Length contraction via Lorentz transformation matrix

Click For Summary
SUMMARY

The discussion focuses on deriving length contraction using the Lorentz transformation matrix, specifically for a ruler of length L along the x-axis as observed from two different inertial frames: K (at rest) and K' (moving with speed v). The transformation equations are established using the Lorentz matrix, yielding results that confirm the length contraction to L/γ. The key insight is that the observer at rest measures the length as L regardless of the timing of measurements, affirming the consistency of the length measurement in their frame.

PREREQUISITES
  • Understanding of Lorentz transformation matrices
  • Familiarity with the concept of inertial reference frames
  • Knowledge of the relativistic factor γ (gamma)
  • Basic principles of special relativity
NEXT STEPS
  • Study the derivation of the Lorentz transformation equations in detail
  • Explore the implications of time dilation in special relativity
  • Learn about the inverse Lorentz transformation for coordinate conversions
  • Investigate practical applications of length contraction in physics experiments
USEFUL FOR

Students of physics, particularly those studying special relativity, educators teaching the principles of relativistic mechanics, and researchers exploring the implications of Lorentz transformations in theoretical physics.

fluidistic
Gold Member
Messages
3,932
Reaction score
283
1,2,3. Homework Statement
I tried to derive the length contraction using the Lorentz transformation matrix and considering 2 events. I reached the correct result but there's a step that I had to assume that I don't understand.

Consider a ruler of length L along the x-axis for an observer at rest with respect to the ruler. The inertial reference frame of that observer is K.

Consider an observer moving alongside the x-axis with speed v. The inertial reference frame of that system is K'.

I considered the events (omitting y and z): ##(t_1,x_1)## and ##(t_2,x_2)##. They are worth ##(t_1,0)## and ##(t_2,L)##, respectively. (*)

Then ##\begin{bmatrix} t_1'\\x_1' \end{bmatrix}=\begin{bmatrix} \gamma & - \beta \gamma \\ -\beta \gamma & \gamma \end{bmatrix}\begin{bmatrix} t_1\\x_1 \end{bmatrix}##. This yields the two equations ##t_1'=\gamma t_1## and ##x_1'=-\beta\gamma t_1##.

While ##\begin{bmatrix} t_2'\\x_2' \end{bmatrix}=\begin{bmatrix} \gamma & - \beta \gamma \\ -\beta \gamma & \gamma \end{bmatrix}\begin{bmatrix} t_2\\L \end{bmatrix}## yields ##t_2'=\gamma t_2 -\beta \gamma L## and ##x_2'=-\beta\gamma t_2 +\gamma L##.

Now since I want to obtain a distance measurement in K', I set ##t_1'=t_2'## and then I solve for ##x_2'-x_1'## and I indeed reach that it's worth ##L/\gamma##.

What didn't like/understand about my own derivation is that for it to work, I had to assume that even though ##t_1 \neq t_2##, ##x_1=0## and ##x_2=L## (see (*)). To me, this is equivalent to say that the length of the ruler is NOT worth L for an observer at rest with respect to it.

Oh wait, actually it could well be... since the observer is at rest with respect to the ruler, it doesn't matter when it measures the spatial distance between the 2 extrema of the ruler, it will always be L no matter when each measurement was performed. Is this reasoning correct?
 
Physics news on Phys.org
Yes, sounds good!

Note, you might try using the inverse transformation matrix for going from primed to unprimed coordinates and seeing if it's any easier. You can let the observation time in the primed frame be ## t_1' = t_2' = 0 ##.
 
Last edited:
  • Like
Likes   Reactions: fluidistic

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
Replies
9
Views
3K
Replies
9
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
26
Views
5K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
1
Views
2K