Lorentz vectors orthogonality

Question

I am reading through the book "Electrodynamics and Classical Theory of Fields and Particles" by A.O. Barut, and I'm still currently in the first chapter of the book in which he is discussing Lorentz vectors and their orthogonality properties. At one point he makes the statement that for given light-like vector, there are only two space-like vectors that are mutually orthogonal to one another and orthogonal to the light-like vector. He then goes on to state that there are three mutually orthogonal space-like vectors. The latter statement seems a lot more intuitive, but both go unproven in the book. So I am pretty curious about how to prove these things, and I tried for a little while today and last night, but I sadly made no progress. Does anyone here know how to prove these things? I would greatly appreciate it!

Answer 1

He then goes on to state that there are three mutually orthogonal space-like vectors.

Sure, the proof is by example: $$\epsilon_1^\mu = (0, 1, 0, 0), \quad \epsilon_2^\mu = (0, 0, 1, 0), \quad \epsilon_3^\mu = (0, 0, 0, 1)$$ where the time component is on the left.

At one point he makes the statement that for given light-like vector, there are only two space-like vectors that are mutually orthogonal to one another and orthogonal to the light-like vector.

The proof is again by construction. Without loss of generality we can take the light-like vector to be $$k^\mu = (1, 0, 0, 1).$$ Now consider a general space-like other vector, $$\epsilon^\mu = (\epsilon^0, \epsilon^1, \epsilon^2, \epsilon^3).$$ Orthogonality to the light-like vector implies $\epsilon^0 - \epsilon^3 = 0$. Since the vector is spacelike, $$(\epsilon^0)^2 < (\epsilon^1)^2 + (\epsilon^2)^2 + (\epsilon^3)^2 = (\epsilon^1)^2 + (\epsilon^2)^2 + (\epsilon^0)^2$$ so the only constraint is that $\epsilon^1$ and $\epsilon^2$ can't both be zero. Now if we consider two such vectors and demand they be orthogonal to each other, we have $$\epsilon_i \cdot \epsilon_j = \epsilon_i^\mu \epsilon_j^\nu \eta_{\mu\nu} = \epsilon_i^1 \epsilon_j^1 + \epsilon_i^2 \epsilon_j^2 = 0.$$ In other words, the two-component vectors $(\epsilon_i^1, \epsilon_i^2)$ have to be nonzero and orthogonal to each other, so clearly at most two vectors can satisfy this. Explicitly, we can take $$\epsilon_1^\mu = (0, 1, 0, 0), \quad \epsilon_2^\mu = (0, 0, 1, 0).$$

Answer 2

Let $\{ \gamma_0, \gamma_1, \gamma_2, \gamma_3 \}$ be an orthonormal basis for Minkowski spacetime, with inner product $$\gamma_\mu \cdot \gamma_\nu = \eta_{\mu\nu}$$ and timelike basis vector $\gamma_0$.

Every lightlike vector can be expressed as a Lorentz transformation of $\gamma_0 + \gamma_3$, to which only two spacelike vectors are orthogonal: $\gamma_1$ and $\gamma_2$.

We can confirm this using the inner product above, $\gamma_1 \cdot (\gamma_0 + \gamma_3) = 0$ and $\gamma_2 \cdot (\gamma_0 + \gamma_3) = 0$, while $\gamma_3 \cdot (\gamma_0 + \gamma_3) = \eta_{33} \not= 0$.