Conservation of Energy

Some of the great tools in physics are so-called "conservation laws" that buttress the laws of motion with certain quantities that remain the same throughout time. Among these great laws is the conservation of energy which states that while energy can change forms, it cannot be created or destroyed.

Here we'll explore the partition between kinetic energy and potential energy, and how energy can in some sense replace forces in our calculations. We conclude with the introduction of formal methods of physics that analyze the dynamic behavior of systems solely in terms of energy, thus replacing the necessity of having to analyze forces at all. Such methods play a dominant role in systems of condensed matter, quantum field theory, and other problems far beyond elementary classical mechanics.

For a person riding a bicycle, their kinetic energy $E$ is equal to the amount of heat that would be dissipated from the moment they pull the brake to the moment they come to rest. If they don't skid, most of this heat will go into heating the brake pad and the metal rim of the wheel. If they do skid, most of the heat will go to the rubber in the tire and to the road. Before they pull the break, this energy can be observed as the ongoing motion of the parts of the bicycle, i.e. the net forward movement of the bicycle and the spinning of the wheels.

There are other forms of energy that a particle can have such as gravitational potential energy, chemical potential energy, electrical potential energy, spring potential energy, et cetera, but the kinetic energy is the portion of an object's energy that is due explicitly to its ongoing motion. For a point particle of mass $\delta m$, moving at speed $v$, the kinetic energy is given by the formula $\frac12 \delta mv^2$, and the kinetic energy of any extended object can be built up from this. If a moving object collides with another object, and there is no dissipation, (no heat is given off, no chemical bonds are rearranged, etc.) then the total kinetic energy of the objects after the collision must equal the kinetic energy of the moving object before the collision.

In Isaac Newton's Principia Mathematica, momentum is defined and is said to be conserved, but there is no mention of energy anywhere in his treatise. Newton was able to work out the elliptical orbits of planets without any reference to energy at all. However, if we start with the assumption that momentum is conserved, we can derive the conservation of kinetic energy, and thus obtain the structure that lay hidden from Newton in his own theory.

Colliding balls on a table

Consider a ball of mass $m_{1}$ moving with velocity $v_{0}$ in the direction of a stationary ball of mass $m_{2}$. Let the velocities of the balls after the collision be $v_{a}$ and $v_{b} ,$ respectively. Moreover, suppose the collision is elastic, i.e. the relative velocity of approach is equal and opposite to the relative velocity of separation, i.e. $\dfrac{\Delta v_i}{\Delta v_f} = -1$.

Therefore, along with the conservation of momentum, we have

$$\begin{aligned} v_{0} &= v_{b} - v_{a} \\ \\ m_{1}v_{0} &= m_{1}v_{a} + m_{2}v_{b}. \end{aligned}$$

Solving these two equations for $v_a$ and $v_b ,$ we find

$$\begin{aligned} v_a &= \dfrac{m_1 - m_2}{m_1 + m_2}v_0 \\\\ v_b &= \dfrac{2m_1}{m_1 + m_2}v_0. \end{aligned}$$

Let us now calculate the initial and final kinetic energies $\big(\frac12 mv^2$ for each ball$\big):$

$$\begin{aligned} \textrm{KE} _{i} &= \dfrac{1}{2}m_{1}v_{0}^2 \\\\ \textrm{KE} _{f} &= \dfrac{1}{2}m_{1}v_{a}^2 + \dfrac{1}{2}m_{2}v_{b}^2. \end{aligned}$$

Using the values of $v_a$ and $v_b$ from above we find (after some tedious algebra)

$$\textrm{KE} _{initial} = \textrm{KE} _{final} .$$

Hence, the total kinetic energy is conserved in elastic collisions.

What about during the collision? Imagine we have two identical balls speeding with equal but opposite velocities to a collision. For an instant, neither ball is moving. So where did the kinetic energy of the balls go during this instant?

The collision of the balls is not instantaneous. For a short interval, the kinetic energy of the balls is stored as elastic potential energy, after which it is reallocated to restore the original kinetic energy of the balls. Elastic energy is the potential energy stored in a deformed body, as for example with the compression of an elastic spring. At all times, the combined kinetic and elastic potential energy of the balls is constant. For more information on elastic collisions, see Analyzing Elastic Collisions.

