# 5.9 Relativistic energy  (Page 6/16)

 Page 6 / 16

## Relativistic energy and momentum

We know classically that kinetic energy and momentum are related to each other, because:

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

Relativistically, we can obtain a relationship between energy and momentum by algebraically manipulating their defining equations. This yields:

${E}^{2}={\left(pc\right)}^{2}+{\left(m{c}^{2}\right)}^{2},$

where E is the relativistic total energy, $E=m{c}^{2}\text{/}\sqrt{1-{u}^{2}\text{/}{c}^{2}},$ and p is the relativistic momentum. This relationship between relativistic energy and relativistic momentum is more complicated than the classical version, but we can gain some interesting new insights by examining it. First, total energy is related to momentum and rest mass. At rest, momentum is zero, and the equation gives the total energy to be the rest energy $m{c}^{2}$ (so this equation is consistent with the discussion of rest energy above). However, as the mass is accelerated, its momentum p increases, thus increasing the total energy. At sufficiently high velocities, the rest energy term ${\left(m{c}^{2}\right)}^{2}$ becomes negligible compared with the momentum term ${\left(pc\right)}^{2};$ thus, $E=pc$ at extremely relativistic velocities.

If we consider momentum p to be distinct from mass, we can determine the implications of the equation ${E}^{2}={\left(pc\right)}^{2}+{\left(m{c}^{2}\right)}^{2},$ for a particle that has no mass. If we take m to be zero in this equation, then $E=pc,\phantom{\rule{0.2em}{0ex}}\text{or}p=E\text{/}c.$ Massless particles have this momentum. There are several massless particles found in nature, including photons (which are packets of electromagnetic radiation). Another implication is that a massless particle must travel at speed c and only at speed c . It is beyond the scope of this text to examine the relationship in the equation ${E}^{2}={\left(pc\right)}^{2}+{\left(m{c}^{2}\right)}^{2}$ in detail, but you can see that the relationship has important implications in special relativity.

Check Your Understanding What is the kinetic energy of an electron if its speed is 0.992 c ?

$\begin{array}{cc}\hfill {K}_{\text{rel}}& =\left(\gamma -1\right)m{c}^{2}=\left(\frac{1}{\sqrt{1-\frac{{u}^{2}}{{c}^{2}}}}-1\right)m{c}^{2}\hfill \\ & =\left(\frac{1}{\sqrt{1-\frac{{\left(0.992c\right)}^{2}}{{c}^{2}}}}-1\right)\left(9.11\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-31}\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}^{2}=5.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}\phantom{\rule{0.2em}{0ex}}\text{J}\hfill \end{array}$

## Summary

• The relativistic work-energy theorem is ${W}_{\text{net}}=E-{E}_{0}=\gamma m{c}^{2}-m{c}^{2}=\left(\gamma -1\right)m{c}^{2}.$
• Relativistically, ${W}_{\text{net}}={K}_{\text{rel}}$ where ${K}_{\text{rel}}$ is the relativistic kinetic energy.
• An object of mass m at velocity u has kinetic energy ${K}_{\text{rel}}=\left(\gamma -1\right)m{c}^{2},$ where $\gamma =\frac{1}{\sqrt{1-\frac{{u}^{2}}{{c}^{2}}}}.$
• At low velocities, relativistic kinetic energy reduces to classical kinetic energy.
• No object with mass can attain the speed of light, because an infinite amount of work and an infinite amount of energy input is required to accelerate a mass to the speed of light.
• Relativistic energy is conserved as long as we define it to include the possibility of mass changing to energy.
• The total energy of a particle with mass m traveling at speed u is defined as $E=\gamma m{c}^{2},$ where $\gamma =\frac{1}{\sqrt{1-\frac{{u}^{2}}{{c}^{2}}}}$ and u denotes the velocity of the particle.
• The rest energy of an object of mass m is ${E}_{0}=m{c}^{2},$ meaning that mass is a form of energy. If energy is stored in an object, its mass increases. Mass can be destroyed to release energy.
• We do not ordinarily notice the increase or decrease in mass of an object because the change in mass is so small for a large increase in energy. The equation ${E}^{2}={\left(pc\right)}^{2}+{\left(m{c}^{2}\right)}^{2}$ relates the relativistic total energy E and the relativistic momentum p . At extremely high velocities, the rest energy $m{c}^{2}$ becomes negligible, and $E=pc.$

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Hi
Hi
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ALFRED
how are you?
hi
asif
hi
Imran
I'm fine
ALFRED
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Anything which changes itself with respect to time or surrounding
Sushant
good
Chemist
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Chemist
No
Sushant
how can u say that
Chemist
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Not so more
Sushant
DHEERAJ
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But ask anything changes itself with respect to time or surrounding A Not any harmful radiation
DHEERAJ
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40 cm into change mm
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
Prema
this msg is out of mistake. sorry friends​.
Prema
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because is the study of mater and natural world
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tahreem
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tahreem
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