
#235
Oct1008, 05:32 PM

Sci Advisor
P: 8,470

You didn't add this until after I had already responded:




#236
Oct1008, 06:39 PM

Mentor
P: 16,477





#237
Oct1008, 08:29 PM

P: 801

Al 



#238
Oct1008, 08:32 PM

P: 801

Al 



#239
Oct1008, 08:33 PM

Sci Advisor
P: 8,470





#240
Oct1008, 08:40 PM

P: 801

Al 



#241
Oct1008, 08:45 PM

Sci Advisor
P: 8,470





#242
Oct1008, 08:54 PM

P: 801

Al 



#243
Oct1008, 09:32 PM

Sci Advisor
P: 8,470





#244
Oct1008, 09:45 PM

P: 801

Al 



#245
Oct1008, 09:52 PM

Sci Advisor
P: 8,470





#246
Oct1008, 09:56 PM

P: 4

You are reffering to time dilation in an inertial reference frame, but the rocket twin is undergoing acceleration and thus is not in an inertial reference frame.




#247
Oct1008, 10:05 PM

P: 801

The question "why can't you consider things from the traveling twin's perspective instead of the Earthtwin's perspective?" has the same answer either way. Al 



#248
Oct1008, 10:18 PM

P: 645

As I read my paragraph again, I see I left ambiguity. I did indeed mean four simultaneous events, in two pairs. I should have written "two pairs of simultaneous events". My error. Also, to remove less obvious ambiguity, I should point out that the coordinate distance in the other inertial frame is according to the first mentioned "taken to be at rest" frame, and the simultaneous events are simultaneous in the "taken to be at rest frame". I took these latter two to be obvious, but I shouldn't do that. So, rephrasing: Is this less ambiguous and, being less ambiguous, correct? If correct, could we be less exacting in order to convey the concept to the student, and then slowly build up the understanding of the conditions under which the equation is correct? (Sort of how one would explain a rainbow in steps, without at first talking about suspension of droplets of water of the correct size to refract rays of light, or the relative placement of those droplets, or how the image of the rainbow is formed in the eye and is not something external to our perceptions of it.) cheers, neopolitan 



#249
Oct1008, 10:44 PM

Mentor
P: 16,477

I'm too tired to untangle your rephrasing.
Given two (straight) worldlines in some inertial coordinate system the spacetime interval of those lines can be written (using my convention above and units where c=1): St² = Tt²  Xt² Sh² = Th²  Xh² If Tt = Th (e.g. the coordinate time between two pairs of simultaneous events) then St²  Sh² = Xh²  Xt² or St² + Xt² = Sh² + Xh² That is all. Just a little algebra on the spacetime interval formula in the special case where the coordinate times are equal for two straight worldlines. So what? What is the value? How is it in any way preferable to the more general spacetime interval formula? 



#250
Oct1108, 03:08 AM

P: 645

St² = Tt²  Xt² (where Xh is zero and Tt = Th, which is all my convoluted phrasing said) And, yes, it is in the third and fourth equations as well. The point is, I guess, that we are talking about spacetime intervals which are the same, so long as you use a consistent inertial frame. Which gets us back to the original point, if you don't use a consistent inertial frame it doesn't work. This seems to be the one factor common to all approaches for explaining the twin paradox. If you use a consistent inertial frame it works. If you don't, you need to do something extra (which can be to consider simultaneity, or to note that there is a missing amount of time equal in magnitude to the disagreed distance travelled, which can be calculated as shown in an earlier post, or you can show that the line is bent in a spacetime vector diagram). I don't think we are getting any further than that, and I don't really think there is any further to go than that. cheers, neopolitan 



#251
Oct1108, 07:21 AM

Mentor
P: 16,477




Register to reply 
Related Discussions  
twin paradox  Special & General Relativity  43  
Twin Paradox  Special & General Relativity  5  
twin paradox  Special & General Relativity  1  
Twin Paradox  Introductory Physics Homework  39 