Looking to understand time dilation

[hprog]

Hi, I am learning SR and I need help to get the idea of relativity with two clocks.

Yet I can understand that two different frames of reference can each one claim to be at rest, since this is just a logical argument.
But I am not getting the point how they can each claim the other ones clock is the one who slows down, after all this is physical question and it is like two people arguing whether the Earth is flat or round in which case only one can be right.

To show an example of what bothers me, let's say that I and another person have synchronized clocks.
Now when it is 12:00 on both of our clocks this person takes off in a linear motion and will never return.
so when my clock will show 5:00, then if the other person is the one who moves then his clock will show 4:00, and if I am the one who moves then the other person's clock will show 6:00.
So the person's clock can be either 4:00 or 6:00 but not both, yet we don't know what it is, but this is like if we don't know if the Earth is flat or not and it is a physical question, and can have only one answer, even if we don't know what the answer is.

It is clear to me that I am missing something, so what is it?

 

[Dale]

I am not getting the point how they can each claim the other ones clock is the one who slows down, after all this is physical question and it is like two people arguing whether the Earth is flat or round in which case only one can be right.

To show an example of what bothers me, let's say that I and another person have synchronized clocks...

Hi hprog, welcome to PF!

The key point is the last two words you used. "Synchronized clocks". How do you synchronize two distant clocks that are moving wrt each other, are you familiar with the Einstein synchronization convention?

The reason that they can each say that the other is going slow is that they disagree on the synchronization of distant clocks. So when one says "my clock reads 5:00 at the same time yours reads 4:00" the other says "no, those two events did not happen at the same time".

 

[hprog]

Hi hprog, welcome to PF!

The key point is the last two words you used. "Synchronized clocks". How do you synchronize two distant clocks that are moving wrt each other, are you familiar with the Einstein synchronization convention?

The reason that they can each say that the other is going slow is that they disagree on the synchronization of distant clocks. So when one says "my clock reads 5:00 at the same time yours reads 4:00" the other says "no, those two events did not happen at the same time".

So what does it clock actually read?
In other words if they would meet there would be some value on both clcoks, what is this value?
Also can this also hold true when both are at rest?

 

[kamenjar]

This question has been bogging me down too, so I could use some help too.

It is claimed that relativistic velocities "slow aging". It is my understanding that even when motion is not towards or away (say accelerated orbital path), there this "effect" is still present. If the object in such orbital motion is moving with a clock made on Earth it will receive signals from Earth at higher frequency than it would generate them. If Earth could display a clock it is obvious that it would appear to run faster, where if it could display it, it would appear to be slower.

 

[Dale]

Now when it is 12:00 on both of our clocks this person takes off in a linear motion and will never return.

In other words if they would meet there would be some value on both clcoks, what is this value?

Do you see how these two scenarios are different scenarios? In the first scenario after 12:00 both observers can be considered inertial and the situation is completely symmetric. In the second scenario at least one of the observers must accelerate after 12:00 in order to re-unite, they cannot both be inertial and the situation is no longer symmetric. The second scenario is the famous twin-paradox: http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/TwinParadox/twin_paradox.html

 

[Dale]

It is claimed that relativistic velocities "slow aging". It is my understanding that even when motion is not towards or away (say accelerated orbital path), there this "effect" is still present.

Yes, in fact exactly this scenario that you describe has been experimentally verified with muons in a ring accelerator very close to c. http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Twin_paradox

 

[kamenjar]

Yes, in fact exactly this scenario that you describe has been experimentally verified with muons in a ring accelerator very close to c. http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#Twin_paradox

But then why does the wikipedia article on time dilation try to imply that "Time dilation due to relative velocity symmetric between observers"? How can it be symmetric when one clock is simply slowed and the subjects "see" them as such. Or do they not?

 

[ghwellsjr]

Hi, I am learning SR and I need help to get the idea of relativity with two clocks.

Yet I can understand that two different frames of reference can each one claim to be at rest, since this is just a logical argument.
But I am not getting the point how they can each claim the other ones clock is the one who slows down, after all this is physical question and it is like two people arguing whether the Earth is flat or round in which case only one can be right.

Actually, you are jumping to a conclusion when you say only one can be right (what if the Earth were a cube?) because we could have any number of other defined reference frames and conclude that both of them are wrong.

If we knew how to determine an absolute rest frame where the speed of light is c in all directions, we could legitimately answer the issue of how to assign times to all events and we could determine the apparent timing rate of all moving clocks. But we can't do that. So we do that next best thing which is to arbitrarily pick any reference frame we want and use that to assign times to all events and use its clocks to define the rate of the passage of time. But we are only allowed to do this for the entire scenario for one frame at at time. It is not legitimate to do this for two frames at the same time and complain about apparent discrepancies. But what is important is that no matter which frame we choose, we will get the same answers for comparing clocks when those clocks are brought together. That's all we can know.

 

[ghwellsjr]

But then why does the wikipedia article on time dilation try to imply that "Time dilation due to relative velocity symmetric between observers"? How can it be symmetric when one clock is simply slowed and the subjects "see" them as such. Or do they not?

