Obtain The Lorentz Transformation in which the velocity is at an infinitesimal angle $d\theta$ counter clockwise from the $x$ axis, by means of a similarity transformation applied to $$L=\begin{bmatrix}\gamma & -\gamma \beta & 0& 0 \\ -\gamma \beta & \gamma & 0 &0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix} \tag{1} $$

show directly that the resulting matrix is orthogonal and that the inverse matrix is obtained by substituting $-v $ for $v$

So I thought this was a pretty straight forward question, Since the vector is counter clockwise from the $x$ axis I thought I would first apply a counterclockwise infinitesimal rotation about the $z$ axis to match the $x$ axis with the vector then apply $L$ and then rotate the resulting transformation clockwise round the $z$ axis to get back to the original basis.

$$\begin{align}T =RLR^{-1} &= \begin{bmatrix}1&0&0&0 \\ 0&1&-d \theta &0 \\ 0&d \theta&1&0 \\ 0&0&0&1 \end{bmatrix} \begin{bmatrix}\gamma & -\gamma \beta & 0& 0 \\ -\gamma \beta & \gamma & 0 &0 \\ 0&0&1&0 \\ 0&0&0&1 \end{bmatrix}\begin{bmatrix}1&0&0&0 \\ 0&1&d \theta &0 \\ 0& -d \theta&1&0 \\ 0&0&0&1 \end{bmatrix} \\ \\&=\begin{bmatrix}\gamma&-\gamma \beta& -\gamma\beta d\theta &0 \\ -\gamma \beta & (\gamma+d \theta)&(\gamma d \theta - d \theta)&0 \\-\gamma \beta d \theta& (\gamma d \theta -d \theta)& 1& 0 \\ 0&0&0&1\end{bmatrix} \end{align} $$ Where I have ignored second order terms of $d \theta$. However, this $T$ is not orthogonal ($T^{T}T \ne I$). Furthermore, replacing $v$ with $-v$ I got the following

$$\begin{align}M=T(v)T(-v) &= \begin{bmatrix}\gamma&-\gamma \beta& -\gamma\beta d\theta &0 \\ -\gamma \beta & (\gamma+d \theta)&(\gamma d \theta - d \theta)&0 \\-\gamma \beta d \theta& (\gamma d \theta -d \theta)& 1& 0 \\ 0&0&0&1\end{bmatrix} \begin{bmatrix}\gamma&\gamma \beta& \gamma\beta d\theta &0 \\ \gamma \beta & (\gamma+d \theta)&(\gamma d \theta - d \theta)&0 \\\gamma \beta d \theta& (\gamma d \theta -d \theta)& 1& 0 \\ 0&0&0&1\end{bmatrix} \\ \\ &= \begin{bmatrix} 1& -\gamma \beta d \theta &0&0 \\ \gamma \beta d \theta&1&0&0\\0&0&1&0 \\ 0&0&0&1\end{bmatrix} \end{align}$$

Which doesnt give me the inverse i.e $T(v)T(-v) \ne I$ therefore $T(-v)\ne T(v)^{-1}$. However, the matrix $M$ does look similar to a Thomas Precession. So I am a bit confused; I'm not sure where I went wrong in calculating the similarity transformation $T$.


1 Answer 1


The reason your final T is not orthogonal is that your original L is not orthogonal ($L^T=L \neq L^{-1}$). In fact Lorentz boosts of $(ct,x)$ really are not orthogonal, and the matrix you have written for L is correct. If L were orthogonal it would leave $t^2+x^2$ invariant. Instead L leaves $t^2-x^2$ invariant.

The request to prove orthogonality in your question doesn't go with the real matrix L that you wrote. Instead orthogonality goes with $L_{Historical}$ rotating by an imaginary angle.

$$ \begin{bmatrix} x'\\ ict'\\ \end{bmatrix} =L_{Historical}\begin{bmatrix} x\\ ict\\ \end{bmatrix} = \begin{bmatrix} cos(i\lambda) & -sin(i\lambda)\\ sin(i\lambda) & cos(i\lambda)\\ \end{bmatrix} \begin{bmatrix} x\\ ict\\ \end{bmatrix} = \begin{bmatrix} cosh(\lambda) & -i\ sinh(\lambda)\\ i\ sinh(\lambda) & cosh(\lambda)\\ \end{bmatrix} \begin{bmatrix} x\\ ict\\ \end{bmatrix} $$ $$ \begin{bmatrix} x'\\ ict'\\ \end{bmatrix} = \begin{bmatrix} \gamma & -i\beta\gamma\\ i\beta\gamma & \gamma\\ \end{bmatrix} \begin{bmatrix} x\\ ict\\ \end{bmatrix} $$

The reason we have heard that boosts are somehow orthogonal rotations is that old time physicists made boosts look like familiar rotations by using imaginary angles and making t imaginary. Please see this Physics Stack answer for a little more explanation.


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