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In his book on relativity, Albert Einstein states in chapter 15 that we now don't write kinetic energy as $K_e = \frac{1}{2}mv^2$ but as $$K_e = \frac{mc^2}{\sqrt{1- v^2/c^2}}-mc^2.$$

How is this claim proven?

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    $\begingroup$ One way would be to write down the ``obvious'' four-momentum $$m\frac{dx^{\mu}}{d\tau}$$ and identify the zero component as the energy. A slightly more satisfying way might apply Noether's theorem for spacetime translations. $\endgroup$ – d_b Sep 16 '19 at 16:57
  • $\begingroup$ There is not just one way to do this. There are many. Some examples are given in answers to this question: physics.stackexchange.com/questions/34008/… $\endgroup$ – user4552 Sep 16 '19 at 21:02
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One method is given on Wikipedia's page on Kinetic Energy and I'll give that here.

You start from the definition of work done :

$$dW = \mathbf{F\cdot}d\mathbf{x}=\frac{d\mathbf{p}}{dt}\cdot \mathbf{v}dt=\mathbf{v}\cdot d\mathbf{p}$$

So we now have an expression and we use the relativistic form for momentum $\mathbf{p}=m\gamma\mathbf{v}$ where $\gamma=\left(1-\frac{v^2}{c^2}\right)^{-\frac 1 2}$

You now integrate to find an expression for kinetic energy :

$$W=\int \mathbf{v}\cdot d(m\gamma\mathbf{v})$$

Integration by parts gets you :

$$W=m\gamma\mathbf{v}\cdot\mathbf{v}-\int m\gamma \mathbf{v}\cdot d\mathbf{v}$$

This resolves down to :

$$W=m\gamma v^2 - \frac m 2 \int \gamma d(v^2)$$

After some maths we get :

$$W=m\gamma v^2+\frac{mc^2}{\gamma} - E_0$$

for some constant of integration $E_0$.

For $\mathbf{v=0}$, $\gamma=1$ and $W=0$ so we can find $E_0$ and get :

$$W=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2$$

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The expression for kinetic energy in relativity is:

$$E=\gamma mc^2-mc^2$$

where $m$ is the object's rest mass and $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ is the Lorentz factor. Substituting, we get an expression for energy as a function of velocity:

$$E(v)=\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}}-mc^2$$

We can also calculate its derivatives:

$$E'(v)=\frac{v}{c^2}\frac{mc^2}{\left(1-\frac{v^2}{c^2}\right)^{3/2}}$$ $$E''(v)=\frac{m}{\left(1-\frac{v^2}{c^2}\right)^{3/2}}+\frac{3v^2}{c^4}\frac{mc^2}{\left(1-\frac{v^2}{c^2}\right)^{5/2}}$$

Using a Taylor series approximation when $v\ll c$, we write:

\begin{align*} E(v)&=E(0)+E'(0)v+\frac{1}{2}E''(0)v^2+...\\ &=0+0\times v+\frac{1}{2}(m)v^2\\ &=\frac{1}{2}mv^2 \end{align*}

which is the classical kinetic energy.

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  • $\begingroup$ This doesn’t answer the question as it currently appears, asking how to derive your first equation. $\endgroup$ – G. Smith Sep 16 '19 at 17:10
  • $\begingroup$ @G.Smith Fair enough; I was going off of a less-clear early edit. $\endgroup$ – probably_someone Sep 16 '19 at 17:11
  • $\begingroup$ In my opinion, ACuriousMind changed the OP’s question, but since the original post was very unclear I’m not sure about that. $\endgroup$ – G. Smith Sep 16 '19 at 17:13
  • $\begingroup$ I think using $E$ to denote kinetic energy rather than kinetic energy plus rest energy is confusing. Normally $E=\gamma mc^2$. $\endgroup$ – G. Smith Sep 16 '19 at 17:21
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according to NEWTON :

$$F=\frac{d}{dt}(m\,v)$$

analog with Minkowski force:

$$K^\mu=\frac{d}{d\tau}(m\,u^\mu)=\frac{d}{d\tau}\,p^\mu\tag 1$$

with:

$$u^\mu=\frac{dx^\mu}{d\tau}=\frac{dx^\mu}{dt}\frac{dt}{d\tau}=\gamma\,\begin{bmatrix} c\\ v\\ \end{bmatrix}$$

$\Rightarrow\quad$ euquation (1)

$$\frac{d}{d\tau}\,p^\mu=\gamma\,\frac{d}{dt}\begin{bmatrix} \gamma\,m\,c\\ \gamma\,m\,v\\ \end{bmatrix}=\begin{bmatrix} K^0\\ \gamma\,F\\ \end{bmatrix}$$

with :

$$u_\mu\,\frac{d}{d\tau}(m\,u^\mu)=\frac{d}{d\tau}\left(\frac{m}{2}\,u_\mu\,u^\mu\right)=0=u_\mu\,K^\mu$$

we obtain $K^0=\frac{\gamma\,v^T\,F}{c}$

$\Rightarrow$

$$\gamma\,\frac{d}{dt}(\gamma\,m\,c)=\frac{\gamma\,v^T\,F}{c}$$

$$\frac{d}{dt}\underbrace{(\gamma\,m\,c^2)}_{E}=\frac{v^T\,F}{c}$$

so:

$$E=\gamma\,m\,c^2=m\,c^2\left(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)$$

and with a Taylor series ( $v^2/c^2 \ll 1) $ you get:

$$E\approx m\,c^2+\frac{m\,v^2}{2}$$

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