Lorenz transformations for two parallel inertial systems

In summary, the conversation discusses the task of generalizing the Lorentz transformations for parallel systems with a general velocity vector. The suggested approach is to express the transformation as a rotation to align the velocity with the x-axis, followed by a boost in the x-direction, and then rotating back to the original axes. Another approach is to resolve the position vector into components parallel and perpendicular to the velocity, where the boost only affects the parallel component. The equations being derived may need to be corrected for the presence of the factor gamma.
  • #1
Tomer
202
0
Thanks for reading!

Homework Statement


I have been given a proof for the lorentz transformations (which I only partly understand) that relied on the two relativity postulates (equivalence of inertial systems and the speed of light being constant) for the case of two standard inertial systems: the system S' moved with velocity v along the x axis, and the systems are parallel.
Now I've been asked to generalize the transformation for the case the the systems are parallel, but the velocity isn't necessarily in the x axis. (a general v vector).
I need to show, that in this case, this equation holds:

1. [itex]\vec{r'}[/itex] = [itex]\vec{r}[/itex] + ([itex]\frac{\vec{v}\cdot\vec{r}}{v^{2}}[/itex]([itex]\beta[/itex] - 1) - [itex]\beta[/itex]t)[itex]\vec{v}[/itex]

2. t' = [itex]\beta[/itex](t - ([itex]\frac{\vec{v}\cdot\vec{r}}{v^{2}}[/itex]))

Homework Equations



The development of the transformation for the simpler case (where v = (v,0,0) ) arrives to the point where it shows that the size x2 + y2 + z2 - c2t2 is invariant. It doesn't to that point use the assumption that the systems are parallel.
It is then sugested that from the fact that the space is isotropic y = y' and z = z'.
This is however no longer the case. So that's where, I guess, I need to work with new tools.

The Attempt at a Solution



By using [itex]\vec{r}[/itex]2 = x2 + y2 + z2 I've tried to make an analogical development, treating |[itex]\vec{r}[/itex]| as x was. I just don't seem to derive the equation, and I don't know how to use the fact that the systems are parallel.

I'm pretty rusty with algebra, I must say, haven't touched it for almost 2 years.

I'd appreciate any help!

Tomer.
 
Physics news on Phys.org
  • #2
Rather than trying to derive the general transformation from scratch, I would try expressing the transformation as a rotation to align the velocity with the +x axis, followed by a boost in the +x direction, followed by a rotation back to the original axes.
 
  • #3
vela said:
Rather than trying to derive the general transformation from scratch, I would try expressing the transformation as a rotation to align the velocity with the +x axis, followed by a boost in the +x direction, followed by a rotation back to the original axes.

Thanks for the answer.

When you say "expressing the transformation", do you mean the expression I've written above? Or the transformation for the standard systems?

I am really rusty :-)
 
  • #4
The one you're trying to derive now. The idea is that you know how to calculate r' and t' when the velocity v lines up with the x-axis, so you use a spatial rotation to bring the original system into that case, do the boost, and then rotate back to the original axes because you want the answer in terms of those axes. That said, I haven't actually worked the problem out, so there might be a simpler way to solve it. I'm just saying this is what I'd try first.
 
  • #5
Actually, now that I've looked at the problem a bit more, it's probably easier to resolve [itex]\vec{r}[/itex] into a component parallel to the velocity and one perpendicular to the velocity:[tex]\vec{r} = \vec{r}_\parallel + \vec{r}_\perp[/tex]The boost will mix t and [itex]r_\parallel[/itex] but will leave [itex]\vec{r}_\perp[/itex] unchanged, so you'll have
\begin{align*}
\vec{r_\perp} &\to \vec{r}_\perp \\
\vec{r_\parallel} &\to \vec{r'}_\parallel \\
t &\to t'
\end{align*}
and [itex]\vec{r}' = \vec{r}_\perp + \vec{r}'_\parallel[/itex].

By the way, are you sure those equations you're trying to derive are correct? You should have factors of [itex]\gamma[/itex] in there somewhere.
 
Last edited:

Related to Lorenz transformations for two parallel inertial systems

1. What are Lorenz transformations for two parallel inertial systems?

Lorenz transformations are a set of equations that describe the relationship between space and time in two parallel inertial systems. These systems are moving at constant velocities relative to each other and have no acceleration or external forces acting upon them.

2. Why are Lorenz transformations important in science?

Lorenz transformations are an essential tool in Einstein's theory of relativity and have many practical applications in modern physics, such as in the study of high-speed particles and the behavior of light. They also help to explain phenomena like time dilation and length contraction.

3. How do Lorenz transformations relate to the concept of simultaneity?

The concept of simultaneity, or the idea that two events occur at the same time, is relative in the theory of relativity. Lorenz transformations show that the time of an event can be perceived differently by observers in different inertial frames of reference, depending on their relative velocity.

4. Can Lorenz transformations be used in non-inertial reference frames?

No, Lorenz transformations are only applicable in inertial reference frames where there is no acceleration or external forces acting upon the system. In non-inertial frames, more complex equations, such as those in general relativity, are needed to describe the relationship between space and time.

5. How are Lorenz transformations different from Galilean transformations?

Lorenz transformations take into account the constancy of the speed of light and the relativity of simultaneity, while Galilean transformations assume that time and space are absolute. This makes Lorenz transformations more accurate and applicable in situations involving high speeds or strong gravitational fields.

Similar threads

  • Advanced Physics Homework Help
Replies
3
Views
488
  • Advanced Physics Homework Help
Replies
2
Views
1K
Replies
16
Views
1K
  • Advanced Physics Homework Help
Replies
15
Views
3K
  • Advanced Physics Homework Help
Replies
8
Views
861
  • Advanced Physics Homework Help
Replies
26
Views
4K
  • Introductory Physics Homework Help
Replies
3
Views
295
  • Advanced Physics Homework Help
Replies
7
Views
1K
  • Advanced Physics Homework Help
Replies
16
Views
3K
Back
Top