How Does the Twin Paradox Affect Aging in Relativity?

[Fredrik]

So the twin in the INERTIAL frame is the twin that is not moving? I must be confused on inertia then. (Visits wiki)

O.K. I looked it up and inertia is resistance to change of motion, so that means that it IS the twin that is not moving in my opinion (hopefully everyones).

"Inertial" means "not accelerating". In the twin "paradox", the twin that stays on Earth can be described as "inertial". The astronaut twin can't, because he's accelerating when he turns around to go back home. The twin on Earth will be older than the astronaut twin when they meet again.

So Fredrik, when you say you are not a big fan of "that way of thinking", exactly what do you mean by that?

I just don't see how the concept of "speed through time" can help anyone understand anything, but I do see how it can confuse people. If someone tells you that an object X with velocity v=0.8c relative to you is moving slower through time than you are, do you even know what that means?* Is it clear to you if the word "space" refers to what you consider space, or what X considers space?** Did you know that you and X won't agree about which slice of spacetime is space?

*) It means that the invariant square of the the projection of the normalized tangent vector of the curve that represents's X's motion onto the normalized tangent vector of the curve that represents your motion is greater than -1. And that's really just a complicated way of saying that it doesn't have the same velocity (through space) as you.

**) It has to be a reference to what you consider space. Any object has velocity 0 through what it considers space.

The funny thing is, when they proceed to say that a photon moving at c is only moving through space, the word "space" is now referring to what it would consider space if it made any sense to define the photon's point of view by considering speed v and taking the limit v→c. And it doesn't make sense to do that, so that statement is not only misleading in at least two different ways, it's also wrong.

This is why:

The reason why we associate a specific inertial coordinate system with the motion of an inertial observer is that there's a clock synchronization procedure that makes that the natural choice. All the statements about Lorentz contracton, time dilation, etc., are consequences of that choice. The claim that massless particles experience no time comes from applying the usual time dilation formula for speed v and taking the limit v→c, but there's no reason why we should think of the result of that procedure as "a photon's point of view". There is however a good reason not to: The clock synchronization procedure doesn't work for massless particles. See my posts in this thread (at Physics Forums) for more about this.

Is there a different/better way to simplify it? The X and Y graph was the best thing I could find on google and youtube. No anger here, just wondering if there's something better.

Nothing is better than spacetime diagrams. (OK, I can think of a few things, but none of them have anything to do with physics).

the video says as long as you exist in the universe and are not moving, you are traveling at the maximum speed through time.

Don't worry about it. It's a useless idea anyway. I would recommend that you get some practice drawing spacetime diagrams, and pay extra attention to the concept of simultaneity. A spacetime diagram has an x and a t axis, with the t axis drawn in the up direction. The most important thing to understand is that if the world line of another inertial observer has slope v in the diagram, his simultaneity lines have slope 1/v. This is the result of the synchronization procedure that I described briefly in the post I linked to in the quote above.

For reference, here is a link to the video.

You didn't actually post the link.

Not a clue about that.

The quote below links to a spacetime diagram for the twin paradox, with comments about how the twins would describe things at different points on their world lines, in terms of the coordinates of their momentarily co-moving inertial frames.

Check out http://web.comhem.se/~u87325397/Twins.PNG .

I'm calling the twin on Earth "A" and the twin in the rocket "B".
Blue lines: Events that are simultaneous in the rocket's frame when it's moving away from Earth.
Red lines: Events that are simultaneous in the rocket's frame when it's moving back towards Earth.
Cyan (light blue) lines: Events that are simultaneous in Earth's frame.
Dotted lines: World lines of light rays.
Vertical line in the upper half: The world line of the position (in Earth's frame) where the rocket turns around.
Green curves in the lower half: Curves of constant -t^2+x^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket left Earth.
Green curves in the upper half: Curves of constant -(t-20)^2+(x-16)^2. Points on the two world lines that touch the same green curve have experienced the same time since the rocket turned around.

The black vertical line serves no purpose at all. I should have deleted it a long time ago, but I've been lazy.

 

[stevmg]

R Shakar at Yale University has a series of physics lectures and he goes into that very question. In essence, the guy standing still ages while the guy moving ages more slowly. That's true for clocks and all systems that work on time.

I can't quote him, Google it and you will see. His lectures 12, 13, 14 and 15 cover simple relativity quite well.

 

[W.RonG]

R Shakar at Yale University has a series of physics lectures and he goes into that very question. In essence, the guy standing still ages while the guy moving ages more slowly. That's true for clocks and all systems that work on time.

I can't quote him, Google it and you will see. His lectures 12, 13, 14 and 15 cover simple relativity quite well.

But in Relativity, nobody is standing still. Everybody is 'the guy moving'.
Ron

 

[W.RonG]

Funny that the name Brian Greene came up. I remembered another example of the aging difference fallacy. I was perusing the physics section of a large book store about a year ago, happened on Mr. Greene's book more recent than the Elegant Universe. Not sure the title, but I opened it to the page where he makes the same mistake of confusing the appearance of a clock ticking more slowly to an observer in a different inertial frame, to make the statement that there is a 'moving' clock ticking more slowly than a 'stationary' clock. Oops.

 

[Fredrik]

Ron, I think you're just misunderstanding what other people are saying. You're right that nobody is standing still in an objective sense, but once you have picked a coordinate system, then it's possible to really stand still in that coordinate system. Is it possible that most of the people that you think are wrong are just describing things in terms of a coordinate system?

 

[sylas]

Ron, it will help clarify your position if you could answer kev, in msg #43.

Can you make it clear if you are stating that in the classic 'twins paradox' you are claiming that when the traveling twin returns home and is once again united with his sibling, that they will not have aged by different amounts in a way that is not only measured by their respective clocks but also manifested physically in how they aged biologically?

If that is what you are claiming, then it contradicts your claim that you 'believe in SR", because anyone that understands SR will know the twins age differentially and that experiments with particles such as muons support that prediction.

