Looking to understand time dilation

[ghwellsjr]

Yes, of course, and when they read each others clock's as a passes A on the return journey, or if b and A read each others clocks in my scenario, then they will each find the other's clocks will be time dilated and read 6 yrs instead of 10 yrs. But that is reading each other's clock.
At the same time each reading their own clock's would read 10yrs.

(Actually they would all read 10 yrs but the time dilated clocks would just be ticking quicker; as each time dilated second would be only 0.6 of a proper second so 10 time dilated(co-ordinate) years would be equal to only 6 proper years).

And as b's speed according to a will be 2v/1+(0.8)² = 1.6/1.64 c = 0.975 c, b's clock when passing A would show 2.195 years had passed according to a.

It's obviously easy to get mixed up as I did a couple posts ago.

But here's what I see you have said: As a goes from A to C and back to a (in my modification), a's clock accumulates 10 years. During that same time, A's clock has also accumulated 10 years. And yet you say that when a and A look at each other's clocks at the end, they see that the other one's clock has accumulated 6 years, correct?

 

[Grimble]

I don't like that diagram, but not for any of the reasons you specify here.

There is no scale indicated, but the drawing is essentially correct for one specific kind of non-inertial reference frame.

No scale needs to be drawn, the angles of the lines show it is not to scale.

There is no requirement that it be rotated about the origin.

How can you say that? If the velocity is shown by rotation then all lines showing a velocity Must pass through the origin. Even if one were to continue the second part of the line back to ct = 0 . Because it then shows the true correspondence to the ct axis.

How can you possibly have lines running back from the deceleration line running the wrong way? They have to slope down to the =ve x origin.

That diagram is just scientifically impossible! Think about it.

No, this is the whole meaning of the relativity of simultaneity. The planes of simultaneity will NOT be parallel in general. If they were always parallel then simultaneity would be absolute, not relative.

Oh yes, so it would! It makes one think doesn't it.

If they rotated as you describe then the speed of light would not be invariant. The rotation is not an ordinary circular rotation, but a hyperbolic rotation. It looks more like a shear along the x=ct line. What is drawn is correct.

I have no idea what that means. I can picture an hyperbolic rotation but I cannot envisage any sort of 'shear'.
The simple amswer is that the rotation is not in the ct,x plane. And why should it be?

You should verify this yourself simply by drawing the lines of constant time and position indicated by the Lorentz transform. I strongly recommend that you go through this exercise. For convenience use c=1 and v=0.6. Simply take the Lorentz transform equation for x' and set x' to 0, this will give you the equation of a line in x and t, plot that line. Then set x' to 1. This will give you the equation of a different line in x and t, plot that line. And so forth for x' = 2 and then t' = 0, 1, 2. That will show you what the Lorentz transform looks like.

Yes absolutely but please, please, please draw it to scale - I will in my next post.

I must apologise if I seem to be getting a bit excited but this is fascinating and I don't mean to be a pain but these things just don't fit as everyone thinks they do.

If you would like I can show you one step at a time dealing with each of my 'problems' as I go. The outcome will surprise you ...

No, you need to learn a little more, the diagram is essentially correct.[/QUOTE]

No, I'm sorry for I know that is what you have been taught, that is what everyone is taught, and so no-one bothers to work it out and see the errors.

If the diagram is drawn correctly all the errors disappear and it all starts to make sense.

 

[Grimble]

It's obviously easy to get mixed up as I did a couple posts ago.

But here's what I see you have said: As a goes from A to C and back to a (in my modification), a's clock accumulates 10 years. During that same time, A's clock has also accumulated 10 years. And yet you say that when a and A look at each other's clocks at the end, they see that the other one's clock has accumulated 6 years, correct?

Yes that is correct, it is reciprocal and it is the effect of taking measurements of a moving object that distorts the readings. This is essentially the effect of c being constant. Any measurement where there is a relative velocity between the frames has to cater for that velocity in the measurement or c would not be constant.

In effect both measurements are correct.

Take a simple case of the light clock where the light pulse takes one second for the resting FoR where it is occurring. For an observer moving at v it would take gamma seconds; and the way that SR resolves this is recognising that gamma coordinate seconds has the same duration, is equal to, the one second proper time.

