Evidence for a Young Earth (Not Billions of Years Old)

[Justatruthseeker]

When he measures A coming toward him at 60% of c.

Lol..... The the best one I have heard so far......

At the distance we are talking about and the velocities, by the time he gets the measurement back he will have already exceeded that velocity..... Not to mention that would require a laser that is capable of reaching light years without diffraction or scattering..... By the time the first beam gets back he will be close to 5/8 to 3/4 of the trip home and still accelerating..... Which of course will tell him he was stationary since he would have to send that beam out at the start....

I've already been generous and given you instantaneous acceleration and deceleration so you could count the entire outbound and return legs as inertial. Don't press my patience by asking for faster than c detection....

 

[Polymath257]

So twin B can not perceive reality. Understood.

There is no velocity without first there being acceleration....

*sigh* I was thinking in terms of twin A going past twin A, continuing on for 10years (in A's frame) or 8 years (in B's frame), then applying rockets, and then coasting back (in A's frame) or having A coast back (in B's frame). Only one period of acceleration is then required.

I am also assuming (to simplify matters) that the acceleration is instantaneous. If you want to add in a period of acceleration, then a much more complicated coordinate frame is required for B. That acceleration means that no single inertial frame will work.

Except why would twin B do an LT for himself? He can't decipher his speed, his speedometer says zero.

And it *is* zero with respect to himself. But he also changes inertial frames.

Every reference point he sees is moving away or towards him at different velocities, so can not decipher it from that. Twin A's clocks are slow, even when they never changed, so he can't use their clock. His clock hasn't changed, even when it has, so there is no need to adjust for time dilation, not that he would know how much has ocurred since he can't decipher his velocity......

He can pick a single inertial frame (he goes through two, at least) and do all measurements and calculations in that inertial frame.

twin B will feel an acceleration as he fires his engines to actually slow down from his first acceleration. then another acceleration as he accelerates back home. then another as he accelerates to decelerate. To him there is no difference between accelerating by firing his engines as there is to decelerate by firing his engines. Both feel as accelerations....

I was thinking of only the turn-around acceleration, putting the speed up and slow down to before and after the 'clocks start'.

no twin B can not. He thinks twin A's clocks run slower.... Only by doing the calculations from Twin A's frame can you di this. Which is exactly why you did do that. You can't help yourself of thinking of it as an absolute frame....

I did an example of just that when I did everything from the outgoing frame. Again, in that, B is at rest for 8 years and then accelerates to 88.2% of c to catch up with A, who is moving away at 60% of c.

In that frame, A has aged 6.4 years at the point that B moves out of that frame and accelerates. Catching up with A takes 17 years (in the outgoing frame), which ages A (by time dilation) by 13.6 years and B (again, by a different time dilation) by 8 years. So when they finally meet again, A has aged by 20 years and B by 16 years total.

Now, where in this did I assume A was

Twin B experiences one at the start and one when he decelerates at the turn around. Then another as he decelerates to land on earth.....

Why would they have different velocities.Their speed (velocity is the same). Is he for some reason incapable of obtaining the same speed on his return half that he did on his outgoing half????? Direction is irrelevant..... Your contradictions are clear in that they can't have different velocities and yet have the same speed.

No, direction is NOT irrelevant! Different directions lead to different frames.

We will ignore the time needed for acceleration and deceleration.....

I was also. A LOT of acceleration, to be sure, but instantaneous.

So we instantly accelerated to 60% of c????

Well, I was having him go past Earth on the way out and on the way back, taking the times when he is at the same position as twin A. For the turn around, I was thinking instantaneous acceleration from 60% outgoing to 605 return (in A's frame) or from 0 to 88.2% B's outward frame).

GR is irrelevant except when the twin lands on earth and we consider the minuscule slowing of his clock due to gravity. At great speeds 60% of light and in the micro-gravity of space, it can be virtually ignored...

Special relativity - Wikipedia

"As of today, special relativity is the most accurate model of motion at any speed when gravitational effects are negligible."

