Looking to understand time dilation

[Grimble]

Grimble;

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

Yes.

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

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

thus it's only applicable to that observer.

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

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

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

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

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

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

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

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

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

Persevere, it's not that complicated.

 

[Dale]

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

Yes. This is called the relativity of simultaneity.

 

[ghwellsjr]

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

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

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

And which conclusion is that?

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

 

[Mentz114]

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

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

 

[ghwellsjr]

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

 

[Passionflower]

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

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

 

[Mentz114]

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

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

PassionFlower, you have introduced the 'quintuplet paradox'

 

[ghwellsjr]

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

 

[Mentz114]

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

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

 

[ghwellsjr]

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

 

[Mentz114]

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

The meaning of "frame dependent" is the same as "dependent on relative velocity". The observations of the twins depend on their relative velocity at the time of the observation.

 

[ghwellsjr]

It doesn't matter which frame you use, you will determine that each twin makes the same measurements.

 

[Mentz114]

It doesn't matter which frame you use, you will determine that each twin makes the same measurements.

The time-dilation between the two frames is the result of the LT of coordinates between frames. But each twin realizes that the others clock is actually working properly in its rest frame, so how can the time-dilation be given any credence ? What is indisputable is the fact that they will have different ages.

I don't see the paradox.

 

[ghwellsjr]

What you need to do is pick ONE frame. Use time dilation and length contraction as appropriate for each twin as they measure the other one's clock, then you'll see the paradox.

 

[Mentz114]

What you need to do is pick ONE frame. Use time dilation and length contraction as appropriate for each twin as they measure the other one's clock, then you'll see the paradox.

No, I won't see a paradox because there isn't one. It sounds as if you're mixing measurements between frames.

I think it's time to bring on the muon. As you know, this particle decays after a short time and it's life-span can be estimated. When the life-span of a muon is measured in the lab, it is found to be much larger than calculated. This is a space-time diagram of a muon being created from a collision, then decaying a short time later into two other particles. Read the diagram from the bottom up which is how time is flowing. All the spatial movement is on the horizontal line, i.e. left-to-right or right-to-left.

In the lab frame the time measured is 8.9 units, but along the muons worldline the interval is 2.9 units. So what is the life-span of the muon ? I go for the time along the muons line for a very good reason, which I will now give, if you're still following me.

The case I've illustrated shows the muon traveling at 0.8c. Suppose in another lab a similar experiment is done where the muon travels at a different speed. The lab clocks now will give something different from the 8.9 I got but they will get the same value along the muons line. So, do we argue about the lab clock reading, or agree on the muons own time, which will be the same ?

 

[ghwellsjr]

You are the one who is mixing measurements between frames. If you won't do what I'm asking you to do, your eyes will never be open.

 

[matheinste]

What you need to do is pick ONE frame. Use time dilation and length contraction as appropriate for each twin as they measure the other one's clock, then you'll see the paradox.

I don't think this is the twin (non) paradox that we all love so much and never get tired of.

Matheinste.

 

[Mentz114]

GHWells, I think we've said all that can be said and we're not getting through to each other so it's probably best to leave it there.

 

[phyti]

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

That's the way you set it up, with A,B, & C in the reference (zero) frame, and a & b moving at .8c relative to it!
We're not dealing with frames in general, but this specific one which you proposed.

 

[granpa]

a and b are stationary and located at the same point and their clocks are synchronized.
b accelerates away at gamma=2
each sees the others clock as time dilated up to the point (call it 'x') that b starts to decelerate.
b reaches 'x' and decelerates to a complete stop
after b stops a perceives that b's clock has ticked half as many times as a's clock.
b agrees.

 

[Mike_Fontenot]

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

It's not a paradox, because the traveler (he) concludes the home twin (she) suddenly ages during his turnaround. When he adds her ageing during his inertial outbound leg, to her ageing during his turnaround, and then to her ageing during his inbound inertial leg, he gets her correct total ageing at the end of his trip (which of course must agree with HER conclusions about her own ageing during his entire trip).

