Relativistic kinetic energy derivation (from Work expended)

  • #1
Hi,

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!
 

Answers and Replies

  • #2
2,793
594
[itex] \gamma m v^2-mc^2+\frac{mc^2}{\gamma}=mc^2(\gamma \frac{v^2}{c^1}-1)+\frac{mc^2}{\gamma}=mc^2 (\gamma (1+1-\frac{v^2}{c^2})-1)+\frac{mc^2}{\gamma} [/itex].
I think you can continue yourself.
 
  • Like
Likes freddie_mclair
  • #3
Jonathan Scott
Gold Member
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997
Just try substituting for ##\gamma## and shuffling that last expression around a bit.
 
  • #4
[itex] \gamma m v^2-mc^2+\frac{mc^2}{\gamma}=mc^2(\gamma \frac{v^2}{c^1}-1)+\frac{mc^2}{\gamma}=mc^2 (\gamma (1+1-\frac{v^2}{c^2})-1)+\frac{mc^2}{\gamma} [/itex].
I think you can continue yourself.
Oh, cool!
Then from your expression (with a tiny corrections on the exponential factor of ## c^1 ## and a change of sign in the ## 1 + 1 ## trick), it follows:

\begin{align}
\gamma m v^2-mc^2+\frac{mc^2}{\gamma}\\
m c^2 \Big(\gamma \frac{v^2}{c^2}-1 \Big)+\frac{m c^2}{\gamma} \\
m c^2 \Big[\gamma \Big( 1 - 1 + \frac{v^2}{c^2}\Big) - 1 \Big]+\frac{m c^2}{\gamma} \\
m c^2 \Big[\gamma \Big( 1 - \gamma^{-2} \Big) - 1 \Big]+\frac{m c^2}{\gamma}\\
m c^2 \Big( \gamma - \frac{1}{\gamma} -1 \Big)+\frac{m c^2}{\gamma}\\
\gamma m c^2 - \frac{m c^2}{\gamma} - m c^2 +\frac{m c^2}{\gamma}\\
\gamma m c^2 - m c^2 = (\gamma - 1)mc^2
\end{align}

Thanks!
 

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