I am reading about SR. The book starts by going over the ordinary rotation matrix, e.g.

$\left( {\begin{array}{cc} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\ \end{array} } \right)$

which takes $x\rightarrow x'$, $y\rightarrow y'$.

Then it goes on to say that if the rotation involves an imaginary axis (like the imaginary time axis $ict$) then the angle of rotation needs to be imaginary as well, like

$\left( {\begin{array}{c} ict' \\ x' \\ \end{array} } \right)=\left( {\begin{array}{cc} \cos i\theta & -\sin i\theta \\ \sin i\theta & \cos i\theta \\ \end{array} } \right)\left( {\begin{array}{c} ict \\ x \\ \end{array} } \right)$

which can then be written with hyperbolic functions instead of ordinary trig. Could somebody explain why the angle needs to be imaginary? Of course a complex number can be rotated by multiplying by $e^{i\theta}=\cos\theta + i\sin\theta$ but I don't understand why here the angle itself is imaginary. Some mathematical explanation would be very helpful. thanks


Note first that working with imaginary time to teach special relativity has fallen way out of favor, it's not considered good pedagogy! And I don't think "hey plug in an imaginary axis" is a very good physical or mathematical argument. But with that out of the way...

The only thing I'm going to do is motivate why when you plug in an imaginary axis, the $i$ hits the angle.

You're right that it has to do with $e^{i\theta}$. It is more enlightening to work with matrices. Denote the identity matrix ${\bf 1}=\begin{bmatrix} 1 & 0 \\ 0 & 1\end{bmatrix}$ and denote another matrix $ i=\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$. Note that $I^2=-\bf 1$, so the matrices of the form $a {\bf 1}+b I$ behave exactly like the complex numbers. So we can write $e^{I \theta}=\cos{\theta}{\bf 1}+\sin(\theta) I=\begin{bmatrix} \cos(\theta) & -\sin(\theta) \\ \sin(\theta) & \cos(\theta)\end{bmatrix}$, where this is the infinite matrix sum ${\bf 1}+I\theta+\frac{1}{2}I^2 \theta^2+\cdots$

In the same way, if you're in three dimensions and you have a vector $\vec{\theta}$, you can create the rotation by an amount $\|\vec{\theta}\|$ about the $\hat{\theta}$ axis, by calculating

$$R=\exp\left( \begin{bmatrix} 0 & -\theta_z & \theta_y \\ \theta_z & 0 & -\theta_x \\ -\theta_y & \theta_x & 0\end{bmatrix} \right)$$

(To prove that it rotates about that axis, note that the vector $(\theta_x,\theta_y,\theta_z)$ times the matrix in the exponential equals zero. So in the Taylor series for exp, identity leaves the vector alone, and every other term just gives zero).

Plugging in $\theta_z=i s$ (in a completely ad hoc manner!) gives:

$$R=\exp\left( i s \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix} \right)$$

ignoring the third axis gives $R=\exp(i s I)$. And according to our previous formula for $e^{I \theta}$, this is just equal to $R=\begin{bmatrix} \cos(is) & -\sin(is) \\ \sin(is) & \cos(is) \end{bmatrix}$, as the book says.

I don't think the author is trying to state a theorem though, I think they're just trying to motivate hyperbolic rotations. Their argument sounds like it's on the basis of "this sounds reasonable, right?"

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  • $\begingroup$ I was just about to say the same thing. $x_4 = ict$ is obsolete notation. $\endgroup$ – DanielC Sep 19 '17 at 8:10

This is simply an application of the identities $i\,\sin z = \sinh(i\,z);\;\cos z=\cosh(i\,z)$. Don't try to read too much more into it than that. I'm not sure what book you're reading, but my personal opinion is that the definition of imaginary time so that Minkowski spacetime "looks Euclidean", which looks very powerful and promising at first, is mostly highly unhelpful to the intuition in relativity. Imaginary time has analytical uses such as Wick rotation, but this is really a way of making certain calculations easier through the techniques of analytic continuation rather than being physically highly meaningful.

This is because the complexification of spacetime co-ordinates destroys the crucial, and physically highly meaningful, notion of metric signature.

Rotations really are different to boosts, or at least we need to think of them thus in relativity. Boosts conserve the direction of the timelike component of a vector - rotations do not. Therefore, in relativity, although simultaneity of events is observer dependent, the order of events is not as long as we consider only less-than-lightspeed relatively moving observers. The fact of my needing to boil eggs before I can have boiled eggs for breakfast (yum!) is a Lorentz invariant fact and this fact is down to the signature of the metric in our universe. Causality is a notion that is probably impossible to make sense of in a universe where the signature were all positive. The compatibility of the experimentally universally treu notion of causality with relativity as long as we stick to subluminal boosts is probably the deepest theoretical argument against faster than light relative motion there is.

Another crucial difference, also down to signature, between rotations and boosts is that the former form a group, the latter do not. Two rotations compose to a rotation (Euler's theorem) but two non collinear boosts compose to a boost together with a rotation. This is something genuinely new and truly quite weird if you're only used to rotation composition; this rotation through the composition of boosts is called Wigner Rotation (not to be confused with the Wick rotation I spoke of earlier) or Thomas Precession. This fact is obscured through the example you give: two boosts in the same direction do indeed compose to a boosts, and there is a one parameter boost group ${\rm SO}(1,1)$.

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  • $\begingroup$ This is a popular interpretation, but to me it sounds blown out of proportions. IMHO the complexification of spacetime makes more physical sense than the "metric signature", as long as time is real while space is imaginary. In either interpretation the geometry is hyperbolic anyway, but imaginary space is a deeper concept than the metric signature. No real object can take a spacelike path. Even the length standard we use (meter) is defined as a fixed fraction of the light-second. All this is way beyond the scope of the OP's question while he has no votes to even post a comment in response :) $\endgroup$ – safesphere Sep 18 '17 at 0:52

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