Correction to Physics Simplified

There is a derivation of the coordinate transformation for the Christoffel symbol ##\mathrm{\Gamma }## (or the connection) on Physics Simplified (here: http://www.physicsimplified.com/2014/06/transformation-of-christoffel-symbol.html)

It is very wrong! One obvious clue is that it twists the indices giving ##{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}## in terms of ##{\mathrm{\Gamma }}^{\mu }_{\nu \lambda }##.

It is also assuming that the connection is torsion free and metric compatible. That is that

\begin{align}

{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu } & \phantom {10000}(1) \\

\end{align}and\begin{align}

\mathrm{\nabla }g_{\mu \nu }=0 & \phantom {10000}(2) \\

\end{align}These are not necessary and I will show why in my next post (I hope.)

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{1}{2}g^{\lambda {'}\rho {'}}\left({\partial }_{\mu {'}}g_{\nu {'}\rho {'}}+{\partial }_{\nu {'}}g_{\rho {'}\mu {'}}-{\partial }_{\rho {'}}g_{\mu {'}\nu {'}}\right) & \phantom {10000}(3) \\

\end{align}If we take the first term in the brackets and use the tensor transformation law we get\begin{align}

{\partial }_{\mu {'}}g_{\nu {'}\rho {'}} & =\frac{\partial }{\partial x^{\mu {'}}}\left(\frac{\partial x^{\nu {'}}}{\partial x^{\nu }}\frac{\partial x^{\rho {'}}}{\partial x^{\rho }}g_{\nu \rho }\right) & \phantom {10000}(4) \\

& =\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial }{\partial x^{\mu }}\left(\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}g_{\nu \rho }\right) & \phantom {10000}(5) \\

& =\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}{\partial }_{\mu }g_{\nu \rho }+g_{\nu \rho }\left(\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\mu }\partial x^{\nu {'}}}+\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\mu }\partial x^{\rho {'}}}\right) & \phantom {10000}(6) \\

\therefore {\partial }_{\mu {'}}g_{\nu {'}\rho {'}} & =\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}{\partial }_{\mu }g_{\nu \rho }+g_{\nu \rho }\left(\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\mu {'}}\partial x^{\nu {'}}}+\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\mu {'}}\partial x^{\rho {'}}}\right) & \phantom {10000}(7) \\

\end{align}That was the first error in Physics Simplified. The indices ##\mu ,\nu ## in the denominator of its last term of its fourth equation were the wrong way round. That would be like swapping ##\nu ,\rho ## in the last term of (7).

We can use (7) to get the other two terms in (3). We start with the first part, ##{\partial }_{\mu }g_{\nu \rho }##, of (7):

Second term (3), change first term of (7) with ##\mu \to \nu ,\ \ \nu \to \rho ,\ \ \rho \to \mu ## \begin{align}

T21=\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}{\partial }_{\nu }g_{\rho \mu } & \phantom {10000}(8) \\

\end{align}Third term (3), change first term of (7) with ##\mu \to \rho ,\ \ \nu \to \mu ,\ \ \rho \to \nu ## and negate

\begin{align}

T31=-\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}{\partial }_{\rho }g_{\mu \nu } & \phantom {10000}(9) \\

\end{align}If we added up those three we get the first part of the fifth equation in Physics Simplified with ##\mu \to \lambda ##

In the above we had really substituted the primed ##\mu ,\nu ,\rho ## (which are the free indices) and then changed the dummy variables ##\mu ,\nu ,\rho##. For the next step we will have to be more careful. Following the same procedure only on the primed indices we get

\begin{align}

T22=g_{\nu \rho }\left(\frac{\partial x^{\rho }}{\partial x^{\mu {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\nu {'}}\partial x^{\rho {'}}}+\frac{\partial x^{\nu }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}}\right) & \phantom {10000}(10) \\

