Looking to understand time dilation

[Grimble]

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.

 

[Dale]

using some sort of accelerometer it is possible to determine what acceleration each one is experiencing but that has NOTHING to do with relativity

This statement is also incorrect. The accelerometer has everything to do with relativity, it is fundamental to both of the two postulates. The Lorentz transform is a transform between two inertial frames. I.e. given one inertial frame you can find any other inertial frame via the Poincare group (which includes the Lorentz transform). But how do you know that your first frame is inertial? It cannot be determined by virtue of the Lorentz transform alone, it can only be determined by some physical experiment, such as using an accelerometer.

Only once you have performed such an experiment can you label a given coordinate system as inertial. Once you have done so you automatically know the metric in that coordinate system, and you can then determine the metric in any arbitrary non-inertial coordinate system also. Evaluated at some point the non-inertial metric will not necessarily reduce to the inertial metric.

 

[ghwellsjr]

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.

Everything was fine up to this point and now it gets confusing. It's OK for a and b to have their own clocks at rest with them but what are you saying they are synchronized to? They cannot be synchronized to each other's clocks or the A, B, and C's clocks. If you want to use a second FoR (the one in which a is at rest) you need to transform the entire scenario into that FoR and you have to define when t0 is relative to the original t0. (It looks like you have done a big shift--which is OK--because in the orginal FoR, t0 was when a was at A and b was a B but now it appears that you have put t0 for this second FoR when a and b are at C). Is that what you really meant to do?

Finally, if you want to consider a third FoR, the one in which b is at rest, you have to do another transform.

I'm not going to comment on the rest of your post until you clarify this issue.

 

[Grimble]

Everything was fine up to this point and now it gets confusing. It's OK for a and b to have their own clocks at rest with them but what are you saying they are synchronized to? They cannot be synchronized to each other's clocks or the A, B, and C's clocks. If you want to use a second FoR (the one in which a is at rest) you need to transform the entire scenario into that FoR and you have to define when t0 is relative to the original t0. (It looks like you have done a big shift--which is OK--because in the orginal FoR, t0 was when a was at A and b was a B but now it appears that you have put t0 for this second FoR when a and b are at C). Is that what you really meant to do?

Finally, if you want to consider a third FoR, the one in which b is at rest, you have to do another transform.

I'm not going to comment on the rest of your post until you clarify this issue.

Then let me repeat and clarify for you:

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.

Let us take the inertial frame of observer a.
If it had an imaginary clock synchronised with the clock adjacent to observer a (why would it - or even how could it - be synchronised with any thing else?) and this imaginary clock were to be passing C at time t0, then it would be exactly 4 ly ahead of a in a's FoR.

And, I thought it quite apparent that we would be considering a similar arrangement in observer b's FoR.

This part is really redundant but just emphasises that in the inertial frames of a and b, the measurements by a and b would be exactly equal to those in our rest frame or AB & C. i.e. that a and b, by their own measurements within their own FoRs, would calculate they had each traveled 4 proper lys in 5 proper years.

 

[ghwellsjr]

OK, I think I see what you are proposing, tell me if this is what you're thinking:

In a's FoR, a has a clock adjacent to himself and another synchronized clock 4 ly ahead of himself. Then when a passes next to A at t0, the other clock is passing by C, again at t0 in a's FoR.

 

[Grimble]

OK, I think I see what you are proposing, tell me if this is what you're thinking:

In a's FoR, a has a clock adjacent to himself and another synchronized clock 4 ly ahead of himself. Then when a passes next to A at t0, the other clock is passing by C, again at t0 in a's FoR.

Precisely, as both a's measurement of a distance of 4 light years and A,B and C's measurement of 4 light years are each made at rest in their individual FoRs, between two clocks that are at rest in their individual FoRs; Thus each measurement is a measurement of Proper distance. And each measurement is made wholly within a single FoR with no reference to the other FoR.

Now as each measurement is 4 Proper light years, at time t0 when a is passing A, a's second clock MUST be passing C.

