One problem Einstein had to face while formulating his special relativity theory is that the first principles of physics still comes first. We must remember that the principles in Newtonian physics are not wrong; the details are just an approximation. His first argument is that objects moving in an inertial reference frame (not accelerating and in a straight line) must corroborate with the relations found in Newtonian mechanics (i.e. \(F = \frac{dp}{dt}\)). Hence, for the kinetic energy of a moving object, it must be shown that \[{E}_{k} = \int _{ 0 }^{ v }{ Fdx } .\]
Also, the kinetic energy \({E}_{k}\) will approach the classical value when the speed of the system approaches zero. **Here we will derive the relativistic energy with first principles and calculus.**

\[ \begin{align*} {E}_{k} &= \int _{ 0 }^{ v }{ Fdx } \\ &= \int _{ 0 }^{ v }{ \frac{dp}{dt}dx } \\ &=\int _{ 0 }^{ v }{d(\gamma mv) \frac{dx}{dt}} \\ &=m\int _{ 0 }^{ v }{{\gamma}^{3} vdv}\\ &=m\int _{ 0 }^{ v }{{\left(1-\frac{{v}^{2}}{{c}^{2}}\right)}^{-3/2} vdv} \\ &=m{c}^{2}\left(\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}} -1\right) \end{align*}\]

Note: It is easy to show that \(\frac{d(\gamma v)}{dv} = {\gamma}^{3}\). Click Here

The total energy of the object, if it were moving would be \({E}_{k} + m{c}^{2}\). Even when you have a perfectly stationary massive object, there is still a lot of energy that is associated with its **rest mass**. Hence, if we let \({E}_{k}\) equal zero, we get \(E = m{c}^{2}\). Therefore, \[E = \frac{m{c}^{2}}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}.\]

Finally we must test that the derived equation for \({E}_{k}\) recovers the kinetic energy in classical mechanics at non-relativistic speeds. By Taylor expansion, \[m{c}^{2}\left(\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}} -1\right) = m{c}^{2}\left(1+ \frac{1}{2}\frac{{v}^{2}}{{c}^{2}}+\frac{3}{8}\frac{{v}^{4}}{{c}^{4}}+ ... -1 \right) \]

which the non-quadratic terms vanish for \(v<<c\). Therefore, \[{E}_{k} = \frac{1}{2}m{v}^{2}.\]

Check out my other notes at Proof, Disproof, and Derivation

## Comments

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TopNewestBut how does one discover that one of the "first principles" should be that \(p=\gamma mv\), and not the Newtonian \(p=mv\)? Some "first principles" are just more "first-er" than other "first principles". Einstein didn't start there. But it's a good exposition, once we've accepted that \(p=\gamma mv\) – Michael Mendrin · 2 years, 8 months ago

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– Steven Zheng · 2 years, 8 months ago

True, the original Newtonian equation for momentum is "first-er" but we must know that we are now working in the relativistic framework. In relativity, if we increase velocity, the mass gets heavier. In many introductory textbooks, they refer the rest mass as \({m}_{0}\) and \(m = \gamma {m}_{0}\). I just dropped the subscript.Log in to reply

Wonderful – Abhijeeth Babu · 2 years, 8 months ago

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how e(k)=mc^2 when v=0 ??? – Danny Kills · 2 years, 8 months ago

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– Steven Zheng · 2 years, 8 months ago

I said it doesn't make sense (linguistically) but there is a tremendous amount of rest mass in you if you are stationary.Log in to reply

– Danny Kills · 2 years, 8 months ago

At the first place, it doesnt make sense mathematically. Setting v=0 in the formula for E(k), i get e(k)=0. Am i wrong???Log in to reply

– Steven Zheng · 2 years, 8 months ago

I see. I should have wrote that if \({E}_{k}\) is zero, the energy equals \(m{c}^{2}\).Log in to reply

– Danny Kills · 2 years, 8 months ago

Yes, absolutely! When i saw that, i totally didnt follow the rest of the text.Log in to reply

i have a question, how did you get gamma^3? Can you explain it to me,Bu what a great note!! – Mardokay Mosazghi · 2 years, 8 months ago

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gamma factor identities – Steven Zheng · 2 years, 8 months ago

Sure. Here you goLog in to reply