Understanding the Twin Paradox: Exploring the Effects of Relativity on Aging

[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:
$$\tau = \int \sqrt {\left (\frac{dt}{d\lambda}\right)^2 - \frac{1}{c^2} \left ( \left (\frac{dx}{d\lambda}\right)^2 + \left (\frac{dy}{d\lambda}\right)^2 + \left ( \frac{dz}{d\lambda}\right)^2 \right) } \,d\lambda$$

 

[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.

 

[JesseM]

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.)

What do you mean by t_B in this case? The spaceship does not remain at rest in any inertial frame, so there's no inertial frame whose time coordinate always keeps pace with the age of the twin on the spaceship (i.e. that twin's proper time). You could construct a non-inertial frame where the spaceship remains at rest and the coordinate time keeps pace with the ship's proper time, but then the normal formulas of SR (like the time dilation formula) would not apply to this frame.

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

If t_A and t_B represented the time coordinates of two inertial frames this would not be correct. Remember, in relativity the situation between two inertial frames is always perfectly symmetric! If you and I are at rest in two different inertial frames, then if it is true in your frame that my clock is running ten times as slow as yours, it must be true in my frame that your clock is running ten times as slow as mine. This symmetry between frames is exactly where the idea for the twin paradox comes from--people think that you should be able to consider things from the perspective of the traveling twin's frame, and in this frame it would be the Earth twin's clock that's running slow, so that this frame would predict that the Earth twin is the younger one when they reunite. Fortunately that is not actually what relativity says, since the traveling twin does not remain at rest in a single inertial frame, and the time dilation formula only applies in inertial frames.

Here's when it would be valid to use the time dilation formula. Suppose you have two inertial frames A and B, and you look at two events that happen at the same position in the A frame, like two events on the worldline of a clock at rest in the A frame. If you want to know the time t_B between these events in the B frame, compared with the time t_A between them in the A frame (where they would just be equal to the proper time measured by a clock at rest in the A frame that's at the same position as both events), then the time dilation formula would say the answer is t_B = t_A*gamma, which means with a velocity of 0.9949874371 you have a ratio of t_B/t_A = 10. On the other hand, if you have two events that happen at the same position in the B frame, like events on the worldline of a clock at rest in this frame, then if the time between these events in the A frame is t_A and in the B frame it's t_B, then the time dilation formula would tell you t_A = t_B*gamma, so t_A/t_B = 10. In general, if you have a clock at rest in a given frame and you want to know the time between two of its readings in another frame, the time dilation formula has the following form:

(coordinate time in frame where clock is moving) = (time as measured by clock itself)*gamma

If you're dealing with two events that don't both happen at a single position coordinate in either of the two frames, then you can't use the time dilation formula at all! Instead you should use the more general time conversion formula from the Lorentz transformation, which works for time and distance intervals as well as time and distance coordinates:

t' = gamma*(t - vx/c^2)

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.

I think the reasons above show why it's a good idea to talk about the proper time $$\tau$$ along a given twin's worldline, rather than just coordinate times, since the twin that turns around doesn't have any single inertial rest frame.

 

[simonh]

OK, so I'm a complete physics wannabe, an armchair one even - considering getting involved in this wonderful world and the concepts do appear to be making some sort of sense. Now while I can quite happily wrap my head around the concepts, I don't understand the why and as such a visual aid would help me.

Now slap me if this is a theoretical impossibility, but it would make my understanding so much clearer!

If A has a telescope and is watching B throughout the entire 40 years (with perhaps the exception of restroom breaks), does B appear to be moving uniformly slower throughout the entire process, or would there be a difference between the departing speed and the return speed from A's perspective.

Or in essence, is the time dilation as a result of B's actions or B's actions comparative to A.

I have a sense of dread as I hit the reply button lol! (Be kind!)

 

[simonh]

In fact scratch that, i read the diagram incorrectly. B does appear to move slower to A than B does to B is what I get from that. But it does however raise another question, is the transition on A of 25.3 years visible to B as one massive time 'burst'?

 

[Dale]

Hi simonh, welcome to PF!

As far as what is actually visible to each twin, that is usually called the "Doppler shift" analysis. There is a good write up about it here:
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html
and please also see figure 2 here:
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#doppler

 

[sylas]

In fact scratch that, i read the diagram incorrectly. B does appear to move slower to A than B does to B is what I get from that. But it does however raise another question, is the transition on A of 25.3 years visible to B as one massive time 'burst'?

Not really. There is rather a massive distance burst that is actually "visible".

Consider this. Twin A stays at rest. From the perspective of twin A, moves out 6 light years at 60% lightspeed, then reverses direction and returns at 60% light speed. This takes 20 years total. I'm picking numbers here to give nice round calculations. The gamma factor is 1.25

What twin A actually SEES, however, is a bit different because of the time delay in light signals.

Twin A sees twin B reverse direction at a point 6 light years distance, and this is seen after 16 years. That's the 10 years B takes for the trip, plus the 6 years it takes for the light to get back. B gets back in another 4 years after that.

So for 16 years, A can see B receding into the distances, and for 4 years A can see B coming back. A knows the turn around was 6 light years away, and so knows that B took 10 years out and 10 years coming back.

What does B see? B sees A receding into the background at 60% light speed. Furthermore, B experiences only 8 years until the turn around. Just before turn around, B is receiving signals from A that are (from B's perspective) coming from a distance of 3 light years away.

How do we get this? Well, the light from A left 3 years previously, which is 5 years after they separated. In five years, at 60% light speed, A has moved 3 lightyears to the rear.

Now B turns around. What changes in what B SEES? B is now in a new different rest frame. In this frame, the signals currently being received from A are not from 3 light years distance. They are from 12 light years distance! And what B sees is that A suddenly becomes much smaller in the sky (being further away). If A sent a signal to B which was received at the moment B turns around, the event of sending that signal is something that occurred 3 years ago in one frame, and 12 years ago in another frame. But it is the same event for A. A is seen to be 4.8 years older than at departure (calculate it) and this event is now 12 years ago rather than 3 years ago. You do not suddenly see A aging. You rather see A at the same age, but that was longer ago, in your new rest frame.

Cheers -- sylas