So both readings are correct from where they are read.

 

[Mentz114]

How can you possibly have lines running back from the deceleration line running the wrong way? They have to slope down to the =ve x origin.

There's only one acceleration in that diagram, the unphysical instant turnaround. In my opinion you're talking rot and I really look forward to your diagram.

That diagram is just scientifically impossible! Think about it.

It is impossible because of the instantaneous turnaround, but otherwise it's fine. I have thought about it and it still makes perfect sense.

 

[ghwellsjr]

Ok, so what happens if a comes to a stop at A at the end of the scenario instead of continuing on? Will both a and A say that 10 years have transpired since the beginning of the scenario?

 

[Dale]

No scale needs to be drawn, the angles of the lines show it is not to scale.

Angles can't show scale, by definition.

How can you say that? If the velocity is shown by rotation then all lines showing a velocity Must pass through the origin.

No, the worldline of an object which is not at the origin at t=0 will not pass through the origin, regardless of velocity. There is certainly no requirement that all worldlines pass through the origin.

That diagram is just scientifically impossible! Think about it.

I have thought about it. The diagram is fine except for subtle issues about diffeomorphism that aren't really important here and impulsive forces which can be a reasonable approximation in appropriate circumstances.

I have no idea what that means. I can picture an hyperbolic rotation but I cannot envisage any sort of 'shear'.
The simple amswer is that the rotation is not in the ct,x plane. And why should it be?

The simpler answer is that it is not a Euclidean rotation.

No, I'm sorry for I know that is what you have been taught, that is what everyone is taught, and so no-one bothers to work it out and see the errors.

This is a very typical crackpot comment. I had not pegged you as a crackpot earlier, but you are rapidly leaning that way. I will advise you to re-read the rules link at the top of the page and specifically recommend that you pay attention to the section on overly-speculative posts. This forum is not a pulpit for crackpots to preach their nonsense, it is dedicated to teaching and learning mainstream science. If you want to "fix" science then there are other venues which are more appropriate.

 

[Grimble]

Ok, so what happens if a comes to a stop at A at the end of the scenario instead of continuing on? Will both a and A say that 10 years have transpired since the beginning of the scenario?

When the traveller decelerates at the end of his journey, the rotation that denotes their relative velocity decreases to zero. And as LT calculates, when the relative velocity is zero then the length contraction and time dilation disappear, so yes they will all read the same, local and remote measurements will agree.

 

[Grimble]

Angles can't show scale, by definition.

No, you are quite right, except that drawn to scale the planes of simultaneity would be parallel to the x axis, as they are not it shews that the ct and ct' axis are not drawn to the correct scales

No, the worldline of an object which is not at the origin at t=0 will not pass through the origin, regardless of velocity. There is certainly no requirement that all worldlines pass through the origin.

But the worldline of an object that has passed through the origin must continue to pass through the origin.

When the velocity of the traveller changes, the rotation that depicts that velocity changes.
As the twin slows down the angle of rotation reduces until as he comes to rest it is along the ct axis of the resting twin. You are confusing the path of the traveller which has a kink in it with the rotation that denotes velocity. They are not the same.[/QUOTE]

I have thought about it. The diagram is fine except for subtle issues about diffeomorphism that aren't really important here and impulsive forces which can be a reasonable approximation in appropriate circumstances.

I have no idea what that is about. The diagram is very simple and drawn correctly there no subtle issues.

The simpler answer is that it is not a Euclidean rotation.

not in the ct,x plane certainly.

This is a very typical crackpot comment. I had not pegged you as a crackpot earlier, but you are rapidly leaning that way. I will advise you to re-read the rules link at the top of the page and specifically recommend that you pay attention to the section on overly-speculative posts. This forum is not a pulpit for crackpots to preach their nonsense, it is dedicated to teaching and learning mainstream science. If you want to "fix" science then there are other venues which are more appropriate.

No, I'm sorry that was a very silly and inappropriate comment of mine.
I am not trying to rewrite Special Relativity but to understand something that doesn't seem to fit logically, when there is a much simpler way of seeing it - drawing the diagram - so everything fits.

It is not the science I have a problem with but the depiction of it.

