Hyperbolic relations in deriving Lorentz transformations

Muphrid

I'm speaking in a more fundamental sense. When you speak of the unit circle, you already have an idea that there is a set of points equidistant from the origin. You need to already have a distance formula--a metric--to do this, and where in Euclidean geometry you get a circle, in Minkowski space you get a hyperbola.

But you're at that point already (I wasn't sure that you were), so that's fine. This is the point where I think a matter of interpretation is in order. Let's say you have a plane. You can talk about components of vectors in this plane by choosing a basis.

i.e. say you have a vector [itex]A[/itex]. One extracts components by taking taking dot products. [itex]A^x = A \cdot e^x[/itex], for example. But you can always choose a different basis. You can choose a new [itex]{e^x}'[/itex]. With a metric--a notion of distance--you can ensure that this vector is normalized, and you can compute its components based on the unit set (the unit circle in Euclidean space, the unit hyperbola in Minkowski).

I feel this is really important: when you choose a new basis, you can express those vectors in terms of the old basis, and in Minkowski space, this is what introduces the hyperbolic trig functions characteristic of the Lorentz transformations.


Let me work the example: one chooses the basis [itex]e_t, e_x[/itex] to describe the xt plane in a Minkowski space--in particular, [itex]e_t \cdot e_t = -1[/itex] and [itex]e_x \cdot e_x = 1[/itex].

Now, we can change the basis describing this plane. Let our new time vector [itex]{e_t}'[/itex] be proportional to [itex]e_t + \beta e_x[/itex]. This should describe someone whose worldline moves in the positive x direction. When we normalize this vector, we get factors of gamma: [itex]{e_t}' = \gamma(e_t + \beta e_x)[/itex].

Now, we need to find a corresponding [itex]{e_x}'[/itex] to go with this vector. To do that, we can invoke a process like the Gram-Schmidt procedure. [itex]{e_x}'[/itex] should be in the direction of [itex]e_x - (e_x \cdot e^t)e_t[/itex]. Consider this an exercise to show that [itex]{e_x}' = \gamma(e_x + \beta e_t)[/itex].

Now then, take a position vector [itex]x e_x + t e_t[/itex]. Evaluate its components in the x't' frame:

[tex](x e_x + t e_t) \cdot {e^x}' = (x e_x + t e_t) \cdot (\gamma e^x - \gamma \beta e^t) = \gamma (x - \beta t) \\
(x e_x + t e_t) \cdot {e^t}' = (x e_x + t e_t) \cdot (\gamma e^t - \gamma \beta e^x) = \gamma (t - \beta x)[/tex]

This is a critical point: while [itex]{e_x}' = \gamma e_x + \gamma \beta e_t[/itex], the reciprocal basis vector [itex]{e^x}' = \gamma e^x - \gamma \beta e^t[/itex], and the same for the time vectors. This follows directly from the metric (but if you like, this can be more explicitly shown as well).

The components are now computed, and all that remains is to make the connection between [itex]\gamma, \beta[/itex] and the hyperbolic trig functions.