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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?

Thanks

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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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