Grimble

 

[ghwellsjr]

Ok, so what happens if a comes to a stop at A at the end of the scenario instead of continuing on? Will both a and A say that 10 years have transpired since the beginning of the scenario?

When the traveller decelerates at the end of his journey, the rotation that denotes their relative velocity decreases to zero. And as LT calculates, when the relative velocity is zero then the length contraction and time dilation disappear, so yes they will all read the same, local and remote measurements will agree.

I am not trying to rewrite Special Relativity...

You have stated that when a starts out at the same place as A and then travels at 0.8c for 5 years, turns around and comes back to A, both a and A will have aged by ten years. Do you not see that this is a statement of the twin paradox in which the traveling twin "a" ages 6 years while the stationary twin "A" ages 10 years?

 

[Grimble]

You have stated that when a starts out at the same place as A and then travels at 0.8c for 5 years, turns around and comes back to A, both a and A will have aged by ten years. Do you not see that this is a statement of the twin paradox in which the traveling twin "a" ages 6 years while the stationary twin "A" ages 10 years?

Not exactly, I see two scenarios.

1. While a is traveling past A at 0.8c, on his return, then a and A will each have aged 10 years, as they can read by their own clocks, but when each examines the others clock they will conclude that only 6 years will have passed for the other and that is the real and original twin paradox.

2. In the second case however, if a were to slow down and come to rest in A's FoR then each will still read 10 years on his own clock and will also conclude that 10years has passed for his brother and that they are, once again the same age.
Which is what one would calculate using the LT equations: when v = 0 t' = t

Grimble

 

[Dale]

No, you are quite right, except that drawn to scale the planes of simultaneity would be parallel to the x axis, as they are not it shews that the ct and ct' axis are not drawn to the correct scales

Please do some research on "relativity of simultaneity". What you are saying here shows that you are still thinking non-relativistically and have not grasped the relativity of simultaneity. It turns out to be the most difficult concept to grasp.

It is not the science I have a problem with but the depiction of it.

I think you don't understand the science or the depiction, because they agree. Please draw your proposed drawing including the scale and I will either show how it matches the above or how it does not match the Lorentz transform.

 

[ghwellsjr]

Not exactly, I see two scenarios.

1. While a is traveling past A at 0.8c, on his return, then a and A will each have aged 10 years, as they can read by their own clocks, but when each examines the others clock they will conclude that only 6 years will have passed for the other and that is the real and original twin paradox.

2. In the second case however, if a were to slow down and come to rest in A's FoR then each will still read 10 years on his own clock and will also conclude that 10years has passed for his brother and that they are, once again the same age.
Which is what one would calculate using the LT equations: when v = 0 t' = t

Grimble

So you think that when two observers are in relative motion but happen to be located at the same place, they cannot correctly see each other's clocks?

But you haven't answered my question about the Twin Paradox: It says that the traveling twin will have aged 6 years while the stationary twin ages 10 years and both of them will agree on this fact at the end when they are stationary once again in the same location, but you are saying that they both age the same 10 years and there is no paradox. What happened to the Twin Paradox?

 

[Grimble]

Please do some research on "relativity of simultaneity". What you are saying here shows that you are still thinking non-relativistically and have not grasped the relativity of simultaneity. It turns out to be the most difficult concept to grasp.

I think you don't understand the science or the depiction, because they agree. Please draw your proposed drawing including the scale and I will either show how it matches the above or how it does not match the Lorentz transform.

Thank you I will do that

 

[Grimble]

So you think that when two observers are in relative motion but happen to be located at the same place, they cannot correctly see each other's clocks?

But you haven't answered my question about the Twin Paradox: It says that the traveling twin will have aged 6 years while the stationary twin ages 10 years and both of them will agree on this fact at the end when they are stationary once again in the same location, but you are saying that they both age the same 10 years and there is no paradox. What happened to the Twin Paradox?

It only exists while they are moving and then each will agree that the other has only aged 6 years.

 

[Mike_Fontenot]

It only exists while they are moving and then each will agree that the other has only aged 6 years.

At any instant when two observers are co-located, they must completely agree about the readings on each of their clocks, whether they are in relative motion or not.

