Understanding the Relationship Between Force and Momentum: F=dp/dt Explained

[John_Doe]

$$F = \frac{dp}{dt}$$
If this is indeed a definition, for what reason is it defined the way it is, and not some other way?

 

[mathphys]

When you give to it a physical interpretation to it, in the Newtonian mechanics context or philosophy, what does it tells you?

F=dp/dt, you may look at it as a simple definition of force and nothing more, a notion who relates two quantities. That it holds for describing natural phenomena is an axiom of Newtonian mechanics. Its acceptance and 'validity' rest on experimental verification, which gives it the character of '(empirical) law'.

Think again in 'given two points there is only one straight line that contains them'. You are assuming a priori the existence of objects called points and straight lines, and the later ennunciate gives a property or relation between this objects as an axiom. And there is no point in asking prove that points & straight lines exist, and that the previous sentence holds, it is an axiom or postulate for eucledian geometry.

When you associate Newton's second law to physical reality, you're axiomatizing that there is a quantity called momenta which characterizes the state of motion of and object (and that may be associated with another characteristic called mass, and that there exist entities called inertial frames, and that space is associated to R^3 and t to R^{+}, etc) and that the rate of change of this quantity in time, depends on another quantity called force which acts on the object , and that may be interpreted (naively) as a 'pull' or a 'push'. \{naive mode on} The stronger the pull the stronger the change in the state of motion\{naive mode off}.


This are the 'physical assumptions' behind the definition, in this context. And the question arises again, can you give a more primitive definition of force or assumptions in the context of classical physics?

 

[John_Doe]

To add another way of looking at it is from the point of view of the action: $$S$$
The action is the generating function from taking the system from one moment in time to another or the time integral of the Lagrangian.
$$\vec{F} = \frac{d \vec{p}}{dt}$$, is the only value which allows $$\delta{S}$$ to be zero.
In other words it is the only value for which the action is stationary, which allows classical physics to be same through out time.*
*I'm only an undergraduate, so I could be incorrect in my assessment of this.

In this thread, thus far, I think that Son Goku is the most accurate in his response.

 

[mathphys]

Sorry, but that is false. The function that extremizes the action S is the one satisfying the corresponding euler-lagrange equations. You arrive to that system of equations using lagrangian mechanics, not to F=dp/dt.

What happens is that from the Euler lagrange equations you may arrive finally to expresions like

ma=-kx (a spring system)

And you may write inmediately (by definition!) that F=dp/dt=d(mv)/dt =ma (for m a constant), and that F=-kx.

 

[John_Doe]

I have a question. Is $$F = \frac{dp}{dt}$$ related to $$F = -\frac{dV}{dx}$$, $$V$$ potential energy?

 

[mathphys]

As someone has previously mentioned in F=dp/dt F represents the net force acting over a body. Let's label it Fn.


On the other hand, a conservative force (field) can be expressed as the gradient of a potential

F= - del (V) for instance.

See http://scienceworld.wolfram.com/physics/Force.html

If a force of this kind is the only one acting on a body then

Fn=F.

Nevertheless if there are more forces acting on the body, that may or may not be conservative (Electrical forces Fe, magnetic forces Fm, other forces Fx) then

Fn=F+Fe+Fm+Fx= dp/dt

 

[John_Doe]

Is $$F = \frac{dp}{dt}$$ a consequence of $$F = -\frac{dV}{dx}$$?

 

[mathphys]

Nop. F= -del (V) just tells you that this force is conservative.

 

[John_Doe]

$$F = -\frac{dV}{dx}$$
According to this relation, the direction of force is always toward decreasing potential energy. This would explain the null case of $$F = \frac{dp}{dt}$$, i.e. objects follow geodesics.

 

[mathphys]

dp/dt=0 just tells that momentum is conserved when the net force equals 0.

 

[John_Doe]

No the null case of Newton's second law of motion is Newton's first law of motion.

 

[mathphys]

Can you demostrate from the second law solely the first law? Including the geodesic path (straight lines in cartesian space, or in an arbitrary space ;) )?

 

[John_Doe]

Possibly, with Newton's first law of motion as a solution. Taking into account relevant conditions, it may work. I think calculus of variations is the best approach. This would then include the geodesic condition.

 

[mathphys]

By the way, Axiomata sives leges motus ? tells you something? Axiomata?
If not , see

"[URL

http://www.gmu.edu/departments/fld/CLASSICS/Newton.leges.html
So do you see why i refer to it as an axiom?
Or Newton liked methapora also? ;) If so, what a beautiful book of poetry he wrote.

