On 30 January 2014 14:17, meekerdb <meeke...@verizon.net> wrote: > On 1/29/2014 5:19 AM, Edgar L. Owen wrote: > >> Brent, >> >> Here's another relativity question I'd like to get your explanation for >> if I may... >> >> In Thorne's 'Black Holes and Time Warps' he gives the following example. >> >> Two observers A and B. >> >> A leaves earth orbit to travel to the center of the galaxy, 30,100 light >> year away, using a constant 1g acceleration to the midpoint and a constant >> 1g decelleration on the second half of the journey to arrive stationary at >> the galactic center, >> >> Thorne tells us that the 30,100 light year trip takes 30,102 years on B's >> clock back on earth but only 20 years on A's clock aboard the spaceship. >> >> Now my question is what causes the extreme slowing of A's clock? >> >> It can't be the acceleration as both A and B experience the exact same 1g >> acceleration for the duration of the trip. >> >> I can understand that during the trip B will observe A's clock to be >> greatly slowed due to the extreme relative motion, but since the motion IS >> relative wouldn't A also observe B's clock to be slowed by the same amount >> during the trip? >> >> And since the time dilation of relative motion is relative then how does >> it actually produce a real objective slowing of A's clock that both >> observers can agree upon? >> >> You had said yesterday that "geometry doesn't cause clocks to slow" but >> other than the trivial 1g acceleration isn't all the rest just geometry in >> this case? >> >> What's the proper way to analyze this to get Thorne's result? >> > > A rough way to see it is right is to note that c/g = 3e7sec ~ 1year << > 30,000yr. So the spaceship spends essentially the whole flight at very > near c. So the trip takes 30,100+ years in the frame of the galaxy. But > the proper time for the spaceship is very small; if it were actually at > speed c, like a photon, its proper time lapse would be zero. Only, because > it can't quite reach c, the time turns out to be 20 years. To get the exact > values you have to integrate the differential equations: > > dt/dtau = 1/gamma > dv/dtau = accel/gamma^2 > dx/dtau = v/gamma > > where gamma=sqrt(1-v^2) >
The equivalence principle indicates that both A and B are in a 1g gravitational field throughout the exercise, hence the time dilation experienced by A can't be gravitational. All that leaves is the different distances they travel through space-time to reach their final meeting, which is indeed down to "geometry" (in this case involving curves rather the straight lines - but that is minor detail, and can be solved by integrating the relevant equations, as indicated). So I assume the overall geometry of their paths through space-time *is*responsible for the final mismatch between their clocks. I'm not sure whether that contradicts "geometry doesn't cause clocks to slow" - probably not. PS I would instruct A to fly above the plane of the galaxy. There is a lot of stuff between the Earth and the galactic centre and I suspect that even a dust grain would hit a relativistic spacecraft like a nuclear bomb once it was near peak velocity, which according to my calculations is 0.9999995c (or in any case p.d.q.) -- You received this message because you are subscribed to the Google Groups "Everything List" group. To unsubscribe from this group and stop receiving emails from it, send an email to everything-list+unsubscr...@googlegroups.com. To post to this group, send email to everything-list@googlegroups.com. Visit this group at http://groups.google.com/group/everything-list. For more options, visit https://groups.google.com/groups/opt_out.
