Register to reply 
WIKI and Time Dilation 
Share this thread: 
#73
Jan211, 11:39 PM

P: 182

Would you please use the experiment in the article to prove reciprocal time dilation? 


#74
Jan211, 11:40 PM

Sci Advisor
P: 8,470

You have to define the precise meanings of t and t' in your equations. t' doesn't deal with any single "moving clock", rather it is the time in the primed frame between the event of the light being emitted from the bottom of the light clock (which is at rest in the unprimed frame and therefore moving relative to the primed frame) and the event of it hitting the top of the light clock (this event will occur at a different position in the primed frame). To measure the time of these two events in the primed frame using clocks, you would need a pair of clocks at different positions in the primed frame. As I explained in #62, each of these primed clocks is individually running slow as measured in the unprimed frame, but the unprimed frame nevertheless agrees that t' is greater than t because of the way the two primed clocks are outofsync in the unprimed frame. 


#75
Jan211, 11:41 PM

P: 140




#76
Jan211, 11:41 PM

P: 424




#77
Jan211, 11:45 PM

P: 182

Since LT matrix invertible, then t' = 25 and t = 20 regardless of which frame is at rest. So if unprimed at rest, t' > t. If primed frame at rest, then t' > t. Have you figured this out yet? 


#78
Jan211, 11:46 PM

P: 182

I am discussing this same subject with Jesse right now. 


#79
Jan211, 11:49 PM

Sci Advisor
P: 8,470




#80
Jan211, 11:49 PM

P: 182

That means, the clock at rest with the light source will measure t=d/c and the frame that views that clock as moving will also conclude it calculates t=d/c and it will view a light beam at an angle with a longer distance to travel. Light abberation is not reciprocal. It is only true for observers moving relative to the light source. 


#81
Jan211, 11:57 PM

P: 182

So, in my mind, I see the two as two different objects. There is the stationary space which has Euclidian properties. There is the Minkowsky space which has odd properties. The light sphere in Minkowsky is off center and not spherical. It is my view these are different. If you have your view fine. This is simply my view. 


#82
Jan211, 11:59 PM

Sci Advisor
P: 8,470

(time interval in frame of observer who sees clock in motion) = (time elapsed on clock)*gamma Do you agree that this always works, regardless of what frame the clock is in or which frame we choose to label "stationary"? 


#83
Jan311, 12:01 AM

P: 424




#84
Jan311, 12:07 AM

Sci Advisor
P: 8,470




#85
Jan311, 12:30 AM

P: 3,967

If you take the primed frame as at rest with respect to the unprimed frame, then t' = t. The v in the gamma factor is the velocity of the two frames relative to each other. If the two frames are at rest with respect to each other then v=0 and gamma=1 and t'=t. [tex]t' = t \gamma [/tex] the elapsed time t' is greater than t for any none zero v. Agree? t is the time measured by a single clock (A) at rest in frame S. Let us say t= 10 seconds. In frame S' clock A appears to be moving at 0.8c which is the velocity of frame S' relative to S. Let us say we have two clocks (B' and C') that are at rest in frame S' and that clock A is initially adjacent to clock B' and ends up adjacent to clock C'. The time measured by the synchronised clocks (B' and C') at rest in S' is 16.666 seconds. So the time measured in frame S' by the moving clock (A) is less than the time measured by clocks B' and C' that at rest in frame S'. This is in agreement with what you say should happen. Note that the time that elapses on clock A is the same (10 seconds) according to observers in frame S AND frame S'. The time interval of 16.666 seconds measured in S' is measured by two spatially separated clocks and is NOT the time measured by clock A according to any observer. Clock A is only running slow in a relative sense compared to the coordinate time difference measured by separate clocks. 


#86
Jan311, 01:55 AM

P: 177

Thus, the unprimed clock ticks fewer times than an equivalent clock at rest in the primed frame for an arbitrary time period measured in the primed frame. So, a clock that ticks once per second in its rest frame would be measured to tick 4 times when traveling at 0.6c relative to a frame in which an equivalent clock at rest would tick 5 times. 