If you pick anyone frame and define two clocks in relative motion (one can be stationary and one moving or they can both be moving), and you analyze how each clock ticks and how each clock measures the ticks that are transmitted at the speed of light from the other clock, you will see that each of them will measure the tick rate of the other clock to be going slower than their own. Neither one of them has to be at rest in the reference frame for this to happen. They can both be traveling in opposite directions at the same speed and the effect will be the same.

 

[HallsofIvy]

So what does it clock actually read?

What do you mean by "actually read"? You seem to be asserting that there is some "correct value". If you really are then you are denying that relativity is true and there is no point in continuing this! (And there is pretty strong experimental evidence in favor of relativity.)

In other words if they would meet there would be some value on both clcoks, what is this value?

What, exactly, do you mean by "meet"? If you mean "are close enough together to read each others clocks, after having, at some point in the past, sychronized their clocks", but still moving relative to one another, the each would observe the others clock as running slower than his own.

Also can this also hold true when both are at rest?

Special relativity talks about the situation when two frames of reference are in motion, relative to one another, at a constant speed. That means that this question as to be dealt with in two separate parts:

a) The "constant speed" is 0. The two frames of reference were at rest relative to one another when the clocks were synchronized and still are. The two observer's will observe both clocks to still be reading the same time.

b) At least one of the frames of reference was in motion relative to the other at the time the clocks were synchronized and has since been accelerated (or decelerated) to match the other's speed. In that case, "all bets are off"- you have now moved outside of special relativity.

 

[Dale]

But then why does the wikipedia article on time dilation try to imply that "Time dilation due to relative velocity symmetric between observers"? How can it be symmetric when one clock is simply slowed and the subjects "see" them as such. Or do they not?

The case of uniform circular motion, e.g. in a muon storage ring, is not symmetric. One observer will measure centrifugal and Coriolis forces in their reference frame and the other will not.

This is a very important point. Inertial motion is relative, accelerated motion is not. Do you understand what an inertial reference frame is?

 

[granpa]

time dilation and length contraction are easy.
its relativity of simultaneity (google it) that confuse all beginners.

 

[kamenjar]

If you pick anyone frame and define two clocks in relative motion (one can be stationary and one moving or they can both be moving), and you analyze how each clock ticks and how each clock measures the ticks that are transmitted at the speed of light from the other clock, you will see that each of them will measure the tick rate of the other clock to be going slower than their own. Neither one of them has to be at rest in the reference frame for this to happen. They can both be traveling in opposite directions at the same speed and the effect will be the same.

Maybe your pulses at "clock rate" and "the clock" is not the same thing.
A clock is a big fat blueshifted Earth doing 5 revolutions around the sun for every second of my clock and a big bluieshifted lightshow above it UTC display spinning days like nuts while I fly by it (or even perpendicularly far away) at .9999 light speed -- and my nice redshifted light show display and showing "barely moving" while you are looking at it when I am redshifted "frozen in time"? What part of clocks appearing to be slowed don't I understand? Is this, rather colorful, statement wrong?

Simplified/Edited:
I understand that while looking at slow moving (static/earth) objects, and moving near light speed, those objects age faster.
I understand that while looking at fast moving objects, subjects in a slow moving (static/earth) reference frames observe those faster objects to be "slowed".
I understand that if I was inside a fast moving ship, and I wave to an "earthling" though a window of a that ship, the earthling would see my hand move slow.
I understand that if I was inside a fast moving ship, and an "earthling" waves to me, his waving would appear to be fast.
One of the ways to observe the aging is to observe Earth rotations.
I understand that while I am in a fast moving spaceship, Earth performs X rotations around the sun while I blink.
For the above cases, assume that the motion is perpendicular to the line of sight to avoid the effects of speed of propagation of light. Imagine me looking at the "earthling" thru the side window.

Which of these statements are wrong?
Am I thinking in terms of "thermodynamic time"?
Am I thinking in terms of "mechanical clocks"?

I believe there is some discrepancy in thinking and it could be related to atomic clock time vs two kinds of clocks mentioned above.

 

[ghwellsjr]

OK, so I forgot to mention that the two clock have to be identical. Com'on now, can't anything be obvious?

I can't tell if your statement is wrong because it is so full of typos and grammar errors (or whatever) I can't figure out what you are asking about. When you read it, does it confuse you? Why don't you edit it so it makes sense and then I will tell you if it is wrong.

 

[Hellabyte]

Now when it is 12:00 on both of our clocks this person takes off in a linear motion and will never return.
so when my clock will show 5:00, then if the other person is the one who moves then his clock will show 4:00, and if I am the one who moves then the other person's clock will show 6:00.
So the person's clock can be either 4:00 or 6:00 but not both, yet we don't know what it is, but this is like if we don't know if the Earth is flat or not and it is a physical question, and can have only one answer, even if we don't know what the answer is.

I believe the misunderstanding you have is from your assumption that when the first clock shows 4:00, you can somehow instantly know what the other guys clock will read. When you look out into space when your clock reads 4:00, the light from when his clock was 5:00 will now be reaching you. His clock must actually be at some later time (Such as 6:00) when this "5:00" light reaches you. It's like looking out at the stars and realizing that we are seeing very old images of them. Those stars could actually have burnt out in exactly the same way that the moving guys clock could actually be at 6:00 when we see the light from when his clock was at 5:00. Also, when his clock reads 6:00 or whatever the light from when your clock read 4:00 will just be reaching him. Please correct me if my understanding is wrong. I am trying to learn this stuff too.

