Friday, 18 December 2015

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 $k^a=(B,B,0,0)$.


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