I am reading a paper that makes the claim that $$ \Lambda^\mu {}_\nu = g^\mu{}_\nu - 2 \frac{(V + W)^\mu (V + W)_\nu}{(V + W)^2} + 2 \frac{W^\mu V_\nu}{V^2} $$ is a proper Lorentz transformation, where $V$ and $W$ are four-vectors such that $V^2 = W^2$.

This transformation is of interest because $$ \Lambda^\mu {}_\nu \, V^\nu = V^\mu - 2 (V + W)^\mu \frac{(V^2 + V \cdot W)}{V^2 + 2V \cdot W + W^2} + 2 W^\mu = V^\mu - (V^\mu + W^\mu) + 2 W^\mu = W^\mu. $$ Thus $\Lambda (V) = W$.

If instead, placing no restrictions on $X$ and $Y$ except that $Y^2 \neq 0$, we define: $$ \Lambda'^\mu {}_\nu = g^\mu{}_\nu - \frac{(X + Y)^\mu (X + Y)_\nu}{X \cdot Y + Y^2} + 2 \frac{X^\mu Y_\nu}{Y^2}, $$ it is easy to see similarly that $$ \Lambda'^\mu {}_\nu \, Y^\nu = Y^\mu - (X + Y)^\mu + 2 X^\mu = X^\mu $$ and so $\Lambda' (Y) = X$.

Clearly this defines a Lorentz-transformation-like $4\times 4$ matrix that transforms one four-vector $Y$ into another of our choosing $X$.

  1. I can see that for $\Lambda'$ to be a Lorentz transformation, it is a necessary condition that $X^2 = Y^2$. Is this also a sufficient condition, and is there an easy way to see why?

  2. Where $X^0 = Y^0$ and $X^2 = Y^2$, I assume this has to reduce to a spatial rotation in order to be a Lorentz transformation - but I can't see how. Does it?

  3. Since the Lorentz group has six degrees of freedom, and we have presumably exhausted three of them in fixing $\Lambda(Y) = X$, there should be three left. What do these correspond to?

  4. Is there any other (perhaps more general) standard way to construct a Lorentz transformation that gives $\Lambda(Y) = X$? I assume you can do it by composing boosts into/out of rest frames but perhaps there's a more direct approach.

  • $\begingroup$ is it obvious how the part of middle term with the -2 factor simplifies as suggested? This would entail $2\frac{V^2+V\cdot W}{V^2+ 2V\cdot W+W^2}=1$… $\endgroup$ Jan 6 at 5:36
  • 3
    $\begingroup$ Which paper? Which page? $\endgroup$
    – Qmechanic
    Jan 6 at 5:36
  • $\begingroup$ @ZeroTheHero $V^2=W^2$ so that fraction does simplify to 1, essentially by construction. $\endgroup$
    – JCW
    Jan 6 at 12:16
  • $\begingroup$ @Qmechanic Page 40 and eq. 5.144 of arxiv.org/abs/hep-ph/9605323 $\endgroup$
    – JCW
    Jan 6 at 12:19
  • 1
    $\begingroup$ @JCW duh! :( It would have helped if I had used this relation rather than being hypnotized by the original ratio. $\endgroup$ Jan 6 at 12:37

1 Answer 1


Possibly interesting reading:

(From my answer at https://www.physicsforums.com/threads/lorentz-boost-matrix-in-terms-of-four-velocity.727933/post-4600482)

  • Fahnline "A covariant four‐dimensional expression for Lorentz transformations",
    Am. J. Phy. 50, 818 (1982); https://doi.org/10.1119/1.12748 $$T^\mu{ }_\nu=g^\mu{ }_\nu-\frac{2 v_B{ }^\mu v_{A\nu}}{c^2}+\frac{\left(v_A{ }^\mu+v_B{ }^\mu\right)\left(v_{A \nu}+v_{B \nu}\right)}{c^2-v_A \cdot v_B}\qquad (16)$$
  • Fahnline "Manifestly covariant, coordinate‐free dyadic expression for planar homogeneous Lorentz transformations"
    J. Math. Phys. 24, 1080 (1983); https://doi.org/10.1063/1.525833
  • Celakoska "On Isometry Links between 4-Vectors of Velocity"
    Novi Sad J. Math. 38(3), 2008, 165-172; http://www.emis.de/journals/NSJOM/Papers/38_3/NSJOM_38_3_165_172.pdf $$(2)\qquad B(U, V)=\delta-\frac{(U+V) \otimes \eta(U+V)}{c^2+U \cdot V}+2 \frac{V \otimes \eta U}{c^2} $$

(In the above references, one should take note of differences in notation and signature-conventions.)

Related: Find boost factor for general Lorentz Transformation


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