So before you say anything, Length Contraction does not come into play for Either of these measurements.

 

[Grimble]

This statement is also incorrect. The accelerometer has everything to do with relativity, it is fundamental to both of the two postulates. The Lorentz transform is a transform between two inertial frames. I.e. given one inertial frame you can find any other inertial frame via the Poincare group (which includes the Lorentz transform). But how do you know that your first frame is inertial? It cannot be determined by virtue of the Lorentz transform alone, it can only be determined by some physical experiment, such as using an accelerometer.

Absolutely! You are right in what you say here, but is it as relevant as it appears to be?
SR, LT and all the rest refer to Inertial FoRs but why are they restricted to Inertial FoRs?
Is it not because it was simpler to deal with them first?
Having said that, what is it about non-inertial frames that precludes their inclusion?

Ah! But of course that is the effects of the extra forces at work that result in the FoR being non-inertial.

But, I say, all we are transforming between frames are coordinates; coordinates at one point in time. It matters not what forces are resulting in those movements and those coordinates, we are not measuring forces only positions, coordinates in 4 dimensional Space-time.

So we are transforming measurements from one FoR to another. Redefining them according to a different set of coordinates. The forces acting upon the objects being measured are not relevant to those measurements.

Or we could say that for any measurement taken the FoR concerned can be considered inertial (from the point of view of those measurements) for the instant that those measurements exist.

And as we are dealing with the movement of one FoR relevant to another we do NOT need to know anything about which if either is inertial, for either one could be, or perhaps neither depending on how one defines them.

Only once you have performed such an experiment can you label a given coordinate system as inertial. Once you have done so you automatically know the metric in that coordinate system, and you can then determine the metric in any arbitrary non-inertial coordinate system also. Evaluated at some point the non-inertial metric will not necessarily reduce to the inertial metric.

I must admit you lose me here, What exactly are you referring to when you talk of the metric? I suspect it is something more than the 4 space-time coordinates and relative velocity that we are dealing with in LT?

 

[Dale]

we are not measuring forces only positions, coordinates in 4 dimensional Space-time.

Do you really believe this? If so, perhaps you can find the x-coordinate-meter in some catalog. I know of many devices that measure distances, but I don't know of any that measure coordinates. Do you see the distinction?

The forces acting upon the objects being measured are not relevant to those measurements.

They are relevant if they affect the outcome of the measurements. When you make a measurement you are not simply mathematically determining a coordinate, you are performing a physical experiment of some sort. The outcome of that physical experiment is generally affected by the forces you refer to. A clock at rest in an rotating reference frame behaves differently from a clock at rest in an inertial frame.

I must admit you lose me here, What exactly are you referring to when you talk of the metric? I suspect it is something more than the 4 space-time coordinates and relative velocity that we are dealing with in LT?

The metric is the geometric object which connects a coordinate system to the physics. It encodes all of the information about distances, angles, durations, relative velocities, and the causal structure of spacetime. For instance, if you measure a small distance the result of the measurement is
ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}
Where ds is the distance, g is the metric, dx is the change in the coordinates, and the Einstein summation convention is used. So even a simple measurement of distance depends on the metric.

http://en.wikipedia.org/wiki/Metric_tensor_(general_relativity)
http://www.phy.duke.edu/~rgb/Class/phy319/phy319/node131.html
http://www.mathpages.com/rr/s5-02/5-02.htm

 

[JesseM]

Absolutely! You are right in what you say here, but is it as relevant as it appears to be?
SR, LT and all the rest refer to Inertial FoRs but why are they restricted to Inertial FoRs?
Is it not because it was simpler to deal with them first?
Having said that, what is it about non-inertial frames that precludes their inclusion?

Nothing precludes their inclusion, but the "twin paradox" is based on the idea that a moving clock runs slower than a clock at rest, with the "paradox" arising because people falsely think this should be true in the non-inertial frame of the traveling twin; in fact, the idea that a moving clock runs slower than a clock at rest is only guaranteed to hold in an inertial frame.

