# Lorentz transformation for 3 frames (2 dimensions)

• I
• chaksome
In summary, In summary, The speaker wants to explore the Lorentz transformation between two frames, and has found two different solutions. They seek to understand the discrepancy between the two solutions and its potential connection to general relativity. The other person provides a mathematical verification for the first solution and explains that the phenomenon is known as Wigner or Thomas rotation. The speaker also asks for analysis of their second solution. The other person suggests that there may be no relative velocity in the x-direction, causing the discrepancy.
chaksome
TL;DR Summary
I want to know why an else solution can not get the right answer. And want to know the way to correct this solution.Supposed that a frame S'' is moving in the lab frame at ##\beta_x## in the x-direction, ##\beta_y## in the y-direction, now I want to find out the Lorentz transformation between these frames.

Applying the vector transformation in special relative, I've already worked out the solution like this(and I think it is the correct answer):$$\begin{cases}x''=\frac{\gamma \beta_x^2+\beta_y^2}{\beta_x^2+\beta_y^2}x+\frac{(\gamma-1)\beta_x\beta_y}{\beta_x^2+\beta_y^2}y-\gamma\beta_x ct\\y''=\frac{\gamma \beta_y^2+\beta_x^2}{\beta_x^2+\beta_y^2}y+\frac{(\gamma-1)\beta_x\beta_y}{\beta_x^2+\beta_y^2}x-\gamma\beta_y ct，(1)\\ct''=\gamma ct-\gamma \beta_x x-\gamma\beta_y y\end{cases}$$(and## \gamma=\frac{1}{\sqrt{1-(\beta_x^2+\beta_y^2)}}##)(if there is anything wrong, please point it out)Now I want to solve this problem in a different way (which is valid if the 3 frames are moving in one direction, but fail to work out the right answer in this scene)First, I make a Lorentz transformation between S and S'(which is moving in the frame S at##\beta_x##in the x-direction),and I get the equation$$\begin{cases}x'=\gamma_x(x-\beta_x ct)\\y'=y\\ct'=\gamma_x(ct-\beta_xx)\end{cases}$$Then,I make a Lorentz transformation between S' and S''(according to the velocity-addition formula,S'' is moving in S' at ##\beta_y'=\gamma_x\beta_y##in the y-direction),and I get the equation$$\begin{cases}x''=x'\\y''=\gamma_y'(y'-\beta_y' ct')\\t''=\gamma_y'(ct'-\beta_y'y')\end{cases}$$OK,now we solve these equation, and find that$$\begin{cases}x''=\gamma_xx+0y-\gamma_x\beta_xct\\y''=\frac{\gamma}{\gamma_x}y+\gamma\beta_x\beta_yx-\gamma\gamma_x\beta_yct, (2)\\ct''=\gamma ct-\gamma\beta_x x-\gamma\beta_y y\end{cases}$$We can see that (1) is obviously different from (2)I hope to find out the essential cause of this phenomenon(maybe it will refer to the general relativity):What causes the discrepancy between (1) and (2)?Why the ct'' stay consistence?Thanks a million!

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Hi.
I would like to give a verification to your (1).

Rotation in xy plane and x-Lorentz transformation are expressed as matrices,
$$R \equiv \begin{pmatrix} cos\phi & sin\phi & 0 \\ -sin\phi & cos\phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ L \equiv \begin{pmatrix} cosh\theta & 0 & sinh\theta \\ 0 & 1 & 0 \\ sinh\theta & 0 & cosh\theta \\ \end{pmatrix}$$
where cosh ##\theta## = ##\gamma##, sinh ##\theta## = - ##\gamma##v/c.

