At the next meeting, which will be "the young" and which will be "the old" ?
Neither. They'll have the same exact age. Is this your "paradox" you were talking about?
At the next meeting, which will be "the young" and which will be "the old" ?
One question: If circular paths are big enough where curvature is no longer noticed; that could be considered a straight path, no?
Is your intent that the debate is restricted to the conditions of SRT alone or will GR come into the debate?
The reason I ask is that in the absence of gravitational fields, the circular paths involve not only speed or velocity, but a constant acceleration defined by the curved paths. While SRT can deal with acceleration, it is far more easily dealt with, from the context of GR and the equivalence principle.
Ok, let's make it asymmetric.
The diameter of trajectory a is 10 light years.
The diameter of trajectory b is 20 light years.
Speed of a 0.98 c
Speed of b 0.49 c
I concur, as this is a simplification of $$\tau = \int_A^B \sqrt{1 - \left( \frac{\vec{v}(\lambda)}{c} \right)^2} \frac{dt}{d\lambda} d\lambda$$ for the case that $$\left| \vec{v}(\lambda) \right| = v_0 = \omega R$$ and $$\int_A^B \frac{dt}{d\lambda} d\lambda = \frac{2 \pi R}{v_0} = \frac{2 \pi}{\omega}$$.$$\tau_A=\tau_B=\frac{2 \pi}{\omega} \sqrt{1-(\frac{\omega R}{c})^2}$$
Ok, let's make it asymmetric.
The diameter of trajectory a is 10 light years.
The diameter of trajectory b is 20 light years.
Speed of a 0.98 c
Speed of b 0.49 c
1. You don't get to move the goalposts.
2. Try to learn a little physics:
$$\tau=\frac{2 \pi}{\omega} \sqrt{1-(\frac{\omega R}{c})^2}=\frac{2 \pi R}{v} \sqrt{1-(\frac{v}{c})^2}$$
3. Where is the "paradox"?
I want to show that there is time paradox in SRT. I will present a concrete situation where is time paradox.
I would like to accept, but you haven't met the threshold of the "Formal Debates" forum.
The case X I will present in the debate.
At this time, X is completely mysterious. It may involve special relativity or it may be off that intended topic. So while it is improper to discuss X in this thread prior to debate, defining X I think is essential if you want anyone to accept your challenge. And in every case of anti-relativity claims being made before, it will be essential for me to invest time explaining the math and/or logic of special relativity for a true meeting of the minds to occur.
Emil, I agree. It looks like you are in the process of formulating a question, not a debate topic.As far as I can see, you have not suggested a topic for debate at this stage.
Please post one, or I'll close this thread.
No -- there are two travelers, a and b. The only reference frame (coordinate system for space and time) used was that where the speed of the travelers and the diameters of their circular trajectories was described.But there are only two reference frames a and b.
That is not a debate, that is asking a question.But it is the responsibility the one who responds to the challenge, and say who is "old" and who is "young", a or b.
Try to understand a little physics.
Physics is much more than SR.
But it is the responsibility the one who responds to the challenge, and say who is "old" and who is "young", a or b.
I concur, as this is a simplification of $$\tau = \int_A^B \sqrt{1 - \left( \frac{\vec{v}(\lambda)}{c} \right)^2} \frac{dt}{d\lambda} d\lambda$$ for the case that $$\left| \vec{v}(\lambda) \right| = v_0 = \omega R$$ and $$\int_A^B \frac{dt}{d\lambda} d\lambda = \frac{2 \pi R}{v_0} = \frac{2 \pi}{\omega}$$.
For one cycle each we have:
$$\tau_A = \frac{\pi \cdot 10 \, \textrm{a} \cdot c}{0.98 \, c} \sqrt{ 1 - \left( \frac{0.98 \, c}{c} \right)^2 } = \frac{500 \pi}{49} \times \frac { 3 \sqrt{11}}{50} \, \textrm{a} = \frac{ 30 \pi \sqrt{11}}{49} \, \textrm{a} \approx 6.4 \, \textrm{a} $$
$$\tau_B = \frac{\pi \cdot 20 \, \textrm{a} \cdot c}{0.49 \, c} \sqrt{ 1 - \left( \frac{0.49 \, c}{c} \right)^2 } = \frac{2000 \pi}{49} \times \frac {\sqrt{7599}}{100} \, \textrm{a} = \frac{ 20 \pi \sqrt{7599}}{49} \, \textrm{a} \approx 110 \, \textrm{a} $$
But it takes 4 cycles around A to catch up to the same position and time of 1 cycle around B, so compare:
$$4 \tau_A = \frac{ 120 \pi \sqrt{11}}{49} \, \textrm{a} < \frac{ 20 \pi \sqrt{7599}}{49} \, \textrm{a} = \tau_B$$.
So B, which travelled a path through space-time that was significantly closer to inertial motion than A, is about 86 years (annum = $$\textrm{a}$$) older than A when they meet again.
Try to understand. In the case presented by me are only two objects a and b and their corresponding reference systems.No -- there are two travelers, a and b. The only reference frame (coordinate system for space and time) used was that where the speed of the travelers and the diameters of their circular trajectories was described.
It is not a question. It is a case with which I prove that the time paradox in SR is a real paradox.That is not a debate, that is asking a question.
It is not a question. It is a case with which I prove that the time paradox in SR is a real paradox.
For that I did not want to present the case, it will not go to ridiculous the discussion.
If you are willing, then formally respond to the debate.
But it takes 4 cycles around A to catch up to the same position and time of 1 cycle around B, so compare:
$$4 \tau_A = \frac{ 120 \pi \sqrt{11}}{49} \, \textrm{a} < \frac{ 20 \pi \sqrt{7599}}{49} \, \textrm{a} = \tau_B$$.
So B, which travelled a path through space-time that was significantly closer to inertial motion than A, is about 86 years (annum = $$\textrm{a}$$) older than A when they meet again.