Mike Fontenot

 

[ghwellsjr]

It only exists while they are moving and then each will agree that the other has only aged 6 years.

So, you think that the Twin Paradox is that at the end, prior to the traveling twin stopping,
each one thinks that they have aged 10 years but their twin has aged only 6 years, and then when the traveling twin stops, they both agree that both of them have aged 10 years, correct?

If this is how you see it, then the two twins are always symmetrical, correct? And it doesn't matter which one takes the trip, correct?

Can you find a reference that describes the Twin Paradox like this? I'm interested in knowing where you learned this.

 

[phyti]

Grimble;

I've read all your posts on this thread, and the same problem keeps showing.
You use the same 'proper time' for all observers. Proper time is indicated on a clock moving with the observer, and its rate is effected by the observers motion in space, thus it's only applicable to that observer. A moving observer and his clock run slower as they move faster.
In the example, the a and b clocks are running slower than the A, B, and C clocks, but a and b can't detect this because their perception is altered to the same degree as the clocks. Upon arriving at C, their own clocks read (.6*5 =) 3 yr, so they conclude the distance is (.6*4 =) 2.4 l yr.
Even though a calculates b's speed as -.976, b's clock must read 3 yr, the same as his. That speed is the relative speed of b by a which is always greater than that measured by the 'chosen' or zero frame. If A uses the speed composition equation with (-.976, .8) the corrected speed for b is .8c.

Persevere, it's not that complicated.

 

[Grimble]

At any instant when two observers are co-located, they must completely agree about the readings on each of their clocks, whether they are in relative motion or not.

Mike Fontenot

Yes of course if they are co-located and reading the same clock they will read the same time but each second for the traveling observer will be equal to 1/γ seconds for the resting observer.

 

[Grimble]

So, you think that the Twin Paradox is that at the end, prior to the traveling twin stopping,
each one thinks that they have aged 10 years but their twin has aged only 6 years, and then when the traveling twin stops, they both agree that both of them have aged 10 years, correct?

If this is how you see it, then the two twins are always symmetrical, correct? And it doesn't matter which one takes the trip, correct?

Yes.

Can you find a reference that describes the Twin Paradox like this? I'm interested in knowing where you learned this.[/QUOTE]

I was aware of, and intrigued by, the idea of each twin appearing to be younger than his sibling in my youth. I subsequently investigated to determine how such a paradox could be explained.

I have followed several different lines and the one that it is due to acceleration doesn't work in any of them. Hence my need for guidance.

Einstein, Minkowski, Lorentz which ever I follow all come to the same conclusion.

 

[granpa]

relativity has 3 parts:
1. time dilation
2. length contraction
3. "relativity of simultaneity"

1 and 2 are easy

your confusion comes from a poor understanding of 3

 

[Grimble]

Grimble;

I've read all your posts on this thread, and the same problem keeps showing.
You use the same 'proper time' for all observers. Proper time is indicated on a clock moving with the observer,

Yes.

and its rate is effected by the observers motion in space,

relative to what? There is no absolute in space for it to move relative to

thus it's only applicable to that observer.

Obviously as it is the clock that is moving with him.

A moving observer and his clock run slower as they move faster.
In the example, the a and b clocks are running slower than the A, B, and C clocks, but a and b can't detect this because their perception is altered to the same degree as the clocks. Upon arriving at C, their own clocks read (.6*5 =) 3 yr, so they conclude the distance is (.6*4 =) 2.4 l yr.

But why do you say that a and b are moving? It is just as true to say that A,B & C are moving at 0.8c relative to a or b.

Even though a calculates b's speed as -.976, b's clock must read 3 yr, the same as his. That speed is the relative speed of b by a which is always greater than that measured by the 'chosen' or zero frame. If A uses the speed composition equation with (-.976, .8) the corrected speed for b is .8c.

Yes I understand all that, and can see how the relative speed can affect the way that clocks et.al. can be measured differently because ofthe difference in the way that the measuring is done. BUT I cannot see how anyone can Possibly apply that to the actual times shown on those clocks.
I couls just as easily say that a is also traveling at O.6c relative to another FoR D and that, by your argument that his clock should read (.8*5 = ) 4 yr! or that there are an infinite number of FoRs that have different speeds relative to A and that his clock should read an infinite number of different times simultaneously!

a's clock will only read one time, the hands will only show one relationship, it will only show one set of numbers at anyone time.

