Is the Twin Paradox Real or Just Apparent in Relativity?

[ghwellsjr]

Yes.

 

[PAllen]

In my variant it happens that all the acceleration intervals are identical between the twins, not just similar, and the two coasting parts are different between the twins and that is why I say that it is exclusively during the coasting intervals that the aging difference between the twins accumulates.

This is a meaningless statement. Apply it to lines on a plane, and you see it is meaningless. It reflects one completely arbitrary way of matching two paths against each other. There are infinitely many other ways of matching, all equally justifiable.

Of course, all intervals contribute to their total ages, including the ones before the scenario began (from the birth of the twins up to the beginning of the scenario and the ones after the scenario ended up to the present moment). Why don't you say they all need to be included, too, to explain the age difference?

Because they are irrelevant. We start clocks syncrhonized at zero, for example, when the twins first diverge in path. Again, looking at lines in a plane shows exactly what is meaningful and what isn't.

 

[ghwellsjr]

If you follow my variant, you will see that the first acceleration, by your definition, is irrelevant, because the twins have not yet diverged in path (as you say), and the time during which the first twin coasts, by your definition, is also irrelevant for the same reason, so why do you insist that these two intervals need to be included in order avoid the "meaningfulness" label?

 

[PAllen]

If you follow my variant, you will see that the first acceleration, by your definition, is irrelevant, because the twins have not yet diverged in path (as you say), and the time during which the first twin coasts, by your definition, is also irrelevant for the same reason, so why do you insist that these two intervals need to be included in order avoid the "meaningfulness" label?

If you are examining path difference between these two twins, you can and should ignore coincident parts before the first divergence. Then, about the rest, all you can say is their total proper time is different and there is absolutely nothing meaningful about how you match them up.

If you are comparing both twin paths to an equivalent inertial path, then it makes sense to start the comparison with the first divergence from inertial path by either twin.

Here are some possibilities to consider, using JesseM's diagram as a reference. Suppose instead of path A, we just repeat path B twice to get the the younger twin. Now instead of extra coasting, you may say we have extra acceleration and coasting. Or construct path A purely of repeats of the 'red' parts, in a sinusoidal path. Now all 'extra' is identical accelerations. Or suppose no parts of the path are similar, but one has very different total interval. Which part is the 'extra'? I claim all of this is nonsense. The total deviation from geodesic is the only thing that is objective; you can match diffferent paths against each other in infinite number of equally meaningless ways.

 

[ghwellsjr]

PAllen, I do not say that your method of determining the age difference of the twins has anything wrong with it or that it will not always work in every situation. I do not say that my explanation is one that will work in all situations. What I am trying to do is help the beginners who posted on this thread understand something about the Twin Paradox. I think they have a long ways to go before they can understand what an expert like you is talking about. Otherwise, they would be answering the questions instead of asking the questions.

The usual variant of the Twin Paradox is that one twin stays at home while the other one accelerates, then coasts away from his brother for a long time, then accelerates in the opposite direction, then coasts back toward his brother for a long time, and finally decelerates to a stop to meet up with his brother where they compare ages. Beginners usually have learned that when two people are in relative motion, they can each view the other one as aging at a slower rate than themselves and the problem appears symmetrical so why does one brother actually end up younger when they finally meet? The usual explanation is that the traveling brother experienced acceleration while his stay-at-home brother does not and so it is the acceleration that determines which one ages less. This, of course, is true, but it implies that it is only the acceleration that causes the younger age, which, of course, is not true. One post (#7) on this thread actually has a method that suggests "during the turnaround, the traveler will conclude that the home twin quickly ages, with very little ageing of the traveler".

So all I'm trying to do is illustrate by using a different variant of the Twin Paradox, that the accelerations, by themselves, do not account for the age difference. There must also be a difference in the coasting intervals for the two twins to accumulate an age difference.

Now, of course, there are other variants of the Twin Paradox where the entire trip of the traveling twin is constantly accelerating or where both twins could accelerate at different rates, and your analysis would work correctly for them and my analysis would not make any sense (it would be meaningless and ludicrous, as you have said), but I believe my variant and explanation can help a beginner understand the Twin Paradox a little better and has nothing wrong with it.

 

[Mike_Fontenot]

[...]
Beginners usually have learned that when two people are in relative motion, they can each view the other one as aging at a slower rate than themselves ...

