# Calculate a specific boost and rotation

• Frostman
In summary, the conversation discusses a method for applying a generic boost in the y-z plane and then making a rotation to realign the photon moment along the abscissa axis. The conversation presents the equations for the boosts and rotations and discusses the simplifications that can be made by choosing a direction for the boost in the y-z plane. It then goes on to solve for the x-component of the boost, which turns out to not be necessary.
Frostman
Homework Statement
In an inertial reference system S, we have a photon with spatial components only along the x axis, its quadrimoment is ##p = E (1, 1, 0, 0)##. There is a special class of Lorentz transformations, called "small group of ##p##", which leaves the components of ##p## unchanged. For example, any rotation around the x axis belongs to this class.
a) Calculate a sequence of boosts and rotations whose product belongs to the small group of p, but which is not a simple rotation around the x axis. To do this, follow the suggestions below.
I) Apply a generic boost in the y-z plane (take advantage of the arbitrariness in deciding the alignment of the y and z axes).
II) Make an appropriate rotation that realigns the photon moment along the abscissa axis.
III) Determine a new boost that brings the photon back to having the initial components.
b) Write explicitly the Lorentz transformation found and verify that it is not a pure rotation.
Relevant Equations
\begin{equation*}
B =
\begin{pmatrix}
\gamma & -\gamma v_x & -\gamma v_y & -\gamma v_z \\
-\gamma v_x & \frac{v_x^2(\gamma - 1)}{|v|^2}+1 & \frac{v_xv_y(\gamma -1)}{|v|^2} & \frac{v_xv_z(\gamma -1)}{|v|^2} \\
-\gamma v_y & \frac{v_xv_y(\gamma -1)}{|v|^2} & \frac{v_y^2(\gamma - 1)}{|v|^2}+1 & \frac{v_yv_z(\gamma -1)}{|v|^2}\\
-\gamma v_z & \frac{v_xv_z(\gamma -1)}{|v|^2} & \frac{v_yv_z(\gamma -1)}{|v|^2} & \frac{v_z^2(\gamma - 1)}{|v|^2}+1
\end{pmatrix}
\end{equation*}
Let's begin with the first point.
a.I) Apply a generic boost in the y-z plane (take advantage of the arbitrariness in deciding the alignment of the y and z axes).
\begin{equation*}
B_{yz} =
\begin{pmatrix}
\gamma & 0 & -\gamma v_y & -\gamma v_z \\
0 & 1 & 0 & 0 \\
-\gamma v_y & 0 & \frac{v_y^2(\gamma - 1)}{|v|^2}+1 & \frac{v_yv_z(\gamma -1)}{|v|^2}\\
-\gamma v_z & 0 & \frac{v_yv_z(\gamma -1)}{|v|^2} & \frac{v_z^2(\gamma - 1)}{|v|^2}+1
\end{pmatrix}
\end{equation*}

I thought that for a generic boost in the yz plane we can consider a boost along the velocity vector ##v=(0, v_y, v_z)##
Applying to our moment vector we have the following result
$$B_{yz} p = E \begin{pmatrix} \gamma \\ 1\\ -\gamma v_y \\ -\gamma v_z \end{pmatrix}$$

a.II) Make an appropriate rotation that realigns the photon moment along the abscissa axis.
Now to realign the moment to the x-axis I have to reset the components ## p_y ## and ## p_z ## with a rotation. But with ##R_x## or ##R_y## or ##R_z## I can't find a right rotation that can make ## p_y = p_z = 0##. How I can find this rotation? Thanks for the help!

Look at the hint about the "arbitrariness" of the boost in the y-z plane. This means you could take the boost to be along the y-axis. That may simplify things.

Frostman
So I can take a ##B_y## or a ##B_z## satisfying the request that the boost take place in the yz plane? Instead to take a combination of boost along ##y## and ##z##?

Frostman said:
So I can take a ##B_y## or a ##B_z## satisfying the request that the boost take place in the yz plane? Instead to take a combination of boost along ##y## and ##z##?
Yes, whatever direction you choose in the y-z plane, I can take that to be my y-axis. We then have the same physical scenario, but my equations will be simpler.

