I was wondering if there is a way to re-write a general Lorentz transformation

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in a Lorentz covariant way. Since one index is up while the other is down, I assumed that it could be written as the tensor product of a 4-vector with a 4-covector. But the only covariant quantities I can think of using are the identity tensor, the 4-velocity and perhaps the levi-civita thensor?

Maybe Lorentz transformations are not tensors. The trace of the matrix is equal to $2(1+\gamma)$ which does not look invariant.

  • $\begingroup$ hyperphysics.phy-astr.gsu.edu/hbase/Relativ/vec4.html#c2 $\endgroup$
    – anna v
    Commented Sep 27, 2023 at 13:36
  • 2
    $\begingroup$ Lorentz transformations transform coordinates from a frame into a primed frame and are often denoted as ${\Lambda_{\mu'}}^\nu\,.$ In contrast, components of a tensor are meaningful only in the same frame: $T_{\mu\nu}$ or $T_{\mu'\nu'}\,.$ Therefore I am not quite sure why that matrix above should be a tensor, and even if it were one, why this is physically meaningful. $\endgroup$
    – Kurt G.
    Commented Sep 27, 2023 at 13:56
  • $\begingroup$ Note that a Lorentz Transformation has to be invertible. $\endgroup$
    – robphy
    Commented Sep 27, 2023 at 14:11
  • $\begingroup$ Quantities with indices are not always tensors. $\endgroup$
    – Ghoster
    Commented Sep 27, 2023 at 16:51
  • $\begingroup$ Since one index is up while the other is down You didn’t show any indices. $\endgroup$
    – Ghoster
    Commented Sep 27, 2023 at 16:52

1 Answer 1


Short answer. You only have to change the sign to the elements on the first row and column, except for the element (1,1).

Some details. Lorentz transformation allows you to find the rule of transformation of the coordinates of a vector $\mathbf{v}$ as seen by two inertial observers $O$ and $O'$ in their own bases, i.e. \begin{equation} \mathbf{v} = v^i \mathbf{b}_i = v'^j \mathbf{b}'_j \ , \end{equation} providing the rule of transformation \begin{equation} v'^j = T^j_i v^i \ , \end{equation} so that \begin{equation} \mathbf{v} = v^i \mathbf{b}_i = v^i T^j_i \mathbf{b}'_j \qquad \rightarrow \qquad \mathbf{b}_i = T^j_i \mathbf{b}'_j \ . \end{equation}

If you want to find the relation between the covariant coordinates as seen by the two observer, you first need to use the metric tensor $g \hspace{-4pt}g$ to transform from contravariant to covariant components \begin{equation} v^i = g^{ik} v_k \qquad , \qquad v'^j = g'^{il} v'_l \ , \end{equation} and insert in the relation above \begin{equation} g'^{jl}v'_l = T^j_i g^{ik} v_k \qquad \rightarrow \qquad v'_l = \underbrace{g'_{jl} T^j_i g^{ik}}_{={T^{cov \ }}^k_l} v_k \ , \end{equation} being the components of the metric tensor (special relativity, Minkowski metric) equal to $diag(-1,1,1,1)$ or $diag(1,-1,-1,-1)$ depending on the metric signature you're using.

This results in the result written in the short answer, i.e. changing the sign of the elements [1,2:4], [2:4,1].


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