Existence proof of Lorentz transformation from lightlike to lightlike vectors

Question

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 1

(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.)

Answer 2

Proposition 1. The Lorentz group $O(3,1)$ acts transitively on non-zero null-vectors.

Since the time reversal transformation $T:t\mapsto -t$ is a Lorentz transformation, it is enough to show the following.

Proposition 2. The restricted Lorentz group $SO^+(1,3;\mathbb{R})\cong SL(2,\mathbb{C})/\mathbb{Z}_2$ acts transitively on future-directed non-zero null-vectors.

Twistor-inspired sketched proof of Proposition 2: Recall first the following facts:

1. Minkowski 4-vectors $\tilde{x}~=~(x^0,x^1,x^2,x^3)$ can be identified with $2\times 2$ hermitian matrices $\sigma$, see e.g. twistor59's Phys.SE answer here or my Phys.SE answer here.

2. The determinant $\det(\sigma)=||\tilde{x}||^2$ is the quadratic Minkowski form.

3. The trace ${\rm tr}(\sigma)=2x^0$ is twice the time coordinate.

4. A group element $g\in SL(2,\mathbb{C})$ in the double cover of the restricted Lorentz group acts on Hermitian $2\times 2$ matrices $\sigma$ via $\rho(g)\sigma= g\sigma g^{\dagger}$.

Now let's return to OP's setup. Light-like means that $\sigma$ has zero determinant, i.e. it has at most rank 1. In other words, there exist two $2\times 1$ column vectors $\lambda,\mu\in \mathbb{C}^2$ such that the $2\times 2$ matrix $$\sigma~=~\lambda\mu^{\dagger}.$$ Hermiticity of $\sigma$ shows that $\lambda || \mu$. Assume that $\sigma$ is not zero. Then it has precisely rank 1. Future-directed means that the remaining non-zero eigenvalue of $\sigma$ is positive, cf. point 3. ($\lambda$ is the corresponding eigenvector.) We can w.l.o.g. assume that $\lambda=\mu$, possibly after a rescaling.

In summary, there exists a non-zero left Weyl spinor $$\lambda~\in~ (\mathbb{C}^2)^{\times}~:=~\mathbb{C}^2\backslash\{(0,0)\},$$ so that the $2\times 2$ matrix $\sigma$ can be written as $$\sigma~=~ \lambda \lambda^{\dagger}.$$

There is a $U(1)$ little group phase-ambiguity in the choice of $\lambda$. The double cover $SL(2,\mathbb{C})$ acts transitively $\rho(g)\lambda=g\lambda$ on $(\mathbb{C}^2)^{\times}$ via regular $2\times 2$ matrices $g$: For all pairs $\lambda, \mu\in (\mathbb{C}^2)^{\times}$ we can find a $2\times 2$ matrix $g\in SL(2,\mathbb{C})$ with determinant 1 such that $\mu= g\lambda$. Altogether, this shows Proposition 2. $\Box$