Frostman
PeroK said:
Yes, whatever direction you choose in the y-z plane, I can take that to be my y-axis. We then have the same physical scenario, but my equations will be simpler.
For the first point I can use this boost.

\begin{equation*}
B_{y} =
\begin{pmatrix}
\gamma & 0 & -\gamma v_y & 0 \\
0 & 1 & 0 & 0 \\
-\gamma v_y & 0 & \frac{v_y^2(\gamma - 1)}{|v|^2}+1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\end{equation*}

Which gives me this new quadrimoment ##p=E(\gamma, 1, -\gamma v_y, 0)##.
Now I have to do a rotation to reset ##p_y = p_z = 0## and I can use the only rotation along z-axis, because\begin{equation*}
R_{x}(\varphi)p =
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \cos\varphi & -\sin\varphi \\
0 & 0 & \sin\varphi & \cos\varphi \\
\end{pmatrix}
p=E
\begin{pmatrix}
\gamma \\
1 \\
-\gamma v_y \cos \varphi \\
-\gamma v_y \sin \varphi \\
\end{pmatrix}
\end{equation*}

But ##\cos## and ##\sin## are never ##0## at the same angle.

\begin{equation*}
R_{y}(\varphi)p =
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \cos\varphi & 0 & \sin\varphi \\
0 & 0 & 1 & 0 \\
0 & -\sin\varphi & 0 & \cos\varphi \\
\end{pmatrix}
p=E
\begin{pmatrix}
\gamma \\
\cos \varphi \\
-\gamma v_y \\
- \sin \varphi \\
\end{pmatrix}
\end{equation*}

By hypothesis ##-\gamma v_y ≠ 0 ##

In the end we have

\begin{equation*}
R_{z}(\varphi)p =
\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \cos\varphi & -\sin\varphi & 0 \\
0 & \sin\varphi & \cos\varphi & 0 \\
0 & 0 & 0 & 1 \\
\end{pmatrix}
p=E
\begin{pmatrix}
\gamma \\
\cos \varphi + \gamma v_y \sin \varphi \\
\sin \varphi - \gamma v_y \cos \varphi \\
0 \\
\end{pmatrix}
\end{equation*}

##p_y## is ##0## for ##\sin \varphi - \gamma v_y \cos \varphi = 0## which is true for ##\varphi = \arctan( \gamma v_y)##

etotheipi
Looks good so far.

Frostman
PeroK said:
Looks good so far.

Calling ## \varphi_z = \arctan( \gamma v_y ) ## and ##\gamma_1## the first boost. We have the last and new quadrimoment:

##p'' = E(\gamma_1, \cos \varphi_z + \gamma_1 v_y \sin \varphi_z, 0, 0) ##
Now let's make the second boost along x-axis:

\begin{equation*}
B_{x} =
\begin{pmatrix}
\gamma_2 & -\gamma_2 v_x & 0 & 0 \\
-\gamma_2 v_x & \frac{v_x^2(\gamma - 1)}{|v|^2}+1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\end{equation*}

\begin{equation*}
B_{x}p'' = E
\begin{pmatrix}
\gamma_1 \gamma_2 - \gamma_2 v_x (\cos \varphi_z + \gamma_1 v_y \sin \varphi_z) \\
-\gamma_1 \gamma_2 v_x + (\frac{v_x^2(\gamma_2 - 1)}{|v|^2}+1)(\cos \varphi_z + \gamma_1 v_y \sin \varphi_z) \\
0 \\
0 \\
\end{pmatrix}
= E
\begin{pmatrix}
1\\
1\\
0\\
0\\
\end{pmatrix}
\end{equation*}

I have to solve this 2 equations in 1 variable (##v_x##)
\begin{cases}
\gamma_1 \gamma_2 - \gamma_2 v_x \cos \varphi_z - \gamma_2 v_x \gamma_1 v_y \sin \varphi_z) = 1 \\
-\gamma_1 \gamma_2 v_x + \gamma_2 \cos \varphi_z + \gamma_2 \gamma_1 v_y \sin \varphi_z = 1
\end{cases}

