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)

Brent


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