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GR and Curvatureby Phymath
Tags: curvature 
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#1
Jan2408, 06:10 PM

P: 189

obviously the equations of tidal forces and geodesic deviation are very similar to lead one to motivate yourself to explore gravity as not a field but as a curved geometry, Einstein also said that if each accelerated frame is locally an inertial one the euclidean geometry of Lorentz can not hold, which also motivated him to explore the geometry instead of a field. can some one explain/give me an example of Lorentz geometry not holding in all accelerated frames, basically prove the second statement.
I am trying to understand why Einstein abandons a field for geometry yes the equations are similar very similar still that is not enough to say it is proof that they are the same, E&M and classic gravity have similar mathematical qualities but they are (as far as we know) not the same. please help me out here proof or giving an example of the second statement, thank you. 


#2
Jan2408, 11:32 PM

P: 443

An excerpt from "Of pots and holes: Einstein’s bumpy road to general relativity", Michel Janssen, http://www.tc.umn.edu/~janss011/



#3
Jan2608, 03:49 PM

P: 189

Thank you that does show me a good example however I would like to check if i fully understand it.
the two observers both measure the diameter of the disk to be the same, now using the method i am about to describe, this is how they determine the circumference of the disk. As the observers pass each other at closest approach they both start their individual clocks, and as the closest approach occurs again the both stop their clocks. The observer not on the disk will say that the circumference of the disk must be: C = v*t where t is the time measured for one full revolution. Both should agree on the velocity because they both could say the other is moving with respect to each other but still both would observe the same velocity of each other. using this idea the time on the clock of the observer not on the disk would be: t = pi d/v where d is the diameter of the disk and v is the velocity of the person on the disk. since the person on the disk would experience time dilation they would report a time for one full revolution of t' = gamma*t where the gamma is the typical Lorentz transformation factor therefore the person on the disk would observe a circumference of the disk they were on of C' = d*t' > which gives pi' d/v = gamma* d*pi/v > pi' = gamma*pi so the person on the disk would find that his constant of pi is smaller then 3.14159... and thus noneuclidean, and also would say that gravity must be deforming the geometry because of the equivalence. is this a correct analysis? 


#4
Jan2708, 08:17 PM

P: 443

GR and Curvature
That problem contains the twin paradox in disguise. Each observer can claim he is stationary and the other observer is rotating around some center. So it's 'paradoxical' why the final result picks out one of them versus the other, why 'pi' is smaller for one of them. The resolution is that the situation just like in the twin paradox is not symmetrical, the observer not on the disk, is implicitly assumed to be really inertial, while the rotating observer is accelerating with respect to the inertial frame so he is never inertial.
I think the correct formula is t (time interval in the inertial frame) = gamma t' (proper time interval measured by the accelerated clock on the disk), where the Lorentz factor gamma >=1. When you apply the time dilation formula, the proper time interval t' (measured by two events happening at the same place of the measuring clock frame) is dilated in any other frame. The above formula is also in agreement with the twin paradox: the nonaccelerating twin measures bigger time t (ages more), than the accelerating twin t'. Since the above is the opposite of your formula, the final answer is reversed: the rotating observer will measure smaller circumference/radius ratio: pi' = pi / gamma. That is the opposite of what the citation is claiming. 


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