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If I take the (corrected) formula : $$R= 2M + \sqrt{r(r-2M)}+2M\ln \left[ \sqrt{\frac{r}{2M} -1} + \sqrt{\frac{r}{2M}} \right] \tag{1}$$ You have (if no error) : $dR = \dfrac{dr}{\sqrt{ 1 - \dfrac{2M}{r}}}$, so $dR^2 = \dfrac{dr^2}{f(r)}$, and this simplifies your metrics. However, you cannot invert the formula $(1)$ to get $r$ as an explicit function of ...


2

First of all, the first equation for $ds^2$ is only valid if $f$ is nothing else than the azimuthal angle $\phi$. Second, if you are evaluating $X_i X^i$, the squared distance from the origin without any infinitesimals, then it is exactly equal to $-t^2+r^2$ and nothing else. The polar coordinate $r$ is chosen as $\sqrt{x^2+y^2}$ so its square already ...


1

In these particular cases, the authors are interested in the conformal structure, i.e. lightcone structure, of the manifold. A conformal structure can be defined by an equivalence class of metrics, all of which are related to each other by a conformal transformation, $$g_{ab}\sim e^{\omega(x)}\bar{g}_{ab}$$ A nice way to characterize a conformal structure ...


1

Everything is standard here. Think simply to a classical field on a Minkowski space $M = R_t \times R^3$. The time translation generator, the hamiltonian, is defined by $H_t = P_0=\int T_{0i} d\sigma^i$, where $d\sigma^i = \epsilon^{ijkl} dx^j \wedge dx^k \wedge dx^l$. Now, we may, in fact, consider only the $d\sigma^o= d\sigma^t$ component which is equals ...


1

I have spent at least 5 minutes decoding what the question could be. Finally, I realized that the question says "When $\mu=1,2$..." and it writes an expression in which sometimes the value $1$ is substituted for $\mu$ in the first expression (not "equation"), sometimes the value $2$. You can't have both of them. If the free index $\mu=1$, then it cannot be ...


2

$$\nabla_{\mu}\left[U,V\right]_{\nu} = \nabla_{\mu}\left(g_{\nu\lambda}\left[U,V\right]^{\lambda}\right)$$ You have two terms inside the parentheses, and you have to apply the derivative to both of them. Myself, I'd just remember that: $$\left[U,V\right]^{a} = U^{b}\nabla_{b}V^{a} - V^{b}\nabla_{b}U^{a}$$, so, we have: $$\begin{align} ...



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