 

[ghwellsjr]

It's not just that it takes time for information to travel to you from a distant clock and so there is a sense in which the information is old but in addition the traveling clock will appear to be ticking at a slower rate. Consider a clock that is traveling toward you at a very high speed. Let's envision a clock that flashes a bright light once a second. As it is moving toward you, you will see the flashed more often than once per second but as you take into account the time it takes for those flashes to get to you, the actual tick rate will be less than once per second. Then as the clock suddenly passes by you and starts moving away from you, the flashes will be less than once per second, but again, when you take into account the travel time, you will conclude that the clock is ticking more slowly by the same amount as when it was traveling toward you. So the moving clock is ticking at a slower rate than your stationary clock. This means that if while your clock read 4:00 his clock read 5:00, then later when your clock reads 5:00 his clock will not be up to 6:00 yet, it will be running slower.

 

[kamenjar]

...

I feel that I am hijacking the thread so I should post a separate thread...I do understand the effect of approaching and "leaving". I do understand the perception of relative speeds for moving objects. For that purpose I will try to come up with an example where these "effects" will be minimal.

 

[jbar18]

Hi, I am learning SR and I need help to get the idea of relativity with two clocks.

Yet I can understand that two different frames of reference can each one claim to be at rest, since this is just a logical argument.
But I am not getting the point how they can each claim the other ones clock is the one who slows down, after all this is physical question and it is like two people arguing whether the Earth is flat or round in which case only one can be right.

To show an example of what bothers me, let's say that I and another person have synchronized clocks.
Now when it is 12:00 on both of our clocks this person takes off in a linear motion and will never return.
so when my clock will show 5:00, then if the other person is the one who moves then his clock will show 4:00, and if I am the one who moves then the other person's clock will show 6:00.
So the person's clock can be either 4:00 or 6:00 but not both, yet we don't know what it is, but this is like if we don't know if the Earth is flat or not and it is a physical question, and can have only one answer, even if we don't know what the answer is.

It is clear to me that I am missing something, so what is it?

First of all, I'm new to these forums as well. So I don't have much experience answering these questions, but I'm pretty sure I'm right here, and I think I understand what you are getting confused about. It seems you are wondering about how both people can see the other person with a slower clock, instead of one person having a slower clock and the other having a faster clock.

The way I think about it is to imagine that you are in space, with your clock, and your friend has their clock some distance away from you. Apart from you two with your clocks, there is literally nothing else around. It might also help if you think of the actual rate of ticking, instead of the actual time shown on the clocks.

So let's pretend that you are passing each other, neither of you know which of you is actually moving. You could both be moving a fraction of the total speed, or one of you could be moving the whole speed and the other could be stationary. You will both agree on the total speed. You've said that you understand that in your question.

Well I think that's your answer - if you pass each other with a given velocity v, you could each argue that you are stationary, which means that you are both arguing that the other person is doing all of the moving. So if you were to pass your friend, you would see him moving at speed v, and his clock would be ticking 1/1-(v^2/c^2) times slower. But then, he also observes you moving at this speed v (since both of you appear stationary to yourselves) and so he would also see your clock ticking 1/1-(v^2/c^2) times slower. You both observe each other's clocks moving slower because as far as you are concerned, you are never moving, everything else is.

Was this what you were getting confused about? The more you think about it the more logical it gets.

I hope that answered your question.

 

[Dale]

hprog, are you comfortable with the concept now or are you still confused?

 

[Grimble]

Well I think that's your answer - if you pass each other with a given velocity v, you could each argue that you are stationary, which means that you are both arguing that the other person is doing all of the moving. So if you were to pass your friend, you would see him moving at speed v, and his clock would be ticking 1/1-(v^2/c^2) times slower. But then, he also observes you moving at this speed v (since both of you appear stationary to yourselves) and so he would also see your clock ticking 1/1-(v^2/c^2) times slower. You both observe each other's clocks moving slower because as far as you are concerned, you are never moving, everything else is.

Was this what you were getting confused about? The more you think about it the more logical it gets.

I hope that answered your question.

You have the explanation in a nut shell there.

And if you take it a step further, with each person slowing an equal amount with respect to the other (to get rid of any arguments about the acceleration) then when they have slowed to be at rest with one another their relative velocity will be zero and their clocks slowing, as shewn by your formula above, will also be zero.

The clocks slowing is due to the relative speeds.
Neither clock will appear to be slowed by the person adjacent to that clock.
So does either clock actually slow, or is the slowing merely an effect of measuring a moving object?

 

[jbar18]

So does either clock actually slow, or is the slowing merely an effect of measuring a moving object?

I think that this is a question I was having trouble with as well. Eventually I came to the conclusion that the question isn't really answerable, it's one of those philosophical questions in disguise as a physics question, kind of like "when you see an object, do you see the object or do you see the light that the object has reflected?". Either way, you know beyond reasonable doubt that the object was there at at least one point in time.