They each conclude that the other twin is ageing more slowly, but ONLY during periods of time when the observer isn't accelerating. Whenever the observer IS accelerating, the ageing rate of the other twin (according to the observer) can be higher or lower than the observer's rate. In fact, the other twin can even be getting younger (according to the observer), for scenarios where the observer is accelerating in the direction away from the home twin.

Mike Fontenot

 

[JJRittenhouse]

They each conclude that the other twin is ageing more slowly, but ONLY during periods of time when the observer isn't accelerating. Whenever the observer IS accelerating, the ageing rate of the other twin (according to the observer) can be higher or lower than the observer's rate. In fact, the other twin can even be getting younger (according to the observer), for scenarios where the observer is accelerating in the direction away from the home twin.

Mike Fontenot

...because acceleration means more time dilation, and can be attributed solely to the twin in acceleration?

Does the twin paradox ignore the effects of gravity on the twin at earth, btw?

 

[Mike_Fontenot]

[...]
...because acceleration means more time dilation, and can be attributed solely to the twin in acceleration?
[...]

I'm not sure what you meant by that. During periods when the observer's acceleration is zero, he will conclude that is twin's rate of ageing is less than his own. That's what is normally meant by "the time dilation result".

But if the observer is accelerating in the direction TOWARD his twin, then he will conclude that her rate of ageing is GREATER than it would be if he weren't accelerating. Depending on how large his acceleration is, and how far apart they are, her rate of ageing can be less than his own rate (but greater than the zero acceleration rate), OR it can be greater (sometimes MUCH greater) than his own rate.

And if the observer is accelerating in the direction AWAY FROM his twin, then he will conclude that her rate of ageing is LESS than it would be if he weren't accelerating. Depending on how large his acceleration is, and how far apart they are, her rate of ageing can be only slightly less than the zero acceleration rate, OR it can even be negative ... i.e., she can be getting younger as he is getting older.

For the simple limiting cases of instantaneous speed changes, you can easily calculate these sorts of results yourself, with the CADO equation I described in my earlier post in this thread (post #7, I think). That post has a simple example of the calculation.

Does the twin paradox ignore the effects of gravity on the twin at earth, btw?

Yes. The standard twin "paradox" has nothing to do with gravitation, or with general relativity.

 

[GODISMYSHADOW]

Friends,
I will be happy if anybody throws light on the following concepts:

(1). In the case of time dilation it is said that a clock in a moving frame appears to go slow to an observer in a resting frame.It leads to the famous twin paradox in which 'A' who spends some time in a spaceship and returns to the Earth appears younger to 'B' who is his twin and stays on Earth at the time A leaves earth. With respect to B , say, 20 years had passed on the Earth but for A, say, only one year has passed in his ship. Here I don't understand one point.
If the theory of relativity says that A will appear 20 years younger than B, is it just an apparent one or real one? During the journey in the spaceship had the body cells of A really slowed down in its metabolism of aging? This point is highly confusing.

(2). According the theory of relativity, the mass of a body increases with the increase in its velocity. In that case, since a photon behaves like a particle, does it acquire mass because of its velocity?

(3). When the velocity of a body increases its mass increases while its length should decrease. Let us suppose that a rod approaches the velocity of light ( not become equal to it ). Then while its mass approaches infinity its length should approach zero! How is it possible for a body to shrink to an almost zero size but at the same time possesses nearly infinite mass?

I know the theory of relativity is a highly a complex one to comprehend. Still if anybody has simple explanations for my above doubts I will be very happy.

In the derivation of special relativistic formulas in college textbooks they show a photon bouncing back and forth between two mirrors in a moving spaceship. Then they apply geometry to the photon's path relative to an observer to come up with the formula. Well, I don't like it! As you know, the double slit experiment shows a photon doesn't have path. Hence, the argument must be invalid. Whether the formula is right or wrong I cannnot say. They just used an invalid argument to reach it.

 

[matheinste]

As you know, the double slit experiment shows a photon doesn't have path. Hence, the argument must be invalid. Whether the formula is right or wrong I cannnot say. They just used an invalid argument to reach it.

If you have an experiment designed to detect the wave nature of EM that is what you will detect. If have an experiment designed to detect the particle nature of EM that is what you will detect.

Matheinste.

 

[Mentz114]

In the derivation of special relativistic formulas in college textbooks they show a photon bouncing back and forth between two mirrors in a moving spaceship. Then they apply geometry to the photon's path relative to an observer to come up with the formula. Well, I don't like it! As you know, the double slit experiment shows a photon doesn't have path. Hence, the argument must be invalid. Whether the formula is right or wrong I cannnot say. They just used an invalid argument to reach it.

