Recognitions:

## Need help understanding the twins

 Quote by cfrogue What was wrong with my original method of the D/c business?
The fact that it doesn't actually give you a way to calculate twin1's final age when twin2 stops accelerating (i.e. the age he is when he receives the light signal, minus D/c) given known values for the variables BT, a, and t that determine twin2's final age. If you think it does, then please show me how you'd calculate a numerical value for twin1's final age given the example values I gave, namely BT=0.5 years, a=2 light years/year^2, and t=5 years.

Again, my method does allow you to calculate twin1's final age given a set of values like this. I can show you the details of the calculation if you're interested.

 At Johns Hopkins as a Jr. Instructor the "twins" paradox was never aproached. Likely there has been two generations of teachers who think there is an "answer" to the paradox. Einstein gave it in 1915; GR. He termed the paradox as "a failure of epistemology." Western 'Greek think" begins with postulates land thru a system of logic arrives at solutions. An unspoken postulate in the twins thing is that the mass environment of neither twin is necessary for fiddling with the SR transformation formulas. Wrong so he thought up GR. If one wants to map wrinkles onto faces of the "twins" one must use the GR equations to the travels of both. Einstein was largely ignored by the media after he kept repeating that immigrants to palestine must get along with Palestinians and that an international police force must have the power to insure the planet that no nuclear weapons existed anywhere. He also insisted that infinities in formulas weren't physical simply imperfect postulates producing math artifacts. He died before the media was allowed to accept infinite densities here and there. I expect to be told I am wrong. Could someone carry this century old "twin" thought to a physicist with whom I can discuss this. Thanks

 Quote by marxmarvelous At Johns Hopkins as a Jr. Instructor the "twins" paradox was never aproached. Likely there has been two generations of teachers who think there is an "answer" to the paradox. Einstein gave it in 1915; GR. He termed the paradox as "a failure of epistemology." Western 'Greek think" begins with postulates land thru a system of logic arrives at solutions. An unspoken postulate in the twins thing is that the mass environment of neither twin is necessary for fiddling with the SR transformation formulas. Wrong so he thought up GR. If one wants to map wrinkles onto faces of the "twins" one must use the GR equations to the travels of both. Einstein was largely ignored by the media after he kept repeating that immigrants to palestine must get along with Palestinians and that an international police force must have the power to insure the planet that no nuclear weapons existed anywhere. He also insisted that infinities in formulas weren't physical simply imperfect postulates producing math artifacts. He died before the media was allowed to accept infinite densities here and there. I expect to be told I am wrong. Could someone carry this century old "twin" thought to a physicist with whom I can discuss this. Thanks
Good, this is not the normal twins paradox.

 Quote by JesseM The fact that it doesn't actually give you a way to calculate twin1's final age when twin2 stops accelerating (i.e. the age he is when he receives the light signal, minus D/c) given known values for the variables BT, a, and t that determine twin2's final age. If you think it does, then please show me how you'd calculate a numerical value for twin1's final age given the example values I gave, namely BT=0.5 years, a=2 light years/year^2, and t=5 years. Again, my method does allow you to calculate twin1's final age given a set of values like this. I can show you the details of the calculation if you're interested.
Yes, I am interested.

No, I cannot give a specific example and you know this.

I would need to actually know the distance between the two when twin2 entered the frame.

For me this does not matter. But for some it does.

So, I am interested how you decide t'.

Recognitions:
 Quote by cfrogue Yes, I am interested.
OK, I'll get to this soon, but first I want to ask a few more questions about how your own approach is supposed to work:
 Quote by cfrogue No, I cannot give a specific example and you know this. I would need to actually know the distance between the two when twin2 entered the frame.
You can calculate the distance between the two in the final rest frame when twin2 finishes his acceleration given BT, a, and t, but it seems to me that even with this information your approach does not tell us who is older. As I said in post #186, if we are analyzing things from the perspective of the inertial frame where they are at rest after acceleration, then in this frame twin1 moves a distance of (c/a)*(cosh(a*BT/c) - 1) during his acceleration phase, after which he is at rest in this frame. Meanwhile we know that twin2 moves inertially for a proper time of (c/a)*sinh(a*BT/c) + t before beginning to accelerate, and in this frame twin2 is moving at constant velocity v=c*tanh(a*BT/c) before beginning to accelerate, so according to the time dilation equation the coordinate time in this frame before twin2 begins to accelerate must be gamma times the proper time, i.e. gamma*[(c/a)*sinh(a*BT/c) + t], and the coordinate distance twin2 covers in this time is just given by velocity*coordinate time, or gamma*c*tanh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]. Then during twin2's acceleration phase, twin2 will cover an additional distance of (c/a)*(cosh(a*BT/c) - 1). So, the total distance twin2 travels from the origin (the point where the two twins first separated) in this frame is:

gamma*c*tanh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] + (c/a)*(cosh(a*BT/c) - 1)

Whereas the total distance twin1 travels from the origin in this frame is:

(c/a)*(cosh(a*BT/c) - 1)

So, subtracting the second from the first shows that the final distance between them will be:

gamma*c*tanh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]

Since gamma is 1/sqrt(1 - v^2/c^2), we can plug in v=c*tanh(a*BT/c), showing that gamma = 1/sqrt(1 - tanh^2(a*BT/c)). And making use of the hyperbolic trig identities here, we known tanh(x) = sinh(x)/cosh(x), so gamma = 1/sqrt(1 - sinh^2/cosh^2) = 1/[sqrt(1/cosh^2)*sqrt(cosh^2 - sinh^2)], and since another identity says that cosh^2 - sinh^2 = 1, this reduces to 1/sqrt(1/cosh^2) = cosh(a*BT/c). So with gamma = cosh(a*BT/c), the final distance between them can be rewritten as:

c*cosh(a*BT/c)*tanh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]

And again making use of the fact that tanh(x) = sinh(x)/cosh(x), this reduces to:

c*sinh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]

So if you plug in BT=0.5 years, a=2 light years/year^2, and t=5 years, this becomes:

sinh(1)*[sinh(1) + 5] = 1.1752011936438014*[1.1752011936438014 + 5] = 7.25710381376082 light years.

But even with the final distance known, can you calculate a numerical value for the final age of twin1?

 Quote by JesseM OK, I'll get to this soon, but first I want to ask a few more questions about how your own approach is supposed to work: You can calculate the distance between the two in the final rest frame when twin2 finishes his acceleration given BT, a, and t, but it seems to me that even with this information your approach does not tell us who is older. As I said in post #186, if we are analyzing things from the perspective of the inertial frame where they are at rest after acceleration, then in this frame twin1 moves a distance of (c/a)*(cosh(a*BT/c) - 1) during his acceleration phase, after which he is at rest in this frame. Meanwhile we know that twin2 moves inertially for a proper time of (c/a)*sinh(a*BT/c) + t before beginning to accelerate, and in this frame twin2 is moving at constant velocity v=c*tanh(a*BT/c) before beginning to accelerate, so the time in this frame before twin2 begins to accelerate must be gamma times the proper time, i.e. gamma*[(c/a)*sinh(a*BT/c) + t], and the distance twin2 covers in this time is just given by velocity*time, or gamma*c*tanh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]. Then during twin2's acceleration phase, twin2 will cover an additional distance of (c/a)*(cosh(a*BT/c) - 1). So, the total distance twin2 travels from the origin (the point where the two twins first separated) in this frame is: gamma*c*tanh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] + (c/a)*(cosh(a*BT/c) - 1) Whereas the total distance twin1 travels from the origin in this frame is: (c/a)*(cosh(a*BT/c) - 1) So, subtracting the second from the first shows that the final distance between them will be: gamma*c*tanh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] Since gamma is 1/sqrt(1 - v^2/c^2), we can plug in v=c*tanh(a*BT/c), showing that gamma = 1/sqrt(1 - tanh^2(a*BT/c)). And making use of the hyperbolic trig identities here, we known tanh(x) = sinh(x)/cosh(x), so gamma = 1/sqrt(1 - sinh^2/cosh^2) = 1/[sqrt(1/cosh^2)*sqrt(cosh^2 - sinh^2)], and since another identity says that cosh^2 - sinh^2 = 1, this reduces to 1/sqrt(1/cosh^2) = cosh(a*BT/c). So with gamma = cosh(a*BT/c), the final distance between them can be rewritten as: c*cosh(a*BT/c)*tanh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] And again making use of the fact that tanh(x) = sinh(x)/cosh(x), this reduces to: c*sinh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] So if you plug in BT=0.5 years, a=2 light years/year^2, and t=5 years, this becomes: sinh(1)*[sinh(1) + 5] = 1.1752011936438014*[1.1752011936438014 + 5] = 7.25710381376082 light years. But even with the final distance known, can you calculate a numerical value for the final age of twin1?

But even with the final distance known, can you calculate a numerical value for the final age of twin1?

Yes, why not?

It is the same type of problem.

I sub off the distance of the twin1 accel phase and then I sub the distance of the twin2 accel phase.

I am left with the distance of the relative motion phase and thus, d/v = t'.

No?

Recognitions:
 Quote by cfrogue I am going to need some time to follow your argument. But even with the final distance known, can you calculate a numerical value for the final age of twin1? Yes, why not? It is the same type of problem. I sub off the distance of the twin1 accel phase and then I sub the distance of the twin2 accel phase. I am left with the distance of the relative motion phase and thus, d/v = t'. No?
Sure, that works. But I thought your calculation method was supposed to involve subtracting D/c from the time twin1 receives the signal--that doesn't appear to happen anywhere in the method above.

