Kinetic Energy: July 2011

Thursday, 14 July 2011

Kinetic energy

The cars of a roller coaster reach their maximum kinetic energy when at the bottom of their path. When they start rising, the kinetic energy begins to be converted to gravitational potential energy. The sum of kinetic and potential energy in the system remains constant, ignoring losses to friction.

The kinetic energy of an object is the energy which it possesses due to its motion.^[1] It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest.
The speed, and thus the kinetic energy of a single object is frame-dependent (relative): it can take any non-negative value, by choosing a suitable inertial frame of reference. For example, a bullet passing an observer has kinetic energy in the reference frame of this observer, but the same bullet is stationary, and so has zero kinetic energy, from the point of view of an observer moving with the same velocity as the bullet.^[2] By contrast, the total kinetic energy of a system of objects cannot be reduced to zero by a suitable choice of the inertial reference frame, unless all the objects have the same velocity. In any other case the total kinetic energy has a non-zero minimum, as no inertial reference frame can be chosen in which all the objects are stationary. This minimum kinetic energy contributes to the system's invariant mass, which is independent of the reference frame.
In classical mechanics, the kinetic energy of a non-rotating object of mass m traveling at a speed v is mv²/2. In relativistic mechanics, this is only a good approximation when v is much less than the speed of light.

History and etymology

The adjective kinetic has its roots in the Greek word κίνησις (kinesis) meaning motion, which is the same root as in the word cinema, referring to motion pictures.

The principle in classical mechanics that E ∝ mv² was first developed by Gottfried Leibniz and Johann Bernoulli, who described kinetic energy as the living force, vis viva. Willem 's Gravesande of the Netherlands provided experimental evidence of this relationship. By dropping weights from different heights into a block of clay, 's Gravesande determined that their penetration depth was proportional to the square of their impact speed. Émilie du Châtelet recognized the implications of the experiment and published an explanation.^[3]
The terms kinetic energy and work in their present scientific meanings date back to the mid-19th century. Early understandings of these ideas can be attributed to Gaspard-Gustave Coriolis, who in 1829 published the paper titled Du Calcul de l'Effet des Machines outlining the mathematics of kinetic energy. William Thomson, later Lord Kelvin, is given the credit for coining the term "kinetic energy" c. 1849–51

Introduction

Energy occurs in many forms, including chemical energy, thermal energy, electromagnetic radiation, gravitational energy, electric energy, elastic energy, nuclear energy, rest energy. These can be categorized in two main classes: potential energy and kinetic energy.
Kinetic energy may be best understood by examples that demonstrate how it is transformed to and from other forms of energy. For example, a cyclist uses chemical energy provided by food to accelerate a bicycle to a chosen speed. On a level surface, this speed can be maintained without further work, except to overcome air resistance and friction. The chemical energy has been converted into kinetic energy, the energy of motion, but the process is not completely efficient and produces heat within the cyclist.
The kinetic energy in the moving cyclist and the bicycle can be converted to other forms. For example, the cyclist could encounter a hill just high enough to coast up, so that the bicycle comes to a complete halt at the top. The kinetic energy has now largely been converted to gravitational potential energy that can be released by freewheeling down the other side of the hill. Since the bicycle lost some of its energy to friction, it never regains all of its speed without additional pedaling. The energy is not destroyed; it has only been converted to another form by friction. Alternatively the cyclist could connect a dynamo to one of the wheels and generate some electrical energy on the descent. The bicycle would be traveling slower at the bottom of the hill than without the generator because some of the energy has been diverted into electrical energy. Another possibility would be for the cyclist to apply the brakes, in which case the kinetic energy would be dissipated through friction as heat.

Newtonian kinetic energy

Kinetic energy of rigid bodies

In classical mechanics, the kinetic energy of a point object (an object so small that its mass can be assumed to exist at one point), or a non-rotating rigid body, is given by the equation

$E_k =\tfrac{1}{2} mv^2$

where

m

is the mass and

v

is the speed (or the velocity) of the body. In SI units (used for most modern scientific work), mass is measured in kilograms, speed in metres per second, and the resulting kinetic energy is in joules.
For example, one would calculate the kinetic energy of an 80 kg mass (about 180 lbs) traveling at 18 metres per second (about 40 mph, or 65 km/h) as

E_k = (1/2) · 80 · 18² J = 12.96 kJ

Since the kinetic energy increases with the square of the speed, an object doubling its speed has four times as much kinetic energy. For example, a car traveling twice as fast as another requires four times as much distance to stop, assuming a constant braking force.
The kinetic energy of an object is related to its momentum by the equation:

$E_k = \frac{p^2}{2m}$

where:

$p\;$ is momentum

$m\;$ is mass of the body

For the translational kinetic energy, that is the kinetic energy associated with rectilinear motion, of a body with constant mass $m\;$ , whose center of mass is moving in a straight line with speed $v\;$ , as seen above is equal to

$E_t =\tfrac{1}{2} mv^2$

where:

$m\;$ is the mass of the body

$v\;$ is the speed of the center of mass of the body.