In the previous example, we saw that the conservation of energy falls out in elastic collisions. What about when an object is accelerated by a gravitational field? Without a detailed understanding of potential energy, can we identify quantities that exchange with kinetic energy? Below, we analyze an especially profitable example of a skier on a frictionless slope.

The inclined plane

Under constant acceleration, an object travels a distance $d$ in the time $t = \sqrt{\frac{2d}{a}}$. For a skier on an inclined plane, this time is given by $\sqrt{\frac{2d}{g\sin\theta}}$. Also, starting from rest at the top of the incline, the skier has speed $v = at = g\sin\theta \sqrt{\frac{2d}{g\sin\theta}} = \sqrt{2g\sin\theta d}$ at the bottom of the incline.

Consider the change in kinetic energy $\frac12 mv^2$. At the top of the incline, this quantity is equal to zero, and at the bottom of the incline it's equal to $mgd\sin\theta$. We notice that $mgd\sin\theta$ is simply the strength of the gravitational force times the vertical distance through which the skier has descended. Further, we notice that the quantity $\frac12 mv^2 + mg(h_0-h)$ is a constant throughout the course of the skier's motion, i.e. kinetic energy exchanges with the quantity $mg\Delta h$.

Evidently, by falling through the distance $\Delta h$, the skier has picked up kinetic energy $\text{KE} = mg\Delta h$. This suggests that $mg\Delta h$ is equal to the gravitational potential energy of the skier, which can reallocate to the kinetic energy, based on the skier's vertical position.

Thus, we have stumbled upon a conservation relation between kinetic energy and the gravitational potential energy. As you might imagine, in this example, the conservation is imposed by the fact that the system follows Newton's second law $F = \frac{dp}{dt}$.

In general, we'll see that energy conservation holds in countless other scenarios. Further, in all the centuries of careful experimentation, a violation has never been identified (in macroscopic systems). For this reason, the conservation of energy has been elevated to the status of a law, i.e. a principle that is expected to hold for all processes in the universe.

In plain English, work means to do something productive. In physics, work is said to be done if force acts to displace an object. For example, when you push a sled across the snow, or lift a bucket of water onto a ledge, or pump up the tire on your bicycle, you've done work. This idea is captured in the definition of work which is the applied force along the direction of displacement times the displacement. Mathematically, we can write

$$W=\int \vec F\cdot d\vec s = Fs\cos\theta,$$

where $\theta$ is the angle between the force and displacement vectors.

As you might imagine, performing work on an object cannot only change its displacement but its kinetic energy as well. We return once more to the example of the skier on the slope to see how these ideas are related.

Work on the skier

Recall our results from above regarding the skier on the slope. Now, consider the work $F\cdot s$, where $F$ is the force of gravity along the incline, and $s$ is the distance traveled along the incline. With $F = mg\sin\theta$ and $s=d$, we find $F\cdot s = mg\sin\theta d$.

This is curious. In descending the slope, we showed that the skier gained an amount of kinetic energy $mgd\sin\theta$ which is also equal to the product of the force that acted on the particle, and the distance over which the particle traveled. Moreover, if we were to pull the skier up the slope at constant velocity, he'd have gained potential energy in the amount $mgd\sin\theta$. One might be tempted to hypothesize that work delivers energy to objects.

Indeed, during his descent the gravitational field does work upon the skier to speed him toward Earth, and in pulling him up the hill, we do work against the gravitational field. Work against a field can always be recouped later and thus we say it is stored kinetic energy or, in other words, potential energy.

Mathematically, we've shown $\vec{F}\cdot\vec{d} = \Delta \textrm{KE} +\Delta \textrm{PE}= \frac12mv_f^2-\frac12mv_i^2 + mg\left(h_f-h_i\right)$.

This suggests that the cumulative pull of the skier by gravity has given rise to the kinetic energy of the particle at the expense of his initial potential energy. Thus, we can either look at the exchange as work done on the skier by the gravitational field, or as a reallocation of potential energy to kinetic energy.

This relationship would be quite useful if it holds in general, but as yet, it is just a nice coincidence we've noticed in our calculation. Next we look to Newton's laws for a way to put it on firm ground.