"As Galilean relativity is now considered an approximation of special relativity that is valid for low speeds, special relativity is considered an approximation of general relativity that is valid for weak gravitational fields, i.e. at a sufficiently small scale (for tidal forces) and in conditions of free fall."

I do believe that the second his engines cut off he is in free fall..... Take the excuses elsewhere....

Like I said, he is in a local Lorentz frame then.

Agreed because the free fall is the result of gravity from curved spacetime... not a force as from acceleration....

Which is why GR is required: to deal with curvature.

We've already established he can't do the calculations from his frame as his speedometer reads as zero and he has no references to deduce his actual velocity. he doesn't even know whn to shut off his engines when he reaches 60% of c....

Yes, when you used A as the absolute frame.....

When did I use A's frame as absolute above?

 

[Polymath257]

Lol..... The the best one I have heard so far......

At the distance we are talking about and the velocities, by the time he gets the measurement back he will have already exceeded that velocity..... Not to mention that would require a laser that is capable of reaching light years without diffraction or scattering..... By the time the first beam gets back he will be close to 5/8 to 3/4 of the trip home and still accelerating..... Which of course will tell him he was stationary since he would have to send that beam out at the start....

I've already been generous and given you instantaneous acceleration and deceleration so you could count the entire outbound and return legs as inertial. Don't press my patience by asking for faster than c detection....

OK, so he precomputes how much to fire his engines to go from a speed of 0 (in outgoing frame) to 88.2% of c towards where A is (which is what he needs to catch up with A at a relative velocity of 60% of c--something he can compute because he knows A is moving away from him at 60% of c and he wants a return relative velocity of 60%).

 

[Astrophile]

But we are not stationary in this frame of reference. This frame of reference is spinning around the earth at 1,000 mph, orbiting the sun at 67,000 mph, which is orbiting the galaxy at 514,000 mph, which is itself moving through space....

So you understand your devices do not reflect reality, just your perception of it.....

At present, I am stationary in the frame of reference of the Earth, and the Sun is moving across the sky. If I were sitting in a train, I should be stationary in the frame of reference of the train and the Earth would be moving beneath me. In the frame of reference of the Moon, I am on a planet that is rotating at 1000 mph; in the frame of reference of the Sun I am on a planet that is both rotating on its own axis and in orbit around the Sun; in the frame of reference of the Galaxy, I am on a rotating planet that is orbiting a star that is itself in orbit around the centre of the Galaxy. As Polymath keeps saying, it all depends on one's frame of reference.

 

[Astrophile]

Care to calculate the acceleration for our orbit around the galaxy?

Assuming a mass for the Galaxy of 400 billion solar masses (8×10^41 kg) and a distance from the Galactic centre of 7900 parsecs, I find an acceleration of 0.9 nm/s².

 

[Astrophile]

Yes, B's view is equally valid. But B undergoes an acceleration, meaning there is a frame difference. Twin A does NOT have an acceleration, so stays in the same frame the whole time. Velocity is relative, acceleration is not.



I did NOT say that changing frames was the reason for the time dilation. If that's what you got from what I wrote, read it again. But, you have to either stay in the same frame for the whole calculation, OR do a LT to go from one frame to another.



At the midpoint of the story for B, there is an acceleration. Twin B will feel the effects of that acceleration as a force. Twin A feels no such force, so is not accelerated, so stays in the same frame the whole time. It really is that simple.

The proper time as experienced by either twin is an objective thing. And both twins can correctly determine the proper time elapsed for either twin. This can also be done in any other inertial frame you want.



Twin B only experiences an acceleration at the turn around and is moving at constant velocity otherwise.


Simply false. The outgoing and the return motions are both uniform: they are constant velocities. That means they are both inertial frames. They have different velocities (different directions, same speed), and so different frames.


OK, you clearly don't know what it means to be in an inertial frame. ALL that is required is that it be moving without acceleration.



Yes. But to do it correctly requires GR and not just SR.



