How Does the Twin Paradox Affect Aging in Relativity?

[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.

 

[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

 

[simonh]

Thanks for the replies guys! Will give me some good material to mull on while I try to get to sleep tonight, even if that last paragraph does give me nightmares :) The concepts seem simple enough after reading 7 pages of this, but as is evident, I'm missing key pieces!

 

[stevmg]

Did I start all that again?

Sorry, Sports Fans.

Just answer just the one question... Twin B ages 2 years, Twin A ages 20 years... yes?

 

[sylas]

Did I start all that again?

Sorry, Sports Fans.

Just answer just the one question... Twin B ages 2 years, Twin A ages 20 years... yes?

In your example, with twin B traveling at +/- 0.9949874371 c relative to an inertial twin A: yes.

 

[stevmg]

A) Exactly, what is "proper time?" Keep the definition simple and NOT RECURSIVE, i.e. - non tautological. Does it mean that in the x, y, z and -ct coordinates, all clocks agree (-ct is constant.) If t were a specific value, then all 3-tuples of x, y and z that calculate to this specific t are merely from the infinite number of F.O.R.'s that relate to this particular t?

B) In the example above (we ignore the acceleration/deceleration to 0.9949874371*c) there is no acceleration or deceleration or curvilinear motion, so aren't both F.O.R.'s "inertial?" F.O.R. A (the earth) and F.O.R. B (the spaceship.)

 

[yuiop]

A) Exactly, what is "proper time?" Keep the definition simple and NOT RECURSIVE, i.e. - non tautological. Does it mean that in the x, y, z and -ct coordinates, all clocks agree (-ct is constant.) If t were a specific value, then all 3-tuples of x, y and z that calculate to this specific t are merely from the infinite number of F.O.R.'s that relate to this particular t?

Let us say a test particle moves from event 1 with coordinates (t1,x1,y1,z1) to event 2 with coordinates (t2,x2,y2,z2) then the proper time is the time interval recorded by a clock attached to the test particle.

Note: From here onwards I am using t, x, y and z as shorthand for (t2-t1), (x2-x1), (y2-y1) and (z2-z1) or (\Delta t, \Delta x, \Delta y, \Delta z)

In flat Minkowski space and considering inertial motion only, the proper time (tau or \tau) is defined mathematically (using units of c=1) as:

\tau = \sqrt{(t^2-x^2-y^2-z^2)}

Different observers will disagree on the values of t,x,y and z for a given particle but they all agree on the value of tau.

For example let us say one observer is watching a particle that moves a distance x=0.8 in time t=1.0 (so it has a velocity of 0.8c relative to his rest frame) then the proper time is:

\tau = \sqrt{(t^2-x^2-y^2-z^2)} = \sqrt{(1-0.8^2-0-0)} = 0.6 s

Another observer is moving with the particle (i.e. at rest with the particle so the velocity of the particle in his reference frame is zero) then he calculates the proper time as:

\tau = \sqrt{(t^2-x^2-y^2-z^2)} = \sqrt{(0.6^2-0-0-0)} = 0.6 s

 

[yuiop]

B) In the example above (we ignore the acceleration/deceleration to 0.9949874371*c) there is no acceleration or deceleration or curvilinear motion, so aren't both F.O.R.'s "inertial?" F.O.R. A (the earth) and F.O.R. B (the spaceship.)

Yes, BUT the spaceship does not remain in one inertial reference frame. For part of the journey the spaceship is in an inertial reference frame that a velocity +v reltive to the Earth and on the way back the spaceship is a different inertial reference frame that has a velocity -v relative to the Earth. That is a crucial difference.

 

[stevmg]

Interesting how my calculations still work out - but that's because the paths have the same distance and velocities back and forth (except for the negative (-v) and negative x (-x).

They wouldn't work out if twin B took a different path back [say like a half semicirlce where going out was along the diameter and the return was along the arc, even if you ignore the circular path (General Relativity) and just go by the longer distance.] There's where your \tau comes into play.

I am going to have to read selected texts about this to gain a better feel for what's what as I am out of my league at this point.

Thanks to all, Jesse, Dale, kev and sylas

 

[stevmg]

To DaleSpam or JesseM or anybody–

Help me with this. I will keep it simple. I just finished an excellent short text by Dr. Richard Wolfson of Middlebury College, Simply Einstein – Relativity Demystified (2002) and it was excellent. He discusses the twin paradox.