Similarly, the old classic "if a tree falls down in the woods and nobody is around to hear it, does it make a sound?" These questions are really impossible to answer, anybody who tries to give a definite answer will probably just sound silly. Basically, does something have to be observed in order to exist?

It's not exactly the same question, but it seems like that's what you're getting at. I suppose that if you were only ever observing yourself, the clocks would never ever slow. So then, if an observer sees them slow, are the clocks actually slowing? Well the clock can be ticking at any speed really, it only ever depends on the observer, so the observer's frame is the important one. And so if we measure something, we have to do it knowing that it might only be correct in our frame. But is that really a problem? Especially now that with relativity we can correct our observations for any frame.

I guess for all practical intents and purposes, yes, it does actually happen.

 

[Dale]

And if you take it a step further, with each person slowing an equal amount with respect to the other (to get rid of any arguments about the acceleration) then when they have slowed to be at rest with one another their relative velocity will be zero and their clocks slowing, as shewn by your formula above, will also be zero.

Careful here. What you describe does not "get rid of any arguments about the acceleration", but it restores the symmetry and gets rid of any argument about asymmetry. Usually the only point of the acceleration is that it breaks the symmetry.

 

[Grimble]

I think that this is a question I was having trouble with as well. Eventually I came to the conclusion that the question isn't really answerable, it's one of those philosophical questions in disguise as a physics question, kind of like "when you see an object, do you see the object or do you see the light that the object has reflected?". Either way, you know beyond reasonable doubt that the object was there at at least one point in time.

Similarly, the old classic "if a tree falls down in the woods and nobody is around to hear it, does it make a sound?" These questions are really impossible to answer, anybody who tries to give a definite answer will probably just sound silly. Basically, does something have to be observed in order to exist?

It's not exactly the same question, but it seems like that's what you're getting at. I suppose that if you were only ever observing yourself, the clocks would never ever slow. So then, if an observer sees them slow, are the clocks actually slowing? Well the clock can be ticking at any speed really, it only ever depends on the observer, so the observer's frame is the important one. And so if we measure something, we have to do it knowing that it might only be correct in our frame. But is that really a problem? Especially now that with relativity we can correct our observations for any frame.

I guess for all practical intents and purposes, yes, it does actually happen.

I think we need to steer a course here that sticks to science rather than philosophy, for scientifically the tree falling causes vibrations in the air at audible frequencies whether there is anyone to hear them or not.

In SR for an observer at rest in the FoR all measurements are taken with rulers and clocks in that FoR.
In SR for an observer moving relative to the body being measured the measurements are taken with the rulers and clocks moving with that body i.e. in that body's FoR. Those measurements are then transformed using the Lorentz equations to give what the moving observer would measure and we see the clock slowing.

But if we were to take those measurements using the rulers and clocks of the moving FoR what would we find?
I have heard it said that we cannot do that, yet it seems to me very simple and it confirms that the clock does not slow, which is not surprising as it can only run at one rate.

 

[Grimble]

Careful here. What you describe does not "get rid of any arguments about the acceleration", but it restores the symmetry and gets rid of any argument about asymmetry. Usually the only point of the acceleration is that it breaks the symmetry.

But where does acceleration occur as a term in any SR formulae?

 

[maxmaxus]

it's just...nature - it sounds contraintuitive because this is just your everyday experience in life, but it's "logical" in terms that there is nothing contradictive in it. For instance...factorial of 2 is 2 but factorial just 4 gitits above is a few miliions...yes - it's "controversial" at first sight - but it is again logical. You can't really find a "rational explation" of relativity (both types...) unless you own a car ever since you were 5 years old and that car is able to accelerate closely to c (which i doubt).

 

[Mike_Fontenot]

But where does acceleration occur as a term in any SR formulae?

That's a very good question.

If you look at the Lorentz equations, you won't find any parameter there for acceleration. But the Lorentz equations directly concern only uniform motion (constant relative velocity) between two separated clocks (or people(!)).

Even in my CADO equation, and also in my delta(CADO_T) equation, which DO concern the situation where the "traveling twin" sometimes accelerates (but the "home twin" does not), there is no parameter there that DIRECTLY refers to acceleration.

So how does acceleration influence the traveler's conclusions about the current age of the home twin?

How this happens can be seen most easily by looking at the delta(CADO_T) equation. That equation is:

delta(CADO_T) = -L * delta(v) ,

where delta(CADO_T) is the CHANGE in the home twin's age, according to the traveling twin, when the traveler does an instantaneous velocity change of delta(v). The quantity L is the separation between the twins (according to the home twin), when the traveler makes the instantaneous velocity change.

Now, when we say that the traveler does an instantaneous velocity change, we really mean that the velocity change occurs over an infinitesimally small elapsed time for the traveler ... call that elapsed time delta(t).

The traveler's acceleration, during that tiny time interval, is just the velocity change, divided by the elapsed time:

a = delta(v) / delta(t) ,

(when delta(t) is sufficiently small).

In the idealized (unrealizable) case of an instantaneous velocity change, delta(v) has a magnitude that can be up to (but not including) a value of 2, whereas delta(t) is infinitesimal. So the acceleration, a, in that case, is extremely large ... so large that we call it infinite. (Mathematically, the acceleration in that case is a Dirac delta function, which is a pulse of infinite height, but with infinitesimal width, and finite area ... if you don't know what that is, don't worry about it).