No, you just used an invalid argument. Of course light has a path. Shine a laser beam through some smoke or dust and you'll see it with your own eyes. The light clock argument is fine, there are no slits or other weird quantum effects involved.

 

[Passionflower]

There seems to be a lot of confusion about the twin experiment from the perspective of the accelerating observer.

For an accelerating observer the total time recorded by the inertial observer can be expressed completely in terms of the moments of acceleration and duration, there is no necessity to express it in terms of velocity, although that is of course possible too.

 

[lj19]

I know about the doppler effect of flying back, where events on Earth would speed up drastically by his perception

Could you explain this to me?

 

[GODISMYSHADOW]

No, you just used an invalid argument. Of course light has a path. Shine a laser beam through some smoke or dust and you'll see it with your own eyes. The light clock argument is fine, there are no slits or other weird quantum effects involved.

With the double-slit experiment, they can reduce the intensity of the light to only one photon at a time. Yet they still get an interference pattern on the screen. Could that mean a photon doesn't have path?

 

[Mentz114]

With the double-slit experiment, they can reduce the intensity of the light to only one photon at a time. Yet they still get an interference pattern on the screen. Could that mean a photon doesn't have path?

This is not relevant to the light clock experiment.

Even in the double slit case the light still gets from A to B, does it not ?

[edit] I should have added that in the two slit experiment, the interference pattern is wiped out if 'which path' information is available in principle. But the light clock doesn't depend on interference.

 

[JesseM]

The light clock thought experiment doesn't require that any actual particle move at c, even if we come up with some arbitrary function x(t) for position as a function of time that no actual object follows, we can still figure out the velocity of a hypothetical entity moving along this path, and if its coordinate speed is c in one frame it must be c in every other frame.

 

[GODISMYSHADOW]

The light clock thought experiment doesn't require that any actual particle move at c, even if we come up with some arbitrary function x(t) for position as a function of time that no actual object follows, we can still figure out the velocity of a hypothetical entity moving along this path, and if its coordinate speed is c in one frame it must be c in every other frame.

The postulate Einstein used was that in any inertial reference frame, the observer will always come up with the same value for the speed of light when he tries to measure it.

That being the case, shouldn't the light clock thought experiment be about an actual physical measurement if it's going to use Einstein's postulate in order to prove something?

 

[JesseM]

The postulate Einstein used was that in any inertial reference frame, the observer will always come up with the same value for the speed of light when he tries to measure it.

Yes, but scientific theories aren't like religious revelations, no special precedence is given to the wording of the original paper that stated some theory, later theorists may come to see the original statement as less than ideal and develop more precise ways of defining the "same theory". For example, nowadays I don't think anyone would say SR was falsified if it turned out that photons actually have a tiny mass and move at less than c, provided it was still true that all the fundamental laws of physics are Lorentz-invariant (meaning they obey the same equations in all the different inertial coordinate systems given by the Lorentz transformation). And the Lorentz transformation still says that any path through spacetime with a coordinate speed of c in one frame will have a coordinate speed of c in other frames, regardless of whether any physical object is actually following that path.

 

[ghwellsjr]

The postulate Einstein used was that in any inertial reference frame, the observer will always come up with the same value for the speed of light when he tries to measure it.

That being the case, shouldn't the light clock thought experiment be about an actual physical measurement if it's going to use Einstein's postulate in order to prove something?

Einstein stated that you cannot measure the one-way speed of light. He postulated that in any inertial reference frame we can define the two one-way times that it takes for light to traverse in each direction between two points are equal. He also affirmed that the round-trip speed of light can be measured and was found to be a constant and he predicted that it will always be measured as the same constant and that has been confirmed over and over again, so much so that now the speed of light has a defined value, but that is not what the theory of Special Relativity is about.

So your first statement is not what Einstein said. No observer can measure what he postulated. Einstein never proposed a thought experiment about an actual physical measurement to prove or disprove his postulate. His postulate cannot be proven to be true and if anyone could ever measure the one-way speed of light, that would be the end of Special Relativity because it would mean that someone had discovered a way to determine an absolute reference frame.

 

[ananthu]

During that first leg of the trip, the two twins are COMPLETELY equivalent. Each can make elementary measurements and simple, first-principle calculations about the current age of the other twin. Each will conclude that the other twin is ageing more slowly. And each of them is correct in their conclusion. Each of their conclusions is as real as anything can be ... it is NOT some kind of illusion.

Mike Fontenot

My confusion has only increased now.

If each looks young to the other, then it means no body has actually aged.