T32=-g_{\nu \rho }\left(\frac{\partial x^{\rho }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\rho {'}}\partial x^{\mu {'}}}+\frac{\partial x^{\nu }}{\partial x^{\mu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\rho {'}}\partial x^{\nu {'}}}\right) & \phantom {10000}(11) \\

\end{align}We now swap the dummy variables ##\nu ,\rho ## in (11) and because of the symmetry of ##g_{\nu \rho }## we can swap them again on that and it becomes\begin{align}

T32=-g_{\nu \rho }\left(\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\rho {'}}\partial x^{\mu {'}}}+\frac{\partial x^{\rho }}{\partial x^{\mu {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\rho {'}}\partial x^{\nu {'}}}\right) & \phantom {10000}(12) \\

\end{align}exploiting the rule that partial derivatives commute we can see that the first term of (12) cancels the last term of (7). Likewise the second term of (12) cancels the first of (10). After swapping dummy variables ##\mu,\rho## again the second term of (10) is the same as its equivalent in (7). So adding up (7), (8), (9) (10) and (12) and putting them into (3) we get

\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{1}{2}g^{\lambda {'}\rho {'}}\left(\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)+2g_{\nu \rho }\frac{\partial x^{\nu }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}}\right) & \phantom {10000}(13) \\

\end{align}If we now apply the tensor transformation law to the ##g^{\lambda {'}\rho {'}}## at the front of that we get

\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{1}{2}\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}\frac{\partial x^{\rho {'}}}{\partial x^{\rho }}g^{\lambda \rho }\left(\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)+2g_{\nu \rho }\frac{\partial x^{\nu }}{\partial x^{\rho{'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}}\right) & \phantom {10000}(14) \\

\end{align}

which is similar to, but not I think the same as the seventh equation in Physics Simplified. Nevertheless simplifying:

\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{1}{2}g^{\lambda \rho }\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)+\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}{\delta }^{\lambda }_{\nu }{\delta }^{\nu }_{\rho }\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}} & \phantom {10000}(15) \\

\end{align}The ##g^{\lambda \rho }\left(\dots \right)/2## part of that is of course ##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }##, so we get

\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }+\frac{\partial x^{\lambda {'}}}{\partial x^{\nu }}\frac{{\partial }^2x^{\nu }}{\partial x^{\nu {'}}\partial x^{\mu {'}}} & \phantom {10000}(16) \\

\end{align}

This is a correct transformation for the connection.

It is definitely not the same as the one given Physics Simplified which was\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{\partial x^{\mu {'}}}{\partial x^{\mu }}\frac{\partial x^{\lambda }}{\partial x^{\lambda {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}{\mathrm{\Gamma }}^{\mu }_{\nu \lambda }+\frac{\partial x^{\mu {'}}}{\partial x^{\mu }}\frac{{\partial }^2x^{\mu }}{\partial x^{\lambda {'}}\partial x^{\nu {'}}} & \phantom {10000}(17) \\

\end{align}

Pdf and docx files at Commentary 3.2 Correction to Physics Simplified.pdf

There is a derivation of the coordinate transformation for the Christoffel symbol ##\mathrm{\Gamma }## (or the connection) on Physics Simplified (here: http://www.physicsimplified.com/2014/06/transformation-of-christoffel-symbol.html)

It is very wrong! One obvious clue is that it twists the indices giving ##{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}## in terms of ##{\mathrm{\Gamma }}^{\mu }_{\nu \lambda }##.

It is also assuming that the connection is torsion free and metric compatible. That is that

\begin{align}

{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }={\mathrm{\Gamma }}^{\lambda }_{\nu \mu } & \phantom {10000}(1) \\

\end{align}and\begin{align}

\mathrm{\nabla }g_{\mu \nu }=0 & \phantom {10000}(2) \\

\end{align}These are not necessary and I will show why in my next post (I hope.)

