I am trying to understand the process of transforming to comoving frame, as outlined in Appendix B1 of this paper.

We can transform some contravariant 4-vector $X^{\nu}$ to the comoving frame as,

$$ X^{(\mu)} = e^{(\mu)}_{\nu} X^{\nu}$$

where the braces denote the comoving frame, the frame moves with 4-velocity $u^{\mu}$, and $e^{(\mu)}_{\nu}$ are the basis 4-vectors of the transformation.

Now, it seems to my intuition that if the frame is at rest such that $u^{\mu} = (1,0,0,0)$ then the vector $X^{\nu}$ should be the same in both frames, but this doesn't seem to be the case. Considering just the $t$ component then

$$ X^{(t)} = e^{(t)}_{\nu} X^{\nu}$$

and from the paper $e^{(t)}_{\nu} = (-u_t, -u_r, -u_{\theta}, -u_{\phi})$. Since $u^{\mu} = (1,0,0,0)$, then,

$$e^{(t)}_{\nu} = (-g_{t t} u^{\mu}, 0, 0, 0)= (-g_{t t} , 0, 0, 0)$$.

Therefore $$X^{(t)} = -g_{tt} X^{t} $$

and since $g_{tt} \ne 1$, then how can the two vectors be equal?



First things first. Vectors have the property that they are objects that are independent of the choice of the basis. Just the components depend on the basis. $V = V^a e_a = V^b e_b$ where $e_a$ and $e_b$ are two different bases and $V^{a}\neq V^b$ (generally).

In the paper they use two different bases. The coordinate basis $e_{\mu}$ and the comoving one $e_{(\mu)}$. For the innerproduct we have $g(e_{(\mu)},e_{(\nu)})=\eta_{(\mu)(\nu)}$ and $g(e_{\mu},e_{\nu})=g_{\mu\nu}$. So $e_{\mu}$ are not orthonormal and $e_{(\mu)}$ are are orthonormal. It immeditly follows that the components of the Vector $X$ should be different.

$X = X^{(\mu)}e_{(\mu)} = X^{\mu}\underbrace{e^{(\mu)}_{\mu} e_{(\mu)}}_{=e_{\mu}}= X^{\mu}e_{\mu}$

So in essence you are right. Both vectors should be the same and they are. And not only for a resting frame but for all moving frames. But the two bases are different and that's why the components are different too.


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