And yes, that would be a locally Lorentz frame. The frame 'at rest' on the surface of the Earth is NOT a Lorentz inertial frame, although it is a good approximation for many situations. But, again, that is a topic for GR, not for SR.

Once again, get off the PR of frame switching. Experiments have already falsified that as being the cause of clocks slowing.

I never said it was the cause of clocks slowing. I said the change of frame is required to be able to do certain versions of the calculations correctly.

I showed how to find the aging of both twins from the outgoing frame (which twin B is in for a while and that is the difference between the two that highlights that frames are changing for B but not for A.

I once asked 'justatruthseeker' a question about the twin paradox, which he never answered. I think that I know the answer, but perhaps you can confirm that I am right or correct me if I am wrong.

Suppose that both twins have a set of clocks, including radiometric clocks using chromium-51 (half-life of 27.5 days) and cobalt-57 (half-life of 270 days). Will twin B's menstrual cycles and pregnancies last the same time according to her clocks as twin A's will according to her Earthbound clocks? In particular, will twin B's menstrual cycles last one half-life of the chromium-51 on the spacecraft, and will her pregnancies last one half-life of the cobalt-57? As I understand it, the answer to both questions is yes, but I know little about the mathematics of relativity, and I should like to have an answer from somebody with qualifications. Also, what will happen at the turn-around point, when B (call her Barbara) changes inertial frames?

 

[tas8831]

You didn't ask if it was "provable", you asked if it was "scientific". Those are not the same.

Goal post shifting? I'm shocked...

 

[tas8831]

My position is that there's ample evidence that the big bang model of the early history of the universe is accurate.


No, an event does not have to be directly observed before it can be scientifically investigated. Scientists investigate all sorts of events that aren't directly observed, e.g., those that are in the distant past, those that take place over long periods of time, or very large-scale events.

Even 'one-off' events that were not observed directly, like volcanic eruptions.

 

[Polymath257]

I once asked 'justatruthseeker' a question about the twin paradox, which he never answered. I think that I know the answer, but perhaps you can confirm that I am right or correct me if I am wrong.

Suppose that both twins have a set of clocks, including radiometric clocks using chromium-51 (half-life of 27.5 days) and cobalt-57 (half-life of 270 days). Will twin B's menstrual cycles and pregnancies last the same time according to her clocks as twin A's will according to her Earthbound clocks?

yes. Both will have the usual duration of menstrual cycle and pregnancy as measured by their own clocks.

In particular, will twin B's menstrual cycles last one half-life of the chromium-51 on the spacecraft, and will her pregnancies last one half-life of the cobalt-57?

yes, one half-life.

As I understand it, the answer to both questions is yes, but I know little about the mathematics of relativity, and I should like to have an answer from somebody with qualifications. Also, what will happen at the turn-around point, when B (call her Barbara) changes inertial frames?

That depends one the exact mechanics of the turn-around. How much acceleration and for how long.

A good way to view a lot of this is by analogy with rotations in the plane. Imagine a xy grid. This grid (coordinate system) can describe every point in the plane by two numbers: an x-coordinate and a y-coordinate. If you have the x and y coordinates of two points, you can compute the distance between then using the Pyhtagorean formula.

In this analogy, the x coordinate is analogous to space and the y coordinate is analogous to time. The distance between points is analogous to 'proper time'.

Now, imagine a second grid, centered at the same place, but rotated with respect to the first. Call the first grid A and the second grid B. The coordinates for most points (all expect the center) will be different in the two grids. The grid B coordinates and the grid A coordinates will often be quite t.

But, you can still find the distance between any two points using the grid B coordinates and the Pythagorean formula.

Now, suppose you take a horizontal line segment in the A grid. Say it has length L, so the difference in the x coordinates is L and there is no difference in the y coordinates. Now look at that same line segment in the B grid. In *that* graid, the difference in x coordinates is smaller than L---there is a length contraction. And now, there is a difference in the y coordinates: there is a violation of simultaneity.

