Proposal: Time paradox in Special Relativity Theory.

[Tach]

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?

 

[tashja]

One question: If circular paths are big enough where curvature is no longer noticed; that could be considered a straight path, no?

 

[OnlyMe]

One question: If circular paths are big enough where curvature is no longer noticed; that could be considered a straight path, no?

How would a curved path or acceleration that is small over a long period of time, be different than a curved path or large acceleration over a short period of time?

To return to the same point, both accelerations should return the same time dilations, as far as the accelerations are concerned.

The time dilation relative to velocity alone may and likely would be different, since the velocity could remain equivalent between the two examples while the time changes.

Tach's response is correct, in either case. The example offered looks like the two paths are equivalent so all time dilations would also be equivalent.

 

[Emil]

Quantum Quack, the trajectories diameter is ..hmm...10 light years.
OnlyMe, is discussed only SR and nothing else.

Objects speed is 0.98c.

 

[Tach]

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.

Nothing to do with GR. Nothing to do with equivalence principle. Nothing to do with acceleration, the elapsed time is a function of velocity only, acceleration does not affect it. The simple answer, using SR , is that the proper times for the twins are:

$$\tau_A=\tau_B=\frac{2 \pi}{\omega} \sqrt{1-(\frac{\omega R}{c})^2}$$

 

[Emil]

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

 

[Tach]

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"?

 

[rpenner]

$$\tau_A=\tau_B=\frac{2 \pi}{\omega} \sqrt{1-(\frac{\omega R}{c})^2}$$

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 $$\int_A^B \frac{dt}{d\lambda} d\lambda = \frac{2 \pi R}{v_0} = \frac{2 \pi}{\omega}$$.

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

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.

 

[Emil]

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"?

Try to understand a little physics. Physics is much more than SR.
I used a helper reference system that everyone understand the case.
But there are only two reference frames a and b.
In these reference frames the trajectory and the speed are changed. (must be translate)
But it is the responsibility the one who responds to the challenge, and say who is "old" and who is "young", a or b.

 

[rpenner]

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.

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.

Emil, I agree. It looks like you are in the process of formulating a question, not a debate topic.

 

[rpenner]

But there are only two reference frames a and b.

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 it is the responsibility the one who responds to the challenge, and say who is "old" and who is "young", a or b.

That is not a debate, that is asking a question.

 

[Emil]

rpenner, you made the same mistake with reference frames.
The speed of a relative to b is the same as the speed of b relative to a.

 

[Tach]

Try to understand a little physics.

It is clear that you are the one who doesn't.

Physics is much more than SR.

Sure, but you are a denier of mainstream physics, obsessed with "refuting" relativity.

But it is the responsibility the one who responds to the challenge, and say who is "old" and who is "young", a or b.

It has been explained to you. The fact that you don't understand doesn't come as a surprise.

 

[Tach]

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.

Throwing pearls in front of .....

 

[Emil]

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.

Try to understand. In the case presented by me are only two objects a and b and their corresponding reference systems.

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.

 

[Tach]

It is not a question. It is a case with which I prove that the time paradox in SR is a real paradox.

All you have proven is your crass ignorance. Again.

 

[Emil]

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.

 

[Tach]

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.

Both rpenner and I responded giving you detailed, well documented scientific answers. The fact that you don't understand the answer is not our fault. Your crass ignorance and gross prejudice is at fault.

 

[Tach]

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.

You need to compare $$\sqrt{7599}=87$$ and $$6 \sqrt{11}=19.8$$, so B is only 67 (a) older than A.

 

[RealityCheck]

Hi everyone.

Like OnlyMe, I only wish to make an observation.

In other threads here, posters have distinguished the SR and the GR 'component contributions' to effects on orbiting satellite clocks/time etc.

Since these two 'components' CAN be distinguished, we then can just treat emil's scenario as two well separated EQUALLY MASSED gravitational bodies such that their respective gravity wells do not much affect the 'other one' of the two (a & b) orbiters (in equally periodic/shaped orbits) around their respective bodies, which respective orbits tangentially momentarily coincide as emil illustrated.

Now, if we separate the GR effects from the purely SR effects, since these two effects can apparently BE separated and allowed for accordingly, the scenario can be analyzed in that manner for the SR component effects only?

Just a suggestion. Cheers!