
#73
Dec1212, 12:49 AM

P: 52

And you were right about Δt' = γ(Δt(vΔx/c^2)) but wrong about Δt' = Δt/γ. When Δx=0, Δt'=γΔt not Δt/γ. I don't understand where you got Δt' = Δt/γ. I've never seen this equation used and it seems to be what is getting in your way. 



#74
Dec1212, 01:05 AM

Sci Advisor
P: 8,470

Look at section 4 of Einstein's 1905 paper, where he uses t for the time elapsed on the clock in the "stationary system" and τ for the time elapsed in the "moving system" when the clock itself is at rest relative to the "moving system", i.e. the time between the two readings in clock's own rest frame, introducing the setup like this: So, after having looking at that section of the 1905 paper, would you agree that Einstein was not talking about the "speed of time" but rather meant the equation Δt' = Δt/γ to apply to the time elapsed between two readings of a physical clock, as measured both in its own rest frame (where Δx'=0) and in the frame where it's moving? Would you agree that given those assumptions about what the equation is supposed to represent, Δt' = Δt/γ can be derived directly from the LT? It's easy to show why, since one of the LT equations is Δt = γ(Δt' + vΔx'/c^{2}), so if you set Δx'=0 you are left with Δt = γΔt' or Δt' = Δt/γ. If you disagree with any of this, please explain why you think I am mistaken about what Einstein meant, preferably pointing to a specific quote from the paper to make your case. 



#75
Dec1212, 01:09 AM

Sci Advisor
P: 8,470





#76
Dec1212, 01:15 AM

P: 52





#77
Dec1212, 01:22 AM

P: 52





#78
Dec1212, 01:24 AM

Sci Advisor
P: 8,470





#79
Dec1212, 01:25 AM

Sci Advisor
P: 8,470





#80
Dec1212, 03:22 AM

P: 3,178

Do you remember where you read that wrong explanation of SR that you keep hanging on? 



#81
Dec1212, 04:12 AM

PF Gold
P: 4,528

Now we use the LT to transform the coordinates of all the events (the 1second ticks represented by blue dots) into the coordinates of another frame moving at 0.866c with respect to the first frame. When we do the LT, we always call the original frame the unprimed frame and the new frame the primed frame: Note that since the speed is 0.866c, gamma is equal to two. You can see that the first event which occurred at the time coordinate of 1 in the first frame occurs now at the time coordinate of 2 in the second frame, just like the LT formula indicates. So far so good. Now we want to see how to use the time dilation formula. Unlike the LT formula which applies to the coordinate times in two different frames, the time dilation formula applies to the relationship between the time (primed) on a clock compared to the coordinate time (unprimed), all in a single frame. So let's apply the time dilation formula to the first graph where the speed of the clock is zero and gamma equals one. The formula says that the time on the clock is the same as the coordinate time and we see that because the spacing of the dots is the same as the spacing of the grid lines for the graph. Now let's apply the time dilation formula to the second graph where the speed of the clock is 0.866 and gamma equals two. The formula says that the time on the clock will be one half of the coordinate time and we see that because the spacing of the dots is twice the spacing of the grid lines for the graph. In other words, the first dot after the origin, representing a clock time of one second occurs at a coordinate time of two seconds. Got it? 



#82
Dec1212, 08:12 AM

P: 52

In the SPECIFIC situation, SR states based on the relative speed "v", the relation of the speed of time should be Δt' = Δt/γ, but regarding the "event period", we should go by Δt' = γ(Δt(vΔx/c^2)). The "event period" reflects the speed of time, isn't it? 



#83
Dec1212, 08:18 AM

Mentor
P: 40,890

Everyone in this thread besides you understands that SR means Special Relativity. And of course Special Relativity includes the Lorentz transformations. Any claim otherwise is just silly. Please give your definition of "SR" in your very next post. (I am giving you the benefit of the doubt here.) 



#84
Dec1212, 08:30 AM

Mentor
P: 40,890

The time dilation formula Δt' = Δt/γ is only true for the special case where the events happen at the same position in the primed frame and thus Δx' = 0. In words, the time dilation formula says that 'moving clocks run slow'. 



#85
Dec1212, 08:56 AM

Mentor
P: 11,239





#86
Dec1212, 10:17 AM

Mentor
P: 16,481

For pedagogical reasons my recommendation is always to follow your last suggestion which I highlighted in bold. It automatically simplifies to the time dilation equation whenever it is appropriate and it avoids accidentally using it when it is not appropriate. 



#87
Dec1212, 10:20 AM

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P: 16,481





#88
Dec1212, 11:39 PM

P: 52

How do observers everywhere in the stationary system S measure the "time period" Δt of ONE SPECIFIC SECTION of continuous events from event 1 at point A to event 2 at point B? They use the synchronized stationary clocks in S. For the observers everywhere in the moving system S', they will use the synchronized moving clocks to measure the "time period" Δt' from event 1 to event 2. Now, we have a better defined case. There are two ways to do the measure jobs. The easy way is to measure event time t1 and t1' for event 1 at point A and measure event time t2 and t2' for event 2 at point B. We will have Δt' = t2't1' and Δt = t2t1. The difficult way is to measure event time by assigned observers, one in S and one in S'. The simplest assginment is let them stay at origin points O and O'. Let us look at this simplest situation. Since the distances of each pair of four points A, B, O and O' could be different, we must adjust the influence of the distance. When A=B=O', it will be the situation arranged by Einstein. We will have Δt' = t2't1' and Δt = t2t1. For this SPECIFIC SET of t1,t2, t1' and t2', SR states based on the relative speed "v", the relation of the speed of time in S' and S should be Δt' = Δt/γ, but regarding the "time period", we should go by Δt' = γ(Δt(vΔx/c^2)). Am I correct? In Δt' = Δt/γ, Δt' is for the time period in the moving system S' of a given SECTION OF EVENTS and Δt is for the stationary system S of the same SECTION OF EVENTS. 



#89
Dec1212, 11:41 PM

P: 52





#90
Dec1312, 05:02 AM

P: 3,178

t1................t2 O S A.................B A,B O' S' t1' t2' 1. For x1'=x2' (Δx'=0, clock at rest in S', moving in S): Δt' = Δt/γ 2. For x1=x2 (Δx=0, clock at rest in S, moving in S'): Δt' = γΔt Once more: you selected the time period of situation 1, with x1'=x2' and thus Δx≠0. You should find from the LT that the time period Δt' = Δt/γ for that situation, based on the relative speed v. If that is not clear to you, please ask. 


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