The successive actions,
1. Rotate the axes so that the S" moves along x axis
2. Do Lorentz transformation for x-moving system
3. Rotate back to original axes
are presented as products of matrices
$$R^{-1}LR= \begin{pmatrix} cosh\theta cos^2\phi +sin^2\phi & (cosh\theta -1)cos\phi sin\phi & sinh\theta cos\phi \\ (cosh\theta -1)cos\phi sin\phi & cosh\theta sin^2\phi + cos^2\phi & sinh\theta sin\phi \\ sinh\theta cos\phi & sinh\theta sin\phi & cosh\theta \\ \end{pmatrix}$$
where matrices product is calculated with above R, L and
$$R^{-1}= \begin{pmatrix} cos\phi & -sin\phi & 0 \\ sin\phi & cos\phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \\$$

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Dale, PeroK and chaksome
If I'm following correctly what you are doing, the phenomenon is called Wigner rotation, or Thomas rotation. Or Wigner-Thomas rotation. Mathematically, it's a consequence of Lorentz transforms not commuting in general (if the velocities are colinear, they do). Physically, it follows from the fact that something moving along the ##y'## axis has ##x'=0##, implying that ##x=x(t)##. So "moving along the ##y##-axis" and "moving along the ##y'##-axis" are two different things - the former implies constant ##x##, the latter implies constant ##x'##. So compositing movement along ##x## with movement along ##y'## is different from compositing movement along ##y## with movement along ##x'##.

vanhees71 and chaksome
mitochan said:
Hi.
I would like to give a verification to your (1).

Rotation in xy plane and x-Lorentz transformation are expressed as matrices,
$$R \equiv \begin{pmatrix} cos\phi & sin\phi & 0 \\ -sin\phi & cos\phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ L \equiv \begin{pmatrix} cosh\theta & 0 & sinh\theta \\ 0 & 1 & 0 \\ sinh\theta & 0 & cosh\theta \\ \end{pmatrix}$$
where cosh ##\theta## = ##\gamma##, sinh ##\theta## = - ##\gamma##v/c.

The successive actions,
1. Rotate the axes so that the S" moves along x axis
2. Do Lorentz transformation for x-moving system
3. Rotate back to original axes
are presented as products of matrices
$$R^{-1}LR= \begin{pmatrix} cosh\theta cos^2\phi +sin^2\phi & (cosh\theta -1)cos\phi sin\phi & sinh\theta cos\phi \\ (cosh\theta -1)cos\phi sin\phi & cosh\theta sin^2\phi + cos^2\phi & sinh\theta sin\phi \\ sinh\theta cos\phi & sinh\theta sin\phi & cosh\theta \\ \end{pmatrix}$$
where matrices product is calculated with above R, L and
$$R^{-1}= \begin{pmatrix} cos\phi & -sin\phi & 0 \\ sin\phi & cos\phi & 0 \\ 0 & 0 & 1 \end{pmatrix} \\$$
Thanks a lot! It’s absolutely a nice way to get the correct answer. And you have inspired me to some extent that maybe the divergence between (1) and (2) is due to the the effect of the rotation. Do you agree?
(For the doubt you bring forward in your second message, I think it is actually no relative velocity in the x-direction because:$$v_x'=\frac{v_x-v_x}{1-\frac{v_x^2}{c^2}}=0$$)

Ibix said:
If I'm following correctly what you are doing, the phenomenon is called Wigner rotation, or Thomas rotation. Or Wigner-Thomas rotation. Mathematically, it's a consequence of Lorentz transforms not commuting in general (if the velocities are colinear, they do). Physically, it follows from the fact that something moving along the ##y'## axis has ##x'=0##, implying that ##x=x(t)##. So "moving along the ##y##-axis" and "moving along the ##y'##-axis" are two different things - the former implies constant ##x##, the latter implies constant ##x'##. So compositing movement along ##x## with movement along ##y'## is different from compositing movement along ##y## with movement along ##x'##.
Though I don't understand the 'Wigner/Tomas/Wigner-Tomas rotation' completely now, but I think you show me a clear road to go by myself. I am so grateful to you for answering my question.

chaksome said:
Thanks a lot! It’s absolutely a nice way to get the correct answer. And you have inspired me to some extent that maybe the divergence between (1) and (2) is due to the the effect of the rotation. Do you agree?
(For the doubt you bring forward in your second message, I think it is actually no relative velocity in the x-direction because:$$v_x'=\frac{v_x-v_x}{1-\frac{v_x^2}{c^2}}=0$$)

If you look at the solution from @mitochan, it doesn't actually say explicitly how the axes in S'' are defined. And, in fact, in order to do the inverse rotation ##R^{-1}## you must orient the axes in S'' so that the motion of S (in S'') is at the same angle as the motion of S'' in S. In this case ##\phi##. (Although, of course, in the opposite direction.)