#87
Jan311, 08:09 AM

Mentor
P: 16,951

VELOCITIES ARE RELATIVE 


#88
Jan311, 08:33 AM

Mentor
P: 16,951




#89
Jan311, 09:39 AM

P: 140

To chinglu1998
Trying to get an example that might make sense to see what they are talking about. If we have two frames of reference the stationary frame and the moving frame. moving frame is moving away from stationary frame at .6c. In the "moving frame" frame of reference the "stationary frame" is moving at .6c, and the "moving frame" is at rest. Then in the "stationary frame" frame of reference the "moving frame" is moving at .6c, and the "stationary frame" is at rest. This also means that which ever frame you chose the other frame's clock is going to be ticking slower. So if you ask which clock is moving slower you have to ask in which frame of reference. Otherwise the question is meaningless. So lets say you have two clocks t and t'. Clock t is in the "stationary frame", t' is in the "moving frame" Now if I put the observer in the "moving frame", to this observer clock t is ticking slower than t'. This is basicly what they are talking about on wiki 


#90
Jan311, 10:44 AM

P: 3,967

Let me try again with a simple example and maybe you can let me know which parts you disagree with.
Anne is at rest in frame S. Anne considers herself stationary. Bob is at rest in frame S'. Bob considers himself stationary. According to Bob, Anne is moving at 0.6c in the x direction. According to Anne, Bob is moving at 0.6c in the x direction. The fact that Bob thinks Anne's velocity is +0.6c and Anne thinks Bob's velocity is 0.6c is not important because the velocity is squared in the gamma factor and the sign of the velocity is "lost". The gamma factor in this case is [tex]\gamma = 1/\sqrt{1(\pm 0.6)^2} = 1.25[/tex] from either point of view. From Bob's point of view: Anne passes Bob at time zero on both their clocks. (Event 1) Bob has a brother Bob2 who is at rest with Bob in frame S'. Bob and Bob2's clocks are synchronised. Anne passes Bob2 when 25 seconds are showing on Bob2's clock. (Event 2) When Anne passes Bob2, 20 seconds have elapsed on her clock. Since Bob and Bob2's clcoks are synchronised, Bob concludes that 25 seconds elapses on his clock between events 1 and 2, when 20 seconds elapses on Anne's clock between those same two events. Therefore according to Bob: [tex]t_{Bob} = \gamma ( t_{Anne}) [/tex] From Anne's point of view: Bob is moving and passes Anne at time zero on both their clocks. (Event 1) Anne has a sister Anne2 who is at rest with Anne in frame S. Anne and Anne2's clocks are synchronised. Bob passes Anne2 when 25 seconds are showing on Anne2's clock. (Event 3) When Bob passes Anne2, 20 seconds have elapsed on his clock. Since Anne and Anne2's clocks are synchronised, Anne concludes that 25 seconds elapses on her clock between events 1 and 3, when 20 seconds elapses on Bob's clock between those same two events. Therefore according to Anne: [tex]t_{Anne} = \gamma ( t_{Bob}) [/tex] ============================== This appears to be a contradiction to the earlier conclusion that according to Bob: [tex]t_{Bob} = \gamma ( t_{Anne}) [/tex] but there is a slight "decepton" going on here. In Anne's rest frame, the time [itex]t_{Anne}[/itex] in the equation [itex]t_{Anne} = \gamma ( t_{Bob} )[/itex] is really the difference between [itex]t_{Anne}[/itex] and [itex]t_{Anne2}[/itex]'s clocks while [itex]t_{Bob}[/itex] is the time measured by a single clock. In Bob's rest frame, the time [itex]t_{Bob}[/tex] in the equation [itex]t_{Bob} = \gamma ( t_{Anne}) [/itex] is really the difference between [itex]t_{Bob}[/itex] and [itex]t_{Bob2}[/itex]'s clocks while [itex]t_{Anne}[/itex] is the time measured by a single clock. In other words, the [itex]t_{Bob}[/itex] in [itex]t_{Bob} = \gamma ( t_{Anne}) [/itex] is not the same as the [itex]t_{Bob}[/itex] in [itex]t_{Anne} = \gamma ( t_{Bob}) [/itex]. [itex]t_{Bob}[/itex] in the first equation is measured by two spatially separated clocks (coordinate time) and [itex]t_{Bob}[/itex] in the second equation is measured by a single clock (proper time). [itex]t_{Bob}[/itex] in the first equation is the difference between events 1 and 2 while [itex]t_{Bob}[/itex] in the second equation is the difference between events 1 and 3. This is the "deception" that is hidden when using the primed notation that gives the appearance of a reciprocal time dilation relationship. Note that the time on the right hand side of both equations is measured by a single clock that is present at both the start and finish events. This is the proper time and I have mentioned it several times before, but I repeat it again, because it is an important concept to understand. All observers agree on the proper time of a clock between two events. 


Register to reply 
Related Discussions  
Gravitational time dilation vs velocity time dilation  Special & General Relativity  22  
Relating SR time dilation and GR timepassage variation  Special & General Relativity  0  
Black Hole Time Machine (time dilation)  Special & General Relativity  13  
Calculating Gravitational Time Dilation in black hole/Future Time Travel  Special & General Relativity  5 