Let us assume A moves away from B and and according to time dilation formula, the clock in the frame of A should appear slower to B and similarly the clock in the frame of B should appear slower to A by the same amount. In this case A and B are moving away from each other. Now take the case of return journey of A. Here both A and B will be approaching each other. Here only I want clarification. When they apporach each other should not the opposite happen ie. the clock the reference frame of A who is actually returning should appear to go fast with respect to that of B and similarly the clock of B should appear faster to A?

So when they meet the time differences that occurred during the forward and return journeys should cancel each other effect and only same time should have elapsed for both of them.
Then it is absurd to say tha when A returns to Earth several years have passed on Earth and B looks older to A and so on.

Can anybody explain this in simple terms?

 

[Passionflower]

Einstein stated that you cannot measure the one-way speed of light.

Where did he state that, in another topic member JesseM stated he implied just the opposite.

 

[ghwellsjr]

Where did he state that, in another topic member JesseM stated he implied just the opposite.

It's in his 1905 paper. In my copy, it's on the second page under the heading "Definition of Simultaneity". He's discussing two observers, A and B separated by a constant distance (they are in the same inertial frame) and how they can each have an identical clock "but it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an 'A time' and a 'B time.' We have not defined a common 'time' for A and B, for the latter cannot be defined at all unless we establish by definition that the 'time' required by light to travel from A to B equals the 'time' it requires to travel from B to A."

 

[ghwellsjr]

My confusion has only increased now.

If each looks young to the other, then it means no body has actually aged.

Let us assume A moves away from B and and according to time dilation formula, the clock in the frame of A should appear slower to B and similarly the clock in the frame of B should appear slower to A by the same amount. In this case A and B are moving away from each other. Now take the case of return journey of A. Here both A and B will be approaching each other. Here only I want clarification. When they apporach each other should not the opposite happen ie. the clock the reference frame of A who is actually returning should appear to go fast with respect to that of B and similarly the clock of B should appear faster to A?

So when they meet the time differences that occurred during the forward and return journeys should cancel each other effect and only same time should have elapsed for both of them.
Then it is absurd to say tha when A returns to Earth several years have passed on Earth and B looks older to A and so on.

Can anybody explain this in simple terms?

I'll try. You need to use just one frame of reference to analyze a particular scenario. Consider two clocks traveling with respect to each other, let's say they are far apart and coming towards each other and eventually meet and then continue on their way getting farther apart again. You can examine this from the point of view of one of the clocks in which case the other clock will actually be running slower. It will also appear to be running slower but that's because it is actually running slower. Remember, the first clock is stationary in the frame of reference that we are considering and the second clock runs slower because it is traveling at a speed from the point of view of the frame of reference. You also need to be aware that the faster the clock travels, the slower it runs. There's no confusion about that, is there? Or you can examine it from the point of view of the other clock in which case the first clock is running slower. That by itself is understandable, too, correct? Or you can examine it from the point of view that is always halfway between them where the two clocks will now be traveling at the same speed and the two clocks will be running at the same speed. Again, no problem, correct?

In all these cases, as long as the clocks never change the speed at which they travel, they each run at a constant rate. So your idea that the clocks run at one speed while approaching each other and at a different speed while retreating from each other is not correct. All that matters is the actual speed in a particular reference frame.

Now if you want to select a frame of reference to examine the entire twin "paradox", the easiest one to select is the frame where one of the clocks, "A", remains stationary. The other clock, "B", travels away from the stationary one, "A", and so it runs slower. Eventually, "B" turns around and comes back, let's assume at the same traveling speed as before and so the clock continues to run slow at the same rate as before. Eventually, when "B" gets back to "A", it has an earlier time on it. Perfectly understandable, correct?

Now if you want to torture yourself, you can select some other frame of reference such as the one where the first clock "B" travels away from "A" and eventually comes back. But in this frame, "B" will be stationary and the other one, "A" will be traveling in the other direction. But near halfway into the scenario, "B" is going to start moving toward "A, but it obviously has to go faster than "A" in order to catch up with it and so it will be running even slower than "A". When "B" finally catches up to "A" it will have an earlier time on it because it was running even slower than "A". Now this "explanation" is not complete for many reasons but it gives the general idea. I would have to torture myself and you to provide all the gory details. I'm merely trying to illustrate how it is possible to get the same answer as to which twin is younger in the end, even if we start out in the reference frame of the other twin.

