Twins Paradox Paradox

Question

I've recently has special-relativity explained to be in a rather elegant way. All objects travel at the speed of light in space time. Thus, when you travel faster through the three dimensions of space, the speed at which you travel through time decreases. Photons do not experience time because all their velocity is in the spatial directions and none of it is in time.

Considering the 4-dimensional universe this invokes, how can I ever interact with anything that has traveled at a different velocity than me unless it changes direction to meet me?

In the terms of the twin paradox, how is it possible for the twins to meet? Twin A, stays on earth traveling at a fixed speed in spacetime. Twin B leaves earth traveling faster in space and slower in time but still traveling at c in spacetime.

Some basic geometry tells me that if two objects forced to travel at a fixed speed from the same origin in changeable directions (in any number of dimensions) that there is no way their future locations can be the same unless they both change direction to provide the possibility of intersection.

If Earth is traveling at a fixed speed and Twin B is traveling at a fixed speed in a space (no matter the dimension) then there is no way they could ever be in the same location in spacetime again unless the earth changes direction and meets Twin B in the middle.

Wouldn't they be forever doomed to be in different places along the axis of time unless Twin A goes on a relativistic voyage to allow Twin B to catch up? Should we not expect time to behave the same way as the spacial dimensions?

Answer 1

That way of thinking about this is a nice one. From “The Elegant Universe”?

Anyway, mathematically, you can define this four-velocity like the normal velocity, the time derivative of the position. However, in special relativity, the position is space and time $(t, x, y, z)$. And the derivative has to be with respect to the proper time (or curve parameter) $\tau$:

$$ u = \frac{\mathrm d}{\mathrm d\tau} x $$

The velocities that you normally measure are calculated with respect to $t$. So we need the chain rule here: $$ \frac{\mathrm d}{\mathrm d\tau} = \gamma \frac{\mathrm d}{\mathrm dt}$$

So the four-velocity is, taking $c = 1$: $$ u = \gamma \frac{\mathrm d}{\mathrm dt} (t, x, y, z) = \gamma (1, v_x, v_y, v_z) $$

As you said:

All objects travel at the speed of light in space time.

That means that $|u| = 1$. With the Minkowsi metric, this is: $$ 1 = \gamma^2 - \gamma^2( v_x^2 + v_y^2 + v_z^2) $$

This means that you travel less through time when you travel through space. But since the equation are for the squares of the velocities, you can indeed have $v_x < 0$.

So the twin that is on the journey can have $v_x > 0$ for the first half and then $v_x < 0$ for the second part and go back to earth. That is not a problem.

If you look at a space time diagram of this, you can construct the time intervals for each party. The observer on the earth will have gone through a couple of time steps. When you draw in the planes of simultaneity of the observer on earth, you will see the time intervals for the traveling observer. There, you can see that the earth time intervals are much longer for the traveler. He will only experience a bit more than two of those intervals during his journey. Therefore, he is younger.

So, when they meet at a space time point again, their age is different. They cannot meet at the space time point where earth is the same age as the traveler, though.