Take the gamma = 2 case, and suppose he ages 20 years on each of his inertial legs, for a total of 40 years. Then she (according to her) ages 40 years during each of his inertial legs, with no ageing during his turnaround, for a total of 80 years.

According to him, she ages 10 years during each of his inertial legs, for a total on the inertial legs of 20 years, plus 60 years during his turnaround, for a total of 80 years during his entire trip (which agrees with her conclusion, as it of course must). No paradox, no inconsistency.

Anyone who insists that there can be no sudden ageing of the home twin, according to the traveler, during his turnaround, must then conclude that the traveler in NOT actually inertial during his constant-velocity legs. I.e., they must conclude that he isn't allowed to use the time-dilation result during his constant-velocity legs.

Mike Fontenot

 

[phyti]

a and b are stationary and located at the same point and their clocks are synchronized.
b accelerates away at gamma=2
each sees the others clock as time dilated up to the point (call it 'x') that b starts to decelerate.
b reaches 'x' and decelerates to a complete stop
after b stops a perceives that b's clock has ticked half as many times as a's clock.
b agrees.

Not if b uses the SR clock synch method, which results in reciprocal measurements, i.e., b will conclude the a clock is running at half rate. SR is symmetrical/reciprocal by design.

 

[ghwellsjr]

Anyone who insists that there can be no sudden ageing of the home twin, according to the traveler, during his turnaround, must then conclude that the traveler in NOT actually inertial during his constant-velocity legs. I.e., they must conclude that he isn't allowed to use the time-dilation result during his constant-velocity legs.

I insist that there can be no sudden aging of anyone and I do not come to either of your other two conclusions. If you will pay attention to what I am going to explain to you here, then you will also be able to understand what is now evading you. OK? Please pay close attention.

First off, you have to understand how an observer in relative motion to another observer measures the time-dilation of the other observer. It has to do with relativistic doppler. It simply means that each twin has an identical clock that the other one can observe. An easy way to make this happen is for each of them to have a clock that flashes at some interval, say, once per second or once per minute. Each observer will count their own flashes and will count their twin's flashes when they see them. They will also calculate the ratio of the clock rate of their twin's clock compared to their own. As long as they are stationary with respect to each other, their own outgoing flash will occur as often as their twin's incoming flash. When they see the flash rates being identical, they can conclude that there is no relative motion between them and there is no time dilation.

Now as soon as the traveling twin accelerates away and achieves a terminal speed, the ratio of the incoming flashes will be lower than the outgoing flashes. From this ratio, they can each calculate the relative speed between them and from that they can calculate the time-dilation factor. It will be symmetrical, they will both measure the same ratio and detemine the same speed and the same time dilation. While they are moving apart, this ratio will be less than one. At the turn around point, assuming the traveling twin achieves the same speed on the inbound leg as the outbound leg, the traveling twin will immediately see an increase in the doppler frequency, in fact, it will be the reciprocal of what it was before, but this new ratio will determine the same speed as before and the same time dilation as before.

However, the home twin will not see the shift in the doppler frequency (from a ratio of less than one to greater than one) until some time later because it takes time for the distant increased frequency of the flashes of light to reach him. But when it does, he will still determine that the speed of his twin is the same as it was before, just in the opposite direction (approaching instead of retreating) and the time-dilation factor is the same as it was before.

Now what accounts for the difference in the aging of the twins? It is simply that they are counting the low rate doppler versus the high rate doppler for different lengths of time. The traveling twin counts the low rate and high rate coming from the home twin for an equal amount of time (corresponding to the outgoing and incoming legs) but the home twin counts the low rate doppler for much longer than the high rate doppler of the traveling twin so his sum total will be much smaller. They both agree on the final age difference of themselves and their twins and they both agree that the traveling twin aged less.