 

[John_Doe]

Yes, but more importantly, can Hamilton's principle be applied to derive $$F = \frac{dp}{dt}$$?

I think that Lagrange's equations are a concequence of Newton's second law of motion. Lagrange's equations have been prooven mathematically. Thus, it may be possible to run the argument backwards and proove Newton's second law of motion as a concequence of Lagrange's equations.

 

[ahrkron]

$$F = \frac{dp}{dt}$$
If this is indeed a definition, for what reason is it defined the way it is, and not some other way?

Because it makes sense. When you try to change the state of motion of a large body, both its speed and its mass come into play.

To quantify the "amount of motion" you can use the product of both (the sum would not do, since the units don't match; you would need to do some trickery to have make that work), and then you can assign a number to the "effort" you need by determining how much you would need to alter that "amount of motion".

Once you have ways to measure lengths, time intervals and mass, you can get numbers for "force", and calibrate devices to produce given amounts of force (or torque, in the case of engines). You can then predict the effect that such devices will have on other objects.

For this to work, the definition of Force could have been different.

 

[ahrkron]

Thus, it may be possible to run the argument backwards and proove Newton's second law of motion as a concequence of Lagrange's equations.

Sure, but that just means that they are equivalent representations of the same behavior. On each, there is some quantity that has a nice feature (action, force), and it is only natural to use it as the starting point.

The situation is similar to the use of different bases for the description of a given set of objects in a plane.

 

[ahrkron]

So do you see why i refer to it as an axiom?
Or Newton liked methapora also? ;) If so, what a beautiful book of poetry he wrote.

The problem with the word "axiom" is that its meaning in the context of mathematical logic is much narrower than its meaning for Newton's work, so it can bring some confusion.

I don't know if, using the current definitions, the Principia could be regarded as a successful attempt at axiomatizing classical mechanics. I think not, but I'm definitely not an expert on the issue... which is also, in my opinion, not the essence of this thread.

 

[ahrkron]

$$F = -\frac{dV}{dx}$$
According to this relation, the direction of force is always toward decreasing potential energy.

Again, that is part of the definition of a "potential" V for a conservative force (for non-conservative forces, a potential cannot be defined).

This would explain the null case of $$F = \frac{dp}{dt}$$, i.e. objects follow geodesics.

Not so. When p is constant, there is no need of a potential to show that the object would move in a straight line, with constant speed. It is just a result of the definition of momentum.

Also, there is no representation of geodesics in Newtonian mechanics. The trayectories implied by p=constant are indeed geodesics in a flat spacetime, but that has no representation in the formalism. The machinery for modelling that (curvature tensors and the like) is not contained, implied or hinted at in the three laws.

 

[mathphys]

Not so. When p is constant, there is no need of a potential to show that the object would move in a straight line, with constant speed. It is just a result of the definition of momentum.
QUOTE]

the Newton first law you mean?. dp/dt=0 is the 'law of conservation of momentum', right?, and you need this one and the definition of momentum and the knowledge of geodesics in cartesian space and that the 'body' will follow one of these to stablish the conclusions, which are summarized in the first law(altough Newton could had not be aware of geodesics). And becase of this, and by the reasons you have exposed in your last paragraph, it is clear that the first law is not redundant for his theory.

 

[Physics Monkey]

Yes, but more importantly, can Hamilton's principle be applied to derive $$F = \frac{dp}{dt}$$?
I think that Lagrange's equations are a concequence of Newton's second law of motion. Lagrange's equations have been prooven mathematically. Thus, it may be possible to run the argument backwards and proove Newton's second law of motion as a concequence of Lagrange's equations.

In the case of both Lagrange's equations and Hamilton's equations, you must still start from a basic principle. For Lagrange's equations you may take that principle to be Hamilton's Principle: the action should be extremized by the classical path. For Hamilton's equations, you can obtain them from Lagrange's equations. You might also posit that the Hamiltonian is the generator of time translations (which can be proven if you start with the Lagrangian) and take the Poisson bracket as fundamental. The point is that no matter which route you take, you have to start somewhere and they are all equivalent (see exceptions below) to each other as you will learn in any good course on classical mechanics. I think it's fair to say that Newton's Laws came first because they were the most intuitive induction from the experimental data, in other words I don't think you would just hit upon Hamilton's Principle by rolling balls down slopes like old Galileo or performing projectile experiments.