If on the other hand you are talking about two twins with purely inertial motion relative to each other, with no turning around and no acceleration, then I would agree with you that it is impossible to define which twin is (really) ageing slower, but that wouldn't be the classic 'twins paradox' would it?

Can you respond to this please Ron? It would help clarify things, I think.

Cheers -- sylas

 

[W.RonG]

Recall that the 'Twins Paradox' arose specifically out of Special Relativity as an effect of Time Dilation. It is quite simple; does Time Dilation cause one clock to tick more slowly than another? the consequence of this phenomenom is the paradox, that one clock will count less time than another if they meet up after 'one of them' has traveled for a while. Answer: no.
Thanks,
Ron

 

[sylas]

Recall that the 'Twins Paradox' arose specifically out of Special Relativity as an effect of Time Dilation. It is quite simple; does Time Dilation cause one clock to tick more slowly than another? the consequence of this phenomenom is the paradox, that one clock will count less time than another if they meet up after 'one of them' has traveled for a while. Answer: no.
Thanks,
Ron

That clarifies. And I mean no offense, but this means you don't believe SR... and if you think you do, then you don't understand it either.

In special relativity, the non-inertial clock will count less time, and this is confirmed when they are brought back together. The amount of time difference between the two clocks can be calculated, and it is confirmed by experiment that the calculations do give the amount of time difference experienced between the two clocks.

It was important that you answered this plainly, though, so we could understand your position more clearly. Thanks.

But you are wrong.

Cheers -- sylas

 

[atyy]

Recall that the 'Twins Paradox' arose specifically out of Special Relativity as an effect of Time Dilation. It is quite simple; does Time Dilation cause one clock to tick more slowly than another? the consequence of this phenomenom is the paradox, that one clock will count less time than another if they meet up after 'one of them' has traveled for a while. Answer: no.
Thanks,
Ron

This is wrong. It can be correct, depending on the exact definition of terms, to say that no ideal clock ever ticks more quickly or more slowly depending on its motion. However, the elapsed proper time will be different for an inertial clock and a non-inertial clock that meet up at the start and end of their spacetime trajectories.

 

[W.RonG]

To 'atyy' and 'sylas': you are assigning preference to one inertial frame over another.
Ron

 

[sylas]

To 'atyy' and 'sylas': you are assigning preference to one inertial frame over another.
Ron

Ron, this is not up for debate. It is not a case of two inertial frames.

If you want to understand this better, we can help; but in my experience it is almost impossible to help someone with this if they insist they already understand it but continue to make all the same errors. As you are doing.

The twin that turns around is not inertial. It is in a different frame on the outward journey, and on the inward journey. As I said previously, it is the non-inertial twin which experiences less elapsed time between departure and when they meet up again.

Cheers -- sylas

 

[Fredrik]

Recall that the 'Twins Paradox' arose specifically out of Special Relativity as an effect of Time Dilation. It is quite simple; does Time Dilation cause one clock to tick more slowly than another? the consequence of this phenomenom is the paradox, that one clock will count less time than another if they meet up after 'one of them' has traveled for a while. Answer: no.

Ah, then I was wrong about what you had misunderstood. SR definitely predicts that the twin that goes on the trip is younger when they meet again. This is an immediate consequence of the axiom that says that a clock measures the proper time of the curve that represents its motion.

That's the coordinate independent resolution of the paradox. If you'd like to see it resolved completely in terms of co-moving inertial frames, then see the spacetime diagram that I linked to in #51 (in the last quote box).

 

[JesseM]

To 'atyy' and 'sylas': you are assigning preference to one inertial frame over another.
Ron

No, you can analyze the problem from any inertial frame and all will have the same answer about the age of the inertial twin and the age of the non-inertial twin when they reunite. Let's call the inertial (Earth-bound) twin "Terence" and the traveling twin "Stella", following the Twin Paradox FAQ. First let's look at the numbers in Terence's rest frame. Suppose that in this frame, Stella travels away from Terence inertially at 0.6c for 10 years, at which point she is at a distance of 0.6*10 = 6 light-years from Earth in this frame, then she turns around (i.e. she accelerates, a non-inertial motion which will cause her to experience G-forces that show objectively that she wasn't moving inertially) and heads back towards Terence at 0.6c, finally reuniting with Terence after 20 years have passed since her departure in this frame. Since Terence is at rest in this frame, he has aged 20 years. But since Stella was moving at 0.6c in this frame, the time dilation formula tells us her aging was slowed down by a factor of \sqrt{1 - 0.6^2} = 0.8, so she only aged 0.8*10 = 8 years during the outbound leg of her trip, and another 0.8*10 = during the inbound leg, so she has only aged 16 years between leaving Earth and returning.

Now let's analyze the same situation in a different inertial frame--namely, the frame where Stella was at rest during the outbound leg of her trip (she can't also be at rest during the inbound leg in this frame, since this is an inertial frame while Stella accelerated between the two legs of the trip). In this frame, Terence on Earth is initially moving away from Stella at 0.6c while she remains at rest. In Terence's frame, remember that Stella accelerated when she was 6 light-years away from Earth, so we can imagine she turns around when she reaches the far end of a measuring-rod at rest in Terence's frame and 6 light-years long in that frame, with Terence sitting on the near end; in the frame we're dealing with now, the measuring-rod will therefore be moving along with Terence at 0.6c, so it'll be shrunk via length contraction to a length of only 0.8*6 = 4.8 light-years. So, Stella accelerates when the distance between her and Terence is 4.8 light-years in this frame, and since Terence as moving away from her at 0.6c in this frame, they will be 4.8 light-years apart after 4.8/0.6 = 8 years have passed. During these 8 years, it is Terence's aging that is slowed down by a factor of 0.8, so while Stella ages 8 years during this leg, Terence only ages 0.8*8 = 6.4 years. Then Stella accelerates to catch up with Terence, while Terence continues to move inertially at 0.6c. Using the relativistic velocity addition formula, if Stella was moving at 0.6c in Terence's frame and Terence is moving at 0.6c in the same direction in this frame, then in this frame Stella must be moving at (0.6c + 0.6c)/(1 + 0.6*0.6) = 0.88235c during the inbound leg. And since Terence is still moving at 0.6c in the same direction, the distance between Stella and Terence will be closing at a "closing speed" of 0.88235c - 0.6c = 0.28235c. Since the distance was initially 4.8 light years at the moment Stella accelerated, in this frame it will take 4.8/0.28235 = 17 years for Stella to catch up with Terence on Earth. During this time Terence has aged another 0.8*17 = 13.6 years, so if you add that to the 6.4 years he had aged during the outbound leg, this frame predicts he has aged 20 years between Stella leaving and Stella returning, same as in Terence's frame. And since Stella is traveling at 0.88235c her aging is slowed by a factor of \sqrt{1 - 0.88235^2} = 0.4706, so during those 17 years in this frame she only ages 0.4706*17 = 8 years during the inbound leg. If you add that to the 8 years she aged during the outbound leg, you find that this frame predicts she has aged 16 years between departing and returning, which again is the same as what was predicted in Terence's frame.

