Changing our Frame to Make Things Easier
In the realm of general relativity, it can be challenging to navigate the vast array of tools and techniques available to us in order to simplify calculations. To demonstrate this, let's take a look at the concept of space-like and time-like vectors. Imagine we are given a metric and two orthogonal vectors. Can we deduce anything about one vector if we know that the other is time-like? One crucial piece of information to keep in mind is that orthogonal vectors, by definition, have a dot product of zero. But what does this tell us about the nature of these vectors in the context of time and space? It's a question that will lead to a deeper understanding of the topic and will be exciting to solve.
For two vectors X and Y, orthogonality with respect to a particular metric g is defined as
$$ g _ { ab } X ^ a Y ^ b =X \cdot Y = X^{\mu}Y_{\mu} = 0 $$
In this situation, it is not straightforward at all to make any deductions about the quantity we need to tell us whether the vector is time-like, null or space-like.
$$ \text{What is the sign of } g _ { a b } Y ^ a Y ^b \text{ ? } $$
However, we can simplify this problem a bit a smart choice of coordinates. At any given point, we can chose the metric to look like Minkowski space. Why is this possible? Well, the metric is defined to be a quadratic form, and Sylvester’s theorem states that we can pick a basis so that the metric is diagonal, but preserves the signature of the original metric. Sylvester’s theorem is important here - it means we can’t just transform this into Euclidean space, for example. With this, we can pick the metric to look like
$$ g = \text{diag} ( - 1, 1 , 1, 1 ) $$
In addition, our whole set-up was completely agnostic to what the x-axis was, and the velocity in which our frame is in. So we can arbitrarily rotate this frame such that the X vector in components looks like the first transform below. We can also do an explicit Lorentz transformation, to show that we can boost our frame such that the spatial component also vanishes.
$$ X \mapsto ( X ^ 0 , X ^ 1, 0 , 0 ) \mapsto ( X^ 0 , 0 , 0 , 0 ) $$
And with this, the rest of the problem is easy to solve!