# Definition of force, kinetic energy and momentum

I've edited the post. Q1 and Q4 are the important ones but I didn't delete Q2 and Q3 since some older answers would not make sense anymore.

To begin with, the formula of the kinetic energy $T$ is $\frac{mv^2}{2}$. Furthermore momentum is conserved $\Sigma m_{i}\vec{v_{i}}=const.$ Then you have the definition that force is the change of momentum with respect to time $\vec{F}=\dfrac{d(m\vec{v})}{dt}$. I've read the chapters concerning mechanics of Physics for scientists and engineers by Giancoli and the Feynman Lectures. Giancoli introduces the arbitrarily work as $W=\int\vec{F} \cdot d \vec{s}$. From this definition of work he derives the kinetic energy to be $\frac{mv^2}{2}$. In contrast to that, in the Feynman lectures you never get a derivation of $\frac{mv^2}{2}$, but it is shown that $\dfrac{dT}{dt}=\vec{F}\cdot\vec{v} = \vec{F}\cdot\dfrac{d\vec{s}}{dt}$. Then it is shown that $dT=\vec{F}\cdot d \vec{s}$ and as a consequence $\Delta T = \vec{F} \cdot \vec{s}$ which is called work. http://www.feynmanlectures.caltech.edu/I_13.html#Ch13-S1

Now I've got some questions:

Q1 Is $F=\dfrac{d(mv)}{dt}$ just an arbitrary definition or is there something "more" behind the formula for force?

Q2 Is $W=\int\vec{F} \cdot d \vec{s}$ just a definition or is there something more behind? I mean, can you derive the formula for work not by taking the formula for the kinetic energy as given.

Q3 How to derive the formula for the kinetic energy and work only form the conservation of momentum $\Sigma m_{i}\vec{v_{i}}=const.$?

Q4 How are work and kinetic energy defined? I have found both: A) Kinetic energy is doable work and work is $F=\dfrac{d(mv)}{dt}$. B) The formula for kinetic energy is: $\frac{mv^2}{2}$. Then, after some maths it follows that $\Delta T = \int \vec{F} \cdot d\vec{s}$. It doesn't make sense to define 2 things in that way. This would be circular logic.

P.S.: I am not a native English Speaker so feel free to edit.

• A word on the place of "derivations" in physics: physics is not a branch of math and some things are accepted because that is the way thing are. We do derivations, because that is how we show that some allegedly fundamental inputs agree with observations and rules from a less fine-grained understanding; but the complex made up of the notion of energy, the work-energy theorem, the concept of momentum, the impulse-momentum theory, and the definitions of force and mass are in a real sense observationally founded. They can be checked for internal consistency, but they are what they are. – dmckee --- ex-moderator kitten May 26 '14 at 22:41

Feynman makes a point of stating explicitly, in vol. 1 of his Lectures on Physics, that $F = \frac{d(mv)}{dt}$ is not the definition of force. In section 12-1 he states

If we have discovered a fundamental law, which asserts that the force is equal to the mass times the acceleration, and then define the force to be the mass times the acceleration, we have found out nothing.

A bit later he states

The real content of Newton's laws is this: that the force is supposed to have some independent properties, in addition to the law $F = ma$; but the specific independent properties that the force has were not completely described by Newton or by anybody else, and therefore the physical law $F = ma$ is an incomplete law. It implies that if we study the mass times the acceleration and call the product the force, i.e., if we study the characteristics of force as a program of interest, then we shall find that forces have some simplicity; the law is a good program for analyzing nature, it is a suggestion that the forces will be simple.

I found the following comments from Terence Tao, on the topic of how physics models work, to be enlightening:

Terence Tao - @Pietro: the way mathematical or physical models work, one assumes the existence of a variety of mathematical quantities (e.g. forces, masses, and accelerations associated to each physical object) that obey a number of mathematical equations (such as F=ma), and one also assumes that the result of various physical measurements can be computed in terms of these quantities. For instance, two physical objects A_1, A_2 will be in the same location if and only if their displacements x_1, x_2 are equal.