The kinetic energy of any entity depends on the reference frame in which it is measured. However the total energy of an isolated system, i.e. one which energy can neither enter nor leave, does not change in whatever reference frame it is measured. Thus, the chemical energy converted to kinetic energy by a rocket engine is divided differently between the rocket ship and its exhaust stream depending upon the chosen reference frame. This is called the Oberth effect. But the total energy of the system, including kinetic energy, fuel chemical energy, heat, etc., is conserved over time, regardless of the choice of reference frame. Different observers moving with different reference frames disagree on the value of this conserved energy.

Quantum mechanical kinetic energy of rigid bodies

In the realm of quantum mechanics, the expectation value of the electron kinetic energy, $\langle\hat{T}\rangle$ , for a system of electrons described by the wavefunction $\vert\psi\rangle$ is a sum of 1-electron operator expectation values:

$\langle\hat{T}\rangle = -\frac{\hbar^2}{2 m_e}\bigg\langle\psi \bigg\vert \sum_{i=1}^N \nabla^2_i \bigg\vert \psi \bigg\rangle$

where

m e

is the mass of the electron and $\nabla^2_i$ is the Laplacian operator acting upon the coordinates of the i^th electron and the summation runs over all electrons. Notice that this is the quantized version of the non-relativistic expression for kinetic energy in terms of momentum:

$E_k = \frac{p^2}{2m}.$

The density functional formalism of quantum mechanics requires knowledge of the electron density only, i.e., it formally does not require knowledge of the wavefunction. Given an electron density $\rho(\mathbf{r})$ , the exact N-electron kinetic energy functional is unknown; however, for the specific case of a 1-electron system, the kinetic energy can be written as

$T[\rho] = \frac{1}{8} \int \frac{ \nabla \rho(\mathbf{r}) \cdot \nabla \rho(\mathbf{r}) }{ \rho(\mathbf{r}) } d^3r$

where

T [ρ]

is known as the von Weizsäcker kinetic energy functional.

Relativistic kinetic energy of rigid bodies

In special relativity, we must change the expression for linear momentum.
Using m for rest mass, v and v for the object's velocity and speed respectively, and c for the speed of light in vacuum, we assume for linear momentum that $\mathbf{p}=m\gamma \mathbf{v}$ , where $\gamma = 1/\sqrt{1-v^2/c^2}$ .
Integrating by parts gives

$E_k = \int \mathbf{v} \cdot d \mathbf{p}= \int \mathbf{v} \cdot d (m \gamma \mathbf{v}) = m \gamma \mathbf{v} \cdot \mathbf{v} - \int m \gamma \mathbf{v} \cdot d \mathbf{v} = m \gamma v^2 - \frac{m}{2} \int \gamma d (v^2)$

Remembering that $\gamma = (1 - v^2/c^2)^{-1/2}\!$ , we get:

$\begin{align} E_k &= m \gamma v^2 - \frac{- m c^2}{2} \int \gamma d (1 - v^2/c^2) \\ &= m \gamma v^2 + m c^2 (1 - v^2/c^2)^{1/2} - E_0 \end{align}$

where E₀ serves as an integration constant. Thus:

$\begin{align} E_k &= m \gamma (v^2 + c^2 (1 - v^2/c^2)) - E_0 \\ &= m \gamma (v^2 + c^2 - v^2) - E_0 \\ &= m \gamma c^2 - E_0 \end{align}$

The constant of integration E₀ is found by observing that, when $\mathbf{v }= 0 , \ \gamma = 1\!$ and $E_k = 0 \!$ , giving

$E_0 = m c^2 \,$

and giving the usual formula:

$E_k = m \gamma c^2 - m c^2 = \frac{m c^2}{\sqrt{1 - v^2/c^2}} - m c^2$

If a body's speed is a significant fraction of the speed of light, it is necessary to use relativistic mechanics (the theory of relativity as developed by Albert Einstein) to calculate its kinetic energy.
For a relativistic object the momentum p is equal to:

$p = \frac{m v}{\sqrt{1 - (v/c)^2}}$ .

Thus the work expended accelerating an object from rest to a relativistic speed is:

$E_k = \frac{m c^2}{\sqrt{1 - (v/c)^2}} - m c^2$ .

The equation shows that the energy of an object approaches infinity as the velocity v approaches the speed of light c, thus it is impossible to accelerate an object across this boundary.
The mathematical by-product of this calculation is the mass-energy equivalence formula—the body at rest must have energy content equal to:

$E_\text{rest} = E_0 = m c^2 \!$

At a low speed (v<<c), the relativistic kinetic energy may be approximated well by the classical kinetic energy. This is done by binomial approximation. Indeed, taking Taylor expansion for the reciprocal square root and keeping first two terms we get:

$E_k \approx m c^2 \left(1 + \frac{1}{2} v^2/c^2\right) - m c^2 = \frac{1}{2} m v^2$