Time Dilation vs Differential Aging

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.

But as I pointed out earlier, we can resolve the twin scenario without resorting to frames or coordinates or Time Dilation or anything else associated with Special Relativity, but that would not be teaching Special Relativity, it would be explaining the problem that Special Relativity addresses, that is, how do we make sense of the differing Proper Times on clocks that accelerate differently?

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.

I don't understand; this is fully addressed in SR.

PAllen

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?

tom.stoer

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

[tex]\tau_A = T[/tex]
[tex]\tau_B = \sqrt{1-v^2}\,T = \sqrt{1-v^2}\,\tau[/tex]

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".

arindamsinha

Didn't get this part. First, are you taking c = 1 in this case?

Secondly, do you actually mean the below?
[tex]\tau_B = \sqrt{1-v^2}\,T = \tau[/tex]

tom.stoer

Sorry; yes, c = 1; and
[tex]\tau_B = \sqrt{1-v^2}\,T = \sqrt{1-v^2}\, \tau_A[/tex]