Register to reply

Simultaneous events - special relativity

by jaumzaum
Tags: events, relativity, simultaneous, special
Share this thread:
jaumzaum
#1
Feb13-13, 10:07 AM
jaumzaum's Avatar
P: 248
I'm studying special relativity and I'm having difficulty in understanding somethings. I have 2 questions I'm confused about.

1) If I am moving in relation to you, when you observe me, I should be in slow-motion. but you are moving in relation to me, so when I observe you, will you be in slow motion too?

2) Consider the following experiment. Pam and Jim are moving in a 1-dimensional space frame. Pam moves to the right with velocity v (in relation to me, I'm at rest at the origin), and Jim moves to the left with velocity v too. Both are hanging a 1 meter rule in the hands, pointing forward. They have always the same distance in relation to the origin.
Consider the following events
Event 1: the front of the Pam's rule touches Jim
Event 2: the back of the Pam's rule passes Jim
Event 3: the front of the Jim's rule touches Pam
Event 4: the back of the Jim's rule passes Pam

Δt(x,y) = time interval between event x and y measured by Pam
Δt'(x,y) = time interval between event x and y measured by Jim
Δt''(x,y) = time interval between event x and y measured by me

For me, events 1 and 3, 2 and 4 should be simultaneous. But let's analyze Jim's frame of reference. For Jim, event 1 and 2 occur at the same place, so, Jim measures the proper time. Thereby Δt(1,2) > Δt'(1,2)
Also, Δt'(3,4) > Δt(3,4)

For symmetry, Δt(3,4)=Δt'(1,2), Δt(1,2)=Δt'(3,4)
So Δt(1,2)>Δt(3,4)
and the events should not be simultaneous to Jim, the same for Pam
Is this right?
Phys.Org News Partner Science news on Phys.org
Experts defend operational earthquake forecasting, counter critiques
EU urged to convert TV frequencies to mobile broadband
Sierra Nevada freshwater runoff could drop 26 percent by 2100
robphy
#2
Feb13-13, 10:18 AM
Sci Advisor
HW Helper
PF Gold
robphy's Avatar
P: 4,137
Try drawing the diagrams that I suggested in this recent post
http://www.physicsforums.com/showpos...1&postcount=13

Often a geometric argument is more tangible than an algebraic calculation.
[please report back if it is helpful]
Mentz114
#3
Feb13-13, 10:48 AM
PF Gold
P: 4,087
These diagrams show how the simultaneity is different when events are seen from three frames. I'm not sure if the labelling corresponds to yours.
Attached Thumbnails
P-G-3.png   P-G-2.png   P-G-1.png  

HallsofIvy
#4
Feb13-13, 11:30 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,568
Simultaneous events - special relativity

Quote Quote by jaumzaum View Post
I'm studying special relativity and I'm having difficulty in understanding somethings. I have 2 questions I'm confused about.

1) If I am moving in relation to you, when you observe me, I should be in slow-motion. but you are moving in relation to me, so when I observe you, will you be in slow motion too?
I will see you "in slow motion" and you will see me "in slow motion", yes.

2) Consider the following experiment. Pam and Jim are moving in a 1-dimensional space frame. Pam moves to the right with velocity v (in relation to me, I'm at rest at the origin), and Jim moves to the left with velocity v too. Both are hanging a 1 meter rule in the hands, pointing forward. They have always the same distance in relation to the origin.
Consider the following events
Event 1: the front of the Pam's rule touches Jim
Event 2: the back of the Pam's rule passes Jim
Event 3: the front of the Jim's rule touches Pam
Event 4: the back of the Jim's rule passes Pam

Δt(x,y) = time interval between event x and y measured by Pam
Δt'(x,y) = time interval between event x and y measured by Jim
Δt''(x,y) = time interval between event x and y measured by me

For me, events 1 and 3, 2 and 4 should be simultaneous.
Well, that's a problem. "Simultaneity" is also relative to the frame of reference. Two events that are seen to be simultaneous from one frame of reference are not necessarily simultaneous from another. In fact, if there exist a frame of reference in which events A and B are seen to be simultaneous, there will exist a frame of reference in which A occurs before B and another in which B occurs before A.

