Thursday, 14 February 2019

Lorentz transformation for velocity

There's an equation on Wikipedia for the Lorentz transformation for velocity at
https://en.wikipedia.org/wiki/Lorentz_transformation#Transformation_of_velocities
which states
\begin{align} \large
{\boldsymbol{\mathrm{u}}}^{\boldsymbol{\mathrm{{'}}}}\boldsymbol{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{1}\boldsymbol{\mathrm{-}}\frac{\boldsymbol{\mathrm{u}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}}{c^{\boldsymbol{\mathrm{2}}}}}\left[\frac{\boldsymbol{\mathrm{u}}}{{\gamma }_{\mathrm{v}}}\boldsymbol{\mathrm{-}}\boldsymbol{\mathrm{v}}\boldsymbol{\mathrm{+}}\frac{1}{c^2}\frac{{\gamma }_{\mathrm{v}}}{{\gamma }_{\mathrm{v}}+1}\left(\boldsymbol{\mathrm{u}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}\right)\boldsymbol{\mathrm{v}}\right]& \phantom {10000}(1)  \\
\end{align}
where the coordinate velocities and Lorentz factor are
\begin{align} \large
\boldsymbol{\mathrm{u}}=\frac{d\boldsymbol{\mathrm{r}}}{dt}\ ,\ \ \ {\boldsymbol{\mathrm{u}}}^{\boldsymbol{\mathrm{{'}}}}=\frac{d{\boldsymbol{\mathrm{r}}}^{\boldsymbol{\mathrm{{'}}}}}{dt^{'}}\ \ ,\ \ \ \ {\gamma }_{\mathrm{v}}=\frac{1}{\sqrt{1-\frac{\boldsymbol{\mathrm{v}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}}{c^2}}} & \phantom {10000}(2) \\
\end{align}
##\boldsymbol{\mathrm{v}}## is the relative velocity of the primed frame of reference. Instead of (1), it is sometimes useful to have the equation
\begin{align} \large
{\boldsymbol{\mathrm{u}}}^{\boldsymbol{\mathrm{{'}}}}\boldsymbol{\mathrm{=}}\frac{\mathrm{1}}{\mathrm{1}\boldsymbol{\mathrm{-}}\frac{\boldsymbol{\mathrm{u}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}}{c^{\boldsymbol{\mathrm{2}}}}}\left[\frac{\boldsymbol{\mathrm{u}}}{{\gamma }_{\mathrm{v}}}\boldsymbol{\mathrm{-}}\boldsymbol{\mathrm{v}}\boldsymbol{\mathrm{+}}\frac{1}{v^2}\left(1-\frac{1}{{\gamma }_{\mathrm{v}}}\right)\left(\boldsymbol{\mathrm{u}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}\right)\boldsymbol{\mathrm{v}}\right] & \phantom {10000}(3) \\
\end{align}
where ##v^2=\boldsymbol{\mathrm{v}}\boldsymbol{\mathrm{\cdot }}\boldsymbol{\mathrm{v}}##.
This form of the equation was useful in the first exercise in Sean M Carroll's Spacetime and Geometry : An Introduction to General Relativity. To get it we just have to show that
\begin{align} \large
\frac{1}{c^2}\frac{{\gamma }_{\mathrm{v}}}{{\gamma }_{\mathrm{v}}+1}=\frac{1}{v^2}\left(1-\frac{1}{{\gamma }_{\mathrm{v}}}\right) & \phantom {10000}(4) \\
\end{align}
We use ##\gamma ## instead of ##{\gamma }_{\nu }## because have terms like ##{\gamma }^2## where the 2 is an exponent.
\begin{align} \large
\frac{1}{c^2}\frac{\gamma }{\gamma +1} & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{{\gamma }^2}{\gamma }\right)=\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\frac{1}{1-{v^2}/{c^2}}}{\gamma }\right) & \phantom {10000}(6) \\
 & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\left[\frac{1-1+{v^2}/{c^2}}{1-{v^2}/{c^2}}\right]c^2}{\gamma v^2}\right) & \phantom {10000}(7) \\
 & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\left[\frac{1}{1-{v^2}/{c^2}}-1\right]c^2}{\gamma v^2}\right) & \phantom {10000}(8) \\
 & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\left[{\gamma }^2-1\right]c^2}{\gamma v^2}\right) & \phantom {10000}(9) \\
 & =\frac{1}{c^2}\frac{1}{\gamma +1}\left(\frac{\left[\gamma +1\right]\left[\gamma -1\right]c^2}{\gamma v^2}\right) & \phantom {10000}(10) \\
 & =\frac{\gamma -1}{\gamma v^2} & \phantom {10000}(11) \\
\therefore \frac{1}{c^2}\frac{\gamma }{\gamma +1} & =\frac{1}{v^2}\left(1-\frac{1}{\gamma }\right) & \phantom {10000}(12) \\
\end{align}
                                         QED
I tried to post this on the Wikimedia page, but it was not allowed :-(
This proof is also available as a pdf here.

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