He posits a scenario of twins being born and one being rushed away to a star some 20 light years away. He treats the problem as if there were two spaceships. One blasting past the Earth at 0.8c when the twins were born (and immediately carrying, say, twin B) and a second spaceship returning to Earth from the star which is 20 light years away from Earth in the Earth-star time frame at 0.8c immediately after twin B arrives. The trip, looking from the Earth resting frame takes 25 light years. The trip from the outgoing spaceship time frame takes 15 light-years. Gamma is, of course SQRT(1 – 0.8)^2 = 0.6. He adds the two times up, the 25 years out and 25 years back and gets 50 years. He adds the two 15 years in spaceship time and gets 30 years – a 20 year difference in age.

He also states that the distance via the spaceship frame is 12 light years and notes that 12 light years/15 years = 4/5 = 0.8c, which agrees with the velocity of the spaceship.

He states because of the supposed symmetry (as one of you once mentioned in a prior post) that there is the paradox of the twin B being younger from the Earth frame and twin A being younger from the spaceship frame. He then goes on to say that, in fact, there is NO symmetry as twin B has to accelerate, fly at 0.8c, slow down, stop, turn around, accelerate again to -0.8c and come back and slow down again.

He later in his book uses the space-time diagram to go into this again and it looks somewhat like this (we use only one dimension – x for distance and t for time):
..t
..|
C|\ ... you’ll have to imagine the 45-degree light line as it is hard to draw with typographical characters
..|. \
..|. . \D
..|. . /
..|. /
A|/___________x

A is the event of the birth of the twins. AC represents the world-line of twin A who just sits there in her own frame of reference. AD represents twin B’s outward journey to the star (20 light years) and DC represents twin B’s return journey to Earth.

This is as simple a diagram that can be presented.

Can you go through the math of the “proper time” [is that “ds” in the equation ds/dt = SQRT[c^2 – (dx/dt)^2]?]. In this case I assumed the “tau” is t (thus dt/dtau = 1 so the first term in this “reverse” Pythagorean expression is c^2) and I ignored the y and z axes as we were only discussing the x-axis (and the t axis.) I cannot understand how, geometrically, ADC (the path of the rocket in space-time) is shorter than AC (sitting still.)

I’ve looked in different books and cannot find it anywhere, just allusions to it, with no specific examples or calculations.

 

[JesseM]

Can you go through the math of the “proper time” [is that “ds” in the equation ds/dt = SQRT[c^2 – (dx/dt)^2]?]. In this case I assumed the “tau” is t (thus dt/dtau = 1 so the first term in this “reverse” Pythagorean expression is c^2) and I ignored the y and z axes as we were only discussing the x-axis (and the t axis.) I cannot understand how, geometrically, ADC (the path of the rocket in space-time) is shorter than AC (sitting still.)

It's because proper time in spacetime is not calculated the same way as length in space. In a 2D space, if the endpoints of a straight segment of a path are at coordinates (x₁, y₁) and (x₂, y₂), then according to the pythagorean theorem, the length of that segment would be \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} (if you don't want to worry about specific coordinates and you know the distance along the x-axis \Delta x and the distance along the y-axis \Delta y between the two endpoints, the formula becomes \sqrt{\Delta y^2 + \Delta x^2 }). On the other hand, in a 2D spacetime with a spatial dimension x and a time dimension t, if you have a constant-velocity segment (like a straight-line segment in your diagram) with endpoints (x₁, t₁) and (x₂, t₂), then the proper time along this segment is \sqrt{(t_2 - t_1)^2 - (1/c^2)*(x_2 - x_1)^2}, or if we use units where c=1 (like light-years and years), it'd be \sqrt{(t_2 - t_1)^2 - (x^2 - x_1)^2} = \sqrt{\Delta t^2 - \Delta x^2 }. You can see that this is different from the pythagorean formula because it has a minus in the middle rather than a plus. That change means that unlike in 2D geometry where a straight-line path is always the shortest distance between two points, in spacetime a constant-velocity path between two events (like the event of the twins departing from one another and the event of the twins reuniting) is always the one with the largest proper time.

Doing the calculations for your specific example:

He posits a scenario of twins being born and one being rushed away to a star some 20 light years away. He treats the problem as if there were two spaceships. One blasting past the Earth at 0.8c when the twins were born (and immediately carrying, say, twin B) and a second spaceship returning to Earth from the star which is 20 light years away from Earth in the Earth-star time frame at 0.8c immediately after twin B arrives. The trip, looking from the Earth resting frame takes 25 light years. The trip from the outgoing spaceship time frame takes 15 light-years. Gamma is, of course SQRT(1 – 0.8)^2 = 0.6. He adds the two times up, the 25 years out and 25 years back and gets 50 years. He adds the two 15 years in spaceship time and gets 30 years – a 20 year difference in age.