With this definition for the axes in S'' you get the "correct" transformation.

If you take your second approach and you map the y'' axis: i.e. the points that have x-y coordinates ##(0, y'')## at some fixed time ##t''##, you will see that this does not map to a vertical line in S. (This is the basis of the Wigner rotation.)

You've effectively done a valid two-stage transformation: but you have transformed to a coordinate system in S'' which is different from the one you got by the first method.

You could work out the angle at which S in moving in S'' in this case. Then, if you really wanted to, you could add a final rotation to rotate the axes in S'' so that S is moving at angle ##\phi## and this should bring you back to the first solution.

Ibix, chaksome and mitochan
PeroK said:
If you look at the solution from @mitochan, it doesn't actually say explicitly how the axes in S'' are defined. And, in fact, in order to do the inverse rotation ##R^{-1}## you must orient the axes in S'' so that the motion of S (in S'') is at the same angle as the motion of S'' in S. In this case ##\phi##. (Although, of course, in the opposite direction.)

With this definition for the axes in S'' you get the "correct" transformation.

If you take your second approach and you map the y'' axis: i.e. the points that have x-y coordinates ##(0, y'')## at some fixed time ##t''##, you will see that this does not map to a vertical line in S. (This is the basis of the Wigner rotation.)

You've effectively done a valid two-stage transformation: but you have transformed to a coordinate system in S'' which is different from the one you got by the first method.

You could work out the angle at which S in moving in S'' in this case. Then, if you really wanted to, you could add a final rotation to rotate the axes in S'' so that S is moving at angle ##\phi## and this should bring you back to the first solution.
I gradually come to understand, the what you say also explain the phenomenon that expression of ct” in (1)and(2) are the same.
Thanks a lot!

PeroK
WOW! I worked out the matrix of the rotation$$R=\begin{pmatrix}cos\varepsilon&sin\varepsilon&0\\-sin\varepsilon&cos\varepsilon&0\\0&0&1\end{pmatrix}$$
and$$sin\varepsilon=\frac{\gamma-1}{\gamma}\frac{\beta_x\beta_y\gamma_x}{\beta_x^2+\beta_y^2}$$
And I've read some paper about the Thomas precession, and I find it more and more interesting to learn more things in this field.
Thank you all again~

Dale, Ibix and PeroK
As for your (2) I write down corresponding matrices.
$$L_1\equiv \begin{pmatrix} c_1 & 0 & s_1 \\ 0 & 1 & 0 \\ s_1 & 0 & c_1 \\ \end{pmatrix}$$

$$L_2\equiv \begin{pmatrix} 1 & 0 & 0 \\ 0 & c_2 & s_2 \\ 0 & s_2 & c_2 \\ \end{pmatrix}$$
where ##c_i\equiv cosh\theta_i##, ##s_i \equiv sinh\theta_i,i=1,2 ##
index i=1 correspond to x-axis Lorentz transformation and i=2 corresponds to that of y axis.

Matrices for successive Lorentz transformations are
$$L_2 L_1 = \begin{pmatrix} c_1 & 0 & s_1 \\ s_1s_2 & c_2 & s_2c_1 \\ s_1c_2 & s_2 & c_1c_2 \\ \end{pmatrix}$$ and $$L_1 L_2 = \begin{pmatrix} c_1 & s_1s_2 & s_1c_2 \\ 0 & c_2 & s_2 \\ s_1 & c_1s_2 & c_1c_2 \\ \end{pmatrix}$$
We see ,if my calculation right, $$L_2 L_1 = (L_1 L_2)^{T}$$

Dale
(contd.)
$$L_1L_2-L_2L_1= \begin{pmatrix} 0 & s_1s_2 & s_1(c_2-1) \\ -s_1s_2 & 0 & s_2(1-c_1) \\ s_1(1-c_2) & s_2(c_1-1) & 0 \\ \end{pmatrix}$$
Say ##\theta_i<<1## leaving only order 2 and neglecting highers, approximately,
$$L_1L_2-L_2L_1= \begin{pmatrix} 0 & \theta_1\theta_2 & 0 \\ -\theta_1\theta_2 & 0 & 0 \\ 0 & 0 & 0 \\ \end{pmatrix}$$
Only rotation components survive. So application order of infinitely small Lorentz transformation, i.e. x first then y or y first then x, results difference of rotation.