Newton's laws: a basis for work

According to the second law, we have the following relationship between changes in velocity and a net applied force (for simplicity we work in one dimension, though the result easily generalizes):

$$\begin{aligned}F_\textrm{net}&=ma\\ &=m\frac{\Delta v}{\Delta t}.\end{aligned}$$

Rearranging, we have

$$\begin{aligned} F_\textrm{net} \Delta t &= m \Delta v \\ &= \Delta p.\end{aligned}$$

Because $\displaystyle\Delta x = v \Delta t$, we can write $\displaystyle F_\textrm{net} \Delta x / v = m \Delta v$, or

$$F_\textrm{net} \Delta x = m v\Delta v.$$

This relation shows that if the object is traveling with velocity $v$ and it is pushed through some small distance $\Delta x$ parallel to the force $\displaystyle F_\textrm{net}$, it will pick up the additional velocity $\displaystyle \Delta v = \frac{F_\textrm{net}\Delta x}{mv}$.

In three dimensions, our result is $\displaystyle \vec{F}_\textrm{net}\cdot\Delta\vec{x} = m\vec{v}\cdot\Delta\vec{v}$.

This relation provides the basis for what we suspected above, that forces can do work to endow particles with kinetic energy. We can now exploit the relation to prove the work-kinetic energy theorem.

Relation between work and kinetic energy

Work-Kinetic Energy Theorem

The net work (i.e. work minus work performed against any external fields) done by the force on an object is equal to its change in kinetic energy:

$$\Delta \text{KE} = \int \vec F \cdot d\vec s.$$

Now think about the ways in which you can prove this theorem before you reveal the proof, try proving it yourself, and don't worry if you weren't able to crack it because most of the people can't do it in a single go.

Proof 1:

If we add up all of the incremental pushes $F_\textrm{net}\Delta s$ that the particle receives over the distance $d$, we get $F_\textrm{net}\sum \Delta s = F_\textrm{net}d$. However, we showed that $F_\textrm{net}\Delta s = mv\Delta v$, so we have $F_\textrm{net}d = m\sum v\Delta v$, a sum over the incremental increases in velocity.

Let us now perform the sum of the $mv\Delta v$. To start, when $v$ is zero, $mv$ is zero; at the end, it is equal to $mv_f$. If we divide the increases up into $n$ equal pieces, the velocity increases $\Delta v$ are each given by $\frac{1}{n}v_f$, which we can pull out of the sum so that the sum becomes $\dfrac{v_f}{n}\sum v$. Suppose we divide the velocity increase over many moments so that $n$ is large and the changes are small.

Now, the sum of $v$ from $0$ to $v_f$ in $n$ equally sized chunks is simply $v$ times the average value of $v$ over the range:

$$\sum\limits_{v=0}^{v_f} v= n\frac{v_f}{2}.$$

Therefore, $\sum mv\Delta v$ is equal to

$$\dfrac{mv_f}{n}\sum\limits_{v=0}^{v_f}\Delta v = \frac{mv_f}{n}n\dfrac{v_f}{2} = \dfrac12 mv_f^2,$$

which implies $F_\textrm{net}d=\frac12 mv_f^2.$

This proves that if we act on an object of mass $m$ with a force $F_\textrm{net}$ over a distance $d,$ it ends up with a kinetic energy $\frac12 mv_f^2$, where the velocity $v_f$ is given by $\sqrt{2F_\textrm{net}d/m}$. $_\square$

$$$$ Proof 2:

We know

$$W = \int \vec F \cdot d\vec s.$$

Since $\vec F =m \vec a=m\dfrac{d\vec v}{dt},$ it follows that

$$\begin{aligned} W &= \int \vec F \cdot d\vec s \\ &=\int m\dfrac{d\vec v}{dt}\cdot d\vec s =\int\limits_{\vec {v_0}}^{\vec{v_f}}m\dfrac{d\vec s}{dt} d\vec v \\ &=m\int\limits_{\vec {v_0}}^{\vec{v_f}} \vec v \cdot d\vec v =\Delta \textrm{KE}, \end{aligned}$$

which proves $\Delta \textrm{KE} = W. \ _\square$

Thus, a force acting on an object through a distance transfers energy to the object, and we call this quantity, $W_\textrm{net} = \displaystyle\int \vec{F}\cdot d\vec{s}$, the work. In a frictionless system, the work is equal to the change in energy caused by the force:

$$W_\textrm{net} = \Delta \textrm{KE} + \Delta \textrm{PE}.$$

More generally, the work is equal to the change in energy plus whatever heat is dissipated to the environment.

