% this tex file was generated by the Physics Derivation Graph 
\documentclass[12pt]{article}
\usepackage{amsmath,amssymb,amsfonts}
\usepackage[dvipdfmx,colorlinks=true,pdfkeywords={physics derivation graph}]{hyperref}
\usepackage{graphicx} % for including PNG files
% inference rules as newcommand for use in the body
\newcommand\LHSofexprequalsLHSofexpr[3]{LHS of Eq.~\ref{eq:#1} is equal to LHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\RHSofexprequalsRHSofexpr[3]{RHS of Eq.~\ref{eq:#1} is equal to RHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\Xcrossbothsidesby[3]{Take cross product of $#1$ and Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\Xdotbothsides[3]{Take inner product of $#1$ with Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\addXtobothsides[3]{Add $#1$ to both sides of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\addexprtoexpr[3]{Add Eq.~\ref{eq:#1} to Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#2}.}
\newcommand\addzerotoLHS[3]{Add zero to LHS of Eq.~\ref{eq:#2}, where $0=#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\addzerotoRHS[3]{Add zero to RHS of Eq.~\ref{eq:#2}, where $0=#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\applydivergence[2]{Apply divergence to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\applyfunctiontobothsidesofexpression[4]{Apply function $#1$ with argument $#2$ to Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}}
\newcommand\applygradienttoscalarfunction[2]{Apply gradient to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\applyoperatortobra[2]{Apply operator in Eq.~\ref{eq:#1} to bra; yields Eq.~\ref{eq:#2}.}
\newcommand\applyoperatortoket[2]{Apply operator in Eq.~\ref{eq:#1} to ket; yields Eq.~\ref{eq:#2}.}
\newcommand\assumeNdimensions[2]{Assume $#1$ dimensions; decompose vector to be Eq.~\ref{eq:#2}.}
\newcommand\bothsidescrossX[3]{Take cross product of Eq.~\ref{eq:#2} and $#1$; yields Eq.~\ref{eq:#3}}
\newcommand\bothsidesdotX[3]{Take inner product of Eq.~\ref{eq:#2} with $#1$; yields Eq.~\ref{eq:#3}}
\newcommand\boundarycondition[2]{Boundary condition: Eq.~\ref{eq:#2} when Eq.~\ref{eq#1}.}
\newcommand\boundaryconditionforexpr[2]{A boundary condition for Eq.~\ref{eq:#1} is Eq.~\ref{eq:#2}}
\newcommand\changefivevariablesinexpr[9]{\def\ArgOne{{#1}}
\def\ArgTwo{{#2}}
\def\ArgThree{{#3}}
\def\ArgFour{{#4}}
\def\ArgFive{{#5}}
\def\ArgSix{{#6}}
\def\ArgSeven{{#7}}
\def\ArgEight{{#8}}
\def\ArgNine{{#9}}
\changefivevariablesinexprRelay
}
\newcommand\changefivevariablesinexprRelay[3]{Change of variable $ArgOne$ to $ArgTwo$ and $ArgThree$ to $ArgFour$ and $ArgFive$ to $ArgSix$ and $ArgSeven$ to $ArgEight$ and $ArgNine$ to $ArgOne0$ in Eq.~\ref{eq:ArgOne1}; yields Eq.~\ref{eq:ArgOne2}.}
\newcommand\changefourvariablesinexpr[9]{\def\ArgOne{{#1}}
\def\ArgTwo{{#2}}
\def\ArgThree{{#3}}
\def\ArgFour{{#4}}
\def\ArgFive{{#5}}
\def\ArgSix{{#6}}
\def\ArgSeven{{#7}}
\def\ArgEight{{#8}}
\def\ArgNine{{#9}}
\changefourvariablesinexprRelay
}
\newcommand\changefourvariablesinexprRelay[1]{Change of variable $ArgOne$ to $ArgTwo$ and $ArgThree$ to $ArgFour$ and $ArgFive$ to $ArgSix$ and $ArgSeven$ to $ArgEight$ in Eq.~\ref{eq:ArgNine}; yields Eq.~\ref{eq:ArgOne0}.}
\newcommand\changesixvariablesinexpr[9]{\def\ArgOne{{#1}}
\def\ArgTwo{{#2}}
\def\ArgThree{{#3}}
\def\ArgFour{{#4}}
\def\ArgFive{{#5}}
\def\ArgSix{{#6}}
\def\ArgSeven{{#7}}
\def\ArgEight{{#8}}
\def\ArgNine{{#9}}
\changesixvariablesinexprRelay
}
\newcommand\changesixvariablesinexprRelay[5]{Change of variable $ArgOne$ to $ArgTwo$ and $ArgThree$ to $ArgFour$ and $ArgFive$ to $ArgSix$ and $ArgSeven$ to $ArgEight$ and $ArgNine$ to $ArgOne0$ and $ArgOne1$ to $ArgOne2$ in Eq.~\ref{eq:ArgOne3}; yields Eq.~\ref{eq:ArgOne4}.}
\newcommand\changethreevariablesinexpr[8]{Change of variable $#1$ to $#2$ and $#3$ to $#4$ and $#5$ to $#6$ in Eq.~\ref{eq:#7}; yields Eq.~\ref{eq:#8}.}
\newcommand\changetwovariablesinexpr[6]{Change variable $#1$ to $#2$ and $#3$ to $#4$ in Eq.~\ref{eq:#5}; yields Eq.~\ref{eq:#6}.}
\newcommand\changevariableXtoY[4]{Change variable $#1$ to $#2$ in Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}.}
\newcommand\claimLHSequalsRHS[1]{Thus we see that LHS of Eq.~\ref{eq:#1} is equal to RHS.}