 

[stevmg]

Sports Fans:

Here is that Yale lecture series I spoke of. I think it is lecture 13 or 14. Start with lecture 12 and step forward.

http://www.youtube.com/watch?v=pHfFSQ6pLGU&feature=SeriesPlayList&p=FE3074A4CB751B2B

Yea, everyone is moving BUT start with twin A on the Earth and look at the second twin B in mtion in space relative to A and it is B who age slower than A. all clocks slow down under time dilation including biological clocks.

Listen to this guy - he explains it a lot better than I can and a lot better than I saw on the earlier posts on this blog.

 

[JesseM]

Yea, everyone is moving BUT start with twin A on the Earth and look at the second twin B in mtion in space relative to A and it is B who age slower than A. all clocks slow down under time dilation including biological clocks.

Only in A's rest frame. In B's rest frame, it is A who ages slower than B (assuming both are moving inertially).

 

[stevmg]

JesseM -

You are the person who straightened my brain out before..
Oh well,

Maybe it has something to do with General Relativity...

If you think I knew nothing about Special Relativity, I won't disappoint you with my less than zero knowledge of GR.

Under GR, time slows down by acceleration and gravity. The guy on Earth is attached to something with more mass than a spaceship.. The B guy has to accelerate and turn (centripetal force), so I guess that is the explanation.

I read Einstein's book "Relativity" and if I wasn't lost and confused after SR, I was clueless and pulseless after reading GR.

Loved that bending the light demonstration and the perihelon{?sp) of Mercury.They even did a movie about a group of guys in 1919 who went to Brazil and did some picture taking of an ecclipse and made it with some local Brazilian girls. Kept my interest up.

Too bad that they didn't believe the 1919 pictures and Uncle Al had to wait until 1923 when a more "official version" was taken and Einstein became God on Earth.

 

[stevmg]

The name of that movie is "House of Sand" - great flick!

 

[Dale]

Not a clue about that. But my mind is sufficiently blown for the night. Google is my friend tomorrow.

Hope you had happy Googling Sorry about my terse response earlier, I was in a rush.

Anyway, a spacetime diagram is simply a graphical plot of the path of a particle. By convention, it is drawn so that the time axis is vertical and the space axis is horizontal. Since paper is traditionally 2D usually we just include one spatial axis and politely ignore the other two, but if you had a 3D graphics system you could easily add a second spatial axis. Also by convention, the time axis is scaled by c (i.e. ct instead of just t) so that even though it is "time" the units have dimensions of length (ct -> L/T T = L).

Using these conventions you can think of the path of particles as geometric figures. A particle at rest would be represented by a vertical "worldline". A pulse of light would be represented by a line at 45 degrees. A particle moving inertially at some v<c would be a straight line with some slope in between. An accelerating particle would have a curved worldline. Etc.

In such a diagram a clock measures the length of the worldline, except instead of using the normal Euclidean metric, the metric that a clock measures is called the Minkowski metric ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2. Then it is no longer very surprising that one clock measures a different time than another since they are measuring the lengths of different lines.

 

[Jackslap]

HA! I'm such a moron for not posting that link. Can't believe I made that mistake. Here it is:

http://www.youtube.com/watch?v=ksbgDJF9bsM&feature=related

Anyway, now that I've officially ruined relationships on this board I suppose I'll continue this pattern by stating that I was confused by the difference of opinions, and quite frankly still am a little bit.

I am thinking that Ron's input must be taking something into consideration. Something that I'm not yet capable of understanding. I see that some of you have suggested such a thing. I shall try to wade through the wealth of information you all have left me so far today. Specifically the Yale lectures. I'll probably only be able to get through one part tonight though.

You guys are really awesome for helping. This site is such a resource for me lately. Such a convenient way to share knowledge and I'm humbled that some of you would share your learning which probably cost a ton of cash with some dude like me that you don't know and probably will never meet.

 

[JesseM]

JesseM -

You are the person who straightened my brain out before..
Oh well,

Maybe it has something to do with General Relativity...

If you think I knew nothing about Special Relativity, I won't disappoint you with my less than zero knowledge of GR.

Under GR, time slows down by acceleration and gravity. The guy on Earth is attached to something with more mass than a spaceship.. The B guy has to accelerate and turn (centripetal force), so I guess that is the explanation.

Well, usually when people discuss the twin paradox they ignore GR--you can imagine a world where all objects, including the Earth, have negligible mass (or you can imagine that the inertial twin is not an Earth but on a space station deep in interstellar space, with too little mass of its own to cause any measurable gravitational time dilation).

 

[stevmg]

You are right, Jesse M. If one did a circular route 10 light years in diameter (30 light years in circumference) it would take only 1/10 g to create the circle to get back to the origin and that won't slow anything down to speak of.

 

[stevmg]

Using the 4-space equation, a different path would change the t although the sum of the x^2, -(ct^2) = sum of the (x')^2, (-ct')^2 is is invariant.