Potential energy is a form of energy due to the configuration or position of an object in a field. It can be seen as a stored form on kinetic energy in that it can be converted to kinetic energy by allowing an object to relax its position or configuration in a field. For example, if we release an electric dipole previously restrained at an angle in an electric field, or drop a massive object in a gravitational field, potential energy is converted to movement through relaxation.

Dams contain massive amount of water stored at a considerable height. When this water is released, the gravitational potential energy is converted to kinetic energy and the water flows, which in turn spins the turbines to generate electricity which can be stored as electric potential energy in batteries. Thus the water which was stored at a height is used to generate electricity in a time of scarcity. Thus kinetic energy can be captured and stored in various forms of potential energy to be extracted later.

Gottfried Wilhelm Leibniz in 1676-1689 was the first to attempt a definition of kinetic energy, and noticed that for some mechanical systems it seemed to be conserved. But remarkably, it was a French mathematician, Emilie du Chatelet, who was the first to propose and derive the conservation of total kinetic and potential energy. She was aware of Isaac Newton's Principia Mathematica and undertook its translation into French, which she completed in 1749. During those years of translation, she included her derivation of the conservation law as a supplement to Newton's treatise. Soon afterwards, mathematicians Leonhard Euler and Joseph-Louis Lagrange went on to develop a mathematical formalism of classical mechanics, based on her works.

The Ball

In this example, we see the interplay of kinetic and potential energies of a falling ball such that their sum is constant, when air resistance is ignored.

Consider a ball of mass $m$ falling from rest from a height $H$ above the ground. Initially, since the ball is at rest its kinetic energy and potential energy, respectively, are

$$\begin{array}{c}&\textrm{KE}_i = 0, &\textrm{PE}_i = mgH. \end{array}$$

Let $v$ be the speed of the ball when it is at height $h$ above the ground. From the second kinematic equation we have $v^2 = 2g(H-h)$. Hence, the kinetic energy of the ball at height $h$ is

$$\textrm{KE}_h = \dfrac{1}{2}mv^2 = mg(H-h).$$

From the expression of potential energy, the potential energy at height $h$ is

$$\textrm{PE}_h = mgh.$$

Thus, the sum of energies is $\textrm{KE}_h + \textrm{PE}_h = mgH$, which is the same as the initial sum of energies.

Spring With Oscillating Mass

When an ideal spring is either compressed or stretched a distance $x$ from the origin where the spring force $=0$, the potential energy stored in the spring with a spring constant $k$ $($which comes from using force to compress the spring against the spring force $F=kx)$ is

$$PE = \dfrac{1}{2}kx^2,$$

while kinetic energy of the oscillating mass $m$ moving at velocity $v$ is

$$KE = \dfrac{1}{2}mv^2,$$

the sum of which is the total energy $E$. $\;$ Since total energy is conserved, $E$ is a constant over time, so differentiating both sides by time $t$, we have

$$\dfrac { d }{ dt } E=0 =\dfrac {dx}{dt} \dfrac { d }{ dx } \left( \dfrac { 1 }{ 2 } k{ x }^{ 2 } \right) +\dfrac {dv}{dt} \dfrac { d }{ dv } \left( \dfrac { 1 }{ 2 } m{ v }^{ 2 } \right) \implies vkx+amv=0.$$

Simplifying, we have

$$-kx = ma = F,$$

which is Hooke's Law stating that the spring force exerted on mass $m$ at any point is equal to $-kx$, a linear function of the distance from the origin where the force $= 0,$ in the direction towards the origin. This can be restated in terms of time $t$

$$\dfrac { m }{ k } \dfrac { { d }^{ 2 }x }{ d{ t }^{ 2 } } +x =0,$$

a $2^{\text{nd}}$ order differential equation which has the solution, where $b$ is a constant:

$$x \left( t \right) =b\sin\left( \sqrt { \dfrac { k }{m } }\, t \right).$$

Thus, the spring with oscillating mass has a period $T,$ which is

$$T=2\pi \sqrt { \dfrac { m }{ k } }.$$

The pendulum

When a pendulum of massless arm length $L$ with mass $m$ at the end of it is at angle $\theta$ from vertical, the gravitational potential energy is