Dale
mitochan said:
As for your (2) I write down corresponding matrices.
$$L_1\equiv \begin{pmatrix} c_1 & 0 & s_1 \\ 0 & 1 & 0 \\ s_1 & 0 & c_1 \\ \end{pmatrix}$$

$$L_2\equiv \begin{pmatrix} 1 & 0 & 0 \\ 0 & c_2 & s_2 \\ 0 & s_2 & c_2 \\ \end{pmatrix}$$
where ##c_i\equiv cosh\theta_i##, ##s_i \equiv sinh\theta_i,i=1,2 ##
index i=1 correspond to x-axis Lorentz transformation and i=2 corresponds to that of y axis.

Matrices for successive Lorentz transformations are
$$L_2 L_1 = \begin{pmatrix} c_1 & 0 & s_1 \\ s_1s_2 & c_2 & s_2c_1 \\ s_1c_2 & s_2 & c_1c_2 \\ \end{pmatrix}$$ and $$L_1 L_2 = \begin{pmatrix} c_1 & s_1s_2 & s_1c_2 \\ 0 & c_2 & s_2 \\ s_1 & c_1s_2 & c_1c_2 \\ \end{pmatrix}$$
We see ,if my calculation right, $$L_2 L_1 = (L_1 L_2)^{T}$$
wah！So many interesting things can be found through your calculation.Yesterday, I was thinking about the physical meaning of the equation,$$L_1L_2=(L_2L_1)^T$$
But I failed to find a clear meaning in physics.
Is that mean a kind of symmetry?
And~according to your theory, the transformation of time is also different if we change the application order of the Lorentz Transformation when the ##\theta _i## is larger.How do we think about it?

Hi.
How about challenging finding the components of the following L and R?

$$L_2L_1=LR$$
where R is mentioned in my post #2 and L is a symmetric matrix representing Lorentz transformation, or boost.

If you succeed, you can learn that two successive boosts are interpreted as successive rotation and a boost.

chaksome
mitochan said:
Hi.
How about challenging finding the components of the following L and R?

$$L_2L_1=LR$$
where R is mentioned in my post #2 and L is a symmetric matrix representing Lorentz transformation, or boost.

If you succeed, you can learn that two successive boosts are interpreted as successive rotation and a boost.
Hello,thank you for your patience and the guidance~
Do you mean that:
$$R^{-1}LR=(LR^{-1})^TR=L^TR^2$$
Then we can define that
$$R^*\equiv R, L^*\equiv L^T=L$$
and we can interpret the successive boosts as##L^*R^*R^*##,or ##LRR##
$$R^*=\begin{pmatrix}cos\epsilon&sin\epsilon&0\\-sin\epsilon&cos\epsilon&0\\0&0&1\end{pmatrix}$$
$$L^*=\begin{pmatrix}cosh\theta&0&sinh\theta\\0&1&0\\sinh\theta&0&cosh\theta\end{pmatrix}$$
$$LRR= \begin{pmatrix} cosh\theta cos^2\phi +sin^2\phi & (cosh\theta -1)cos\phi sin\phi & sinh\theta cos\phi \\ (cosh\theta -1)cos\phi sin\phi & cosh\theta sin^2\phi + cos^2\phi & sinh\theta sin\phi \\ sinh\theta cos\phi & sinh\theta sin\phi & cosh\theta \\ \end{pmatrix}$$
which meets my (1)

(correction and supplement to my post #12)
Hypothesis : First x- then y- boosts is interpreted as first boost in one direction then rotation, i.e.
$$L_2 L_1 =RL$$
If so
$$R^{-1} L_2 L_1 =L$$
with ##R^{-1}## in post #2, ##L_2L_1## in post #9 and careful choice of angle ##\phi##, RHS L would become a symmetric matrix representing boost.