Similarly, if we take a line segment in the B grid and look at it in the A grid, there is also length contraction and lack of simultaneity. And, we can do the same with veritcal line segments to show time dillation (although in the analogy, it is contraction--the specific formulas used in SR are different than euclidean geometry).

Now, a 'frame' is simply a grid-like coordinate system, possibly rotated with respect to A or B or both. All frames give different results for x and y coordinates, but all also give the same results for distance between points. All are related by a rotation and maybe a translation.

Now, suppose in this plane, we have two twins. One twin, A, has a path in the plane that is a single vertical line in the A grid. The other has a path that consists of a broken line starting at the same place A starts going a bit to the side and then returning back to where A stops.

Several questions:
1. Is the path length for A longer or shorter than that for B?
2. Does the length of the two paths depend on which grid is used?
3. Is there any one, single grid in which B is always moving in the y direction?

I think it is easy enough to see that the answers are:
1. A's path is shorter than B's.
2. This doesn't depend on any particular grid used to describe the paths.
3. No single grid will work for all of B's path: you can get grids for the first part or the second part, but none for both.

Now, this is an analogy to what happens with the relativistic twins. There are some major differences:
1. instead of distance, the 'invariant' is proper time. ALL frames will compute the same proper time.
2. Instead of the Pythagorean formula, there is a Minkowskian formula, where the square of the time minus the square of the distance is the square of the proper time.
3. because of this, longer paths in the usual distance typically have *shorter* proper times.

So, we have two twins: one (A) has a path always in the y direction for some grid (his frame of reference). The other (B) moves away and then returns. The difference is that no single frame will work for B: that break in the path prevents a single grid from always having B's path in the y direction. Instead, B need two different grids for that.

None the less, the path lengths (proper times) can be computed correctly in any grid *as long as we use coordinates from that grid). There is a basic assymetry between the two twins because A's path is always in the y direction of some grid, but that is not the case for B. This is partly why the euclidean distance for B is longer than that for A. In the analogy, the proper time for B is smaller than the proper time for A. In other words, B ages less than A.

B 'changes frames' while A does not.

 

[Justatruthseeker]

OK, so he precomputes how much to fire his engines to go from a speed of 0 (in outgoing frame) to 88.2% of c towards where A is (which is what he needs to catch up with A at a relative velocity of 60% of c--something he can compute because he knows A is moving away from him at 60% of c and he wants a return relative velocity of 60%).

No he doesn't. He doesn't know how fast he is going in the least. If he starts at 0 (stationary to A) then he can not compute how long to fire hist rockets because A is going 0 relative to him. It's even worse in the outgoing frame as by the time light catches up to him he will have again surpassed 60% of c and still be accelerating on the outward journey. Then he won't know when to turn around because by the time he does figure out his velocity with respect to A, he will have overshot his intended distance....

Traveling at anything in fractions of c is quite beyond our ability to hit any distance accurately. Trying to travel to the nearest star at 60% of c (even if we could reach that velocity) would most likely result in our smashing right into the star. We would have to slow to but fractions of a percent before we even got close, so we could get distance measurements correct.

It is clear people on here do not have the faintest clue as to why light remains c regardless of velocity....

 

[Subduction Zone]

No he doesn't. He doesn't know how fast he is going in the least. If he starts at 0 (stationary to A) then he can not compute how long to fire hist rockets because A is going 0 relative to him. It's even worse in the outgoing frame as by the time light catches up to him he will have again surpassed 60% of c and still be accelerating on the outward journey. Then he won't know when to turn around because by the time he does figure out his velocity with respect to A, he will have overshot his intended distance....

Traveling at anything in fractions of c is quite beyond our ability to hit any distance accurately. Trying to travel to the nearest star at 60% of c (even if we could reach that velocity) would most likely result in our smashing right into the star. We would have to slow to but fractions of a percent before we even got close, so we could get distance measurements correct.

It is clear people on here do not have the faintest clue as to why light remains c regardless of velocity....