Or we could say that for any measurement taken the FoR concerned can be considered inertial (from the point of view of those measurements) for the instant that those measurements exist.

What do you mean by "the FoR concerned"? A measurement doesn't come attached to a frame of reference, the reading on any physical measuring instrument is a frame-independent fact which all frames should make the same prediction about.

And as we are dealing with the movement of one FoR relevant to another we do NOT need to know anything about which if either is inertial, for either one could be, or perhaps neither depending on how one defines them.

You do if you want to use them to make physical predictions about things like the elapsed time on a clock with known velocity as a function of time in some coordinate system.

I must admit you lose me here, What exactly are you referring to when you talk of the metric? I suspect it is something more than the 4 space-time coordinates and relative velocity that we are dealing with in LT?

In any coordinate system, you want to be able to calculate the frame-independent proper time along any timelike worldline (along with a frame-independent 'proper distance' along any spacelike worldline). For any given coordinate system, there is an associated metric which gives the line element which you can integrate along a worldline with known coordinates in your coordinate system and get the correct proper time/proper distance for that worldline. For example, in an inertial coordinate system in flat spacetime the line element could be written as ds^2 = dt^2 - (1/c^2)(dx^2 + dy^2 + dz^2) (this is just the infinitesimal form of the spacetime interval which gives proper time/proper distance along a constant velocity-path with known endpoints (t1, x1, y1, z1) and (t2, x2, y2, z2), in which case the proper time could be calculated by \sqrt{(t_2 - t_1)^2 - (1/c^2)[(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2]})

 

[ghwellsjr]

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.

OK, I think I see what you are proposing, tell me if this is what you're thinking:

In a's FoR, a has a clock adjacent to himself and another synchronized clock 4 ly ahead of himself. Then when a passes next to A at t0, the other clock is passing by C, again at t0 in a's FoR.

Precisely, as both a's measurement of a distance of 4 light years and A,B and C's measurement of 4 light years are each made at rest in their individual FoRs, between two clocks that are at rest in their individual FoRs; Thus each measurement is a measurement of Proper distance. And each measurement is made wholly within a single FoR with no reference to the other FoR.

Now as each measurement is 4 Proper light years, at time t0 when a is passing A, a's second clock MUST be passing C.

So before you say anything, Length Contraction does not come into play for Either of these measurements.

I'm afraid that you've done a Galilean Transformation which is only applicable at low speeds, certainly not at the 0.8c you have proposed in your scenario. You need to perform a valid Lorentz Transformation from your orginal scenario where A, B, and C are at rest to this one where a is at rest. If you had done so, you would understand that all dimensions along the direction of a's motion are contracted, so that the distances from A to C and from C to B are not 4 ly. Also, the clocks that were synchronized in the original FoR or no longer synchronized in this new FoR and they are time dilated.

Your concept of LT as you indicated at the end of the first quote is all mixed up.

But I'm wondering, why are you interested in doing this transformation? What added information do you think it is going to provide?

 

[Grimble]

I'm afraid that you've done a Galilean Transformation which is only applicable at low speeds, certainly not at the 0.8c you have proposed in your scenario. You need to perform a valid Lorentz Transformation from your orginal scenario where A, B, and C are at rest to this one where a is at rest. If you had done so, you would understand that all dimensions along the direction of a's motion are contracted, so that the distances from A to C and from C to B are not 4 ly. Also, the clocks that were synchronized in the original FoR or no longer synchronized in this new FoR and they are time dilated.

Your concept of LT as you indicated at the end of the first quote is all mixed up.

But I'm wondering, why are you interested in doing this transformation? What added information do you think it is going to provide?

I am sorry but you are still not grasping what I am saying.
I am not doing any form of transformation, not even a Galilean one.
I am dealing with two separate sets of coordinates.
In one case A,B & C are making measurements entirely within their own set of coordinates ...
And, completely independently a is making measurements according to his own set of coordinates ...
And according to the observations made in the first FoR an event occurs where a passes A at time t0.
Then a second event occurs where a passes C at time t1.