Initially, the numerical quantities in these models (such as F, m, a) are unknown. However, because of their relationships to each other and tophysical observables, one can in many cases derive their values from physical measurement, followed by mathematical computation. Using rulers, one can compute displacements; using clocks, one can compute times; using displacements and times, one can compute velocities and accelerations; by measuring the amount of acceleration caused by the application of a standard amount of force, one can compute masses; and so forth. Note that in many cases one needs to use the equations of the model (such as F=ma) to derive these mathematical quantities. (The use of such equations to compute these quantities however does not necessarily render such equations tautological. If, for instance, one defines a Newton to be the amount of force required to accelerate one kilogram by one meter per second squared, it is a non-tautological fact that the same Newton of force will also accelerate a two-kilogram mass by only one half of a meter per second squared.)

If one has found a standard procedure to compute one of these quantities via a physical measurement, then one can, if one wishes, take this to be the definition of that quantity, but there are multiple definitions available for any given quantity, and which one one chooses is a matter of convention. (For instance, the definition of a metre has changed over time, to make it less susceptible to artefacts.)

In some cases, it is not possible to measure a parameter in the model through physical observation, in which case the parameter is called "unphysical". For instance, in classical mechanics the potential energy of a system is only determined up to an unspecified constant, and is thus unphysical; only the difference in potential energies between two different states of the system is physical. However, unphysical quantities are still useful mathematical conveniences to have in a model, as they can assist in deriving conclusions about other, more physical, parameters in the model. As such, it is not necessary that every quantity in a model come with a physical definition in order for the model to have useful physical predictive power.

• Would it be possible to tell me the source, where you found this explanation of Tao ? I'd be very (!) grateful, as I would be interested in reading what other things Tao had to say about this subjects, but googling I couldn't find this commentary on his blog - or elsewhere... – l7ll7 Sep 24 '15 at 14:39
• @user10324 Tao made this comment on his google buzz, back when that was a thing, and I copied the comment to my email when I read it. I'm not sure if Tao's buzz posts are still online -- if I find out more I'll let you know. – littleO Sep 24 '15 at 18:05
• I just went through Taos complete Google+ account, but couldn't find an entry, where this comment would have fit in. If you saved this comment of Tao's to your email - which is not surprising, I would have done the same, since this comment of Tao's is so supremely clear and enlightening - I image you have other nuggets of wisdom from Tao (on physics). Couldn't you post them somewhere online, so that the rest of the world can be enlightened by Tao too ? Your karma might improve significantly ;) – l7ll7 Sep 25 '15 at 20:15

Q1 Is $F=\frac{d(mv)}{dt}$ just a definition or is there something "more" behind the formula for force?

$\vec{p} = m\vec{v}$ is a definition! $\vec{F} = m\vec{a}$ is not! Pratically speaking, you can measure both Force (using an elastic material and its deformation) and acceleration (double derivative of space respect to time), Newton discovered they're proportional so, since they're definite indipendently, $\vec{F} = m\vec{a}$ is the relation between $\vec{F}$ and $\vec{a}$ not the definition of force! You can't measure momentum, so you'll need to define it! The definition is $\vec{p} = m\vec{v}$ and the reasons are perfectly explained on Feynman books.

$$\vec{F} = m\vec{a} = m\frac{d\vec{v}}{dt} = \frac{d(m\vec{v})}{dt}$$

Q2 Is $W=\int\vec{F}.d\vec{S}$ just a definition or is there something more behind? I mean, can you derive the formula for work not by taking the formula for the kinetic energy as given.

$$W = \int \vec{F} \cdot d\vec{s}$$ is the definition of work done by a force $\vec{F}$ along a path, you do not need kinetic energy to define work. Energy is the possible work doable by a moving body.

Q3 How to derive the formula for the kinetic energy and work only from the conservation of momentum $\sum_i\vec{v_i}$=const.?

You know that $\vec{F} = \frac{d(m\vec{v})}{dt} = \frac{d\vec{p}}{dt}$ but since $\vec{p}$ is costant, is derivative will be zero. In case it is moving with costant speed $\vec{v_0}$. Now, imagine a costant force applied to the mass, you'll have that $v^2 - v_{0}^2 = 2as$ (from cinematics) $Fs = W = \Delta E$, $mas = W \Rightarrow$ $$v^2 - v_0^2 = 2W/m$$ $$W = \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2$$ The applied force is going to stop the mass, so the mass reacts with an opposite force, and since energy is the possible work done by the moving mass. Since it is opposed and $v= 0$ we obtain: $$W = \frac{1}{2}mv_0^2$$ Energy depends only on the initial velocity because momentum (before the costant force applies) is conservating.

