Time Effects in Separate Frames of Ref.

[Qoz]

Hi, before I question let me say I'm new both to this site and Physics in general, as such please forgive any ignorance on my part.

I was given an example of time difference in regards to a 'my twin traveling at near the speed of light' scenario and I'm having trouble reconciling some of the logic.

In the example my twin leaves Earth traveling at .8c the twin travels 12.5 years out, turns around and returns. For me, from the Earth frame or reference 25 years has passed. For my twin 15 years has passed (from the ships frame of reference). I'm fine up to this point. Where I start falling short is the explanation as to why this is. I was told that it's because Special Relativity requires consistent movement and my twin has had to turn around, therefor no Special Relativity.

This doesn't make sense to me in as much as raises the question, "If I send my twin on a circular course, where the arc of that course will take 25 years at .8c (from the Earth frame of reference) to loop back to it's origin point. Wouldn't the same thing happen?". In this case isn't it still 15 years from my twins frame of reference, though the aspect of turn around and return has been eliminated?

In essence instead of a "out and back" my twin is being sent on a continual curve, under which Special Relativity would still apply.

 

[WannabeNewton]

Your disagreement with the given explanation is certainly justified because of course special relativity still applies! Indeed in your scenario there would still be a discrepancy amongst the twins. Instead of reexplaining the twin paradox here, let me link you to a resource that can do it much more coherently: http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

Welcome to the forums by the way!

 

[ghwellsjr]

Hi, before I question let me say I'm new both to this site and Physics in general, as such please forgive any ignorance on my part.

No forgiveness is necessary. The stated purpose of this forum is to remove ignorance. You've come to the right place.

I was given an example of time difference in regards to a 'my twin traveling at near the speed of light' scenario and I'm having trouble reconciling some of the logic.

In the example my twin leaves Earth traveling at .8c the twin travels 12.5 years out, turns around and returns. For me, from the Earth frame or reference 25 years has passed. For my twin 15 years has passed (from the ships frame of reference). I'm fine up to this point. Where I start falling short is the explanation as to why this is. I was told that it's because Special Relativity requires consistent movement and my twin has had to turn around, therefor no Special Relativity.

What you were told is incorrect, Special Relativity can handle any situation except gravity. Your analysis is perfectly correct.

This doesn't make sense to me in as much as raises the question, "If I send my twin on a circular course, where the arc of that course will take 25 years at .8c (from the Earth frame of reference) to loop back to it's origin point. Wouldn't the same thing happen?".

Yes.

In this case isn't it still 15 years from my twins frame of reference, though the aspect of turn around and return has been eliminated?

Yes, except the turn around hasn't been eliminated, it's just been applied throughout the entire trip.

In essence instead of a "out and back" my twin is being sent on a continual curve, under which Special Relativity would still apply.

Special Relativity applies equally well in both cases.

Here's the general rule: use an Inertial Reference Frame (IRF) to describe your scenario. Start with both twins at the same place. Have one twin travel at a constant speed in any direction until he returns. You divide however long he has been gone according to the IRF by gamma (calculated based on his speed) and you get how much he has aged. It's real simple.