0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This makes totally sense, thank you! $\endgroup$ – Nabla Jul 2 at 13:18

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.