Working in GR (3+1 dimensions) with a certain metric, 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, which is 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 lightlike 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.

  • 1
    $\begingroup$ The sign of the contraction will tell you whether or not the two vectors have the same time-orientation (pointing towards the past/future) $\endgroup$
    – Slereah
    Mar 9 at 16:32

1 Answer 1


Since the contraction of two vectors need them both to be defined at the same point $p$ and we are free to choose whichever coordinates we prefer on the manifold, we can always choose normal coordinates at $p$, so that the problem ends up reducing to the same thing in Minkowski spacetime.

In a more physics-like language, we can do the computation in a free-falling frame so that the computation is done as in Minkowski spacetime.

Let us choose the null vector $n^a = \left(\frac{\partial}{\partial t}\right)^a + \left(\frac{\partial}{\partial x}\right)^a$ (you can check it is indeed lightlike). The vectors $t^a_\pm = \pm \left(\frac{\partial}{\partial t}\right)^a$ are timelike. Notice now that, under the $-+++$ convention, \begin{align} n_a t^a_\pm &= \pm \left(\frac{\partial}{\partial t}\right)^a \left(\frac{\partial}{\partial t}\right)_a + \pm \left(\frac{\partial}{\partial t}\right)^a \left(\frac{\partial}{\partial x}\right)_a, \\ &= \mp 1 + 0, \\ &= \mp 1. \end{align} Hence, the contraction of a timelike vector with a null vector can be either positive or negative. A similar result works for spacelike vectors, and can be proven in a similar way.

In the $-+++$ convention, it does hold that the contraction of a timelike and a null vector will be negative if they both have the same time-orientation (i.e., if both point to the future or both point to the past). To see this, pick coordinates such that the timelike vector is of the form $\left(\frac{\partial}{\partial t}\right)^a$ (I'll assume the timelike vector to be normalized for simplicity). Completing the coordinate system so that it is normal, we can write the most general vector as $$n^a = \alpha \left(\frac{\partial}{\partial t}\right)^a + \beta \left(\frac{\partial}{\partial x}\right)^a + \gamma \left(\frac{\partial}{\partial y}\right)^a + \delta \left(\frac{\partial}{\partial z}\right)^a.$$

Imposing $n^a n_a = 0$ now leads to $$-\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = 0,$$ and hence $n^a$ is given by $$n^a = \pm \sqrt{\beta^2 + \gamma^2 + \delta^2} \left(\frac{\partial}{\partial t}\right)^a + \beta \left(\frac{\partial}{\partial x}\right)^a + \gamma \left(\frac{\partial}{\partial y}\right)^a + \delta \left(\frac{\partial}{\partial z}\right)^a.$$

If $n^a$ and $\left(\frac{\partial}{\partial t}\right)^a$ both point to the same time direction, then the $+$ sign is to be chosen. Otherwise, the $-$ sign. Assuming $+$, we see that $$n_a \left(\frac{\partial}{\partial t}\right)^a = - \sqrt{\beta^2 + \gamma^2 + \delta^2} < 0,$$ where the strict negativity comes from the fact that if $\sqrt{\beta^2 + \gamma^2 + \delta^2} = 0$, then $n^a$ would be timelike instead of null.


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