ขอต้อนรับ ผู้มาเยือน กรุณา ล็อกอิน หรือ สมัครสมาชิก

ล็อกอินด้วยชื่อผู้ใช้ รหัสผ่่าน และระยะเวลาใช้งาน

 
Advanced search

41330 Posts in 6202 Topics- by 8771 Members - Latest Member: ChinNaaek
Pages: 1   Go Down
Print
Author Topic: GR8677.021  (Read 3002 times)
0 Members and 1 Guest are viewing this topic.
conantee
SuperHelper
*****
Offline Offline

Posts: 1400

เราเป็นอย่างไร สังคมเป็นอย่างนั้น


« on: December 31, 2009, 11:06:44 AM »

21. Two observers O and O^\prime observes two events, A and B. The observers have a constant relative speed of 0.8 c. In units such that the speed of light is 1, observer O obtained the following coordinates:
      Event A: x = 3, y = 3, z = 3, t= 3
      Event B: x = 5, y = 3, z = 1, t= 5
What is the length of the space-time interval between these two events, as measured by O^\prime?
(A) 1
(B) \sqrt{2}
(C) 2
(D) 3
(E) 2 \sqrt{3}
Logged
conantee
SuperHelper
*****
Offline Offline

Posts: 1400

เราเป็นอย่างไร สังคมเป็นอย่างนั้น


« Reply #1 on: December 31, 2009, 11:53:43 AM »

First, the space-time interval is independent of observer. So, don't worry about the speed of observer O^\prime.
Next, let's calculate the space-time interval between two events. You should know that space and time are treated differently. Instead of usual cartezian coordinates, time is treated in imaginary axis. The "distance" in space-time is then can be calculated from:
(\Delta D) ^2 = (\Delta x)^2 + (\Delta y) ^2 + (\Delta z) ^2 - (\Delta t)^2 where speed of light is 1

In our problem \Delta x = 5-3 = 2, \Delta y = 3-3 = 0, \Delta z = 3-1 =2, \Delta t = 3-5 = -2
Hence, (\Delta D)^2 = 2^2 +0^2+ 2^2 - (-2)^2 = 4. So, \Delta D = 2.

The answer is (C). 24 of 100 people got the right answer.
Logged
Pages: 1   Go Up
Print
Jump to: