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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?

Spine Feast
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  • 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. – Ben Grossmann Nov 16 '15 at 13:51
  • 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! – Spine Feast Nov 16 '15 at 14:03
  • It helps to note that the "contravariant" vectors are column-vectors, whereas the "covariant" vectors are row-vectors. – Ben Grossmann Nov 16 '15 at 14:07
  • I don't see how to generalize this to tensors though..perhaps you could elaborate? – Spine Feast Nov 16 '15 at 14:41
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    it is better to write $\Lambda^{\alpha}{}_{\beta}$ to keep track all the way of which index is row and which index is column – janmarqz Nov 17 '15 at 01:50
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    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 $\Lambda_\beta{}^\alpha$ ?? – Spine Feast Nov 17 '15 at 19:17
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    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. – amcalde Nov 18 '15 at 14:08
  • @DepeHb, how do you multiply matrices? If $A=[a_{ij}]$ and $B=[b_{ij}]$ are some matrices that can be multiplied then their multiplication would be $AB=[c_{ij}]$ where $c_{ij}=a_{is}b_{sj}$, which its indentation version also is there to help – janmarqz Nov 19 '15 at 15:52

2 Answers2

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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.

amcalde
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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 $$T'=\Lambda T\Lambda^{\top}.$$

janmarqz
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  • 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. – janmarqz Nov 19 '15 at 18:06
  • I don't see how you conclude $T'=\Lambda T\Lambda^{\top}$ from $T'^{\alpha\beta} = \Lambda^\alpha{}\gamma T^{\gamma \delta}(\Lambda^{\top})\delta{}^\beta{}$ ... – Spine Feast Nov 20 '15 at 14:48
  • first: you got to see that $\Lambda T=[\Lambda^{\alpha}{}_{\gamma}T^{\gamma\delta}]$ – janmarqz Nov 20 '15 at 17:29
  • for the component $[\Lambda T]^{\alpha\delta}$ – janmarqz Nov 20 '15 at 17:32
  • 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? – Spine Feast Nov 21 '15 at 20:53
  • let me refrain how anyone multiply matrices: $[a_{ij}][b_{ij}]=[a_{is}b_{sj}]$, the $s$ index means summation – janmarqz Nov 21 '15 at 21:07
  • Sure, I know that, but why the transpose? Why do you switch to $\Lambda^\alpha{}\gamma T^{\gamma \delta}(\Lambda^{\top})\delta{}^\beta{}$ before concluding the final form? – Spine Feast Nov 21 '15 at 21:15
  • $T^{\gamma \delta}\Lambda^\beta{}\delta=T^{\gamma \delta}(\Lambda^{\top})\delta{}^{\gamma}$ means $T\Lambda^{\top}$ – janmarqz Nov 21 '15 at 21:25
  • the position of $\delta$: second in $T$ and first in $\Lambda^{\top}$ indicates multiplication – janmarqz Nov 21 '15 at 21:27
  • I think I understand now. So the point is to have the summed over index as the "rows", right? – Spine Feast Nov 21 '15 at 22:12
  • do you know that any rank-2 tensor can be consider as a bilinear map between 4 flavors? Take a look onto a picture used in wikipedia's "bilinear maps" https://en.wikipedia.org/wiki/Bilinear_map#/media/File:FourfacesofBilinaearmaps.PNG which settles the right, useful indexation convention to find how a rank 2 tensor changes components – janmarqz Nov 22 '15 at 00:02
  • also look at the 4 basis changes induced by a change $V_e\dashrightarrow V_b$ from old basis $e_i$ to new one $b_i$ and the other three: $V_b\dashrightarrow V_e$, $V^_{\varepsilon}\dashrightarrow V^{\beta}$ and $V^*{\beta}\dashrightarrow V^*_{\varepsilon}$, in a photo https://www.facebook.com/AlgebraMultilineal/photos/pb.341191865975214.-2207520000.1448150674./832025246891871/?type=3&theater. Here $\dashrightarrow$ for a change of basis, $\varepsilon^i$ dual to $e_j$ and $\beta^i$ dual to $b_j$. – janmarqz Nov 22 '15 at 00:09