Inner product and projection in pseudo-riemannian manifolds

Question

My question is about how to properly compute the projection of a tensor in a given direction using inner product product in a pseudo-riemannian manifold, since inner product isn't defined positive.

For simplicity, suppose we have a 2-dimensional curved manifold with signature $ (-, +) $ and a generic metric tensor $ g_ {\mu \nu} $. I want to project a time-like vector $ t ^ \mu = (t ^ 0, t ^ 1) $ on another time-like vector $ n ^ \mu = (n ^ 0, n ^ 1) $ and suppose to know that these two vector form an acute angle, so we expect that the projected component $ t ^ \mu_p $ will be something like

$ t ^ \mu_p = An ^ \mu $ with $A> 0$

because the projected component of $ t ^ \mu $ has the same direction of $ n ^ \mu $. To determine $ A $ I suppose I have to use the inner product $ t_ \mu n ^ \mu $, but this of course is negative, since both vectors are time-like. How do I justify mathematically the use of the minus sign that I want in the projection in order to obtain the correct result?

Then, if I want to do the same with a p-rank tensor $ T $, do I have to consider any minus sign somewhere? For example, for a 2-rank tensor the projected component

$ T ^ {\mu \nu} _p = (T _ {\alpha \beta} n ^ \alpha n ^ \beta) n ^ \mu n ^ \nu $

or do I need a minus sign?

Answer 1

The product $n_{\mu}n^{\mu}$ takes care of the sign without having to write it by hand, whatever the nature of $n$ is that is what your $t$ inherits. Multiply, better still, contract your first expression with $n_{\mu}$ then since $n_{\mu}n^{\mu}\neq 0$ you can get the $A$ straight away. Just be careful with notation:

$$ n_{\mu} t_p^{\mu} = A n_{\mu}n^{\mu}\\[10pt] A = \frac{n_{\mu} t_p^{\mu}}{n_{\nu}n^{\nu}} = \frac{n_{\mu} t_p^{\mu}}{||n||^2}.$$ The sign of $A$ is given by this, not because you set it up front. If you force this sign you might be forcing a contradiction.