Proving Orthogonality of θ_{ik}V^k to U_i

[devd]

Homework Statement

Show that the tensor
θ$_{ik}$ = $g_{ik} - U_{i}U_{k}$
projects any vector, $V^{k}$, into a 3-surface orthogonal to the unit time-like
vector $U_{i}$ (By a projection, the vector $θ_{ik}V_{k}$, is implied).

Homework Equations

The Attempt at a Solution

The projection should be,
$θ_{ik} V^k = g_{ik} V^k - U_i U_k V^k<br /> \Rightarrow θ_{ik} V^k U_i = g_{ik} V^k U_i - U_i U_k V^k U_i$. This should equal zero, for the projection to be orthogonal. But, I'm not being able to proceed.
By timelike, the problem means $U_i U^i \ge 0$, right? But, i don't see how that helps me.

 

[strangerep]

Show that the tensor
θ$_{ik}$ = $g_{ik} - U_{i}U_{k}$
projects any vector, $V^{k}$, into a 3-surface orthogonal to the unit time-like
vector $U_{i}$ (By a projection, the vector $θ_{ik}V_{k}$, is implied).

The phrase "unit time-like vector $U_i$" is important. What do you think "unit" means here?

[...] By timelike, the problem means $U_i U^i \ge 0$, right? But, i don't see how that helps me.

Yes, but (again), what does "unit" mean here?

The projection should be,
$θ_{ik} V^k = g_{ik} V^k - U_i U_k V^k<br /> \Rightarrow θ_{ik} V^k U_i = g_{ik} V^k U_i - U_i U_k V^k U_i$.

Whenever you're contracting indices in such contexts, it should be between an upstairs and a downstairs pair of indices. So your last part should be
$$\theta_{ik} V^k U^i ~=~ g_{ik} V^k U^i - U_i U_k V^k U^i $$Now, can you simplify these subexpressions:
$$g_{ik} U^i ~~~\text{and}~~~ U_i U^i ~~~?$$Then consider further my remarks about the meaning of "unit"...

 

[devd]

Ohho, sorry! I missed the 'unit' bit. $U_i U_k$ can be written as $g_{ik} U_i U^i$. Now, $U_i$ is a unit timelike vector. By definition, then, $U_i U^i =$ 1. Then, the terms cancel and the result follows, right! Thanks for pointing that out! :)