Is this a proof for energy conservation?

Question

I recently watched this video which proves energy conservation by defining kinetic energy as $\frac{1}{2}mv^2$ and potential energy as the negative integral of force with respect to position. The proof is derived from basic Newtonian mechanics. Here it is:

$$v^2 = u^2+2a(s_f-s_i)$$ $$v^2=u^2+2\left(\frac{F}{m}\right)(s_f-s_i)$$ $$\frac{1}{2}mv^2-\frac{1}{2}mu^2 = F(s_f-s_i)$$

Here, the assumption is that force is constant from $s_i$ to $s_f$. However, this assumption can be eliminated by considering a small change in position $\Delta s$ such that $F$ is constant within this interval, even though it may be changing over the entire distance. Then:

$$\frac{1}{2}mv^2-\frac{1}{2}mu^2=\int_{s_i}^{s_f}F\cdot ds=\int_{0}^{s_f}F\cdot ds-\int_{0}^{s_i}F\cdot ds$$ $$\frac{1}{2}mv^2-\int_{0}^{s_f}F\cdot ds=\frac{1}{2}mu^2-\int_{0}^{s_i}F\cdot ds$$

Now, if you define the potential energy to be the negative integral of force with respect to distance, we get:

$$\frac{1}{2}mv^2+U_f=\frac{1}{2}mu^2+U_i$$

But I thought that energy conservation was an experimentally deduced fact. So I'm guessing there are some additional assumptions associated with this derivation, which makes this proof applicable to only a special subset of systems. What are these assumptions? And how far can you get with such an approach? In other words, how much can you generalize the energy conservation law using Newtonian mechanics?

Answer 1

It may have started as an observation stemming from Newtonian mechanics, but energy conservation has become a far more general principle.

The most general or fundamental derivation I know of is using Noether's Theorem. Discovered by mathematician Emmy Noether in the context of Lagrangian Mechanics, it can be shown that the postulate of Time Symmetry – that a physical experiment will have the same result regardless of being performed yesterday, today, or 100 years from now – directly implies conservation of energy. Likewise, linear physical symmetry (an experiment at my location has the same result as the same one performed 100 km to my right) directly implies conservation of momentum. And many other conserved quantities can be derived from a corresponding symmetry.

Importantly, it can be shown that in systems without Time Symmetry, such as the large scale structure of our universe, energy conservation does not hold. The universe is constantly expanding and changing such that an experiment 2 hours after the Big Bang would not give the same result as the same experiment now. The microwave frequency light we observe in all directions has a peak frequency in the range of $10^{11} ~\mathrm{Hz}$ after 14 billion years of recession and redshift during the expansion. When it was emitted it had a frequency of about $10^{14} ~ \mathrm{Hz}$. That 1000-fold reduction in energy, by $E=hf$ is an example of energy non-conservation in a system without Time Symmetry.

Answer 2

But I thought that energy conservation was an experimentally deduced fact.

It is. That does not mean it cannot be derived mathematically. After all, it is agreed by many that mathematics is the language of physics ("Only through mathematics can one achieve lasting truth in physics. Those who neglect mathematics wander endlessly in a dark labyrinth." - Galileo).

So I'm guessing there are some additional assumptions associated with this derivation, which makes this proof applicable to only a special subset of systems.

As discussed by RC_23 his answer, there is the more fundamental concept of time-translational invariance. Another condition for energy conservation is that the force(s) involved must be conservative. In fact, Noether's theorem states that for a physical system with conservative forces, each symmetry of the action has a corresponding conservation law.

how much can you generalize the energy conservation law using Newtonian mechanics?

As much you like (in non-relativistic, classical systems as pointed out above by RC_23), provided all the forces involved are conservative.

Now, if you define the potential energy to be the negative integral of force with respect to distance

It's not a question of "if". It is necessary so that your result

$\frac{1}{2}mv^2+U_f=\frac{1}{2}mu^2+U_i$

naturally follows.

That is, for conservation of mechanical energy to hold true, the force(s) that bring about the changes in potential or kinetic energy, necessarily have the property that they can be written as the gradient of some potential energy function. This also means that the force is irrotational (zero curl) and that the work done is path independent (which also means that the line integral of the force over a closed loop will give you zero).

Answer 3

What that derivation shows is that $F=ma$ and conservation of the sum of potential and kinetic energy are related.

Those two are not independent cases: they are two ways to express the same thing.

Here's the thing: have you ever wondered why the expression for kinetic energy $\tfrac{1}{2}mv^2$ features that factor $\tfrac{1}{2}$?

What we have there is an expression with a squared quantity, and a factor $\tfrac{1}{2}$.