Why do you think that distances cannot be calculated? To keep it simpler let's assume that acceleration takes a very small part of the time of a trip so it can be treated as being instantaneous. Why could a person not measure distances and relative velocities and make calculations?

 

[Justatruthseeker]

yes. Both will have the usual duration of menstrual cycle and pregnancy as measured by their own clocks.


yes, one half-life.



That depends one the exact mechanics of the turn-around. How much acceleration and for how long.

A good way to view a lot of this is by analogy with rotations in the plane. Imagine a xy grid. This grid (coordinate system) can describe every point in the plane by two numbers: an x-coordinate and a y-coordinate. If you have the x and y coordinates of two points, you can compute the distance between then using the Pyhtagorean formula.

In this analogy, the x coordinate is analogous to space and the y coordinate is analogous to time. The distance between points is analogous to 'proper time'.

Now, imagine a second grid, centered at the same place, but rotated with respect to the first. Call the first grid A and the second grid B. The coordinates for most points (all expect the center) will be different in the two grids. The grid B coordinates and the grid A coordinates will often be quite t.

But, you can still find the distance between any two points using the grid B coordinates and the Pythagorean formula.

Now, suppose you take a horizontal line segment in the A grid. Say it has length L, so the difference in the x coordinates is L and there is no difference in the y coordinates. Now look at that same line segment in the B grid. In *that* graid, the difference in x coordinates is smaller than L---there is a length contraction. And now, there is a difference in the y coordinates: there is a violation of simultaneity.

Similarly, if we take a line segment in the B grid and look at it in the A grid, there is also length contraction and lack of simultaneity. And, we can do the same with veritcal line segments to show time dillation (although in the analogy, it is contraction--the specific formulas used in SR are different than euclidean geometry).

Now, a 'frame' is simply a grid-like coordinate system, possibly rotated with respect to A or B or both. All frames give different results for x and y coordinates, but all also give the same results for distance between points. All are related by a rotation and maybe a translation.

Now, suppose in this plane, we have two twins. One twin, A, has a path in the plane that is a single vertical line in the A grid. The other has a path that consists of a broken line starting at the same place A starts going a bit to the side and then returning back to where A stops.

Several questions:
1. Is the path length for A longer or shorter than that for B?
2. Does the length of the two paths depend on which grid is used?
3. Is there any one, single grid in which B is always moving in the y direction?

I think it is easy enough to see that the answers are:
1. A's path is shorter than B's.
2. This doesn't depend on any particular grid used to describe the paths.
3. No single grid will work for all of B's path: you can get grids for the first part or the second part, but none for both.

Now, this is an analogy to what happens with the relativistic twins. There are some major differences:
1. instead of distance, the 'invariant' is proper time. ALL frames will compute the same proper time.
2. Instead of the Pythagorean formula, there is a Minkowskian formula, where the square of the time minus the square of the distance is the square of the proper time.
3. because of this, longer paths in the usual distance typically have *shorter* proper times.

So, we have two twins: one (A) has a path always in the y direction for some grid (his frame of reference). The other (B) moves away and then returns. The difference is that no single frame will work for B: that break in the path prevents a single grid from always having B's path in the y direction. Instead, B need two different grids for that.

None the less, the path lengths (proper times) can be computed correctly in any grid *as long as we use coordinates from that grid). There is a basic assymetry between the two twins because A's path is always in the y direction of some grid, but that is not the case for B. This is partly why the euclidean distance for B is longer than that for A. In the analogy, the proper time for B is smaller than the proper time for A. In other words, B ages less than A.

B 'changes frames' while A does not.

Distance is quite relevant for the twins. B's ruler is shorter than A's. B sees a greater distance than does A. So what is 1 light year in respect to A, is greater than 1 light year to B. This is why both A and B would see light reach the same location at the same time. B does not find it surprising that it takes longer than 1 light year to reach the target by his slower clock, because to B the target is greater than 1 light year.... This is simplified and does not take into account the shifting of our zero points.....

