# Generalized Lorentz Transformation

• unified
In summary: Lorentz transformation from ##S##.How do you define the coordinates in ##\bar S##? There's no single orientation of axes. The axes can be... well, pretty much anything, as long as it's a Lorentz transformation from ##S##.I see what you're saying. I just assumed that since the problem doesn't give any other information, we can assume the orientation of the axes to be the one given by the transformation. But, I guess that's not the case. Thank you for your help!I see what you're saying. I just assumed that since the problem doesn't give any other information, we can assume the orientation of the axes
unified
Summary: The problem is to generalize the Lorentz transformation to two dimensions.

Relevant Equations

Lorentz Transformation along the positive x-axis:
$$\begin{pmatrix} \bar{x^0} \\ \bar{x^1} \\ \bar{x^2} \\ \bar{x^3} \\ \end{pmatrix} = \begin{pmatrix} \gamma & -\gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} {x^0} \\ {x^1} \\ {x^2} \\ {x^3} \\ \end{pmatrix}$$
Lorentz Transformation along the positive y-axis:
$$\begin{pmatrix} \bar{x^0} \\ \bar{x^1} \\ \bar{x^2} \\ \bar{x^3} \\ \end{pmatrix} = \begin{pmatrix} \gamma & 0 & -\gamma \beta & 0 \\ 0 & 1 & 0 & 0 \\ -\gamma \beta & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} {x^0} \\ {x^1} \\ {x^2} \\ {x^3} \\ \end{pmatrix}$$
Velocity transformations from S to S' where S' moves along the positive x-axis at speed v relative to S:
$$u'_x = \frac{u_x - v}{1 - \frac{u_xv}{c^2}} \\ u'_y = \frac{u_y}{\gamma(1 - \frac{u_xv}{c^2})}$$

Problem Statement

$$\bar{S} \text{ moves at velocity } \\ \vec{v} = \beta c(Cos\phi \hat{x} + Sin\phi \hat{y}) \text{ relative to S, with axes parallel and origins coinciding at time } \\ t = \bar{t} = 0. \text{ Find the Lorentz transformation from S to } \bar{S}$$

Attempt at a solution
$$\text{Let S' move at velocity } \beta cCos\phi\hat{x} \text{ relative to S. We can find the velocity of } \bar{S} \text { relative to S' using the velocity transformations. } \\ u'_x = 0 \\ u'_y = \gamma_x\beta cSin\phi \\ where \, \gamma_x = \frac{1}{(1-\beta^2Cos^2\phi)^\frac{1}{2}}$$

$$\text{We can express the coordinates in } \bar{S} \text{ given the coordinates in S' by the y-axis Lorentz transformation, with } \\ \bar{\beta} = \beta Sin\phi \gamma_x \, \\ \bar{\gamma} = \frac{\gamma}{\gamma_x} \\ where \, \gamma = \frac{1}{(1 - \beta^2)^\frac{1}{2}}$$$$\text{And we can express the coordinates in S' given the coordinates in S by the x-axis transformation with } \\ \beta' = \beta Cos\phi \\ \gamma' = \gamma_x$$

$$\text{ Therefore we can express the coordinates in } \bar{S} \text { given the coordinates in S as follows }$$

$$\begin{pmatrix} \bar{x^0} \\ \bar{x^1} \\ \bar{x^2} \\ \bar{x^3} \\ \end{pmatrix} = \begin{pmatrix} \frac{\gamma}{\gamma_x} & 0 & -\frac{\gamma}{\gamma_x}\beta Sin\phi \gamma_x & 0 \\ 0 & 1 & 0 & 0 \\ -\frac{\gamma}{\gamma_x}\beta Sin\phi \gamma_x & 0 & \frac{\gamma}{\gamma_x} & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \gamma_x & -\gamma_x\beta Cos\phi & 0 & 0 \\ -\gamma_x \beta Cos\phi & \gamma_x & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{pmatrix} =$$
\begin{pmatrix}
\gamma & -\gamma \beta Cos\phi & -\gamma \beta Sin\phi & 0 \\
-\gamma_x\beta Cos\phi & \gamma_x & 0 & 0 \\
-\gamma \gamma_x\beta Sin\phi & \gamma \gamma_x\beta^2Sin\phi Cos\phi & \frac{\gamma}{\gamma_x} & 0 \\
0 & 0 & 0 & 1 \end{pmatrix}
\begin{pmatrix}
x^0 \\
x^1 \\
x^2 \\
x^3
\end{pmatrix}