\newcommand\claimexprequalsexpr[2]{Thus we see that Eq.~\ref{eq:#1} is equivalent to Eq.~\ref{eq:#2}.}
\newcommand\combineliketerms[2]{Combine like terms in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\conjugatebothsides[2]{Conjugate both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\conjugatefunctionX[3]{Conjugate $#1$ in Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\conjugatetransposebothsides[2]{Conjugate transpose of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\declareassumption[1]{Eq.~\ref{eq:#1} is an assumption.}
\newcommand\declarefinalexpr[1]{Eq.~\ref{eq:#1} is one of the final equations.}
\newcommand\declareguesssolution[2]{Judicious choice as a guessed solution to Eq.~\ref{eq:#1} is Eq.~\ref{eq:#2},}
\newcommand\declareidentity[1]{Eq.~\ref{eq:#1} is an identity.}
\newcommand\declareinitialexpr[1]{Eq.~\ref{eq:#1} is an initial equation.}
\newcommand\differentiatewithrespectto[3]{Differentiate Eq.~\ref{eq:#2} with respect to $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\distributeconjugatetofactors[2]{Distribute conjugate to factors in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\distributeconjugatetransposetofactors[2]{Distribute conjugate transpose to factors in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\dividebothsidesby[3]{Divide both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\divideexprbyexpr[3]{Divide Eq.~\ref{eq:#1} by Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\dropnondominantterm[3]{Based on the assumption $#1$, drop non-dominant term in Eq.~\ref{#2}; yeilds Eq.~\ref{#3}}
\newcommand\evaluatedefiniteintegral[2]{Evaluate definite integral Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\expandLHS[2]{Expand the LHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\expandRHS[2]{Expand the RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\expandintegrand[2]{Expand integrand of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\expandmagnitudetoconjugate[3]{Expand $#1$ in Eq.~\ref{eq:#2} with conjugate; yields Eq.~\ref{eq:#3}.}
\newcommand\exprisequivalenttoexprunderthecondition[3]{Eq.~\ref{eq:#1} is equivalent to Eq.~\ref{eq:#2} under the condition in Eq.~\ref{eq:#3}.}
\newcommand\expristrueunderconditionexpr[3]{Eq.~\ref{eq:#1} is valid when Eq.~\ref{eq:#2} occurs; yields Eq.~\ref{eq:#3}.}
\newcommand\factoroutX[3]{Factor $#1$ from Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\factoroutXfromLHS[3]{Factor $#1$ from the LHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\factoroutXfromRHS[3]{Factor $#1$ from the RHS of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\functioniseven[5]{$#1$ is even with respect to $#2$, so replace $#1$ with $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}.}
\newcommand\functionisodd[5]{$#1$ is odd with respect to $#2$, so replace $#1$ with $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}.}
\newcommand\indefiniteintegralover[3]{Indefinite integral of both sides of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\indefiniteintegrateLHSover[3]{Indefinite integral of LHS of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\indefiniteintegrateRHSover[3]{Indefinite integral of RHS of Eq.~\ref{eq:#2} over $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\indefiniteintegration[2]{Indefinite integral of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\integrate[2]{Integrate Eq.~ref{eq:#1}; yields Eq.~ref{eq:#2}.}
\newcommand\integrateoverfromto[5]{Integrate Eq.~\ref{eq:#4} over $#1$ from lower limit $#2$ to upper limit $#3$; yields Eq.~\ref{eq:#5}.}
\newcommand\makeexprpower[3]{Make Eq.~\ref{eq:#2} the power of $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\maximumofexpr[3]{The maximum of Eq.~\ref{eq:#2} with respect to $#1$ is Eq.~\ref{eq:#3}}
\newcommand\multiplyLHSbyunity[3]{Multiply LHS of Eq.~\ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~\ref{eq:#3}}
\newcommand\multiplyRHSbyunity[3]{Multiply RHS of Eq.~\ref{eq:#2} by 1, which in this case is $#1$; yields Eq.~\ref{eq:#3}}