My method for calculating twin1's time from the perspective of the final rest frame was basically similar but without the need to consider distances. I would just say that since we know twin2 moved inertially at v=c*tanh(a*BT/c) before beginning to accelerate, and we know twin2 experienced a proper time of [(c/a)*sinh(a*BT/c) + t] before beginning to accelerate, the according to the time dilation equation the coordinate time in this frame before twin2 accelerates would just be gamma times the proper time, i.e. gamma*[(c/a)*sinh(a*BT/c) + t]. As I showed in my previous post, if v=c*tanh(a*BT/c) then gamma=cosh(a*BT/c), so the coordinate time before twin2 accelerates can also be written as cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]. Then the coordinate time for twin2 to accelerate in this frame is (c/a)*sinh(a*BT/c), so the total coordinate time from start to finish is cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] + (c/a)*sinh(a*BT/c).

Now, we know that in this frame it took twin1 a coordinate time of (c/a)*sinh(a*BT/c) to do his own acceleration, so the coordinate time from the end of twin1's acceleration to the end of twin2's acceleration must be:

[cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] + (c/a)*sinh(a*BT/c)] - (c/a)*sinh(a*BT/c)

Which just reduces to cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]...since twin1 was at rest throughout this period, this would also be the proper time for twin1 from the end of his acceleration to the moment when twin2 stopped his own acceleration. And we know the proper time for twin1 during his acceleration was BT, so add them together and we have twin1's total proper time from start to finish:

BT + cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]

Since gamma = cosh(a*BT/c) is greater than 1, this total proper time will naturally be larger than twin2's total proper time from start to finish, i.e. BT + (c/a)*sinh(a*BT/c) + t. So, twin1 will be older than twin2 at the moment twin2 stops his acceleration and we compare their ages in the final rest frame.

 Quote by JesseM Sure, that works. But I thought your calculation method was supposed to involve subtracting D/c from the time twin1 receives the signal--that doesn't appear to happen anywhere in the method above. My method for calculating twin1's time from the perspective of the final rest frame was basically similar but without the need to consider distances. I would just say that since we know twin2 moved inertially at v=c*tanh(a*BT/c) before beginning to accelerate, and we know twin2 experienced a proper time of [(c/a)*sinh(a*BT/c) + t] before beginning to accelerate, the according to the time dilation equation the coordinate time in this frame before twin2 accelerates would just be gamma times the proper time, i.e. gamma*[(c/a)*sinh(a*BT/c) + t]. As I showed in my previous post, if v=c*tanh(a*BT/c) then gamma=cosh(a*BT/c), so the coordinate time before twin2 accelerates can also be written as cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]. Then the coordinate time for twin2 to accelerate in this frame is (c/a)*sinh(a*BT/c), so the total coordinate time from start to finish is cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] + (c/a)*sinh(a*BT/c). Now, we know that in this frame it took twin1 a coordinate time of (c/a)*sinh(a*BT/c) to do his own acceleration, so the coordinate time from the end of twin1's acceleration to the end of twin2's acceleration must be: [cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] + (c/a)*sinh(a*BT/c)] - (c/a)*sinh(a*BT/c) Which just reduces to cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t]...since twin1 was at rest throughout this period, this would also be the proper time for twin1 from the end of his acceleration to the moment when twin2 stopped his own acceleration. And we know the proper time for twin1 during his acceleration was BT, so add them together and we have twin1's total proper time from start to finish: BT + cosh(a*BT/c)*[(c/a)*sinh(a*BT/c) + t] Since gamma = cosh(a*BT/c) is greater than 1, this total proper time will naturally be larger than twin2's total proper time from start to finish, i.e. BT + (c/a)*sinh(a*BT/c) + t. So, twin1 will be older than twin2 at the moment twin2 stops his acceleration and we compare their ages in the final rest frame.
Since you said I am stupid in the other thread, what would make you think I could follow this unless you are stupid.

You are inconsistent.

Recognitions:
 Quote by cfrogue Since you said I am stupid in the other thread, what would make you think I could follow this unless you are stupid. You are inconsistent.
I didn't say you were stupid, I said your argument was stupid. An intelligent person can make a stupid argument if they are too confident of their own correctness and are too quick to be snidely dismissive of the counterarguments made by others and not really pay attention to what people tell them, as seems to be the case with you on all of these threads.

 Quote by JesseM I didn't say you were stupid, I said your argument was stupid. An intelligent person can make a stupid argument if they are too confident of their own correctness and are too quick to be snidely dismissive of the counterarguments made by others and not really pay attention to what people tell them, as seems to be the case with you on all of these threads.
Yes, maybe read this to yourself.

You can already tell I understand SR.

Have you considered yet I might be seeing something?

I think you can see this though.

Anyway, you said I was stupid. No matter, I am not.

But, I am going to look through your length argument here.

I will confess, you seem to have something with it.

Note how I am not so arrogant as to assume I know everything.

 Quote by cfrogue You can already tell I understand SR.
LOL. Are you aware that length contraction is a major part of SR?

 Quote by Al68 LOL. Are you aware that length contraction is a major part of SR?
LOL, what is that?

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