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

[John_Doe]

Why does F=dp/dt?

 

[da_willem]

I'm not sure that is a question you're posing, but..

With \vec{p}=m\vec{v}[/tex] the chain rule gives <br /> <br /> \frac{d\vec{p}}{dt}=\frac{dm}{dt}\vec{v}+\frac{d\vec{v}}{dt}m=m\vec{a}+\frac{d\vec{v}}{dt}m<br /> <br /> For objects of constant mass the last term vanishes and you remain with \vec{F}=m\vec{a}[/tex]...

 

[Pengwuino]

Well I'll give the non-mathematical proof for it.

momentum = mass * velocity right?

Force = mass * acceleration as well.

Well if you have a 1kg mass initially moving at 20m/s being pushed by a 10N force, it's going to accelerate at 10m/s right? Well let's say it only acts for 1 second. That means it should accelerate 10m/s and end at 30m/s. It's initial momentum would be 1*20 = 20kg*m/s. Its final momentum at 30m/s would be 1*30=30kg *m/s. So through this, you can see that the momentum changed 10kg*m/s in that 1 thus you can see the change in momentum over the change in time is (10kg*m/s)/1s which is equal to 1kg *m/s^2 which is equal to 10 N which is equal to your original force!

 

[The Bob]

the chain rule gives

I don't know if this is what you call it but I think it is the product rule.

Also, it would appear that you have answered John Doe's question. He seemed to want to know why the derivative of momentum is equal to ma. That is would you have found.

The Bob (2004 ©)

 

[dx]

Right now, there's no answer to why F equals \dot{p}. It is a basic law deduced from experiment. In fact, this is the way Newton originally gave his law, sometimes also
\frac{F}{m}\delta{T}=\delta{v}
All these are equivalent to
F = ma
when the mass is constant. Differentiating p with respect to time assuming mass is constant gives ma

 

[ahrkron]

The expression you quote is a definition for force. It has the nice feature of sharing some properties with our intuitive idea of "force", but the quantity dp/dt is what you have to think about when using the concept in the context of physics.

 

[arildno]

As ahrkon says, F=dp/dt is pretty much a definitional equation in MODERN physics, rather than a derived one.

Quite differently, from a historical point of view, the definitional equations within classical mechanics (for the closed system) was F=ma, and mass conservation.

The subtle shift to the view in modern physics consists of replacing the velocity vector&mass with the momentum vector&energy as the primary dynamical quantities, whereas the velocity vector&mass were the primary dynamical quantities in classical mechanics.

 

[Physics Monkey]

I would just point out that force has always been
<br /> \vec{F} = \frac{d \vec{p}}{dt},<br />
this is the way Newton wrote it in the Principia. The simplification \vec{F} = m \vec{a} can be made for most systems, but Newton did have it right from the beginning.

Also, the equation
<br /> \vec{F} = \frac{d \vec{p}}{d t}<br />
cannot be just differentiated to obtain
<br /> \vec{F} = \frac{dm}{dt} \vec{v} + m \frac{d \vec{v}}{dt},<br />
one simple way to see this is that the right hand side of the above equation isn't Galilean invariant! The error comes because the above equation has an implicit redefinition of the system which is invalid, in plain words you have to think about the momentum change of the little mass element dm because it makes a finite contribution to the right hand side. A more careful analysis indicates that what multiplies the \frac{d m}{dt} term is basically the speed of the mass dm relative to the main body. Clearly this modification produces a Galilean invariant result since it is the relative velocity that enters. This result is quite general in non-relativistic classical mechanics since bodies lose mass by shedding some of their material (think a rocket) and gain mass by acquiring extra bulk from their surroundings (think a snowball).

Note that these considerations are modified in a relativstic setting where the "relativistic mass" can change without the body shedding any material, and the desired invariance is Lorentz invariance. In this case, it turns out that the naive differentiation is essentially ok.

 

[arildno]

I would just point out that force has always been
<br /> \vec{F} = \frac{d \vec{p}}{dt},<br />
this is the way Newton wrote it in the Principia. The simplification that \vec{F} = m \vec{a} can of course be made for most systems, but Newton did have it right from the beginning.

Also, the equation
<br /> \vec{F} = \frac{d \vec{p}}{d t}<br />
cannot be just simply differentiated to obtain
<br /> \vec{F} = \frac{dm}{dt} \vec{v} + m \frac{d \vec{v}}{dt},<br />
one simple way to see this is that the right hand side of the above equation isn't Galiliean invariant!

Which is why the basic equations in classical mechanics PROPERLY stated (for the material system) should be be F=ma and conservation of mass (from which F=dp/dt trivially follows).

 

[Physics Monkey]

arildno,

I have to disagree, the basic equation as formulated by Newton is properly stated as
<br /> \vec{F} = \frac{d \vec{p}}{dt},<br />
not \vec{F} = m \vec{a}. Why are you arguing with this statement?

