Orthogonality of null vectors in Minkowski Spacetime

Question

In Minkowski spacetime, is it true to say that a null vector is orthogonal to itself? Why or why not? Can a timelike vector be orthogonal to a null vector? Can a timelike vector be orthogonal to another timelike vector?

Answer 1

Short answer:

"In Minkowski spacetime, is it true to say that a null vector is orthogonal to itself?" -- Yes

"Can a timelike vector be orthogonal to a null vector?" -- Yes, but only if the null vector is the zero vector

"Can a timelike vector be orthogonal to another timelike vector?" -- No

Long answer:

Definition: Given two four vectors $\mathbf U = U^\mu \mathbf e_\mu = \left(U^0,\vec u\right)$ and $\mathbf V = V^\mu \mathbf e_\mu = \left(V^0,\vec v\right)$ in Minkowski spacetime, the inner product $\mathbf U·\mathbf V$ is defined as

$$\mathbf U·\mathbf V = g_{\mu\nu}U^\mu V^\nu = -U^0V^0 + \vec u·\vec v$$

where

$$g_{\mu\nu} = \left(\begin{matrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{matrix}\right)$$

is the covariant Minkowski metric, and $U^\mu$ and $V^\mu$ are respectively the contravariant components of $\mathbf U$ and $\mathbf V$ expressed in a covariant basis $\{\mathbf e_\mu\}$ that satisfies $\mathbf e_\mu· \mathbf e_\nu = g_{\mu\nu}$; the indices $\mu$ and $\nu$ each range from 0 to 3. The scalars $U^0$ and $V^0$ are respectively referred to as the temporal components of $\mathbf U$ and $\mathbf V$ while the three-vectors $\vec u$ and $\vec v$ are respectively referred to as the spatial components of $\mathbf U$ and $\mathbf V$. The product $\vec u·\vec v$ has the same form as the familiar inner product defined for vectors existing in a 3-dimensional Euclidean space.

Let us consider a second covariant basis $\{\mathbf e_\mu'\}$ that satisfies $\mathbf e_\mu'· \mathbf e_\nu' = g_{\mu\nu}$ and a Lorentz transformation $\mathbf \Lambda = \mathbf e_\mu' \otimes \mathbf e^\mu = \left(\Lambda_\mu\,^\nu \mathbf e_\nu \right)\otimes \mathbf e^\mu$ that transforms the basis $\{\mathbf e_\mu\}$ into $\{\mathbf e_\mu'\}$. From the expression for the transformation, we see that $\mathbf e_\mu' = \Lambda_\mu\,^\nu \mathbf e_\nu$, thus

$$\mathbf e_\mu'· \mathbf e_\nu' = \Lambda_\mu\,^\kappa \mathbf e_\kappa· \Lambda_\nu\,^\lambda \mathbf e_\lambda = \Lambda_\mu\,^\kappa \Lambda_\nu\,^\lambda g_{\kappa\lambda}$$

and so $\Lambda_\mu\,^\kappa \Lambda_\nu\,^\lambda g_{\kappa\lambda} = g_{\mu\nu}$.

If we express $\mathbf U = U'^\mu \mathbf e_\mu'$ and $\mathbf V = V'^\mu \mathbf e_\mu'$ in the basis $\{\mathbf e_\mu'\}$, we see that

$$\mathbf U = U'^\mu \mathbf e_\mu' = U'^\mu \Lambda_\mu\,^\nu \mathbf e_\nu$$ $$\mathbf V = V'^\mu \mathbf e_\mu' = V'^\mu \Lambda_\mu\,^\nu \mathbf e_\nu$$

and so $U^\nu = U'^\mu \Lambda_\mu\,^\nu$ and $V^\nu = V'^\mu \Lambda_\mu\,^\nu$.

Evaluating $\mathbf U·\mathbf V$ in the basis $\{\mathbf e_\mu\}$ yields

$$\mathbf U·\mathbf V = g_{\mu\nu}U^\mu V^\nu = g_{\mu\nu}U'^\kappa \Lambda_\kappa\,^\mu V'^\lambda \Lambda_\lambda\,^\nu = g_{\kappa\lambda}U'^\kappa V'^\lambda$$

thus, for an arbitrary basis $\{\mathbf e_\mu\}$ that satisfies $\mathbf e_\mu· \mathbf e_\nu = g_{\mu\nu}$, the value of the inner product $\mathbf U·\mathbf V$ does not depend on the basis $\{\mathbf e_\mu\}$ specified, and the inner product is said to be invariant under a Lorentz transformation.

