In special relativity one looks at coordinate transformations that consist of combinations of Lorentz boosts, rotations and reflections - members of the Lorentz group. Under an arbitrary transformation like that, a 4-vector $\vec{x}$ transforms as:

$$x'^\alpha = \Lambda^\alpha{}_\beta x^\beta$$

Where $\Lambda^\alpha{}_\beta$ is represents this transformation (is this a $(1,1)$ tensor itself?). This can be written in matrix form

$$x' = \Lambda x$$

Now, the basis vectors transform in another way:

$$\vec{e'_\alpha} = (\Lambda^{-1})^\beta{}_\alpha\vec{e_\beta}$$

A 1-form $\tilde{p}$ transforms like this too:

$$p'_\alpha = (\Lambda^{-1})^\beta{}_\alpha p_\beta$$

while the basis 1-forms obey

$$\omega'^\alpha = \Lambda^\alpha{}_\beta \omega^\beta$$

Anyway, for a tensor representable by a matrix, that is a $(0,2)$, $(2,0)$ or $(1,1)$ tensor, there are different transformation properties:

$$T'^{\alpha\beta} = \Lambda^\alpha{}_\gamma \Lambda^\beta{}_\delta T^{\gamma \delta}$$ $$T'_{\alpha\beta} = (\Lambda^{-1})^\gamma{}_\alpha (\Lambda^{-1})^\delta{}_\beta T_{\gamma \delta}$$ $$T'^{\alpha}{}_\beta = \Lambda^\alpha{}_\gamma (\Lambda^{-1})^\delta{}_\beta T^\gamma{} _\delta$$

My question is: how to translate these rules into matrix equations? I've seen formulas involving the transpose (no idea how that would even come about) of $\Lambda$ but never with an explanation where this came from. So how do I come up with the right order and transposes/inverses on the matrices?

  • $\begingroup$ I've also seen these referred to as "covariant and contravariant transformations". All of these rules are simply a change of basis under different guises. I'll write something up if I have the chance, but it tends to be very difficult to discuss these things in a non-confusing way. $\endgroup$ Nov 16, 2015 at 13:51
  • $\begingroup$ Indeed. Yes, from what I understand covariant means the same transformation as the basis, inverse to a "normal" vector, which in physics is the more natural quantity to consider. The transformations are straightforward to write down following this index notation rules, but I'm having a lot of trouble "translating" it into matrix notation! $\endgroup$ Nov 16, 2015 at 14:03
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    $\begingroup$ it is better to write $\Lambda^{\alpha}{}_{\beta}$ to keep track all the way of which index is row and which index is column $\endgroup$
    – janmarqz
    Nov 17, 2015 at 1:50
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    $\begingroup$ I added the indentations, however to be honest I don't know what they mean. What is the difference between $\Lambda^{\alpha}{}_{\beta}$ and $\Lambda^{\alpha}_{\beta}$ and $\Lambda_\beta{}^\alpha$ ?? $\endgroup$ Nov 17, 2015 at 19:17
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    $\begingroup$ Well, $\Lambda$ is a matrix. In general if you switch the indices (Latin or Greek) on a matrix you mean the transpose operation. So without knowing if the matrix is symmetric, you should not do that willy-nilly. $\endgroup$
    – amcalde
    Nov 18, 2015 at 14:08

2 Answers 2


Let us rearrange your $$T'^{\alpha\beta} = \Lambda^\alpha{}_\gamma \Lambda^\beta{}_\delta T^{\gamma \delta},$$ into $$T'^{\alpha\beta} = \Lambda^\alpha{}_\gamma T^{\gamma \delta}\Lambda^\beta{}_\delta,$$ and one step more $$T'^{\alpha\beta} = \Lambda^\alpha{}_\gamma T^{\gamma \delta}(\Lambda^{\top})_\delta{}^\beta{}.$$

In this last equation anyone could clearly see that to get the components of $T'$ one needs to multiply accordingly to $$T'=\Lambda T\Lambda^{\top}.$$

