# [Special Relativity] Lorentz Transformation and Boosts

## Homework Statement:

Find the matrix for Lorentz transformation consisting of a boost of speed ##v## in the ##x##-direction followed by a boost of speed ##w## in the ##y'## direction. Show that the boosts performed in the reverse order would give a different transformation.

## Relevant Equations:

Refer to the below calculations ##\longrightarrow##
[BEGINNGING NOTICE]
Before I begin showing my attempted solution, I would just like to quickly mention that this is a "repost" of the same question I had around a week ago. While I would usually use the "reply" function on the same thread, I believe that thread is getting pretty messy (sometimes with wrong LaTeX typing conventions) and would only ultimately serve to confuse the readers that truly want to help me - this is especially the case since I am no longer able to edit my post. Therefore, rather than having a bunch of misinformation and facts being jumbled together in disarray in the previous thread, I decided to create this new one to properly show what I have gathered and learned regarding how to (attempt to) solve this problem.

Of course, for any administrator(s) of the forum reading this, please feel free to delete the past thread as I'm sure duplicate posts that may serve to confuse future readers are not allowed on this site. Here is the past thread: https://www.physicsforums.com/threa...ity-lorentz-transformation-and-boosts.980562/.
*Note that I did mention all of the provided information a user gave in the past thread here.

Without further due, below is my attempted solution. I would sincerely appreciate any amount of help from the community. Thank you for reading this as well as providing your kind assistance!

[Attempted Solution]
Below are the equations I used to set up - and will hopefully solve - for the above question.

For Boost:
A Lorentz boost in the ##x##-direction would look like this below:
$$\begin{bmatrix} \gamma(v) & -\beta(v) \gamma(v) & 0 & 0 \\ - \beta(v) \gamma(v) & \gamma(v) & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

Or, the same Lorentz boost of speed ##v## in the ##x##-direction could be written in this way as well:
$$\begin{cases}t' = \gamma (t- \frac{vx}{c^2}) \\ x' = \gamma(x-vt) \\ y' = y \\ z' = z\end{cases}$$

As for my Lorentz boost in the ##y'##-direction with speed ##w##, I used the following matrix:
$$\begin{bmatrix} \gamma(w) & 0 & -\beta(w) \gamma(w) & 0 \\ 0 & 1 & 0 & 0 \\ -\beta(w) \gamma(w) & 0 & \gamma(w) & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$

*Please note that $$\beta = \frac{v}{c}$$ and $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$

Beyond setting up these equations, I honestly don't know how to continue ...

To begin with, @mitochan did kindly provided me a related website that may be able to help me with this question: https://www.physicsforums.com/threads/lorentz-transformation-for-3-frames-2-dimensions.978937/
However, as much as I would like to properly understand its content, I - unfortunately - do admit that I am having a hard time wrapping my head around the provided thread in relation (and application) to my question here.

Furthermore, the user also stated that "two successive perpendicular boosts equals a rotation after a boost". However, how is this fact able to show (or answer) that the boosts (in the question) performed in the reverse order would give a different transformation?

From another user (not from here), he suggested that I should multiply the matrices I found in one order and then in the opposite order afterward to show for the different transformations. While the user did not specify what matrices I should use, I believe I should use the above matrices I found here in the ##x## and ##y'## direction. But, what made me falter in continuing the step is because the user did specifically stated that for my second matrix (I assume the one in the "##y'##"-direction) needs to have ##B##, ##C##, and ##D## in the second row or column. However, I fail to understand or know how to accomplish that - let alone how to find the correct values for the supposed ##B##, ##C##, and ##D## variables.

As I've said, I am not entirely sure how the above equations can help solve the question - if at all. Therefore, any help toward answering the question or help solidify my concept regarding boosts and Lorentz transformations will be much appreciated. Thank you very much for the community's assistance!

Reference Material for the above-attempted solution: https://physics.stackexchange.com/questions/30166/what-is-a-lorentz-boost-and-how-to-calculate-it

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PeroK
Homework Helper
Gold Member
A transformation is a change in coordinates. The first matrix represents the relationship between unprimed and primed coordinates. The second matrix represents the relationship between the primed and double primed coordinates.

If you look at the maths behind matrix multiplication you'll see that the product of the two matrices represents the relationship between the unprimed and double primed coordinates.

• Athenian
A transformation is a change in coordinates. The first matrix represents the relationship between unprimed and primed coordinates. The second matrix represents the relationship between the primed and double primed coordinates.

If you look at the maths behind matrix multiplication you'll see that the product of the two matrices represents the relationship between the unprimed and double primed coordinates.

