into the system, but is transferred within the system. The overall mass of the system remains constant at (m + delta m).

Δp=p(t+Δt)−p(t)=(m+Δm)⋅(v+Δv)−m⋅v=F⋅Δt

In the last but one 'side' of your equation, it's not just mv that you should be subtracting but also the initial momentum of Delta m, which (see post 15) is (Delta m)w. That way we get the change in momentum of the system.

If we put F = 0 into your equation as quoted then we find that delta v is independent of w. This can't be right, as, with F = 0, we are dealing with a simple case of a two body collision. Note, too, that in terms of orders of magnitude (delta m) w is no more negligible than (delta m)v.

 

[DrStupid]

As I said above, my system is the accreting body and the stuff it accretes in time delta t. So mass delta m isn't transferred into the system, but is transferred within the system. The overall mass of the system remains constant at (m + delta m).

Than we do not talk about the force acting on a system with variable mass.

If we put F = 0 into your equation as quoted then we find that delta v is independent of w.

That's correct.

$F_A = m_A \cdot \dot v_A + v_A \cdot \dot m_A = 0$

results in

$\dot v_A = - \frac{{v_A \cdot \dot m_A }}{{m_A }}$

This can't be right, as, with F = 0, we are dealing with a simple case of a two body collision.

It's a special case of a two body collision.

 

[Philip Wood]

Than we do not talk about the force acting on a system with variable mass.

No, but I do talk about the force on a BODY of varying mass. My point is that in Galilean/Newtonian Physics a body can't acquire mass without the mass coming from somewhere outside the original body. And as the body acquires mass it also acquires momentum, dependent on the velocity of the acquired mass before it was acquired. You haven't (yet) convinced me that the equation I gave originally is wrong.

 

[DrStupid]

[

No, but I do talk about the force on a BODY of varying mass.

I still don't get it. What systems and what forced/interactions between which of the systems are you talking about exactly? I counted four systems

A: the body of initial mass m
B: the additional mass Δm
C: the common system of the body and the additional mass
D: an external system exerting a force the body

and two interactions

a: the collision of A and B
b: the force F between A and D

Is this correct?

If yes, than force F is not related to any kind of mass transfer. In that case your scenario is not suitable for the discussion of forces between open systems.

If not, please give a more detailed explanation of your scenario.

My point is that in Galilean/Newtonian Physics a body can't acquire mass without the mass coming from somewhere outside the original body. And as the body acquires mass it also acquires momentum, dependent on the velocity of the acquired mass before it was acquired.

That doesn't justify a modification of the second law.

You haven't (yet) convinced me that the equation I gave originally is wrong.

As it is your equation, you have to convince me that it is correct.

However, I can show you, that your force violates the second and the third law. The violation of the second law is obvious. Your force differs by the additional term $-w \cdot \dot m$ from the change of momentum. In order to show the violation of the second law, I repeat my calculation from #24 with your force. The forces between the interacting open systems are

$F_A = m_A \cdot \dot v_A + \left( {v_A - v_B } \right) \cdot \dot m_A$
$F_B = m_B \cdot \dot v_B + \left( {v_B - v_A } \right) \cdot \dot m_B$

with the third law and conservation of mass (you already agreed that the mass can change by mass transfer between the systems only) this results in

$m_A \cdot \dot v_A = 2 \cdot \left( {v_B - v_A } \right) \cdot \dot m_A - m_B \cdot \dot v_B$

This is obviously different from my result in #24. Thus at least one concept of force must be wrong. In order to find the correct result independent from the definition of force, I again repeat the calculation with momentum only. The changes of momentum are

$\dot p_A = m_A \cdot \dot v_A + v_A \cdot \dot m_A$
$\dot p_B = m_B \cdot \dot v_B + v_B \cdot \dot m_B$

with conservation of momentum

$\dot p_A + \dot p_B = 0$

and conservation of mass this results in.

$m_A \cdot \dot v_A = \left( {v_B - v_A } \right) \cdot \dot m_A - m_B \cdot \dot v_B$

That mean there is something wrong with your force.

 

[Philip Wood]

I counted four systems

A: the body of initial mass m
B: the additional mass Δm
C: the common system of the body and the additional mass
D: an external system exerting a force the body

and two interactions

a: the collision of A and B
b: the force F between A and D

Is this correct?

If yes, than force F is not related to any kind of mass transfer. In that case your scenario is not suitable for the discussion of forces between open systems.

I agree with your summary (A, B, C, D, a, b). I agree, too, that the force F_AD is not related to any kind of mass transfer, in that I'm trying to deal with cases where the force and the mass-accretion are independent, in the sense that we could conceive of either being turned off independently of the other e.g mass-accreting body experiencing a force (say from an electric field) unrelated to the mass accretion.

I repeat my calculation from #24 with your force. The forces between the interacting open systems are

FA=mA⋅v˙A+(vA−vB)⋅m˙A

But my force wasn't the force of B on A, but was a force from outside (D in your notation).

At this point I have to leave this discussion for a few hours. I have to say that I am finding it very interesting indeed, and I must thank you for engaging in it. The difference of opinion clearly centres on the treatment of open systems.

May I please check that we're in agreement on the following two things…

(1) For a body of mass m moving at velocity v accreting mass at a rate dm/dt from a 'mist' of particles with (pre-capture) velocity w, then if no external force (FAD) acts, from momentum conservation we have

(m dv/dt + v dm/dt) = w dm/dt

(2) the force on an open system is equal to the limit as delta t approaches zero of (final p – initial p)/delta t.