# Lorentz vectors orthogonality

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!

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 $$\epsilon_1^\mu = (0, 1, 0, 0), \quad \epsilon_2^\mu = (0, 0, 1, 0).$$

• Okay I don't have a problem with these specific cases, I was wondering how to prove these statements starting from general time-like or space-like vectors. – user132849 Apr 1 at 1:54
• @user132849 Are you looking for a proof that every light-like vector can be Lorentz transformed to (1,0,0,1)? Can you see how the rest follows from that (Minkowski dot products are invariant under a Lorentz transform of both vectors). – Subhaneil Lahiri Apr 1 at 2:17
• If I could see that all light-like vectors can somehow be transformed to (1,0,0,1), then yes, I would understand the rest of the answer. – user132849 Apr 1 at 2:19
• @user132849 First, orient your spatial axes so the spatial part of the vector is along the $z$-axis. Then the vector must be of the form $(k, 0, 0, k)$. Then perform a boost along the $-z$ direction until you get to $(1, 0, 0, 1)$. – knzhou Apr 1 at 2:27
• Yeah I that makes sense I guess. Thanks for the answers! – user132849 Apr 1 at 2:30

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 $$\gamma_3 \cdot (\gamma_0 + \gamma_3) = \eta_{33} \not= 0$$.

• Okay! How can we construct this orthonormal basis w.r.t the given inner product and prove that it indeed satisfies the inner product? Additionally, how can we show that every light-like vector is a transformation of \gamma_0 and \gamma_3? – user132849 Apr 1 at 2:22
• There are three transitive surfaces of the Lorentz group acting on Minkowski space, corresponding to the timelike, spacelike, and lightlike surfaces. What this means is that any two points on the same surfaces are related by a Lorentz transformation. In particular, there exists a Lorentz transformation mapping $\gamma_0 + \gamma_3 \mapsto k$ under the group action, if $k$ is also lightlike. E.g. see en.m.wikipedia.org/wiki/Lorentz_group#Restricted_Lorentz_group – Luke Burns Apr 6 at 2:47