Tme dilation and inertial reference In answering the standard 'twin paradox', the reason for the reduced age of the 'traveling' twin is given as due to the change in acceleration and direction relative to the earthbound twin. This is done to resolve the question of the 'relative motion' of the earthbound twin from the perspective of the traveling twin (since the earthbound twin is moving, relative to the traveling twin, why doesn't that twin also have a reduced age).
Isn't this bugaboo an unnecessary avoidance of inertial frames of reference?
Muons created at the earth's upper atmosphere should not survive the distance traveling at near-light speed to the earth's surface, since they should decay due to their halflife prior to reaching the surface.
But they don't due to their relativistic time dilation - without any change in direction or acceleration.
 A: The time dilation is directly connected to acceleration. The best way to see this is as follows. The action principle for inertial motion involves an action whose minimization yields that inertial motion. For motion that deviates from it, and involves acceleration, the action will have a larger value.
And the punch-line is about to follow ... as a double-whammy.
More precisely, given two space-time points $A$ and $B$, with $A$ occurring before (i.e. in the causal past of) $B$, the motion which minimizes the action integral $S(A,B)$ is the inertial motion. If $A$ and $B$ coincide, as positions in space (with respect to a given frame of reference, that is) then $S(A,B) = 0$ precisely for inertial motion, otherwise $S(A,B) > 0$.
So, what is that integral? The time-dilation!
(More precisely: the time-dilation up to a fixed proportionality with the sign reversed.)
QED
And now for the second punchline... if you look at this in greater depth, you're going to get an even bigger surprise: Relativity is a red-herring. That is:
there is also a non-relativistic version of both time-dilation and the twin paradox!
And they are each the non-relativistic limits of their respective relativistic versions; with both limits being non-zero and non-trivial.
The difference between the two paradigms (relativistic versus non-relativistic) is not a difference of kind, but a difference of degree. The relativistic version of the time-dilation presents itself as a small relativistic correction of the non-relativistic version.
It's all a matter of how you take the non-relativistic limit.
For a body of mass $m$, the action integral governing its motion has the following form: $S(A, B) = \int_A^B L dt$, involving a Lagrangian that is usually expressed as $L = -mc^2 ds$, where $s$ is the proper time. But equivalently, and more traditionally (and more correctly) it can also be expressed in the form $L = mc^2 (dt - ds)$, that is: as the difference between the proper time $s$ and real time $t$. The proportionality factor is $mc^2$, where $c$ is the speed of light.
In the non-relativistic limit, i.e. as $c → ∞$, proper and real time coincide: $t - s → 0$. However, the Lagrangian itself continues to have meaning as a non-trivial non-relativistic limit. This puts the spotlight on the time-dilation, itself, but scaled up by a factor of $c^2$ as: $u = c^2(s - t)$. The difference $s - t$ is the one I use out of habit, rather than $t - s$, and is the one that corresponds to time-dilation with the correct sign.
The Lagrangian can, thus, be written as $L = -m du$, and the action is just $S(A,B) = -mu$, where $u$ is evaluated from $A$ to $B$.
For acceleration-free motion, the scaled-up time-dilation $u$ is as large as it can be (and -$u$ is minimized, as is the action $-mu$, itself). With accelerated non-inertial motion of any kind, the action $-mu$ is larger, and scaled-up time-dilation $u$ is smaller. The time difference between the two will simply be the difference of their respective $u$ coordinates ... scaled back down by dividing out by $c^2$.
If the spatial locations of points $A$ and $B$ are the same, then the inertial twin was actually at rest the whole time and $u$ will be 0, while for the other twin, it will be negative (and the action positive).
There is a geometry associated with the $u$ coordinate. This is best seen by writing the quadratic form representing the Minkowski metric out explicitly, as well as the linear form representing the proper time:
$$ds = dt + \frac{1}{c^2} du, \hspace 1em c^2 ds^2 = c^2 dt^2 - (dx^2 + dy^2 + dz^2).$$
Substituting $u$ into the quadratic form for $s$ and bringing everything over to one side, the result is:
$$dx^2 + dy^2 + dz^2 + 2 dt du + \frac{1}{c^2} du^2 = 0.$$
Minkowski space embeds, as a 4-dimensional subspace, into a 5-dimensional geometry, with the extra coordinate $u$.