Now I have to remember that
##\gamma_1 = \frac{1}{\sqrt{1-v_y^2}} ##
##\gamma_2 = \frac{1}{\sqrt{1-v_x^2}} ##
##\cos(\arctan(\gamma_1 v_y)) = \frac{1}{\sqrt{1+\gamma_1^2 v_y^2}} = \sqrt{1-v_y^2} = \gamma_1^{-1}##
##\sin(\arctan(\gamma_1 v_y)) = \frac{\gamma_1 v_y}{\sqrt{1+\gamma_1^2 v_y^2}}=v_y ##
And simplify these equations.
\begin{cases}
\gamma_1 \gamma_2 (1-v_x)=1 \\
\gamma_1 \gamma_2 (1-v_x)=1 \\
\end{cases}

\begin{cases}
\frac{ 1 }{ \sqrt{1-v_y^2}} \frac{1}{\sqrt{1-v_x^2}} (1-v_x)=1 \\
\frac{ 1 }{ \sqrt{1-v_y^2}} \frac{1}{\sqrt{1-v_x^2}} (1-v_x)=1 \\
\end{cases}

Solving them I find out:

$$v_{x_{1,2}} = \frac{1 \pm \sqrt{1 - (2-v_y^2) v_y^2}}{ 2 - v_y^2}$$

That gives me a condition another condition for ##v_y##: ## v_y < -1 \vee v_y > 1 ##
Can you confirm?

And this gives me the sequence of the Lorentz's transformation that I should apply:
$$B_xR_z(\varphi_z)B_yp=p$$

So, if is it all correct ##B_xR_z(\varphi_z)B_y=\mathbb{1}##.
It seemed like a very long exercise full of calculus. Isn't it that I did something wrong or is there a much faster way? It is an exam topic and it is 1 of the 4 exercises to do.

If ##v## is the first boost in the y direction, with gamma factor ##\gamma##, then try ##u = \frac{\gamma^2 -1}{\gamma^2 + 1}## for the boost in the x-direction.

There are a number of ways to simplify what you've done. If I have time, I'll post this later.

Frostman
I don't understand why did you choose this boost ##u = \frac{\gamma^2-1}{\gamma^2+1}##.
I'm interested and I want to try understand on this ways to simplify my solution.

Frostman said:
I don't understand why did you choose this boost ##u = \frac{\gamma^2-1}{\gamma^2+1}##.
I'm interested and I want to try understand on this ways to simplify my solution.
After the second transformation, the problem reduces to one spatial dimension. You want:
$$\begin{pmatrix} \gamma_u &-\gamma_u u \\ -\gamma_u u & \gamma_u \\ \end{pmatrix} \begin{pmatrix} \gamma \\ \gamma \\ \end{pmatrix} = \begin{pmatrix} 1 \\ 1 \\ \end{pmatrix}$$
Solving that using ##\gamma_u^2 = \frac 1 {1-u^2}## gives you that equation for ##u## in terms of ##\gamma##.

Frostman
Oh... I should have notice this before...

##\cos(\arctan(\gamma_1 v_y)) = \frac{1}{\sqrt{1+\gamma_1^2 v_y^2}} = \sqrt{1-v_y^2} = \gamma_1^{-1}##
##\sin(\arctan(\gamma_1 v_y)) = \frac{\gamma_1 v_y}{\sqrt{1+\gamma_1^2 v_y^2}}=v_y ##

With this last step is the most simplified solution or did I miss something again?

Frostman said:
Oh... I should have notice this before...

##\cos(\arctan(\gamma_1 v_y)) = \frac{1}{\sqrt{1+\gamma_1^2 v_y^2}} = \sqrt{1-v_y^2} = \gamma_1^{-1}##
##\sin(\arctan(\gamma_1 v_y)) = \frac{\gamma_1 v_y}{\sqrt{1+\gamma_1^2 v_y^2}}=v_y ##

With this last step is the most simplified solution or did I miss something again?
That's messy. Just let ##\cos \theta = \frac 1 \gamma## and ##\sin \theta = -v##.

In general, you don't need to solve all these intermediate equations. You can simply use the relationship between variables.

Also, using both ##v## and ##\gamma## is useful. You can express some things in terms of ##v## and some in terms of ##\gamma##.

Frostman
Here's a simple approach. First, we have a boost in the y-direction:
$$B_y(v) = \begin{pmatrix} \gamma & 0 & -\gamma v & 0 \\ 0 & 1 & 0 & 0 \\ -\gamma v & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$
Then, we have a rotation about the z-axis:
$$R_z(\theta) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta & 0 \\ 0 & \sin \theta & \cos \theta & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$
And, if we have ##\sin \theta = v## and ##\cos \theta = \frac 1 \gamma## (check that this is valid), then we get:
$$R_z(\theta)B_y(v) \begin{pmatrix} 1 \\ 1 \\ 0 \\ 0 \\ \end{pmatrix} = \begin{pmatrix} \gamma \\ \gamma \\ 0 \\ 0 \\ \end{pmatrix}$$
Then we have a boost in the x-direction, with ##u = \frac{\gamma^2 -1}{\gamma^2 + 1}##.