What I am showing is that the measurements taken within each FoR are equal as absolute proper measurements.

They obviously will not be equal in either FoR where LT means that the 'remote/moving' measurement observed from each frame WILL be length contract/time dilated.
But, nevertheless the measurements of time and distance made by A of the distance AC and the time t1-t0, will by necessity have to be equal to the measurements made by a of the distance between his clocks and the time of the two events t1-t0 as these measurements are all entirely local within each FoR and NO TRANSFORMATION is involved.

 

[Grimble]

Do you really believe this? If so, perhaps you can find the x-coordinate-meter in some catalog. I know of many devices that measure distances, but I don't know of any that measure coordinates. Do you see the distinction?

No. I'm sorry but you have me there.
My understanding is that the coordinates place an event at a place and time according to the Frame of Reference and that the x coordinate is the measure of the displacement parallel to the x axis. But apparently that is not correct?

They are relevant if they affect the outcome of the measurements. When you make a measurement you are not simply mathematically determining a coordinate, you are performing a physical experiment of some sort.

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?

 

[Grimble]

Nothing precludes their inclusion, but the "twin paradox" is based on the idea that a moving clock runs slower than a clock at rest,

But it doesn't does it? It only appears to because of the distortion caused by taking measurements at speed.

with the "paradox" arising because people falsely think this should be true in the non-inertial frame of the traveling twin;

How can you say that this isn't true? Special relativity - the movement of one relevant to the other - reciprocal movement.

in fact, the idea that a moving clock runs slower than a clock at rest is only guaranteed to hold in an inertial frame.

Yet it doesn't does it? There is nothing in Einstein's SR that says that it would, only that the observer at rest would see it run slow.

 

[JesseM]

Nothing precludes their inclusion, but the "twin paradox" is based on the idea that a moving clock runs slower than a clock at rest,

But it doesn't does it? It only appears to because of the distortion caused by taking measurements at speed.

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.

with the "paradox" arising because people falsely think this should be true in the non-inertial frame of the traveling twin;

How can you say that this isn't true? Special relativity - the movement of one relevant to the other - reciprocal movement.

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.

in fact, the idea that a moving clock runs slower than a clock at rest is only guaranteed to hold in an inertial frame.

Yet it doesn't does it? There is nothing in Einstein's SR that says that it would, only that the observer at rest would see it run slow.

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.

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

 

[ghwellsjr]

I'm afraid that you've done a Galilean Transformation which is only applicable at low speeds, certainly not at the 0.8c you have proposed in your scenario. You need to perform a valid Lorentz Transformation from your orginal scenario where A, B, and C are at rest to this one where a is at rest. If you had done so, you would understand that all dimensions along the direction of a's motion are contracted, so that the distances from A to C and from C to B are not 4 ly. Also, the clocks that were synchronized in the original FoR or no longer synchronized in this new FoR and they are time dilated.

Your concept of LT as you indicated at the end of the first quote is all mixed up.

But I'm wondering, why are you interested in doing this transformation? What added information do you think it is going to provide?

I am sorry but you are still not grasping what I am saying.
I am not doing any form of transformation, not even a Galilean one.
I am dealing with two separate sets of coordinates.
In one case A,B & C are making measurements entirely within their own set of coordinates ...
And, completely independently a is making measurements according to his own set of coordinates ...
And according to the observations made in the first FoR an event occurs where a passes A at time t0.
Then a second event occurs where a passes C at time t1.

What I am showing is that the measurements taken within each FoR are equal as absolute proper measurements.

They obviously will not be equal in either FoR where LT means that the 'remote/moving' measurement observed from each frame WILL be length contract/time dilated.
But, nevertheless the measurements of time and distance made by A of the distance AC and the time t1-t0, will by necessity have to be equal to the measurements made by a of the distance between his clocks and the time of the two events t1-t0 as these measurements are all entirely local within each FoR and NO TRANSFORMATION is involved.

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:

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.

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

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.

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.

 

[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.