This final matrix represents the solution to the problem. But, this solution is wrong, as the correct matrix is given an follows.

\begin{pmatrix}
\gamma & -\gamma \beta Cos\phi & -\gamma \beta Sin\phi & 0 \\
-\gamma \beta Cos\phi & (\gamma Cos^2\phi + Sin^2\phi) & (\gamma - 1)Sin\phi Cos\phi & 0 \\
-\gamma \beta Sin\phi & (\gamma - 1)Sin\phi Cos\phi & (\gamma Sin^2\phi + Cos^2\phi) & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}

I am unable to determine what is wrong with this solution, any suggestions would be appreciated.

Moderator's note: Moved to advanced physics homework forum.

The product of two pure boost in different directions wii not be a pure boost anymore.
$$\exp(\zeta^1X_1)\cdot\exp(\zeta^2X_2)\neq\exp(\zeta^1X_1+\zeta^2X_2)$$
Because the boosts in different directions do not commute with each other.

Jianbing_Shao said:
The product of two pure boost in different directions wii not be a pure boost anymore.
$$\exp(\zeta^1X_1)\cdot\exp(\zeta^2X_2)\neq\exp(\zeta^1X_1+\zeta^2X_2)$$
Because the boosts in different directions do not commute with each other.

I noticed that of course. But, isn't it true that the matrix gives S(bar) in terms of S? I don't see how the logic could be wrong. We give S(bar) in terms of S' and S' in terms of S, thus S(bar) in terms of S. I agree the formula is not a boost, but isn't it still correct?

unified said:
I noticed that of course. But, isn't it true that the matrix gives S(bar) in terms of S? I don't see how the logic could be wrong. We give S(bar) in terms of S' and S' in terms of S, thus S(bar) in terms of S. I agree the formula is not a boost, but isn't it still correct?

What you have is a valid transformation. For example, the time dilation effect will be correct.

But, if you assume that the frame ##\bar S## is moving with velocity ##\vec v## in frame ##S##, then frame ##S## is not moving with velocity ##-\vec v## in your frame ##\bar S##.

In fact, that requirement is usually an unstated assumption in this problem. The assumption that you want symmetry of relative velocity between ##S## and ##\bar S##.

Your approach does not achieve this, but transforms to a rotated version of the "required" coordinates for ##\bar S##.

PeroK said:

What you have is a valid transformation. For example, the time dilation effect will be correct.

But, if you assume that the frame ##\bar S## is moving with velocity ##\vec v## in frame ##S##, then frame ##S## is not moving with velocity ##-\vec v## in your frame ##\bar S##.

In fact, that requirement is usually an unstated assumption in this problem. The assumption that you want symmetry of relative velocity between ##S## and ##\bar S##.

Your approach does not achieve this, but transforms to a rotated version of the "required" coordinates for ##\bar S##.

If what I have is a valid transformation, then I'm confused about what is calculating. Given the coordinates in S, does it give the coordinates in S(bar)?

unified said:
If what I have is a valid transformation, then I'm confused about what is calculating. Given the coordinates in S, does it give the coordinates in S(bar)?
How do you define the coordinates in ##\bar S##? There's no single orientation of axes. The axes can be in any directions you want.

Take a simple boost in the x-direction from ##S## to ##S'##. There's nothing to say that ##S'## must take the same x-axis as ##S##. It's perfectly valid for ##S'## to define the direction of motion as the ##y## direction. Or, to rotate its coordinates in any way.

It's by convention that the coordinates in ##S'## are defined in a certain way with respect to the ##S## coordinates.

