# 5.9 Relativistic energy  (Page 2/16)

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## Relativistic kinetic energy

Relativistic kinetic energy of any particle of mass m is

${K}_{\text{rel}}=\left(\gamma -1\right)m{c}^{2}.$

When an object is motionless, its speed is $u=0$ and

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

so that ${K}_{\text{rel}}=0$ 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 $\frac{1}{2}\phantom{\rule{0.2em}{0ex}}m{u}^{2}.$ To show that the expression for ${K}_{\text{rel}}$ reduces to the classical expression for kinetic energy at low speeds, we use the binomial expansion to obtain an approximation for ${\left(1+\epsilon \right)}^{n}$ valid for small $\epsilon$ :

${\left(1+\epsilon \right)}^{n}=1+n\epsilon +\text{​}\frac{n\left(n-1\right)}{2!}{\epsilon }^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{\epsilon }^{3}+\cdots \approx 1+n\epsilon$

by neglecting the very small terms in ${\epsilon }^{2}$ and higher powers of $\epsilon .$ Choosing $\epsilon =-{u}^{2}\text{/}{c}^{2}$ and $n=-\frac{1}{2}$ leads to the conclusion that γ at nonrelativistic speeds, where $\epsilon =u\text{/}c$ is small, satisfies

$\gamma ={\left(1-{u}^{2}\text{/}{c}^{2}\right)}^{-1\text{/}2}\approx 1+\frac{1}{2}\left(\frac{{u}^{2}}{{c}^{2}}\right).$

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 speed here, most terms are very small. Thus, the expression derived here for $\gamma$ is not exact, but it is a very accurate approximation. Therefore, at low speed:

$\gamma -1=\frac{1}{2}\left(\frac{{u}^{2}}{{c}^{2}}\right).$

Entering this into the expression for relativistic kinetic energy gives

${K}_{\text{rel}}=\left[\frac{1}{2}\left(\frac{{u}^{2}}{{c}^{2}}\right)\right]m{c}^{2}=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}m{u}^{2}={K}_{\text{class}}.$

That is, relativistic kinetic energy becomes the same as classical kinetic energy when $u\text{<}\text{<}c.$

It is even more interesting to investigate what happens to kinetic energy when the speed of an object approaches the speed of light. We know that $\gamma$ becomes infinite as u approaches c , so that ${K}_{\text{rel}}$ also becomes infinite as the velocity approaches the speed of light ( [link] ). The increase in ${K}_{\text{rel}}$ is far larger than in ${K}_{\text{class}}$ as v approaches c. 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    .

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

An electron has a velocity $v=0.990c.$ (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 $9.11\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-31}\text{kg}.$ )

## Strategy

The expression for relativistic kinetic energy is always correct, but for (a), it must be used because the velocity is highly relativistic (close to c ). First, we calculate the relativistic factor $\gamma ,$ and then use it to determine the relativistic kinetic energy. For (b), we calculate the classical kinetic energy (which would be close to the relativistic value if v were less than a few percent of c ) and see that it is not the same.

in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
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there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
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40cm=40.0×10^-2m =400.0×10^-3m =400mm.
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