Yes, of course an infinite number of people can read an infinite number of different times according to their relative velocities, BUT ONLY if it is the measuring of the time under the condition of moving at that velocity that affects the time read.

This is just simple logic. Moving with respect to the object measured changes the SCALE upon which that measurement is read.

And if velocity is shewn by the angle of rotation between two FoR than a change of velocity, i.e. an acceleration MUST be shewn by a change in that angle of rotation it must increase or decrease, that is the angle of rotation from the origin. We can't have the path of a body in one frame being shown by a dogleg in the other...

Persevere, it's not that complicated.

 

[Dale]

there are an infinite number of FoRs that have different speeds relative to A and that his clock should read an infinite number of different times simultaneously!

Yes. This is called the relativity of simultaneity.

 

[ghwellsjr]

I was aware of, and intrigued by, the idea of each twin appearing to be younger than his sibling in my youth. I subsequently investigated to determine how such a paradox could be explained.

I have followed several different lines and the one that it is due to acceleration doesn't work in any of them. Hence my need for guidance.

Einstein, Minkowski, Lorentz which ever I follow all come to the same conclusion.

And which conclusion is that?

Before we can come to any resolution of the Twin Paradox, we have to understand what it is. The two twins start out at the same age in the same location. One of them accelerates away to some high speed. During this time, as they each observe the other's clock, they are each running slower than their own. This part is symmetrical. Eventually, the traveling twin decelerates and comes back at the same high speed. During this time, as they each observe the other's clock, they are each running slower than their own, just like before. The only time when this is not true is during the brief period of time when they observe the process of acceleration/deceleration. Eventually the traveling twin comes to a stop at the starting point and when they compare the actual times on their clocks, the traveling one has less time on it. Do you agree with this as a statement of the Twin Paradox?

 

[Mentz114]

I think I should point out that the classical twin paradox where one remains inertial is just a special case of two worldlines starting at the same point and rejoining after they have done some travelling. In all cases, regardless of how the travellers moved, the elapsed time on their clock will be the Lorentzian length between the parting and meeting of the 4D curves that describe their journeys.

That's it. All observers agree on those elapsed times because they are the geometric invariant of Minkowski spacetime. What more can be said ? There is no paradox or confusion possible about this, surely ?

 

[ghwellsjr]

Mentz, how does your explanation demonstrate that there is a paradox that needs to be resolved?

 

[Passionflower]

I think I should point out that the classical twin paradox where one remains inertial is just a special case of two worldlines starting at the same point and rejoining after they have done some travelling. In all cases, regardless of how the travellers moved, the elapsed time on their clock will be the Lorentzian length between the parting and meeting of the 4D curves that describe their journeys.

Indeed, take several observers traveling between event A and B, one non-accelerating and others accelerating different ways. The observer who does not accelerate has the maximum elapsed time while the other observers have less elapsed times possibly at varying degrees.

 

[Mentz114]

Mentz, how does your explanation demonstrate that there is a paradox that needs to be resolved?

There is no paradox to be resolved. I thought I said that pretty clearly.

PassionFlower, you have introduced the 'quintuplet paradox'

 

[ghwellsjr]

Mentz, you don't think that each twin observing and measuring the other one's clock as running slower than his own during both legs of the trip but yet only the traveler's clock ends up with a lower time is a paradox?

 

[Mentz114]

Mentz, you don't think that each twin observing and measuring the other one's clock as running slower than his own during both legs of the trip but yet only the traveler's clock ends up with a lower time is a paradox?

No. A paradox is when observers disagree about an invariant, not a frame dependent measurement. If we label an oberver as A, then asking a number of other inertial observers what the rate of A's clock is will yield a lot of different answers depending on relative velocities at the times when the measurements are made. Which one is correct ?

 

[ghwellsjr]

But there are no frame dependent measurements in the Twin Paradox, just like in the real world. In fact, there can be no frame dependent measurements in any scenario, or else we would have a preferred frame and relativity would not be a viable theory about reality.