## Here is the correct derivation.

We have the (torsion free , metric compatible) equation for the connection, which can be written with or without primes\begin{align}{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{1}{2}g^{\lambda {'}\rho {'}}\left({\partial }_{\mu {'}}g_{\nu {'}\rho {'}}+{\partial }_{\nu {'}}g_{\rho {'}\mu {'}}-{\partial }_{\rho {'}}g_{\mu {'}\nu {'}}\right) & \phantom {10000}(3) \\

\end{align}If we take the first term in the brackets and use the tensor transformation law we get\begin{align}

{\partial }_{\mu {'}}g_{\nu {'}\rho {'}} & =\frac{\partial }{\partial x^{\mu {'}}}\left(\frac{\partial x^{\nu {'}}}{\partial x^{\nu }}\frac{\partial x^{\rho {'}}}{\partial x^{\rho }}g_{\nu \rho }\right) & \phantom {10000}(4) \\

& =\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial }{\partial x^{\mu }}\left(\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}g_{\nu \rho }\right) & \phantom {10000}(5) \\

& =\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}{\partial }_{\mu }g_{\nu \rho }+g_{\nu \rho }\left(\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\mu }\partial x^{\nu {'}}}+\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\mu }\partial x^{\rho {'}}}\right) & \phantom {10000}(6) \\

\therefore {\partial }_{\mu {'}}g_{\nu {'}\rho {'}} & =\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}{\partial }_{\mu }g_{\nu \rho }+g_{\nu \rho }\left(\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\mu {'}}\partial x^{\nu {'}}}+\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\mu {'}}\partial x^{\rho {'}}}\right) & \phantom {10000}(7) \\

\end{align}That was the first error in Physics Simplified. The indices ##\mu ,\nu ## in the denominator of its last term of its fourth equation were the wrong way round. That would be like swapping ##\nu ,\rho ## in the last term of (7).

We can use (7) to get the other two terms in (3). We start with the first part, ##{\partial }_{\mu }g_{\nu \rho }##, of (7):

Second term (3), change first term of (7) with ##\mu \to \nu ,\ \ \nu \to \rho ,\ \ \rho \to \mu ## \begin{align}

T21=\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}{\partial }_{\nu }g_{\rho \mu } & \phantom {10000}(8) \\

\end{align}Third term (3), change first term of (7) with ##\mu \to \rho ,\ \ \nu \to \mu ,\ \ \rho \to \nu ## and negate

\begin{align}

T31=-\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}{\partial }_{\rho }g_{\mu \nu } & \phantom {10000}(9) \\

\end{align}If we added up those three we get the first part of the fifth equation in Physics Simplified with ##\mu \to \lambda ##

In the above we had really substituted the primed ##\mu ,\nu ,\rho ## (which are the free indices) and then changed the dummy variables ##\mu ,\nu ,\rho##. For the next step we will have to be more careful. Following the same procedure only on the primed indices we get

\begin{align}

T22=g_{\nu \rho }\left(\frac{\partial x^{\rho }}{\partial x^{\mu {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\nu {'}}\partial x^{\rho {'}}}+\frac{\partial x^{\nu }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}}\right) & \phantom {10000}(10) \\

T32=-g_{\nu \rho }\left(\frac{\partial x^{\rho }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\rho {'}}\partial x^{\mu {'}}}+\frac{\partial x^{\nu }}{\partial x^{\mu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\rho {'}}\partial x^{\nu {'}}}\right) & \phantom {10000}(11) \\

\end{align}We now swap the dummy variables ##\nu ,\rho ## in (11) and because of the symmetry of ##g_{\nu \rho }## we can swap them again on that and it becomes\begin{align}

T32=-g_{\nu \rho }\left(\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\rho {'}}\partial x^{\mu {'}}}+\frac{\partial x^{\rho }}{\partial x^{\mu {'}}}\frac{{\partial }^2x^{\nu }}{\partial x^{\rho {'}}\partial x^{\nu {'}}}\right) & \phantom {10000}(12) \\