If you have ##\bar S## moving in both the x and y directions relative to ##S##, then it's not so obvious how you want to define the x,y axes in ##\bar S## with respect to the x, y axes in ##S##.

The only thing that must be true is that the magnitude of the velocity of ##S## must be correct. But, any orientation of axes where this holds is a valid coordinate system for ## \bar S##.

However, the coordinate system where the velocity of ##S## is ##- \vec v## is big convention the one given for the 2D boost.

unified said:
If what I have is a valid transformation, then I'm confused about what is calculating. Given the coordinates in S, does it give the coordinates in S(bar)?
PS note that for a boost along the x-axis, all the axes in ##S'## can be taken to be parallel to those in ##S##. I.e. the two sets of axes coincide when the origins coincide.

But, if the boost is not along a single coordinate axis, then the mutually orthogonal x, y axes in one frame cannot coincide with mutually orthogonal x', y' axes in the other.

It's a good exercise to check this out.

unified said:
I noticed that of course. But, isn't it true that the matrix gives S(bar) in terms of S? I don't see how the logic could be wrong. We give S(bar) in terms of S' and S' in terms of S, thus S(bar) in terms of S. I agree the formula is not a boost, but isn't it still correct?
Perhaps you can think of the problem in such a way:
$$\exp(\zeta^1 X_1)\exp(\zeta^2 K_2)=\lim_{n\rightarrow \infty}\prod^{n}_{i=1}\exp\left(\frac{\zeta^1 X_1}{i}\right) \prod^{n}_{j=1}\exp\left(\frac{\zeta^2 X_2}{j}\right)$$
and
$$\exp(\zeta^1 X_1+\zeta^2 K_2)=\lim_{n\rightarrow \infty}\prod^{n}_{i=1}\exp\left(\frac{\zeta^1 X_1}{i}\right)\exp\left(\frac{\zeta^2 X_2}{i}\right)$$
Obviously the two formulas describe different physical processes

I'm not sure whether the problem is well posed. I interpreted it differently than what the discussion suggest:

I think what's asked for is the rotation free Lorentz boost along an arbitrary velocity ##\vec{v}## between the two frames. This will give a symmetric Lorentz-transformation matrix. It's not difficult to derive from the Lorentz transformation in one direction by writing it in 3D covariant terms.

Another understanding of the question seems to be that it's ask to do the transformation in two steps, i.e., first a boost in 1-direction and then a boost in 2-direction. This leads, however, not to a rotation free boost for the reasons stated above, i.e., the composition of two rotation-free boosts in different direction leads to a Lorentz boost followed by a rotation.

For the rotation-free boost, see

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf

## 1. What is the Generalized Lorentz Transformation?

The Generalized Lorentz Transformation is a mathematical formula that describes the relationship between space and time in the theory of special relativity. It is used to calculate how measurements of space and time change between two different reference frames that are moving at a constant velocity relative to each other.

## 2. How is the Generalized Lorentz Transformation different from the original Lorentz Transformation?

The original Lorentz Transformation only applies to objects moving at a constant velocity along one axis. The Generalized Lorentz Transformation extends this concept to include objects moving in any direction and at any velocity. It also takes into account the effects of acceleration.

## 3. What are the key components of the Generalized Lorentz Transformation?

The Generalized Lorentz Transformation involves four main components: time dilation, length contraction, velocity addition, and the transformation of coordinates. These components work together to describe how measurements of space and time change between different reference frames.

## 4. How is the Generalized Lorentz Transformation used in practical applications?

The Generalized Lorentz Transformation is used in many practical applications, including GPS systems, particle accelerators, and satellite communication. It is also used in the calculations of relativistic effects in high-speed travel, such as in space travel and airplanes.

## 5. Are there any limitations to the Generalized Lorentz Transformation?

While the Generalized Lorentz Transformation is a very accurate and useful tool, it does have some limitations. It only applies to objects moving at a constant velocity in a vacuum and does not take into account the effects of gravity. It also cannot be used to describe the behavior of objects at the quantum level.

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