Brent, PS on quoted post:
If, as you say, A's proper time (comoving clock) is running much slower all during (most of) the trip to the center of the galaxy, then doesn't that mean A would observe all the intergalactic stuff passing by him at MUCH greater than the speed of light? How would that work? Stay at home observer B sees him traveling at just under the speed of light, but if A's proper time is actually slowing by a factor of ~31000/20=1550 he must see himself traveling at much GREATER than the speed of light relative to the intergalactic stuff he is passing. Is that right? Second, I seem to recall you saying that in the case of an observer falling through the event horizon of a black hole his own comoving clock (proper time) does NOT slow and he just observes himself falling right through it and reaching the singularity only 20 minutes later on his clock. So why is the comoving clock traveling to the center of the galaxy slowed but the comoving clock falling into a black hole isn't? In both cases we agree that a stationary observer watching back on earth sees these clocks both slowing don't we? Thanks, Edgar On Thursday, January 30, 2014 5:11:28 PM UTC-5, Edgar L. Owen wrote: > > Brent, > > So what you are saying is that because everything travels through > spacetime at the speed of light in all frames (my STc Principle) and A's > path through SPACE is much longer than B's (which is zero) that A's path > through time must be correspondingly shorter? > > At least that's my understanding and the way I'd express it..... > > However according to what you said yesterday that the time slowing effect > is due to the longer travel time of photons due to relative motion away > from each other, wouldn't A see B's clock slow by the SAME amount that B > see's A's clock slow DURING the trip due to the equal and opposite relative > motion and the equally longer and longer time photons from each take to > reach the other? > > But that seems to contradict the first result which implies A and B should > observe each other's clocks NOT slowing by the same rate DURING the trip > because A is actually moving in space and B isn't. > > So what's your explanation for the apparent contradiction? > > Thanks, > Edgar > > > > On Thursday, January 30, 2014 12:25:09 AM UTC-5, Brent wrote: >> >> On 1/29/2014 5:39 PM, LizR wrote: >> >> On 30 January 2014 14:17, meekerdb <meek...@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. >> >> >> Exactly. The clocks faithfully measure the interval along their >> respective paths. It's the difference in the paths, the geometry, that is >> the difference in duration. >> >> >> 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.) >> >> >> Even without dust the intergalactic hydrogen atoms would make it similar >> to standing in the LHC beam - but with a lot more luminosity. >> >> Brent >> > -- 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.