# Lorentz group, boost and indices

• I
• TimeRip496
In summary, the inverse matrix of the Lorentz transformation is obtained by contracting the two matrices between alternate indices.
TimeRip496
Compare this with the definition of the inverse transformation Λ-1:

Λ-1Λ = I or (Λ−1)ανΛνβ = δαβ,...(1.33)
where I is the 4×4 indentity matrix. The indexes of Λ−1 are superscript for the first and subscript for the second as before, and the matrix product is formed as usual by summing over the second index of the first matrix and the first index of the second matrix. We see that the inverse matrix of Λ is obtained by

−1)αν = Λνα,.....(1.34)
which means that one simply has to change the sign of the components for which only one of the indices is zero (namely, Λ0i and Λi0) and then transpose it:

Source: http://epx.phys.tohoku.ac.jp/~yhitoshi/particleweb/ptest-1.pdf - Page 12

I understand everything except the bolded part. How does the author know to do that? Even if he meant to inverse the matrix in the conventional sense, it doesn't seems like it.

Last edited by a moderator:
Try raising and lowering the indices ##\nu## and ##\alpha## on the Lorentz transformation with the metric tensor.

TeethWhitener said:
Try raising and lowering the indices ##\nu## and ##\alpha## on the Lorentz transformation with the metric tensor.
You mean like
−1)αν = Λνανβgvvgβα

whereby $$g_{vv}=g^{βα}= \begin{pmatrix} 1 & 0 & 0&0 \\ 0 & -1 & 0&0 \\ 0 & 0 & -1&0 \\ 0 & 0 & 0&-1 \\ \end{pmatrix}$$

Just to confirm, the contraction between the two matrices can only be between alternate indices(as well as superscript and subscript) but not between the same indices of the matrices.

Try ##\Lambda_{\nu}{}^{\alpha}=g^{\alpha\beta}g_{\mu\nu}\Lambda^{\mu}{}_{\beta}##
TimeRip496 said:
Just to confirm, the contraction between the two matrices can only be between alternate indices(as well as superscript and subscript) but not between the same indices of the matrices.
I'm not sure what this means. You can always sum over shared indices as long as one is up and the other is down.

TimeRip496
TeethWhitener said:
Try ##\Lambda_{\nu}{}^{\alpha}=g^{\alpha\beta}g_{\mu\nu}\Lambda^{\mu}{}_{\beta}##

I'm not sure what this means. You can always sum over shared indices as long as one is up and the other is down.
$$g_{μv}=g^{αβ}= \begin{pmatrix} 1 & 0 & 0&0 \\ 0 & -1 & 0&0 \\ 0 & 0 & -1&0 \\ 0 & 0 & 0&-1 \\ \end{pmatrix}$$
$$g_{μv}.g^{αβ}=I$$
$$I.\Lambda^{\mu}{}_{\beta}=\Lambda^{\mu}{}_{\beta}$$
I know how the contraction in terms of Einstein notation works but I still can't get to its inverse in terms of the matrix. I try matrix product them but I still get back the original Lambda instead of its inverse. Did I get somewhere wrong here?

##g_{\mu\nu}g^{\alpha\beta}\neq I##. On the contrary, ##g_{\mu\nu}g^{\nu\beta}=I##. That's the definition of matrix multiplication (which in this case is a contraction of a (2,2)-tensor over its inner indices to give a (1,1)-tensor).

Edit: it's probably easiest to just plug in a few numbers for the indices. For instance, try to evaluate ##(\Lambda^{-1})^0{}_1## using the equation for ##\Lambda_{\nu}{}^{\alpha}## that I mentioned in post #4.

TeethWhitener said:
##g_{\mu\nu}g^{\alpha\beta}\neq I##. On the contrary, ##g_{\mu\nu}g^{\nu\beta}=I##. That's the definition of matrix multiplication (which in this case is a contraction of a (2,2)-tensor over its inner indices to give a (1,1)-tensor).