But let's analyze Jim's frame of reference. For Jim, event 1 and 2 occur at the same place, so, Jim measures the proper time. Thereby Δt(1,2) > Δt'(1,2)
Also, Δt'(3,4) > Δt(3,4)

For symmetry, Δt(3,4)=Δt'(1,2), Δt(1,2)=Δt'(3,4)
So Δt(1,2)>Δt(3,4)
and the events should not be simultaneous to Jim, the same for Pam
Is this right?
ghwellsjr
#5
Feb13-13, 02:55 PM
PF Gold
P: 4,745
Quote Quote by jaumzaum View Post
I'm studying special relativity and I'm having difficulty in understanding somethings. I have 2 questions I'm confused about.

1) If I am moving in relation to you, when you observe me, I should be in slow-motion. but you are moving in relation to me, so when I observe you, will you be in slow motion too?

2) Consider the following experiment. Pam and Jim are moving in a 1-dimensional space frame. Pam moves to the right with velocity v (in relation to me, I'm at rest at the origin), and Jim moves to the left with velocity v too. Both are hanging a 1 meter rule in the hands, pointing forward. They have always the same distance in relation to the origin.
Consider the following events
Event 1: the front of the Pam's rule touches Jim
Event 2: the back of the Pam's rule passes Jim
Event 3: the front of the Jim's rule touches Pam
Event 4: the back of the Jim's rule passes Pam

Δt(x,y) = time interval between event x and y measured by Pam
Δt'(x,y) = time interval between event x and y measured by Jim
Δt''(x,y) = time interval between event x and y measured by me

For me, events 1 and 3, 2 and 4 should be simultaneous.
Yes, in your rest frame, events 1 and 3 are simultaneous. But events 2 and 4 are simultaneous in every frame since they are the same event.
Quote Quote by jaumzaum View Post
But let's analyze Jim's frame of reference. For Jim, event 1 and 2 occur at the same place, so, Jim measures the proper time. Thereby Δt(1,2) > Δt'(1,2)
It's not a good idea to compare Coordinate Times or Coordinate Time intervals between frames. But the way you expressed it is good because you referred to the Proper Time. However the best way to say it is: in Pam's rest frame, the Coordinate Time interval between events 1 and 2 is greater than the corresponding Proper Time interval on Jim's clock.
Quote Quote by jaumzaum View Post
Also, Δt'(3,4) > Δt(3,4)
Similarly, in Jim's rest frame, the Coordinate Time interval between events 3 and 4 is greater than the corresponding Proper Time interval on Pam's clock.

I'm not just nitpicking-these are important distinctions. Jim and Pam can actually measure the Proper Times on each of their own clocks but they cannot measure the Coordinate Times or intervals of remote events. Rather, we calculate those times based on the selected Inertial Reference Frame.
Quote Quote by jaumzaum View Post
For symmetry, Δt(3,4)=Δt'(1,2),
Two Proper Time intervals will have the same value-no problem.
Quote Quote by jaumzaum View Post
Δt(1,2)=Δt'(3,4)
Two Coordinate Time intervals will have the same value but why are you comparing them?
Quote Quote by jaumzaum View Post
So Δt(1,2)>Δt(3,4)
Now you are back to one IRF and your conclusion is probably correct, but I don't know the why you are doing this.
Quote Quote by jaumzaum View Post
and the events should not be simultaneous to Jim, the same for Pam
Is this right?
What events are you referring to? You've been dealing with time intervals and time intervals are not events so I don't know what you are asking.


Register to reply

Related Discussions
Two Events in Special Relativity with on Time Dialation Introductory Physics Homework 9
[Special Relativity] Find frame K' in which events in K appear to be simultaneous Introductory Physics Homework 4
Special Relativity: Find the reference frame in which two events are Simultaneous Introductory Physics Homework 10
Special Relativity calculating a time period so simultaneous in all reference frames Advanced Physics Homework 4
Simultaneous events in SR...? Special & General Relativity 12