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


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...