Looking to understand time dilation

[Grimble]

It does run slower relative to a given inertial frame, and that isn't a "distortion" any more than any other frame-dependent observation like the observation that one object has a greater velocity than another. My point was that the twin paradox is based on falsely thinking that the frame-dependent truth "a moving clock runs slower than a clock at rest" would still work in a non-inertial frame.

Agreed.

Nope, that's an entirely false statement if you are talking about non-inertial movement. Special relativity does not claim there is the least bit of reciprocity if one object is moving inertially and the other is moving non-inertially. Acceleration is absolute in relativity.

But how can you have one inertial and one non-inertial if their movement is only relevant, one to the other?
There is no inertial absolute frame everything is relevant.
If one takes a body alone in space it is, it has to be, at rest. There is nothing for it to be moving relative to.

If one then adds a second body then any movement is relative to the first. That is the only thing there for it to be relative to.
If observers on those two bodies measure that the distance between them is increasing non-linearly then each is accelerating because its movement is relative to the other.
There could be any number of forces acting upon those bodies, acting equally on those bodies.
For example they could be at rest one relative to the other yet both be subject to an enormous force yet that would have nothing to do with them being at rest.

Relativity is about movement of one body relative to another, how changes in coordinates may be mapped from one to the other and yes, they may show that when mapped the traveling clock runs slow as measured by the moving observer; but this is reciprocal and each would see the other's clock slow, as both are subject to the same effects.
BUT is those two observers were to compare the measurements taken only within their own FoRs they would agree about the results.

The rate that a clock runs relative to coordinate time (i.e. $$d\tau / dt = \sqrt{1 - v^2/c^2}$$) is frame-dependent, but Einstein's SR says that if you have two specific events A and B on its worldline such that it moved inertially between those events, then in any inertial frame, if the coordinate time between A and B is $$\Delta t$$ and the speed of the clock as it moved between them in that frame is v, the time $$\Delta \tau$$ elapsed on the clock is $$\Delta \tau = \Delta t \sqrt{1 - v^2/c^2}$$. The time elapsed on a given worldline between two specific events on that worldline is frame-independent, so all inertial frames will agree on the value even though they disagree on the value of $$\Delta t$$ and v.

Agreed.

Did you read my post on the [post=2972720]geometric analogy[/post]?

yes

Do you agree that if we have two dots A and B on a 2D plane, with a straight line segment joining them, then in any Cartesian coordinate system where the difference in x-coordinate of A and B is $$\Delta x$$ and the slope $$\Delta y / \Delta x$$ is S, then the length of the line segment is $$\sqrt{\Delta x^2 + \Delta y^2} = \sqrt{\Delta x^2}\sqrt{1 + (\Delta y^2 / \Delta x^2)} = \Delta x \sqrt{1 + S^2}$$? And that this length is coordinate-independent, so different Cartesian coordinate systems with their x-y axes will all agree on the value of $$\Delta x \sqrt{1 + S^2}$$ even though they disagree on the value of $$\Delta x$$ and S? If you agree with that,

yes

perhaps you can see that this formula can also be applied to a polygonal path made up of multiple straight segments, like a V-shaped path, so for example if the path consists of a straight segment going from A to B and another straight segment with a different slope going from B to C, the total length of this path would be $$\Delta x_{AB} \sqrt{1 + S^2_{AB}} + \Delta x_{BC} \sqrt{1 + S^2_{BC}}$$.

Yes

You can do the same sort of thing in SR if you want to figure out a clock's elapsed time (a frame-independent quantity) along a polygonal worldline consisting of multiple inertial segments joined by instantaneous acceleration, like the V-shaped worldline of the traveling twin in the simplest version of the twin paradox.

Yes - are you referring to the Minkowski diagram?

 

[Grimble]

A Frame of Reference is the same thing as a Set of Coordinates, so when you talk about different observers making measurements with their own set of coordinates it's the same thing as saying they are making measurements within the Frame of Reference in which they are at rest. I'm sure you know this.

