Too Simple & Short Solution of Twin Paradox

[vidwansoni]

Consider two points A and B (Fig.1) at rest in S frame having a distance L0 between them (a long scale may be imagined as placed between A and B). ------ (1)
The ‘traveler’ twin (S-frame) starts from A at time t = 0 and suddenly accelerates to attain a constant velocity v in the view point of the S frame. After reaching B it suddenly decelerates, comes to rest, and returns back. Neglecting the acceleration and deceleration times, the ‘stay-at-home’ twin will check that the ‘traveler’ twin has reached the point B in a time (two synchronized clocks can be placed respectively at points A & B to check this time in S frame; the stay-at home's own frame)
t = L0 /γ (1)
Further in the view point of the ‘traveler’ twin (S1 frame), a ‘Lorentz contracted’ length L0 / (see Figure 2) passes by him (which he can check by comparing with a scale longitudinally placed in his own frame) taking a time shown on his (S1) clock as
t1= L0 /γ v (2)

where γ = (1- v2/c2)-1/2

From equations (1) and (2), we get
t1= t (1- v2/c2)1/2 (3)
Eq. (3) shows that t1 is less than t.
After completing the to and fro journey, the ‘traveler’ twin has spent a total time

T1= 2 t1 = 2 t (1- v2/c2)1/2 = T (1- v2/c2)1/2 (4)

or T1 = T (1- v2/c2)1/2 .
This shows that the ‘traveler twin’ has aged less than the ‘stay-at-home’.

Traveler twin (S1 frame) also checks that he has aged less
In the above discussion the time readings on the clocks positioned at A and B of the relatively moving S frame can also be checked from the view point of the S1 frame. Due to the time dilation effect in the view point of S1 frame, the time t1 on the S clock at A ought to be

t2= t1/γ = L0 /γ2v (5)
Now according to S1 criteria of simultaneity, the S clocks will show a progressive difference of readings (see Fig.2) so that in the view point of S1 frame, the time at the point B1 of the relatively moving S frame comes out to be
t2 = t1 + L0 v/c2 = L0 /γ2v + L0 v/c2 = L0/v = t (6)
which is the same value of time as is read in S frame. So the traveler twin also finds that the time passed in S frame is more than that passed in his own frame.
Hence Proved.
(FOR FIGURES SeeIAPT Physics Bulletin (Kanpur India) Feb 2014; p.32.

 

[ghwellsjr]

Consider two points A and B (Fig.1) at rest in S frame having a distance L0 between them (a long scale may be imagined as placed between A and B). ------ (1)
The ‘traveler’ twin (S-frame) starts from A at time t = 0 and suddenly accelerates to attain a constant velocity v in the view point of the S frame. After reaching B it suddenly decelerates, comes to rest, and returns back. Neglecting the acceleration and deceleration times, the ‘stay-at-home’ twin will check that the ‘traveler’ twin has reached the point B in a time (two synchronized clocks can be placed respectively at points A & B to check this time in S frame; the stay-at home's own frame)
t = L0 /γ (1)

I think you meant:

t = L0 /v

Further in the view point of the ‘traveler’ twin (S1 frame), a ‘Lorentz contracted’ length L0 / (see Figure 2) passes by him (which he can check by comparing with a scale longitudinally placed in his own frame) taking a time shown on his (S1) clock as
t1= L0 /γ v (2)

where γ = (1- v2/c2)-1/2

From equations (1) and (2), we get
t1= t (1- v2/c2)1/2 (3)
Eq. (3) shows that t1 is less than t.
After completing the to and fro journey, the ‘traveler’ twin has spent a total time

T1= 2 t1 = 2 t (1- v2/c2)1/2 = T (1- v2/c2)1/2 (4)

or T1 = T (1- v2/c2)1/2 .
This shows that the ‘traveler twin’ has aged less than the ‘stay-at-home’.

Aside from your one typo, the above looks fine.

Traveler twin (S1 frame) also checks that he has aged less
In the above discussion the time readings on the clocks positioned at A and B of the relatively moving S frame can also be checked from the view point of the S1 frame.

What do you mean by the S1 frame? Is it the inertial frame in which the traveler twin is at rest during the "to" part of the trip? Or is it the inertial frame in which the traveler twin is at rest during the "fro" part of the trip? Or is it a non-inertial frame in which the traveler twin is at rest during the entire trip? Or something else?

Due to the time dilation effect in the view point of S1 frame, the time t1 on the S clock at A ought to be

t2= t1/γ = L0 /γ2v (5)

It would really be helpful if you would use the superscript function in the advanced editing window. Then I think your equation would look like this:

t2= t1/γ = L0 /γ²v

But I don't think this is a good thing to do. You are applying shortcut formulas for Time Dilation when you really need to apply the full Lorentz Transformation formulas.

