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For a diagonal metric $g_{\mu\nu}$ and an orthonormal tetrad with metric $\gamma_{mn}=\eta_{mn}=diag(-1,1,1,1)$, it can be proven by $$ g_{\mu\nu}dx^\mu dx^\nu=\eta_{mn}{e^{m}}_{\mu}{e^{n}}_{\nu}dx^\mu dx^\nu$$ that $${e_m}^\mu=diag((-g_{00})^{-1/2},g_{11}^{-1/2},g_{22}^{-1/2},g_{33}^{-1/2})$$

Consider now the classic Schwarzschild metric with signature (-,+,+,+) and the vierbein: $${e_0}^0=\left(1-\dfrac{r_s}{r}\right)^{-1/2}$$ Apart from the $r=r_s$ case which is proven to be a simple coordinate failure, solved by proper transformation, Hamilton in his GR book imposes another question. Indeed, for $r<r_s$ we have that $r_s/r>1$ and so $$1-\dfrac{r_s}{r}<0$$ which leads to the conclusion that ${e_0}^0\in\mathbb C$. Same holds for ${e_1}^1$. Does this mean that the vierbein fails inside the horizon, is the question imposed.

My first response was to try other coordinates and I think if one chooses the Lemaître coordinates: $$-dT^{2}+{\frac {r_{\mathrm {s} }}{r}}\,dR^{2}+r^{2}\,d\Omega ^{2}$$ one solves the problem, proving that this must be a coordinate failure, because ${e_0}^0=1$ and ${e_1}^1=(r/r_s)^{1/2}$. Is this true? or is there another meaning for the vierbein being not real inside the horizon for many coordinate choices?

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  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please add this tag to this type of problem. This is one of the things that we ask you to do in our homework policy: physics.meta.stackexchange.com/questions/714/… $\endgroup$ – user4552 Jan 5 '20 at 15:44
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Some physical intuition is helpful here: recall orthonormal tetrads correspond to timelike observers. In Schwarzschild-Droste coordinates, your definition leads to $$(e_0)^\mu=\big((1-2M/r)^{-1/2},0,0,0\big)$$ Note this is the 4-velocity of a static observer, meaning one with constant $\theta$, constant $\phi$, and constant $r$. However a static observer can only exist at $r>2M$, so the tetrad (as interpreted this way...) is only meaningful there.

Now for $r<2M$, we can still use your method of rescaling coordinate vectors to construct an orthonormal tetrad, but this time the timelike vector (4-velocity of the corresponding observer) is: $$(e'_0)^\mu=\Big(0,\sqrt{2M/r-1},0,0\Big)$$ These observers have $t=\textrm{const}$, and may only exist for $r<2M$. They have "energy per mass" $0$.

Yes, you can apply the same procedure for Lemaitre coordinates, and this also yields an orthonormal tetrad, however a different one. In particular, $$(e^\textrm{Lemaitre}_0)^\mu=(1,0,0,0)$$ in these coordinates, but the same vector has components $$(e^\textrm{Lemaitre}_0)^\mu=\Big((1-2M/r)^{-1},-\sqrt{2M/r},0,0\Big)$$ in Schwarzschild-Droste coordinates. You may recognise these observers (?) This velocity field can be extended to $r=2M$ (I suggest working in Gullstrand-Painleve coordinates), and thus exists for all $r>0$.

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Note that in Schwartschild metric $$ ds^2 = - (1-r_s/r) dt^2 + (1-r_s/r)^{-1} dr^2 + r^2 d \Omega_2^2 $$ for $r>r_s$ we have that $t$ is a timelike coordinate and $r$ is a spacelike one; by this I mean that the vectors $\partial_t$ and $\partial_r$ are, respectively, timelike and spacetlike. In this region of spacetime is therefore meaningful to consider $g_{tt} \sim {e_0}^0 = (1-r_s/r)^{1/2}$ and $g_{rr} \sim {e_1}^1 = (1-r_s/r)^{-1/2}$.

Then at $r=r_s$ we have the event horizon, and the metric is singular.

Inside the horizon, namely for $r<r_s$, note that $g_{tt}$ and $g_{rr}$ change sign, now $g_{rr}<0$ and $g_{tt}>0$: $t$ is therefore a space coordinate ($\partial_t$ is spacelike) and $r$ can be thought as a time coordinate ($\partial_r$ is timelike). We'd better, therefore, to relate $g_{rr} \sim {e_0}^0 = (r_s/r - 1)^{-1/2}$ and $g_{tt} \sim {e_1}^1 = (r_s/r - 1)^{1/2} $.

What we learn from this example is that we had a metric defined in two separate regions of spacetime, and therefore a priori we cannot expect it to have a uniform behaviour. In this case, changing region, we have to change expression to the vierbeins and to change their connection with metric as well.

Note that, having the metric constant signature, the vierbeins always exist: what is to be considered with care is their expression in coordinates.

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    $\begingroup$ First of all, I think the quantities you are equating with the vierbeins actually correspond to the inverse vierbeins used to relate the coordinate metric with the tetrad one, i.e $g_{\mu\nu}={e^m}_\mu{e^n}_\nu\gamma_{mn}$. Would it be abuse of notation if I just chose the inverse vierbeins to be ${e^0}_0=\sqrt{|-g_{00}|}$ and ${e^1}_1=\sqrt{|g_{11}|}$? This choice seems valid for both areas ($r>r_s$ and $r<r_s$) and the coordinate frame stays unchanged in this manner. $\endgroup$ – kospall Jan 30 '17 at 2:06
  • $\begingroup$ But if the coordinate frame does not change, then the tetrad frame must change, which means a tetrad transformation (should be a Lorentz one) must take place that has to leave the tetrad metric unchanged (as we're talking about orthonormal tetrads). I'm getting really confused here, because I'm new to this. Could you explain your thought in a more detailed way please? $\endgroup$ – kospall Jan 30 '17 at 2:07
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    $\begingroup$ Sorry I was sloppy with the powers $\pm 1/2$, I will double check. The coordinate frame is divided in three regions: $r>r_s$ (where $g_{rr}>0$ and $g_{tt}<0$), $r=r_s$ (metric is degenerate) and $r<r_s$ ($g_{rr}<0$ and $g_{tt}>0$); this is not a good patch for spacetime. These coordinates work inside and outside the horizon, but fail on it. I insist that it is not the vierbeins to breack down, but the expression of the metric $g$ in these coordinates: these coordinates are not valid at the horizon: you should treat them as defined in two disconnected regions of spacetime $\endgroup$ – user139175 Jan 30 '17 at 8:30
  • $\begingroup$ Also your identification with absolute values is incorrect because ${e^0}_0$ [${e^1}_1$] is thought to be related with the negative [positive] eigenvalue, but $g_{tt}$ [$g_{rr}$] is positive [negative] inside the horizon $\endgroup$ – user139175 Jan 30 '17 at 8:34

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