| … | … | @@ -7099,6 +7099,28 @@ The hedgehog is |
| 7099 | 7099 | \end_inset |
| 7100 | 7100 | |
| 7101 | 7101 | . |
| 7102 | We'll also take |
| 7103 | \begin_inset Formula $\hat{n}=\hat{r}$ |
| 7104 | \end_inset |
| 7105 | |
| 7106 | viz |
| 7107 | \begin_inset Formula $n_{\mu}=x_{\mu}/r$ |
| 7108 | \end_inset |
| 7109 | |
| 7110 | and thus |
| 7111 | \begin_inset Formula $\gamma^{\mu}$ |
| 7112 | \end_inset |
| 7113 | |
| 7114 | is proportional to |
| 7115 | \begin_inset Formula $\gamma^{a}$ |
| 7116 | \end_inset |
| 7117 | |
| 7118 | and the only question here is, |
| 7119 | do I need to have a factor of |
| 7120 | \begin_inset Formula $\lambda$ |
| 7121 | \end_inset |
| 7122 | |
| 7123 | on the left hand side? |
| 7102 | 7124 | \end_layout |
| 7103 | 7125 | |
| 7104 | 7126 | \begin_layout Paragraph* |
| … | … | @@ -7144,6 +7166,121 @@ e_{3} & =\frac{1}{\lambda r\sin\theta}\partial_{\phi} |
| 7144 | 7166 | The spin connection |
| 7145 | 7167 | \end_layout |
| 7146 | 7168 | |
| 7169 | \begin_layout Standard |
| 7170 | Let assume the earlier expressions for the spin connection were correct. |
| 7171 | Well, |
| 7172 | lets assume they are not, |
| 7173 | and restart from first principles. |
| 7174 | So |
| 7175 | \begin_inset Formula |
| 7176 | \begin{align*} |
| 7177 | d\lambda & =d\frac{2A}{1-r^{2}}\\ |
| 7178 | & =\frac{2A}{\left(1-r^{2}\right)^{2}}2rdr\\ |
| 7179 | & =\frac{\lambda^{2}}{A}rdr |
| 7180 | \end{align*} |
| 7181 | |
| 7182 | \end_inset |
| 7183 | |
| 7184 | Right? |
| 7185 | I think that's right. |
| 7186 | So |
| 7187 | \begin_inset Formula |
| 7188 | \begin{align*} |
| 7189 | de^{0} & =0\\ |
| 7190 | de^{1} & =0\\ |
| 7191 | de^{2} & =d\left(\lambda r\right)\wedge d\theta\\ |
| 7192 | de^{3} & =d\left(\lambda r\sin\theta\right)\wedge d\phi |
| 7193 | \end{align*} |
| 7194 | |
| 7195 | \end_inset |
| 7196 | |
| 7197 | So I crank this by hand. |
| 7198 | Which is tedious, |
| 7199 | but so it goes. |
| 7200 | First, |
| 7201 | |
| 7202 | \begin_inset Formula |
| 7203 | \begin{align*} |
| 7204 | d\left(\lambda r\right) & =rd\lambda+\lambda dr\\ |
| 7205 | & =\lambda\left(\frac{\lambda}{A}r^{2}+1\right)dr\\ |
| 7206 | & =\lambda\left(\frac{2}{1-r^{2}}r^{2}+\frac{1-r^{2}}{1-r^{2}}\right)dr\\ |
| 7207 | & =\lambda\left(\frac{1+r^{2}}{1-r^{2}}\right)dr\\ |
| 7208 | & =\left(\frac{1+r^{2}}{1-r^{2}}\right)e^{1} |
| 7209 | \end{align*} |
| 7210 | |
| 7211 | \end_inset |
| 7212 | |
| 7213 | so that |
| 7214 | \begin_inset Formula |
| 7215 | \[ |
| 7216 | de^{2}=\frac{1}{\lambda r}\left(\frac{1+r^{2}}{1-r^{2}}\right)e^{1}\wedge e^{2} |
| 7217 | \] |
| 7218 | |
| 7219 | \end_inset |
| 7220 | |
| 7221 | The next one is |
| 7222 | \begin_inset Formula |
| 7223 | \begin{align*} |
| 7224 | d\left(\lambda r\sin\theta\right) & =\sin\theta\;d\left(\lambda r\right)+\lambda r\cos\theta\;d\theta\\ |
| 7225 | & =\lambda\left[\left(\frac{1+r^{2}}{1-r^{2}}\right)\sin\theta\;dr+r\cos\theta\;d\theta\right]\\ |
| 7226 | & =\left(\frac{1+r^{2}}{1-r^{2}}\right)\sin\theta\;e^{1}+\cos\theta\;e^{2} |
| 7227 | \end{align*} |
| 7228 | |
| 7229 | \end_inset |
| 7230 | |
| 7231 | so |
| 7232 | \begin_inset Formula |
| 7233 | \[ |
| 7234 | de^{3}=\frac{1}{\lambda r}\left[\left(\frac{1+r^{2}}{1-r^{2}}\right)\;e^{1}\wedge e^{3}+\frac{\cos\theta}{\sin\theta}\;e^{2}\wedge e^{3}\right] |
| 7235 | \] |
| 7236 | |
| 7237 | \end_inset |
| 7238 | |
| 7239 | Next, |
| 7240 | go back to the torsion–free kinematic equation for the spin connection |
| 7241 | \begin_inset Formula |
| 7242 | \[ |
| 7243 | de^{a}+\omega_{\;b}^{a}\wedge e^{b}=0 |
| 7244 | \] |
| 7245 | |
| 7246 | \end_inset |
| 7247 | |
| 7248 | and solve it component by component. |
| 7249 | Is there an easier way? |
| 7250 | Not obviously so. |
| 7251 | Moving through the Ricci components also seems to need a lot of algebra. |
| 7252 | I could use that to double check my results, |
| 7253 | which would be cool, |
| 7254 | if I could trick Claude into not lying to me. |
| 7255 | But tricking Claude into doing the right thing takes about as much time as not using Claude at all... |
| 7256 | Fuck me. |
| 7257 | Whatever. |
| 7258 | Lets blast ahead, |
| 7259 | perhaps stupidly, |
| 7260 | but lets do it anyway. |
| 7261 | \end_layout |
| 7262 | |
| 7263 | \begin_layout Standard |
| 7264 | So |
| 7265 | \begin_inset Formula $0=\omega_{\;b}^{1}\wedge e^{b}$ |
| 7266 | \end_inset |
| 7267 | |
| 7268 | means that |
| 7269 | \begin_inset Formula |
| 7270 | \begin{align*} |
| 7271 | 0 & =\omega_{\;1}^{1}\wedge e^{1}+\omega_{\;2}^{1}\wedge e^{2}+\omega_{\;3}^{1}\wedge e^{3}\\ |
| 7272 | & = |
| 7273 | \end{align*} |
| 7274 | |
| 7275 | \end_inset |
| 7276 | |
| 7277 | |
| 7278 | \end_layout |
| 7279 | |
| 7280 | \begin_layout Paragraph* |
| 7281 | Punt |
| 7282 | \end_layout |
| 7283 | |
| 7147 | 7284 | \begin_layout Standard |
| 7148 | 7285 | Lets assume the earlier expressions for the spin connection were correct. |
| 7149 | 7286 | \end_layout |