
#19
Feb413, 08:55 AM

P: 1,657





#20
Feb413, 09:02 AM

P: 1,657




#21
Feb413, 10:07 AM

Physics
Sci Advisor
PF Gold
P: 5,518

Of course, it's not *necessary* to analyze the twin paradox this way, since spacetime is flat in the standard twin paradox, and there is never any need to introduce a "gravitational field" in flat spacetime. But conceptually, allowing for the possibility of a "gravitational field" as a coordinatedependent thing (as Einstein was using the term, it basically means nonzero connection coefficients) was a way of getting to the principle of general covariancethat in fact you *can* use any coordinates you like; you're not limited to inertial coordinates. But if you use noninertial coordinates, you're going to find these "gravitational fields" present, even if no gravitating masses are present. In other words, it was a way of expressing the unity of SR and GR, that they are really just one single theory, as I said above. 



#22
Feb413, 02:13 PM

PF Gold
P: 4,081

Thanks to PeterDonis and stevendaryl for those posts and links. I was always a bit confused by Einsteins 'dialog' ( Ialog ?) but the Usenet Physics article explains how to use a gravitational field to unkink the worldlines. It is an ingenious construction.




#23
Feb413, 03:56 PM

P: 1,657





#24
Feb413, 05:00 PM

P: 1,098





#25
Feb413, 05:04 PM

P: 1,098





#26
Feb413, 05:26 PM

Sci Advisor
Thanks
P: 2,973

[Of course stevendaryl knows this already. I'm just trying to stop someone else who doesn't know this from being confused by this agingfaster/aging slower right now thing] 



#27
Feb413, 05:36 PM

PF Gold
P: 4,081





#28
Feb413, 05:44 PM

Physics
Sci Advisor
PF Gold
P: 5,518

http://math.ucr.edu/home/baez/physic.../SR/clock.html Basically, it's the postulate that a clock's "rate of time flow", as seen by an observer, is not affected by its acceleration; it's only affected by the clock's velocity relative to the observer. 



#29
Feb413, 05:59 PM

P: 1,098





#30
Feb613, 02:33 PM

PF Gold
P: 4,081

This diagram is the scenario where the twins T1 (blue) and T2 (green) comove, then part company. The measurements are made between times t_{0} and t_{2}.
The proper times of interest are AB, along T2's worldline, and CD+DE along T1's worldline. Without doing any calculations ( I think these proper times are stevendaryl's τ_{2} and τ_{1}), it is obvious that whether AB is greater than or less than CD+DE, this relationship will be true from any inertial frame. The reason being, that proper time is invariant under LT. Definition of proper time [tex] d\tau^2 = dt^2dx^2 [/tex] Transform the intervals dx>dX, dt>dT with a Lorentz transformation [tex] dT=\gamma dt + \gamma\beta dx,\ \ dX=\gamma dx + \gamma\beta dt [/tex] Calculate new proper time [tex] \begin{align} dT^2dX^2 &= (\gamma dt + \gamma\beta dx)^2  (\gamma dx + \gamma\beta dt)^2\\ &= \gamma^2(1\beta^2)dt^2 \gamma^2(1\beta^2)dx^2\\ &= dt^2  dx^2 \end{align} [/tex] Any decent, simple book on relativity tells us this. 



#31
Feb613, 03:23 PM

Physics
Sci Advisor
PF Gold
P: 5,518





#32
Feb613, 03:41 PM

Physics
Sci Advisor
PF Gold
P: 5,518

We have the following events (coordinates t, x, y are given relative to Frame A): #1: [itex](0, 0, 0)[/itex] T1 starts the experiment, moving in the x direction at velocity v. #2: [itex](0, 0, 1)[/itex] T2 starts the experiment, moving in the x direction at velocity v. #3: [itex](t_1, v t_1, 0)[/itex] T1 stops moving. #4: [itex](t_2, v t_2, 1)[/itex] T2 stops moving and ends the experiment. #5: [itex](t_2, v t_1, 0)[/itex] T1 ends the experiment. We have, by hypothesis, [itex]t_2 > t_1[/itex], and for convenience I will define [itex]\delta t = t_2  t_1[/itex]. The proper times in Frame A are then: [tex]\tau_1 = \frac{t_1}{\gamma} + \left( t_2  t_1 \right) = \frac{t_1}{\gamma} + \delta t[/tex] [tex]\tau_2 = \frac{t_2}{\gamma} = \frac{t_1 + \delta t}{\gamma}[/tex] This makes it obvious that [itex]\tau_1 > \tau_2[/itex]. Now let's look at things in Frame B. Here are the event coordinates t', x', y' in that frame, obtained by Lorentz transforming the coordinates given above (note that we have assumed the origins of both frames are the same, at event #1): #1: [itex](0, 0, 0)[/itex] T1 starts moving in the x direction at velocity v. #2: [itex](0, 0, 1)[/itex] T2 starts moving in the x direction at velocity v. #3: [itex](t_1 / \gamma, 0, 0)[/itex] T1 stops moving. #4: [itex](t_2 / \gamma, 0, 1)[/itex] T2 stops moving. #5: [itex](t_1 / \gamma + \gamma \delta t,  \gamma v \delta t, 0)[/itex] T1 ends the experiment. The proper times in this frame are then: [tex]\tau_1 = \frac{t_1}{\gamma} + \frac{t_1 / \gamma + \gamma \delta t  t_1 / \gamma}{\gamma} = \frac{t_1}{\gamma} + \delta t[/tex] [tex]\tau_2 = \frac{t_2}{\gamma} = \frac{t_1 + \delta t}{\gamma}[/tex] In other words, the proper times are the same in both frames, as they should be. The key thing to note, of course, is that in Frame B, event #5 happens *later* than event #4, and the additional coordinate time that this adds to T1's "moving" segment in that frame more than compensates for the fact that T1 is moving while T2 is at rest. This is basically the same resolution as the previous scenario; the y coordinate drops out of the analysis since all the motion is in the x direction, but there is still separation in the x direction at the end of the experiment (even though there isn't at the start), so relativity of simultaneity still comes into play in making event #5 later than event #4 in Frame B. 



#33
Feb713, 05:32 AM

Mentor
P: 14,483

Thread locked pending cleanup.
Metz114 et al: Please remember to use the report button to let the mentors know about nonsense such as that which you highlighted. OK. Cleanup complete. I deleted 50 posts. That's a bit much, perhaps too much. Those posts are still here; I softdeleted them. Let me know if there's anything that you members strongly feel needs to be restored. 


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