For finite accelerations, there is also an effect on the current age of the home twin (according to the traveler). In fact, when the separation is large enough, even +-1g accelerations produce effects that are qualitatively quite similar to the effects of instantaneous accelerations.

The way finite accelerations enter into the CADO equation is fairly involved to explain. I'll only be able here to give an idea of how it enters. The CADO equation is:

CADO_T = CADO_H - L*v,

where

CADO_T is the current age of the home twin, ACCORDING TO THE TRAVELER, at any given instant t in the traveler's life,

CADO_H is the current age of the home twin, ACCORDING TO THE HOME TWIN, at any given instant t in the traveler's life,

L is their current distance apart, in lightyears,
according to the home twin,

and

v is their current relative speed, in lightyears/year,
according to the home twin. v is positive
when the twins are moving apart.

Although it's not shown explicitly, all of the quantities in the CADO equation are functions of the traveler's age, t. To use the CADO equation when the traveler undergoes a finite acceleration, a(t), we have to determine how the velocity, v(t), varies as a function of the traveler's age. When you make that computation, you have to use the acceleration a(t). That's how the acceleration comes into it.

Here's a link to a previous posting that gives an example of how the CADO equation is used, when the velocity changes are instantaneous:

https://www.physicsforums.com/showpost.php?p=2934906&postcount=7

And here's a link to a previous posting that gives an example of how the delta(CADO_T) equation is used, when the velocity changes are instantaneous:

https://www.physicsforums.com/showpost.php?p=2923277&postcount=1

There is an example with +-1g accelerations on my webpage:

http://home.comcast.net/~mlfasf

The CADO equation is derived in my paper:

"Accelerated Observers in Special Relativity",
PHYSICS ESSAYS, December 1999, p629. Mike Fontenot

 

[Grimble]

I'm sorry Mike but I cannot follow how acceleration can have ANY effect upon the ages of the twins.

Maybe I see it too simply but the Lorentz Transformation Formulae are showing how the age of one twin is measured differently by the other twin as a factor of their current velocity. It is purely an instantaneous observation and has nothing to do with what the apparent ages were even a fraction of a second before the current observation.

I would give the analogy of measuring the apparent height of a man walking away from an observer. His apparent height is solely due to their current separation.

So a change in the twins relative velocity could only affect the current measurements.
And if the twins come back together and come to rest with respect to one another their respective measurements as given by LT will be identical. It is only while they have a relative velocity that the LT gives a difference to their ages.

i.e. When the traveller turns round he first comes to rest with respect to the home twin. At this moment their ages, as shown by LT, will once again be equal. Then the traveller returns and during that return journey their ages and apparent ages will differ once again but when the traveller has returned and come to rest once again they will have identical ages once again.

It is only if the traveller keeps on traveling past the home twin on his return that the age difference is apparent and it will be apparent to each of them.

Their relative velocity is a rotation. a change of velocity means that the angle of rotation changes; therefore acceleration (change of velocity) is shown by rotation, this cannot lead to a wiggly line! It means that the whole line remains straight. it is only the angle between the lines that changes. When deceleration to rest occurs the angle between the lines becomes zero and the lines become one.

How else can it work?

Sorry to go on but I am confused at it all seems so simple and straightforward. It is all about which rulers and clocks are being used to take the measurements.

Grimble

 

[matheinste]

I would give the analogy of measuring the apparent height of a man walking away from an observer. His apparent height is solely due to their current separation.

So a change in the twins relative velocity could only affect the current measurements.
And if the twins come back together and come to rest with respect to one another their respective measurements as given by LT will be identical. It is only while they have a relative velocity that the LT gives a difference to their ages.

i.e. When the traveller turns round he first comes to rest with respect to the home twin. At this moment their ages, as shown by LT, will once again be equal. Then the traveller returns and during that return journey their ages and apparent ages will differ once again but when the traveller has returned and come to rest once again they will have identical ages once again.

It is only if the traveller keeps on traveling past the home twin on his return that the age difference is apparent and it will be apparent to each of them.

Grimble

This maybe where you are going wrong. When the traveller reaches the turnaround point he is momentarily, or for any desired length of time, at rest with respect to the stay at home and their clocks will be ticking at the same rate as each others as they are at rest with respect to each other.

The same is true when the traveller returns and halts next to the stay at home, they are at rest in the same inertial frame and so their clocks tick at the same rate as each other. And of course at that stage it is meaningful to compare clocks, and although they are now once again ticking at the same rate, the traveller's clock shows less accumulated time.

Matheinste.

 

[Mike_Fontenot]

[...]
When the traveller turns round he first comes to rest with respect to the home twin. At this moment their ages, as shown by LT, will once again be equal.
[...]

No, that's incorrect. Whenever their relative velocity is zero, their ageing RATES will indeed be equal. But their AGES will generally NOT be equal. However, they WILL each agree about the correspondence between their ages (whereas, when their relative velocity is non-zero, they will generally disagree about that correspondence).

Mike Fontenot

 

[Dale]

But where does acceleration occur as a term in any SR formulae?