Definition: The zero four-vector, denoted as $\mathbf 0 = 0\mathbf e_\mu = \left(0, \vec 0\right)$, is the unique four-vector such that, for an arbitrary four-vector $\mathbf U$, $\mathbf U·\mathbf 0 = 0$, and for an arbitrary Lorentz transformation, $\mathbf \Lambda\mathbf 0 = \mathbf 0$.

Definition: Given a four-vector $\mathbf U$, if $\mathbf U·\mathbf U < 0$, then $\mathbf U$ is said to be timelike. For timelike vectors, a basis $\{\mathbf e_\mu\}$ exists such that $\mathbf U = \left(U^0,\vec 0\right)$.

Definition: Given a four-vector $\mathbf U$, if $\mathbf U·\mathbf U > 0$, then $\mathbf U$ is said to be spacelike. For spacelike vectors, a basis $\{\mathbf e_\mu\}$ exists such that $\mathbf U = \left(0,\vec u\right)$.

Definition: Given a four-vector $\mathbf U$, if $\mathbf U·\mathbf U = 0$, then $\mathbf U$ is said to be null.

Definition: Given two four-vectors $\mathbf U$ and $\mathbf V$ , if $\mathbf U·\mathbf V = 0$, then $\mathbf U$ and $\mathbf V$ are said to be orthogonal to each other.

Proposition 1: If $\mathbf U$ is a null vector, then $\mathbf U$ is orthogonal to itself.

Proof: If $\mathbf U$ is a null vector, then $\mathbf U·\mathbf U = 0$. Therefore, if $\mathbf U$ is a null vector, then $\mathbf U$ is orthogonal to itself.

Proposition 2: Given two four-vectors $\mathbf U$ and $\mathbf V$, if $\mathbf U·\mathbf V = 0$ and $\mathbf U$ is timelike and $\mathbf V$ is null, then $\mathbf V = \mathbf 0$

Proof: If $\mathbf U$ is timelike, then in an arbitrary basis $\{\mathbf e_\mu\}$, $\mathbf U·\mathbf U = -\left(U^0\right)^2 + \vec u·\vec u < 0$. Thus, $\left(U^0\right)^2 > \vec u·\vec u$, and since $\vec u·\vec u \ge 0$, $\left(U^0\right)^2 > 0$ and thus $U^0\ \neq 0$ in the arbitrary basis $\{\mathbf e_\mu\}$.

If $\mathbf V$ is null, then in an arbitrary basis $\{\mathbf e_\mu\}$, $\mathbf V·\mathbf V = -\left(V^0\right)^2 + \vec v·\vec v = 0$. Thus, $\left(V^0\right)^2 = \vec v·\vec v$ in the arbitrary basis $\{\mathbf e_\mu\}$.

Apply a Lorentz transformation $\mathbf \Lambda$ to transform into a basis $\{\mathbf e_\mu\}$ in which $\mathbf U = \left(U^0,\vec 0\right)$. The inner product $\mathbf U·\mathbf V$ evaluated in this basis is $\mathbf U·\mathbf V = -U^0 V^0 = 0$. But since $U^0\ \neq 0$, $V^0 = 0$, which implies that $\vec v·\vec v = 0$ and thus $\vec v = \vec 0$

Therefore, if $\mathbf U·\mathbf V = 0$ and $\mathbf U$ is timelike and $\mathbf V$ is null, then $\mathbf V = \mathbf 0$.

Proposition 3: Two timelike vectors $\mathbf U$ and $\mathbf V$ cannot be orthogonal to each other.

Proof: Assume that there exists a pair of timelike vectors $\mathbf U$ and $\mathbf V$ which are orthogonal to each other; then $\mathbf U·\mathbf V = 0$.

From the previous proof, we know that if $\mathbf U$ is timelike, then $U^0 \neq 0$ in an arbitrary basis $\{\mathbf e_\mu\}$. Similarly, if $\mathbf V$ is timelike, then $V^0 \neq 0$ in the same arbitrary basis $\{\mathbf e_\mu\}$.

Apply a Lorentz transformation $\mathbf \Lambda$ to transform into a basis $\{\mathbf e_\mu\}$ in which $\mathbf U = \left(U^0,\vec 0\right)$. The inner product $\mathbf U·\mathbf V$ evaluated in this basis is $\mathbf U·\mathbf V = -U^0 V^0$. But since $U^0 \neq 0$ and $V^0 \neq 0$, $\mathbf U·\mathbf V \neq 0$, which contradicts our original assumption.

Therefore, two timelike vectors $\mathbf U$ and $\mathbf V$ cannot be orthogonal to each other.