# Time Dilation vs Differential Aging

by arindamsinha
Tags: aging, differential, dilation, time
P: 5,435
 Quote by ghwellsjr But in post #4 you said, "one introduces an inertial frame S with coordinate time t" and proceeded to use an equation with t in it, so I'm confused.
Yes, in order to make contact with well-known formulas. The invariant length of a curve C can be defined w/o referring to coordinates. And to explain how the 'would-be paradox is resolved' one only needs a), i.e. no coordinates.

 Quote by ghwellsjr ... how do we make sense of the differing Proper Times on clocks that accelerate differently?
I don't understand; this is fully addressed in SR.
PF Gold
P: 5,059
 Quote by tom.stoer Yes, in order to make contact with well-known formulas. The invariant length of a curve C can be defined w/o referring to coordinates. And to explain how the 'would-be paradox is resolved' one only needs a), i.e. no coordinates.
You can define it without coordinates, and derive properties without coordinates, but I'm not clear how you could compute anything without coordinates. Is there a way?
P: 5,435
 Quote by PAllen ?... but I'm not clear how you could compute anything without coordinates. Is there a way?
It depends what you mean.

If observer A stays at rest whereas B moves along a circular curve with constant speed v, then one immediately finds the well-known result

$$\tau_A = T$$
$$\tau_B = \sqrt{1-v^2}\,T = \sqrt{1-v^2}\,\tau$$

Note that I haven't introduced any specific spatial coordinates. But I had to introduce at least time t, and I don't see a way to derive this result w/o using any coordinate at all. So strictly speaking my answer to your question is "no".

I would say that this holds even in GR; one can derive many results w/o specific charts, but (in the framewok of differential geometry) one always uses the existence of charts, so again strictly speaking the answer is "no".
P: 181
 Quote by tom.stoer ... $$\tau_A = T$$ $$\tau_B = \sqrt{1-v^2}\,T = \sqrt{1-v^2}\,\tau$$ ...
Didn't get this part. First, are you taking c = 1 in this case?

Secondly, do you actually mean the below?
$$\tau_B = \sqrt{1-v^2}\,T = \tau$$
 Sci Advisor P: 5,435 Sorry; yes, c = 1; and $$\tau_B = \sqrt{1-v^2}\,T = \sqrt{1-v^2}\, \tau_A$$

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