Proper acceleration and coordinate acceleration are essential to the definition of an inertial reference frame. I.e. an inertial frame is an orthonormal coordinate system in which any object with a proper acceleration of 0 also has a coordinate acceleration of 0.

 

[ghwellsjr]

...
When the traveller reaches the turnaround point he is momentarily, or for any desired length of time, at rest with respect to the stay at home and their clocks will be ticking at the same rate as each others as they are at rest with respect to each other.

The same is true when the traveller returns and halts next to the stay at home, they are at rest in the same inertial frame and so their clocks tick at the same rate as each other. And of course at that stage it is meaningful to compare clocks, and although they are now once again ticking at the same rate, the traveller's clock shows less accumulated time.

Matheinste.

All that you say here is true, but it is also true for any other reference frame that you want to analyze the situation in. It's not because two separated clocks are at rest in the same reference frame that makes them tick at the same rate, it's that they are at rest with respect to each other that makes them tick at the same rate, even if they are not at rest in the reference frame. And, as you pointed out, the two clocks need to be brought back together to compare their times before any conclusion can be made about the difference in their accumulated times. But the same conclusion will be achieved if you analyze the whole situation in a different reference frame where they are not at rest at the beginning, middle and end. Reference frames are human conventions and have nothing to do with what is happening in reality.

My complaint with what Mike is proposing with his CADO formula is that he keeps changing reference frames during the course of his analysis and that can lead to any kind of conclusion. I have also stated to him that when an object experiences acceleration, it changes its aging rate, not its age, but rather it ages differently compared to what it was doing before it accelerated. But he wants to attribute huge age differences to objects that are not accelerating just because some other object is accelerating (or something like that). If he would just say that he has discovered a formula that correctly calculates the final age difference between the two twins without any interpretation of what is happening while they are separated, then I could recognize his achievement, but he insists that his interpretation has some legitimate real meaning all during the trip.

 

[Mike_Fontenot]

[...]
My complaint with what Mike is proposing with his CADO formula is that he keeps changing reference frames during the course of his analysis ...
[...]

No, I use a SINGLE reference frame for the traveler during his entire voyage. I call that single reference frame the "CADO" or "{MSIRF(t)}" reference frame ... it is NOT an inertial frame.

Mike Fontenot

 

[matheinste]

It's not because two separated clocks are at rest in the same reference frame that makes them tick at the same rate, it's that they are at rest with respect to each other that makes them tick at the same rate

Aren't these two statements saying the same thing.

Matheinste.

 

[ghwellsjr]

No, I use a SINGLE reference frame for the traveler during his entire voyage. I call that single reference frame the "CADO" or "{MSIRF(t)}" reference frame ... it is NOT an inertial frame.

Mike Fontenot

But even if a single inertial reference frame is used, no conclusion can be made about the age difference between two clocks while they are separated until they come back together. You will come to different conclusions about their relative ages while separated just by selecting different inertial reference frames. If no inertial reference frame can be considered "preferred" according to SR, how can you claim that a non-inertial reference frame is "preferred" just because it is the one a particular observer is at rest in?

 

[ghwellsjr]

Aren't these two statements saying the same thing.

Matheinste.

Yes, I'm not disagreeing with anything you said, I'm merely pointing out that we shouldn't think that it's necessary to say that the two "are at rest in the same inertial frame" because when they are at rest with each other in one inertial frame, they will be at rest with each other in all inertial frames.

 

[matheinste]

Yes, I'm not disagreeing with anything you said, I'm merely pointing out that we shouldn't think that it's necessary to say that the two "are at rest in the same inertial frame" because when they are at rest with each other in one inertial frame, they will be at rest with each other in all inertial frames.

That's fine.

Matheinste.

 

[Mike_Fontenot]

[...]
If no inertial reference frame can be considered "preferred" according to SR, [...]
[...]

If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame. There is only one inertial frame that agrees with the elementary measurements and elementary calculations that he makes, using clocks and rulers with which he is stationary. That preferred inertial frame is the one in which he is stationary.

Of course, other inertial observers (who are moving at some constant velocity with respect to the above inertial observer) will prefer a different inertial frame (the one in which THEY are stationary).

What special relativity says is that there is no single inertial frame that ALL inertial observers prefer. There is no single inertial reference frame that is UNIVERSALLY special.

Mike Fontenot

 

[Dale]

If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame.

Only as a matter of personal and computational convenience, not as a matter of physical necessity. And there may be some situations where a different frame is more convenient, e.g. the center of momentum frame for a collision.

There is only one inertial frame that agrees with the elementary measurements and elementary calculations that he makes, using clocks and rulers with which he is stationary.

You have never provided any substantiation for this, and it is only even possibly true if you define "elementary calculations" such that it is a tautology. Otherwise the principle of relativity ensures that all inertial reference frames will give the same predictions for the result of any given measurement and your statement is wrong.

 

[ghwellsjr]

If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame. There is only one inertial frame that agrees with the elementary measurements and elementary calculations that he makes, using clocks and rulers with which he is stationary. That preferred inertial frame is the one in which he is stationary.

Of course, other inertial observers (who are moving at some constant velocity with respect to the above inertial observer) will prefer a different inertial frame (the one in which THEY are stationary).

