TL;DR

For a ##\binom{1}{1}## tensor ##L^{\alpha}_{\phantom{\alpha}\mu'}## should the upper index ##\alpha## refer to the row or does the lower index ##\mu'## when representing as a matrix?

I have been following the 8.962 class on OCW and I was thinking I was writing out the components correctly with first index row, second index column like matrices have been written in every other course I have taken, and pretty much every example we have gone through in the class in the first few lectures was symmetric so it didn't matter. But now when discussing from changing from primed cylindrical coordinates ##(t,r,\theta,z)## to unprimed cartesian coordinates ##(t,x,y,z)## in the fifth lecture the professor represented the transformation matrix $$L^\alpha_{\phantom{\alpha}\mu'} = \frac{\partial x^\alpha}{\partial x^{\mu'}}$$ between representations in components as

$$
L^{\alpha}_{\phantom{\alpha}\mu'} \to
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \cos{\theta} & \sin{\theta} & 0 \\
0 & -r\sin{\theta} & r\cos{\theta} & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
$$

so that the transformation matrix was
$$
L^{\alpha}_{\phantom{\alpha}\mu'} \to
\begin{pmatrix}
\frac{\partial t}{\partial t} & \frac{\partial x}{\partial t} & \frac{\partial y}{\partial t} & \frac{\partial z}{\partial t} \\
\frac{\partial t}{\partial r} & \frac{\partial x}{\partial r} & \frac{\partial y}{\partial r} & \frac{\partial z}{\partial r} \\
\frac{\partial t}{\partial \theta} & \frac{\partial x}{\partial \theta} & \frac{\partial y}{\partial \theta} & \frac{\partial z}{\partial \theta} \\
\frac{\partial t}{\partial z} & \frac{\partial x}{\partial z} & \frac{\partial y}{\partial z} & \frac{\partial z}{\partial z} \\
\end{pmatrix}
$$

or equivalently

$$
L^{\alpha}_{\phantom{\alpha}\mu'} \to
\begin{pmatrix}
\frac{\partial x^0}{\partial x^{0'}} & \frac{\partial x^1}{\partial x^{0'}} & \frac{\partial x^2}{\partial x^{0'}} & \frac{\partial x^3}{\partial x^{0'}} \\
\frac{\partial x^0}{\partial x^{1'}} & \frac{\partial x^1}{\partial x^{1'}} & \frac{\partial x^2}{\partial x^{1'}} & \frac{\partial x^3}{\partial x^{1'}} \\
\frac{\partial x^0}{\partial x^{2'}} & \frac{\partial x^1}{\partial x^{2'}} & \frac{\partial x^2}{\partial x^{2'}} & \frac{\partial x^3}{\partial x^{2'}} \\
\frac{\partial x^0}{\partial x^{3'}} & \frac{\partial x^1}{\partial x^{3'}} & \frac{\partial x^2}{\partial x^{3'}} & \frac{\partial x^3}{\partial x^{3'}} \\
\end{pmatrix}
$$

But why are rows being identified with the second ##\mu'## index now instead of the first ##\alpha## index like in every other matrix I have ever written? Did the professor accidentally write the transpose of the matrix or am I missing something important?

Prof discusses this at 1:16:17 of the lecture:

Kick-Stand said:

Or does this even really matter since the formulas used in this course will be in terms of components instead of matrices so who cares whether it's ##\alpha## or ##\mu'## that gives the row index of the matrix representation of the ##\binom{1}{1}## tensor ##L^\alpha_{\phantom{\alpha}\mu'}##?

A matrix (specifically a square matrix) is a decent way to express all of the components of a (1, 1) tensor. The problem is that the traditional convention for index placement in linear algebra has, as you mentioned, the first index representing row, and second index representing column. However, a rank (1, 0) tensor is, by convention, a vector (typically represented as a single column), while a rank (0, 1) tensor is a co-vector (typically represented as a single row). So if you wanted to extend this convention to a rank (1, 1) tensor, then the upper index would represent columns and the lower index would represent rows.

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Kick-Stand said:

Did the professor accidentally write the transpose of the matrix or am I missing something important?

Let's consider a simple example: transformation of ##~\eta_{\alpha\beta}~## to cylindrical coordinates. Einstein's summation convention is unambiguous:$$g'_{\mu'\nu'}=\frac{\partial x^\alpha}{\partial x'^{\mu'}}\eta_{\alpha\beta}\frac{\partial x^\beta}{\partial x'^{\nu'}}\quad.\tag{1}$$Each component of ##~g'_{\mu'\nu'}~## is a sum of products, and we know exactly which components participate in each product. In many cases it is convenient to use matrices for calculations of this sort. So (1) is written as$$g'_{\mu'\nu'}=L_{\mu'}{}^\alpha ~\eta_{\alpha\beta}~ L^\beta{}_{\nu'} \tag{2}$$(rows - left index, columns - right). According to my calculation, the professor wrote the transpose (check whether I'm correct).

Note that in (1) we have terms of the form ##~\partial x^\gamma/\partial x'^{\rho'}~## on both sides of ##\eta## , but the rule of matrix multiplication causes us to interpret the left one as the transpose of the right in (2) (suggesting that the definition in the first equation in post #1 may be too restrictive).

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