@PeroK, thank you for your reply! By stating "the first and second matrix" in your response post, do you refer specifically to the two matrices (i.e. for the ##x## and ##y'## directions) I set up in my first post here?
Furthermore, when you say that once I perform matrix multiplication, I'll see that the product of the two matrices represents the relationship between the unprimed and double primed coordinates, does this "relationship" essentially equate to a transformation?

If so, below are my calculations. However, if you did not mean this in your post, please feel free to ignore the below matrix multiplication.
Multiplying the boost with speed ##v## in the ##x##-direction by the boost with speed ##w## in the ##y'##-drection ##\longrightarrow##
$$\begin{pmatrix}\gamma(v) &-\beta(v) \gamma(v) &0&0\\ -\beta(v) \gamma(v) & \gamma(v) &0&0\\ 0&0&1&0\\ 0&0&0&1\end{pmatrix}\begin{pmatrix}\gamma(w) & 0& -\beta(w) \gamma (w)&0\\ 0&1&0&0\\ -\beta(w) \gamma (w) &0&\gamma (w)&0\\ 0&0&0&1\end{pmatrix}=\begin{pmatrix}\gamma (v) \gamma (w) & -\gamma(v) \beta (v) & -\beta (w) \gamma (v) \gamma (w) & 0 \\ -\gamma (v) \beta(v) \gamma (w) & \gamma (v) & \beta(w) \gamma(v) \beta(v) \gamma (w) &0\\ -\beta(w) \gamma (w) &0& \gamma (w) &0\\ 0&0&0&1\end{pmatrix}$$

Assuming I am in the right direction, I then later swap the order of multiplication between the "##x##" and "##y'##" boosts to show for the "different transformation" as the question asked?

I apologize for my confusion. However, any amount of help is welcomed. Thank you for your assistance!

PeroK
Homework Helper
Gold Member
@PeroK, thank you for your reply! By stating "the first and second matrix" in your response post, do you refer specifically to the two matrices (i.e. for the ##x## and ##y'## directions) I set up in my first post here?
Furthermore, when you say that once I perform matrix multiplication, I'll see that the product of the two matrices represents the relationship between the unprimed and double primed coordinates, does this "relationship" essentially equate to a transformation?

If so, below are my calculations. However, if you did not mean this in your post, please feel free to ignore the below matrix multiplication.
Multiplying the boost with speed ##v## in the ##x##-direction by the boost with speed ##w## in the ##y'##-drection ##\longrightarrow##
$$\begin{pmatrix}\gamma(v) &-\beta(v) \gamma(v) &0&0\\ -\beta(v) \gamma(v) & \gamma(v) &0&0\\ 0&0&1&0\\ 0&0&0&1\end{pmatrix}\begin{pmatrix}\gamma(w) & 0& -\beta(w) \gamma (w)&0\\ 0&1&0&0\\ -\beta(w) \gamma (w) &0&\gamma (w)&0\\ 0&0&0&1\end{pmatrix}=\begin{pmatrix}\gamma (v) \gamma (w) & -\gamma(v) \beta (v) & -\beta (w) \gamma (v) \gamma (w) & 0 \\ -\gamma (v) \beta(v) \gamma (w) & \gamma (v) & \beta(w) \gamma(v) \beta(v) \gamma (w) &0\\ -\beta(w) \gamma (w) &0& \gamma (w) &0\\ 0&0&0&1\end{pmatrix}$$

Assuming I am in the right direction, I then later swap the order of multiplication between the "##x##" and "##y'##" boosts to show for the "different transformation" as the question asked?

I apologize for my confusion. However, any amount of help is welcomed. Thank you for your assistance!
That looks right. Although, actually, that is doing the ##w## boost first in the ##y## direction, followed by the ##v## boost in the x-direction. Including the vectors, you have:

$$\begin{pmatrix} t'' \\ x'' \\ y'' \\ z'' \end{pmatrix} = \begin{pmatrix}\gamma (v) \gamma (w) & -\gamma(v) \beta (v) & -\beta (w) \gamma (v) \gamma (w) & 0 \\ -\gamma (v) \beta(v) \gamma (w) & \gamma (v) & \beta(w) \gamma(v) \beta(v) \gamma (w) &0\\ -\beta(w) \gamma (w) &0& \gamma (w) &0\\ 0&0&0&1\end{pmatrix} \begin{pmatrix} t \\ x \\ y \\ z \end{pmatrix}$$

Last edited:
• Athenian
Awesome! Thank you very much for your help!