# In special relativity, why is the time dilation factor different from different points in space?

So I look at the Lorentz transformation in the usual one dimensional case. Let's say the transformations are:

$$t' = \gamma(t-vx/c^2) \\ x'=\gamma(x-vt)$$

If I want to "track" the origin of the prime frame in terms of non-primed coordinates, then I need to ensure that $$x=vt$$ which gives me $$t'=t/\gamma$$. We can see that the primed clock is slowing down.

However, if I want to target some fixed non-zero location in the primed frame, say $$x_0'$$, then I need to have $$x=vt+x_0'/\gamma$$. That means the transformed time of the clock standing at the $$x_0'$$ position in primed frame has reading $$t'=t/\gamma - \frac{v}{c^2\gamma^2}x_0'$$

There could be some further simplifications probably. But why is it that, depending on the fixed position in the primed frame, the time in the unprimed frame has a different offset?

I understand this has no affect on the time "ticking rate" of the primed frame, from the POV of the unprimed frame, because the ticking rate is not affected by the time offset.

In some sense, if you freeze time "t" in the unprimed frame, and "walk" around the universe, you would actually be moving through time in the primed frame!

I drew a x-t plot and understand that this non-uniform (in space) offsetting of $$t'$$ helps to keep the $$x=ct$$ line invariant in both frames. But is there a more intuitive physics explanation?

• don't you mean $x'_0$? – JEB Jun 5 at 4:20
• Yes. Thanks, fixed. – Shuheng Zheng Jun 5 at 5:52

You have already stated the intuitive reason: simultaneity is relative, so by shifting $$x'_0$$ (I refuse to call an $$x'$$-coordinate $$x$$-anything without a prime) at fixed $$t'=0$$, the time in the $$x$$-frame changes. Or, moving observers have different hyperplanes of simultaneity, which in layman's terms is: their definitions of "now" differ, and the size of disagreement depends on position.