# Is the rotation angle in Minkowski's diagram real or imaginary?

1. May 13, 2009

### gene1721

In Figure_1(b) I have depicted a simplified version of Minkowski's diagram, where

$$\beta = \frac{v}{c}= \tanh \psi= - i \tan i \psi= \frac{( e^{\psi} - e^{-\psi})}{( e^{\psi} + e^{-\psi})}$$,

the rotation of $$(x',t')$$-axes being defined as imaginary (considering x and t real, and c=1).

However, I have found books where this rotation is considered as real, while the rotation in Figure_1(a) is shown as imaginary (considering x real, t imaginary, and c=1).

Could you explain which rotation is real and which imaginary?

2. May 13, 2009

### gene1721

Figure_1 is here!

#### Attached Files:

• ###### Figure_1.jpg
File size:
12.8 KB
Views:
165
3. May 13, 2009

### kuon

Doing a Lorentz transformation (boost) by a velocity v to some 4-vector is equivalent to rotate it an imaginary angle $$\chi$$ such that:

$$\chi = arctan( v )$$

4. May 13, 2009

### kuon

5. May 13, 2009

### gene1721

Thanks Kuon!

If I understand correctly, you come with a third possibility, right?

6. May 14, 2009

### Fredrik

Staff Emeritus
That depends on how you define "rotation". I wouldn't call a Lorentz boost a "rotation". I don't mind calling it a "hyperbolic rotation" or a "Minkowski orthogonal transformation" but calling it a "rotation" is a bit weird in my opinion.

There's no need to bring imaginary numbers into this.

The angle of a rotation is always real and can be defined by writing the matrix as

$$\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}$$

A 2×2 matrix represents a proper rotation if and only if it's orthogonal and has determinant 1. Alll 2×2 matrices with those properties can be put in the form above, with different values of $\theta$.

The rapidity of a Lorentz boost in 1+1 dimensions is always real and can be defined by writing the matrix as

$$\begin{pmatrix}\cosh\theta & -\sinh\theta\\ -\sinh\theta & \cosh\theta\end{pmatrix}$$

A 2×2 matrix $\Lambda$ represents a proper orthochronous Lorentz boost if and only if it satisfies $\Lambda^T\eta\Lambda=\eta$, $\det\Lambda=1$, and $\Lambda_{00}\geq 1$. (That's just its upper left component). All 2×2 matrices with those properties can be put in the form above, with different values of $\theta$. This $\theta$ is usually referred to as the "rapidity" of the Lorentz transformation, not the "angle", but I wouldn't mind calling it the "hyperbolic angle" or even the "angle of a hyperbolic rotation". I just don't want to call it a "rotation angle", as you did in the thread title.

Note that you can't turn one of these matrices into the other just by substituting $\theta\rightarrow i\theta$. That's why it doesn't quite make sense to describe a Lorentz transformation as a rotation by an imaginary angle.

Last edited: May 14, 2009
7. May 14, 2009

### gene1721

Scott Walter (http://www.univ-nancy2.fr/DepPhilo/walter/), who published several papers on Minkowski and his diagram, wrote on page 9 of this article http://www.univ-nancy2.fr/DepPhilo/walter/papers/nes.pdf

We can see now not only that it does make sense to describe a Lorentz transformation as a rotation by an imaginary angle, but to also inquire why would Minkowski really want to use the imaginary rotation. (I have to say that I buy only partially into Walter's explanation!)

Trying to understand why wanted Minkowski to use the imaginary angle, I found in different sources three ways (including Koun's version!) to explain the rotation angle in Minkowski's diagram:

1. x real, t real and the rotation angle real;
2. x real, t real and the rotation angle imaginary;
3. x real, t imaginary and the rotation angle imaginary.

So, given that Minkowski didn't want to use version 1, which between version 2 and version 3 is the correct one?

Last edited by a moderator: Apr 24, 2017