I have to accept the 4-space equation on faith alone.

 

[Dale]

I have to accept the 4-space equation on faith alone.

In science the unproven assumptions of a theory are called postulates. They are not accepted on faith alone, but they are verified experimentally. When a particular experiment agrees with the logical consequences of a postulate or set of postulates then the theory based on those postulates is said to be experimentally verified.

In the case of the Minkowski metric, the experimental evidence is ample and such statements about faith are rather absurd.
http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html

 

[TcheQ]

Using the 4-space equation, a different path would change the t although the sum of the x^2, -(ct^2) = sum of the (x')^2, (-ct')^2 is is invariant.

Keep reviewing relevant literature, various lectures on youtube etc etc.

Alternatively you can always take a postgrad theoretical physics class

 

[stevmg]

With all due respect, it is this blog that is absurd...

I am using ordinary figures of speech and not literal meanings...

Am signing off, folks...

Hasta la vista...[figure of speech-wise, not literally Spanish]

Steve G

 

[JesseM]

You are right, Jesse M. If one did a circular route 10 light years in diameter (30 light years in circumference) it would take only 1/10 g to create the circle to get back to the origin and that won't slow anything down to speak of.

Actually gravitational time dilation only occurs in spacetime with "real" gravity due to mass and energy curving spacetime. The G-forces you feel when accelerating in flat spacetime (like taking a circular path in deep space far from any massive objects) don't cause any gravitational time dilation, at least not if you analyze things from the perspective of an inertial frame (if you analyze things in a non-inertial frame you can have a pseudo-gravitational field which leads to pseudo-gravitational time dilation, as discussed in this section of the twin paradox page, but when dealing with accelerating objects in flat spacetime it's much easier to just use inertial frames to calculate how much time goes by on their clocks).

 

[sylas]

Actually gravitational time dilation only occurs in spacetime with "real" gravity due to mass and energy curving spacetime. The G-forces you feel when accelerating in flat spacetime (like taking a circular path in deep space far from any massive objects) don't cause any gravitational time dilation, at least not if you analyze things from the perspective of an inertial frame (if you analyze things in a non-inertial frame you can have a pseudo-gravitational field which leads to pseudo-gravitational time dilation, as discussed in this section of the twin paradox page, but when dealing with accelerating objects in flat spacetime it's much easier to just use inertial frames to calculate how much time goes by on their clocks).

Actually, JesseM, you do get time dilation in a pseudogravitational field.

You can show this by consideration of a long spaceship which is experiencing a constant acceleration. First, take a single particle (the spaceship captain) and work out their world line so that they experience a constant acceleration. Then get another particle representing the rear of the spaceship, which has a world line defined so that the pilot can send a light signal from their location in the cockpit to a mirror at the back of the ship, and receive a reflection, and this is always a constant time by the pilot's clock.

This will give you a world line for the rear of the space ship, so that everyone at any point on the ship is experiencing constant acceleration, and everyone on board agrees that the ship is not stretching or doing anything weird like that.

What you will find is that different parts of the ship experience a different acceleration! And there is a measurable time dilation difference all along the ship, exactly analogous to the time dilation differences you find in a gravitational field.

Cheers -- sylas

 

[JesseM]

Actually, JesseM, you do get time dilation in a pseudogravitational field.

I agree, that's what I meant when I said "if you analyze things in a non-inertial frame you can have a pseudo-gravitational field which leads to pseudo-gravitational time dilation". My point was that both the pseudo-gravitational field and the pseudo-gravitational time dilation are absent if you analyze the accelerating object from the point of view of an inertial frame, which is always possible in flat spacetime; from the perspective of the inertial frame, all the differential aging of the accelerating object can be accounted for in terms of velocity-based time dilation, even though the non-inertial frame accounts for the differential aging differently (but as always, all frames agree on local predictions like what ages two observers will be when they meet at a single point in spacetime).

 

[sylas]

... all the differential aging of the accelerating object can be accounted for in terms of velocity-based time dilation ...

Ah! Glad we are on the same page, sorry I misunderstood you.

The above comment is a hugely important point, IMO, which lots of people get wrong. Your comment is succinct and spot on. I'm going to steal your phrasing when I am explaining it for others! Kudos.

 

[Jackslap]

lol, this has gone WAY beyond my ability to compute. I watched the first half of the Yale lecture number 12 that was linked to, but it froze up at 29 minutes in. I found myself understanding the instructors hypothetical situations and things that he said in ENGLISH. But he quickly starts jotting down formulas and using tons of variables for which I have no place for right now in my physics infancy.

I think I'd need to start at his lecture number 1, Newtonian Mechanics. However, it's freaking Yale and I'm thinking even then I don't have enough background to begin there. For your info, the highest level math class I've ever taken was an Algebra 2 class in college as a pre-req for the x-ray program. That's it. However I was encouraged when the formulas he was using in lecture 12 were fairly simply manipulated using algebra anyway. But at some point he started using sin, cos, which is trigonometry stuff. I aint's got a clue 'bout none o' that.

Not your guys job to teach me though, I must begin at the beginning if I am going to learn this. And my beginning will be quite slow due to work and family schedules. Baby steps as Bob would say.

 

[TcheQ]

I think I'd need to start at his lecture number 1, Newtonian Mechanics. However, it's freaking Yale and I'm thinking even then I don't have enough background to begin there. For your info, the highest level math class I've ever taken was an Algebra 2 class in college as a pre-req for the x-ray program. That's it. However I was encouraged when the formulas he was using in lecture 12 were fairly simply manipulated using algebra anyway. But at some point he started using sin, cos, which is trigonometry stuff.

It doesn't matter what university it is, they all teach the same thing (assuming it's maths or physics :P). The Yale guy is particularly good. As for maths, you need to learn calculus, differentials and matrices for relativity. If you want to learn quantum mechanics you need some form of statistics and computing, signal interpretation (such as Fourier transforms) and a background in computing. Nothing that's beyond a normal human being. It's always best to start at the beginning :)

 

[stevmg]

Cut me a break DaleSpam as you have been most infolrmative otherwise...

