1
$\begingroup$

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.

$\endgroup$
1
  • 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

2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.