Sunday, November 5, 2017

Lowering the Index on the Riemann-Christoffel Tensor

Lowering the Index on the Riemann-Christoffel Tensor For the past few months I've been learning about Tensor Calculus. My original goal was to learn about Relativity but there was too much I didn't understand because of the use of tensors. I've always been intrigued by the word "tensors" but not having the foggiest idea what they were I had no ability to go forward.

So, I found a lecture series on YouTube called Tensor Calculus and the Calculus of Moving Surfaces taught by Professor Pavel Grinfeld. I was so intrigued that I bought the text book Introduction to Tensor Analysis and the Calculus of Moving Surfaces and proceeded to do the problems. I've just gotten to the chapter on curvature and figured I'd attempt solutions to some of the problems on-line. (I find the answer key somewhat lacking at times. Dr. Grinfeld has links on the YouTube channel for various resources.) My only quibble about the book is that there are a number of typographical errors. I hope it goes through a second printing.

The other thing I should note before I get started is that there is a lot of stuff on the web about differential geometry and tensors and the notation is all over the map. One of the pleasing things about Grinfeld's treatment is that the notation is fairly simple and consistent. So, I'll be using that notation because of this and also because it's the only one I know!

So, our problem today is to lower the index on the Riemann-Christoffel tensor. I don't pretend that this is the only or most affective or efficient way of doing this. It's the machinery that I employed because of the level of my understanding.

The Riemann-Christoffel tensor, \(R^\gamma_{\cdot\delta\alpha\beta}\) is the interesting components of the following expression; the commutator applied to a contravariant tensor. $$ \begin{equation} (\nabla_\alpha\nabla_\beta - \nabla_\beta\nabla_\alpha)T^\gamma = R^\gamma_{\cdot\delta\alpha\beta}T^\delta \label{eq:commutator} \end{equation} $$ Working out all the covariant derivatives you get the following famous expression for the Riemann-Christoffel tensor: $$ \begin{equation} R^\gamma_{\cdot\delta\alpha\beta} = \pard{\Gamma^\gamma_{\beta\delta}}{S^\alpha} - \pard{\Gamma^\gamma_{\alpha\delta}}{S^\beta} + \Gamma^\gamma_{\alpha\omega}\Gamma^\omega_{\beta\delta} - \Gamma^\gamma_{\beta\omega}\Gamma^\omega_{\alpha\delta} \label{eq:rup} \end{equation} $$ Lowering of the contravariant index, \(\gamma\), is achieved by multiplying by the metric tensor. $$ \begin{equation} \eqalign{ R_{\gamma\delta\alpha\beta} &= S_{\gamma\epsilon}R^\epsilon_{\cdot\delta\alpha\beta} \cr &= S_{\gamma\epsilon}\left(\pard{\Gamma^\epsilon_{\beta\delta}}{S^\alpha} - \pard{\Gamma^\epsilon_{\alpha\delta}}{S^\beta} + \Gamma^\epsilon_{\alpha\omega}\Gamma^\omega_{\beta\delta} - \Gamma^\epsilon_{\beta\omega}\Gamma^\omega_{\alpha\delta}\right) } \label{eq:rsx} \end{equation} $$ So, the interesting part of this is that we get accustomed to just dropping the indices because that is part of the beauty of tensor notation. The problem comes when you can't do that. The issue here is that multiplication of the metric tensor does not penetrate the partial derivative unscathed as it would for a covariant derivative. We have to be somewhat careful.

Here is what I did. Consider this expression which we can evaluate by the produce rule $$ \begin{equation} \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} = \pard{\left(S_{\gamma\epsilon}\Gamma^\epsilon_{\beta\delta}\right)}{S^\alpha} = S_{\gamma\epsilon}\pard{\Gamma^\epsilon_{\beta\delta}}{S^\alpha} + \Gamma^\epsilon_{\beta\delta}\pard{S_{\gamma\epsilon}}{S^\alpha} \label{eq:lc2} \end{equation} $$ Note also that the partial derivative of the metric tensor can be found to be $$ \begin{equation} \pard{S_{\gamma\epsilon}}{S^\alpha} = \Gamma_{\gamma,\epsilon\alpha} + \Gamma_{\epsilon,\gamma\alpha} \label{eq:lc3} \end{equation} $$ Giving $$ \begin{equation} \eqalign{ S_{\gamma\epsilon}\pard{\Gamma^\epsilon_{\beta\delta}}{S^\alpha} &= \pard{\Gamma_{\gamma,\alpha\beta}}{S^\alpha} - \Gamma^\epsilon_{\beta\delta}\pard{S_{\gamma\epsilon}}{S^\alpha} \cr &= \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \left(\Gamma_{\gamma,\epsilon\alpha} + \Gamma_{\epsilon,\gamma\alpha}\right)\Gamma^\epsilon_{\beta\delta} \cr &= \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \Gamma_{\gamma,\epsilon\alpha}\Gamma^\epsilon_{\beta\delta} - \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} } \label{eq:lc4} \end{equation} $$ If you look at equation \ref{eq:lc4}, it gives us the value of the first term of equation \ref{eq:rsx} after you multiply out the metric tensor. Doing the same thing to the second term gives us $$ \begin{equation} \eqalign{ S_{\gamma\epsilon}\pard{\Gamma^\epsilon_{\alpha\delta}}{S^\beta} &= \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} - \Gamma_{\gamma,\epsilon\beta}\Gamma^\epsilon_{\alpha\delta} - \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} } \label{eq:lc5} \end{equation} $$ The 3rd and 4th terms of equation \ref{eq:rsx} are just the Cristoffels with the index lowered. Putting everything together gives us $$ \begin{equation} \eqalign{ R_{\gamma\delta\alpha\beta} &= \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \bbox[5px,border:2px solid red]{\Gamma_{\gamma,\epsilon\alpha}\Gamma^\epsilon_{\beta\delta}} - \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} - \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} + \bbox[5px,border:2px solid blue]{\Gamma_{\gamma,\epsilon\beta}\Gamma^\epsilon_{\alpha\delta}} + \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} + \bbox[5px,border:2px solid red]{\Gamma_{\gamma,\alpha\omega}\Gamma^\omega_{\beta\delta}} - \bbox[5px,border:2px solid blue]{\Gamma_{\gamma,\beta\omega}\Gamma^\omega_{\alpha\delta}} } \label{eq:rl1} \end{equation} $$ Careful inspection will reveal that the red boxed terms are the same because the Cristoffel symbols are symmetric in the last two indices and the name of the contracted index doesn't matter. And the same goes for the terms in the blue boxes. It's interesting to note that the original Cristoffel products are disposed of and replaced with a different set. I don't know if there is some deeper meaning there.

The final result is (rearranging to get a more aesthetically pleasing result) $$ \begin{equation} \eqalign{ R_{\gamma\delta\alpha\beta} &= \pard{\Gamma_{\gamma,\beta\delta}}{S^\alpha} - \pard{\Gamma_{\gamma,\alpha\delta}}{S^\beta} + \Gamma_{\epsilon,\gamma\beta}\Gamma^\epsilon_{\alpha\delta} - \Gamma_{\epsilon,\gamma\alpha}\Gamma^\epsilon_{\beta\delta} } \label{eq:final} \end{equation} $$ More coming on the this topic.

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