Both of these invariants continue to have meaning in the non-relativistic limit. As $c → ∞$, they approach the following:
$$ds = dt, \hspace 1em dx^2 + dy^2 + dz^2 + 2 dt du = 0.$$
That's the geometry of non-relativistic theory. It's called the Bargmann geometry.
So, now you know what the fifth coordinate for Bargmann geometry is. It's just time-dilation, up to rescaling. The moorings of time-dilation may have gotten cut loose from $s$ and $t$, because those two now coincide, but time-dilation is still there, hiding where it always was.
For a body in motion, with its velocity given by
$$ = \left(\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}\right),$$
if it is subject to the quadratic constraint:
$$
0 = \left(\frac{dx}{dt}\right)^2 + \left(\frac{dx}{dt}\right)^2 + \left(\frac{dx}{dt}\right)^2 + 2 \frac{dt}{dt} \frac{du}{dt} = v^2 + 2 \frac{du}{dt},$$
then the following will be the equation of motion for its $u$ coordinate:
$$\frac{du}{dt} = -\frac{v^2}{2}.$$
The action is just $S(A,B) = \int_A^B ½ m v^2 dt = -m u$. So, when you scale back down $u$ by dividing by $c^2$, you get the time dilation - in time units.
If you compare the non-relativistic "time dilations" for the two twins to their respective relativistic versions, you will find that the latter are each virtually identical to the former. They will each differ from their non-relativistic versions by small corrections on the order of $1/c^2$.
Thus, the twin paradox survives the passage into non-relativistic theory. You just needed to know where to look for it, and how to take the non-relativistic limit correctly, to discover where it was hidden all along.
A: 
In answering the standard 'twin paradox', the reason for the reduced age of the 'traveling' twin is given as due to the change in acceleration and direction relative to the earthbound twin.

This is not correct. The reason why traveling twin is younger is NOT because it accelerates. It's because it is flying fast. Nothing interesting happened to him during acceleration. Theoretically the traveling twin can change his velocity within 1 second by Earth time (and even faster by his own time), and in this case he will not age for more than 1 second during acceleration.
Still acceleration is very important for twin paradox. Because during the acceleration period weird things happen to the Earth twin as seen by traveling twin!
According to traveling twin while he is flying away from Earht the Earth one is younger. Then traveling twin stops. Within this split of second the Earth twin becomes much older! Traveling twin accelerates to go back - and Earth twin becomes even more old! From now on traveling twin moves back home with constant speed, the Earth twin is aging slower then himself, but the net effect is that when he reaches Earth, the Earth twin is older.
The Earth twin suddenly aged during acceleration not because he was moving fast/slow. This effect can't be described by usual time-dilation formula. This happens because the frame of reference is not inertial.
UPDATE
Special relativity theory describes the world using inertial frames of reference. All the famous formulae are written for inertial frames of reference.
Special relativity theory states that time is going slower on moving objects. To be precise: if in some inertial frame of reference an object has velocity $v$, than time on this object is going slower by factor $\sqrt{1-v^2/c^2}$. There is not a single word about acceleration here. If traveling twin started his journey having speed 0.8$c$ he is aging 0.6 times slower. In 1 year (according to home-sitting-twin) the traveling twin will be (1 - 0.6) = 0.4 years younger.
Now traveling twin bounces from a metal wall and is moving back home. His speed is still 0.8 $c$ and he is still aging 0.6 times slower. According to home-sitting-twin the traveling one will be back home in 1 more year and by that time he will be 0.8 years younger than himself.
Again: there is nothing about acceleration in all these calculations. And this IS NOT a twin paradox yet.
Twin paradox happens when we try to describe what was happening using the frame of reference "attached" to a traveling twin. At a first glance it looks like the situation is symmetric: the second twin is staying still all the time, the first one moves back and forth, and the final result should be that the first twin (one which was sitting all the time on the traveling Earth) should be younger. And this IS a paradox, because we can just compare both twins in the end of experiment and it can not happen that each of them is younger than the other.
The explanation of this paradox is that we mistakenly used "usual" special relativity formulae in not-inertial frame of reference. If one really wants to use the frame of reference "attached" to accelerating twin he should use not Special, but General relativity theory. This exercise is not for faint-hearted! But the final result will be consistent: the twin sitting on Earth will be older than the other one.
