Lorentz Boosts: Finding Speed, Coordinates & Rotation w/ Matrix Multiply

[Athenian]

TL;DR

QUESTION:
How can one tell (or figure out) the boost speed(s), direction coordinates, and rotation through matrix multiplication?

Recently, I've been studying about Lorentz boosts and found out that two perpendicular Lorentz boosts equal to a rotation after a boost. Below is an example matrix multiplication of this happening:

$$
\left(
\begin{array}{cccc}
\frac{2}{\sqrt{3}} & 0 & -\frac{1}{\sqrt{3}} & 0 \\
0 & 1 & 0 & 0 \\
-\frac{1}{\sqrt{3}} & 0 & \frac{2}{\sqrt{3}} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
\frac{2}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & 0 & 0 \\
-\frac{1}{\sqrt{3}} & \frac{2}{\sqrt{3}} & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
=
\left(
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \frac{4 \sqrt{3}}{7} & -\frac{1}{7} & 0 \\
0 & \frac{1}{7} & \frac{4 \sqrt{3}}{7} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
\left(
\begin{array}{cccc}
\frac{4}{3} & -\frac{2}{3} & -\frac{1}{\sqrt{3}} & 0 \\
-\frac{2}{3} & \frac{25}{21} & \frac{2}{7 \sqrt{3}} & 0 \\
-\frac{1}{\sqrt{3}} & \frac{2}{7 \sqrt{3}} & \frac{8}{7} & 0 \\
0 & 0 & 0 & 1 \\
\end{array}
\right)
$$

Source: https://physics.stackexchange.com/q...ing-to-a-rotation-after-a-boost/515690#515690

According to the expert (on the website above) providing the example matrix multiplication, he stated that, "the left-hand-side represents a boost by $c/2$ along the $x$-direction followed by a boost by $c/2$ along the $y$-direction, and that the right-hand-side represents a boost by $\sqrt{7}c/4$ in the direction $(2/\sqrt{7},\sqrt{3/7},0)$ followed by rotation around the $z$-axis by $\cos^{-1}(4\sqrt{3}/7)$ or $8.21$ degrees".

However, my question is how was he able to know the boosts for the $x$ and the $y$ directions are both $c/2$? Furthermore, how was he able to know what are the magnitude, direction, and rotation about the $z$-axis of the matrices after the matrix multiplication? For some reason, I am having a hard time figuring out how he was able to find these numerical values.

Finally, a last yet not-so-bright question, how was the author able to get the answer yet split the matrix in two in that specific manner provided in the example above? Does this help with the solution process?

Any help to provide any amount of insight to answering my question will be much appreciated. Thank you very much for your time and assistance!

 

[PeroK]

Let me approach this issue in a different way. First, let's assume that reference frame S' is moving at some velocity $\vec v$ in the x-y plane, relative to frame S.

Now, the time on a clock in S' (relative to a clock in S) is determined by this. But, an observer in frame S' could set up their x'-y'-z' axes in any way they like. There is nothing that compels them to do it any particular way.

Let's suppose, however, that this observer wants to align their x'-y' axes in some way with the x-y axes in frame S.

(Note that if the relative motion is along the x-axis, then the x-y axes in S are orthogonal in S' when the origins coincide. So, S' can choose these direction for the x'-y' axis. And, of course, this is what the Lorentz boost assumes.)

However, if the relative velocity is not along either the x or y-axis, then when the origins coincide the x-y axes are not orthogonal in S'. Exercise: check this out. So, there is no immediate way for S' to align its axes with those in S.

Then S' has an idea! If the velocity of S' relative to S is $\vec v$, then why not choose the x'-y' axes so that the velocity of S in S' is $-\vec v$? That creates a symmetry between the frames.

This convention is used to define the x'-y'-z' coordinate system in S' when dealing with boosts along an arbitrary direction. All the formulas for a general boost assume this convention. This uniquely defines an x'-y'-z' coordinate system for S'.

Now, suppose instead we simply apply a boost in the x-direction of $v_x$, to define an intermediate frame, T say. Then define another boost in T of $v_y$. This defines a reference frame, T', which is similar to S', except we are not sure that the choice of x'-y'z' axes that arises from this matches the ones we choose above. And, in fact, a closer look reveals that, in fact, the velocity of S in frame T' is not $-\vec v$.

So, what coordinates have arisen for the frame T' from this process? The answer is that the coordinates in T' are rotated compared to the coordinates in S'. By doing two boosts we lost the symmetry in the relative velocity between the frames. And, it's not too hard to calculate the rotation between these coordinate systems.

Note that the relative speed between S and T' is, correctly, $|\vec v|$, so the time coordinate is correct.

Once you have all that it is not too hard to find the coordinates for frames S' and T' and to define the rotation needed to go from one to the other.

 

[Ibix]

I see @PeroK covered the right hand side. The left hand side is simply two Lorentz transforms in matrix form. The diagonal elements are $\gamma$ and the off diagonals are $-v\gamma$.

 

[PeterDonis]

how was he able to know the boosts for the $x$ and the $y$ directions are both $c/2$?

Try calculating $\gamma = 1 / \sqrt{1 - v^2}$ and $\gamma v$ for $v = 1/2$. Then think about what the matrix elements of a matrix that describes a pure boost look like in terms of $\gamma$ and $\gamma v$.

how was he able to know what are the magnitude, direction, and rotation about the $z$-axis of the matrices after the matrix multiplication?

Think about what the $tt$ and $zz$ matrix elements need to be for a matrix that describes no boost and a rotation about the $z$ axis.

how was the author able to get the answer yet split the matrix in two in that specific manner provided in the example above?

By understanding particular facts about matrices that describe pure rotations and pure boosts, and using them to factor the product of the two boost matrices into a pure rotation matrix and a pure boost matrix.

Does this help with the solution process?

Most definitely.

 

[Athenian]

Thank you @PeterDonis, @Ibix, and @PeroK for all your kind assistance and explanations! I sincerely appreciate the help and I now have a much better understanding with regards to my original question.

However, if the relative velocity is not along either the x or y-axis, then when the origins coincide the x-y axes are not orthogonal in S'. Exercise: check this out. So, there is no immediate way for S' to align its axes with those in S.

Think about what the $tt$ and $zz$ matrix elements need to be for a matrix that describes no boost and a rotation about the $z$ axis.

And yes, I will be sure to do the proposed exercise to reinforce the validity of your statement, PeroK. In addition, I'll be sure to think about "what the $tt$ and $zz$ matrix elements need to be for a matrix that describes no boost and a rotation about the $z$ axis".

Of course, if I have any questions, I'll be sure to update this thread afterward. In the meantime, thank you all for your kind and helpful responses!