Now according to S1 criteria of simultaneity, the S clocks will show a progressive difference of readings (see Fig.2) so that in the view point of S1 frame, the time at the point B1 of the relatively moving S frame comes out to be
t2 = t1 + L0 v/c2 = L0 /γ2v + L0 v/c2 = L0/v = t (6)
which is the same value of time as is read in S frame.

I see a problem here. If we look at one of your intermediate terms, we get:

t2 = L0/v

But previously you said:

t2= L0 /γ²v

Which means that they can only both be true when γ=1 which is when v=0.

So the traveler twin also finds that the time passed in S frame is more than that passed in his own frame.
Hence Proved.
(FOR FIGURES SeeIAPT Physics Bulletin (Kanpur India) Feb 2014; p.32.

It's very difficult to understand what you are saying when you refer to figures that are not available. I found this link to your bulletin:

http://indapt.org/index.php/publications

But the latest one is from last year. If you have a hard copy of this publication, I would suggest that you redraw the figures and upload them so that we can see what you are talking about.

 

[vidwansoni]

Dear Sir
I could not put the figures in this (do not know how to do). Please send me your email address so that I send you the article published in IAPT Bulletin.
Thanks
Vidwan Singh Soni (Dr)
444/1, Urban Estate Patiala 147002, India.

 

[vidwansoni]

When you get my article, please upload it on my behalf.
Thanks
V S Soni

 

[ghwellsjr]

Dear Sir
I could not put the figures in this (do not know how to do). Please send me your email address so that I send you the article published in IAPT Bulletin.
Thanks
Vidwan Singh Soni (Dr)
444/1, Urban Estate Patiala 147002, India.

When you get my article, please upload it on my behalf.
Thanks
V S Soni

It's not hard for you to do it yourself. Just click on the "Go Advanced" button below this post and scroll down to the "Manage Attachments" button. After you click the button, it will take you to a dialog box that will allow you to Browse to your file and then Upload it. If you can't do that, then I won't be able to do it either.

 

[vidwansoni]

Too short solution of Twin Paradox

Please see in attachment 2 pages which simply solve the paradox.
V S Soni
Patiala India

 

[ghwellsjr]

Please see in attachment 2 pages which simply solve the paradox.
V S Soni
Patiala India

Thanks for uploading it. I knew you could do it.

Now that I see the original paper, I see what the problem was that I earlier referred to. The problem was that you didn't correctly transcribe from the paper to your post.

Starting with equation (2), you changed t' to t1 and then when you got down to equation (5) where there is a t1, you changed it to t2 which then led to my observation that equation (6) couldn't be right. So your post is all messed up but your paper is consistent.

However, I don't know where you got one of the terms in equation (6):

t₂ = t₁ + L₀v/c² = L₀/γ²v + L₀v/c² = L₀v = t

Where did you get L₀v/c² from?

 

[georgir]

You've not proven anything, because you've not looked at the "paradox" part, namely the supposed symmetry of the situation. That is to say, the traveling twin may believe that he did not in fact move at all, and instead A and B and the other twin did the acceleration/deceleration and movement. He may carry the giant ruler with himself, and consider himself the older and wiser twin of the two :p

 

[vidwansoni]

You have not studied my article properly. See carefully the Thumb nails above.
Vidwan Singh Soni

 

[vidwansoni]

Your contention that the 'traveler' thinks that he is does not move at all is wrong. An accelerated frame always feel his motion!
V S Soni

 

[ghwellsjr]

You have not studied my article properly. See carefully the Thumb nails above.
Vidwan Singh Soni

Maybe not but I'm asking for help. I cannot see any prior reference to your term L₀v/c² until it pops up in equation (6). I don't doubt that the equation is valid, I just don't understand where you got that term that makes it valid.

 

[Nugatory]

Your contention that the 'traveler' thinks that he is does not move at all is wrong. An accelerated frame always feel his motion!
V S Soni

If you haven't already seen http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html, you may want to give it a quick read - you're misunderstanding the point of the paradox.

There are acceleration-free ways of constructing twin paradox situations: For example, you can have the traveler turn around by doing a tight hyperbolic orbit around a massive object; the traveler will remain in free fall for the entire journey, just as does the stay-at-home twin. In that case, there is absolutely no basis for claiming that one of the twins is moving and one isn't.

For that matter, there doesn't have to be any turnaround at all to construct the paradox. Suppose that the outgoing traveler never turns around, keeps on flying away forever. However, at some point on his journey he encounters an inbound spaceship traveling in the opposite direction; and as they pass the inbound traveler sets his clocks to whatever the outbound traveler's watch reads. When inbound meets up with stay-at-home, how do their clocks compare? And how do you reconcile this result with the fact that at every point in the journey, the travelers correctly found that stay-at-home's clock was slow because of time dilation; yet stay-at-home correctly says the same thing about the traveling clocks?