# 28.6 Relativistic energy  (Page 4/12)

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## Kinetic energy and the ultimate speed limit

Kinetic energy is energy of motion. Classically, kinetic energy has the familiar expression $\frac{1}{2}{\mathrm{mv}}^{2}$ . The relativistic expression for kinetic energy is obtained from the work-energy theorem. This theorem states that the net work on a system goes into kinetic energy. If our system starts from rest, then the work-energy theorem is

${W}_{\text{net}}=\text{KE}.$

Relativistically, at rest we have rest energy ${E}_{0}={\mathrm{mc}}^{2}$ . The work increases this to the total energy $E={\mathrm{\gamma mc}}^{2}$ . Thus,

${W}_{\text{net}}=E-{E}_{0}=\gamma {\mathrm{mc}}^{2}-{\mathrm{mc}}^{2}=\left(\gamma -1\right){\mathrm{mc}}^{2}.$

Relativistically, we have ${W}_{\text{net}}={\text{KE}}_{\text{rel}}$ .

## Relativistic kinetic energy

Relativistic kinetic energy is

${\text{KE}}_{\text{rel}}=\left(\gamma -1\right){\mathrm{mc}}^{2}.$

When motionless, we have $v=0$ and

$\gamma =\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}=1,$

so that ${\text{KE}}_{\text{rel}}=0$ at rest, as expected. But the expression for relativistic kinetic energy (such as total energy and rest energy) does not look much like the classical $\frac{1}{2}{\mathrm{mv}}^{2}$ . To show that the classical expression for kinetic energy is obtained at low velocities, we note that the binomial expansion for $\gamma$ at low velocities gives

$\gamma =1+\frac{1}{2}\frac{{v}^{2}}{{c}^{2}}.$

A binomial expansion is a way of expressing an algebraic quantity as a sum of an infinite series of terms. In some cases, as in the limit of small velocity here, most terms are very small. Thus the expression derived for $\gamma$ here is not exact, but it is a very accurate approximation. Thus, at low velocities,

$\gamma -1=\frac{1}{2}\frac{{v}^{2}}{{c}^{2}}.$

Entering this into the expression for relativistic kinetic energy gives

${\text{KE}}_{\text{rel}}=\left[\frac{1}{2}\frac{{v}^{2}}{{c}^{2}}\right]{\mathrm{mc}}^{2}=\frac{1}{2}{\mathrm{mv}}^{2}={\text{KE}}_{\text{class}}.$

So, in fact, relativistic kinetic energy does become the same as classical kinetic energy when $v\text{<<}c$ .

It is even more interesting to investigate what happens to kinetic energy when the velocity of an object approaches the speed of light. We know that $\gamma$ becomes infinite as $v$ approaches $c$ , so that KE rel also becomes infinite as the velocity approaches the speed of light. (See [link] .) An infinite amount of work (and, hence, an infinite amount of energy input) is required to accelerate a mass to the speed of light.

## The speed of light

No object with mass can attain the speed of light.

So the speed of light is the ultimate speed limit for any particle having mass. All of this is consistent with the fact that velocities less than $c$ always add to less than $c$ . Both the relativistic form for kinetic energy and the ultimate speed limit being $c$ have been confirmed in detail in numerous experiments. No matter how much energy is put into accelerating a mass, its velocity can only approach—not reach—the speed of light.

## Comparing kinetic energy: relativistic energy versus classical kinetic energy

An electron has a velocity $v=0\text{.}\text{990}c$ . (a) Calculate the kinetic energy in MeV of the electron. (b) Compare this with the classical value for kinetic energy at this velocity. (The mass of an electron is $9\text{.}\text{11}×{\text{10}}^{-\text{31}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ .)

Determine the total force and the absolute pressure on the bottom of a swimming pool 28.0m by 8.5m whose uniform depth is 1 .8m.
for the answer to complete, the units need specified why
That's just how the AP grades. Otherwise, you could be talking about m/s when the answer requires m/s^2. They need to know what you are referring to.
Kyle
Suppose a speck of dust in an electrostatic precipitator has 1.0000×1012 protons in it and has a net charge of –5.00 nC (a very large charge for a small speck). How many electrons does it have?
how would I work this problem
Alexia
how can you have not an integer number of protons? If, on the other hand it supposed to be 1e12, then 1.6e-19C/proton • 1e12 protons=1.6e-7 C is the charge of the protons in the speck, so the difference between this and 5e-9C is made up by electrons
Igor
what is angular velocity
angular velocity can be defined as the rate of change in radian over seconds.
Fidelis
Why does earth exert only a tiny downward pull?
hello
Islam
Why is light bright?
an 8.0 capacitor is connected by to the terminals of 60Hz whoes rms voltage is 150v. a.find the capacity reactance and rms to the circuit
thanks so much. i undersooth well
what is physics
is the study of matter in relation to energy
Kintu
physics can be defined as the natural science that deals with the study of motion through space,time along with its related concepts which are energy and force
Fidelis
a submersible pump is dropped a borehole and hits the level of water at the bottom of the borehole 5 seconds later.determine the level of water in the borehole
what is power?
power P = Work done per second W/ t. It means the more power, the stronger machine
Sphere
e.g. heart Uses 2 W per beat.
Rohit
A spherica, concave shaving mirror has a radius of curvature of 32 cm .what is the magnification of a persons face. when it is 12cm to the left of the vertex of the mirror
did you solve?
Shii
1.75cm
Ridwan
my name is Abu m.konnek I am a student of a electrical engineer and I want you to help me
Abu
the magnification k = f/(f-d) with focus f = R/2 =16 cm; d =12 cm k = 16/4 =4
Sphere
what do we call velocity
Kings
A weather vane is some sort of directional arrow parallel to the ground that may rotate freely in a horizontal plane. A typical weather vane has a large cross-sectional area perpendicular to the direction the arrow is pointing, like a “One Way” street sign. The purpose of the weather vane is to indicate the direction of the wind. As wind blows pa
hi
Godfred
Godfred
If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag