Indices of the Minkowski metric and Lorentz transformation

I am currently in the process of studying special Relativity and I keep stumbling over a concept I can't make consistent for myself. It is about the fact which index of a Lorentz transform and the Minkowski metric denotes a column and which one denotes the row.

My thoughts so far: If I take a look at the Matrix multiplication $$\mathbf{x}^{'\nu}=\mathbf{\Lambda}^\nu\,_\mu\cdot\mathbf{x}^\mu$$, then the upper index $$\mu$$ must be the Index indicating a row as $$\mathbf{x}$$ is a column vector and therefor it's Index must specify the entry (row). Same holds true for the upper index $$\nu$$ of $$\mathbf{x}^{' \nu}$$. Additionally I know that the product describes a matrix multiplication with a vector and due to the Einstein sum rule the expression is summed over $$\mu$$. So if I am interested in the first entry of $$\mathbf{x}^{'}=\mathbf{x}^{'0}$$ I have to multiply the first row of $$\mathbf{\Lambda}=\mathbf{\Lambda}^0$$ with the column of $$\mathbf{x}$$. Therefor the upper index $$\nu$$ of the Lorentz transformation describes the row of the matrix and the lower index $$\mu$$ the column. So far so clear.\

Now I encounter some difficulties. For example our professor writes the following: $$x_\nu=\eta_{\mu\nu}x^\mu$$. (first question: strictly speaking and also written here https://en.wikipedia.org/wiki/Raising_and_lowering_indices (example in Minkowski spacetime): is $$x_\nu$$ a row vector?). The same reasoning as above cannot be used here as we have no upper/lower index. But after the same logic in $$\mathbf{x}^\mu$$, $$\mu$$ describes a row index and therefor in $$\eta_{\mu\nu}$$ $$\nu$$ must describe a row again and $$\mu$$ a column. So now the latter index is the row index (after my understanding). Now the inconsistencies start: In the book here equation (5.12)+(5.13) the authors say that $$\Lambda^\mu\,_\alpha \eta_{\mu\nu}\Lambda^\nu\,_\beta$$ is not a matrix multiplication as the index $$\mu$$ is a column index in the first Lorentz transformation as well as in the Minkowski metric.

I also know (very little to be honest) about Tensors and that they play an important role in the lowering and raising of indexes, but there has to be a self consistent answer to my Index problem somewhere. I have yet to find a satisfying answer to my problem and would be grateful for any help you can provide.

Edit: I have found a post where in one answer a reference to another question is given where the author says the left most index indicates the row. That would at least support my claim for the index label of the Lorentz transformation, yet the problem with the Minkowski metric remains.

When tensor notation is first introduced, it can help the beginner student to show matrix equations that do the same thing, so that they can see it is just a linear transformation. A common convention is to have column matrices representing contravariant vectors (the ones with with upper indices), and write a chain of matrix multiplications in tensor notation as $$x^\alpha=A^\alpha_\beta B^\beta_\gamma C^\gamma_\delta\dots X^\mu_\nu x^\nu$$ sorted into an order where the lower index of each term is the same as the upper index of the following term. But matrices can only handle a handful of cases - where you have vectors or 1-forms that can be written $$x^\mu$$ or $$x_\mu$$ respectively, and $$\left(\begin{smallmatrix}1\\1\end{smallmatrix}\right)$$-tensors $$X^\mu_\nu$$. Any other sort of tensor, and the analogy breaks. And with matrices you have to get the order right, or again it breaks. A row-vector times a column vector is not the same thing as a column-vector times a row-vector.

So the best thing to do, once the idea has been introduced, is to emphasise to the student that tensors are not the same thing as matrices. After the first week, don't try to interpret everything as matrices or row/column vectors, because in a lot of cases it doesn't work, and it's likely to mislead you.

A tensor, in coordinate form, is better thought of as an $$n$$-dimensional array of numbers, without worrying about what direction each axis is in. The dimensions are better thought of as first, second, third, ... (upper or lower) index rather than row, column, rising-up-out-of-the-page, ... so we don't get stuck once we get beyond two or maybe three dimensions. The rule for combining them is the Einstein summation convention, which says you sum over each repeated pair of upper/lower indices. That means you can write them in any order, because the index labels will tell you which dimensions to combine. It means you can have arrays of three or four or five dimensions, and not have to worry about how to extend the convention "rows in the first matrix are multiplied by columns in the second matrix" to something that won't fit into a flat 2D page. It means you can combine dimensions in tensors that are not sat next to one another. It's far more powerful and general, and in some ways simpler.

• Thank you for the answer. So I guess you both say the same thing. Not to worry about rows and collumns but to think about tensors as a generalisation of matrices to be able to use them to solve more complicated stuctures. In that case I should get myseld familiar with tensors. Jul 16, 2022 at 14:38
• To get the proper alignment of indices on tensors, you can use braces. For instance, {X^\mu}_\nu for ${X^\mu}_\nu$ or {X_\nu}^\mu for ${X_\nu}^\mu$. Jul 16, 2022 at 20:31

First, it should probably be said that the notation $$\mathbf{x}^{'\nu}=\mathbf{\Lambda}^\nu\,_\mu\cdot\mathbf{x}^\mu$$ is a kind of overkill since one should either use the index notation $$x'^\nu = \Lambda^\nu_{\,\mu} \,x^\mu$$ or matrix notation (e.g., $$\bf{x'}=\bf{\Lambda} \cdot \bf{x}$$).

If you don't want to learn about tensors at the moment (though you will have to do it sooner or later in this branch of physics), then the most fair answer is: there's not any equivalence between matrix notation and index notation. Think of $$V^\mu$$ and $$V_\mu$$ as two slightly different objects, both of which can be represented by a quarduplet of numbers. But don't wonder too much if they are ordered in a row or in a column. It's up to you how you write them down on the paper - if you want, you can transpose them later if you change your mind.
As an analogy, consider an array (field) of several numbers in you favorite programming language. I guess you access its elements by calling something like V[1] or V[2], you can even have multi-dimensional arrays which you access by something like L[1,2]. But you didn't have to specify to the computer which index denotes a column and which is a row.

I some sense it makes some sense to think of $$x^\mu$$ as a column and $$x_\mu$$ as a row, but in general the matrix notation is not flexible enough to capture the more complex cases, such as objects with more than two indices (since we only have 2_dimensional papers, right?).
Similar problem arises when you want to write $$\eta_{\mu\nu}$$ as a matrix: of course you cast it as $$\eta = \begin{pmatrix} -1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$$ but you can't express that both indices are lower, which should mean that they are column-type. In this case, you can still use the matrix notation using transpositions: the "Minkowski scalar product" $$x^\mu \eta_{\mu\nu} y^\mu$$ can be written in matrix notation as $$\bf{x}^T \bf{\eta} \,\bf{y}$$, similarly you can obtain $$x_\mu$$ in matrix notation from $$x^\mu$$ by $$(\eta \,\bf{x})^T$$. Nevertheless, as you can see, it is slightly cumbersome and this is the reason why we usually prefer using the index notation and say that there is no strict correspondence between lower/upper indices and rows/colums.