 

[arildno]

Because:
1. It is readily verified that, within classical limits, mass does not vary with velocity.
2. Therefore, F=ma is NECESSARILY Galilean invariant (whether or not we have verified that a closed system has conserved mass) , whereas F=dp/dt requires the ADDITIONAL assumption for the closed system that mass is conserved in order to be Galilean invariant.
3. Thus, to regard F=ma and mass conservation as the basic equations for the closed /material system is simpler than F=dp/dt and mass conservation.

 

[Physics Monkey]

The assumption of Galilean invariance and the equation F = dp/dt together require that mass be conserved and that mass not vary with velocity. I do not have to assume that mass is conserved. Moreover, in both the Lagrangian and Hamiltonian frameworks it is dp/dt that enters (this is especially important when considering angular momentum). Furthermore, in relativity the correct expression is dp/dt which only reduces to ma in the limit of small velocities. Clearly the expression dp/dt is more fundamental.

 

[Son Goku]

Why does F=dp/dt?

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.

 

[D H]

Because:
1. It is readily verified that, within classical limits, mass does not vary with velocity.
2. Therefore, F=ma is NECESSARILY Galilean invariant (whether or not we have verified that a closed system has conserved mass) , whereas F=dp/dt requires the ADDITIONAL assumption for the closed system that mass is conserved in order to be Galilean invariant.
3. Thus, to regard F=ma and mass conservation as the basic equations for the closed /material system is simpler than F=dp/dt and mass conservation.

I disagree. F=dp/dt describes the basic equations of motion. F=ma is valid for only constant mass system only. F=dp/dt is the general form. Newtons Laws apply to much more than simple closed systems.

While mass does not vary with velocity in classical mechanics, it does vary with time in some systems. Watch the next liftoff of the Space Shuttle to see such a system in action. One had better use F=dp/dt to model rocket dynamics. F=ma leads to invalid results.

 

[John_Doe]

I would like to see a solid proof for this statement:

F=dp/dt

Modern physics is dependant on this premise. This cannot be a definition given that a definition still requires justification for it's usage, and it's validity as a scientific statement is not readily obvious. I accept the fact that science is based upon experiment, however any theory should be logically consistent.

Could someone please explain?

 

[da_willem]

F=dp/dt is the definition of force. It's usefulness stems from the observation that in nature there exist all sorts of phenomena that can elegantly described by this definition: -GMm/r^2, -kx, q_1q_2/4\pi\epsilon[/tex]...

 

[robphy]

It should be noted that F=dp/dt (or F=ma) is NOT a definition of force. The left hand side is the NET FORCE (i.e. the vector sum of the external forces on the object in the question)... and the equality is not a definition, but a physical law.

 

[pmb_phy]

Why does F=dp/dt?

The relation F=dp/dt is not an equality but an identity. I.e. one does not derive this relation. It is a definition of F in terms of p. It used to be that it was a postulate and referred to as Newton's second law of motion but physicists have a difficult time when its referred to as a law since it is unclear how to measure F unless it is first defined. However this in itself opens up a whole can of worms since different people have very strong feelings about what I just said on both sides of the issue.

Pete

 

[John_Doe]

The relation
F = \frac{dp}{dt}
can be integrated and rewritten as
p = \int{Fdt}
does this help?

 

[mathphys]

hi

As most people here has said, F=\dot p is a definition of force(theoretical) and a physical law (fact, empirical). This Newton's Second Law is a statement from the theoretical point of view (one of the 3 Newton defined) that can be verified experimentally. Of course I'm talking in the reing of classical mechanics.


If make no sense to ask about a proof that F=\dot p. p=int(\F dt) dt is the same but expressed in different language.

Furthermore, Asking ' prove that F=\dot p ' is like asking: Prove that t is time and that R^3 is space. Those are also basic definitions in the classical mechanics reign, since The Principia, they're taken as 'granted' from our 'expirience'. They are the analogs of point, line , plane, etc. of Euclid's geometry. Abstract but 'well known' (supposely) objects (in the practical perspective), whose existence and properties are axioms (in the theoretical sense).

Regards

 

[John_Doe]

I repeat, it's validity as a scientific statement is not readily obvious.

Besides, all experiment does is provide support for a theory; it doesn't proove it.

Without any justification or proof, I see no reason whatsoever for this relation to be correct:
F = \frac{dp}{dt}

However, all of modern physics is based upon this premise.

Again, why?

 

[mathphys]

hi

Define 'proof', or 'proves it'. In the mathematical sense?

In the mathematical sense F=\dot p is an axiom (in Newtonian mechanics). Just like in Euclidian Geometry ' given to points, there is just one straight line that contains them'. It is an axiom, no proof for it. It is a premise, and the theory is constructed over it. You may read the initial chapter of Einstein's Relativity if my explication does not satisfy you.