• So, according to your answer, force is defined as the Deformation of elastic material?? You won't find any material that will deform exact linearly, when a force is applied, so I personally think your definition of force as a Deformation is wrong, but I am not sure. Secondly, I Interpret your answer as follows: You define work arbitrarily as F*s and energy is doable work. Then you have to get 0,5mv^2. Is that right? – user50224 May 26 '14 at 21:16

How to derive the formula for the kinetic energy and work only form the conservation of momentum

Here's an approach for the kinetic energy using the Lagrangian mechanics.

The Lagrangian for a free particle is simply the kinetic energy $T$ of the particle and the conservation of momentum is expressed by Euler-Lagrange equation (working in one dimension)

$$\frac{d}{dt}\frac{\partial L}{\partial \dot q} = \frac{d}{dt}\frac{\partial T}{\partial v} = \frac{dp}{dt} = 0$$

Thus, we have

$$\frac{\partial T}{\partial v} = p = mv$$

which implies

$$T = \frac{mv^2}{2}$$

(We could have included constants of integration and then set them to zero on physical grounds.)

• I don't understand the Lagrangian approach. I think I have a lot to learn ;). Is there any other way to derive the formula for kinetic energy? Furthermore, when you have the formula for kinetic energy the formula of work is a consequence, right? And: The concept of force is just an arbitrary definition, right? – user50224 May 26 '14 at 19:13
• @user50224, there a numerous paths to deriving the formula for KE and learning each one will deepen your understanding of the nature of KE. By the way, keep in mind that the $F$ in $F=ma$ is the net force so it isn't really a definition of force. For example, in statics, we set the sum of forces to be zero (so everything is static). en.wikipedia.org/wiki/Statics – Alfred Centauri May 26 '14 at 22:39

Q1 Is $F=\frac{d(mv)}{dt}$ just a definition or is there something "more" behind the formula for force?

In the absence of forces the momentum $P$ is conserved so $\frac{dP}{dt}=0$. So, if we observe the momentum changing, we may as well define it as being due to a force $F=\frac{dP}{dt}$. So, in my view, it's a definition.

Q2 Is $W=\int F\bullet ds$ just a definition or is there something more behind? I mean, can you derive the formula for work not by taking the formula for the kinetic energy as given?

Suppose that the kinetic energy $T$ is some function of the momentum, so $T=f(P)$. In the absence of forces the kinetic energy is conserved $\frac{dT}{dt}=0$. However, some force is causing the kinetic energy to change, $$dT=\frac{df(P)}{dP}dP=\frac{df(P)}{dP}Fdt$$ and we'd like to have the usual formula for work $dT=Fdx$ so this means $\frac{df(P)}{dP}dt=dx$ and so, $$\frac{df}{dP}=v=\frac{P}{m}$$ which implies, $$T=\frac{P^{2}}{2m}$$ So, in my view, the formula for work implies that we already have the formula for kinetic energy.

Q3 How to derive the formula for the kinetic energy and work only form the conservation of momentum $\frac{dP}{dt}=0$?

It's only necessary to get the formula for the energy because the answer to question two shows that the formula for work follows from the formula for energy.

In order to get the formula for energy one needs to study the group for Galilean space-time. It has the Lie algebra, $[K,P]=m$; $[K,H]=P$; $[P,H]=0$ where $P$ is momentum, $H$ is energy and $K$ is the velocity boost. The conservation of momentum is implied by the third Lie bracket which says that momentum commutes with energy. Now, one uses the fact that the Lie algebra has a representation by the Poisson brackets (PBs) of Hamiltonian mechanics. The first PB is, $$m=[K,P]=\frac{\partial K}{\partial x}\frac{\partial P}{\partial P} - \frac{\partial K}{\partial P}\frac{\partial P}{\partial x}=\frac{\partial K}{\partial x}$$ This implies the generator of boosts is $K=mx$. Now use this formula for the boost in the second PB. $$[K,H]=\frac{\partial K}{\partial x}\frac{\partial H}{\partial P}=m\frac{\partial H}{\partial P}=P$$ and the last equality implies, $$\frac{\partial H}{\partial P}=\frac{P}{m}$$ which, upon integrating wrt $P$ gives the usual formula for the kinetic energy $H=\frac{P^{2}}{2m}$ and the usual symbol $H$ (Hamiltonian) is being used for kinetic energy $T$ in this case.

