You're making a mistake 1st years, learning special relativity for the first time, shouldn't even be making. The 'twin paradox' is not an actual paradox, it is called it because it runs counter to our intuition but it isn't logically inconsistent. If you could do the special relativity calculations you'd find that you cannot have both twins meet up again and them both be younger than the other. One and only one is always younger than the other. Which one is the younger one is also simple to work out. Saying "The twin paradox invalidates special relativity!" is just plain ignorant.
OK
Discuss it.
Which of the two of the twins will be younger than?
And - why?
Hi Masterov;
That's a question I too have always wondered about. When (if ?) Alpha Numeric replies, I hope he also includes a simplistic answer, to cater for those who are quite untrained / ignorant in these areas, though nonetheless interested (me for instance).
Special relativity teaches us that the elapsed time experienced by an object moving inertially between events A and B is, in any inertial coordinate system:
$$\tau = \frac{1}{c} \sqrt{c^2 (t_B - t_A)^2 - (x_B - x_A)^2 - (y_B - y_A)^2 - (z_B - z_A)^2} \equiv \frac{1}{c} \sqrt{c^2 (\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2} $$
So it's a good thing that all inertial coordinate systems agree on the value of this number because any two events described in any two coordinate systems:
$$c^2 (\Delta t)^2 - (\Delta x)^2 - (\Delta y)^2 - (\Delta z)^2 = c^2 (\Delta t')^2 - (\Delta x')^2 - (\Delta y')^2 - (\Delta z')^2$$
So the reason "why" the twin paradox always has one and only one answer of which twin is younger and by how much is one of space-time (hyperbolic) geometry.
Imagine a triangle of slower-than-light paths through space time. You can get from A to B via an inertial path, or you can get from A to C and then from C to B.
So the twin paradox is equivalent to saying that the inertial path has the largest proper time of any slower-than-light path from A to B.
Or "the straight line is the longest path between to points in space-time."
Or $$\tau_{AB} \ge \tau_{AC} + \tau_{CB}$$ which I shall demonstrate in a coordinate system.
Let
$$U = \sqrt{u_x^2 + u_y^2 + u_z^2} < c \\ V = \sqrt{v_x^2 + v_y^2 + v_z^2} < c \\ \begin{pmatrix} t_C - t_A \\ x_C - x_A \\ y_C - y_A \\ z_C - z_A \end{pmatrix} = \begin{pmatrix} 1 \\ u_x \\ u_y \\ u_z \end{pmatrix} \Delta t_{AC} \\ \begin{pmatrix} t_B - t_C \\ x_B - x_C \\ y_B - y_C \\ z_B - z_C \end{pmatrix} = \begin{pmatrix} 1 \\ v_x \\ v_y \\ v_z \end{pmatrix} \Delta t_{CB}$$
Then it follows that:
$$ -c^2 \lt - U V \leq \vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z \leq U V \lt c^2
\\ \begin{pmatrix} t_B - t_A \\ x_B - x_A \\ y_B - y_A \\ z_B - z_A \end{pmatrix} = \begin{pmatrix} \Delta t_{AC} + \Delta t_{CB} \\ u_x \Delta t_{AC} + v_x \Delta t_{CB} \\ u_y \Delta t_{AC} + v_y \Delta t_{CB} \\ u_z \Delta t_{AC} + v_z \Delta t_{CB} \end{pmatrix}
\\ \tau_{AC} = \sqrt{1 - \frac{u_x^2 + u_y^2 + u_z^2}{c^2}} \Delta t_{AC} = \sqrt{1 - \frac{U^2}{c^2}} \Delta t_{AC}
\\ \tau_{CB} = \sqrt{1 - \frac{v_x^2 + v_y^2 + v_z^2}{c^2}} \Delta t_{CB} = \sqrt{1 - \frac{V^2}{c^2}} \Delta t_{CB}
\\ \tau_{AB} = \sqrt{( \Delta t_{AC} + \Delta t_{CB} )^2 - ( \frac{u_x}{c} \Delta t_{AC} + \frac{v_x}{c} \Delta t_{CB} )^2 - ( \frac{u_y}{c} \Delta t_{AC} + \frac{v_y}{c} \Delta t_{CB} )^2 - ( \frac{u_z}{c} \Delta t_{AC} + \frac{v_z}{c} \Delta t_{CB} )^2 }
\\ \tau_{AB} = \sqrt{(1 - \frac{u_x^2+u_y^2+u_z^2}{c^2})(\Delta t_{AC})^2 + 2 (1 - \frac{u_x v_x + u_y v_y + u_z v_z }{c^2}) \Delta t_{AC} \Delta t_{CB} + (1 - \frac{v_x^2+v_y^2+v_z^2}{c^2})( \Delta t_{CB} )^2 }
\\ \tau_{AB} = \sqrt{(1 - \frac{U^2}{c^2})(\Delta t_{AC})^2 + (1 - \frac{V^2}{c^2})( \Delta t_{CB} )^2 + 2 (1 - \frac{\vec{u} \cdot \vec{v}}{c^2}) \Delta t_{AC} \Delta t_{CB} }
$$
So
$$ \tau_{AC} + \tau_{CB} = \sqrt{(\tau_{AC} + \tau_{CB})^2}
\\ \tau_{AC} + \tau_{CB} = \sqrt{ (1 - \frac{U^2}{c^2})(\Delta t_{AC})^2 + (1 - \frac{V^2}{c^2})( \Delta t_{CB} )^2 + 2 \sqrt{1 - \frac{U^2}{c^2}} \sqrt{1 - \frac{V^2}{c^2}} \Delta t_{AC} \Delta t_{CB} }
\\ \tau_{AB} = \sqrt{ ( \tau_{AC} + \tau_{CB} )^2 + 2 \Delta t_{AC} \Delta t_{CB} ( 1 - \sqrt{1 - \frac{U^2}{c^2}} \sqrt{1 - \frac{V^2}{c^2}} - \frac{\vec{u} \cdot \vec{v}}{c^2} ) }$$
Observe that $$ \tau_{AB} \lt \tau_{AC} + \tau_{CB} $$ (which is the opposite of what we wish to show) requires:
$$0 \leq 1 - \frac{\vec{u} \cdot \vec{v}}{c^2} \lt \sqrt{1 - \frac{U^2}{c^2}} \sqrt{1 - \frac{V^2}{c^2}} $$
but
$$ 0 \leq 1 - \frac{UV}{c^2} \le 1 - \frac{\vec{u} \cdot \vec{v}}{c^2}$$
and
$$ ( 1 - \frac{UV}{c^2} )^2 = 1 + \frac{U^2V^2}{c^4} - \frac{2 U V}{c^2} \ge 1 + \frac{U^2V^2}{c^4} - \frac{U^2 + V^2}{c^2} = \left( \sqrt{1 - \frac{U^2}{c^2}} \sqrt{1 - \frac{V^2}{c^2}} \right)^2$$
because $$0 \ge - \frac{ (U - V)^2 }{c^2} $$
Therefore $$ \tau_{AB} \ge \tau_{AC} + \tau_{CB} $$.
And so the statement "An inertial path between two causality related events is the path with the longest elapsed proper time" does not depend on any of the three legs of the triangle being at rest in the chosen coordinate system. It is a matter of simple (hyperbolic) geometry and doesn't have anything to do with coordinates at all. Coordinates only help in specifying precisely which events we are talking about so we can calculate exactly how different the twin's ages are when they reunites at the same event in space-time (B).