My preliminary calculation suggests such a coordinated rotation angle be
$$tan\phi=\frac{s_1 s_2 }{c_1+c_2}$$
and then L becomes boost matrix for boost velocity $$(c\beta_x,\ c\gamma_x\beta_y)$$, which is a kind you introduced in post #1, but not successive boosts but a single boost.
Further, formula
$$R=L_2L_1L^{-1}$$
suggests that a rotation is interpreted and represented as three successive boosts of first ##(-c\beta_x,-\ c\gamma_x\beta_y)##, then ##c\beta_x## and at last ##c\beta_y##.

I should appreciate it if you check calculations and hopefully verify it.

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mitochan said:
$$tan\phi=\frac{s_1 s_2 }{c_1+c_2}$$
I finally found it exactly meet what I work out in my #8, and I try to follow your thoughts and verify that the angle you show is right. Thank you for giving me another opinion of this question ~

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$$sin\phi=-\frac{s_1s_2}{c_1c_2+1}$$
$$cos\phi=\frac{c_1+c_2}{c_1c_2+1}$$
$$tan\phi=-\frac{s_1s_2}{c_1+c_2}$$

$$L=\begin{pmatrix} 1+\frac{s_1^2c_2^2}{c_1c_2+1} & \frac{s_1s_2c_2}{c_1c_2+1} & s_1c_2 \\ \frac{s_1s_2c_2}{c_1c_2+1} & 1+\frac{s_2^2}{c_1c_2+1} & s_2 \\ s_1c_2 & s_2 & c_1c_2 \\ \end{pmatrix}$$

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(continued)

Indtroducing ##s'_2=s_2/c_1##,

$$L=\begin{pmatrix} 1+\frac{s_1^2 c_2^2}{c_1c_2+1} & \frac{s_1s_2' c_1c_2}{c_1c_2+1} & s_1c_2 \\ \frac{s_1s_2'c_1c_2}{c_1c_2+1} & 1+\frac{s_2'^2c_1^2}{c_1c_2+1} & s_2'c_1 \\ s_1c_2 & s_2'c_1 & c_1c_2 \\ \end{pmatrix}$$

Symmetry between 1 and 2 is apparent.

$$L=\begin{pmatrix} 1+\alpha (v_1/c)^2 & \alpha (v_1/c)(v'_2/c) & -(v_1/c)c_1c_2 \\ \alpha (v_1/c)(v'_2/c) & 1+\alpha (v'_2/c)^2 & -(v'_2/c)c_1c_2 \\ -(v_1/c)c_1c_2 & -(v'_2/c)c_1c_2 & c_1c_2 \\ \end{pmatrix}$$
where
$$\alpha=\frac{c_1^2c_2^2}{c_1c_2+1}$$
$$c_i=\frac{1}{\sqrt{1-v_i^2/c^2}}=\gamma_i$$
$$v_i/c=-tanh\theta_i=-s_i/c_i$$
$$v'_2=v_2/c_1$$

This matrix is expected to represent boost for velocity ##(v_1,v_2/\gamma_1)##.
Factor ##1/\gamma_1## reduces velocity of the second boost in original S-S'-S" procedure. This is interpreted to take place by time dilation of S' ,where ##v_2## is measured, against S.

In this synthesized boost ##s_2## is reduced to ##s'_2=s_2/\gamma_1##. however ##c_2## appears with no modification in the boost matrix. I would like to understand it in depth.

