How to visualize rotation on a hyperbola?

Question

I am studying Lorentz transform and I do not quite get what it means to use the hyperbolic matrix to rotation a point on a hyperbola, mainly it is because the hyperbola consists of two divergent curves (with a huge space in between that nothing can jump across) and the curves don't bend the way that circles do.

If I were to draw the $0$ and $180^{\deg}$ radial plot on top of a hyperbola, where would they be?

Just to make it perfectly elucidating, if I were a red block lying on a curve of the hyperbola, what does it mean for me to rotate? Where would I be after rotation?

Answer 1

When we have a point on a sphere in Euclidean space this means it falls a certain distance from the center of the sphere. This distance is the radius and, assuming the center is the origin, we have $x^2+y^2+z^2=R^2$. If we rotate the point about the origin then it is still on the sphere.

In a similar way, if we take an event in space-time then it has a certain interval from the origin. Depending on what you read, either $-c^2t^2+x^2+y^2+z^2=I$ or $-I$. Let's suppose the signature is $(-1,1,1,1)$ for the Minkowski metric. Then, an operation which leaves the interval invariant is known as a Lorentz transformation. We can move the event around on the hyperboloid $-c^2t^2+x^2+y^2+z^2=I$ in space time by a Lorentz transformation. Suppose we drop $y,z$ and set $c=1$ then $-t^2+x^2=I$ is what is left. If we move the event to $t',x'$ where $$\left[ \begin{array}{c} t' \\ x' \end{array} \right] = \left[ \begin{array}{cc} \gamma & -\gamma \beta \\ -\gamma \beta & \gamma \end{array} \right]\left[ \begin{array}{c} t \\ x \end{array} \right] $$ where $\gamma = \cosh \phi$ and $\beta \gamma = \sinh \phi$ then \begin{align} -(t')^2+(x')^2 &= -(\gamma t-\gamma \beta x)^2+(-\gamma \beta t+\gamma x)^2 \\ &= \gamma^2\left(-t^2+2\beta xt-\beta^2x^2+\beta^2 t^2-2\beta xt + x^2\right) \\ &= \gamma^2 (1-\beta^2) \left( -t^2 +x^2 \right) \end{align} But, $\gamma^2-\beta^2 \gamma^2 = \cosh^2 \phi - \sinh^2 \phi = 1$ hence $$ -(t')^2+(x')^2 = -t^2+x^2 $$ I wouldn't recommend a direct visualization. But, perhaps the analogy to rotations in three dimensions is helpful. The concept here is that the event in space time is geometric, but the coordinates $t,x,y,z$ are merely based on the artificial choice of a reference frame. Einstein told us, the laws of physics should be independent of that choice (at least he told us this in 1905, maybe the story changes in 1916). The reason for these weird Lorentz transformations is that they are necessary for us to keep the speed of light invariant under these mixing transformations between space and time.

Answer 2

Wikipedia has a couple of nice animations showing Lorentz transformations. First there's one that shows how your frame of reference changes continuously as you travel along a world line:

Another one shows a single hyperbolic rotation. It won't upload to Stack Exchange for some reason, but here's a link:

http://upload.wikimedia.org/wikipedia/commons/b/b4/Animated_Lorentz_Transformation.gif

Answer 3

When you take a point on a circle and rotate it in Euclidean space, the point ends up somewhere else on the circle.

When you take a point on a hyperbola and "rotate" it in Lorentzian space, it ends up somewhere else on the hyperbola.

Now, in the Lorentzian case, we have some caveats: while the solution set to $x^2 - y^2 = 1$ has two disjoint curves, there are no rotation operations that can transform a point from one of those curves to the other.

Moreover, while rotation in Euclidean space is periodic (and hence, we can identify $\theta$ with $\theta + 2\pi n$ for integer $n$ for the purposes of rotation), rotation in Lorentzian space is not periodic. Hyperbolic angles are unbounded.

Taking your red block and rotating it hyperbolically through some huge angle will send it down one direction or the other of the hyperbolic curve it sits on. The red block can be rotated to any other point on the same hyperbolic curve, but there is one and only one angle that corresponds to such a rotation--again, unlike Euclidean space, where rotations are periodic.