Convention for writing $\binom{1}{1}$ tensors in matrix form?

[Kick-Stand]

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]

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 $L^\alpha_{\phantom{\alpha}\mu'}$?

 

[Pencilvester]

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.

 

[Ibix]

I would say, in short, tensors aren't matrices and don't follow the same rules. For example, $g_{ab}x^ax^b$ is a perfectly valid tensor product that yields a scalar, as you can see from the lack of free indices, but if you use the "vectors are columns, convectors are rows" rule then it makes no sense. And $g_{ab}x^a$ gives a covector, but a square matrix times a column vector gives a column vector.

It is certainly very common to represent rank-2 tensors as arrays of numbers, but there isn't a one-to-one correspondence between these arrays and matrices. I agree it is helpful to consistently map first indices to rows and second to columns, or vice versa, so it's possible the professor just made a mistake, but note that the $L^a{}_{a'}=\partial x^a/\partial x^{a'}$ is completely unambiguous, and that's the important bit.

I will add that there is a kind of clumsy way to make a sort of matrix notation that's more consistent with tensor rules, wherein you write rank-2 tensors as row vectors of column vectors or column vectors of row vectors depending on index placement. To my way of thinking it's a neat trick rather than really useful, but I mention it for completeness. I don't have a written reference for that, but eigenchris' YouTube Tensors for Beginners covers it - starting around lecture 11, from the thumbnails.



Edit: that link took me to the playlist in preview and takes me to lecture -1 now it's posted. Weird. If you don't get a playlist, go to eigenchris' channel page and he has a playlist called Introduction to Tensors for Beginners, which is what I was trying to link.

Edit 2: And now it shows as a dead YouTube link, but still seems to work...? I give up. If the embedded video doesn't work, go to YouTube and search for eigenchris.

 

[Kick-Stand]

I would say, in short, tensors aren't matrices and don't follow the same rules. For example, $g_{ab}x^ax^b$ is a perfectly valid tensor product that yields a scalar, as you can see from the lack of free indices, but if you use the "vectors are columns, convectors are rows" rule then it makes no sense. And $g_{ab}x^a$ gives a covector, but a square matrix times a column vector gives a column vector.

It is certainly very common to represent rank-2 tensors as arrays of numbers, but there isn't a one-to-one correspondence between these arrays and matrices. I agree it is helpful to consistently map first indices to rows and second to columns, or vice versa, so it's possible the professor just made a mistake, but note that the $L^a{}_{a'}=\partial x^a/\partial x^{a'}$ is completely unambiguous, and that's the important bit.

I will add that there is a kind of clumsy way to make a sort of matrix notation that's more consistent with tensor rules, wherein you write rank-2 tensors as row vectors of column vectors or column vectors of row vectors depending on index placement. To my way of thinking it's a neat trick rather than really useful, but I mention it for completeness. I don't have a written reference for that, but eigenchris' YouTube Tensors for Beginners covers it - starting around lecture 11, from the thumbnails.

It's kind of funny because one of the first things the prof does in Lecture 1 of that class is tell his students that yes you're going to feel in your soul that you should treat these early rank 2 tensors (like the Minkowski metric $\eta_{\alpha\beta}$ or Lorentz transform $\Lambda^\mu_{\phantom{\mu}\bar{\nu}}$) as matrices but don't because for one thing it makes things way more complicated and two you're screwed once you get to tensors with 3,4,5 indices.

 

[JimWhoKnew]

It's kind of funny because one of the first things the prof does in Lecture 1 of that class is tell his students that yes you're going to feel in your soul that you should treat these early rank 2 tensors (like the Minkowski metric $\eta_{\alpha\beta}$ or Lorentz transform $\Lambda^\mu_{\phantom{\mu}\bar{\nu}}$) as matrices but don't because for one thing it makes things way more complicated and two you're screwed once you get to tensors with 3,4,5 indices.

Note that $~\partial x^\alpha/\partial x'^{\mu'}~$ is not a tensor. If you want to transform it to a third coordinate system, it doesn't follow the tensor transformation rule (because it already involves 2 coordinate systems rather than one). Moreover, for raising the lower index you should use $~g'^{\mu'\nu'}~$ , while for lowering the other you should use $~g_{\alpha\beta}~$. The $~\Lambda^\alpha{}_{\mu'}~$ in SR are an exception, due to the invariance of $~\eta_{\mu\nu}~$ under Lorentz transformations.

For true tensors, order of indices (left to right) is important. $~M_{\mu\nu}~$ is the transpose of $~M_{\nu\mu}~$ , and they are not equal in general. You can use that to work out the relation between $~M^\mu{}_\nu~$ and $~M_\nu{}^\mu~$ .

 

[JimWhoKnew]

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).

 

[pervect]

I don't know if this is standard, but I think of a vector, in tensor notation written as $u^a$ as a column vector in matrix notion, represented graphically as a little arrow with magnitude and direction, and a one-form, $u_a$ in tensor notation, as a row vector in matrix notation, represented graphically by a set of parallel plates.

Then a tensor like $L^a{}_b$ would logically be arranged so that $L^a{}_b \cdot u^b$ is like $L \cdot \vec{b}$ in vector notation, where L is a matrix and b is a column vector.

[add]A web seems to confirm my recollection, but I don't have any references that would pass PF standards on this point.

 

[Pencilvester]

I don't know if this is standard, but I think of a vector, in tensor notation written as ua as a column vector in matrix notion, represented graphically as a little arrow with magnitude and direction, and a one-form, ua in tensor notation, as a row vector in matrix notation, represented graphically by a set of parallel plates.

I’ve maintained this visual representation as well ever since learning about it in MTW! I can always visualize a vector as a 1-dimension little arrow, but if the manifold is n-dimensional, where n>3, while I can still only visualize a one-form as a set of parallel, 2-dimensional plates, I have to remember that it’s really parallel, (n-1)-dimensional hyperplanes.