Edit: it's probably easiest to just plug in a few numbers for the indices. For instance, try to evaluate ##(\Lambda^{-1})^0{}_1## using the equation for ##\Lambda_{\nu}{}^{\alpha}## that I mentioned in post #4.
I think I have something wrong with my understanding when it comes to operating them in terms of matrix.
##\Lambda_{\nu}{}^{\alpha}=g^{\alpha\beta}g_{\mu\nu}\Lambda^{\mu}{}_{\beta}=
\begin{pmatrix}
1 & 0 & 0&0 \\
0 & -1 & 0&0 \\
0 & 0 & -1&0 \\
0 & 0 & 0&-1 \\
\end{pmatrix}
\begin{pmatrix}
1 & 0 & 0&0 \\
0 & -1 & 0&0 \\
0 & 0 & -1&0 \\
0 & 0 & 0&-1 \\
\end{pmatrix}
\begin{pmatrix}
Λ^0{}_0 & Λ^0{}_1 & Λ^0{}_2&Λ^0{}_3 \\
Λ^1{}_0 & Λ^1{}_1 & Λ^1{}_2&Λ^1{}_3 \\
Λ^2{}_0 & Λ^2{}_1 & Λ^2{}_2&Λ^2{}_3 \\
Λ^3{}_0 & Λ^3{}_1 & Λ^3{}_2&Λ^3{}_3 \\
\end{pmatrix}=
\begin{pmatrix}
1 & 0 & 0&0 \\
0 & -1 & 0&0 \\
0 & 0 & -1&0 \\
0 & 0 & 0&-1 \\
\end{pmatrix}
\begin{pmatrix}
Λ^0{}_0 & Λ^0{}_1 & Λ^0{}_2&Λ^0{}_3 \\
-Λ^1{}_0 & -Λ^1{}_1 & -Λ^1{}_2&-Λ^1{}_3 \\
-Λ^2{}_0 & -Λ^2{}_1 & -Λ^2{}_2&-Λ^2{}_3 \\
-Λ^3{}_0 & -Λ^3{}_1 & -Λ^3{}_2&-Λ^3{}_3 \\
\end{pmatrix}=
\begin{pmatrix}
Λ^0{}_0 & Λ^0{}_1 & Λ^0{}_2&Λ^0{}_3 \\
Λ^1{}_0 & Λ^1{}_1 & Λ^1{}_2&Λ^1{}_3 \\
Λ^2{}_0 & Λ^2{}_1 & Λ^2{}_2&Λ^2{}_3 \\
Λ^3{}_0 & Λ^3{}_1 & Λ^3{}_2&Λ^3{}_3 \\
\end{pmatrix}
##
I did matrix product in between them as there are same indices and I don't know how to take the trace after tensor product. I think my problem is that I don't know how this works in terms of matrix.

It's probably better not to think of the problem in terms of matrices. Let's consider the specific example of ##(\Lambda^{-1})^1{}_0 = \Lambda_0{}^1##. Lowering and raising indices gives us (with sums written explicitly)
$$\Lambda_0{}^1 = \sum_{\mu ,\beta} g^{1\beta}g_{\mu 0}\Lambda^{\mu}{}_{\beta}$$
But ##g^{1\beta}## will be zero unless ##\beta=1##. Similarly, ##g_{\mu 0}## is only nonzero when ##\mu =0##. This means the sum only has one nonzero term:
$$\Lambda_0{}^1 = g^{11}g_{00}\Lambda^{0}{}_{1}$$
Does this make it clear why 1) the indices flip, and 2) the sign changes when exactly one of the indices is zero?