Here we have three constant-velocity segments as you described:

A is the event of the birth of the twins. AC represents the world-line of twin A who just sits there in her own frame of reference. AD represents twin B’s outward journey to the star (20 light years) and DC represents twin B’s return journey to Earth.

The segments are AC, AD, and DC. If we're calculating things in the rest frame of twin A, then in this frame \Delta x for AC is zero (since twin A doesn't change positions between the two endpoints) and \Delta t for AC is 50 years. So, the proper time on this segment is \sqrt{\Delta t^2 - \Delta x^2 } = \sqrt{50^2 - 0^2 } = 50 years. Meanwhile, in this same frame if we look at segment AD, \Delta x is 20 light-years while \Delta t is 25 years, so the proper time of this segment is \sqrt{\Delta t^2 - \Delta x^2 } = \sqrt{25^2 - 20^2 } = \sqrt{225} = 15 years. And for DC, in this frame \Delta x is again 20 light-years while \Delta t is again 25 years, so the proper time on this segment is 15 years as well. Of course you could repeat the calculations of the proper time in a different frame using x',t' coordinates where the \Delta x&#039; and \Delta t&#039; for each segment would be different, but the proper time on each segment would remain the same.

 

[stevmg]

1. I keep getting gamma screwed up. It is [SQRT(1 - v²/c²)]^-1. I apologize for that. I also don't know how to get all that what I wrote under the square root sign in this blog. I tried what you did but it doesn't come out correctly.

2. I just remember three things:

a - Items are longest" in their "rest" frame (when they are not moving.) That applies to distances as well such as the 20 lt-year distance between the Earth and the star as Dr. Wolfson proposed.

b - Time is most "rapid" in its "rest frame" (where there is no motion of the rest frame.) Hence, in Dr. Wolson's example, the trip is 25 years in the Earth-star frame but only 15 years in either spaceship frame (to and fro) which are moving at 0.8c.

c - Mass (not a player here) is "least" or "smallest" in the rest frame and increases as the frame moves. Dr. Wolfson never goes into this in his book.

Thus \gamma is always a ratio:

\gamma = length. at. rest/length. in. motion
\gamma = duration. at. rest/duration. in. motion
1/(\gamma) = mass. at. rest/mass. in. motion

Sorry for all those "."'s but the words run onto each other unless I insert them - Ooops! Just saw the "white spaces and dots selection in the ?Latex? Reference (what's this, we dealing with rubber gloves?) Oh, well...

The reason why Dr. Wolfson chose 0.8c is that it is a neat square root as 1 - 0.8² = 1 - 0.64 which = 0.36 and \sqrt{0.36} = 0.6. Hey! Look at that! I got the "\sqrt{}" to work - for a simple expression.

Now, the coup de grace

In your example below which I quote from you, you actually show how a world-line is calculated in explainable terms. I see clearly why the AC length (the twin that sits still) is less than the ADC length (the twin that moved) - it is because of the negative sign summation under the square root. This sort of makes the world line diagram counter-intuitive as the "longer" path is the "shorter" path in proper time. That is freakin' weird. Is there a way of looking at that diagram to make it look not so weird?

It's because proper time in spacetime is not calculated the same way as length in space. In a 2D space, if the endpoints of a straight segment of a path are at coordinates (x₁, y₁) and (x₂, y₂), then according to the pythagorean theorem, the length of that segment would be \sqrt{(y_2 - y_1)^2 + (x_2 - x_1)^2} (if you don't want to worry about specific coordinates and you know the distance along the x-axis \Delta x and the distance along the y-axis \Delta y between the two endpoints, the formula becomes \sqrt{\Delta y^2 + \Delta x^2 }). On the other hand, in a 2D spacetime with a spatial dimension x and a time dimension t, if you have a constant-velocity segment (like a straight-line segment in your diagram) with endpoints (x₁, t₁) and (x₂, t₂), then the proper time along this segment is \sqrt{(t_2 - t_1)^2 - (1/c^2)*(x_2 - x_1)^2}, or if we use units where c=1 (like light-years and years), it'd be \sqrt{(t_2 - t_1)^2 - (x^2 - x_1)^2} = \sqrt{\Delta t^2 - \Delta x^2 }. You can see that this is different from the pythagorean formula because it has a minus in the middle rather than a plus. That change means that unlike in 2D geometry where a straight-line path is always the shortest distance between two points, in spacetime a constant-velocity path between two events (like the event of the twins departing from one another and the event of the twins reuniting) is always the one with the largest proper time.

