
#19
Jan1313, 04:01 PM

Sci Advisor
P: 5,307





#20
Jan1313, 06:09 PM

Sci Advisor
PF Gold
P: 4,863





#21
Jan1313, 10:18 PM

Sci Advisor
P: 5,307

If observer A stays at rest whereas B moves along a circular curve with constant speed v, then one immediately finds the wellknown result [tex]\tau_A = T[/tex] [tex]\tau_B = \sqrt{1v^2}\,T = \sqrt{1v^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". 



#22
Jan1413, 12:14 AM

P: 181

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



#23
Jan1413, 03:20 PM

Sci Advisor
P: 5,307

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


Register to reply 
Related Discussions  
Gravitational time dilation and special relativity time dilation  Special & General Relativity  9  
Aging and time dilation  Special & General Relativity  3  
Time Dilation, who is aging slower?  Special & General Relativity  10  
Time Dilation and Aging?  Special & General Relativity  6  
Time Dilation and Differential Aging  Special & General Relativity  31 