TimeRip496
TeethWhitener said:
It's probably better not to think of the problem in terms of matrices. Let's consider the specific example of ##(\Lambda^{-1})^1{}_0 = \Lambda_0{}^1##. Lowering and raising indices gives us (with sums written explicitly)
$$\Lambda_0{}^1 = \sum_{\mu ,\beta} g^{1\beta}g_{\mu 0}\Lambda^{\mu}{}_{\beta}$$
But ##g^{1\beta}## will be zero unless ##\beta=1##. Similarly, ##g_{\mu 0}## is only nonzero when ##\mu =0##. This means the sum only has one nonzero term:
$$\Lambda_0{}^1 = g^{11}g_{00}\Lambda^{0}{}_{1}$$
Does this make it clear why 1) the indices flip, and 2) the sign changes when exactly one of the indices is zero?
I think I get it and I know why my above matrix doesn't work cause my metric did not affect the indices of the original Λ matrix. Does it work if I see it in matrix like this?
##\Lambda_{\nu}{}^{\alpha}=g^{\alpha\beta}g_{\mu\nu}\Lambda^{\mu}{}_{\beta}=
\begin{pmatrix}
g^{00} & g^{01} & g^{02}&g^{03} \\
g^{10} & g^{11} & g^{12}&g^{13} \\
g^{20} & g^{21} & g^{22}&g^{23} \\
g^{30} & g^{31} & g^{32}&g^{33} \\
\end{pmatrix}
\begin{pmatrix}
g_{00} & g_{01} & g_{02}&g_{03} \\
g_{10} & g_{11} & g_{12}&g_{13} \\
g_{20} & g_{21} & g_{22}&g_{23} \\
g_{30} & g_{31} & g_{32}&g_{33} \\
\end{pmatrix}
\begin{pmatrix}
Λ^0{}_0 & Λ^0{}_1 & Λ^0{}_2&Λ^0{}_3 \\
Λ^1{}_0 & Λ^1{}_1 & Λ^1{}_2&Λ^1{}_3 \\
Λ^2{}_0 & Λ^2{}_1 & Λ^2{}_2&Λ^2{}_3 \\
Λ^3{}_0 & Λ^3{}_1 & Λ^3{}_2&Λ^3{}_3 \\
\end{pmatrix}
##

Last edited:
If you do want to think in terms of matrices, make sure you're actually doing matrix multiplication: ## \mathbf{AB} = A_{ij}B^{jk}##. The indices have to be matched up in a way that the summed-over index (j in this case) is on the inside of the products. So ##g^{\alpha\beta}g_{\mu\nu}\Lambda^{\mu}{}_{\beta}## isn't a valid form to represent matrix multiplication. There is a way to arrange this equation (and ONLY this equation) using the properties of the metric tensor so that you can matrix multiply: ##g^{\alpha\beta}g_{\mu\nu}\Lambda^{\mu}{}_{\beta} = g_{\nu\mu}\Lambda^{\mu}{}_{\beta}g^{\beta\alpha}##. This second equation represents a matrix multiplication. You can do this because of the fact that the metric tensor is symmetric.

TimeRip496
TeethWhitener said:
If you do want to think in terms of matrices, make sure you're actually doing matrix multiplication: ## \mathbf{AB} = A_{ij}B^{jk}##. The indices have to be matched up in a way that the summed-over index (j in this case) is on the inside of the products. So ##g^{\alpha\beta}g_{\mu\nu}\Lambda^{\mu}{}_{\beta}## isn't a valid form to represent matrix multiplication. There is a way to arrange this equation (and ONLY this equation) using the properties of the metric tensor so that you can matrix multiply: ##g^{\alpha\beta}g_{\mu\nu}\Lambda^{\mu}{}_{\beta} = g_{\nu\mu}\Lambda^{\mu}{}_{\beta}g^{\beta\alpha}##. This second equation represents a matrix multiplication. You can do this because of the fact that the metric tensor is symmetric.
Thanks a lot!

No problem. A lot of the time, for me at least, explicitly writing out the sums helps me figure out what's going on (even if it's a bit tedious).

## 1. What is the Lorentz group?

The Lorentz group is a mathematical concept used in physics to describe the symmetries of spacetime. It is a group of transformations that preserve the fundamental laws of physics, including the speed of light, in different frames of reference.

## 2. What is a boost in the context of the Lorentz group?

A boost is a transformation within the Lorentz group that changes the coordinates of an object in one frame of reference to its coordinates in another frame of reference that is moving at a constant velocity relative to the first frame. This transformation involves both spatial and temporal components.

## 3. How are boosts represented mathematically?

Boosts are represented using the Lorentz transformation equations, which involve a matrix of coefficients known as the Lorentz matrix. These coefficients depend on the relative velocity between the two frames of reference and can be used to calculate the coordinates of an object in different frames.

## 4. What are indices in the context of the Lorentz group?

Indices refer to the superscripts and subscripts used in the Lorentz transformation equations to represent the components of a vector or tensor in different frames of reference. These indices correspond to the dimensions of spacetime, which are typically denoted as x, y, z, and t.

## 5. How is the Lorentz group related to special relativity?

The Lorentz group is closely related to special relativity, as it provides a mathematical framework for understanding the effects of time and space dilation, length contraction, and the relativity of simultaneity. These concepts are all fundamental principles of special relativity and can be described using the Lorentz group transformations.

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