The Lorentz Transformation

Einstein postulated that the speed of light is the same in any inertial frame of reference. It is not possible to meet this condition if the transformation from one inertial reference frame to another is done with a universal time, that is, \bgroup\color{black}$ t'=t$\egroup. Let us study a transformation from one inertial reference frame to another that is moving with a constant velocity \bgroup\color{black}$ v$\egroup in the \bgroup\color{black}$ x$\egroup direction. Such a transformation is usually referred to as a boost.

Newton's motion with constant velocity \bgroup\color{black}$ x=vt$\egroup transforming to \bgroup\color{black}$ x'=v't'$\egroup requires that the transformation be linear, like a rotation. We therefore try a linear transformation in which both the position and the time transform. Since this linear transformation will mix \bgroup\color{black}$ x$\egroup and \bgroup\color{black}$ t$\egroup, it is reasonable to try to transform quantities that have the same units 2, so we will try a dimensionless transformation of \bgroup\color{black}$ x$\egroup and \bgroup\color{black}$ ct$\egroup.

We will work in just two dimensions, \bgroup\color{black}$ x$\egroup and \bgroup\color{black}$ t$\egroup, like a rotation in the \bgroup\color{black}$ xt$\egroup plane. (For the boost along the \bgroup\color{black}$ x$\egroup direction, \bgroup\color{black}$ y$\egroup and \bgroup\color{black}$ z$\egroup are not changed.)

Assume the origin of a second inertial frame (primed), is moving at velocity \bgroup\color{black}$ v$\egroup in the \bgroup\color{black}$ x$\egroup direction, and corresponds to the (unprimed) origin at \bgroup\color{black}$ t=0$\egroup. All of these choices can be made without loss of generality. The most general linear transformation of \bgroup\color{black}$ x$\egroup and \bgroup\color{black}$ t$\egroup can be written with four dimensionless coefficients. We will call these coefficients \bgroup\color{black}$ \gamma$\egroup, \bgroup\color{black}$ u$\egroup, \bgroup\color{black}$ \Gamma$\egroup, and \bgroup\color{black}$ \alpha$\egroup.

$\displaystyle x'=\gamma(x-{u\over c} ct)$    
$\displaystyle ct'=\Gamma(ct-\alpha x)$    

By the very definition of the transformation we have that the origin of the primed frame is at \bgroup\color{black}$ x=vt$\egroup in the unprimed frame. (This makes the boost clearly different from a rotation.)

  $\displaystyle x'=\gamma(x-{u\over c} ct)$ the given transformation    
  $\displaystyle x'=0=\gamma(vt-{u\over c} ct)$ definition of the boost    
  $\displaystyle u=v$    

So we can identify the constant \bgroup\color{black}$ u$\egroup to be the velocity of the transformation.

  $\displaystyle x'=\gamma(x-{v\over c} ct)$    
  $\displaystyle ct'=\Gamma(ct-\alpha x)$    

The inverse of the transformation must be the same as a transformation with \bgroup\color{black}$ -v$\egroup as the velocity of moving frame. \bgroup\color{black}$ \gamma$\egroup and \bgroup\color{black}$ \Gamma$\egroup are scale factors that depend on the velocity of the transformation. Since there is no difference between the \bgroup\color{black}$ +x$\egroup and \bgroup\color{black}$ -x$\egroup directions in physics, we must use the same \bgroup\color{black}$ \gamma$\egroup and \bgroup\color{black}$ \Gamma$\egroup in the inverse transformation which has the same magnitude of velocity.