\newcommand\multiplybothsidesby[3]{Multiply both sides of Eq.~\ref{eq:#2} by $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\multiplyexprbyexpr[3]{Multiply Eq.~\ref{eq:#1} by Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\normalizationcondition[1]{Normalization condition is Eq.~\ref{eq:#1}.}
\newcommand\partiallydifferentiatewithrespectto[3]{Partially differentiate Eq.~\ref{eq:#2} with respect to $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\raisebothsidestopower[3]{Raise both sides of Eq.~\ref{eq:#2} to $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\replaceconstantwithvalue[5]{Replace constant $#1$ with value $#2$ and units $#3$ in Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}}
\newcommand\replacecurlwithLeviCevitasummationcontravariant[2]{Replace curl in Eq.~\ref{eq:#1} with Levi-Cevita contravariant; yields Eq.~\ref{eq:#2}.}
\newcommand\replacescalarwithvector[2]{Replace scalar variables in Eq.~\ref{eq:#1} with equivalent vector variables; yields Eq.~\ref{eq:#2}.}
\newcommand\replacesummationnotationwithvectornotation[2]{Replace summation notation in Eq.~\ref{eq:#1} with vector notation; yields Eq.~\ref{eq:#2}.}
\newcommand\selectimaginaryparts[2]{Select imaginary parts of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\selectrealparts[2]{Select real parts of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\separatethreevectorcomponents[4]{Separate three vector components in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3} and Eq.~\ref{eq:#4}.}
\newcommand\separatetwovectorcomponents[3]{Separate two vector components in Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3}}
\newcommand\separatevectorintotwotrigonometricratios[4]{Separate vector in Eq.~\ref{eq:#2} into components related by angle $#1$; yields Eq.~\ref{eq:#3} and Eq.~\ref{eq:#4}.}
\newcommand\simplify[2]{Simplify Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\solveforX[3]{Solve Eq.~\ref{eq:#2} for $#1$; yields Eq.~\ref{eq:#3}.}
\newcommand\squarerootbothsides[3]{Take the square root of both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2} and Eq.~\ref{eq:#3}.}
\newcommand\substituteLHSofexprintoexpr[3]{Substitute LHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\substituteLHSoffiveexpressionsintoexpr[7]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} and LHS of Eq.~\ref{eq:#5} into Eq.~\ref{eq:#6}; yields Eq.~\ref{eq:#7}.}
\newcommand\substituteLHSoffourexpressionsintoexpr[6]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} into Eq.~\ref{eq:#5}; yields Eq.~\ref{eq:#6}.}
\newcommand\substituteLHSofsixexpressionsintoexpr[8]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} and LHS of Eq.~\ref{eq:#4} and LHS of Eq.~\ref{eq:#5} and LHS of Eq.~\ref{eq:#6} into Eq.~\ref{eq:#7}; yields Eq.~\ref{eq:#8}.}
\newcommand\substituteLHSofthreeexpressionsintoexpr[5]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} and LHS of Eq.~\ref{eq:#3} into Eq.~\ref{eq:#4}; yields Eq.~\ref{eq:#5}.}
\newcommand\substituteLHSoftwoexpressionsintoexpr[4]{Substitute LHS of Eq.~\ref{eq:#1} and LHS of Eq.~\ref{eq:#2} into Eq.~\ref{eq:#3}; yields Eq.~\ref{eq:#4}.}
\newcommand\substituteRHSofexprintoexpr[3]{Substitute RHS of Eq.~\ref{eq:#1} into Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\subtractXfrombothsides[3]{Subtract $#1$ from both sides of Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#3}.}
\newcommand\subtractexprfromexpr[3]{Subtract Eq.~\ref{eq:#1} from Eq.~\ref{eq:#2}; yields Eq.~\ref{eq:#2}.}
\newcommand\sumexponents[2]{Sum exponents on LHS and RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\sumexponentsLHS[2]{Sum exponents on LHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\sumexponentsRHS[2]{Sum exponents on RHS of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\newcommand\swapLHSwithRHS[2]{Swap LHS of Eq.~\ref{eq:#1} with RHS; yields Eq.~\ref{eq:#2}.}
\newcommand\takecurlofbothsides[2]{Apply curl to both sides of Eq.~\ref{eq:#1}; yields Eq.~\ref{eq:#2}.}
\title{Lorentz transformation}
\date{\today}
\author{none}
\setlength{\topmargin}{-.5in}
\setlength{\textheight}{9in}
\setlength{\oddsidemargin}{0in}
\setlength{\textwidth}{6.5in}
\begin{document}
\maketitle
\begin{abstract}
Generated by the \href{http://derivationmap.net/}{Physics Derivation Graph}.