I got a simplified form for the final Lorentz transformation:
$$L = B_x(u)R_z(\theta)B_y(v) = \begin{pmatrix} 1 + \frac{v^2}{2} & -\frac{v^2}{2} & -v & 0 \\ \frac{v^2}{2} & 1 - \frac{v^2}{2} & -v & 0 \\ -v & v & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$$
That gives you something to shoot for. You can see that it leaves the original vector components unchanged. You could also check that ##L## is a Lorentz Transformation:
$$L^{T}\eta L = \eta$$
Where ##\eta## is the diagonal matrix ##(1, -1, -1, -1)## or ##(-1, 1, 1, 1)##.

etotheipi and Frostman
Really appreciate. To check if is a Lorentz Transformation I agree, I studied this demonstration. The equalities ##\sin \theta = v ## and ##\cos \theta = \frac{1}{\gamma}## come from the observation that I did before about the relation between ##\cos \theta## and ##\sin \theta## with ##\arctan \theta##? If I want to check that ##L## it's not a pure rotation I can use the fact that ##L^TL≠\mathbb{1}## like a pure rotation (##R^TR≠\mathbb{1}##)?

Frostman said:
Really appreciate. To check if is a Lorentz Transformation I agree, I studied this demonstration. The equalities ##\sin \theta = v ## and ##\cos \theta = \frac{1}{\gamma}## come from the observation that I did before about the relation between ##\cos \theta## and ##\sin \theta## with ##\arctan \theta##? If I want to check that ##L## it's not a pure rotation I can use the fact that ##L^TL≠\mathbb{1}## like a pure rotation (##R^TR≠\mathbb{1}##)?
The matrix representing a rotation about the x-axis has a specific form. In particular for the time components. You can see "by inspection" that it's not a pure rotation, for ##v \ne 0##.

Frostman
I understand, that's a good point of view without make any calculus.

Thank you so much!

berkeman
Frostman said:
The equalities ##\sin \theta = v ## and ##\cos \theta = \frac{1}{\gamma}## come from the observation that I did before about the relation between ##\cos \theta## and ##\sin \theta## with ##\arctan \theta##?
Interestingly:
$$\sin^2 \theta + \cos^2 \theta = v^2 + \frac 1 {\gamma^2} = v^2 + (1 - v^2) = 1$$
You can compare that with the rapidity ##\phi##, where:
$$\gamma = \cosh \phi, \ \gamma v = \sinh \phi \ \ \text{and} \ \ v = \tanh \phi$$

Frostman

## 1. How do you calculate a specific boost and rotation?

To calculate a specific boost and rotation, you will need to use the appropriate formulas and equations for the given situation. This may involve using vector calculations, trigonometry, or other mathematical methods. It is important to have a clear understanding of the physical principles involved in order to accurately calculate the boost and rotation.

## 2. What is the difference between boost and rotation?

Boost and rotation are two different types of motion. Boost refers to the change in speed or velocity of an object, while rotation refers to the change in orientation or direction of an object. In other words, boost involves linear motion while rotation involves angular motion.

## 3. How do you determine the magnitude and direction of a boost?

The magnitude and direction of a boost can be determined using vector calculations. This involves breaking down the boost into its components and using trigonometry to determine the magnitude and direction of each component. The magnitude of the boost can then be found by using the Pythagorean theorem and the direction can be determined using inverse trigonometric functions.

## 4. How do you calculate the final position after a boost and rotation?

To calculate the final position after a boost and rotation, you will need to use the initial position, the boost vector, and the rotation angle. First, apply the boost to the initial position to determine the new position. Then, rotate the new position by the given angle to determine the final position. This can be done using vector and trigonometric calculations.

## 5. Can you provide an example of calculating a specific boost and rotation?

Yes, for example, if an object starts at position (2,3) and has a boost of magnitude 5 and direction of 30 degrees, and then rotates by 45 degrees, you can calculate the final position by first applying the boost to (2,3) to get a new position of (6.33, 5.1). Then, rotate this new position by 45 degrees to get a final position of (3.535, 7.778). This calculation involves vector addition and trigonometric functions.

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