"Faith alone" does not refer to a religious or mystical approach. My reference is that I have seen the Minkowski 4-space but I do not understand the mathematical derivation of it. I do not challenge but do accept it. This is allegory.

I do not understand how it is shown that the Twin B does not age as fast as the twin A.
I do understand that if one takes a different course in the (twin B) xyz coordinates then the ct coordinate is different than the original (twin A). It is the Minkowski 4-space that establishes that I am told but I do not understand. Given that the Minkowski 4-space is true ("faith alone" statement) then I accept the outcome.

Don't get too freakin' literal as I do write in conversational language. To wit - when we used to write fitness reports on officers, we would always write "water-walkers" [reference to Jesus] to say that this guy or gal was OK. We weren't really saying they were like Jesus. Again, that's a "figure of speech" or allegory.

Let's get on the same page.

 

[W.RonG]

Well, this has certainly been enlightening. Since I have only been 'dabbling' in Physics for forty years or so, I was aware only of the 'classical' Twins Paradox. That's the one that came out of Special Relativity, where 'B' flies around in space while 'A' watches, and since neither is in a preferred inertial frame neither clock can run slower than any other.
There seems to be a new Twins Paradox. This one allows clocks to be assigned 'inertial' and 'non-inertial' so it exists outside the realm of Special Relativity. A 'GR' neo-twins-paradox perhaps? that's the only reasonable explanation for the dilemma.
Thanks again,
Ron

 

[sylas]

Well, this has certainly been enlightening. Since I have only been 'dabbling' in Physics for forty years or so, I was aware only of the 'classical' Twins Paradox. That's the one that came out of Special Relativity, where 'B' flies around in space while 'A' watches, and since neither is in a preferred inertial frame neither clock can run slower than any other.

As you have been told now many times, you are incorrect. This is not meant to be an insult; it is an attempt to help.

One twin is inertial. The other twin is not. The twin that is in an inertial frame is the one that WILL experience less time, when they have come back together again.

This is the only "twin" paradox. It is the same as explained by Einstein in his original works on the subject. The effect is confirmed by direct measurement.

Your reasoning is incorrect, because you have failed to note that there is a real difference between being in an inertial frame and NOT being in an inertial frame. The two twins are not symmetrical, and the only reason this is a paradox is because some people find it hard to understand; not because there is any actual inconsistency.

You also don't understand this yet -- despite 40 years of dabbling. This, by the way, is not all that unusual, and I mean no offense. But this is something that is taught in first year physics classes or even high school in some cases. You will do better to accept that you might need a teacher rather than rely on your own dabbling.

Cheers -- sylas

 

[Dale]

Cut me a break DaleSpam as you have been most infolrmative otherwise...

"Faith alone" does not refer to a religious or mystical approach.

...

Don't get too freakin' literal as I do write in conversational language. To wit - when we used to write fitness reports on officers, we would always write "water-walkers" [reference to Jesus] to say that this guy or gal was OK. We weren't really saying they were like Jesus. Again, that's a "figure of speech" or allegory.

I will gladly cut you a break if you will stop with the religious and Biblical allusions. They are simply not appropriate to the forum, even in context. Being a person of both science and faith I don't like to see either misunderstood or treated so lightly and dismissively. I know you do not intend your comments to be provocative in this way, but to me they make a real impediment to actually understanding what your scientific or technical question is. I appreciate that to you they are "conversational" expressions, but the internet is a medium where such colloquial expressions do not come through and misunderstandings can easily result. It requires more care in communication than a casual conversation with friends where everyone knows your background, can read your body language, and understand immediately your intent.

 

[yuiop]

Well, this has certainly been enlightening. Since I have only been 'dabbling' in Physics for forty years or so, I was aware only of the 'classical' Twins Paradox. That's the one that came out of Special Relativity, where 'B' flies around in space while 'A' watches, and since neither is in a preferred inertial frame neither clock can run slower than any other.
There seems to be a new Twins Paradox. This one allows clocks to be assigned 'inertial' and 'non-inertial' so it exists outside the realm of Special Relativity. A 'GR' neo-twins-paradox perhaps? that's the only reasonable explanation for the dilemma.
Thanks again,
Ron

Hi Ron,

You are right that if two twins moving relative to each other with purely inertial motion (I.e. in a straight line and never accelerating, decelerating or turning around) then each measures the others clock to be running slower than their own clock and there is no way to determine which twin is "really" ageing less than the other. However, the classic twin's paradox has a non inertial element because one twin has to turn around and this involves acceleration. There is a popular conception that SR can not handle acceleration but this is not true. The introduction of acceleration into purely inertial SR is very easy to do because it turns out that acceleration has no effect on the instantaneous proper rate of a clock which is determined purely by the instantaneous velocity of the moving/accelerating clock. The twin's paradox is called a paradox because of the apparent paradoxical (contradictory) conclusions that inertial considerations say there is no real differential ageing of the twins while the round trip thought experiment says they will age differently. Of course it is not really a paradox because it can be resolved and SR predicts only one outcome (The twin's age differentially). The paradoxical nature is compounded because as I said before, acceleration has no direct effect on the clock rates! The solution and full understanding of the twin's paradox is subtle and involves differences in paths through spacetime that can not be transformed away from any point of view. The subtlety of the resolution to the paradox is the reason that there are literally hundreds of threads and thousands of posts on the subject. Once you accept that differential ageing occurs as has been demonstrated by actual experiments I strongly recommend that you do not jump to the conclusion that acceleration is the cause of differential ageing because that is a false conclusion and this is also a fact proven by experiments.