source: \cite{1999_Tipler_Llewellyn}, page 21; see also https://en.wikipedia.org/wiki/Lorentz\_transformation and https://en.wikipedia.org/wiki/Derivations\_of\_the\_Lorentz\_transformations
\end{abstract}
% step ID = 1337934
\declareinitialexpr{5427510}
equation 1-13 on page 21 in \cite{1999_Tipler_Llewellyn}
\begin{equation}
x' = \gamma (x - v t)
\label{eq:5427510}
\end{equation}
% step ID = 9869359
\declareinitialexpr{2283140}
equation 1-14 on page 21 in \cite{1999_Tipler_Llewellyn}
\begin{equation}
x = \gamma (x' + v t')
\label{eq:2283140}
\end{equation}
% step ID = 6150706
\substituteLHSofexprintoexpr{5427510}{2283140}{4471422}
solve output expr for t'
\begin{equation}
x = \gamma ( \gamma (x - v t) + v t' )
\label{eq:4471422}
\end{equation}
% step ID = 4777578
\simplify{4471422}{7169020}
\begin{equation}
x = \gamma ( \gamma x - \gamma v t + v t' )
\label{eq:7169020}
\end{equation}
% step ID = 5708661
\simplify{7169020}{6463955}
\begin{equation}
x = \gamma^2 x - \gamma^2 v t + \gamma v t'
\label{eq:6463955}
\end{equation}
% step ID = 3244831
\subtractXfrombothsides{\gamma^2 x}{6463955}{8494407}
\begin{equation}
x - \gamma^2 x =  - \gamma^2 v t + \gamma v t'
\label{eq:8494407}
\end{equation}
% step ID = 9520931
\factoroutXfromLHS{x}{8494407}{3992172}
\begin{equation}
x (1 - \gamma^2 ) = - \gamma^2 v t + \gamma v t'
\label{eq:3992172}
\end{equation}
% step ID = 8199201
\addXtobothsides{\gamma^2 v t}{3992172}{6047713}
\begin{equation}
x (1 - \gamma^2 ) + \gamma^2 v t = \gamma v t'
\label{eq:6047713}
\end{equation}
% step ID = 6431475
\dividebothsidesby{\gamma v}{6047713}{5995189}
\begin{equation}
\frac{x (1 - \gamma^2 )}{\gamma v} + \frac{\gamma^2 v t}{\gamma v} = t'
\label{eq:5995189}
\end{equation}
% step ID = 4052521
\simplify{5995189}{7546640}
\begin{equation}
\frac{\gamma x (1 - \gamma^2 )}{\gamma^2 v} + \gamma t = t'
\label{eq:7546640}
\end{equation}
% step ID = 3002960
\swapLHSwithRHS{7546640}{1693888}
\begin{equation}
t' = \frac{\gamma x (1 - \gamma^2 )}{\gamma^2 v} + \gamma t
\label{eq:1693888}
\end{equation}
% step ID = 3151962
\declareinitialexpr{6319661}
\begin{equation}
x^2 + y^2 + z^2 = c^2 t^2
\label{eq:6319661}
\end{equation}
% step ID = 5637413
\declareinitialexpr{5649086}
\begin{equation}
x'^2 + y'^2 + z'^2 = c^2 t'^2
\label{eq:5649086}
\end{equation}
% step ID = 5796383
\declareassumption{6316097}
\begin{equation}
y' = y
\label{eq:6316097}
\end{equation}
% step ID = 2712078
\declareassumption{7666907}
\begin{equation}
z' = z
\label{eq:7666907}
\end{equation}
% step ID = 8057586
\substituteLHSoffourexpressionsintoexpr{7666907}{6316097}{1693888}{5427510}{5649086}{4326342}
\begin{equation}
\gamma^2 (x - v t)^2 + y^2 + z^2 = c^2 \gamma^2 \left( t + \frac{ 1 - \gamma^2 }{ \gamma^2 } \frac{x}{v} \right)^2
\label{eq:4326342}
\end{equation}
% step ID = 5537889
\simplify{4326342}{6066191}
expanded the squared terms
\begin{equation}