Doing the calculations for your specific example:

Here we have three constant-velocity segments as you described:

The segments are AC, AD, and DC. If we're calculating things in the rest frame of twin A, then in this frame \Delta x for AC is zero (since twin A doesn't change positions between the two endpoints) and \Delta t for AC is 50 years. So, the proper time on this segment is \sqrt{\Delta t^2 - \Delta x^2 } = \sqrt{50^2 - 0^2 } = 50 years. Meanwhile, in this same frame if we look at segment AD, \Delta x is 20 light-years while \Delta t is 25 years, so the proper time of this segment is \sqrt{\Delta t^2 - \Delta x^2 } = \sqrt{25^2 - 20^2 } = \sqrt{225} = 15 years. And for DC, in this frame \Delta x is again 20 light-years while \Delta t is again 25 years, so the proper time on this segment is 15 years as well. Of course you could repeat the calculations of the proper time in a different frame using x',t' coordinates where the \Delta x&#039; and \Delta t&#039; for each segment would be different, but the proper time on each segment would remain the same.

"The proper time along this segment is SQRT[(t_2 - t_1)^2 - (1/c^2)*(x_2 - x_1)^2]" Is that the same as s, or, to say it another way:

s² = (t_2 - t_1)² - (1/c²)*(x_2 - x_1)²

I've seen that equation s² = ya-de-da in the past and is "s" proper time?

Can you recommend a (text) book that would go into this for me as I have had no success in finding one that isn't either too simple or too complex. I thought I had struck paydirt with "Relativity Demystified" by David McMahom but - boom! He goes into tensors and matrices and all that chazarei (that's Yiddish for "garbage" - I speak 23 languages superfluously.)

Finally, besides the seemingly "loose" explanation that this "twin paradox" really is not a paradox based on the fact that twin B accelerates to speed, decelerates at the star, turns around, re-accelerates to speed and decelerates at the Earth to meet her sister. Is there a more scientific quantitative explanation that establishes that you really DO use twin A (the lady that "sits") as the ultimate "rest frame?"

I am a retired physician (still have a license) with a math major and Masters that teaches college here and there and have time in my life now to learn about things that I have seen before but never had the time to study - after all - when I was doctoring, wouldn't you have preferred that I read medical articles (which I did) rather than study theoretical physics, which I didn't.

 

[yuiop]

Can you recommend a (text) book that would go into this for me as I have had no success in finding one that isn't either too simple or too complex. I thought I had struck paydirt with "Relativity Demystified" by David McMahom but - boom! He goes into tensors and matrices and all that chazarei (that's Yiddish for "garbage" - I speak 23 languages superfluously.)

I have that book too and the cover and title is misleading. The cartoon ilustration on the front and the promise of "laymen's explanations" on the back lead one to believe it going to be a clearly explained text but not far into it you discover that that author assumes everyone was born knowing what Kronecker delta function is. It certainly isn't a simple introductory text to relativity if you do not have pretty advanced math skills already and pretty much omits the Michelson Morley experiment and discussions of things like the twin's paradox, presumably because the author thinks those subjects are self explanatory.

 

[atyy]

Is there a more scientific quantitative explanation that establishes that you really DO use twin A (the lady that "sits") as the ultimate "rest frame?"

Yes. In physics, you can choose any reference frame you want, no matter how weird. But you pay the price for the weirdness, in that the laws of physics will look weird, even though they haven't changed at all. It's the same as if you define green as red, and red as green. Then grass is red, but the grass hasn't changed, just that you have chosen a weird nomenclature for colours. Of course, once you know the laws of physics in anyone reference frame inertial or not, you know the laws of physics in any other reference frame, just as once you know that grass is green in a particular colour nomenclature, then you know the colour of grass in any colour nomenclature.

In Newtonian physics and special relativity, there are a special class of reference frames called inertial reference frames, in which the laws of physics look nice, or at least have their standard textbook form. The fact that there is a class of frames that are inertial, rather than only one such frame is due to symmetries in the laws of physics. The difference between Newtonian physics and special relativity is that the symmetry is Galilean in the former, but Lorentzian in the latter. In the context of the twin paradox, twin A is special because she is always at rest in a particular inertial frame. In the broader context of special relativity, twin A is not special compared to another person who is moving at constant velocity relative to her, since that person is at rest in another inertial reference frame.

