How Does Relativity Affect Perceptions of Time and Length in Motion?

Click For Summary
SUMMARY

The discussion focuses on the effects of relativity on time and length as experienced by a race horse traveling at 0.9c down a 2000m straightaway. According to a timer in the grandstand, the time taken for the horse to complete the distance is calculated using the formula ∆t = length/speed, resulting in a specific value. The jockey perceives the length of the straightaway as contracted, calculated using the Lorentz contraction formula L = Lo[sqrt(1-(u^2)/(c^2))]. The jockey's perspective also affects the calculation of time and speed, confirming that both the grandstand and the jockey agree on their relative speeds of 0.9c.

PREREQUISITES
  • Understanding of special relativity concepts, including time dilation and length contraction.
  • Familiarity with the Lorentz transformation equations.
  • Basic knowledge of speed calculations in physics.
  • Ability to manipulate algebraic equations for solving physics problems.
NEXT STEPS
  • Study the Lorentz transformation in detail to understand its applications in various scenarios.
  • Learn about time dilation effects in high-speed travel, particularly in the context of special relativity.
  • Explore practical examples of length contraction in real-world physics problems.
  • Investigate the implications of relativity on GPS technology and satellite communications.
USEFUL FOR

Students of physics, educators teaching special relativity, and anyone interested in the practical applications of relativistic effects in high-speed scenarios.

Oijl
Messages
102
Reaction score
0

Homework Statement


A race horse can travel down a 2000m straightaway at a speed of 0.9c.

a) How long does it take the horse to run the straightaway according to a timer sitting in the grandstand?
b) From the jockey's perspective, the length of the straightaway appears to be less than 2000 m. What is the apparent length?
c) How long does it take for the horse to run the straightaway according to the jockey?
d) What would the jockey calculate for the speed of the horse relative to the grandstand?


It seems like I'm doing things right until the last part, but I put up my work for the first three parts so you'll know my thought process that leads me to my problem in part d.

Homework Equations





The Attempt at a Solution



a) ∆t = length/speed
Length of the straightaway according to a timer in the grandstand would be 2000m, because there would be no length contraction for the timer.
length=2000m
The speed is given at 0.9c.
speed=0.9c
Therefore,
∆t = 2000/(0.9c) = some value

b)L = Lo[sqrt(1-(u^2)/(c^2))]
Lo = 2000m
u = 0.9c
So,
L = 2000[sqrt(1-0.9^2)] = some other value

c) Again, ∆t = length/speed (but here we'll call it ∆to)
∆to = L/(0.9c), where L is the L of part b

d) ...
Speed is length over time, and the length according to the jockey is L, and the time according to the jockey is ∆to, which was calculated using 0.9c as the speed, so 0.9c would be the speed of the horse.



Anyway, thanks for looking at this, and I'm sure someone can give me a good, quick help that sets me straight because I know this is a simple problem.
 
Physics news on Phys.org
Looks good to me. The grandstand sees the horse/jockey moving at speed 0.9c, so the horse/jockey sees the grandstand moving at speed 0.9c. (They agree on each other's relative speed.)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 67 ·
3
Replies
67
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
3
Views
2K
  • · Replies 36 ·
2
Replies
36
Views
3K
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K