Understanding Lorentz Transforms: Interpreting Results and Visualizing Answers

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The discussion revolves around the application of Lorentz transforms to analyze the events involving a moving rocket. The user calculated the length of the rocket as observed in frame S, finding it to be 800 meters, and derived spatial and temporal coordinates for four events. Confusion arose regarding the simultaneous nature of events B and C in frame S', leading to questions about their interpretation and the positions of observers. Clarifications indicated that while events B and D occur simultaneously, they do not happen at the same location, and the calculations provided were mostly correct, with minor adjustments suggested for precision. The user expressed a desire to better visualize and understand the results of Lorentz transforms in future problems.
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Ok, this is my last study problem, I think I got it, but my answers seem a little odd...

Homework Statement


A rocket of length 1000 meters is at rest in S'. The nose of the rocket is atx'=0and the tail of the rocket is atx'=-1000 meters. S' is moving with a velocity of v=\frac{3c}{5} in the positive x direction relative to S.

Four events are given:

Event A is the synchronizing event where the nose of the rocket is at the origin in both frames:
x_{A}=x'_{A}=t_{A}=t'_{A}=0

Event B is simultaneous with A in S:
t_{B}=t_{A}=0

Event C is when the tail of the rocket passes the origin as observed in S

Event D is simultaneous with C and is when an observer in S sees the nose of the rocket pass by him.

Homework Equations



x'=\gamma(x-vt)

t'=\gamma(t-\frac{vx}{c^{2}}

I also used,

L=\frac{L_{0}}{\gamma}

The Attempt at a Solution



This is why I'm worried... it seems straight-forward...

All I did was calculate the length of the rocket as observed in S:

L=\frac{1000 meters}{5/4}=800 meters

I believe this gives me spatial coordinates for all of the events...

x_{C}=x_{A}=0

x_{B}=-800 meters

x_{D}=800 meters

as well as temporal coordinates:

t_{A}=t_{B}=0 (A is given in the problem and A,B are simultaneous)

t_{C}=t_{D}=\frac{x_{C}-x_{B}}{v}=\frac{x_{D}-x_{A}}{v}=4.4475 x 10^{-6} seconds

Then I just applied the coordinate transforms, and got:

x'_{A}=0 (given)
x'_{B}=-1000 meters
x'_{C}=-1000 meters
x'_{D}=0

t'_{A}=0
t'_{B}=2.0014 x 10^{-6} seconds
t'_{C}=5.5559 x 10^{-6} seconds
t'_{D}=3.55802 x 10^{-6} seconds

Now, everything here looks a little wierd... events B and C in the S' frame happen in the same place? And the sequence of events in S' is A-B-D-C?

Am I mis-applying the transforms or misinterpreting the problem?

If not, could someone please help me with interpreting the answers?

Much thanks in advance!
 
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What is the question? calculate when all 4 events happen in both frames?
 
yes, sorry, it's basically fill in all the unknown times...
 
strangequark said:
yes, sorry, it's basically fill in all the unknown times...

What is event B?
 
I'm also confused about events C and D... if within the same frame, they happen at the same position at the same time... then they represent the same event. is the observer described in D at the origin in S?
 
Sorry, my description was bad...

Event B is when an observer in S (not at the origin) sees the tail of the rocket pass over his head at the same time that an observer at the origin in S sees the nose of the rocket pass over HIS head...

For events C and D, they happen at the same time, but not the same position... the observer of event C is at the origin, while the observer of D is not (both observers are in the S frame though).
 
Your answers all look right to me. Only thing is the decimal places I think...

tc is exactly 40/9 * 10^-6 = 4.444 * 10^-6

tb' = 2.00*10^-6 exactly
tc' = (50/9)*10^-6 = 5.556 * 10^-6
td' = (32/9)*10^-6 = 3.556 * 10^-6
 
Great, thanks again... I feel like I have a decent understanding of how to set up these problems as I do them, but everytime I come to an answer that I can't visualize it makes me wonder... I suppose I'm used to being able to tell if an answer is reasonable just by looking at the problem and I can't seem to do that yet with the Lorentz transforms... I'll keep at it, eventually it will click.
 
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