 

[matheinste]

1.
Finally, besides the seemingly "loose" explanation that this "twin paradox" really is not a paradox based on the fact that twin B accelerates to speed, decelerates at the star, turns around, re-accelerates to speed and decelerates at the Earth to meet her sister. Is there a more scientific quantitative explanation that establishes that you really DO use twin A (the lady that "sits") as the ultimate "rest frame?"

You can look at it from the point of view of twin A, who is at rest in an inertial reference frame throughout. For the simple case of a constant and equal speed of outward and inward journeys, and assuming instantaneous accelerations at the end points and turning point, the time passed for B is twice the time dilated value of time passed for A during the outward journey of B. If finite accelerations are used, leading to non-linear speed profiles, then integration is required of B's instantaneous co-moving inertial frames.

Matheinste.

 

[JesseM]

In your example below which I quote from you, you actually show how a world-line is calculated in explainable terms. I see clearly why the AC length (the twin that sits still) is less than the ADC length (the twin that moved) - it is because of the negative sign summation under the square root. This sort of makes the world line diagram counter-intuitive as the "longer" path is the "shorter" path in proper time. That is freakin' weird. Is there a way of looking at that diagram to make it look not so weird?

I don't know of any way to represent it so paths with longer proper time actually have a longer spatial length on the diagram...I think you just kind of have to remember that spacetime geometry works differently than spatial geometry.

I've seen that equation s² = ya-de-da in the past and is "s" proper time?

s² is the square of the spacetime interval which is defined a little differently than proper time...whereas proper time squared would be equal to \Delta t^2 - (1/c^2)*\Delta x^2, the square of the spacetime interval s is equal to \Delta x^2 - c^2 \Delta t^2. So s² is basically just -c² times the proper time squared, with s having units of distance rather than time. For any two points in spacetime, the separation between them is either "spacelike" which means s² is positive, "timelike" which means it's negative, or "lightlike" which means it's zero. Any two events on the worldline of a slower-than-light object have a timelike separation, any two events on the worldline of a light ray moving in a single direction in a vacuum have a lightlike separation, and if two events have a spacelike separation then no particle moving at or slower than light can have both events on its worldline (a spacelike separation also means there is one inertial frame where the two events occurred simultaneously at different positions in space, and s would be the spatial distance between them in that frame). A light cone consists of all the points with a timelike or lightlike separation from the event on the "tip".

Can you recommend a (text) book that would go into this for me as I have had no success in finding one that isn't either too simple or too complex. I thought I had struck paydirt with "Relativity Demystified" by David McMahom but - boom! He goes into tensors and matrices and all that chazarei (that's Yiddish for "garbage" - I speak 23 languages superfluously.)

https://www.amazon.com/dp/0716723271/?tag=pfamazon01-20 by A.P. French was my college relativity book, I remember it being pretty clear.

Finally, besides the seemingly "loose" explanation that this "twin paradox" really is not a paradox based on the fact that twin B accelerates to speed, decelerates at the star, turns around, re-accelerates to speed and decelerates at the Earth to meet her sister. Is there a more scientific quantitative explanation that establishes that you really DO use twin A (the lady that "sits") as the ultimate "rest frame?"

You don't need to use twin A's frame to analyze this problem, you can use any inertial frame and you'll still get the same answer. Check out my post #36 on this thread where I talked about how you could analyze a twin paradox scenario from the perspective of two different frames, one where the inertial twin "Stella" was at rest, and another where the non-inertial twin "Terence" was at rest during the outbound leg of his trip (but not the inbound leg).

I am a retired physician (still have a license) with a math major and Masters that teaches college here and there and have time in my life now to learn about things that I have seen before but never had the time to study - after all - when I was doctoring, wouldn't you have preferred that I read medical articles (which I did) rather than study theoretical physics, which I didn't.

Well, if you'd been my doctor I'd have preferred you read medical articles! ;) But yeah, studying things on your own that you didn't get to learn in school can be a great experience...

 

[DrGreg]

Can you recommend a (text) book that would go into this for me as I have had no success in finding one that isn't either too simple or too complex. I thought I had struck paydirt with "Relativity Demystified" by David McMahom but - boom! He goes into tensors and matrices and all that chazarei (that's Yiddish for "garbage" - I speak 23 languages superfluously.)

You might like to try General Relativity from A to B by Robert Geroch, University of Chicago Press, 1978, ISBN 0-226-28864-1. The preface says it is based on lectures given to non-science undergraduates. Despite the name, most of the book is about Special Relativity (i.e. without gravity), but carefully written so that when gravity is introduced at the end, nothing previously said turns out to be incorrect. It's a moderately slim paperback, with only light-weight maths in it.

A substantial number of sample pages are available at Google Books.