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opencog-learn commits / b977cd2

Attempt to fix the bad math
b977cd2398bbf9a5a5d00139547cc56e3d2c42eb
by Linas Vepštas <linasvepstas@gmail.com> · 2026-01-28T00:05:41.000Z

Changed files (1) +54 -72

learn-lang-diary/learn-lang-diary-part-ten-F.lyx +54 -72
……@@ -6966,33 +6966,46 @@ or
69666966
69676967\end_inset
69686968
6969which smells OK.
6970\end_layout
6969which is probably right but smells bad.
6970 It's probably right because I think Claude copied it out of a book.
6971 It smells bad because it is putting tangent–space indexes (the latin letters) on coordinate–space partial derivatives.
6972 So let me see if I can patch this up by hand.
6973 Let me try to patch this by writing
6974\begin_inset Formula $\partial^{a}=e^{a\mu}\partial_{\mu}$
6975\end_inset
69716976
6972\begin_layout Standard
6973But now there is stumbling on notational issues.
6974 The greek letters are for coordinate spaces;
6975 the latin letters are for the tangent spaces.
6976 Lots of abuse of notation ensues.
6977\end_layout
6977.
6978 This is an abuse of notation,
6979 but I think we can get away with it if we are careful.
6980 This gives
6981\begin_inset Formula
6982\[
6983\omega_{\;\;\mu}^{ab}=\frac{1}{\lambda}\left(\delta_{\mu}^{a}e^{b\nu}-\delta_{\mu}^{b}e^{a\nu}\right)\partial_{\nu}\lambda
6984\]
69786985
6979\begin_layout Standard
6980First,
6981
6982\begin_inset Formula $\partial^{a}=e^{a\mu}\partial_{\mu}$
69836986\end_inset
69846987
6985 with
6988Then for
69866989\begin_inset Formula $e^{a\mu}=\delta^{a\mu}/\lambda$
69876990\end_inset
69886991
6989.
6990 The calculation in greek indexes,
6992 (another abuse of notation) this becomes
6993\begin_inset Formula
6994\[
6995\omega_{\;\;\mu}^{ab}=\frac{1}{\lambda^{2}}\left(\delta_{\mu}^{a}\delta^{b\nu}-\delta_{\mu}^{b}\delta^{a\nu}\right)\partial_{\nu}\lambda
6996\]
6997
6998\end_inset
6999
7000The partial derivative gets done in greek indexes,
7001 i.e.
7002 in coordinate space,
69917003 for
69927004\begin_inset Formula $\lambda=2A/\left(1-r^{2}\right)$
69937005\end_inset
69947006
6995 gives
7007.
7008 This gives
69967009\begin_inset Formula
69977010\[
69987011\partial_{\mu}\lambda=\frac{4A}{\left(1-r^{2}\right)^{2}}r\partial_{\mu}r=\frac{\lambda^{2}}{A}r\partial_{\mu}r=\frac{\lambda^{2}}{A}x_{\mu}
……@@ -7000,35 +7013,41 @@ First,
70007013
70017014\end_inset
70027015
7003So converting to latin:
7016So
70047017\begin_inset Formula
70057018\[
7006\partial^{a}\lambda=\frac{\lambda}{A}x^{a}
7019\omega_{\;\;\mu}^{ab}=\frac{1}{A}\left(\delta_{\mu}^{a}\delta^{b\nu}-\delta_{\mu}^{b}\delta^{a\nu}\right)x_{\nu}
70077020\]
70087021
70097022\end_inset
70107023
7011and so now comes more confusion.
7012 Earlier,
7013 Claude claimed,
7014 in Cartesian:
7024Drop mu next,
7025 to get
70157026\begin_inset Formula
7016\[
7017\omega^{ab}=\frac{\lambda}{A}\left(x^{b}dx^{a}-x^{a}dx^{b}\right)
7018\]
7019
7020\end_inset
7027\begin{align*}
7028\omega^{ab} & =\frac{1}{A}x_{\nu}\left(\delta^{b\nu}dx^{a}-\delta^{a\nu}dx^{b}\right)\\
7029 & =\frac{\lambda}{A}x_{\nu}\left(e^{b\nu}dx^{a}-e^{a\nu}dx^{b}\right)\\
7030 & =\frac{\lambda}{A}\left(x^{b}dx^{a}-x^{a}dx^{b}\right)
7031\end{align*}
70217032
7022but this is off by a factor of
7023\begin_inset Formula $\lambda$
70247033\end_inset
70257034
7026 overall.
7027 Fuck me.
7028 It is better to avoid using expressions like this.
7029 I leave this here,
7030 for now,
7031 with a ***caution*** sticker on it.
7035with this third line being the abuse of notion.
7036 This need a big ***caution*** sticker on it:
7037 its dangerous but OK,
7038 as long as one consistently abuses the notation.
7039 So we plow ahead,
7040 damn the torpedoes.
7041 Not sure who ever launched a torpedo towards a plow,
7042 but we're not exactly sailing here,
7043 either.
7044 Claude's wordcel trickery has slowed me down.
7045 That it is not a shape rotator becomes exposed very clearly in exercises like this.
7046 Alas.
7047 For me,
7048 alas.
7049 I can't trust it.
7050 Lets plow some more ravines.
70327051\end_layout
70337052
70347053\begin_layout Paragraph*
……@@ -7130,43 +7149,6 @@ e_{3} & =\frac{1}{\lambda r\sin\theta}\partial_{\phi}
71307149
71317150The spin connection.
71327151
7133\begin_inset Formula
7134\[
7135\omega^{ab}=\frac{1}{\lambda}\left(dx^{a}\,\partial^{b}\lambda-dx^{b}\,\partial^{a}\lambda\right)
7136\]
7137
7138\end_inset
7139
7140so for for
7141\begin_inset Formula $\lambda=2A/\left(1-r^{2}\right)$
7142\end_inset
7143
7144 we get
7145\begin_inset Formula $\partial^{2}\lambda=\partial^{3}\lambda=0$
7146\end_inset
7147
7148 and
7149\begin_inset Formula $\partial^{1}\lambda=\lambda^{-1}\partial_{1}\lambda=\lambda r/A$
7150\end_inset
7151
7152 and so ...
7153\end_layout
7154
7155\begin_layout Standard
7156No mixed time–space:
7157\begin_inset Formula
7158\begin{align*}
7159\omega^{0i} & =0\\
7160\omega^{12} & =-\frac{1}{\lambda r}d\theta\\
7161\omega^{13} & =-\frac{1}{\lambda r}\sin\theta\,d\phi\\
7162\omega^{23} & =-\cos\theta\,d\phi
7163\end{align*}
7164
7165\end_inset
7166
7167I copied these from Claude,
7168 but they look maybe not right.
7169 Let me check.
71707152\end_layout
71717153
71727154\begin_layout Section*