Kinetic Energy equation: Is $K=\frac12mv^2$ a Definition, or a derived Theorem?

Question

I am trying to understand classical physics as a mathematical model. I will first specify the trail of thoughts that led up to this question. (Please correct me if anything is wrong with the reasoning below)

From what I have understood so far, in our mathematical model of the universe, we observed(or rather, derived) that some quantity is always conserved, and we call that quantity energy. In some sense, this conservation of energy is a constraint imposed to our system, and we can exploit this property to derive conclusions about physical phenomena. (What Is Energy? Where did it come from?)

This quantity, energy, can be divided and attributed to several fundamental notions of physics, such as mass, position, velocity, and acceleration.

For example, suppose we have a particle of mass $m$ floating in a vacuum space. We drag it to some constant direction for a few seconds, and the particle ended up moving at a speed of $v$. At this point we say that its kinetic energy is $\frac12mv^2$.

Then, the question: Why is this quantity, $\frac12mv^2$, can be identified as the portion of the conserved quantity, energy?

I looked for the answer on web, but most of them says about energy being the ability to do work, and they propose the typical integration thing. $$\int\vec F\cdot\vec{\mathrm dx}=\int m\frac{\vec{\mathrm dv}}{\mathrm dt}\cdot\vec{\mathrm dx}=\int mv\,\mathrm dv=\frac12mv^2$$

However, in our definition of energy, it has nothing to do with work (which is defined as $\int\vec F\cdot\vec{\mathrm dx}$).

Then I found another concept that might resolve all these mess. The work-energy theorem(?), which says the amount of energy transfered to an object by changing its motion is equal to the work done. However, by saying that it is a theorem, it makes the whole story a complete circular reasoning.

TLDR: Is $K=\frac12mv^2$ itself a postulate? Or is it a consequence of some another postulate?

Apologies if this question is too dumb. Any help would be very much appreciated.

Answer 1

We have the following mathematical property:

with:
$a$ acceleration
$v$ velocity
$s$ position

$$ \int_{s_0}^s a \ ds = \tfrac{1}{2}v^2 - \tfrac{1}{2}v_0^2 \tag{1} $$

(The above relation is derived in my answer to the question What is energy?)

I used $a$, $v$, and $s$ for readability, but this is a general property that is applicable whenever there is a phenonomenon that is described with a state $q$, first time derivative $\frac{dq}{dt}$, and second time derivative $\frac{d^2q}{dt^2}$

In the case of classical mechanics:

Multiply both sides with inertial mass $m$:

$$ \int_{s_0}^s m \ a \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{2} $$

The above is still purely a mathematical statement. Combining (2) with $F=ma$ turns it into a physics statement:

$$ \int_{s_0}^s F \ ds = \tfrac{1}{2}mv^2 - \tfrac{1}{2}mv_0^2 \tag{3} $$

(3) is the work-energy theorem.

The concept of kinetic energy capitalizes on the work-energy theorem. By defining kinetic energy according to the work-energy theorem we get the property that the sum of kinetic energy and potential energy is a constant.

The concept of kinetic energy had a predecessor: Vis Viva, the living force, defined as $mv^2$. Using $mv^2$ as quantity of motion is less practical, because obviously the sum of potential energy and Vis Viva is not a conserved quantity.

Answer 2

TLDR: Is $K=\frac12mv^2$ itself a postulate? Or is it a consequence of some another postulate

It is the definition of "kinetic energy" in the modern formulation of Newtonian mechanics; this term could be defined differently, e.g. as Leibniz' vis viva $mv^2$, used in the past, but already in 18th century Lagrange i his Analytical Mechanics introduced and used the function $T=\frac{1}{2}mv^2$ ($T$ probably from French travaille - work). This function has advantages, and based on later work clearing up things with vis viva/energy and its conservation (e.g. Coriolis Du calcul du l'effet des machines,1829), we call this quantity (with one half) kinetic energy.

The standard definition of kinetic energy $\frac12mv^2$ has the advantage that net work done by all forces acting on a system of particles for motion from state 1 to state 2 equals change of kinetic energy $\sum_a \frac{1}{2}m_av_a^2$: $$ \sum_a \int_1^2 \mathbf F_{-a}\cdot d\mathbf r_a = \sum_a \frac{1}{2}m_av_{a,2}^2 - \frac{1}{2}m_av_{a,1}^2, $$ where $\mathbf F_{-a}$ is net force acting on the particle $a$.

This is the so-called work-energy theorem (where "energy" means kinetic energy); it can be derived from laws of mechanics and definitions of kinetic energy and work, thus it is a theorem. Thus kinetic energy is related to work and its modern definition comes from the observation that work (product of displacement and force in its direction) equals changes of kinetic energy.