Until you understand why c is always c regardless of velocity, such things will always confuse you..... You will confuse both times and distances as being the same proper time and distance when they are not...... and continue to talk about frames when they are really just a distraction....

 

[Subduction Zone]

Distance is quite relevant for the twins. B's ruler is shorter than A's. B sees a greater distance than does A. So what is 1 light year in respect to A, is greater than 1 light year to B. This is why both A and B would see light reach the same location at the same time. B does not find it surprising that it takes longer than 1 light year to reach the target by his slower clock, because to B the target is greater than 1 light year....

Until you understand why c is always c regardless of velocity, such things will always confuse you..... You will confuse both times and distances as being the same proper time and distance when they are not...... and continue to talk about frames when they are really just a distraction....

Inside out and backwards. That you do not understand inertia frames of reference is what is causing your problems understanding this concept. They are hardly a distraction. Why not take some time to learn?

 

[Justatruthseeker]

Why do you think that distances cannot be calculated? To keep it simpler let's assume that acceleration takes a very small part of the time of a trip so it can be treated as being instantaneous. Why could a person not measure distances and relative velocities and make calculations?

When does he reach 60% of c? On the outward trip at 60% of c (to him stationary) he will have sent out a pulse of light. This means he has covered 60% of one light year to frame A when the pulse reaches him. By the time the light pulse catches back up to twin B he has exceeded 60% of c and gone another 45% or so or roughly a little over one light year. But..... that light pulse only carries the information sent when it was sent at 60%, in the meantime he has continued to increase in velocity, waiting for the pulse to tell him his velocity. Since it takes but a short time to accelerate he has probably reached 80% of c by now..... It is only slightly less problematic on the return, since he is decreasing the distance to A not increasing it, therefore light will return slightly sooner....

 

[Subduction Zone]

When does he reach 60% of c? On the outward trip at 60% of c (to him stationary) he will have sent out a pulse of light. This means he has covered 60% of one light year to frame A when the pulse reaches him. By the time the light pulse catches back up to twin B he has exceeded 60% of c and gone another 45% or so or roughly a little over one light year. But..... that light pulse only carries the information sent when it was sent at 60%, in the meantime he has continued to increase in velocity, waiting for the pulse to tell him his velocity. Since it takes but a short time to accelerate he has probably reached 80% of c by now..... It is only slightly less problematic on the return, since he is decreasing the distance to A not increasing it, therefore light will return slightly sooner....

It does not matter "when". We are going to treat this as a case where the time to accelerate is very short to keep it simple. Also one does not measure distance form one frame to another. That makes no sense. One measures distance in one's own frame. One can calculate the distance that another would observe in his frame. For now let's treat velocities as instantaneous. Start with the simple and add complicating factors later. That is the way to understand some problems.

But to make it easy let's say that he is traveling from star to star which have a velocity of zero relative to each other. If he does that he can get his relative velocity to the star that he is approaching instantaneously.

 

[Justatruthseeker]

I once asked 'justatruthseeker' a question about the twin paradox, which he never answered. I think that I know the answer, but perhaps you can confirm that I am right or correct me if I am wrong.

Suppose that both twins have a set of clocks, including radiometric clocks using chromium-51 (half-life of 27.5 days) and cobalt-57 (half-life of 270 days). Will twin B's menstrual cycles and pregnancies last the same time according to her clocks as twin A's will according to her Earthbound clocks? In particular, will twin B's menstrual cycles last one half-life of the chromium-51 on the spacecraft, and will her pregnancies last one half-life of the cobalt-57? As I understand it, the answer to both questions is yes, but I know little about the mathematics of relativity, and I should like to have an answer from somebody with qualifications. Also, what will happen at the turn-around point, when B (call her Barbara) changes inertial frames?

No. B's cycle will last longer because B's radiometric clocks decay slower...... B just still calls a different duration tick of time a second, and so THINKS they are the same duration as they were before.

The twin ages slower not because his clocks tick slower, but because every atom in his body is decaying at a slower rate. Including the chromium-51 and cobalt-57 on his ship.