\end{align}exploiting the rule that partial derivatives commute we can see that the first term of (12) cancels the last term of (7). Likewise the second term of (12) cancels the first of (10). After swapping dummy variables ##\mu,\rho## again the second term of (10) is the same as its equivalent in (7). So adding up (7), (8), (9) (10) and (12) and putting them into (3) we get

\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{1}{2}g^{\lambda {'}\rho {'}}\left(\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)+2g_{\nu \rho }\frac{\partial x^{\nu }}{\partial x^{\rho {'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}}\right) & \phantom {10000}(13) \\

\end{align}If we now apply the tensor transformation law to the ##g^{\lambda {'}\rho {'}}## at the front of that we get

\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{1}{2}\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}\frac{\partial x^{\rho {'}}}{\partial x^{\rho }}g^{\lambda \rho }\left(\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{\partial x^{\rho }}{\partial x^{\rho {'}}}\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)+2g_{\nu \rho }\frac{\partial x^{\nu }}{\partial x^{\rho{'}}}\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}}\right) & \phantom {10000}(14) \\

\end{align}

which is similar to, but not I think the same as the seventh equation in Physics Simplified. Nevertheless simplifying:

\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}\frac{1}{2}g^{\lambda \rho }\left({\partial }_{\mu }g_{\nu \rho }+{\partial }_{\nu }g_{\rho \mu }-{\partial }_{\rho }g_{\mu \nu }\right)+\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}{\delta }^{\lambda }_{\nu }{\delta }^{\nu }_{\rho }\frac{{\partial }^2x^{\rho }}{\partial x^{\nu {'}}\partial x^{\mu {'}}} & \phantom {10000}(15) \\

\end{align}The ##g^{\lambda \rho }\left(\dots \right)/2## part of that is of course ##{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }##, so we get

\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{\partial x^{\lambda {'}}}{\partial x^{\lambda }}\frac{\partial x^{\mu }}{\partial x^{\mu {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}{\mathrm{\Gamma }}^{\lambda }_{\mu \nu }+\frac{\partial x^{\lambda {'}}}{\partial x^{\nu }}\frac{{\partial }^2x^{\nu }}{\partial x^{\nu {'}}\partial x^{\mu {'}}} & \phantom {10000}(16) \\

\end{align}

This is a correct transformation for the connection.

It is definitely not the same as the one given Physics Simplified which was\begin{align}

{\mathrm{\Gamma }}^{\lambda {'}}_{\mu {'}\nu {'}}=\frac{\partial x^{\mu {'}}}{\partial x^{\mu }}\frac{\partial x^{\lambda }}{\partial x^{\lambda {'}}}\frac{\partial x^{\nu }}{\partial x^{\nu {'}}}{\mathrm{\Gamma }}^{\mu }_{\nu \lambda }+\frac{\partial x^{\mu {'}}}{\partial x^{\mu }}\frac{{\partial }^2x^{\mu }}{\partial x^{\lambda {'}}\partial x^{\nu {'}}} & \phantom {10000}(17) \\

\end{align}

Pdf and docx files at Commentary 3.2 Correction to Physics Simplified.pdf

Thanks for the correction to Physics Simplified. I did notice the indices get a bit messy at the end in that work.

ReplyDeleteOne thing to mention, in eqn (14), you have 4 rho's hanging around on the RHS. I think it's relatively clear which pairs you're summing over but it's incorrect notation nonetheless.

Thanks and quite right about the ##\rho##'s. Hope you don't mind if I leave it uncorrected.

DeletePS I use mathjax / Latex in this website and it work's in comments!

DeleteI guess there is an error in equation 4. the transformation of the metric inside the parenthesis seems wrong

ReplyDeleteOops! But it is mysteriously corrected in (5)!

DeletePartial derivatives in different coordinate systems don't commute in general which you seem to have done in the third step. Also I don't see anything wrong in Carroll's derivation except the sign which is obviously a typo. Carrol's result and yours (albeit wrongly derived in my opinion) are in fact equivalent.

ReplyDelete