That combination of a squared quantity and a factor $\tfrac{1}{2}$ is typical for the outcome of evaluating the area between a slanted line and a horizontal line:

With cartesian coordinates, take the following simple graph: a straight line extending from the origin $(0,0)$, according to: $f(x) = x$

Next we give the expression for the area between that diagonal line and the $x$-axis, from the origin $0$ to a final point $x_f.$

The triangle with the base from $(0,0)$ to $(x_f, 0)$ is a right-angled triangle, which gives that the area between that graph and the x-axis is given by $\tfrac{1}{2}{x_f}^2$.

Now we take the following well known expression for uniform acceleration, with:

• $s_f$ final position
• $s_i$ initial position.

\begin{equation}\label{eq:1} s_f - s_i = v_i t + \tfrac{1}{2}at^2. \tag{1} \end{equation}

The factor $\tfrac{1}{2}$ of $\tfrac{1}{2}at^2$ correlates with the area of a triangle: with uniform acceleration the distance covered as a function of time is proportional to half the area of the rectancle with sides $t$ and $at$.

For later use:
\eqref{eq:2} for the relation that with uniform acceleration the velocity increases linear with time:

\begin{equation}\label{eq:2} v_f - v_i = at. \tag{2} \end{equation}

Multiply both sides of \eqref{eq:1} by $a$ for acceleration, and group the factor $(at)$ together:

\begin{equation}\label{eq:3} a(s_f - s_i) = v_i(at) + \tfrac{1}{2}(at)^2. \tag{3} \end{equation}

Use \eqref{eq:2} to substitute the factor $(at)$:

\begin{equation}\label{eq:4} a(s_f - s_i) = v_i (v_f - v_i) + \tfrac{1}{2}(v_f - v_i)^2 \tag{4} \end{equation}

\begin{equation}\label{eq:5} a(s_f - s_i) = v_i v_f - {v_i}^2 + \tfrac{1}{2}{v_f}^2 - v_i v_f + \tfrac{1}{2}{v_i}^2 .\tag{5} \end{equation}

The result is:

\begin{equation}\label{eq:6} a(s_f - s_i) = \tfrac{1}{2}{v_f}^2 - \tfrac{1}{2}{v_i}^2 \tag{6}. \end{equation}

Both sides of \eqref{eq:6} can be interpreted in terms of area:

• The left hand side is the area of a rectangle, $a(s_f - s_i)$
• In the right hand side the factor $\tfrac{1}{2}{v_f}^2$ is the area of a right triangle with $v_f$ for the length of the base and $v_f$ for the height. Likewise for $v_i$.

Next:
To combine $F=ma$ and \eqref{eq:6}:

take $F=ma$, and multiply both sides by $(s_f - s_i)$:

\begin{equation}\label{eq:7} F(s_f - s_i) = ma(s_f - s_i) \tag{7} \end{equation}

\begin{equation}\label{eq:8} F \Delta s = \tfrac{1}{2}m{v_f}^2 - \tfrac{1}{2}m{v_i}^2. \tag{8} \end{equation}

The definitions of potential energy and kinetic energy are specifically chosen to slot in with the above.

Until around the 1850's physicists used a precursor to the kinetic energy concept that was called 'Vis Viva' (The living force), which was defined as $mv^2$.

It was around 1850 that a french physicist by the name of Gustave Gaspard Coriolis proposed a new name and a new definition: 'kinetic energy' and $\tfrac{1}{2}mv^2$. That innovation was adopted rapidly.

(Recognizing the value of defining a quantity $F\cdot \Delta s$ goes back all the way to Huygens. The interpretation of what it stands for changed over time, but not the definition.)

The point is: it is not a remarkable coincidence that $F=ma$ and conservation of the sum of potential and kinetic energy slot together: the definition of kinetic energy was specifically chosen to make that happen. The choice was made because physicists recognized how it makes a lot of things fall into place.

Generalizations

Generalization of \eqref{eq:6} to non-uniform acceleration.

