How can I visualize Lorentz transformations? I know it's four dimensional, but without a visual picture in my mind I can't understand any of this. For example how do:

Time reflection

Space reflection

Proper Lorentz transformations

Improper Lorentz transformations

Homogeneous Lorentz transformations

Orthochronous Lorentz transformations

Antichronous Lorentz transformations

look like? I don't have any idea about how they look like. Maybe this can be explained with Minkowski diagrams or in some other form.

  • $\begingroup$ Yes, of course you can draw them as before-after Minkowski diagrams. $\endgroup$
    – ACuriousMind
    Commented Jan 31, 2016 at 14:55

1 Answer 1


Let the Minkowski spacetime be the set $\mathbb{R}^4$ parametrized by coordinates $\underline x = (t,x_1,x_2,x_3)$.

You can visualize this set with a $4$-dimensional cartesian diagram, with axes labeled $(t, x_1, x_2, x_3)$. You can help your intuition just by drawing the 3-dimensional one (suppressing one of the space dimensions).

A Lorentz transformation is an isometry of the Minkowski spacetime, which means that it preserves the inner product given by the metric $g = -dt^2 + dx_1^2+dx_2^2+dx_3^2$. The associated matrix is $$ g = \left(\begin{array}[cccc] --1 & & & \\ & 1 & & \\ & & 1 & \\ & & & 1 \end{array}\right) $$

One has that all such isometries are of the form $\underline x' =\Lambda \underline x + C$, where $\Lambda$ is a matrix such that $\Lambda^{-1} g \Lambda = g$.

Let's look at the group in detail:

  1. Time reflection just reverses the direction of time: $t'=-t$, $x'_1 = x_1$, $x_2' = x_2$, $x_3' = x_3$. In the Minkowski diagram, this means reflection wrt the $t$-axis.

  2. Space reflection just reverses one space direction. Without loss of generality, let's suppose that this direction is aligned with the $x_1$-axis (you can always assume that). Then, the corresponding transformation is like placing a $3$-d straight mirror $M$ (the analogue of a 2d flat mirror [a plane], but with another dimension). In this case, the mirror will be the set for which $x_1 = 0$.

  3. A proper Lorentz transormation is such that $\det \Lambda = 1$. This means that the transformation is orientation preserving. Roughly, this means that the total number of spacetime ``flippings'' given by the transformation is even. For example, space and time reflections are both improper, in the sense that $\det \Lambda = -1$, and they both flip only one direction. Composing time reflection and space reflection along the $x_1$ direction gives a proper transformation: we have an even number (2) of spacetime flippings. The matrix relative to such transformation is $$ \Lambda = \left(\begin{array}[cccc] --1 & & & \\ & -1 & & \\ & & 1 & \\ & & & 1 \end{array}\right) $$ and can be visualized as a rotation of the spacetime which keeps $x_3$ and $x_4$ fixed, and rotates by $180^°$ the $(t,x_1)$ coordinates. All such rotations are proper (e.g. also only space rotations).

  4. Homogeneous transformations are those that can be written in the form $\underline x' = \Lambda \underline x$, which means that the origin $(\underline 0)$ is kept fixed. They do not translate the spacetime.

  5. Orthochronous transformations are those which preserve the direction of time. In our picture, we know that the time direction is the direction $(1,0,0,0)$. The transformed time direction is $V = \Lambda (1,0,0,0)$. If the $t$-coordinate of $V$ is positive, the transformation is orthochronous, if it is negative it is antichronous. For example, time reflection is antichronous, whereas space rotations and reflections are orthochronous.

Finally, there is an extremely cool video that shows a picture of the only ``nontrivial'' Lorentz transformations, i.e. the Lorentz boosts. Here it is: video.

  • 1
    $\begingroup$ Why -1? I like this answer because you wrote about every transformation I asked about. But I will wait for other answers, but thx so far :) $\endgroup$ Commented Jan 31, 2016 at 20:26
  • $\begingroup$ So inhomogeneous transformations aren't preserving the origin? And what are Lorentz boosts? $\endgroup$ Commented Jan 31, 2016 at 21:01
  • 1
    $\begingroup$ Exactly, inhomogeneous transformations are not preserving the origin. Lorentz boosts are those transformations which are preserving $g$, but are not preserving the Euclidean metric $g'= dt^2 + dx_1^2+dx_2^2+dx_3^2$. They are those responsible for spacetime contraction and dilation. An example of such a transformation is, letting $\alpha \in \mathbb{R}$, $$t' = t \cosh \alpha - x_1\sinh \alpha, \qquad x_1' = -t \sinh \alpha +x_1 \cosh \alpha$$. $\endgroup$
    – snefs
    Commented Feb 1, 2016 at 23:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.