# How does time dilation work without a privileged reference frame?

As I understand special relativity, light travels at the same speed in all reference frames.

What I fail to understand is why time dilation would occur in one reference frame, but not by an equal amount in all other reference frames, at the same time.

To put it another way, if I am sitting on earth, observing a space ship that is flying at a speed near $c$, I should observe him moving through time slower than me. However, looking out his window, he should observe me traveling relative to him at a speed near $c$ and should likewise see me moving through time slower than himself.

What am I missing from my understanding of special relativity?

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possible duplicate of Time Dilation - how does it know which FoR to age slower? – Qmechanic Dec 29 '11 at 16:27

If you are standing up, and your friend is inclined on a tilted incline of slope 45 degrees, and you are both the same height, you would say that your friend is shorter by a factor of .707. But from his tilted point of view, you are also shorter by the same factor. There is no contradiction, and there is no need to invoke an absolute notion of up. This is not confusing, because we know there is such a thing as a rotation.

Similarly in relativity, if something is moving, the graph of its motion in time is a straight line tilted away from the t-axis of a non-moving observer, so it is obvious that the equally spaced clock-ticks on the moving line will not be equally spaced to the stationary observer, and that the effect is symmetric, because from the moving point of view, the stationary frame is tilted.

The relativistic transformations are just (hyperbolic) rotations of space into time. They are no more confusing than rotations (although for me, relativity just served to highlight how counterintuitive rotations really are when you think about them deeply).

This is also explained in these answers: What are the mechanics by which Time Dilation and Length Contraction occur? and Einstein's postulates <==> Minkowski space. (In layman's terms)

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+1 explaining the answer and also including links to the related material FULL and Complete Answer deserved a bump – Argus May 28 '12 at 20:17

I like to think of it this way, say you have a mirror like sphere such as a silicone ball you would find on a bird bath, now you look at the your reflection that's in the middle of the ball (THAT CAN REPRESENT YOUR REFERNCE FRAME) and then you look at the perceived edge/border of the visible side of the reflective surface (THAT CAN REPRESENT THE SPEED OF LIGHT BOUNDARY) now from your point of view, you see a fairly normal reflection, where if another person where standing at an angle from you there reflection would appear distorted (REPRESENTING TIME DILATION) but to that other person, they're reflection would appear normal and yours would look warped to there frame of reference, also; if you spin the ball or change positions from around the sphere, everything always looks the same. This is a 3d representation that can be translated to a space-time scenario.

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ok, please tell me what was wrong with this analogy. – GammaRay Mar 9 '14 at 6:44
as far as I am concerned it is OK. A bit more complicated than the simple rotation argument of Ron's answer above. – anna v Mar 10 '14 at 7:11