Here is a demonstration of how acceleration does not effect clocks with reference to the relativistic rocket equations here: http://www.phys.ncku.edu.tw/mirrors/physicsfaq/Relativity/SR/rocket.html

We see that the instantaneous velocity of the accelerating rocket is:

v = at / sqrt(1 + (at/c)^2)

which can be solved for acceleration (a) to give:

a = v/t*1/sqrt(1 - v^2/c^2 )

The instantaneous time dilation factor (gamma) of a clock on the accelerating rocket (T) relative to the initial inertial clock (t) is given as:

t/T = sqrt(1 + (at/c)^2)

Substitution of the value found for a into the above equation gives:

t/T = sqrt(1 + (v/c*1/sqrt(1 - v^2/c^2 )^2) which simplifies to:

t/T = 1/sqrt(1 - v^2/c^2 )

which is the familiar time dilation gamma factor of SR with no dependence on acceleration whatsoever.

 

[Fredrik]

Well, this has certainly been enlightening. Since I have only been 'dabbling' in Physics for forty years or so, I was aware only of the 'classical' Twins Paradox. That's the one that came out of Special Relativity, where 'B' flies around in space while 'A' watches, and since neither is in a preferred inertial frame neither clock can run slower than any other.
There seems to be a new Twins Paradox. This one allows clocks to be assigned 'inertial' and 'non-inertial' so it exists outside the realm of Special Relativity. A 'GR' neo-twins-paradox perhaps? that's the only reasonable explanation for the dilemma.
Thanks again,
Ron

One of them has to accelerate after they separate. Otherwise the distance between them would just keep increasing, and they would never meet again.

The scenario that goes by the name "the twin paradox" involves two twins that are are both present at two events E and F. One of the twins has been doing inertial motion from E to F, and the other twin hasn't. Special relativity predicts that if they're both the same age at E, the one who stayed inertial is older at F.

You said that "since neither is in a preferred inertial frame neither clock can run slower than any other". This is incorrect. The fact that neither of the frames is preferred implies that if A's clock is slow in B's rest frame, then B's clock is slow in A's rest frame. But this doesn't imply that A's clock can't be slow in B's rest frame or that B's clock can't be slow in A's rest frame.

 

[Fredrik]

One twin is inertial. The other twin is not. The twin that is in an inertial frame...
...
being in an inertial frame and NOT being in an inertial frame.

I'm just going to nitpick your choice of words a bit. (It wouldn't be fair to Mentz114 if I let this one slide ). I think it's fine to say that "one twin is inertial". That just means that he isn't accelerating. But the phrase "is in an inertial frame" doesn't really mean anything. I know that you meant the same thing as when you said that the twin "is inertial", and in this context it's probably clear to everyone else as well, but it's still a meaningless phrase that novices in particular should be discouraged from using.

It's not that much more work to type "X is at rest in an inertial frame" instead of just "X is in an inertial frame".

 

[sylas]

It's not that much more work to type "X is at rest in an inertial frame" instead of just "X is in an inertial frame".

I appreciate a good suggestion and this is a good suggestion and a legitimate clarification of poor phrasing.

Thanks!

 

[stevmg]

I will gladly cut you a break if you will stop with the religious and Biblical allusions. They are simply not appropriate to the forum, even in context. Being a person of both science and faith I don't like to see either misunderstood or treated so lightly and dismissively. I know you do not intend your comments to be provocative in this way, but to me they make a real impediment to actually understanding what your scientific or technical question is. I appreciate that to you they are "conversational" expressions, but the internet is a medium where such colloquial expressions do not come through and misunderstandings can easily result. It requires more care in communication than a casual conversation with friends where everyone knows your background, can read your body language, and understand immediately your intent.

Dale -

Sorry - the military will continue to talk in their superlative and metaphorical terms which involves a whole lot of swearing for years to come. Saying "someone is a genius" just means that "he comes to work on time." "Has a knack for tracking details" means "he follows the rules and is not a serial killer." How about the old Irish phrase - "May the Lord take a liking to you - but not too soon!" I never had anybody ever misunderstand me before - even when I spoke in German and my German wasn't very good. Of course, neither is my English.

But the original statement and question about taking the Minkowski 4-space as it relates to this twin paradox is "If you look at x,y,z,ct coordinants and the x',y',z',ct' coordinates and they are the same at the beginning and end the the journey for the twins - aren't the t and t' going to be the same (i.e. - same age)?

Other posts on this thread all ascribe to supposition that you can't tell which frame of reference (FOR) is the basis so neither twin will be older or younger - or is it that one twin will always see the second twin as younger? (Choose one twin - the other guy is younger. Choose the other twin, and the first guy is younger.)

By the way (an aside) - you can have same embryo twins of different sex - did you know that? You will floor me if you know how that is possible.

I know that the GPS satellites atomic clocks run slower than their Earth counterparts (hence are factored up by the gamma factor to offset time dilation.) Don't ask me where I heard that - I can't remember. Yes, it is only nanoseconds but didn't Eddington (May 1919) and later the Aussies of 1922 "confirm" Einstein's GR with trivial differences in where stars were and where they were supposed to be ("trivial" is not a put-down but is a reference to something really, really, really tiny.)

Seems like if Eddington and those Aussies in 1922 (I don't remember their names) both had the Earth as the FOR, so maybe there is a way to select the proper FOR.

Now, again, I don't know how Minkowski came up with a four-space so I do take it on "faith" (NOT IN A RELIGIOUS SENSE) alone - meaning I trust Minkowski and all you folks who understand it and know the confirmatory evidence to be giving me a "good" equation; good enough that I can plug in numbers and come up with a numeric answer.

The word "faith" has many meanings and not just a religious one, so I am not incorrect in its usage in my statement.