Now here's the scenario as you have defined it:

But then you make a claim about what a and b will measure or see from their Frames of Reference:

But, you don't realize that you cannot just make this claim without doing a Lorentz Transformation to get from your first FoR to your second FoR. You need to understand that in the original FoR, the distances involved for a and b will be contracted to 0.6 and the clocks for a and b will be running at 0.6 of normal, so a and b will measure the 4 ly distances as 2.4 ly and the 5 y times as 3 years, and yes, they will measure their own speeds as 2.4 ly / 3 y or 0.8c. They can have synchronized clocks within their FoR's but they won't be synchronized with the original FoR or with each other.

No, I'm sorry I seem to be confusing you. Let me try again:
Let us take observe a, at rest in his own FoR and as far as he is concerned at rest in the universe.
Now a observes A pass him at 0.8c and decides that this is t0. Some time later he observes C pass him and he declares that this happened at time t1.
Now what is t1 - t0 for a? And remember he has seen neither A's clock nor C's clock nor any measurement made in ABC's FoR.

a has an standard clock at rest in his frame of reference and it is, therefore measuring proper time, just like A,B & C's clocks. so for him t1 - t0 must also = 5 years and AB must = 4 Light years.

It is only if a or A were to try and use the others clock's or rulers that LC and TD would apply.

 

[Grimble]

Gentlemen, allow me to add that the idea that one if one twin could read the other's clock he would read a different time is impossible. He would read whatever time the clock is shewing.
The difference is that the passage of time on that clock would be slower to him. Each second would take less time to pass so the total time ie #of seconds x duration of each second would be less total time.

 

[ghwellsjr]

No, I'm sorry I seem to be confusing you. Let me try again:
Let us take observe a, at rest in his own FoR and as far as he is concerned at rest in the universe.
Now a observes A pass him at 0.8c and decides that this is t0. Some time later he observes C pass him and he declares that this happened at time t1.
Now what is t1 - t0 for a? And remember he has seen neither A's clock nor C's clock nor any measurement made in ABC's FoR.

a has an standard clock at rest in his frame of reference and it is, therefore measuring proper time, just like A,B & C's clocks. so for him t1 - t0 must also = 5 years and AB must = 4 Light years.

It is only if a or A were to try and use the others clock's or rulers that LC and TD would apply.

If it is true (and it is true because you stated it as a given in your original description of this thought experiment) that the distance from A to C in the FoR in which they are at rest is 4 light years, and a is traveling from A to C at 0.8c, and a reads his clock when he passes A and then he reads his clock again when he passes C, then the difference in his readings will be 3 years, not 5. If you are going to claim that it is 5, then you have described a different thought experiment from your original one. Also, based on his measured speed of 0.8c, he will conclude that the distance between A and C is 2.4 light years, not 4.

 

[Grimble]

If it is true (and it is true because you stated it as a given in your original description of this thought experiment) that the distance from A to C in the FoR in which they are at rest is 4 light years, and a is traveling from A to C at 0.8c, and a reads his clock when he passes A and then he reads his clock again when he passes C, then the difference in his readings will be 3 years, not 5. If you are going to claim that it is 5, then you have described a different thought experiment from your original one. Also, based on his measured speed of 0.8c, he will conclude that the distance between A and C is 2.4 light years, not 4.

But no, a's clock,is local to him, at rest with him and with time in his own FoR and so it will read 5 years.
It is only if he were to read A's Clock and C's clock that he would read 3 years.
For a, his clock is not moving, it is the clock's of A and C that are moving. And the distance would only be 2.4 light years if he were to measure the distance AC using A's ruler and then transform the measurement to allow for the speed.
If a calculates the distance from his own measurements of his own clock there is nothing to transform.

 

[ghwellsjr]

But no, a's clock,is local to him, at rest with him and with time in his own FoR and so it will read 5 years.
It is only if he were to read A's Clock and C's clock that he would read 3 years.
For a, his clock is not moving, it is the clock's of A and C that are moving. And the distance would only be 2.4 light years if he were to measure the distance AC using A's ruler and then transform the measurement to allow for the speed.
If a calculates the distance from his own measurements of his own clock there is nothing to transform.