  $\displaystyle x'=\gamma(x-{v\over c} ct)$ the transformation    
  $\displaystyle ct'=\Gamma(ct-\alpha x)$    
  $\displaystyle \begin{pmatrix}x'\cr ct'\end{pmatrix}=\begin{pmatrix}\gamma & -{v...
...gamma\cr -\alpha\Gamma & \Gamma\end{pmatrix}\begin{pmatrix}x\cr ct\end{pmatrix}$ transformation as matrix eq.    
  $\displaystyle \mathbb{B}(v)=\begin{pmatrix}\gamma & -{v\over c}\gamma\cr -\alpha\Gamma & \Gamma\end{pmatrix}$ transformation matrix as function of $ v$    
  $\displaystyle \mathbb{B}(-v)=\mathbb{B}^{-1}(v)$ inverse is $\displaystyle \mathbb{B}(-v)$    
  $\displaystyle \begin{pmatrix}\gamma & +{v\over c}\gamma\cr -\alpha_-\Gamma & \G...
...amma}\begin{pmatrix}\Gamma&\alpha\Gamma\cr {v\over c}\gamma&\gamma\end{pmatrix}$ except $ \alpha$ depends on $ v$    
  $\displaystyle \begin{pmatrix}\gamma & {v\over c}\gamma\cr -\alpha_-\Gamma & \Ga...
...)}\begin{pmatrix}\Gamma & {v\over c}\gamma\cr \alpha\Gamma &\gamma\end{pmatrix}$ rhs is just inverse of 2X2    
  $\displaystyle \gamma={1\over \gamma\Gamma (1-{v\over c}\alpha)}\Gamma={1\over \gamma (1-{v\over c}\alpha)}$ upper left    
  $\displaystyle \Gamma={1\over \gamma\Gamma (1-{v\over c}\alpha)}\gamma={1\over \Gamma (1-{v\over c}\alpha)}$ lower right    
  $\displaystyle \gamma^2={1\over (1-{v\over c}\alpha)}$    
  $\displaystyle \Gamma^2={1\over (1-{v\over c}\alpha)}$    
  $\displaystyle \Gamma=\gamma=\sqrt{1\over 1-{v\over c}\alpha}$ must be $\displaystyle >0$    
  $\displaystyle \begin{pmatrix}\gamma & {v\over c}\gamma\cr -\alpha_-\gamma & \ga...
...=\begin{pmatrix}\gamma& {v\over c}\gamma \cr \alpha\gamma & \gamma\end{pmatrix}$ plug $ \Gamma=\gamma$    
  $\displaystyle \alpha={v\over c}$ upper left    
  $\displaystyle \alpha_-=-{v\over c}$ lower left    
  $\displaystyle \mathbb{B}(v)=\begin{pmatrix}\gamma & -{v\over c}\gamma\cr -{v\ov...
...ix} =\begin{pmatrix}\gamma & -\beta\gamma\cr -\beta\gamma & \gamma\end{pmatrix}$    
  $\displaystyle \begin{pmatrix}x'\cr ct'\end{pmatrix}=\begin{pmatrix}\gamma & -\beta\gamma\cr -\beta\gamma & \gamma\end{pmatrix}\begin{pmatrix}x\cr ct\end{pmatrix}$ the transformation now    

Now, consider a pulse of light moving in the \bgroup\color{black}$ x$\egroup direction emitted at \bgroup\color{black}$ x=0$\egroup and \bgroup\color{black}$ t=0$\egroup in one inertial frame. Since the origins of the two systems coincide at \bgroup\color{black}$ t=0$\egroup, this light is emitted in the primed frame with \bgroup\color{black}$ x'=0$\egroup, and \bgroup\color{black}$ t'=0$\egroup. At a later time, the position of the light pulse will be at \bgroup\color{black}$ x=ct$\egroup. By Einstein's postulate that the speed of light is independent of inertial frame, (and by the Michelson-Morley measurement). The pulse of light should be at \bgroup\color{black}$ x'=ct'$\egroup in the primed frame. Our transformation must give this result, so lets try it. Transforming the later position of the light pulse, we get.

  $\displaystyle x'=\gamma(x-{v\over c} ct)$    
  $\displaystyle ct'=\gamma(ct-{v\over c} x)$    
  $\displaystyle x'=\gamma(ct-{v\over c} ct)$ plug in $ x=ct$    
  $\displaystyle ct'=\gamma(ct-{v\over c} ct)$ plug in $ x=ct$    
  $\displaystyle x'=ct'$ from the 2 eq.    

The condition that the speed of light is the same in the two reference frames is met.

We have shown that the most general transformation to a frame moving with a velocity \bgroup\color{black}$ v$\egroup, that is consistent with Newton's laws and the isotropy of space, and that satisfies the condition that the inverse of the transformation is a transformation with velocity \bgroup\color{black}$ -v$\egroup, is given by:

\bgroup\color{black}$ \displaystyle \begin{pmatrix}x'\cr ct'\end{pmatrix}=\begin... -\beta\gamma & \gamma\end{pmatrix}\begin{pmatrix}x\cr ct\end{pmatrix}$\egroup
\bgroup\color{black}$\displaystyle \beta={v\over c} $\egroup
\bgroup\color{black}$\displaystyle \gamma={1\over\sqrt{1-\beta^2}} $\egroup
\bgroup\color{black}$ \displaystyle y'=y$\egroup
\bgroup\color{black}$ \displaystyle z'=z$\egroup
This transformation also gives the result that the speed of light is independent of which inertial frame of reference we use.

It took surprisingly little physics input to derive the Lorentz transformation for the space-time coordinates.

Jim Branson 2012-10-21