Nice graphical presentation, Jorrie. So, I conclude that the apparent "time loop" that seems to be (and conceived by Pam) in the Jim's worldline (at the midpoint) is because of the distance (6ly) between Pam and Jim. If Jim was near Pam his apparent worldline will be the green one (the star's worldline). But I assume that Pam will not actually see any "time inversion" concerning Jim (in other words if she could see Jim she will not observe his life "going backwards" for a time interval). Unlike it seems that all these happen just in a moment. It seems that in the midpoint the Jim's life makes a "jump" (from the Pam's perspective) and she has lose the part of his life which is in this "loop"... Am I right???... If so, maybe, would it be better to represent this part of the red line as an horizontal part???... ...

Hi George, you wrote: "But I assume that Pam will not actually see any "time inversion" concerning Jim (in other words if she could see Jim she will not observe his life "going backwards" for a time interval)."

Correct. Pam could possible read a series of clocks that Jim laid out and synchronized in his frame, but they will surely not show any time reversal. It is only when she measures her instantaneous speed relative to Jim's frame and considers his definition of simultaneity that she can come to the conclusion of time reversal for Jim. It is hence only apparent.

You wrote: "It seems that in the midpoint the Jim's life makes a "jump" (from the Pam's perspective) and she has lose the part of his life which is in this "loop"... Am I right???"

No, because of the smooth deceleration and turnaround (not a jerk), there is no sudden jump in the apparent age of Jim.

Jorrie

Hi Jorrie,

If we assume that Pam is on a rigid platform as long as the distance from earth to Alpha proxima. Then suppose Pam accelerates rapidly to a velocity of c/2 in 1 sec. then during the one sec. due to lorentz contraction the star would appear to come closer to Pam by a distance of L(1-(1-1/4)^1/2) = 0.134 L where L is 4 light-years which is an enormous distance. This shows that the apparent velocity of the star w.r.t. Pam will be much higher than velocity of light. Also the other end of Pam's platform which is rigid w.r.t. Pam will shoot off away from the star at incredible speed far beyond 'c' . How do these facts reconcile with Special theory of Relativity.

Hi kasharma, you wrote: "... then during the one sec. due to lorentz contraction the star would appear to come closer to Pam by a distance of L(1-(1-1/4)^1/2) = 0.134 L where L is 4 light-years which is an enormous distance. This shows that the apparent velocity of the star w.r.t. Pam will be much higher than velocity of light."

Yep, apparent velocities can exceed c, but never properly observed velocities, e.g., the blueshift of the star must still be finite as measured by Pam; hence the observed relative velocity cannot exceed c.

You further wrote: "Also the other end of Pam's platform which is rigid w.r.t. Pam will shoot off away from the star at incredible speed far beyond 'c' . How do these facts reconcile with Special theory of Relativity."

No, the star will not observe the "other end of Pam's platform" to shoot off at a speed exceeding c. How did you get to this conclusion?

Jorrie

Hi Jorrie,

I agree the Star will not observe the other end of Pam's Platform moving at incredible speed. With respect to the Star the other end of Pam's Platform will appear to move away slower at some smaller acceleration of C² / (C²/a + L) where 'a' is the acceleration at Pam's end (as per theory). Please correct me if I am wrong. But the movement of time on the Plaform at the other end will also appear much slower as seen by the Star. But, for a person standing on the Platform at the other end the speed would be incredible as his time is the same as Pam's, being rigidly attached to Pam's position and therefore whatever is observed by Pam will also be observed by the person at the other end.

"Do you agree?"

Well, DUH! I'm not likely qualified to disagree, and even if I were, I think this is closer to correct than otherwise. But while I digest it, there's this one little question... Presuming it is possible to measure such a quality in photons, which travel inherently "at" the speed of light (being "light" sort of makes this obligatory, eh?), can photons be said to "age"? In other words, experience duration over the time required to travel for years and years from distant stars? I don't know what changes we could measure to define this, there may be none, but they DO get "older", do they not?

Re: can photons be said to "age"? More 'to-the-point', what you are actually asking is "do photons SHOW their age?" ... because everything that exists is most certainly aging with each passing moment of its existence. Even those miniscule bits of matter just coming into existence as this perceived universe expands begin to tally-up their own time-periods of existence.

