# General Lorentz transformation between two four-vectors

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.

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

Am. J. Phy. 50, 818 (1982); https://doi.org/10.1119/1.12748 $$T^\mu{ }_\nu=g^\mu{ }_\nu-\frac{2 v_B{ }^\mu v_{A\nu}}{c^2}+\frac{\left(v_A{ }^\mu+v_B{ }^\mu\right)\left(v_{A \nu}+v_{B \nu}\right)}{c^2-v_A \cdot v_B}\qquad (16)$$
Novi Sad J. Math. 38(3), 2008, 165-172; http://www.emis.de/journals/NSJOM/Papers/38_3/NSJOM_38_3_165_172.pdf $$(2)\qquad B(U, V)=\delta-\frac{(U+V) \otimes \eta(U+V)}{c^2+U \cdot V}+2 \frac{V \otimes \eta U}{c^2}$$