• So, you are of the opinion that force is just an arbitrary definition. The definition of work is a consequence if we know the formula for kinetic energy. Is this right? I don't understand your idea of Lie algebra. Could you maybe explain how to get the formula for kinetic energy? – user50224 May 26 '14 at 19:09
• @user50224 : Yes, I think force is a definition that just captures the observation that something is changing the momentum. The definition of work is a consequence of the formula for kinetic energy. The background to the answer to your third question is hard work. I just wanted to show how the formulae of mechanics ultimately depend on the group properties of the space-time. – Stephen Blake May 26 '14 at 19:33

Feynman has derived formulae for potential energy, kinetic energy, and work, and the law of conservation of momentum, assuming nothing but the absence of any demonstrable existence of perpetual motion machines and assuming Newton's second and third laws, specifically Newton's definition/formula for force.

Feynman starts by deriving the formula for gravitational potential energy from the basics in Chapter 4 (http://www.feynmanlectures.caltech.edu/I_04.html), assuming only that man has never seen a case of work getting done for free without the expenditure of energy. He shows an experiment where a ball at height h can raise 3 balls of the same individual weights as itself to height h/3 by lowering itself fully. So, by a given gravitational potential energy (obtained because of its height), the height to which a target object can be raised is inversely proportional to that target object's weight. Thus we have:

(i) $${\rm PE} = w h$$

Next, one can conceivably go to different heights above and below the earth's surface and discover, using a sufficiently sensitive instrument, the following:

(ii) g varies with height: an object dropped accelerates at different rates at different heights;

(iii) $$w = m g$$: the force exerted by an object, say in the extent to which it compresses a spring it is placed on, varies with g which varies with height as in (ii).

So we get:

(iv) PE = wh = mgh = ma*h

Next, Feynman shows the pendulum experiment (Ch.4.3) where the potential energy that the pendulum has at height h is completely converted to kinetic energy when it reaches the bottom, and that kinetic energy gets converted completely back to potential energy when the pendulum reaches the top on the other side. So, all loss of potential energy during the pendulum swing downward is being converted to kinetic energy. Therefore:

d(PE)/dt = -d(KE)/dt

Subsuming the minus sign into the direction of change of h and expanding, we get:

d(KE)/dt = d(mgh)/dt = mgv = mav

that is:

d(KE)/dt = mv(dv/dt)

To get the formula for kinetic energy KE we integrate the above:

(v) KE = (1/2)*m**v^2

Next, in Chapter 10, Feynman assumes Newton's second and third laws:

Second Law: force is the rate of change of momentum;

Third Law: every action has an equal and opposite reaction)

Feynman combines the two laws as follows: if we have two objects of momenta p1 and p2 colliding head on, then from Newton's third law they will experience equal and opposite forces, but those forces are given by Newton's second law as d(p1)/dt and d(p2)/dt, so we have:

d(p1)/dt = -d(p2)/dt

i.e.,

d(p1)/dt + d(p2)/dt = 0

i.e.,

d(p1+p2)/dt = 0

integrating which we get:

(vi) (p1+p2)_initial = (p1+p2)_final

which is the law of conservation of momentum.

Next, for work done, consider Feynman's pendulum (Ch.4.3) when it is at the bottom of its swing. There is kinetic energy but no potential energy at that point. Now the kinetic energy does work to push the pendulum back to height h at the opposite end, working against the gravitational force and fully converting itself into potential energy in the process. How much work is done? That is nothing but mgh, the potential energy that Feynman initially proved for a ball at height h! So we have:

(vii) work W = mgh = F*h = force * distance

Thus, Feynman has derived formulae for potential energy, kinetic energy, and work, and the law of conservation of momentum, assuming nothing but the absence of any demonstrable existence of perpetual motion machines and assuming Newton's second and third laws, specifically Newton's definition/formula for force.