$$PE=mgh=mgL(1-\cos\theta),$$

while the rotational kinetic energy of the mass $m$ is

$$KE=\dfrac { 1 }{ 2 } I{ { \omega }^{ 2 } }=\dfrac { 1 }{ 2 } m{ L }^{ 2 }{ \omega }^{ 2 },$$

the sum of which is the total energy $E$. $\;$ Since total energy is conserved, $E$ is a constant over time, so differentiating both sides with respect to time $t$, we have

$$\begin{aligned} \dfrac { d }{ dt } E&=0\\ \dfrac { d\theta }{ dt } \dfrac { d }{ d\theta } [mgL( 1-\cos\theta)] +\dfrac { d\omega }{ dt } \dfrac { d }{ d\omega } \left( \dfrac { 1 }{ 2 } m{ L }^{ 2 }{ \omega }^{ 2 } \right)&=0 \\ \dfrac { d\theta }{ dt } mgL\sin\theta +\dfrac { d\omega }{ dt } m{ L }^{ 2 }\omega &=0\\ \frac { d\theta }{ dt } mgL\sin\theta +\frac { { d }^{ 2 }\theta }{ d{ t }^{ 2 } } m{ L }^{ 2 }\frac { d\theta }{ dt }&=0 \\ \dfrac { L }{ g } \dfrac { { d }^{ 2 }\theta }{ d{ t }^{ 2 } } +\sin\theta &=0. \end{aligned}$$

As an approximation for small $\theta$, we have

$$\dfrac { L }{ g } \dfrac { { d }^{ 2 }\theta }{ d{ t }^{ 2 } } +\theta =0,$$

a $2^{\text{nd}}$ order differential equation which has the solution, where $b$ is a constant:

$$\theta \left( t \right) =b\sin\left( \sqrt { \dfrac { g }{ L } }\, t \right).$$

Thus, the simple pendulum, for small $\theta$, has a period $T$ which is dependent only on $L$ and $g:$

$$T=2\pi \sqrt { \dfrac { L }{ g } }.$$

Or more generally, for small $\theta$ but with a rotational inertia of $I$

$$T=2\pi \sqrt { \dfrac { I }{ mgL } } ,$$

where $L$ is the distance from the pivot to the center of mass of the pendulum.

Compare the equations of both spring oscillation and simple pendulum, and see how there's an analogy of Hooke's Law in the latter.

Throughout the first half of the $19^\text{th}$ century, engineers realized that thermal heat was another form of energy, capable of mechanical work. By 1850, the law of conservation of energy was formally stated for the first time, as the first law of thermodynamics. In the last decades of the $19^\text{th}$ century, Ludwig Boltzmann developed the statistical mechanics theory of heat and entropy, showing the equivalence of kinetic and thermal heat energies, based on the theory that matter was made of atoms. Over time, other forms of energy were recognized such as elastic, electromagnetic, chemical, and nuclear energies, and the law of conservation of energy was generalized to include all such forms of energies.

Firing an Inelastic Shell

A tank fires a special hollow shell that has $N$ neon atoms inside it, so cold (somewhere near absolute zero Kelvin) that relative to the shell the atoms are not moving. The shell has a velocity of $680$ meters per second, and it slams into a dirt wall and suffers a 100% inelastic collision, where the shell comes to a dead stop. But the neon atoms continue to bounce around inside the shell at the same velocity. What is the temperature of the neon gas inside the shell, after the shell has come to a dead stop?

Treat neon as an ideal gas, i.e. point-like particles with mass, and disregard the mass or size of the hollow shell. Neon atoms are $20.1797\text{ amu},$ or $\frac { 20.1797 }{ \text{mole} }$ grams, where $1\text{ mole} = 6.022\times { 10 }^{ 23 }$.

