Relativistic kinetic energy derivation (from Work expended)

[freddie_mclair]

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!

 

[ShayanJ]

$\gamma m v^2-mc^2+\frac{mc^2}{\gamma}=mc^2(\gamma \frac{v^2}{c^1}-1)+\frac{mc^2}{\gamma}=mc^2 (\gamma (1+1-\frac{v^2}{c^2})-1)+\frac{mc^2}{\gamma}$.
I think you can continue yourself.

 

[Jonathan Scott]

Just try substituting for $\gamma$ and shuffling that last expression around a bit.

 

[freddie_mclair]

$\gamma m v^2-mc^2+\frac{mc^2}{\gamma}=mc^2(\gamma \frac{v^2}{c^1}-1)+\frac{mc^2}{\gamma}=mc^2 (\gamma (1+1-\frac{v^2}{c^2})-1)+\frac{mc^2}{\gamma}$.
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!

 

[sardar]

Hello,

Your approach to deriving the relativistic kinetic energy from the work expended is correct. However, you have reached a point where you need to use the relationship between velocity and the Lorentz factor, $\gamma$, to simplify the equation.

Since you are assuming that the body is at rest initially, $u = 0$ and $\gamma = 1$. Substituting these values into your equation, we get:

\begin{align}
T &= \gamma m v^2 + m c^2 \sqrt{1-v^2/c^2} - mc^2 \\
&= m v^2 + m c^2 \sqrt{1-v^2/c^2} - mc^2 \\
&= mc^2 \left(v^2 + \sqrt{1-v^2/c^2} - 1 \right)
\end{align}

Now, using the relationship $v = c\sqrt{1-1/\gamma^2}$, we can simplify the equation further:

\begin{align}
T &= mc^2 \left(c^2 \left(1-\frac{1}{\gamma^2} \right) + \sqrt{1-c^2 \left(1-\frac{1}{\gamma^2} \right)} - 1 \right) \\
&= mc^2 \left(c^2 \left(1-\frac{1}{\gamma^2} \right) + \frac{1}{\gamma} - 1 \right) \\
&= mc^2 \left(\frac{c^2}{\gamma^2} + \frac{1}{\gamma} - 1 \right) \\
&= mc^2 \left(\frac{1}{\gamma} - 1 \right) \\
&= (\gamma - 1) mc^2
\end{align}

Therefore, the relativistic kinetic energy can be written as $T = (\gamma - 1) mc^2$, as desired.

I hope this helps. Keep up the good work in your studies of relativistic physics!