Talk:PlanetPhysics/Spacetime Interval Is Invariant for a Lorentz Transformation

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: spacetime interval is invariant for a Lorentz transformation %%% Primary Category Code: 03.30.+p %%% Filename: SpacetimeIntervalIsInvariantForALorentzTransformation.tex %%% Version: 2 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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The \htmladdnormallink{spacetime}{http://planetphysics.us/encyclopedia/SR.html} interval between two events $E_1( x_1, y_1, z_1, t_1 ) $ and $E_2( x_2, y_2, z_2, t_2 )$ is defined as

$$ (\triangle s)^2 = c^2 \triangle t^2 - (\triangle x)^2 - (\triangle y)^2 - (\triangle z)^2. $$

If $\triangle s$ is in \htmladdnormallink{reference frame}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} $S$, then $\triangle s'$ is in reference frame $S'$ moving at a \htmladdnormallink{velocity}{http://planetphysics.us/encyclopedia/Velocity.html} $u$ along the x-axis. Therefore, to show that the spacetime interval is invariant under a \htmladdnormallink{Lorentz transformation}{http://planetphysics.us/encyclopedia/CosmologicalConstant.html} we must show

$$ (\triangle s)^2 = (\triangle s')^2 $$

with the reference frames related by \htmladdnormallink{The Lorentz transformation}{http://planetphysics.us/encyclopedia/LorentzTransformation.html} $$ x' = \frac{ x - u t }{ \sqrt{1 - u^2/c^2} } $$ $$ y' = y $$ $$ z' = z $$ $$t' = \frac{ t - ux/c^2 }{ \sqrt{ 1 - u^2/c^2 } }.$$

The change in coordinates between events in the $S'$ frame is then given by

$$ \triangle x' = \left ( \frac{ x_2 - u t_2 }{ \sqrt{1 - u^2/c^2} } \right ) - \left ( \frac{ x_1 - u t_1 }{ \sqrt{1 - u^2/c^2} } \right ) = \frac{ \triangle x - u \triangle t } { \sqrt{1 - u^2/c^2} } $$

$$ \triangle y' = y_2 - y_1 = \triangle y $$

$$ \triangle z' = z_2 - z_1 = \triangle z $$

$$ \triangle t' = \left ( \frac{ t_2 - u x_2/c^2 }{ \sqrt{1 - u^2/c^2} } \right ) - \left ( \frac{ t_1 - u x_1 / c^2 }{ \sqrt{1 - u^2/c^2} } \right ) = \frac{ \triangle t - u \triangle t } { \sqrt{1 - u^2/c^2} }. $$

Squaring the terms yield

$$ (\triangle x')^2 = \left ( \frac{ \triangle x - u \triangle t } { \sqrt{1 - u^2/c^2} } \right ) \left ( \frac{ \triangle x - u \triangle t } { \sqrt{1 - u^2/c^2} } \right ) = \frac{ (\triangle x)^2 - 2 u \triangle x \triangle t + u^2 (\triangle t)^2 }{ 1 - u^2/c^2 }$$

$$ (\triangle y')^2 = (\triangle y)^2 $$

$$ (\triangle z')^2 = (\triangle z)^2 $$

$$ (\triangle t')^2 = \left ( \frac{ \triangle t - u \triangle t } { \sqrt{1 - u^2/c^2} } \right ) \left (  \frac{ \triangle t - u \triangle t } { \sqrt{1 - u^2/c^2} } \right ) = \frac{ (\triangle t)^2 - 2u \triangle x \triangle t / c^2 + u^2 (\triangle x)^2 / c^4 }{ 1 - u^2/c^2 }. $$

Substituting these terms into the spacetime interval gives

$$ (\triangle s')^2 = \frac{ c^2 ( (\triangle t)^2 - 2u \triangle x \triangle t /c^2 + u^2 (\triangle x)^2 /c^4 )}{ 1 - u^2/c^2 } - \frac{ ( (\triangle x)^2 - 2 u \triangle x \triangle t + u^2 (\triangle t)^2)}{ 1 - u^2/c^2 } - (\triangle y)^2 - (\triangle z)^2 .$$

Adding the first two terms with common denominators together yields

$$ (\triangle s')^2 = \frac{ c^2 (\triangle t^2) - (\triangle x)^2 - u^2 (\triangle t)^2 + u^2 (\triangle x)^2 / c^2}{ 1 - u^2/c^2 } - (\triangle y)^2 - (\triangle z)^2 .$$

Pulling out a $-u^2/c^2$

$$ (\triangle s')^2 = \frac{ c^2 (\triangle t^2) - (\triangle x)^2 - u^2/c^2 ( c^2(\triangle t)^2 + (\triangle x)^2 )}{ 1 - u^2/c^2 } - (\triangle y)^2 - (\triangle z)^2 .$$

Factoring out a $c^2 (\triangle t)^2 - (\triangle x)^2$ in the numerator

$$ (\triangle s')^2 = \frac{ (c^2 (\triangle t^2) - (\triangle x)^2) (1 - u^2/c^2) }{ 1 - u^2/c^2 } - (\triangle y)^2 - (\triangle z)^2 .$$

Finally, canceling terms gives

$$ (\triangle s')^2 = c^2 (\triangle t^2) - (\triangle x)^2 - (\triangle y)^2 - (\triangle z)^2  = (\triangle s)^2 .$$

Hence, the spacetime interval is invariant under a Lorentz transformation.

\begin{thebibliography}{9} \bibitem{Carroll} Carroll, Bradley, Ostlie, Dale, {\em An Introduction to Modern Astrophysics}. Addison-Wesley Publishing Company, Reading, Massachusetts, 1996. \bibitem{Cheng} Cheng, Ta-Pei, {\em Relativity, Gravitation and Cosmology}. Oxford University Press, Oxford, 2005. \bibitem{Einstein} Einstein, Albert, \htmladdnormallink{http://planetphysics.us/encyclopedia/SpecialTheoryOfRelativity.html}. 1916. \end{thebibliography}

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