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The kinetic energy of $N$ neon atoms travelling at $680$ meters per second is

$$N\frac { 1 }{ 2 } m{ v }^{ 2 }=N\frac { 1 }{ 2 } \left( \frac { 20.1797 }{ 6.022\times { 10 }^{ 23 } } \frac { 1 }{ { 10 }^{ 3 } } \right) { \left( 680 \right) }^{ 2 }\text{ Joules}=N\left( 7.747\times { 10 }^{ -21 } \right)\text{ Joules}.$$

Since total energy is conserved, the kinetic energy of the neon atoms has been converted into thermal heat energy. From statistical mechanics of ideal gases, we then have

$$N\left( 7.747\times { 10 }^{ -23 } \right)\text{ Joules}=N\frac { 3 }{ 2 } kT \text{ Joules},$$

where $k=1.381\times { 10 }^{ -23 }\text{ Joules/Kelvin}$ is the Boltzmann Constant, and $T$ is temperature in $\text{Kelvin}$. From this, we have

$$\left( 7.757\times { 10 }^{ -21 } \right) \frac { 2 }{ 3k } =374.088\text{ Kelvin},$$

which is just above the boiling point of water.

Note that this result does not depend on the number $N$ of neon atoms so that there could be a gram of neon in the shell, or a single atom of neon. The temperature of an ideal gas depends solely on the mass of the atom and the average velocity of the atoms, is not a form of energy, and does not have the physical dimensions of energy, which thermal heat energy does.

The Rocket

Rocket physics describes one kind of rocket that uses fuel as a stored form of chemical energy, which is released in the combustion chamber of the rocket to generate the thrust to propel it. As the fuel is used up, the mass of the rocket drops, while its velocity increases. Using the law of conservation of momentum, velocity of a rocket as a function of mass describes how the following rocket equation can be derived:

$$\begin{aligned} v_f &= v_0 + u\log \frac{M_0}{M_f} , \end{aligned}$$

where $v_f$ and $M_f$ are the final velocity and mass, respectively, $v_0$ and $M_0$ are the initial velocity and mass, respectively, and $u$ is the velocity of the rocket exhaust relative to the rocket.

$$\text{Indian Polar Satellite Launch Vehicle XL which uses HTPB propellant}$$

We can use the law of conservation of energy to derive the same result as follows:

Let $M_0-m$ be the rocket’s mass, where $m$ is the consumed mass of fuel. Also, let $v(m)$ be the velocity of the rocket as a function of $m$, and $u$ the velocity of the exhaust relative to the rocket. Then we have the following as the total energy of the rocket/exhaust gas system:

$$E=\displaystyle \int _{ 0 }^{ m }{ \dfrac { 1 }{ 2 } } \big[ u-v\left( m \right) \big]^2\, dm+\dfrac { 1 }{ 2 } \left( { M }_{ 0 }-m \right) \big[v(m)\big]^2+{ PE }_{ \text{Fuel} }\left( { M }_{ 0 }-m \right),$$

where the first term is the kinetic energy of the rocket exhaust gas, the second term is the kinetic energy of the rocket, and the last term ${ PE }_{\text{Fuel} }\left( { M }_{ 0 }-m \right)$ is the potential energy contained in the unburned fuel. The kinetic potential energy of this fuel when it is converted into exhaust gas is simply the kinetic energy it has at velocity $u$, so we can rewrite this equation as

$$E=\displaystyle \int _{ 0 }^{ m }{ \dfrac { 1 }{ 2 } } \big[u-v(m)\big]^2 \, dm+\dfrac { 1 }{ 2 } \left( { M }_{ 0 }-m \right) \big[v(m)\big]^2+\dfrac { 1 }{ 2 } \left( { M }_{ 0 }-m \right) { u }^{ 2 }.$$

Since total energy is conserved, we can differentiate both sides by $m$ with the result

$$\begin{aligned} \dfrac { dE }{ dm } &=0\\\\ \dfrac { 1 }{ 2 }\big[u-v(m)\big]^2-\dfrac { 1 }{ 2 }\big[v(m)\big]^2-\left( {M}_ {0}-m \right) v\left( m \right) \dfrac { dv }{ dm } -\dfrac { 1 }{ 2 } { u }^{ 2 }&=0\\\\ -u-\left( {M}_ {0}-m \right) { \dfrac { dv }{ dm } }&=0\\\\ -\int _{ 0 }^{ m }{ \frac { u }{ {M}_ {0}-m } } \, dm&=\int dv \\\\ u\log\left( \dfrac { {M}_ {0 }}{ {M}_ {0}-m } \right) &={v}_{m}-{v}_{0}, \end{aligned}$$

which agrees with the rocket equation given above. This equation is the most simplified form of rocket physics, which assumes that all the thrust is provided by the exhaust velocity of the combustion gases alone. A real rocket involves other considerations such as combustion gas pressure inside the nozzle.