Miriam Webster Dictionary (I had to use it since my English is bad)
Main Entry: faith
Pronunciation: \ˈfāth\
Function: noun
Inflected Form(s): plural faiths \ˈfāths, sometimes ˈfāthz\
Etymology: Middle English feith, from Anglo-French feid, fei, from Latin fides; akin to Latin fidere to trust — more at bide
Date: 13th century
1 a : allegiance to duty or a person : loyalty b (1) : fidelity to one's promises (2) : sincerity of intentions
2 a (1) : belief and trust in and loyalty to God (2) : belief in the traditional doctrines of a religion b (1) : firm belief in something for which there is no proof (2) : complete trust
3 : something that is believed especially with strong conviction; especially : a system of religious beliefs <the Protestant faith>

synonyms see belief

— on faith : without question <took everything he said on faith>

On this blog, and since you are a science contributor and because of space limitations, can't I take certain things YOU say "on faith?"

But, please, let's address the scientific question I asked.

 

[Dale]

But the original statement and question about taking the Minkowski 4-space as it relates to this twin paradox is "If you look at x,y,z,ct coordinants and the x',y',z',ct' coordinates and they are the same at the beginning and end the the journey for the twins - aren't the t and t' going to be the same (i.e. - same age)?

Other posts on this thread all ascribe to supposition that you can't tell which frame of reference (FOR) is the basis so neither twin will be older or younger - or is it that one twin will always see the second twin as younger? (Choose one twin - the other guy is younger. Choose the other twin, and the first guy is younger.)

The time measured by a clock is called the "http://en.wikipedia.org/wiki/Proper_time" " and is a function of not only the endpoints, but of clock's entire path. So two clocks that begin at one event and end at another event may record different times depending on the paths that they took between the two events. The proper time is a quantity that is agreed on by all inertial frames, so if two clocks take different paths from one event to another then all reference frames will agree how much proper time elapsed for each and therefore all reference frames will agree which is younger.

By the way (an aside) - you can have same embryo twins of different sex - did you know that? You will floor me if you know how that is possible.

I don't know how it is relevant, and I am certainly no expert in the subject, but my understanding is that an embryo at a very early stage is "genderless" and that the development of male or female gonads and genitalia occurs in response to the amount of specific hormones including estrogen and testosterone (among others that I don't know). The presence of a Y chromosome or an extra X chromosome usually ensures the correct balance, but in certain pathologies the hormone levels can be skewed resulting in atypical development.

 

[Fredrik]

But the original statement and question about taking the Minkowski 4-space as it relates to this twin paradox is "If you look at x,y,z,ct coordinants and the x',y',z',ct' coordinates and they are the same at the beginning and end the the journey for the twins - aren't the t and t' going to be the same (i.e. - same age)?

The twins are not going to be the same age when they meet. The easiest way to see that this is what SR predicts is to use the axiom that says that a clock measures the proper time of the curve that represents its motion. OK, that may not be the easiest way for you, but it is for someone who is familiar with the concept of proper time. If you want an explanation in terms of the coordinates of inertial frames, you should study the spacetime diagram I linked to in #51. (See the last quote box).

Other posts on this thread all ascribe to supposition that you can't tell which frame of reference (FOR) is the basis so neither twin will be older or younger - or is it that one twin will always see the second twin as younger? (Choose one twin - the other guy is younger. Choose the other twin, and the first guy is younger.)

Those are all written by W.RonG, and he's wrong. As you may have noticed at least four people have tried to explain that to him, three of whom are science advisors. Only one of the twins is doing inertial motion, so we can't associate a single inertial frame with each twin. If we insist on describing things in terms of inertial frames, we need at least three of them for the two twins.

 

[stevmg]

Dale -

Has that got something to do with "tau" for each twin (the time on each twin using that particular twin's place as the frame of reference for him/her and for a particular twin there is never a difference in the xyz vector but just the time (ct) vector)? In other words, tau-A and tau-B which, according to you (I will have to do more reading) which will vary according to the path, so different paths mean different "tau's" hence different time vectors or the different ages of the twins. In other words - will the twins differ in age (assuming they do rejoin) - is that true when all is said and done. You don't have to "prove" Minkowski to me, as I will review it again in Einstein's book or other books on the subject.

Quick answer to the different sex, same embryo or same zygote phenomenon. Very very rare. Sometimes (recorded only a few times that I know of) a male zygote (X-Y), when it divides, a Y chromosome is "lost" in one of the daughter cells. If the twinning occurs at that stage (generally the morula stage or 8-cell) that cell which is the Y-deficient cell goes on to become the second baby. That second baby is genotype X-O which is a Turner baby. This is a female but does have some differences from other "normal" (X-X) females (they do not develop ovaries and cannot bear children, have congenital heart disease and are short.) They are not deficient mentally. The original morula goes on to become an X-Y baby or boy.

If that Y-deficient cell is made at a different stage, then it is just lost and does not become a second individual.

This is rare as hens teeth.

 

[Dale]

Has that got something to do with "tau" for each twin (the time on each twin using that particular twin's place as the frame of reference for him/her and for a particular twin there is never a difference in the xyz vector but just the time (ct) vector)? In other words, tau-A and tau-B which, according to you (I will have to do more reading) which will vary according to the path, so different paths mean different "tau's" hence different time vectors or the different ages of the twins. In other words - will the twins differ in age (assuming they do rejoin) - is that true when all is said and done.

Yes, on both counts, the proper time is traditionally identified by the variable \tau, and the twins will in general be different ages when they rejoin.

For a worldline parametrized by \lambda the proper time is given by:
\tau <br /> = \int \sqrt {\left (\frac{dt}{d\lambda}\right)^2 - \frac{1}{c^2} \left ( \left (\frac{dx}{d\lambda}\right)^2 + \left (\frac{dy}{d\lambda}\right)^2 + \left ( \frac{dz}{d\lambda}\right)^2 \right) } \,d\lambda

 

[stevmg]

Thank you particularly to DaleSpam AND Fredrik who gave me a definite answer on this subject. It will take me weeks of review and study to understand the vector calculus that will enable me to understand the integral noted above.

Is that integral "integratable" (in the sense that there is an algebraic expression or "anti-derivative" that does represent it after the proper operations are done?) It is clearly integratable as all tge variables are continuous over the specified interval so the limits do exist.

