# Is this a one dimensional Lorentz Boost? And can you have a 1-d Boost without group structure?

Someone has claimed that he has constructed a quaternion representation of the one dimensional (along the x axis) Lorentz Boost.

His quaternion Lorentz Boost is $v'=hvh^*+ 1/2( [hhv]^*-[h^*h^*v^*]^*)$ where h is (sinh(x),cosh(x),0,0). He derived this odd transform by substituting the hyperbolic sine and cosine for the sine and cosine in the usual unit quaternion rotation $v'=hvh^*$ and then subtracting out unwanted factors. You can see the short "proof" here:

http://visualphysics.org/preprints/qmn10091026

I have argued that, whereas the quaternion rotations form a group, his newly devised transform probably does not. Two transformations of the form $v'=hvh^*+ 1/2( [hhv]^*-[h^*h^*v^*]^*)$ almost certainly do not make another transformation of the form $v'=fvf^*+ 1/2( [ffv]^*-[f^*f^*v^*]^*)$ He has not attempted to prove that and he won't because he thinks it is unnecessary.

He has responded that, even if I'm correct, he still has created a Lorentz Boost despite the fact that he has not created a group. I've argued that that it is essential that two Lorentz Boosts make another Lorentz Boost. Without that group structure there is no Boost. Is this correct?

Can you have a Lorentz Boost along the x axis without having the group structure Boost+Boost=Boost? (for the case of Boosts along the x axis. I know that, for boosts in different directions, Boost+Boost=Rotation)

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Boost along the $x$ direction. The result is $X' = \Lambda(\phi) X$, where $$X = \left( \begin{array}{c} c t \\ x\end{array}\right) \hspace{5ex} \textrm{and} \hspace{5ex} \Lambda(\phi) = \left( \begin{array}{cc} \cosh \phi & -\sinh\phi \\ -\sinh\phi & \cosh \phi \end{array}\right),$$ and where $\phi$ is the rapidity.

Boost again along the $x$ direction, $X'' = \Lambda(\phi')X' = \Lambda(\phi')\Lambda(\phi) X$. Using the properties of the hyperbolic trigonometric functions, we can see that $X'' = \Lambda(\phi'') X$, where $\phi'' = \phi + \phi'$. (Do the matrix multiplication, $\Lambda(\phi')\Lambda(\phi)$.) Thus, the composition of boosts along the $x$ direction is another boost along the $x$ direction. The rest of the group axioms are also satisfied. Thus, the collection of boosts along the $x$ direction form a group.

The group structure is independent of the representation. If the "representation" is not closed under composition, it is definitely not a representation of the group.

Lorentz transformations can be represented by quaternions. See here for example.

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This is really just a matter of deciding if his idea to call this a "Lorentz boost" is justified or not. It's a choice of terminology, not a fixed mathematical idea that needs a mathematician to decide whether it is right or wrong.

For regular Minkowski space, the composition of two boosts need not be a pure boost: the composition generally incorporates a spatial rotation. So I don't see any good reason that boosts should be required to form a group (although all of the Lorentz transformations form a group.)

Edit: After rereading I see that you probably know that, but are just questioning whether or not boosts composing to a nonboost would occur in a 1-dimensional setting. Unless his definition is a special case or version of higher dimensional cases, this could be debatable. Be sure to post in the physics forum too!

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