Mass Energy in Nuclear Reactions

A Deuterium atom, $_1 ^2\ce{H}$, is an isotope of hydrogen that has a nucleus containing one proton and one neutron. Two of them, if collided together at a sufficiently high speed to overcome the Coulomb barrier, can fuse and produce a helium-3 atom $\left(_2 ^3\ce{He}\right)$ and a free neutron $\left(_0 ^1\ce{n}\right)$. $_2 ^3\ce{He}$ is an isotope of helium that has two protons but only one neutron. Excess energy is produced from this fusion, which is carried away as kinetic energy in the products.

In atomic mass units $(\text{amu})$, the masses of the nuclei are

$$\begin{array}{llll} _1 ^2\ce{H}&=&2.01410179 &\text{ (amu)}\\ _2 ^3\ce{He}&=&3.0160293 &\text{ (amu)}\\ _0 ^1\ce{n}&=&1.008664916 &\text{ (amu)}. \end{array}$$

Per Einstein’s $E=m{ c }^{ 2 }$, mass-to-energy conversion from $\text{amu}$ to million electron Volts is

$$1\text{ amu} = 941.494\text{ MeV}.$$

Now, calculate the energy $E$ in $\text{MeV}$, produced as the result of the following Deuterium+Deuterium fusion reaction:

$$_{ 1 } ^{ 2 }\ce{H} + \ _{ 1 } ^{ 2 }\ce{H} \to \ _{ 2 }^{ 3 }\ce{He}+ \ _0^1\ce{n}+{ E }_{ \text{MeV} }.$$

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Two deuterium nuclei have a total mass of $4.02820358\text{ amu},$ but the total mass of $_2 ^3\ce{He}$ and $_0 ^1\ce{n}$ is $4.024694216\text{ amu},$ which is less. The theory of relativity says mass is another form of energy, so since total energy is conserved that missing $0.00350936\text{ amu}$ mass was converted into energy. Using the mass-to-$\text{MeV}$ conversion, this works out to about $E=3.26895\text{ MeV}.$

Note: $\text{eV}$, or electron volt, is not to be confused with $\text{V}$, or volt. $\text{eV}$ has dimensions of energy, or joules, while $\text{V}$ is joules divided by charge.

If the energy produced is carried away by the products, and if both kinetic energy and momentum are conserved, what is the kinetic energy of the neutron after this fusion reaction? (Disregard the initial kinetic energies of the reactant Deuterium nuclei.)

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The conservation equations are

$$\begin{aligned} \dfrac { 1 }{ 2 } m_{(^3\ce{He})} v_{(^3\ce{He})}^2+\dfrac { 1 }{ 2 } m_n v_n^2&=E\\ { m }_{ (^3\ce{He}) }{ { v }_{ (^3\ce{He}) } }+{ m }_{ n }{ { v }_{ n } }&=0. \end{aligned}$$

Letting ${ E }_{(^3\ce{He} ) }$ and ${ E }_{ n }$ be the kinetic energies of the products resulting from the fusion, we can restate the two equations from both conservation laws as follows:

$$\begin{aligned} { E }_{ (^3\ce{He}) }+{ E }_{ n }&=E\\ { E }_{(^3\ce{He})}{ m }_{ (^3\ce{He}) }&={ E }_{ n }{ m }_{ n }. \end{aligned}$$

From these, we can work out the kinetic energy of the neutron

$${ E }_{ n }=\dfrac { { m }_{ (^3\ce{He}) } }{ { m }_{ (^3\ce{He}) }+{ m }_{ n } } E,$$

which works out to approximately $2.44969\text{ MeV}$ or $2.450\text{ MeV}$, which is the usual published value.

