I was deriving the matrix form of Lorentz boosts and I came up with a doubt. I don't think I quite understand hyperbolic rotations.

The stadard basis of the Minkowski space is given by $\{e_0, e_1,e_2,e_3\}$, where the square norm of any vector $x=x_0 e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3$ is $|x|^2 = x_0 ^2 - x_1 ^2 - x_2 ^2 -x_3 ^2$. Note: $x_0$ is the time component ($e_0$ indicates time axis), $x_1, x_2, x_3$ is the spatial component ($e_1 , e_2, e_3$ indicate the $x, y,$ and $z$ axis).

Let $B_i$ be a Lorentz boost in the ith direction. This boost will only modify the time component and the $ith$ component, and like any other lorentz transformation, it will preserve the norm of any vector. Consider $B_i e_0 = a e_0 +b e_i = e_0 '$. Then,

$(B_i e_0)^2 = e_0 ^2 = 1$

$a^2 - b^2 = 1$

The solutions of this lie on a hyperbola along the $e_0$ axis. We can parametrize it and obtain:

$a=cosh \theta$ and $b=sinh\theta$

I do the same procedure for $B_i e_i = a e_0 +b e_i = e_i '$. $a$ and $b$ satisfy the equation: $b^2 -a^2 = 1$ which is a hyperbola along the $e_i$ axis.

My intuition: I imagine the $e_0$ as the horizontal axis in the x-y plane and the $e_i$ as the vertical axis. If I shift $e_0$ by and angle $\theta$ (counterclockwise) over the hyperbola, the resulting vector $e_0'$ will fall on the first quadrant. So $e_0 ' = cosh \theta e_0 + sinh \theta e_i $. Now I need to shift $e_i$ by the same angle (counterclockwise) over the corresponding hyperbola so that I can keep $e_i$ and $e_0$ orthogonal to each other. The resulting vector $e_i'$ will fall over the 2nd quadrant. So $e_i ' = -sinh \theta e_0 + cosh \theta e_i $

For $i=1$, I end up with a boost $B_i$ of the form: $$B_i= \left( \begin{matrix} \cosh \theta & -sinh \theta & 0 & 0\\ sinh \theta & cosh \theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix}\right) $$ when I should end up with:

$$B_i= \left( \begin{matrix} \cosh \theta & sinh \theta & 0 & 0\\ sinh \theta & cosh \theta & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ \end{matrix}\right) $$ Edit: The columns of this matrix are correct because the matrix satisfies Lorentz transformation condition $\eta = B_1 ^T \eta B_1$ (where $\eta$ is the minkowski metric tensor) while my boost matrix doesn't.
Question : Since taking the square might make me lose/add a negative sign, is there a better way to obtain $e_0'$ and $e_1'$ with all the necessary negative signs in its components? Right now I would just be using trial an error, until the boost matrix I obtain satisfies $\eta = B_1 ^T \eta B_1$ I appreciate any insights, thanks!


migrated from math.stackexchange.com Nov 16 '17 at 23:01

This question came from our site for people studying math at any level and professionals in related fields.

  • $\begingroup$ You want "Orthogonal in Minkowski geometry", right? As rapidity varies, both the timelike vector and its spacelike orthogonal either approach or recede from the lightlike direction. $\endgroup$ – robphy Nov 16 '17 at 23:46
  • $\begingroup$ because the product of the e_0 components need to have the same sign as the product of the e_i components right? I thought about orthogonality (with minkowski metric) between column vectors before, but then it seemed that kind of intuition only applied to real orthogonal matrices, when their transpose is the inverse of the matrix. The constraint in Lorentz transformations is not that (although kind of similar), so I thought this intuition was wrong and it only had to do with satisfying \eta = B^T \eta B. I'm new in this I appreciate the insight. $\endgroup$ – Cami77 Nov 17 '17 at 0:12

In your attempted construction, you first use the future unit-hyperbola for the tips of your unit-timelike vectors. Note that this hyperbola has spacelike tangents. In fact, following Minkowski's own construction from "Space and Time"

We decompose any vector, such as that from O to x, y, z, t into four components x, y, z, t. If the directions of two vectors are, respectively, that of a radius vector OR from O to one of the surfaces ∓F = 1, and that of a tangent RS at the point R on the same surface, the vectors are called normal to each other. Accordingly, $$c^2tt_1 − xx_1 − yy_1 − zz_1 = 0$$ is the condition for the vectors with components x, y, z, t and $x_1$, $y_1$, $z_1$, $t_1$ to be normal to each other.

If R is the tip of the unit-timelike vector OR from O, the tangent to the hyperbola at R is [Minkowski-]orthogonal to OR. The "intuition" to have is that the tangent to the "circle" in that geometry is orthogonal to the radius vector.

(You don't need to use the other hyperbola [with timelike tangents].)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.