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?

  • $\begingroup$ The twin situation which is usually called a "paradox" explicitly requires the "travelling" one to change direction to meet their sibling. And (s)he is not travelling at the speed of light, otherwise (s)he would have no mass and would never be able to decrease velocity. Also, for the n'th time, there is no frame of reference for something travelling at the speeed of light. This thing about time not passing when you are at the speed of light needs to stop. $\endgroup$ Jul 18, 2014 at 19:20
  • $\begingroup$ "This thing about time not passing when you are at the speed of light needs to stop." That makes sense but is not necessary for my problem. If the earth is traveling in spacetime at the same speed the "traveler" is traveling in spacetime then would the earth not have moved farther along the axis of time than the traveler and thus not be there when the "traveler" comes back to the earth's old location? $\endgroup$ Jul 18, 2014 at 19:25
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    $\begingroup$ I think this is actually a good question. I'm not quite sure why people are downvoting it. It comes down to the fact that the explanation you're received is not entirely accurate, but it is a common explanation, and I don't fault you for asking about it. $\endgroup$
    – David Z
    Jul 18, 2014 at 19:28
  • $\begingroup$ They meet up again. However, the whole concept of perceived time depends on the path! $\endgroup$ Jul 18, 2014 at 19:28
  • $\begingroup$ The phrase "All objects travel at the speed of light in space time." need to be understood in the correct mathematical context (and with an awareness that the distribution of that motion in "time" and "space" is observer dependent), and you are trying to force and almost-but-not-quite-correct understanding onto it. $\endgroup$ Jul 18, 2014 at 19:28

1 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.

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  • $\begingroup$ All this is true of course, but I think it may miss the key conceptual issue, which is how the other twin can be traveling at $c$ through spacetime and yet still be on Earth to meet the traveling twin when he/she comes back. $\endgroup$
    – David Z
    Jul 18, 2014 at 19:32
  • $\begingroup$ You are right, his comment on the question clarified it. I will draw a picture and write a bit more. $\endgroup$ Jul 18, 2014 at 19:33
  • $\begingroup$ Essentially I interpreted what I understood in the sense of a 4-dimensional metric space where everything is traveling at a fixed velocity and mutable direction. I am a computer scientist who knows enough math to get myself into trouble. $\endgroup$ Jul 18, 2014 at 19:40
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    $\begingroup$ One key point is that minus sign in the metric. All possible four-velocites do not lie on a sphere, but on a hyperboloid. That makes the whole geometry really unintuitive. $\endgroup$ Jul 18, 2014 at 19:44
  • $\begingroup$ So what I am taking out of this is that time does not have the same properties as the other dimensions because you cannot go backward in time (and thus a hyperbolid convex to the axis of time rather than a sphere)? $\endgroup$ Jul 18, 2014 at 19:56

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