ATP synthase, a molecular motor

Spring Pendulum: A Preview of Lagrangian Mechanics

Examples given above included a pendulum in a gravitational field, and an oscillating spring-mass system. What if we were to combine them? A massless pendulum arm of relaxed length $L$ is capable of stretching or compressing lengthwise, and a mass $m$ is attached to the end of it. Then the position of the mass can be described by two functions to be determined:

$$\begin{aligned} \text{Distance from pivot point } &=L+x(t) \\ \text{Angle from vertical } &=\theta(t). \end{aligned}$$

The total kinetic energy is then the sum of radial and tangential components

$$T=\dfrac { 1 }{ 2 } m { \left( \dfrac { dx }{ dt } \right) }^{ 2 }+\dfrac { 1 }{ 2 } m { \left( (L+x)\dfrac { d\theta }{ dt } \right) }^{ 2 }$$

and the total potential energy is the sum of gravitational and spring potentials

$$V=V\left(x,\theta\right)=-mg(L+x)\cos \theta +\dfrac { 1 }{ 2 } k{ x }^{ 2 },$$

where $g$ is the gravitational acceleration and $k$ is the spring constant.

There are two approaches to classical mechanics (see Lagrangian formulation of mechanics), which can be briefly summarized as follows:

$$\begin{array}{rl} \text{Hamiltonian:} &\mathcal{H} = T+V\\ \text{Lagrangian:} &\mathcal{L}=T-V, \end{array}$$

which can be shown to be equivalent through the Legendre transform. In this exercise, we express this spring pendulum by its Lagrangian as follows:

$$\mathcal{L}=T-V= \dfrac { 1 }{ 2 } m{ { \dot { x } }^{ 2 }+ \dfrac { 1 }{ 2 } m { \left( (L+x) \dot {\theta } \right) }^{ 2 } }+mg(L+x)\cos \theta -\dfrac { 1 }{ 2 } k{ x }^{ 2 }.$$

The two Euler-Lagrange equations for these are then

$$\begin{aligned} \dfrac { d }{ dt } \left( \dfrac { \partial \mathcal {L} }{ \partial \dot { x } } \right) &=\dfrac { \partial \mathcal {L} }{ \partial x } \implies m\ddot { x } =m(L+x){ \dot { \theta } }^{ 2 }+mg\cos \theta -kx\\ \dfrac { d }{ dt } \left( \dfrac { \partial \mathcal {L} }{ \partial \dot { \theta } } \right) &=\dfrac { \partial \mathcal {L} }{ \partial \theta } \implies m(L+x)\ddot { \theta } +2m\dot { x } \dot { \theta } =-mg\sin\theta. \end{aligned}$$

The first is the radial force along the length of the pendulum arm while the second is the tangential force. The complete solution to this pair of differential equations is beyond the scope here, but this example is given to illustrate the generalized utility of treating problems of classical mechanics through the use of kinetic and potential energies, rather than the use of forces and momenta.

The LC Circuit: Via the Lagrangian Approach

The classic LC circuit, in the simplest form, consists of an inductor and a capacitor, both of which are capable of storing energy: one electromagnetic and the other electrostatic. This circuit can be analyzed using Kirchhoff’s laws, which is the conventional approach. However, to demonstrate the power of the Lagrangian method, which treats energy as a generalized abstract quantity, the following is a way how the LC circuit can be analyzed in terms of energies alone:

Let $q\left(t\right)$ be the charge as a function of time $t$. Also, let $L$ be the inductance of the inductor, and $C$ the capacitance of the capacitor. Then we can make the analogy of kinetic and potential energies as follows:

$$\begin{aligned} T&=\dfrac { 1 }{ 2 } L{ \left( \dfrac { dq }{ dt } \right) }^{ 2 }\\ V&=\dfrac { 1 }{ 2C } { q }^{ 2 }. \end{aligned}$$

The Lagrangian is then

$$\mathcal{L}=\mathcal{L}\left( \dot { q } ,q \right) =T-V=\dfrac { 1 }{ 2 } L{ \dot { q } }^{ 2 }-\dfrac { 1 }{ 2C } { q }^{ 2 }.$$

The Euler-Lagrange equation is then

$$\dfrac { d }{ dt } \left( \dfrac { \mathcal{L} }{ \partial \dot { q } } \right) -\dfrac { \mathcal{L} }{ \partial q } =0\implies \ddot { q } +\dfrac { q }{ LC } =0,$$

which yields the same solution as can be found by using Kirchhoff’s laws

$$q\left( t \right) ={ q }_{ 0 }\cos\left( \omega t \right)$$

given that $q\left(0\right)=0, \dot{q}\left(0\right)=0$ and $\omega =\frac { 1 }{ \sqrt { LC } } .$