 

[mathphys]

And no, 'modern' physics is not based on this premise. Classical Newtonian Mechanics is based on this and other premises. Alternatively ,you may,must or prefer to use lagrangian or Hamiltonian formalisms or geometric algebra as the "language" for mechanics, i.e., No Newton Laws as axioms.

 

[John_Doe]

'ax·i·om (ăk'sē-əm)
n.
A self-evident or universally recognized truth; a maxim: “It is an economic axiom as old as the hills that goods and services can be paid for only with goods and services” (Albert Jay Nock).
An established rule, principle, or law.
A self-evident principle or one that is accepted as true without proof as the basis for argument; a postulate.
[Middle English, from Old French axiome, from Latin axiōma, axiōmat-, from Greek, from axios, worthy.]'

F = \frac{dp}{dt}
is not self-evident, is not a universally recognized truth. You cannot argue that it is a axiom, for it is not.

Also, Euclid's theorem 'given any line, and any point not on that line, there is a unique line through that point that is parallel to the given line' cannot be proven in terms of the axioms of Euclidean geometry, and it was this that gave way to non-Euclidean geometries.

 

[mathphys]

It is an axiom for NEWTONIAN MECHANICS.

The 2 points-line is an axiom, is one of the axioms of EUCLIDIAN GEOMETRY, not a theorem. Of course that it cannot be proved from the other axioms. Remove it from EUCLIDIAN GEOMETRY and you have another type of Geometry.

Axioms are accepted as valid and accepted as true for the theories that are constructed from them, solely.

Physical Theories are accepted as valid if they sucess in explaning physical phenomena in a consistent way. Newtonian mechanics is a cornerstone of classical physics because it does, has done and will do in many (of course not all) cases, and that is why is teached...it is useful for constructing bridges ;), for instance.

 

[John_Doe]

That doesn't change the fact that it is not self-evident, is not a universally recognized truth. You cannot argue that it is a axiom, for it is not.

 

[mathphys]

:P read my previous post. It is for NEWTONIAN MECHANICS, ...not for life :P


If you think like Newton did, is an axiom,,for his theory...and applications, when it applies

 

[John_Doe]

And so the theories of modern physics, based upon classical physics are to only apply when his theory does?

So, given the results of this 26 reply thread, when does his 'axiomic' theory apply?

 

[ahrkron]

The expression "F=dp/dt" is a definition of what we mean by "force".

Newton did not "discover" that such relation works. There was no definition of "force" at the time, so he could not possibly measure many forces to compare them with the corresponding products ma[/color].

What he did was more subtle, but tremendously important: he identified an important concept to quantify, and provided a definition that allows us to do so, and to work with the concept, separated from circumstancial details that are not relevant for the physics.

 

[John_Doe]

Yet, by reasons already mentioned, it cannot be. If it were a definition, that would also logically imply that it were also an axiom, which it cannot be.

 

[mathphys]

Yes i agree with ahrkron, in the sense that he has written F=dp/dt is a definition of force more than an axiom.

But also, as part of the set of Newton laws, it may be tought as an axiom. It all depends, if there are previous concepts or products of ma then he may define a quantity related to it(;) But now think how to define mass). Or you can start with the set of those 3 laws as axioms and develop the rest of the theory, as an axiomatic one.

On the other hand ,why keep insisting that the whole of physics(mechanics) depends on it? Ask geometric algebra guys if they think so. Classical? Think about lagrangian and hamiltonian formalism. Simpler but nevertheless fundamental? Newtonian Mechanics, it boosted physical sciences and because of this is , sic, tremendously important.


Regards.

quoting Einstein 'God does not care about our mathematical difficulties', nor the way we theoretize. So does Nature.

 

[John_Doe]

Do you deny that for example electromagnetic theory is based upon this premise, and therefore the general theory of relativity and quantum theory? As for classical physics, that is without a doubt based upon it.

 

[mathphys]

EM theory is not consistent completely and by that not at all with Newtonian mechanics theory. Maxwell knew that, Lorentz and Einstein also. That may be thought as one of the principal motivations for the theory of relativity, which has Newtonian mechanics as a particular case, but departs from a completely different perspective, see Einstein's Relativity book. Quantum Mechanics is directly and historically related to the lagrangian and hamiltonian formalisms of mechanics, just see Schrödinger's work.

 

[dx]

Electromagnetic theory is not based upon Newtonian mechanics, nor does the general theory of relativity or quantum mechanics. The latter two include Newtonian mechanics as a special case.

 

[ahrkron]

If it were a definition, that would also logically imply that it were also an axiom

The word "axiom" is technically used in mathematical logic. Here, you can think of it as an axiom only as a metaphor; but, as all metaphors, it is limited.

No matter how you call it, is is an operational definition of the quantity "force".

Can you find another, independent, one in the context of classical physics?

 

[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.