Well if that's true, then wouldn't the same thing be true for b? Here's your original statement of the scenario:

Now for seeing it all in one FoR:

Let us envisage three observers, A,B & C at rest in space.
They are situated in a straight line in the order ACB.
AC = CB = 4 light years, so AB = 8 ly
Also, because they are at rest in an inertial FoR the synchronised clocks that each possesses will be showing Proper time. (clock at rest in an inertial FoR is following its world line and is therefore measuring Proper time)

Imagine another two observers a & b each traveling at 0.8c. a is traveling from A towards C and b is traveling from B towards C.

And let us say that in our rest FoR they simultaneously pass A and B at time t0.

The observer at C will observe them both passing him at time t1 and t1 - t0 = 5 years.

Note that this has nothing to do with LT as all our observations and measurements are
made in the resting FoR of observers A,B & C.
The traveling observers, a and b, are detected at points A and C and points B and C respectively. From these observations A, B & C can determine that their speeds are equal and are each 0.8c.

Now what do a and b observe in their FoRs?
Well they too may consider them selves at rest in inertial FoRs. Therefore their clocks will also be showing proper time. And if they had imaginary synchronised clocks at rest with them and passing observer C at t0 we know that the proper distance between a and C at time t0 measured by a in a's FoR would be 4 proper light years.

Therefore a and b would each observe that they would travel 4 lt in 5 years at a speed of 0.8c.

It is only when the observers A, B and C use the measurements from a and b's FoRs that LT comes into play, for if C were to read a & b's clocks as they passed him at time t1 he would read time dilated times and distances - and calculate that a and b each traveled a distance of 2.4 (contracted) light years in a time of 3 (dilated) years.
Travelling at a speed of 2.4/3 = 0.8c.

So, yes, you do say that what is true for a is true for b, that is, b will go from B to C in the same 5 years as it took for a to go from A to C, and as you said, they both pass C at the same time, and according to C's local clock, it took 5 years.

But let's continue, don't you agree that in another 5 years, according to a's local clock, a will go from C to B and according to b's local clock, b will go from C to A in the same 5 years?

And won't you say that in the original FoR in which A, B, and C are at rest, that according to A's local clock, b will pass A 10 years after a passed A and according to B's local clock, a will pass B 10 years ater b passed B?

Do I have everything correct according to your understanding?

 

[Dale]

Are, you mean like taking measurements of an object that is moving with a high velocity relative to you. Like the way those measurements are different from those taken with rulers and clocks adjacent to the object. The way that those measurements of a moving object are distorted as a function of the relative velocity?

Yes. The fictitious forces in non-inertial frames affect measurements in much a similar way. E.g. in a non-inertial frame like Rindler coordinates two clocks may have time dilation relative to each other even though they are at rest wrt each other. Similarly, in non-inertial frames the coordinate speed of light may not be c. Non-inertial frames have a different metric than inertial frames, so you cannot neglect that fact without getting wrong results.

 

[Grimble]

Well if that's true, then wouldn't the same thing be true for b? Here's your original statement of the scenario:

So, yes, you do say that what is true for a is true for b, that is, b will go from B to C in the same 5 years as it took for a to go from A to C, and as you said, they both pass C at the same time, and according to C's local clock, it took 5 years.

But let's continue, don't you agree that in another 5 years, according to a's local clock, a will go from C to B and according to b's local clock, b will go from C to A in the same 5 years?

And won't you say that in the original FoR in which A, B, and C are at rest, that according to A's local clock, b will pass A 10 years after a passed A and according to B's local clock, a will pass B 10 years ater b passed B?

Do I have everything correct according to your understanding?

Yes, exactly

 

[ghwellsjr]

OK, what if after a travels from A to C, instead of continuing in the same direction when he meets b coming the other way, he turns around (instantly) and follows along with b from C to A. Will both a and b continue to have the same times on their two clocks?

 

[Grimble]

OK, what if after a travels from A to C, instead of continuing in the same direction when he meets b coming the other way, he turns around (instantly) and follows along with b from C to A. Will both a and b continue to have the same times on their two clocks?

OK, what if after a travels from A to C, instead of continuing in the same direction when he meets b coming the other way, he turns around (instantly) and follows along with b from C to A. Will both a and b continue to have the same times on their two clocks?