But, does all 'matter' (as we refer to it), including photons, CHANGE in any measurable or perceptible way? Certainly not in the same fashion as living tissue of any sort that we're aware of!

Dust we once were, and to dust we shall all return. Blessings to all

Not to put too fine a point on it, but as I stated:

"...I don't know what changes we could measure to define this, there may be none, but they DO get "older", don't they?..."

Yes, "matter" ages - granite crumbles, iron rusts, but how about energy? Do photons "age" in any appreciable manner? What about magnetism? Or gravity? Any signs of them getting "older" over time? What would we even look for?

Hi EnviroMan, you asked: "Presuming it is possible to measure such a quality in photons, which travel inherently "at" the speed of light (being "light" sort of makes this obligatory, eh?), can photons be said to "age"?"

The prevailing wisdom is that photons do not "age", meaning that they experience zero proper time as they travel the universe. Anyway, what does "aging" of an elementary particle mean? Unless it decays and has a statistical "half-life", it may be meaningless concept!

Jorrie

Jorrie; Thanks for the reply! I was almost of the opinion that there would be no meaningful measurement to determine "age" in a photon, but thought that worth confirming. So a photon formed from nuclear fusion in our sun today is no more "fresh and new" than one formed at the time of the Big Bang. That almost seems to have some sort of significance, but I'm not sure what, exactly. I shall ponder this for a bit and perhaps return with another question later. Or, I may conclude that it is an inherently meaningless concept after all...

Hi, EnviroMan. That was an interasting question. I would like to approach this issue in a more theoretical (or philosophical) way. If we could just "observe" a photon (whatever this means, without disturbing it in any way) during its travel through the space (with a constant speed = c) we should see it as it's not "aging" because the time on its reference frame is not "passing" (from our point of view). We should see it "frozen" in time. The same happens for a photon. As it sees the universe travels in a speed c (from its point of view) it consider the universe "frozen" (or it sees a "static" universe). The things are not so when we have any change of the photon's speed, i.e. a curved orbit due to an intense gravitational field or because it passes through a material (e.g. glass or water). By the way, when the photon travels through a (e.g.) piece of glass (with a lower speed than c) then we should see a kind of slow "aging" of the photon, as the time (on its frame and relative to us) now seems that is "passing" (although in a very "slow rate"). The same happens to the universe from the photons point of view. I assume that sth like this happens if the photon follows a circular path... ... Jorrie, do you agree???...

(In fact, when we actually see a photon this means that it is the end of its journey (on the retina of our eye) and, hence, the end of its existence. We see it as it is, only in an instant (the end of its life). In order to consider an "age" we should be able to see it in two different instants in order to observe any "changes" of it, thus an "aging". But this is impossible.)

Hi George, you wrote: "By the way, when the photon travels through a (e.g.) piece of glass (with a lower speed than c) then we should see a kind of slow "aging" of the photon, as the time (on its frame and relative to us) now seems that is "passing" (although in a very "slow rate"). The same happens to the universe from the photons point of view. I assume that sth like this happens if the photon follows a circular path..."

I do not think any of these (even in glass) make a "photon age" - for one thing, photons in a circular orbit at r = 1.5R_hor around a black hole are not thought to age!

You are right about the impossibility of measuring a photon's age - they have no clocks and no half-live!

Jorrie

Jorrie, you wrote: "I do not think any of these (even in glass) make a "photon age"..." ... Why not??? It's almost the same thing as with a traveller which is moving with a speed very near to c (e.g. 0.95c). We "see" him aging slower than we do (from our reference frame) as his time "flows" slower than ours. The same should happen with a photon which is travelling through a material and its speed is lower than c (e.g. 0.95c). We should "see" the time (on the photon's reference frame) passing by (much slower than our time, of course). This could be considered as a kind of "aging", with the meaning of the "flow" of time on it (with a philosophical temper). I can't understand your objection... ??? ...

Hi George, you wrote: "The same should happen with a photon which is travelling through a material and its speed is lower than c (e.g. 0.95c). We should "see" the time (on the photon's reference frame) passing by (much slower than our time, of course). ... I can't understand your objection... ???"

