Is this a Lorentz transformation?

In summary: If K'' moves at velocity vy with respect to K', and K' moves at velocity vx with respect to K, then K'' moves at velocity vx+vy with respect to K.In summary, the conversation discusses a coordinate transformation in special relativity that involves two velocity boosts in different directions. The resulting transformation matrix is not the same when the boosts are done in reverse order. However, both transformations are Lorentz transformations and are closely related. The meaning of vy in the assignment is not explicitly stated, but it is assumed to be the velocity of K'' with respect to K'.
  • #1
Libra82
13
0

Homework Statement


Question as stated: In special relativity consider the following coordinate transformation between inertial frames: first make a velocity boost [itex]v_x[/itex] in the x-direction, then make a velocity boost [itex]v_y[/itex] in the y-direction. 1) Is this a Lorentz transformation? 2) Find the matrix of this transformation. 3) Consider the boosts in inverse order - is it the same transformation?


Homework Equations


[itex] \beta_i = \frac{v_i}{c} [/itex]
[itex] \gamma_i = \frac{1}{\sqrt{1-\frac{v_i^2}{c^2}}}[/itex]


The Attempt at a Solution


I use the [itex] c = 1 [/itex] convention.

I wrote down the two transformations as:

x-direction:
[itex]
\begin{pmatrix}
t' \\
x'\\
y' \\
z'
\end{pmatrix} = \begin{pmatrix}
\gamma_x & -\beta_x \gamma_x & 0 & 0 \\
-\beta_x \gamma_x & \gamma_x & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
t\\
x\\
y\\
z
\end{pmatrix}
[/itex]
and for the y-direction:
[itex]
\begin{pmatrix}
t'' \\
x''\\
y'' \\
z''
\end{pmatrix} = \begin{pmatrix}
\gamma_y & 0 & -\beta_y \gamma_y & 0 \\
0 & 1 & 0 & 0 \\
-\beta_y \gamma_y & 0 & \gamma_y & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
t' \\
x' \\
y' \\
z'
\end{pmatrix}
[/itex]
and combined these to get
[itex]
\begin{pmatrix}
t'' \\
x''\\
y'' \\
z''
\end{pmatrix} = \begin{pmatrix}
\gamma_y & 0 & -\beta_y \gamma_y & 0 \\
0 & 1 & 0 & 0 \\
-\beta_y \gamma_y & 0 & \gamma_y & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}\begin{pmatrix}
\gamma_x & -\beta_x \gamma_x & 0 & 0 \\
-\beta_x \gamma_x & \gamma_x & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
t\\
x\\
y\\
z
\end{pmatrix} = \begin{pmatrix}
\gamma_x \gamma_y & -\beta_x \gamma_x \gamma_y & -\beta_y \gamma_y & 0 \\
-\beta_x \gamma_x & \gamma_x & 0 & 0 \\
-\beta_y \gamma_x \gamma_y & \beta_x \beta_y \gamma_x \gamma_y & \gamma_y & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} \begin{pmatrix}
t\\
x\\
y\\
z
\end{pmatrix}
[/itex]
This should be the answer to question 2).

If I inverse the order of the boosts I notice that the resulting transformation matrix is the transpose of the above matrix.

As the transformation matrix for the two cases in question are not equal the two transformations are not the same? (question 3)

I am uncertain on how to explain whether or not the resulting transformations are Lorentz transformations.

why do the question headlines automatically appear each time I preview my post?
 
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  • #2
First of all, I am uncomfortable with the vy part of the question.
What is vy?
Is that the velocity of the K'' frame origin with respect to the initial K frame?
Or is that the velocity of the K'' frame with respect to the K' frame?
Clarifying this may help.

The resulting transformations are each Lorentz transformations.
There are many ways to prove that, depending on which starting point is assumed.
If the Lorentz transform is defined as a linear transformation that keeps the ds² invariant, then the proof is obvious.

A more algebraic would try to indentify the resulting transformation as a boost along a certain resultant velocity.

Finally, note also the relation Lt(v) = L-1(-v) .
The transformations are different, but nevertheless closely related.
 
Last edited:
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  • #3
Thank you for your reply.

I am, however, unsure of the interpretation of the vy part of the assignment. The exercise is written as in the OP. In my solution I have assumed vy was K'' with respect to K'.

Of course I could check whether or not ds² is invariant with respect to the double transformation! Thanks for the hint (and now I'm sad that I didn't figure this out for myself).

The final relation between the two is the same one I noticed while carrying out the calculations. Different transformations though as one matrix entry switches place.
 
  • #4
You made a choice for the meaning of vy.
You simply need to keep that in mind.
Of course if you reverse the order of the boost, you need to switch the meanings too.
 
  • #5
If you learned about velocity composition in SR, then you could make use of of this knowledge.
 

What is a Double Lorentz transformation?

A Double Lorentz transformation is a mathematical equation used in physics to describe how an object's coordinates appear to change when observed from two different frames of reference that are moving with a constant velocity relative to each other. It takes into account the effects of both time dilation and length contraction.

Why is a Double Lorentz transformation important?

A Double Lorentz transformation is important because it helps us understand the effects of relativity on measurements and observations made from different reference frames. It is a crucial tool in the study of special relativity and is used in a wide range of fields such as particle physics, astrophysics, and cosmology.

How is a Double Lorentz transformation calculated?

A Double Lorentz transformation is calculated using two equations: the time dilation equation and the length contraction equation. These equations take into account the relative velocities of the two frames of reference and the speed of light, and they can be combined to determine the transformation of an object's coordinates from one frame to another.

What is the difference between a Double Lorentz transformation and a single Lorentz transformation?

A Double Lorentz transformation takes into account the effects of both time dilation and length contraction, whereas a single Lorentz transformation only considers one of these effects. A single Lorentz transformation is used when the relative velocity between the two frames of reference is parallel to the direction of motion, while a Double Lorentz transformation is used for any relative velocity.

Can a Double Lorentz transformation be applied to any reference frames?

Yes, a Double Lorentz transformation can be applied to any reference frames as long as they are moving with a constant velocity relative to each other. It is a fundamental principle of special relativity that the laws of physics should be the same in all inertial reference frames, and the Double Lorentz transformation allows us to maintain this consistency when making measurements and observations.

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