  • $\begingroup$ in the indentation case $C=AB$ then the entries for $C$ are $c^i{}_j=a^i{}_sb^s{}_j$ and we had used $A^i{}_j=(A^{\top})_j{}{}^i$ for the transpose. $\endgroup$
    – janmarqz
    Nov 19, 2015 at 18:06
  • $\begingroup$ I don't see how you conclude $T'=\Lambda T\Lambda^{\top}$ from $T'^{\alpha\beta} = \Lambda^\alpha{}_\gamma T^{\gamma \delta}(\Lambda^{\top})_\delta{}^\beta{}$ ... $\endgroup$ Nov 20, 2015 at 14:48
  • $\begingroup$ first: you got to see that $\Lambda T=[\Lambda^{\alpha}{}_{\gamma}T^{\gamma\delta}]$ $\endgroup$
    – janmarqz
    Nov 20, 2015 at 17:29
  • $\begingroup$ for the component $[\Lambda T]^{\alpha\delta}$ $\endgroup$
    – janmarqz
    Nov 20, 2015 at 17:32
  • $\begingroup$ Yes, but how does one get from that to the entire product. I mean I could just as well say that without swapping $\beta$ with $\delta$ to get the transpose - why do the indices need to be arranged that way not the other? $\endgroup$ Nov 21, 2015 at 20:53

My old professor has a book that starts out going through this in detail. Look at pages 20 - 22 or so to start. Here's the draft of his book

My Full Answer: If you ask "My question is: how to translate these rules into matrix equations?" I have to say that they already are matrix equations. Let me explain a bit.

In an early course on Special relativity you may actually write out the Lorentz Transformations as a matrix-column vector operation:

$$\tilde{x} = \Lambda x$$

This gets tedious quickly as you consider more general situations, like boosts with rotations, or more general geometry like when there is curvature. So we don't often write out what the components of $\Lambda$ are except in the easy cases.

When multiplying vectors with a matrix you get a vector. But it is often helpful to think about the components one at a time. This is what the indices do. They do more also, because you've chosen some basis to write them down in. When you write down $$\tilde{x}^\alpha = \Lambda^{\alpha}_{\phantom{\alpha}\beta} x^\beta$$

It sometimes help to think of this just like $$ y = Ax \Rightarrow y_i = \sum_j A_{i j} x_j$$

It is essentially the same thing. Seeing this equation should not make you think that it is not a matrix equation. It is! $A_{ i j}$ are the components of the matrix (one at a time!). It is an algebraic expression with indices that represents a real life matrix multiplying a real life vector.

Now one big difference with tensors is that with higher rank you need more multiplies of the $\Lambda$ matrix. This gets confusing because with things like

$$\tilde{R}^{\mu \nu \epsilon \phi} = \Lambda^{\mu}_{\phantom{\mu} \alpha} \Lambda^{\nu}_{\phantom{\nu} \beta} \Lambda^{\epsilon}_{\phantom{\epsilon} \gamma} \Lambda^{\phi}_{\phantom{\phi} \delta } R^{\alpha \beta \gamma \delta}$$

You're sort of still talking about matrix multiply but not with a vector. With a rank 4 tensor. So the algebra works out that you have to do 4 sums. There is little in high school or undergraduate math where this is the case out side of SR or GR. In fact the only thing I've ever seen that even comes close is where you talk about transforming a matrix by right and left multiplication of some other matrix: $$\tilde{M} = U.M.V $$ $$ \tilde{M}_{i l} = \sum_j \sum_k U_{i j} V_{k l} M_{j k}$$

Leaving out the sums really only simplifies things when you already know what you're talking about.

If you are trying to write this down in a familiar way using matrix notation with entries and all of that: GOOD LUCK. This is problematic for many reason but not least of all because usually your higher rank tensors are not really 2D things! So writing them out on paper is hard because you have to somehow represent them and the traditional box shaped matrix won't work (the components are multi-dimensional arrays).

Sorry if my explanation was rambling. I can elaborate but I feel that I may have just made things worse. Let me know what was most confusing about my answer and I'll try to fix it.

Best of luck in a confusing subject.


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