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 $X$.

*

*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?


*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?


*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?


*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.
 A: 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
