I once asked 'justatruthseeker' a question about the twin paradox, which he never answered. I think that I know the answer, but perhaps you can confirm that I am right or correct me if I am wrong.
Suppose that both twins have a set of clocks, including radiometric clocks using chromium-51 (half-life of 27.5 days) and cobalt-57 (half-life of 270 days). Will twin B's menstrual cycles and pregnancies last the same time according to her clocks as twin A's will according to her Earthbound clocks?
yes. Both will have the usual duration of menstrual cycle and pregnancy as measured by their own clocks.
In particular, will twin B's menstrual cycles last one half-life of the chromium-51 on the spacecraft, and will her pregnancies last one half-life of the cobalt-57?
yes, one half-life.
As I understand it, the answer to both questions is yes, but I know little about the mathematics of relativity, and I should like to have an answer from somebody with qualifications. Also, what will happen at the turn-around point, when B (call her Barbara) changes inertial frames?
That depends one the exact mechanics of the turn-around. How much acceleration and for how long.
A good way to view a lot of this is by analogy with rotations in the plane. Imagine a xy grid. This grid (coordinate system) can describe every point in the plane by two numbers: an x-coordinate and a y-coordinate. If you have the x and y coordinates of two points, you can compute the distance between then using the Pyhtagorean formula.
In this analogy, the x coordinate is analogous to space and the y coordinate is analogous to time. The distance between points is analogous to 'proper time'.
Now, imagine a second grid, centered at the same place, but rotated with respect to the first. Call the first grid A and the second grid B. The coordinates for most points (all expect the center) will be different in the two grids. The grid B coordinates and the grid A coordinates will often be quite t.
But, you can still find the distance between any two points using the grid B coordinates and the Pythagorean formula.
Now, suppose you take a horizontal line segment in the A grid. Say it has length L, so the difference in the x coordinates is L and there is no difference in the y coordinates. Now look at that same line segment in the B grid. In *that* graid, the difference in x coordinates is smaller than L---there is a length contraction. And now, there is a difference in the y coordinates: there is a violation of simultaneity.
Similarly, if we take a line segment in the B grid and look at it in the A grid, there is also length contraction and lack of simultaneity. And, we can do the same with veritcal line segments to show time dillation (although in the analogy, it is contraction--the specific formulas used in SR are different than euclidean geometry).
Now, a 'frame' is simply a grid-like coordinate system, possibly rotated with respect to A or B or both. All frames give different results for x and y coordinates, but all also give the same results for distance between points. All are related by a rotation and maybe a translation.
Now, suppose in this plane, we have two twins. One twin, A, has a path in the plane that is a single vertical line in the A grid. The other has a path that consists of a broken line starting at the same place A starts going a bit to the side and then returning back to where A stops.
Several questions:
1. Is the path length for A longer or shorter than that for B?
2. Does the length of the two paths depend on which grid is used?
3. Is there any one, single grid in which B is always moving in the y direction?
I think it is easy enough to see that the answers are:
1. A's path is shorter than B's.
2. This doesn't depend on any particular grid used to describe the paths.
3. No single grid will work for all of B's path: you can get grids for the first part or the second part, but none for both.
Now, this is an analogy to what happens with the relativistic twins. There are some major differences:
1. instead of distance, the 'invariant' is proper time. ALL frames will compute the same proper time.
2. Instead of the Pythagorean formula, there is a Minkowskian formula, where the square of the time minus the square of the distance is the square of the proper time.
3. because of this, longer paths in the usual distance typically have *shorter* proper times.
So, we have two twins: one (A) has a path always in the y direction for some grid (his frame of reference). The other (B) moves away and then returns. The difference is that no single frame will work for B: that break in the path prevents a single grid from always having B's path in the y direction. Instead, B need two different grids for that.
None the less, the path lengths (proper times) can be computed correctly in any grid *as long as we use coordinates from that grid). There is a basic assymetry between the two twins because A's path is always in the y direction of some grid, but that is not the case for B. This is partly why the euclidean distance for B is longer than that for A. In the analogy, the proper time for B is smaller than the proper time for A. In other words, B ages less than A.
B 'changes frames' while A does not.
