How exactly are Lorentz transformations rotations?

Question

I've always seen everywhere that Lorentz transformations are rotations in 4D. Let's stick with 2D (one space axis, one time axis) for simplicity.

Rotations of two dimensional space axes look completely different from the 2D Lorentz transformation. To rotate space axes, we rotate both x and y axes by an angle of same magnitude with same sign. This results in the axes still remaining at 90 degrees after the rotation.

But Lorentz transformations in 2D look like rotating both space and time axes by an angle of same magnitude but opposite signs. The axes don't remain at 90 degrees after Lorentz transformations, instead they form a V-shape. How is this a rotation? Are we using some generalised definition of rotation? Also, why not just rotate both space and time axes in the same direction (so that they remain at 90 degrees), like we do with two space axes? (I've read the reason we don't treat the time-axis like a usual space-axis is that we can't move backwards in time. If this reason is correct, then please elaborate on this. How does rotating both x and t axes in the same direction mean we are going backwards in time?)

EDIT- https://www.mathpages.com/rr/s1-07/1-07.htm I'm coming off this text. Near the end of the first paragraph, it says that the reason we choose to do the 'opposite sign' rotation is that we can't go backwards in time.

I've removed a number of comments that were attempting to answer the question and/or responses to them. Please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. — David Z, Apr 15 '20 at 11:01

score 14 · Answer 1 · edited Apr 15 '20 at 17:12

14

The relevant generalized notion of "rotation" is that a rotation is a transformation that fixes one point and preserves all distances. In euclidean space, this means that if you have two points with x-coordinates differing by $\Delta x$ and y-coordinates ďiffering by $\Delta y$, then the value of $\Delta x^2+\Delta y^2$ is unaffected by a rotation. In Minkowski space it means that $\Delta x^2-\Delta t^2$ is unaffected.

edited Apr 15 '20 at 17:12

John Rennie

355,118

answered Apr 15 '20 at 02:55

WillO

15,072

1

What motivates us to use the minus sign in the latter formula (apart from observing constant speed of light)? Is it that if we kept it positive, then it'd imply we can move backwards in time? – Ryder Rude Apr 15 '20 at 02:58
4

What motivates the minus sign is that we want rotations to keep the speed of light constant, as you surmised. – WillO Apr 15 '20 at 03:02
2

There is some other motivation too. https://www.mathpages.com/rr/s1-07/1-07.htm This text predicts Special Relativity only using pure mathematics (without relying on observing anything to move at the maximum speed). Use 'Find in page : backwards in time' on that page. – Ryder Rude Apr 15 '20 at 03:07
1

@WillO Nice and to-the-point answer, but I think it might benefit from stressing a bit more that the definition of "distance" in this context can vary from one type of rotation to another, and that for the Lorentz transformation, it's a distinctly different notion of "distance" than the one we're used to from everyday life. – David Z Apr 15 '20 at 11:03

score 12 · Accepted Answer · answered Apr 15 '20 at 05:09

A Lorentz boost is not a rotation by a real angle. Instead. it is a strain by a real angle. The transformation of the x,t axis where both move inward by a small angle in radians $d\lambda$ (called the Lorentz boost parameter) is well known to mechanical engineers in the x,y plane. The engineer distorts a square in the x,y plane so that both edges of the square move inward by a small angle in radians $d\epsilon$ (called the strain). The square becomes a parallelepiped. The matrices which do these transformations for non-infinitesimal angles are: $$ \begin{bmatrix} cosh(\lambda) & sinh(\lambda)\\ sinh(\lambda) & cosh(\lambda)\\ \end{bmatrix} \begin{bmatrix} x\\ ct\\ \end{bmatrix} \quad and \quad \begin{bmatrix} cosh(\epsilon) & sinh(\epsilon)\\ sinh(\epsilon) & cosh(\epsilon)\\ \end{bmatrix} \begin{bmatrix} x\\ y\\ \end{bmatrix} $$ The reason you have heard that boosts are somehow rotations is that old time physicists made boosts look like familiar rotations by using imaginary angles and making t imaginary.

$$ \begin{bmatrix} x'\\ ict'\\ \end{bmatrix} = \begin{bmatrix} cos(i\lambda) & -sin(i\lambda)\\ sin(i\lambda) & cos(i\lambda)\\ \end{bmatrix} \begin{bmatrix} x\\ ict\\ \end{bmatrix} $$ $$ \begin{bmatrix} x'\\ ict'\\ \end{bmatrix} = \begin{bmatrix} cosh(\lambda) & -i\ sinh(\lambda)\\ i\ sinh(\lambda) & cosh(\lambda)\\ \end{bmatrix} \begin{bmatrix} x\\ ict\\ \end{bmatrix} $$ $$ \begin{bmatrix} x'\\ ct'\\ \end{bmatrix} = \begin{bmatrix} cosh(\lambda) & sinh(\lambda)\\ sinh(\lambda) & cosh(\lambda)\\ \end{bmatrix} \begin{bmatrix} x\\ ct\\ \end{bmatrix} $$ Space-space parallelepiped strains leave $x^2-y^2$ invariant. Space-time parallelepiped strains leave $x^2-(ct)^2$ invariant. Rotations leave $x^2+(ict)^2$ invariant. Please see my answer to this question if you would like more math.

I don't think "the reason we don't treat the time-axis like a usual space-axis is that we can't move backwards in time" is a good argument. However, if $ct>x$ a boost can not make $ct'<x'$ because $x^2-(ct)^2$ is invariant. Thus if an event is causal in the forward light cone then it is causal in the forward light cone in all boosted frames too. A real rotation of x and real ct could turn $ct>x$ into $ct'<x'$ and screw up causality.

score 2 · Answer 3 · answered Apr 15 '20 at 21:34

2

Wikipedia has some nice animations showing that Lorentz transformations are indeed some sort of rotation, e. g. this one

or this

One might say that they do their best to look like rotation under the constraint of not crossing the lightcone

answered Apr 15 '20 at 21:34

მამუკა ჯიბლაძე

513

How exactly are Lorentz transformations rotations?

3 Answers3

Linked