Relativity - Lorentz Transformation

Click For Summary
SUMMARY

The discussion focuses on solving problems related to Lorentz transformations in the context of special relativity, specifically involving a spacecraft moving at 0.8c. The calculations reveal that when a message is sent from base station A to spacecraft B after 2 years, B receives it at 6 years in its own frame. Additionally, B takes 1 year to respond, leading to a total time of 7 years for B when sending the message back. The confusion arises in determining the correct x' value for A's calculations, with the consensus being that x' should be set to 0 for the photo taken at the spacecraft's location.

PREREQUISITES
  • Understanding of Lorentz transformations and their equations
  • Familiarity with the concept of time dilation in special relativity
  • Knowledge of the speed of light and its significance in relativity (c)
  • Basic algebraic manipulation skills for solving equations
NEXT STEPS
  • Study the derivation and applications of Lorentz transformations in different scenarios
  • Explore time dilation effects in various frames of reference
  • Learn about the implications of relative velocity on simultaneity in special relativity
  • Practice solving problems involving multiple observers and coordinate systems in relativity
USEFUL FOR

Students and educators in physics, particularly those focusing on special relativity, as well as anyone interested in understanding the mathematical framework of Lorentz transformations and their applications in real-world scenarios.

missmaths.
Messages
1
Reaction score
0

Homework Statement


A is at the base station and given in K co-ordinates
B is on a spacecraft and given in K' co-ordinates.
The velocity of the spacecraft is v=0.8c

Question 1
After t = 2y (y = years) A sends a message by radio to B demanding a picture. Which time t' does B have when the signal arrives and what is B's distance x to the base station then?

Question 2
It took B t' = 1y time to take the picture and get the radio transmission on its
way. What time t is it for A and where (x) is the spacecraft when B launches
his message?

Question 3
What time t is it for A when the message from B arrives and where (x) is
B when this happens?


Homework Equations


[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}[/tex]
so in this question [tex]\gamma = \frac{1}{0.6}[/tex]

Lorentz's Transformations
[tex]t' = \gamma(t-\frac{vx}{c^{2}})[/tex],
[tex]x' = \gamma(x-vt)[/tex],
[tex]t = \gamma(t'+\frac{vx'}{c^{2}})[/tex],
[tex]x = \gamma(x'+vt')[/tex].



The Attempt at a Solution



For 1 I have done the following

[tex]x_{B}=vt[/tex],
[tex]x_{message}=c(t-2y)[/tex]

So I need to find when these to meet i.e.
[tex]vt=c(t-2y)[/tex]
which leads to
[tex]t=\frac{2y}{1-\frac{v}{c}} = \frac{2y}{1-0.8} = 10y[/tex]
therefore
[tex]x_{B}=0.8c * 10y = 8Ly[/tex]

then from the Lorentz t' equation I get 6y (just substituting values above in).


For 2 I have done the following

From 1 I saw that t' = 6y when B got the message therefore when B sends the message it must be 7y as tooks 1y to sort it.

Now I need to find t from this which is where I am stuck.

Obviously I use the t Lorentz equation however it is the entry for x' I am confused about.
As I am calculating t for A must I put in [tex]x'_{A} = -vt'[/tex] or should I put in x' = 0 as the photo is at x' = 0.

I have calculated both ways and for x' = -vt' I got 4.2y and for x' = 0, I got 11.67.
The second seems more correct as it is greater then 10 the t found in part 1, but this is relativity so I am confused.

Once I have sorted this bit I am sure I can do part 3.

Any help on my problem would be much appreciated.
Thank You.
 
Physics news on Phys.org
Your instinct and reasoning that x' = 0 is correct. The primed coordinates have nothing to do with observer A; they are the coordinates B uses. You calculate for A by using the Lorentz transformations with the relative velocity of the two frames.
 

Similar threads

Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
3K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
9
Views
2K
Replies
6
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K