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OK, I understand why $KE=\gamma mc^2-mc^2$, but why is it also equal to $E-E_0$?

Qmechanic
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Max
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    have a look at the answers to this question - see if that is helpful.... http://physics.stackexchange.com/q/143652/ – tom Jan 11 '15 at 18:54
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    What is $E$ and what is $E_0$? – Sofia Jan 11 '15 at 18:59
  • @Sofia typically $E$ is total energy and $E_0$ is rest mass energy - the question I linked has a good discussion about different ways of expressing these values and particularly $m$ and $m_0$ – tom Jan 11 '15 at 19:03
  • Welcome @Max - nice question - you might want to look at how to put equations and symbols into your questions with latex type codes MathJax - see for example http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – tom Jan 11 '15 at 19:07

3 Answers3

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I think the answer is that if we start with

$$KE = \gamma mc^2 - m c^2$$

then because $mc^2$ is the rest mass energy. $E_0$

$$ E_0 = m c^2 $$

and the total energy $E$ is given by

$$ E = \gamma m C^2 $$

so substituting into your first equation gives

$$ KE = E - E_0 $$

does this help?

tom
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  • OK, but I think the equation for the rest mass energy is derived from the one I don't understand. If not, then please give me the derivation of the equation you've used – Max Jan 12 '15 at 16:54
  • $E=mc^2$ is pretty fundamental and I really suggest that you look at this question to see where it comes from --- http://physics.stackexchange.com/questions/143652/is-e2-mc22pc2-correct-or-is-e-mc2-the-correct-one -- there are some really good answers there given by people who probably understand it better than I do - I have an answer there too, but it is not as detailed as some of the others. - Comment back here again if that does not help and we can take it from there – tom Jan 12 '15 at 17:11
  • I don't think so as the equation for the relativistic kinetic energy was discovered far earlier than the $E=mc^2$ one. Unfortunately, the question on the link you gave me doesn't provide with the derivation nor any information about $E=mc^2$ – Max Jan 12 '15 at 21:41
  • @Max sorry it did not help - i am not sure if the following link is interesting as a step towards what you are looking for http://physicsworld.com/cws/article/news/2011/aug/23/did-einstein-discover-e-equals-mc-squared ... or perhaps here? https://www.physicsforums.com/threads/einstein-did-not-derive-e-mc2-first.28362/ – tom Jan 12 '15 at 23:29
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If you take the non-(special)-relativistic limit of that expression you find that $$\text{KE} \approx \left(\frac12mv^2 + mc^2\right) - mc^2$$ which is what you would expect from classical mechanics. The other way to see this is that, when a particle is at rest in a certain inertial frame of reference, its relativistic energy is given by its rest mass through $m_0c^2$. This represent a sort of frozen energy contained inside the body. Kinetic energy is then extra energy the body gains when it aquires a certain velocity.

Phoenix87
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Relativistic kinetic energy is defined as $E-E_0$ because this definition naturally follows from the more general definition of relativistic energy as $E = \gamma mc^2$ where $\gamma$ depends upon the velocity.

It then follows to define rest energy for $v=0, \gamma=1$ as $E_0 = mc^2$, and relativistic kinetic energy as the excess energy when its moving over the energy when it's not: $E-E_0 =\gamma mc^2 - mc^2$.

The more fundamental question to then ask is: why is relativistic energy defined as $E = \gamma mc^2$?

Energy is defined via Noether's theorem as that physical quantity that's conserved for a system whose physical laws don't change over time. For a single point particle of mass m

$$E =\gamma mc^2$$

John McAndrew
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