0
$\begingroup$

This question already has an answer here:

How did the universe 'know' to do that? Or is there an intuitive explanation for it?

$\endgroup$

marked as duplicate by John Rennie, Danu, user10851, Kyle Kanos, Ben Crowell Oct 11 '14 at 14:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

0
$\begingroup$

Because space is not Euclidean. It's that simple. IMO This also precludes the mediation of gravity by particles.

$\endgroup$
  • $\begingroup$ Maybe you want to expand a little on the first sentence so that the OP can understand what you mean? Also, the second sentence seems out of place for this question. $\endgroup$ – Martin Oct 11 '14 at 11:20
0
$\begingroup$

The root cause is the fact that the speed of light is the same value for all observers, whether they are moving towards the source or away from it. The consequence is that people travelling at different velocities relative to each other measure things differently. As to why c = const, nobody knows.

$\endgroup$
0
$\begingroup$

The universe doesn't have to "know" that any more than it has to "know" that a tilted ladder should "become" shorter. The key effect is the relativity or simultaneity.

  • The length is defined as the distance of the two end points at the same time. Due to relativity of simultaneity, the same time in the observer's frame corresponds to two different times in the object's frame. Between the two times, the observer and object have moved relative to each other, therefore the distance of the two points changed. A detailed analysis shows that for the observer, the object seems shorter.

  • The "mass gain" is actually just an accounting issue: If $m_0$ is the invariant ("rest") mass of an object, the particle's momentum is $$p = \frac{m_0 v}{\sqrt{1-v^2/c^2}}.$$ Now one way to interpret this equation is to interpret the denominator as belonging to the mass, thus restoring the Newtonian $p=mv$ with a "relativistic mass" $$m = \frac{m_0}{\sqrt{1-v^2/c^2}}.$$ Note however that with this "relativistic mass" the Newtonian equation $F=ma$ does not hold in general.

    Indeed, in theoretical physics these days the concept of a relativistic mass is considered outdated, and the only relevant mass is the invariant ("rest") mass.

    Anyway, the concept of relativistic mass only describes how we account for certain factors in the momentum, which the universe of course doesn't care of.

  • Time dilation can be again understood from relativity of simultaneity by the object moving into regions with "more delayed time" (that is, into regions where we consider earlier times of the object's frame to be simultaneous). Thus the object's time seems to go more slowly.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.