In fact, not all mechanical systems can be easily handled in the Hamiltonian or Lagrangian frameworks, so it does make sense to keep Newton's Laws as the most fundamental statement of the "rules" of classical mechanics. Just to clarify, while it is certainly true that at the basic level (i.e. quantum field theory) our description is entirely in terms of the Lagrangian/Hamiltonian, it is the case at the macroscopic level of description that certain systems are not easily incorporated in the Lagrangian framework. For example, there is no Hamilton's Principle for a system with a retarding force that is proportional to velocity. The Lagrangian may still be written down but it doesn't contain all the information. Lagrange's equations still hold, but only with the modification that they include other generalized forces (drag force) not obtainable from a variational approach.

 

[Son Goku]

Sorry, but that is false. The function that extremizes the action S is the one satisfying the corresponding euler-lagrange equations. You arrive to that system of equations using lagrangian mechanics, not to F=dp/dt.

This is my understanding of it.
(Could be wrong, but anyway)
From what I've seen, when you vary the Action you get the expression:
$$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } \delta{S} = \int_{t1}^{t2} (\pd{L}{q}{} - \frac {d}{dt}\pd{L}{\dot{q}}{}) dt$$
$$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } \pd{L}{\dot{q}}{}$$ is momentum $$p$$
So $$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } \frac {d}{dt}\pd{L}{\dot{q}}{}$$ is $$\frac {dp}{dt}$$

$$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } \pd{L}{q}{}$$ is $$- F$$*

So the formula for the action can be rewritten as
$$\delta{S} = \int_{t1}^{t2} (F - \frac {dp}{dt}) dt$$
Which means $$\delta{S} = 0$$, if $$F = \frac {dp}{dt}$$

$$\delta{S} = 0$$ is used to show that when the action is used as a canonical transformation to bring the system from one time to another, the physics remains unchanged.
So $$F = \frac {dp}{dt}$$ allows the physics to remain unchanged.
It doesn't prove why $$F = \frac {dp}{dt}$$, but I think it shows its necessity.
Again, I could be wrong on any of this, so corrections would be appreciated.
*My gigantic assumption here is a conservative force

 

[mathphys]

This is my understanding of it.
(Could be wrong, but anyway)
From what I've seen, when you vary the Action you get the expression:
$$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } \delta{S} = \int_{t1}^{t2} (\pd{L}{q}{} - \frac {d}{dt}\pd{L}{\dot{q}}{}) dt$$
$$\newcommand{\pd}[3]{ \frac{ \partial^{#3}{#1} }{ \partial {#2}^{#3} } } \pd{L}{\dot{q}}{}$$ is momentum $$p$$

No, it is not momentum p, it is generalized momentum. There is a difference.
It may be linear momentum, angular momentum or another quantity related to the generalized coordinates of the problem.


Nevertheless ,you are right in the sense that IF {qi}={x,y,z} and F=-del (V) the equation of motion that extremizes the action is F=dp/dt. Assuming a conservative force simplifies the problem but restricts is scope, how about not conservative forces (like friction forces)?
As physics_monkey has pointed out you will have to use generalized forces (and restrictions) not obtainable from variational principles in some cases.

 

[Son Goku]

No, it is not momentum p, it is generalized momentum. There is a difference.

Yes, of course you're right, I'm so use to just calculating Langrangians and Hamiltonians (Currently doing Classical Mechanics II) I'd forgotten this.

 

[Physics Monkey]

Hi, Son Goku, you seem to have a good grasp of things but I just wanted to point out a small technical error in your discussion of the variation of the action. You seem to have forgotten to include the $$\delta q$$ in statement of the variation of the action. You should have
$$\delta S = \int \left(\frac{\partial L}{\partial q} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}} \right) \delta q \,dt$$
after the usual integration by parts and ignoring the boundary terms. This factor is important because it allows you to set the rest of the integrand to zero (in general, just because the integral of a function is zero doesn't mean the function is zero). Hope this helps.

By the way, you wouldn't happen to go to Georgia Tech would you?

 

[John_Doe]

What's the lagrangian you're using?

 

[Son Goku]

This factor is important because it allows you to set the rest of the integrand to zero (in general, just because the integral of a function is zero doesn't mean the function is zero). Hope this helps.
By the way, you wouldn't happen to go to Georgia Tech would you?

Thanks for the part in bold, I was actually looking at what I'd written after wondering why it wasn't $$\delta{S} = constant$$.
As for your question, I go to NUI Maynooth in Ireland.

What's the lagrangian you're using?

A general one for a conservative force.