- #1

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I'm trying to get the relativistic kinetic energy, ## T ##, from the work expended, ## W ##, (assuming that the body is at rest initially) and I'm doing it like this (in 1D):

\begin{equation}

W = T = \int F ds = m \int \frac{d(\gamma u)}{dt}u dt = m\int u d(\gamma u)

\end{equation}

Where, ## u ## is the speed, and ## \gamma ## the Lorentz factor.

Now, putting some limits on it, and integrating by parts:

\begin{align}

T &= \gamma m u^2 \Big|_{0}^{v} - m \int_0^v \gamma u du \\

&= \gamma m v^2 - m \int_0^v \gamma u du \\

&= \gamma m v^2 - m \int_0^v \frac{u}{\sqrt{1-u^2/c^2}} du\\

&= \gamma m v^2 + m c^2 \sqrt{1-u^2/c^2} \Big|_0^v\\

&= \gamma m v^2 + m c^2 \sqrt{1-v^2/c^2} - mc^2

\end{align}

I can also write it as:

\begin{align}

T &= \gamma m v^2 + \frac{mc^2}{\gamma} - mc^2

\end{align}

But to me it looks like a dead end here...

Well, what I really want to get is the relativistic kinetic energy written like this: ##T = (\gamma - 1)mc^2 ##.

Could you give me some advice if I'm doing this correctly, and if so, how to proceed from here?

Thanks in advance!