Now if you want to say that each twin can observe the aging of the other twin, you would have to say that each twin always measures a constant aging of the other twin because the time-dilation factors never change except for the brief moments during acceleration.

Now I want to point out for the sake of others on this thread, that I have not declared any frame of reference in the analysis of the scenario. In fact, it doesn't matter which frame of reference you want to use, the analysis will be identical. All measurements and observations made by observers in a scenario will not change just because you analyze it from different reference frames.

 

[Grimble]

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

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

And the time measured along those those world lines, between two points/events on those world lines will be identical. That is how it seems to me. How can it be anything else?

The Lorentzian length, must be the same for each as calculated from the other, they are, after all, reciprocal journeys, one of the other.

 

[granpa]

Not if b uses the SR clock synch method, which results in reciprocal measurements, i.e., b will conclude the a clock is running at half rate. SR is symmetrical/reciprocal by design.

As I clearly stated, a and b are, at that point, stationary with respect to each other.

 

[Grimble]

Yes. This is called the relativity of simultaneity.

The clock can only show one time, that is what clocks do ...

Isn't it just that different observers read/observe/record different times on that same single clock?

And what has that to do with thehttp://www.bartleby.com/173.html" ?

 

[Grimble]

And which conclusion is that?

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

Yes that is the twin paradox, but I don't see any reason at all why the acceleration should have that effect. The paradox to me is that they should each see that the other is younger than they are and that occurs in the outward leg of the journey.

 

[granpa]

acceleration changes the relativity of simultaneity which changes the accelerating twins calculated value of the present time on the stationary twins clock

 

[Grimble]

acceleration changes the relativity of simultaneity which changes the accelerating twins calculated value of the present time on the stationary twins clock

Yes, it changes the angle of rotation between the FoRs?

 

[Dale]

The clock can only show one time, that is what clocks do ...

Isn't it just that different observers read/observe/record different times on that same single clock?

And what has that to do with thehttp://www.bartleby.com/173.html" ?

One clock only reads a single number at any given event, and all observers in all reference frames agree on that value. But that is not the issue here. The issue here is how two different clock's times compare, and that is determined by the relativity of simultaneity.

 

[ghwellsjr]

And which conclusion is that?

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

Yes that is the twin paradox, but I don't see any reason at all why the acceleration should have that effect. The paradox to me is that they should each see that the other is younger than they are and that occurs in the outward leg of the journey.

If you agree that the traveling twin in the Twin Paradox ends up aging less than the home twin, we do you insist that the traveler, a, in your scenario ends up aging the same as the stationary observer, A?

Also, why do you think the time-dilation occurs only on the outward leg of the journey?--it occurs during the entire journey except for the brief intervals of acceleration/deceleration.

Please explain your concern about acceleration, it's just the way one twin gets to a new speed with respect to the other twin. The acceleration itself has no affect on the age, it just changes the aging rates. They have to spend time at the relative speed for the different aging rates to result in a different ages. If you don't understand that, please be more specific in your questions/comments.

 

[Grimble]

If you agree that the traveling twin in the Twin Paradox ends up aging less than the home twin, we do you insist that the traveler, a, in your scenario ends up aging the same as the stationary observer, A?

No he ages the same but that age appears to be less to the other twin.
Each sees the other as younger.

Also, why do you think the time-dilation occurs only on the outward leg of the journey?--it occurs during the entire journey except for the brief intervals of acceleration/deceleration.

Time dilation occurs for the whole time that they are moving one to the other.
The amount of time dilation is a function of the current velocity and exists only while the relative motion exists, and it changes every time their relative velocity changes. But once they come to rest with each other and there is no longer a relative velocity there is, no longer, any time dilation. As an effect of their motion it can only exist while there is motion.

Please explain your concern about acceleration, it's just the way one twin gets to a new speed with respect to the other twin. The acceleration itself has no affect on the age, it just changes the aging rates. They have to spend time at the relative speed for the different aging rates to result in a different ages. If you don't understand that, please be more specific in your questions/comments.