Well, again we have a difference between the accepted wisdom and teachings of the Minkowski diagram showing the dogleg portrayal of the traveling twin's journey.
If one takes that diagram:

There are 4 problems that leap out at one:
1. It is not drawn to a consistent scale;
2. The second part of the traveling twin's journey is not rotated about the origin;
3. The plane of simultaneity must surely also encompass all points on the stationary twin's FoR that are simultaneous; that is parallel to its x axis. We are, after all, viewing the stationary twin's FoR.
4. If the relative velocity is depicted by a clockwise rotation of the traveller's ct' axis, then how can it also be shown by the corresponding anti-clockwise rotation of the x' axis? Surely the ct', x' frame can only rotate one way?


Resolving these issues by drawing diagrams that reflect reality and answer all these points shows that indeed a anb will continue to have the same times on their clocks!

 

[ghwellsjr]

So now let's just look at a as he travels from A to C and back to A (we'll ignore B and B):

First, a description of this part of your scenario--A and C are at rest with respect to each other and are 5 light years apart and have synchronized clocks. a is moving at 0.8c with respect to A and C and travels from A to C and back to A.

In the stationary FoR, a passes by A at t0 and arrives at C at t1, 4 light years later. He instantly turns around and arrives back at A at t2, which takes another 4 light years. The difference, t2 - t0, is the time it took in the stationary frame of reference for a to make the round trip and is 8 light years.

From a's point of view, it also took 4 light years to go from A to C and another 4 light years to go back from C to A for a total of 8 light years.

I know I have modified your original scenario by leaving b and B out and adding t2 but did I get everything correct?

 

[Grimble]

So now let's just look at a as he travels from A to C and back to A (we'll ignore B and B):

First, a description of this part of your scenario--A and C are at rest with respect to each other and are 5 light years apart and have synchronized clocks. a is moving at 0.8c with respect to A and C and travels from A to C and back to A.

In the stationary FoR, a passes by A at t0 and arrives at C at t1, 4 light years later. He instantly turns around and arrives back at A at t2, which takes another 4 light years. The difference, t2 - t0, is the time it took in the stationary frame of reference for a to make the round trip and is 8 light years.

From a's point of view, it also took 4 light years to go from A to C and another 4 light years to go back from C to A for a total of 8 light years.

I know I have modified your original scenario by leaving b and B out and adding t2 but did I get everything correct?

Not quite A and Care 4 light years apart; to and t1 are 5 years apart as they are times. And the time for the round trip would be 10 years.

From a's point of view it also took 5 years to travel the 4 l.y.s from A to C and the same to return.

The important point is that time dilation (and Length contraction) only occurs when one is reading the others clocks.

 

[ghwellsjr]

Sorry for the mixup and thanks for clarification.

The important point is that time dilation (and Length contraction) only occurs when one is reading the others clocks.

But I thought you said that A's clock and a's clock had the same time on them at the beginning and both observers can correctly read the other one's clock. So can't they both correctly read each other's clocks 10 years later at the end?

 

[Grimble]

Sorry for the mixup and thanks for clarification.

But I thought you said that A's clock and a's clock had the same time on them at the beginning and both observers can correctly read the other one's clock. So can't they both correctly read each other's clocks 10 years later at the end?

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.

 

[Dale]

There are 4 problems that leap out at one:

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

1. It is not drawn to a consistent scale;

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

2. The second part of the traveling twin's journey is not rotated about the origin;

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

3. The plane of simultaneity must surely also encompass all points on the stationary twin's FoR that are simultaneous; that is parallel to its x axis. We are, after all, viewing the stationary twin's FoR.

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.

4. If the relative velocity is depicted by a clockwise rotation of the traveller's ct' axis, then how can it also be shown by the corresponding anti-clockwise rotation of the x' axis? Surely the ct', x' frame can only rotate one way?

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.

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.

Resolving these issues by drawing diagrams that reflect reality and answer all these points shows that indeed a anb will continue to have the same times on their clocks!

No, you need to learn a little more, the diagram is essentially correct.

 

[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