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chaksome
(correction)
$$-v_i/c=tahn\theta_i=s_i$$

Following the prescription in post #2, i.e. rotate, x-boost and rotate back, I calculated matrices to be

$$L=R^{-1}(\phi)L_1(V)R(\phi)$$

where

$$V^2/c^2\equiv v_1^2/c^2 + \frac{v_2^2/c^2}{\gamma_1^2}=v_2^2+ \frac {v_1^2/c^2} {\gamma_2^2}$$
,symmetric between 1 and 2, and
$$=1-\frac{1}{\gamma_1^2\gamma_2^2}\equiv 1-\frac{1}{\gamma^2}$$
$$\phi=tan^{-1}\frac{v_2}{v_1\gamma_1}$$
Going backto post #14 with this,

$$L_2(v_2)L_1(v_1)=R(\Phi)R^{-1}(\phi)L_1(V)R(\phi)$$

where

$$\Phi=-tan^{-1}\frac{v_1v_2/c^2}{\gamma_1\gamma_2(\gamma_1+\gamma_2)}$$

I have not calculated on the commutation but hopefully expect

$$L_1(v_1)L_2(v_2)=R(\Phi)R^{-1}(\phi)L_1(V)R(\phi)$$

but where ##\phi## has different value from the above case, so

$$\phi=tan^{-1}\frac{v_2\gamma_2}{v_1}$$

with reminding that suffix 1 denotes x coordinate and 2 denotes y coordinate.

So I summerize here my understanding of making ##L_1L_2## and ##L_1L_2## transformations in S,
- First, boost with velocities of the same magnitude but different directions,
- Then rotate with the same angle.

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chaksome
chaksome said:
And~according to your theory, the transformation of time is also different if we change the application order of the Lorentz Transformation when the ##\theta _i## is larger.How do we think about it?
From the previous post the both undertake the rotation of the same angle so common rotation component coefficicent ##v_1v_2/c^2## as the top order. Difference in boost directions effects in higher orders.

Say "boosts go-around",

$$L_2(-v_2)L_1(-v_1)L_2(v_2)L_1(v_1)$$
$$=R(\Phi)R^{-1}(\phi+\pi/2)L_1(V)R(\phi+\pi/2)R(\Phi)R^{-1}(\phi)L_1(V)R(\phi) \approx R(2\Phi)$$

for infinitesimal boosts. Rotation would be interpreted as integrated infinitesimal boost go around sets.

Obviously,

$$L_1(-v_1)L_2(-v_2)L_2(v_2)L_1(v_1)=I$$

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chaksome
(correction) angle for inverse direction is ##\phi+\pi##, not ##\phi+\pi/2##.

## 1. What is the Lorentz transformation for 3 frames?

The Lorentz transformation for 3 frames is a mathematical formula that describes how measurements of time and space are affected when transitioning between three different inertial frames of reference in two dimensions. It is an extension of the original Lorentz transformation, which was developed by Dutch physicist Hendrik Lorentz in the late 19th century to explain the effects of relativity on the concept of simultaneity.

## 2. How does the Lorentz transformation account for time dilation?

The Lorentz transformation takes into account the fact that time is not absolute and can appear to pass at different rates for observers in different frames of reference. This is known as time dilation and is a fundamental aspect of Einstein's theory of special relativity. The transformation includes a factor known as the time dilation factor, which adjusts for the difference in perceived time between frames.

## 3. Can the Lorentz transformation be applied to any number of dimensions?

Yes, the Lorentz transformation can be applied to any number of dimensions. In its most basic form, it applies to two dimensions, but it can be extended to three or more dimensions by adding additional terms to the transformation equations. This allows for the description of more complex scenarios, such as objects moving in three-dimensional space or in curved spacetime.

## 4. How is the Lorentz transformation related to the concept of spacetime?

The Lorentz transformation is closely related to the concept of spacetime, which combines the three dimensions of space with the dimension of time into a single four-dimensional continuum. In fact, the transformation equations are derived from the equations of spacetime, which describe how events are perceived by observers in different frames of reference.

## 5. Can the Lorentz transformation be used to explain the twin paradox?

Yes, the Lorentz transformation can be used to explain the twin paradox, which is a thought experiment that demonstrates the effects of time dilation. In this scenario, one twin travels at high speed to a distant star and back, while the other twin remains on Earth. When the traveling twin returns, they have aged less than the twin who stayed on Earth, due to the effects of time dilation predicted by the Lorentz transformation.

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