First step.
Let there be two stages of uniform acceleration: from $s_0$ to $s_1$, and from $s_1$ to $s_2$.

\begin{equation}\label{eq:7a} a(s_1 - s_0) = \tfrac{1}{2}{v_1}^2 - \tfrac{1}{2}{v_0}^2 \tag{7} \end{equation}

\begin{equation}\label{eq:8a} a(s_2 - s_1) = \tfrac{1}{2}{v_2}^2 - \tfrac{1}{2}{v_1}^2. \tag{8} \end{equation}

Add \eqref{eq:7a} and \eqref{eq:8a} together:

\begin{equation}\label{eq:9} a(s_1 - s_0) + a(s_2 - s_1) = \tfrac{1}{2}{v_1}^2 - \tfrac{1}{2}{v_0}^2 + \tfrac{1}{2}{v_2}^2 - \tfrac{1}{2}{v_1}^2. \tag{9} \end{equation}

On the left hand side the factor $s_1$ drops out and on the right hand side the factor $\tfrac{1}{2}{v_1}^2$ drops out.

\begin{equation}\label{eq:10} a(s_2 - s_0) = \tfrac{1}{2}{v_2}^2 - \tfrac{1}{2}{v_0}^2. \tag{10} \end{equation}

This dropping out of intermediate terms generalizes to summation of a sequence of infinitesimally narrow rectangles, which of course is evaluation of an integral.

\begin{equation}\label{eq:11} \int_{s_i}^{s_f} a \ \mathrm{d}s = \int_{v_i}^{v_f} v \ \mathrm{d}v \\ = \tfrac{1}{2}{v_f}^2 - \tfrac{1}{2}{v_i}^2 \tag{11} \end{equation}

(In \eqref{eq:11}, to go from the first to the second step:
$$\mathrm{d}s = v \ \mathrm{d}t \quad \text{and} \quad a \ \mathrm{d}t = \mathrm{d}v.$$
We have:
$$\int_0^s a \ \mathrm{d}s = \int_0^t a \ v \ \mathrm{d}t = \int_0^t v \ a \ \mathrm{d}t = \int_0^v v \ \mathrm{d}v$$).

Repeating \eqref{eq:11} with the inbetween step omitted:

\begin{equation}\label{eq:12} \int_{s_i}^{s_f} a \ \mathrm{d}s = \tfrac{1}{2}{v_f}^2 - \tfrac{1}{2}{v_i}^2. \tag{12} \end{equation}

In mechanics the starting point is $F=ma$. Integrating with respect to position coordinate, and using \eqref{eq:12} to process $\int m \ a \ \mathrm{d}s$:

\begin{equation}\label{eq:13} \int_{s_i}^{s_f} F \ \mathrm{d}s \\ = \int_{s_i}^{s_f} ma \ \mathrm{d}s \\ = \tfrac{1}{2}m{v_f}^2 - \tfrac{1}{2}m{v_i}^2 \tag{13} \end{equation}

Generalization to areas other than mechanics

\eqref{eq:12} is not a physical property: it is a mathematical property.

Let me refer to not taking a derivative as 'zeroth derivative'.

If you have occuring together:
zeroth derivative
first derivative
second derivative,
then there is a counterpart of \eqref{eq:12}.

Let $s$ represent any form of state that has a first time derivative and a second time derivative. Then:

\begin{equation}\label{eq:14} \int_{s_i}^{s_f} \frac{\mathrm{d}^2s}{\mathrm{d}t^2} \ \mathrm{d}s = \tfrac{1}{2}{\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)_f}^2 - \tfrac{1}{2}{\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)_i}^2. \tag{14} \end{equation}

Example:
The case of electric current in a circuit:
change of current strength is the second derivative of position of charge. A coil with self-inductance will oppose change of current strength.
Through an inductor: change of current strength is proportional to electromotive force (electric potential), divided by the coefficient of self-induction (inductance).

With:

• $F_e$ electromotive force
• $L$ inductance
• $C$ position of charge
• $I$ first time derivative: current,

\begin{equation}\label{eq:15} \frac{\mathrm{d}^2 C}{\mathrm{d}t^2} = \frac{F_e}{L}, \tag{15} \end{equation}

which can be restated as:

\begin{equation}\label{eq:16} F_e = L \frac{\mathrm{d}^2C}{\mathrm{d}t^2}. \tag{16} \end{equation}

So when a current $I$ is flowing through an inductor, we expect an associated energy of $\tfrac{1}{2}LI^2.$

Generally:

In any area of physics where the dynamics is described with an equation that expresses that the second time derivative of some state relates to a particular change causing factor the relation \eqref{eq:14} can be used to transform the expression to energy form.

Answer 4

There are already great detailed answers, but staying in elementary level physics:

defining [...] potential energy as the negative integral of force with respect to position

The important experimental observation here is that this definition makes sense. You could imagine a universe with only two particles and only gravity, where bringing the two masses of 1 kg from 1 m distance to 10 m distance required a different amount of negative integral of force with respect to position, depending on how you did it. Defining "potential energy" in that way would still give the same result of energy conservation; however, to know the "potential energy" of a system you would need to know the entire history, so the concept wouldn't be very useful!

In other words, it's an experimental observation that the potential energy, defined in this way, is uniquely determined by looking at the current state of the system, regardless of history.