\gamma^2 x^2 - \gamma^2 2 x v t + \gamma^2 v^2 t^2 + y^2 + z^2 = c^2 \gamma^2 \left(\frac{1-\gamma^2}{\gamma^2}\right)\frac{x^2}{\gamma^2} + c^2 \gamma^2 2 t \left(\frac{1-\gamma^2}{\gamma^2}\right)\frac{x}{\gamma} + c^2 \gamma^2 t^2
\label{eq:6066191}
\end{equation}
% step ID = 7160101
\simplify{6066191}{4202425}
grouped by terms for x^2, xt, and t^2
\begin{equation}
\left( \gamma^2 - c^2 \gamma^2 \left( \frac{1-\gamma^2}{\gamma^2} \right)^2 \frac{1}{v^2} \right) x^2 + y^2 + z^2 + \left( -\gamma^2 2 x v t - c^2 \gamma^2 2 t \left( \frac{1-\gamma^2}{\gamma^2} \right) \frac{x}{v} \right) = t^2 \left( c^2 \gamma^2 - \gamma^2 v^2 \right)
\label{eq:4202425}
\end{equation}
% step ID = 4875843
\exprisequivalenttoexprunderthecondition{6319661}{4202425}{2562123}
based on the comparison of the x^2 terms
\begin{equation}
\gamma^2 - c^2 \gamma^2 \frac{(1-\gamma^2)^2}{v^2 \gamma^4} = 1
\label{eq:2562123}
\end{equation}
% step ID = 1336657
\exprisequivalenttoexprunderthecondition{6319661}{4202425}{3640931}
based on the comparison of the t^2 terms
\begin{equation}
-\gamma^2 v^2 + c^2 \gamma^2 = c^2
\label{eq:3640931}
\end{equation}
% step ID = 7832193
\exprisequivalenttoexprunderthecondition{6319661}{4202425}{6685577}
based on the comparison of the (x t) terms
\begin{equation}
-\gamma^2 2 x v t - c^2 \gamma^2 2 t \left( \frac{1-\gamma^2}{\gamma^2} \right) \frac{x}{v} = 0
\label{eq:6685577}
\end{equation}
% step ID = 5029881
\subtractXfrombothsides{\gamma^2}{2562123}{7403799}
solve for \gamma
\begin{equation}
- c^2 \gamma^2 \frac{(1-\gamma^2)^2}{v^2 \gamma^4} = 1 - \gamma^2
\label{eq:7403799}
\end{equation}
% step ID = 3464414
\dividebothsidesby{1 - \gamma^2}{7403799}{4052253}
\begin{equation}
- c^2 \frac{(1-\gamma^2)}{v^2 \gamma^2} = 1
\label{eq:4052253}
\end{equation}
% step ID = 4044426
\multiplybothsidesby{v^2 \gamma^2}{4052253}{8195408}
\begin{equation}
-c^2 + c^2 \gamma^2 = v^2 \gamma^2
\label{eq:8195408}
\end{equation}
% step ID = 3201871
\addXtobothsides{c^2}{8195408}{6913493}
\begin{equation}
c^2 \gamma^2 = v^2 \gamma^2 + c^2
\label{eq:6913493}
\end{equation}
% step ID = 8488825
\subtractXfrombothsides{v^2 \gamma^2}{6913493}{5207615}
\begin{equation}
c^2 \gamma^2 - v^2 \gamma^2 = c^2
\label{eq:5207615}
\end{equation}
% step ID = 7599260
\factoroutXfromLHS{\gamma^2}{5207615}{8842089}
\begin{equation}
\gamma^2 (c^2 - v^2) = c^2
\label{eq:8842089}
\end{equation}
% step ID = 9486255
\dividebothsidesby{c^2 - \gamma^2}{8842089}{7595841}
\begin{equation}
\gamma^2 = \frac{c^2}{c^2 - \gamma^2}
\label{eq:7595841}
\end{equation}
% step ID = 5211114
\squarerootbothsides{7595841}{3040283}{6010461}
\begin{equation}
\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}
\label{eq:3040283}
\end{equation}
\begin{equation}
\gamma = \frac{-1}{\sqrt{1-\frac{v^2}{c^2}}}
\label{eq:6010461}
\end{equation}
% step ID = 5619063
\declarefinalexpr{3040283}
Lorentz factor definition
\bibliographystyle{plain}
\bibliography{pdg.bib}
\end{document}
% EOF