I view this in the light of my black hole example. A distant observer can observe light to travel slower near a black hole, but there is no way to observe that light to age. So, of we stick to observables, light never ages!

Jorrie

PS: Light "travels slower" near a black hole, or is "delayed" from the perspective of a distant observer, because spacetime is "more dense" there. The same should happen in a medium, like air or glass.

"...spacetime is "more dense" there..."

Due to increased gravity from the extra mass, or does it work the other way 'round? That is, there's more mass, hence more gravity, because space/time is more dense? The second way doesn't seem right, but so much of this is not intuitively obvious...

Hi EnviroMan, you asked: "Due to increased gravity from the extra mass, or does it work the other way 'round?"

The first way, I think. Time and space are "contracted", relative to a distant (flat space) observer, meaning clocks tick slower and measuring rods appears shorter, presumably due to the presence of the mass.

This effect is not observable locally, near the mass. A local observer can however observe the distant universe to display the inverse effect - faster clocks and expanded rods.

This is as stated in "ordinary language" - relativistically, it is just the curvature of spacetime that is greater the nearer one gets to a massive body. Light follows the straightest possible path (a spacetime geodesic) through curved spacetime, which appears to be curved when viewed from flat space.

Jorrie

Hi Jorrie, you wrote: "A distant observer can observe light to travel slower near a black hole, but there is no way to observe that light to age."... I agree with you because (as I said in a previous post) we need to be able to observe a photon in two different instants in order observe any "change" of its state thus an "aging", but this is imposible. So in a (practical) way we cannot pronounce on any "aging" of a photon (and we can say that there isn't any "aging").

You wrote: "So, of we stick to observables, light never ages!"... You are right... But what if we don't stick to observables??? (I said that I try to approach this issue in a more philisophical way)... Just replace a real photon with a "magic photon" and imagine that this photon could send clock pulses (magically)... As it travels in the space at speed c we cannot count any ticks of its clock (as an infinite time duration separates one tick from the next tick). We conclude that the time is "frozen" on the photon (there is no "aging"). Then this photon approach a BH and follows into... or enters a material... So its speed becomes smaller than c. This time we can count ticks of its clock. These ticks are far away to each other (more sparse than the ticks of our clock)... But, anyway, there are ticks... So, we conclude that the time on the photon is not "frozen" any more... There is a time flow (although very slow) thus a kind of "aging". If you don't feel convenient with the "photons with clocks" you could replace them with "imaginary travellers carrying clocks" (able to travel at speeds equal and just below c though).

Any comments???

I see how that works, Pam and Jim are two such imaginary clock-toting travelers. But as you already pointed out, photons have no half-life, thus carry no clocks. However, once observed, a photon ends it's service life and is no more, so from the time it was generated in a star or by whatever process until then, a certain amount of measurable (although not by the photon) time has passed. Is there any difference in the photon if it travels a light year as compared to only a light minute?

Hi EnviroMan... Yeap, it seems that it is so... Time doesn't flow for a photon (relative to us)... Your question was "food for thought" though...

Hi again George; you wrote: "But what if we don't stick to observables??? (I said that I try to approach this issue in a more philisophical way)... Just replace a real photon with a "magic photon" and imagine that this photon could send clock pulses (magically)..."

The problem is that when philosophy enters, science leaves...

"Then this photon approach a BH and follows into... or enters a material... So its speed becomes smaller than c."

Careful! A photon's speed does not become smaller than c when it nears a black hole. A local observer will still measure it as c. It is just a distant observer that measures an apparent speed smaller than c. I still maintain that the photon experience no proper time near a black hole and GR supports that.

"If you don't feel convenient with the "photons with clocks" you could replace them with "imaginary travellers carrying clocks" (able to travel at speeds equal and just below c though)."

Again, this is philosophy, not science. I cannot even imagine a traveler moving at c, together with a photon! Irrespective of how close to c he moves relative to you, the photon still travels at c relative to him.

I'll rather stick to the observables - meaning that a 'photon's time' is a meaningless concept.

Jorrie

PS: I'm not qualified to talk about the characteristics of photons inside matter (like glass), but AFAIK, since matter is mostly empty space, photons are believed to still travel at exactly c between molecules/atoms. They are then absorbed and emitted again, causing a delay - this is apparently what refractive index is all about.