Yes acceleration merely changes the apparent ageing rate;
and No, the length of time spent in motion has no effect whatsoever, the ONLY thing that affects the amount of time dilation is the current velocity.

 

[Passionflower]

I include a spacetime diagram of 5 twins, the elapsed time on the twins clocks depends on the length of their spacetime paths, only when the lengths are identical they will record an identical elapsed time.
[PLAIN]http://img713.imageshack.us/img713/9677/event.gif

 

[Grimble]

Diagram 1. The Minkowski diagram of the twin paradox drawn to scale.
The diagram is the FoR of the resting twin. (the floater)
As it is from that FoR, the travellers times and distances are dilated and contracted respectively'

http://img707.imageshack.us/img707/3873/simultaneity.jpg

Uploaded with ImageShack.us

Diagram 2. The path of the traveller shown as rotations, the rotation changing as the traveller decelerates and returns.

http://img17.imageshack.us/img17/9214/figure5z.jpg

Uploaded with ImageShack.us

When the velocity is shown as rotation a change in the velocity will be shown as the rotation increasing or decreasing and the corresponding contracted length changing.

 

[Mike_Fontenot]

Anyone who insists that there can be no sudden ageing of the home twin, according to the traveler, during his turnaround, must then conclude that the traveler in NOT actually inertial during his constant-velocity legs. I.e., they must conclude that he isn't allowed to use the time-dilation result during his constant-velocity legs.

I insist that there can be no sudden aging of anyone and I do not come to either of your other two conclusions.
[...]

Using the gamma = 2 example I gave previously, it is indisputable that the TOTAL ageing by the home twin, during the entire trip, must be 80 years. The home twin AND the traveler can't possibly disagree about that fact.

IF you insist that, ACCORDING TO THE TRAVELER, the home twin doesn't age AT ALL during the turnaround, then he (the traveler) must conclude that ALL of her (the home twin's) ageing must occur during the two constant-velocity legs of the trip. So the sum of her ageing during his two constant-velocity legs MUST be 80 years (while he ages a total of only 40 years).

But if you consider him to be "inertial" on his OUTBOUND leg (and thus justified in using the time-dilation result), he would conclude that she ages only 10 years during his outbound leg.

Similarly, if you consider him to be "inertial" on his INBOUND leg (and thus justified in using the time-dilation result), he would conclude that she ages only 10 years during his inbound leg.

So, if you consider him to be inertial on BOTH of his constant-velocity legs, AND if you Insist that she doesn't age (according to him) during his turnaround, then he MUST conclude that her TOTAL ageing during his entire trip was only 20 years. But it is indisputable that her total ageing during his entire trip is 80 years. There is a missing 60 years. Where does it occur?

ANY proposed alternative frame for the traveler that gives zero ageing for the home twin during the traveler's turnaround, MUST get a total of 80 years for the home twin's ageing during the two constant-velocity legs. There is just no way of avoiding that (given the insistence that no ageing occurs during the turnaround).

If you insist that there is no ageing of the home twin (according to the traveler) during the turnaround, then you have various alternatives for apportioning the required 80 years of home-twin ageing, among the two constant-velocity legs.

You can adopt PassionFlower's frame for the traveler, in which the traveler is inertial (and thus justified in using the time-dilation result) during the outbound leg, but NOT on the inbound leg. So, in that alternative, the total home-twin ageing during the traveler's outbound leg is 10 years, and so her total ageing during his inbound leg must be 70 years.

Or, you can use the Dolby & Gull frame for the traveler, in which the traveler isn't inertial (and thus isn't justified in using the time-dilation result) in EITHER of the constant-velocity legs.

I consider both of those alternatives to be fatally flawed. I have critiqued both of them in another thread, starting with this posting:

https://www.physicsforums.com/showpost.php?p=2983139&postcount=76 ,

and continuing in the following two posts.

Mike Fontenot