What special relativity says is that there is no single inertial frame that ALL inertial observers prefer. There is no single inertial reference frame that is UNIVERSALLY special.

Mike Fontenot

The term "preferrred frame" has a specific meaning. You are making the same mistake that I am dealing with someone else on in this thread:

https://www.physicsforums.com/showthread.php?t=442132&page=2

It doesn't mean a personal preference like "I prefer blondes". Even if everyone in the world preferred blondes, it wouldn't mean they are "UNIVERSALLY special". I hope you don't think the word "special" in Special Relativity has anything to do with the concept of "preferred". This has nothing to do with anyone's preferences. It has to do with whether we base our physics on the concept of an æther.

 

[Mike_Fontenot]

[...]
If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame.
[...]

Only as a matter of personal and computational convenience, not as a matter of physical necessity.
[...]

It's not a matter of physical necessity that you refrain from constantly hitting yourself in the head with a hammer, either.

The well-known results of time-dilation and length-contraction are directly available for your use (when you are an inertial observer), provided that you choose to use the usual Lorentz coordinates, in an inertial frame in which you are stationary. And the time coordinate, with that choice, corresponds to the time shown on your OWN watch. The spatial coordinate, with that choice, correponds to the lengths as reported by your OWN rulers and measuring tapes. There is a REASON why Einstein chose those coordinates, when developing his special theory.

In the spirit of general relativity, you are of course free to choose some other set of coordinates, by transforming those Lorentz coordinates in an almost unlimited number of ways, provided that the eigenvalues of the resulting metric are either {1, -1, -1, -1} or {-1, 1, 1, 1}, assuming that spacetime is everywhere flat.

In those alternative coordinate systems, your "clocks" would behave in very odd ways, compared to ordinary clocks. And likewise for your "rulers". Real work-a-day experimental physicists would seldom, if ever, choose those types of "measuring devices" to do their experiments in their laboratories. There is a REASON that the phrase "in the laboratory frame" is used so often.

[...]
And there may be some situations where a different frame is more convenient, e.g. the center of momentum frame for a collision.
[...]

Not if you're one of the two people involved in the impending collision. I've spent a lot of hours flying airplanes and gliders, constantly needing to avoid mid-air collisions. In all that time, I've NEVER used the center of mass coordinate system to help me avoid mid-air collisions.

Mike Fontenot

 

[Mike_Fontenot]

[...]
I hope you don't think the word "special" in Special Relativity has anything to do with the concept of "preferred". This has nothing to do with anyone's preferences. It has to do with whether we base our physics on the concept of an æther.
[...]

No, the term "special" in "special relativity" refers to the special case of limitless flat spacetime (no gravitational fields).

My other use of the term "special", as in "There is no single inertial reference frame that is UNIVERSALLY special", referred to the fact that in flat spacetime, there is no single inertial reference frame that is preferred by ALL inertial observers.

Mike Fontenot

 

[ghwellsjr]

You got the word "special" right, I just don't know why you can't get the word "preferred" right.

 

[ghwellsjr]

Real work-a-day experimental physicists would seldom, if ever, choose those types of "measuring devices" to do their experiments in their laboratories. There is a REASON that the phrase "in the laboratory frame" is used so often.

I thought your CADO formula was for a traveler going at substantial speeds, traveling incredible distances for almost an entire liftetime to "know" at any instant of time how old his daughter was back at home.

 

[JesseM]

And the time coordinate, with that choice, corresponds to the time shown on your OWN watch.

But your own watch can only be used to assign coordinates to events on your own worldline, an accelerating observer could come up with multiple non-inertial coordinate systems which all agree that the coordinate time between events on his worldline is equal to his own proper time between those events, but which disagree about the time between events far from him or about simultaneity of distant events. There is no physical reason to think any of these non-inertial coordinate systems more accurately represents his "own measurements" of time than any other.

The spatial coordinate, with that choice, correponds to the lengths as reported by your OWN rulers and measuring tapes.

For an inertial observer "their own" ruler presumably means an inertial ruler at rest relative to themselves. But what does it mean for an accelerating observer? Does it mean that they are using an accelerating ruler, and if so how to decide how parts far from them are accelerating (is the ruler accelerating in a Born rigid way for example?) Or are you supposing that at each moment they define "their own" ruler to be a different inertial ruler which is instantaneously at rest relative to themselves at that moment?

In those alternative coordinate systems, your "clocks" would behave in very odd ways, compared to ordinary clocks. And likewise for your "rulers".

Again this is meaningless if you haven't defined precisely what "your clocks" and "your rulers" means for an accelerating observer, and why you think the accelerating observer doesn't have a choice of how clocks far from his own worldline should move or be synchronized with one another, and likewise why he doesn't have a choice about how points on his own (accelerating?) ruler far from his own worldline should themselves be moving (remember that in SR there is no simple notion of how the back end of a 'rigid ruler' should accelerate if we know how the front end is accelerating, since there is no well-defined notion of 'rigidity' for objects undergoing arbitrary acceleration, physical rulers will behave like silly putty or slinkys when you accelerate one end)

 

[Dale]

It's not a matter of physical necessity that you refrain from constantly hitting yourself in the head with a hammer, either.
...
I've spent a lot of hours flying airplanes and gliders, constantly needing to avoid mid-air collisions. In all that time, I've NEVER used the center of mass coordinate system to help me avoid mid-air collisions.

