Harry Veeder wrote:
Stephen A. Lawrence wrote:
Harry Veeder wrote:
Stephen A. Lawrence wrote:
This is not a paradox, and the "paradoxical" nature of the problem was
in fact resolved something on the order of a century ago. The traveling
twin accelerates; the stay-at-home twin does not; thus, the symmetry is
broken.
That works in SR, but the solution is inconsistent with GR.
Wrong. In fact the full solution can only be had using techniques
commonly considered to be part of GR.
Acceleration is acceleration, in either SR or GR. In either case you
integrate the path length, using the pseudo-Riemannian metric of
Minkoski, in order to find the elapsed proper time of either twin, and
in either case the path which includes the acceleration comes out "shorter".
A geodesic in GR is the longest distance between two points.
No, it is the shortest "distance" between two points on a spacetime
manifold.
Work it out or look it up if you don't believe me. It's the path with
maximal magnitude, not minimal. It's commonly referred to as "extremal"
but unlike most "extremal" paths we deal with it's (locally) maximal,
not (locally) minimal.
Take a simple example: The metric, with c=1 and signature
diag(1,-1,-1,-1), in 1 space dimension, can be written as dt^2 - dx^2.
If I go directly from point (0,0), to point (5,3), the squared path
length is
25 - 9 = 16
and elapsed proper time is 4. If, instead, I go from (0,0) to (2,1) and
then from (2,1) to (5,3) (with a kink in the path), the squared path
length is
(4-1) + (9-4) = 8
which is smaller; the elapsed proper time is 2*sqrt(2), which is also
smaller than 4.
Again, the geodesic is the path which (locally) maximizes the proper time.
(I can't cough up the "geodesic equation" to prove this in general
without some time dredging around in my memory or some time looking at a
book; maybe tomorrow.)
Acceleration pushes you off the geodesic, as a result of which you
follow a shorter path. If the two twins have worldlines which cross at
two points, and if one accelerates while the other follows a geodesic,
the one on the geodesic will "age" more. That's straight out of GR ...
or SR, take your pick.
In GR, acceleration due to gravity is treated as indistinguishable from
a manufactured acceleration.
In GR acceleration due to gravity appears as nonzero connection
coefficients in the metric, which is also where you see the effects of a
uniform "acceleration field". In that sense, they're indistinguishable.
However, gravity due to real bodies also has tidal effects, which
manifest themselves as nonzero curvature; that is _not_ the same as
acceleration. Curvature can be tested for locally, and cannot be
transformed away by careful choice of coordinates, unlike uniform
acceleration.
Curvature -- tidal effects -- cannot be handled in special relativity.
If both accelerate, then neither follows a geodesic and you need to know
the details of the problem to determine who ages more (if either).
GR and SR only really differ when you introduce gravity, which doesn't
enter into this problem.
You can't ignore gravity.
You can in the twins problem; there isn't any. But in any case, gravity
doesn't affect the fact that geodesics are paths which maximize the
proper time -- that's true regardless of the presence of gravity.
The raison d'être of GR is to explain gravity.
That's right. But you don't need it to resolve the twins problem, which
takes place in flat space.
Ignore gravity and you are back in the flat spacetime of SR.
Harry
