What happens when contracting a lightlike with spacelike/timelike vectors?

Question

I would like to understand what result I get contracting a lightlike four-vector with a spacelike and/or a timelike four-vectors (I'm interested in both cases) in terms of sign. Is the result of these contractions negative or positive? Can you explain me the thought process behind this?

I tried to imagine the issue graphically, i.e. in terms of the light cone graph. Contracting e.g. a timelike with a lightlike vector would mean to project one on another, right? Hence I would obtain as a result of such contraction a scalar that has the same sign of the norm of a timelike vector, or at most it should be null. Is this right or am I thinking badly?

A little context: I am trying to compute the Null Energy Condition (i.e. contraction of a stress-energy tensor with two lightlike vectors) considering a stress-energy tensor that contains a four-velocity, i.e. a timelike vector. The result is that some terms of this condition will contain contractions like $u^\mu u^\nu k_\mu k_\nu$, with $u^i$ being the four-velocities and $k^i$ being generic lighlike vectors. It is not clear to me what is the contribution in terms of sign of these terms.

I apologize if the question is too naive.

Answer 1

Hint: Note that if $v$ is a space/light/timelike vector, so is $-v$.

Note, on the other hand, that your contribution is $$u^\mu u^\nu k_\mu k_\nu= u^\mu k_\mu \; u^\nu k_\nu=\left(u^\mu k_\mu\right)^2 \geq 0\,.$$

We can look at this in more detail: You say $k^\mu$ is null, so you can choose a basis such that $k^\mu=(1,0,0,1)$. Then $u^\mu k_\mu=u^0-u^3$. We can distinguish two cases:

• $u^\mu u_\mu\leq0$, i.e. $u$ is spacelike or null. Here, $u^\mu k_\mu$ can be anything - consider spacelike examples $\pm(1,0,2,0)$ and $(1,0,1,1)$ or lightlike $(1,0,0,\pm 1)$.
• $u^\mu u_\mu>0$, i.e. $u$ is timelike. In other words, $(u^0)^2>(u^1)^2+(u^2)^2+(u^3)^2\geq (u^3)^2$, so $u^\mu k_\mu$ is never zero. Hence, your contribution for lighlike $k$ and timelike $u$ is positive rather than just nonnegative.