Very cute. Completely irrelevant to the topic, but cute.

So Mike, are you going to continue to dodge the challenge and hide from the issue? After a dozen or so requests you have had plenty of opportunity but you still can't even define your terms let alone demonstrate your claim. I suspect that you know that your claim is wrong.

 

[ghwellsjr]

I think Mike has defined his terms "elementary measurements" and "elementary calculations". Look at this post:

If the traveler is perpetually inertial, then he certainly DOES have a "preference" for one particular inertial frame. There is only one inertial frame that agrees with the elementary measurements and elementary calculations that he makes, using clocks and rulers with which he is stationary. That preferred inertial frame is the one in which he is stationary.

And now this one:

The well-known results of time-dilation and length-contraction are directly available for your use (when you are an inertial observer), provided that you choose to use the usual Lorentz coordinates, in an inertial frame in which you are stationary. And the time coordinate, with that choice, corresponds to the time shown on your OWN watch. The spatial coordinate, with that choice, correponds to the lengths as reported by your OWN rulers and measuring tapes. There is a REASON why Einstein chose those coordinates, when developing his special theory.

It looks like Mike thinks that Einstein was claiming that you have to use the inertial frame in which you are stationary to make SR work.

Then, somehow, he slips from a perpetually inertial frame to a non-inertial frame when the traveler starts his voyage:

No, I use a SINGLE reference frame for the traveler during his entire voyage. I call that single reference frame the "CADO" or "{MSIRF(t)}" reference frame ... it is NOT an inertial frame.

And I'm sure he really believes this is what Einstein was promoting.

 

[Grimble]

This maybe where you are going wrong. When the traveller reaches the turnaround point he is momentarily, or for any desired length of time, at rest with respect to the stay at home and their clocks will be ticking at the same rate as each others as they are at rest with respect to each other.

The same is true when the traveller returns and halts next to the stay at home, they are at rest in the same inertial frame and so their clocks tick at the same rate as each other. And of course at that stage it is meaningful to compare clocks, and although they are now once again ticking at the same rate,

But yes I agree with what you are saying here! So where does this show that I am going wrong?

the traveller's clock shows less accumulated time.

Now this is where you lose me, What has SR and transforming measurements with LT possibly got to do with ACCUMULATED time?

LT is used to transform measurements: what a clock currently reads, not a calculation of the difference between two readings on a clock.

LT is a function of the CURRENT velocity. It has absolutely nothing to do with the history of how their relative speed might have varied over the course of their travel relative to one another.

 

[Dale]

It looks like Mike thinks that Einstein was claiming that you have to use the inertial frame in which you are stationary to make SR work.

Which is a complete misunderstanding of SR, particularly the first postulate. I think it is obvious to everyone besides Mike.

 

[matheinste]

i.e. When the traveller turns round he first comes to rest with respect to the home twin. At this moment their ages, as shown by LT, will once again be equal. Then the traveller returns and during that return journey their ages and apparent ages will differ once again but when the traveller has returned and come to rest once again they will have identical ages once again.

It is only if the traveller keeps on traveling past the home twin on his return that the age difference is apparent and it will be apparent to each of them.

Grimble

"when the traveller has returned and come to rest once again they will have identical ages once again."

"It is only if the traveller keeps on traveling past the home twin on his return that the age difference is apparent and it will be apparent to each of them."

These two statements are incorrect.

Perhaps I have missed something from your previous posts that puts these statements in context, if so forgive me and point them out.

Don't forget you cannot use one LT for the whole of the journey for the travellers non inertial frames.

Matheinste

 

[ghwellsjr]

Now this is where you lose me, What has SR and transforming measurements with LT possibly got to do with ACCUMULATED time?

LT is used to transform measurements: what a clock currently reads, not a calculation of the difference between two readings on a clock.

LT is a function of the CURRENT velocity. It has absolutely nothing to do with the history of how their relative speed might have varied over the course of their travel relative to one another.

The Lorentz Transform assumes that there has only ever been one velocity relative to the frame of reference. In this case, you can use LT to calculate both the difference in clock readings between the stationary one and the moving one, and/or you can calculate the relative tick rate between the two clocks. If you then accelerate the moving clock, you have to be very careful about how you describe what is happening to get a precise unique result. Usually, people just take a short cut and assume that the accelerating clock is gradually changing to its new tick rate as calculated by LT. Then to get an idea of the difference in the clock readings, you have to integrate, or multiply the time interval that a clock was ticking at a particular rate by its tick rate to find the accumulated time at the point of acceleration. And remember, this is all being done from the one frame of reference of the stationary observer (or any other single frame of reference).

In other words, if there were only one velocity involved, you could use LT directly to calculate the difference in clock readings OR you could use the tick rate multiplied by the time interval and you will get the same answer but when you introduce a new velocity, you cannot use the former method to calculate the difference in clock readings, you have to use the later twice, once for each time interval that the clock was moving at each speed and then add them together.