There is another frame of reference we could use and that is the one corresponding to "B's" return trip. In this case, "A" will be traveling at one constant speed the entire time while "B" will be traveling at a higher speed. About halfway into the scenario, "B" stops and eventually "A" approaches until they meet. Again, since "B" was traveling at a higher speed to begin with, it will experience a greater slowing down the "A". This "explanation" is also not complete but is just to give a general idea.

Remember, it's the speed of each clock in the reference frame that determines how fast it runs, not the relative speed between the other clock.

And remember, as long as you analyze the complete scenario in just one reference frame, you will get the same final result as you would in any other reference frame, even ones that are not associated with any objects in the scenario.

 

[yuiop]

My confusion has only increased now.

If each looks young to the other, then it means no body has actually aged.

Note that Mike was talking about the first leg of the journey before either of the twins accelerated or changed direction. While both twins have constant inertial motion relative to each other, a comparison of relative ageing rates is meaningless and it is impossible to determine which twin is really ageing slower than the other, until one of them accelerates and comes to rest in the reference frame of the other twin.

Let us assume A moves away from B and and according to time dilation formula, the clock in the frame of A should appear slower to B and similarly the clock in the frame of B should appear slower to A by the same amount. In this case A and B are moving away from each other. Now take the case of return journey of A. Here both A and B will be approaching each other. Here only I want clarification. When they apporach each other should not the opposite happen ie. the clock the reference frame of A who is actually returning should appear to go fast with respect to that of B and similarly the clock of B should appear faster to A?

I think you are getting Doppler shift mixed up with time dilation. In Newtonian physics a clock that emits a light signal once per second will appear to emit a signal at less than once per second when it is going away from the observer and when the clock is coming towards the observer the clock appear to emit signals are more than one per second. However when the moving clock is compared to the stationary clock of the observer in Newtonian physics, the clocks show no difference in elapsed time and the apparent change in clock rate of the moving clock due to Doppler shift is just an illusion caused by light signal travel times. In Special Relativity a comparison of clock rates normally discounts any Doppler shift effects and in the case of the twin's paradox the difference in ageing rates is real.

You can also note that the time dilation formula is t' = t*sqrt(1-v^2/c^2). The velocity factor is squared and so a positive or negative velocity of the same magnitude yields a time dilation factor that is independent of the direction of the traveling clock.

 

[yuiop]

It's in his 1905 paper. In my copy, it's on the second page under the heading "Definition of Simultaneity". He's discussing two observers, A and B separated by a constant distance (they are in the same inertial frame) and how they can each have an identical clock "but it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an 'A time' and a 'B time.' We have not defined a common 'time' for A and B, for the latter cannot be defined at all unless we establish by definition that the 'time' required by light to travel from A to B equals the 'time' it requires to travel from B to A."

In short, to measure the one-way speed of light we need to synchronise two spatially separated clocks. To synchronise two spatially separated clocks we need to make an assumption about the one-way speed of light. Therefore any attempt to measure the one-way speed of light involves circular reasoning.

 

[Mike_Fontenot]

My confusion has only increased now.

If each looks young to the other, then it means no body has actually aged.

In special relativity, whenever the twins AREN'T co-located, they generally will NOT agree about the correspondence between their current ages, nor about their respective current rates of ageing. And neither of them is "the one who is ACTUALLY correct" ... they are EACH correct: each of their conclusions is CORRECT and REAL, in the sense that each of their conclusions agrees with each of their own elementary measurements and first-principle calculations. This is true no matter WHAT either of them is doing ... whether or not either of them is inertial or accelerating.

[...]
When they approach each other should not the opposite happen, i.e., the clock in the reference frame of A who is actually returning should appear to go fast with respect to that of B and similarly the clock of B should appear faster to A?

Here, I think you are being confused by different meanings and usages of the term "appear". If one of the twins sees a TV image of the other twin (perhaps holding a sign giving her age), then when the twins are approaching each other, the age reported on the sign will be changing faster than the observer's age. That rate of ageing IS appropriately referred to as the "apparent ageing" of the other twin. But it obviously is NOT the true rate of ageing of the other twin, according to the observer. The age reported on the sign is the other twin's age at the instant the image was TRANSMITTED. The observer must determine how much the other twin aged while the message was in transit, and use that additional ageing to calculate the age of the other twin at the instant that the message was RECEIVED. The result of that calculation is the correct current age of the other twin, according to the observing twin. Some people mistakenly call this latter result "the apparent age" of the other twin, but that is very misleading terminology, and should not be used. That (properly) computed result is as real as real can be.

Mike Fontenot