I'm working on an assignment for an intro to relativity class and I've come up against this series of equations.

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The step I'm confused about is going from summation notation to matrix notation where they've taken the transpose of one of the Lorentz transform matrices. I understand that you can do this, but I don't understand why. What's the motivation? Wouldn't the result (det(L)=+-1) be exactly the same if you didn't take the transpose? There's obviously something I'm missing here, but I don't know what.


After reading a few responses I'm editing this to add that what I'm asking is if someone can explain to me (in detail, showing all the steps) how to go from tensor notation to matrix notation. I think (I hope) that will make it clear what's happening between eqns. 1.7 and 1.10.


The author actually explains this in the excerpt you've provided:

(...) in order that we may use matrix multiplication rules.

If you multiply two $N\times N$ matrices $A$ and $B$ with components $a_{ij}$ and $b_{lk}$ together, you write: $$ C = AB = \sum_{m=1}^N a_{im}b_{mk} $$ So as you see, the $(i,k)$th element of the resulting matrix $C$ is the scalar product of the $i$th row of the first matrix with the $k$th column of the second matrix. To ensure this, the indices you sum over have to somehow be "next to each other" in the notation.

If you take your formula $$ \Lambda^{\mu'}_{\;\;\mu}\Lambda^{\nu'}_{\;\;\nu}\,\eta_{\mu'\nu'} = \eta_{\mu\nu} $$ and want to write it in matrix form, you see that after you change the multiplication order $$ \Lambda^{\mu'}_{\;\;\mu}\Lambda^{\nu'}_{\;\;\nu}\,\eta_{\mu'\nu'} = \Lambda^{\mu'}_{\;\;\mu}\,\eta_{\mu'\nu'}\Lambda^{\nu'}_{\;\;\nu} $$ the $\nu'$s are next to each other, but the $\mu'$s aren't. So you have to switch around the indices in the first $\Lambda$, which is done via the transpose.

Only after you did that you have the correct order of the indices and you can write the equation in matrix form.

Addendum: Let me try to explain the process in more detail in a simpler toy model. Although you say you do not understand how to go from tensor notation to matrix notation, this has actually nothing to do with the objects being tensors, so let's just drop the distinction between co- and contravariant indices and see if that way we can clear things up.

In the very simple example of two $2\times2$ matrices $A$ and $B$ $$ A=\left(\begin{matrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{matrix}\right) \qquad B =\left(\begin{matrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{matrix}\right) $$ when we start at calculating $A\cdot B$ it is clear what has to happen: $$ C:= A\cdot B = \begin{pmatrix}a_{11} & a_{12}\end{pmatrix}\cdot\begin{pmatrix}b_{11}\\b_{21}\end{pmatrix}\cdot\left(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix}\right) + \begin{pmatrix}a_{21} & a_{22}\end{pmatrix}\cdot\begin{pmatrix}b_{11}\\b_{21}\end{pmatrix}\cdot\left(\begin{matrix} 0 & 0 \\ 1 & 0 \end{matrix}\right) + \begin{pmatrix}a_{11} & a_{12}\end{pmatrix}\cdot\begin{pmatrix}b_{12}\\b_{22}\end{pmatrix}\cdot\left(\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}\right)+\begin{pmatrix}a_{21} & a_{22}\end{pmatrix}\cdot\begin{pmatrix}b_{12}\\b_{22}\end{pmatrix}\cdot\left(\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix}\right) = \left(\begin{matrix} a_{11}b_{11}+a_{12}b_{21} & a_{11}b_{12}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{21} & a_{21}b_{12}+a_{22}b_{22} \end{matrix}\right) $$ This is what I described above by "scalar product of row with column", but it should be clear from your linear algebra course how to do it.

The point I am trying to stress is when you are given two matrices and are supposed to calculate their product, the procedure is straightforward and it is also immediately clear how to write down the multiplication in indices. You just check the $(1,2)$ element of the resulting matrix $C$ and can simply read off which elements of $A$ you are supposed to combine with $B$.

On the other hand, if you know nothing about matrix multiplication and are just given the two matrices as some sort of "container" with four elements each, you are free to combine those element any way you like.

This is where the formula for the Lorentz transformation comes in: Your desired transformation property just gives you a procedure, written down as a specific summation, without telling you what this procedure means in terms of matrices.

The analogue in our toy model would be to combine the elements of $A$ and $B$ in the following fashion, where $d_{ij}$ is what I call this new combination: $$ d_{ij} := \sum_{m=1}^2 a_{im}b_{jm} $$ Now, what is $d_{11}$? That's easy, just plug the numbers in: $d_{11} = a_{11}b_{11}+a_{12}b_{12}$.

Huh, that's weird. This specific combination of elements does not appear in any of the components of the matrix product, $C$. What about the rest?

$$ D := (d_{ij})_{i,j=1,2} = \left(\begin{matrix} a_{11}b_{11}+a_{12}b_{12} & a_{11}b_{21}+a_{12}b_{22} \\ a_{21}b_{11}+a_{22}b_{12} & a_{21}b_{21}+a_{22}b_{22} \end{matrix}\right) $$

This $D$ is obviously a matrix representation of the above index formula, but here's the thing: As it stands, it has no simple connection with the constituent matrices. When someone asks you: "How did you get matrix $D$?", you have to answer "Well, I did this thing where I took the elements according to the formula and I arrived at this result", which is a perfectly valid response, but you cannot say what you did with the matrices $A$ and $B$ as a whole.

Now if you look close at your matrix $D$ you notice something peculiar: It almost looks like the result of matrix multiplication, just that the elements $b_{12}$ and $b_{21}$ are in the wrong places, they are even exactly in each others' place!

