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As far as I know, a Lorentz transformation is a linear map $\Lambda$ that preserves the Minkowski metric (which, for concreteness, I'll take with signature $(+,+,+,-)$). Then $\Lambda$ is orthochronous if $\lambda_{44} >0$ and is proper if $\det \Lambda =1$. So far, so good, I guess.

Here, for example, a pure boost in the direction of a unit vector (in $\Bbb R^3$) is defined as a proper and orthochronous Lorentz transformation that leaves unchanged the normal plane to the vector (in $\Bbb R^3$, again). In some other sources, such as Wikipedia, we see stuff such as "boosts" and "pure Lorentz transformations".

I'd like to know which precisely is the difference between these concepts. What is the difference between a "boost", a "pure boost", and a "pure Lorentz transformation"? What does the "pure" entails?

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The proper orthochronous Lorentz group is, as the name suggest, a group. It may be defined, as you said, as the set of transformations leaving the metric invariant, with unit determinant and positive time component. A usual $\mathbb R^3$ rotation, without involving time, also preserves $ds^2=d\vec r^2-(c\,dt)^2$. Other known type of transformation that doesn't change $ds^2$ is \begin{equation}t\,\to\,\gamma(t-v\,x),\qquad x\,\to\, \gamma(x-v\,t),\tag{1}\end{equation} with $\gamma\equiv1/\sqrt{1-v^2}$ (all of this with $c=1$).

I'd say that "Lorentz transformation" should be the name for any transformation inside the Lorentz group, which could be any of the previous ones, or a combination of them. "Boost", "pure boost" or "pure Lorentz transformation" are names usually given only to the type of transformation in (1). Notice that this would be a boost along the $x$-axis, but you could have boost defined in any arbitrary direction of your space.

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  • $\begingroup$ Then there's no difference? I'm sorry, this doesn't look precise enough for me... $\endgroup$ – Ivo Terek Feb 7 '17 at 18:07
  • $\begingroup$ May I ask where would you like more precision? $\endgroup$ – anonymous Feb 8 '17 at 3:50
  • $\begingroup$ I read it again after resting a bit and I guess you were explicit enough in saying that there's no difference. I computed a "boost" (according to link in question) in $\Bbb R^n_1$ in an arbitrary direction, and for $n=4$ and ${\bf v}={\bf e}_1$ it coincide with your equation (1), so I'm happy. Thanks! $\endgroup$ – Ivo Terek Feb 8 '17 at 15:37
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    $\begingroup$ Good! I think that these confusions come from historical reasons. Probably at the beginning, transformation (1) was called "Lorentz transformation", but after some the definition of the Lorentz group, that term would be ambiguous and that why they came up with "purity". But that's just my opinion. Glad to help. $\endgroup$ – anonymous Feb 9 '17 at 1:54

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