homework and exercises - Existence proof of Lorentz transformation from lightlike to lightlike vectors

This is a question that might have quiet an easy answer and intuitively has an easy solution, but I struggle a bit with the strict mathematical proof. The statement is relatively simple:

Let $t^\mu, u^\mu$ be two non-zero lightlike four-vectors, so $t_\mu t^\mu = u_\mu u^\mu = 0$. There is always a Lorentz transformation $\Lambda$ so that $\Lambda^\mu_{\nu} t^\nu = u^\mu$.

I assume that the proof might be quite easy and somewhat elegant, but I've looked at this problem for quite a while now and seemed to take the wrong paths which only lead to complications. Maybe someone has got a keyword or a fresh idea.

Answer

(I took a trivial mistake beforehand. Your statement is true, here is the proof.)

I assume $c=1$ throughout the proof.

Consider $t\neq 0$ and $u\neq 0$ both light-like and future-oriented. Fix a Minkowski reference frame with future-oriented temporal coordinate.

In coordinates $t= (t^0, \vec{t})$ and $u= (u^0, \vec{u})$ with $t^0= ||\vec{t}||$ and $u^0= ||\vec{u}||$ because both vectors are future-oriented and light-like.

Exploiting a Lorentz transformation $\Lambda_R$ completely defined by a spatial rotation $R$, we can transform $t= (||\vec{t}||, \vec{t})$ to $\Lambda_R t = (||\vec{t}||, R\vec{t}) = (||R\vec{t}||, R\vec{t})$, where $R\vec{t}$ is parallel to $\vec{u}$.

To conclude, it is enough proving that there exist a boost $\Lambda_b$ such that $\Lambda_b \Lambda_R t =u$.

For our convenience we rotate the spatial axes of our reference frame in order to have $\vec{u}$ and $\vec{\Lambda_Rt}$ directed along $x^3$. In these coordinates

$$\Lambda_R t = (a,0,0,a)\quad u = (b,0,0,b)\:,$$

The action of the boost $\Lambda_b$ along $x^3$ is, for some $\chi \in \mathbb R$,

$$\Lambda_b (a,0,0,a) = (\cosh \chi a + \sinh \chi a, 0,0, \sinh \chi a + \cosh \chi a)$$

The problem reduces to finding $\chi \in \mathbb R$ such that

$$\cosh \chi a + \sinh \chi a =b$$

where $a,b>0$ are given. Namely

$$e^\chi a =b$$

the solution always exists and is, obviously,

$\chi = \ln (b/a)$.

Summing up, if we apply first $\Lambda_R$ and then $\Lambda_p$ to $t$ we obtain $u$ as wanted. The composition $\Lambda_p \Lambda_R$ is the wanted Lorentz transformation.

In case $t$ and $u$ have opposite temporal directions, the reasoning above must be completed by adding a further time reversal operation to $t$ (which is a Lorentz transformation too) before using $\Lambda_R$ and $\Lambda_p$.

COMMENT. This result is interesting because it proves that the action of the Lorentz group is transitive on the boundary of the light cone. This fact is instead false if referring to the interior of the light cone, because Lorentz transformations preserve the Lorentzian length of vectors and thus vectors with different length cannot be mapped to each other by any Lorentz transformation. (The surface of the light cone is the limit case of an internal surface of fixed-length vectors [a mass shell in momentum representation] in agreement with the found result.)