This is where I hope to finalize this addendum: While you can leave your matrix $D$ as is, one final step makes it possible for you to write down the formula for the $d_{ij}$ in such a way that you can completely rely on the matrices $A$ and $B$ without having to address their elements.

This final step is taking the transpose of $B$ and realizing that the matrix $D$ has actually the whole time been $B^T\cdot A$.

Analogous considerations explain how to get the $\Lambda^T$.

Executive summary of the addendum: The formula for the Lorentz transformations, arising from physical considerations and being correct without any conversions of it, is without those conversions just a formulaic procedure prescribing how to combine elements of $\Lambda$ and $\eta$ to get the desired result.

Conversion of one of the factors into its transpose allows not only to drop the summation notation, but also to express the formula in terms of the objects $\Lambda$ and $\eta$ themselves, without having to address their elements.

  • $\begingroup$ Okay, that makes sense, but why was the order changed at all? Couldn't you just leave it the same, take the determinant of both sides and get the same answer? $\endgroup$ – Chris Bennett Mar 1 '17 at 19:50
  • $\begingroup$ @ChrisBennett I've changed the order to make the picture of the indices being next to each other clear, you don't need to do that. I'm not sure what you mean by "take the determinant" - so far, the determinant has nothing to do with the matrix multiplication. $\endgroup$ – Wojciech Morawiec Mar 1 '17 at 20:03
  • $\begingroup$ Is it not valid to just multiply the The two Lorentz transformation matrices without taking the transpose? If not, why not? $\endgroup$ – Chris Bennett Mar 1 '17 at 20:15
  • $\begingroup$ Now that I've stared at what you've written for a while, I actually don't see how the mu-primes aren't "next to" each other. I see no difference between them and the nu's. $\endgroup$ – Chris Bennett Mar 1 '17 at 20:31
  • $\begingroup$ @ChrisBennett Between the $\mu'$ from the $\Lambda$ and the $\mu'$ from the $\eta$ there is a $\mu$ (without the prime). You sum over the primed indices, so the first summand looks like $\Lambda^0_{\;\;\mu}\eta_{00}\Lambda^0_{\;\;\nu}$ an so on. This way you should see that the $\mu'$ is a row index for $\Lambda$, which is the same as a column index for $\Lambda^T$. $\endgroup$ – Wojciech Morawiec Mar 1 '17 at 20:39

I am assuming that when you ask why, you mean you understand the index-to-matrix conversion. That is, I am assuming you understand that $\Lambda^{\mu'}_{\ \ \mu} \eta_{\mu' \nu'} \Lambda^{\nu'}_{\ \ \nu}=\eta_{\mu \nu}$ implies the matrix equation $\Lambda^T \eta \Lambda=\eta$ [square matrices]. If you don't understand this, please refer to the other answers. It's mathematically completely correct!

The question then would be why take the transpose in the first place, and why work with the tensor equation in the first place. Imagine the following:

You are given two column vectors, $x$ and $y$. You have metric $g$, your square matrix. The inner product of $x$ and $y$ is the scalar $x^Tgy$. You demand that the transformation (square matrix) $\Lambda$, upon sending $x$ to $\Lambda x$ and $y$ to $\Lambda y$, leaves inner products unchanged. For example, this is what a rotation does. Then you say that for all vectors $x$ and for all vectors $y$:

$$x^Tgy=(\Lambda x)^Tg(\Lambda y)=x^T\Lambda^Tg\Lambda y$$

You exploit the fact that this holds for all $x$ and $y$, and exploit some properties of $g$ to show that in fact, we must have $g=\Lambda^Tg\Lambda$.

In the case that $g$ is the Minkowski metric, you get the Lorentz transformations. In the case that $g$ is the Euclidean metric, it's the identity matrix, and you get the relation $\Lambda^T=\Lambda^{-1}$ which holds for rotation matrices.

  • $\begingroup$ No, the index to matrix conversion is the thing I don't understand. $\endgroup$ – Chris Bennett Mar 2 '17 at 1:00

There are already several good answers. Here we will just make a simple consistency check why one of the $\Lambda$-matrices in eq. (1.10) should be transposed:

  1. Recall that Lorentz transformations (i.e. between inertial frames) must form a group under composition. In particular, the composition of two Lorentz transformations should again be a Lorentz transformation.

  2. Note that condition (1.10) defines a group under matrix multiplication. The proof is straightforward. It uses in particular the fact that the transposed of a product of matrices is equal to the opposite product of transposed matrices.

  3. Note that condition (1.10) without the transposed does not form a group under matrix multiplication.

  • $\begingroup$ Sorry, I'm not familiar with group theory. $\endgroup$ – Chris Bennett Mar 3 '17 at 0:12
  • $\begingroup$ I updated the answer. $\endgroup$ – Qmechanic Mar 3 '17 at 14:16

$\Lambda^{\mu'}_{\;\;\mu}\Lambda^{\nu'}_{\;\;\nu}\,\eta_{\mu'\nu'} = \Lambda^{\mu'}_{\;\;\mu}\,\eta_{\mu'\nu'}\Lambda^{\nu'}_{\;\;\nu}$

$\quad\quad\quad\quad\quad\quad = (\Lambda^T)_{\;\;\mu}{}^{\mu'}\,\eta_{\mu'\nu'}\Lambda^{\nu'}_{\;\;\nu}$

This is because $\Lambda^{\mu'}_{\;\;\mu}=(\Lambda^T)_{\;\;\mu}{}^{\mu'}\,$. Just a simple matrix algebra.

  • $\begingroup$ But that's the thing I don't understand. How does going from the summation notation to the matrix algebra make that happen. And why does it need to happen? $\endgroup$ – Chris Bennett Mar 2 '17 at 1:00

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