This is basically the result of Michelson-Morley experiment: $$ (x^0)^2 - \sum_{\mu=1}^3 (x^\mu)^2 \stackrel{!}= (\bar x^0)^2 - \sum_{\mu=1}^3 (\bar x^\mu)^2 $$ Why can this be interpreted as a rotation?

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    $\begingroup$ You can think of it as a hyperbolic rotation. $\endgroup$ Oct 18, 2016 at 23:02
  • $\begingroup$ In the book "Geometric algebra for physicists" by Doran and Lasenby, the authors within the formalism of geometric algebra do treat Lorentz transformation as rotation in 4-d space with the peculiar metric it has. $\endgroup$
    – Deep
    Oct 19, 2016 at 4:59
  • $\begingroup$ What does "$\stackrel{!}=$" mean? How is this different from the ordinary "="? $\endgroup$ Jan 25, 2021 at 14:35

3 Answers 3


Short version: The Lorentz transformation is not a rotation. It's just very similar.

Medium version: Look at the similarities!

A 2D rotation matrix can be written as

$$R(\theta)=\left( \begin{matrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{matrix}\right)$$

These rotation matrices satisfy the identity $R(\theta)R(\phi)=R(\theta+\phi)$. This is equivalent to the angle addition formulas for sine/cosine. They're used to rotate the $x$ dimension into the $y$ dimension.

A 1 space + 1 time dimension Lorentz transformation can be written as:

$$L(\alpha)=\left( \begin{matrix} \cosh(\alpha) & \sinh(\alpha) \\ \sinh(\alpha) & \cosh(\alpha) \end{matrix}\right)$$

These rotation matrices satisfy the identity $L(\alpha)L(\phi)=L(\alpha+\phi)$. These are equivalent to the hyperbolic angle addition formulas. They're used to transform the time dimension partway into the space dimension.

So mathematically, aesthetically, and almost every-which-way, the Lorentz matrices behave a lot like rotation matrices. However, even in special relativity, time and space are different. There are three spacelike axes and one timelike axis. The timelike axis stays timelike no matter how you Lorentz transform. But in Euclidean space, you can rotate your x axis into any other axis you so please. So these are NOT rotations in 4-dimensional Euclidean space. They are the analogue of rotations in 4-dimensional Minkowski space.

Long version: generalizing inner products.

If you want to generalize the notion of angles, you can start from the dot product formula: $\vec{a}\cdot \vec{b}=a_x b_x+a_y b_y+a_z b_z=\|a\|\|b\|\cos(\theta)$, with theta being the angle between them. You can write this as a matrix equation: $\vec{a}\cdot \vec{b}=\vec{a}^T I \vec{b}$, with $\vec{a}$ and $\vec{b}$ interpreted as column matrices and $I$ the identity matrix. You can imagine instead of putting an identity matrix in there you stuck in any other matrix, say $A$. So you define $\vec{a}\cdot \vec{b}=\vec{a}^T A \vec{b}$. Usually the symbol $\cdot$ is reserved for the dot product, so mathematicians like to write this as $\langle \vec{a},\vec{b}\rangle$, and physicist notation is $a_\mu b^\mu$. I'll use the matrix notation. In Euclidean geometry, you have: $$A_E=\left( \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}\right)$$

In Minkowski geometry, you have:

$$A_M=\left( \begin{matrix} -1 & 0 & 0 &0 \\ 0 & 1 & 0&0 \\ 0 & 0 & 1&0 \\ 0&0&0&1 \end{matrix}\right)$$

So in this generalization, a generalized rotation $R$ should be a linear transformation that leaves all generalized angles invariant. ie, it should leave all dot products invariant. Writing that out: we send $\vec{a}$ to $R\vec{a}$ and $\vec{b}$ to $R\vec{b}$ with $R$ a linear transformation and $\vec{a}$ and $\vec{b}$ column vectors. We demand that $\vec{a}^TA\vec{b}=(R\vec{a})^T A (R\vec{b})$. If you scratch your head a little bit, and necessitate that this holds for ALL vectors $\vec{a}$ and $\vec{b}$, you'll find that this implies $A=R^T A R$.

If you plug in $A_E$, the only matrices $R$ that satisfy these are the rotation matrices (the group of rotation matrices is denoted $O(3)$). If you plug in $A_M$, the only matrices $R$ that satisfy these are the Lorentz transformations (the group of these matrices is denoted $O(3,1)$ for 3 positive signs and one negative sign in the matrix $A_M$).

Note that you can have the "length" of a vector, $\vec{a}^T A_M \vec{a}$ be negative. This makes the Lorentz inner products useless for a lot of mathematics. For example, in real analysis you don't get anything useful out of this, because the space becomes non-hausdorff, which is a big no-no. (That is, if you define $d(\vec{a},\vec{b})=(\vec{b}-\vec{a})^TA_M(\vec{b}-\vec{a})$ as a distance function, it's pretty useless). Whenever you do a volume integral or try to give a notion of distinct points, you don't use the Minkowski inner product. For this reason, mathematicians define an inner product space and explicitly include the condition of positive definiteness, which precludes using anything like $A_M$. This makes perfect sense in mathematics, but in physics you really do want to keep the Minkowski metric.

  • $\begingroup$ How about mentioning hyperbolic trig? Could help the similarity clearer. $\endgroup$
    – innisfree
    Oct 19, 2016 at 0:17
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    $\begingroup$ @innisfree I added a "medium version". Let me know if it's what you had in mind! $\endgroup$
    – user12029
    Oct 19, 2016 at 2:23
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    $\begingroup$ Nice one, that's what I meant $\endgroup$
    – innisfree
    Oct 19, 2016 at 2:38

Rotation changes direction of a vector without changing its length. Can you define length of a four vector to be invariant under Lorentz transformation?


A rotation matrix $R$ can be defined as a square matrix for which $R^T=R^{-1}$.

For a Lorenz transformation $\Lambda$, if we define its transpose as $(\Lambda^T)_\alpha{}^\beta=\Lambda^\beta{}_\alpha$, then we have $(\Lambda^T)^\alpha{}_\beta=(\Lambda^{-1})^\alpha{}_\beta$.


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