general relativity - Tangent vector of photon

If I had some line element such as minkowski line element:

$ds^2 = -dt^2 + dx^2 + dy^2 + dz^2$

And assuming this photon or beam of photons travels in the x direction, how would one find the components of the tangent vector?

Actual answer should be $k^\mu = B(1,1,0,0)$, but how is this obtained?

Answer

For an inertial observer with 4-velocity $t^a$ such that $g_{ab}t^a t^b=t_bt^b=-1$,
let $x^a$ be the unit spacelike-vector orthogonal to $t^a$.
So, $g_{ab}x^a t^b=x_bt^b=0$ and $g_{ab}x^a x^b=x_b x^b=1$.
(In this frame, $t^a=(1,0,0,0)$ and $x^a=(0,1,0,0)$.)

So, $k^a=A t^a+B x^a$.

Since the photon tangent vector must satisfy $g_{ab}k^a k^b=k_b k^b=0$,
we have:
$\begin{align} 0&=k_b k^b\\ &=(A t_b+B x_b)(A t^b+B x^b)\\ &=A^2t_bt^b+B^2x_bx^b\\ &=-A^2+B^2 \end{align}$

So, $|A|=|B|$.

Thus, for a future-directed, forward-pointing tangent vector, we have $k^a=Bt^a+Bx^a$, where $B>0$. In coordinates, this is $k^a=(B,B,0,0)$.