To wit, an example of an expression that is integratable (the limits do exist) but for which there is no anti-derivative is the probability distribution (normal curve)

The height (ordinate) of a normal curve is defined as:

P(x) = [1/[sigmaSQRT(2pi)]] * [e^ [-(x-mu)^2]/2sigma^2]]

There is no anti-derivative for this equation but it is integratable in the sense that limits do exist. The limit between - infinity and + infinity can be derived by some mathematical trick of doing a rotation of the figure and getting its definite integral and then taking the square root.

 

[Dale]

Is that integral "integratable" (in the sense that there is an algebraic expression or "anti-derivative" that does represent it after the proper operations are done?) It is clearly integratable as all tge variables are continuous over the specified interval so the limits do exist.

That depends on the specifics of the functions t(\lambda), x(\lambda), y(\lambda), z(\lambda). For example, if \lambda = t (i.e. \lambda is just another name for coordinate time), x = v t, y=0, z=0 (i.e. object moving inertially at velocity v along the x axis) then we recover the familiar time dilation formula \tau = t \sqrt{1-\frac{v^2}{c^2}}

But for other expressions it may not be so nice.

 

[stevmg]

A beautiful clarification of what is going on. I'm at my limit now but will study further. For now this is good enough.

I know what parametric equations are and how they can be used to represent a line - straight or curved - in space. (as opposed to a plane or a surface.) Never went that far in math or analytical geometry.

I think I used the wrong word... "integratable" should be "integrable."

When we studied integrals years ago, the distinction between anti-derivative and integral (although equivalent one-to-one in their final meanings) was never made, so we used the word "integrate" to refer to obtaining the anti-derivative, which is not really saying the same thing as there are equations as pointed out that have no obtainable anti-derivative but are integrable and as such do have numeric solutions (such as the normal distribution curve used in statistics.) It took me many years later to figure that out and the sense of the Riemann sum finally came to me on what this was all about. I wish this dual approach [integral = lim (Riemann sum) and anti-derivative ] had been taught and we would not have gotten into circular or tautological logic back then.

That's it for me on this thread.

 

[Dale]

A beautiful clarification of what is going on. I'm at my limit now but will study further. For now this is good enough.

I am glad to have helped. I really think that the concept of proper time and other Lorentz invariants is one of the most important ideas in SR, so please take your time and ask any questions that arise.

 

[stevmg]

Hey, Sports Fans...

A few days off and I'm back. Here is the simpleton's (that's me) approach to the twin paradox problem. We're going to use Simple Relativity and no references to anything else which may detract from this simple illustration which follows. Again, assume there are twins A and B - and believe me, that's how we would label them in the delivery room and nursery, even after Mom would give them their "real" names. Now, take twin B and put him/her on a spaceship at a velocity of 0.9949874371*c. Assume the jump to warp speed is instanenous. Start the two clocks (for A and for B) at that instant) t_A0 = t_B0 = 0.0.

Now send that little guy, B, off, say, to the right at that "warp"* speed: 0.9949874371*c while A remains here going nowhere. We will use the Earth as a frame of refrence F.O.R.. Keep him (B) going for ten (10) years Earth time t_A1.

Now, after 10 years Earth time or t_A1, turn him (B) around and return at the same speed [this time the velocity sign is reversed (from a plus (+) to a minus (-)], so make that -0.9949874371*c and he will get home in twenty years - Earth time t_A2. Now, that isn't too hard to wrap your brains around, is it?

We will proceed using Simple Relativity and will ignore the deceleration and subsequent acceleration back to warp speed for the return trip. Thus, we will not attempt to apply General Relativity. General Relativity would slow down the spaceship time even more because of the forces of acceleration/deceleration, so we can proceed with Simple Relativity and not lose the flavor of what we are trying to illustrate.

Using the time-dilation formula:

t = \gamma*t' or, in this case: t_A = \gamma*t_B

We have two F.O.R.s: The Earth (t_A and the spaceship t_B.) Continuing on with the use of the time-dilation formula:

Remember, \gamma = 1/SQRT[(1 - v²/c²)

We get the ratio of t_A/t_B = 10

DO THE MATH, IT'S GOING TO WORK OUT, just plug in the v that I showed you above.

Since it's 10 Earth time units per Spaceship time units the outbound trip consumed 10 years of Earth time or t_A1 but one year of spaceship time or t_B1. Now:
t_A1 = 10 years (that's Earth time)
t_B1 = 1.0 years (spaceship time)

Even though on the return, the spaceship is traveling at -v, under the square root sign in the expression of gamma this quantity is squared, thus losing the negative sign and square root quantity is the same as on the outbound journey:

SQRT(1 - (-v)²/c²) = SQRT(1 - v²/c²)

As such the 10:1 ratio of Earth time to spaceship time is preserved. Since it takes another ten (10) years for the spaceship to return in Earth time, then t_A2 = 20 years. Likewise, the return trip in the spaceship is another year so we have t_B2 = 2 years.

So now, twin A is twenty (20) years old, while twin A is two (2) years old.

This clearly displays the twin paradox using Simple Relativity by itself without going into "world lines" or \tau time coordinates or whatever. You do have to ignore the acceleratio/deceleration aspect but so did all the other posts on this blog.

Read Section XII. The Behavior of Measuring-Rods and Clocks in Motion in Einstein's book "Relativity." That gives you the inside dope on this problem by the master himself (Einstein, that is.)

By the way, how did I guess at v = 0.9949874371*c for the spaceship so that it would come out as the 10:1 ratio described above? I didn't guess. I worked the arithmetic backwards assuming the 10:1 ratio and wound up obtaining the SQRT (0.99) = 0.9949874371

Life is so full of tricks!

* I know, "warp speed" in sci-fi literature actually refers to busting through light speed but I just wanted to use a phrase that means something really, really, really fast!

 

[stevmg]

P.S.
The baby would "weigh" ten (10) times what he weighed on Earth, although neither he or anyone else in his spaceship would know it.

Oh yes, I know it is the mass that increases, but on Earth that does translate to weight.