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For Lorentz transformations, if we put $x^1=x$ and $x^2=ct$ and restrict ourselves to $2D$ we get $$x'=\gamma(x-\beta ct) \tag{1} $$ $$ct'=\gamma(ct-\beta x) \tag{2} $$ The matrix associated with this is $$L=\gamma\begin{pmatrix} 1 & -\beta\\ -\beta& 1\end{pmatrix}$$ This is a symmetric matrix as expected because of the symmetric nature of equations 1 and 2

However if we define $x^2=ict$ where $i=\sqrt{-1}$, then we get $$x'=\gamma[x+(i\beta)(ict)] \tag{3} $$ $$ict'=\gamma[ict-(i\beta) x] \tag{4} $$ And the matrix becomes $$L=\gamma\begin{pmatrix} 1 & i\beta\\ -i\beta& 1\end{pmatrix}$$ Now I understand that in equation 3 we had a minus sign already, allowing us to split it into $i^2$ and satisfying $x^2=ict$. And because in equation 4 we didn't have this, we had to insert an $i$ on our own causing the asymmetry. However, I cannot understand the cause of this asymmetry at a deeper and more intuitive level.

Qmechanic
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GedankenExperimentalist
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3 Answers3

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The matrix $\,\rm L\,$ of a Lorentz boost is symmetric. The matrix of a general Lorentz transformation is not symmetric. For example, the matrix $\,\Lambda\,$ of a proper homogenous Lorentz transformation is not symmetric. A transformation of this kind could be expressed as product of a Lorentz boost $\,\rm L\,$ and a rotation in space $\,\mathcal R$ : $\Lambda=\rm L\mathcal R$. This rotation breaks the symmetry of the matrix $\,\rm L\,$ producing the asymmetric $\,\Lambda$. See my answer here Show that any proper homogeneous Lorentz transformation may be expressed as the product of a boost times a rotation.

VoulKons
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I believe there is nothing to understand at a deeper level. The asymmetry is just an artifact of the algebra (or notation used). Notice that nowadays the "i" is seldom used. It became obsolete.

By the way, the Lorentz matrix is symmetric only for a boost transformation. When your reference frames are rotated, the matrix is no longer symmetric.

facenian
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Putting aside the issue of whether it’s a good idea to add a factor of $i$ to the time coordinate, a Lorentz transformation is a linear transformation whose corresponding matrix has one upstairs index and one downstairs index. Whether such a matrix is symmetric or not is generally basis-dependent. That is, a linear transformation which is symmetric in one basis is generally not symmetric in another.

If we restrict our attention to orthogonal changes of basis - which have matrices whose transpose is equal to their inverse - then any matrix which is symmetric in one basis remains symmetric in another. However, a generic change of basis is not orthogonal. In this case, your proposed change of basis matrix would be $$\pmatrix{i&0\\0&1}$$ which is not orthogonal.

Albatross
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