Yes, he THINKS it is the same rate, but it isn't. He can not perceive the changes to his clocks or decay rates and so simply continues to call longer ticks of time seconds.... his 86,400 seconds take longer than A's 86,400 seconds.....

No one is arguing B doesn't THINK it remains the same because he can't perceive the changes to his decay rates. He just keeps calling a longer duration tick of time a second, as A calls a shorter tick of time a second. They are not of the same duration..... What A calls a second does not equal what B calls a second....

 

[Subduction Zone]

No. B's cycle will last longer because B's radiometric clocks decay slower...... B just still calls a different duration tick of time a second, and so THINKS they are the same duration as they were before.

The twin ages slower not because his clocks tick slower, but because every atom in his body is decaying at a slower rate. Including the chromium-51 and cobal-57 on his ship.

Yes, he THINKS it is the same rate, but it isn't. He can not perceive the changes to his clocks or decay rates and so simply continues to call longer ticks of time seconds.... his 86,400 seconds take longer than A's 86,400 seconds.....

No one is arguing B doesn't THINK it remains the same because he can't perceive the changes to his decay rates. He just keeps calling a longer duration tick of time a second, as A calls a shorter tick of time a second. They are not of the same duration.....

Now you are back to Flat Earth thinking again. There is no preferred frame of reference. In A's frame of reference B's clock is moving more slowly than his. In B's frame of reference A's clock is moving more slowly than his. Both see their own clocks as moving a regular speed. This is why understanding frames of reference is key to understanding relativity. B's 84,000 seconds take longer for A but at the same time A's 84,000 seconds take more time for B. Time is frame dependent.

 

[Polymath257]

No he doesn't. He doesn't know how fast he is going in the least. If he starts at 0 (stationary to A) then he can not compute how long to fire hist rockets because A is going 0 relative to him.

In the outgoing frame, yes. And he can compute in that frame how long he will have to fire his rockets to get to 88.2% of c. That is how long he does so.

It is clear people on here do not have the faintest clue as to why light remains c regardless of velocity....

It does so because the laws of physics are Lorentz invariant.

 

[Polymath257]

No he doesn't. He doesn't know how fast he is going in the least. If he starts at 0 (stationary to A) then he can not compute how long to fire hist rockets because A is going 0 relative to him. It's even worse in the outgoing frame as by the time light catches up to him he will have again surpassed 60% of c and still be accelerating on the outward journey. Then he won't know when to turn around because by the time he does figure out his velocity with respect to A, he will have overshot his intended distance....

Traveling at anything in fractions of c is quite beyond our ability to hit any distance accurately. Trying to travel to the nearest star at 60% of c (even if we could reach that velocity) would most likely result in our smashing right into the star. We would have to slow to but fractions of a percent before we even got close, so we could get distance measurements correct.

]

Once again, it is clear you don't comprehend what an inertial frame is. Twin B can determine that twin A is moving at 60% of c away from him. he can use that to determine when to fire his rockets (when he has gone 8 years). He can know how long to fire those rockets in the outgoing frame to get to a speed of 88.2% of c heading towards A. Once he has done so, he is in a different frame. It will take him 8 years (in that frame) to catch up with twin A.

At the end, he will have aged 16 years. And, based on his calculations in the outgoing frame, he will correctly predict that twin A will have aged 20 years.

 

[Justatruthseeker]

Now you are back to Flat Earth thinking again. There is no preferred frame of reference. In A's frame of reference B's clock is moving more slowly than his. In B's frame of reference A's clock is moving more slowly than his. Both see their own clocks as moving a regular speed. This is why understanding frames of reference is key to understanding relativity. B's 84,000 seconds take longer for A but at the same time A's 84,000 seconds take more time for B. Time is frame dependent.

A is stationary...... We already know B's observation of A's clocks ticking slower is WRONG. This is why when he returns he is younger, not A...... If B could actually judge reality correctly, A would not be older....