With special relativity, we have discovered that this definition of kinetic energy is deficient when speeds approach the speed of light, because then it does not obey the work-energy theorem exactly anymore. So we have fixed the definition, in special relativity, into $$ K = \left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} - 1\right) mc^2 $$ which does obey the work-energy theorem for all speeds. This special relativistic definition gives very close value to the Newtonian definition for low speeds, but diverges from it dramatically for high speeds.

Then, the question: Why is this quantity, $\frac12mv^2$, can be identified as the portion of the conserved quantity, energy?

Because in mechanics of systems free of friction, in the work-energy theorem, the forces are such that increase of kinetic energy $T$ has the same value as decrease of certain potential energy expression $V$, which is a function of particle positions only:

$$ \Delta T = - \Delta V $$ so change of their sum $$ \Delta (T+V) = 0. $$

Thus it makes sense to introduce their sum $E=T+V$, which we call total energy, because it does not change in time:

$$ \Delta E = 0. $$

Answer 3

TLDR: Is $K=\frac12mv^2$ itself a postulate? Or is it a consequence of some another postulate?

$K=\frac{1}{2}mv^2$ is a definition used in classical non-relativistic mechanics. It is used because it is useful.

By definition, the kinetic energy of a system of $N$ particles whose positions, with respect to three cartesian axes $x$, $y$, and $z$ are given by $x_i(t)$, $y_i(t)$ and $z_i(t)$, where $t$ is time and where $i$ indexes the particle, and where $m_i$ is the mass of the ith particle, is: $$ K \equiv \sum_{i=1}^N \frac{1}{2}m_i\left( \dot x_i^2 + \dot y_i^2 + \dot z_i^2\;, \right) $$ where the dot symbol denotes differentiation with respect to time.

At least since the 19th century it has been well known that such a definition of kinetic energy allows one to straightforwardly arrive at the Lagrange equations of motion (see, for example, the 1904 textbook of Whittaker "A Treatise on the Analytical Dynamics of Particles and Rigid Bodies" at page 37): $$ \frac{d}{dt}\frac{\partial K}{\partial \dot q_i} - \frac{\partial K}{\partial q_i} = Q_i\;, $$ where $q_i$ are the generalized coordinates and $Q_i$ are the generalized forces.

Alternatively, in the case of conservative forces, where the $Q_i$ can be written as $-\frac{\partial U}{\partial q_i}$, the Lagrange equations of motion can be written as: $$ \frac{d}{dt}\frac{\partial L}{\partial \dot q_i} - \frac{\partial L}{\partial q_i} = 0\;, $$ where $L=K-U$ is the celebrated Lagrangian function.

Answer 4

You are correct in noting that, as Marius Meyer also correctly points out in a comment, that work is really defined in terms of energy, and that the work-energy theorem is actually pretty circular. It is just bad teaching.

However, while you are trying to restrict the complexity and work solely in terms of non-relativistic Newtonian mechanics, this is actually bad for understanding deeper of what it is you are doing.

We cannot define energy as $\frac12mv^2$ for the extremely simple reason that the kinetic energy in Special Theory of Relativity is something else. Similarly, we cannot define momentum as $m\vec v$ because in Special Theory of Relativity it is also redefined.

The Work-Energy Theorem is celebrated in standard textbooks. However, it is completely missing in modern physics, because if we want to generalise it to Special Theory of Relativity, it becomes obvious that such an endeavour is theoretically nonsense: it would pretty much assume the form of the energy as defined in Special Theory of Relativity, and thus be of no use whatsoever at all.

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This is where it is quite necessary to realise that Physics is an experimental science. Nobody actually knows exactly what it is that momentum is, nor what it is that energy is. However, we know a lot of facts surrounding momentum and energy, and that they are the relevant conserved quantities that are suitable to build a theoretical understanding of our universe. We can only motivate certain definitions, but in the end, it is the experimental evidence that convinces us of their usefulness.

To that end, for the beginner being introduced to non-relativisitic Newtonian mechanics for the first time, it is understood by educators everywhere that the imparting of knowledge is smooth if we point out to students that

1. We live on the surface of a planet that has an approximately constant acceleration due to gravity $g$
2. We live in a universe whereby momentum is easy to motivate
3. When acceleration is constant, a relevant important quantity to scrutinise, even from the basic kinematics equations, is $\frac12mv^2$
4. This quantity is known to be fruitful to study because, when extended to include many different kinds of potential energy, its conservation allows, with momentum, a solution to a great many problems, at least approximately.