Hi Jorrie. You said: "The problem is that when philosophy enters, science leaves..."... I agree with you. But sometimes it's o.k. to wonder "what if...". (This may gives you a little push further in comprehension of the reality)...

You said: "... photons are believed to still travel at exactly c between molecules/atoms. They are then absorbed and emitted again, causing a delay - this is apparently what refractive index is all about." ... After a second thought this is correct. The apparent reduction of the photon's velocity is because of this.

About the fall of the photon in the BH I have to agree with you (you know better afterall). (...but, still, if I was a photon falling into the BH I would perceive the near space around me "frozen" in time, but what about the distant space???... oups... philosophy again... ... pardon... ...I'll just accept what you said: "I still maintain that the photon experience no proper time near a black hole and GR supports that." ... and I stop it right here... ...)

So it seems that, anyway, a photon doesn't get any speed lower than c in any case. So it's impossible to be any "time flow" on it according to my scenario (even in a general philosophical meaning... oups... sorry again... ...) Since there is no other way to observe any "time flow" on it, we have to admit that there isn't such a thing like "aging".

I did a little research elsewhere, and found what appears to me to be a good non-mathematical explanation. Copied herewith:

Suppose I ask you "what is the speed of the Mississippi between St.
Louis and New Orleans?". Well, what speed? If you're talking about
the river as a whole, there is no speed. It is just there, extended
between those two points (and beyond). So, you need some marker. You
may ask "if I drop a buoy in the river, in St. Louis, how long before
it'll get to New Orleans?" This time it can be measured and dividing the
distance by this time will give you some speed. Or, you can ask "If I
create a flow disturbance near St. Louis (say, by opening or closing
a dam), how long before the results will be detected in New Orleans?"
This'll give some other speed (different from the first). And neither
of these two speeds has much to do with the speed of the individual
water molecules which make up the river.

The situation with light in matter is similar. You can talk about
phase velocity, which is the velocity at which a constant phase
surface propagates (that's the one to which most simple-minded popular
writeups refer). This one depends on the light's frequency and can be
either smaller or *larger* than c. No contradiction to relativity
since no information is conveyed. A more realistic measure may be the
so called "group velocity" which is the velocity at which a pulse of
light (which contains many frequencies) propagates. This one is in
general different than the phase velocity and is usually <=c, but in
some pathological situations may be > c as well. Then you can get to
"energy flow velocity" (self explanatory, I hope) or "signal
velocity", which is the velocity in which a disturbance in the EM
field propagates. These last two are always <= c. The point is,
you've a bunch of different "velocities", all of which may be
different, in general, and none of which must equal the speed of the
photons (which is always c).

In any case, my original question was: Does the photon, or what ever
is representing it as it travels through the medium that makes it
slow, experience time differently than a photon not going through the
medium? And if so, what effects would a person see?

And the answer is that the medium *does not* make the photon slow. A
photon always propagates at c.

How much confusion has been either added or subtracted by this?

Hi Enviroman. You said: "How much confusion has been either added or subtracted by this?"... Not confusion at all... This was interesting...

"And the answer is that the medium *does not* make the photon slow. A
photon always propagates at c."... This is exactly what Jorrie pointed out (about the propagation of the photons inside a material and their apparent velocity reduction) and I agree with him.

But I'm still thinking about the case of the photon's fall inside a Black Hole. A distant observer (not a local one) consider an apparent velocity reduction of the photon as it falls inside the BH. What's hapenning from the photon's perspective??? 1) The photon should consider (respectively) an increment of the time flow of the distant universe, just because of the extreme gravity. 2) But, on the other hand, we have the removal of the photon (away from everything else) at speed c. Because of this, the photon considers the sorrounding universe to be frozen in time. What's the "combination" of (1) and (2)??? I think the movement of the photon at c must overcome the effect of gravity. And I think that the movement at c is sth so absolute "extreme" that the final result is (essentially) only because of this. And the result must be "frozen time of the sorrounding universe" from the photon's perspective.

I'll put it in another way: In another forum Jorrie and I had a discussion concerning what happens to a traveller falling in a BH. Initially, I had the wrong conception that the traveller will experience an extreme acceleration of the time flow of the sorrounding universe (he should experience an infinite blue shift, hence he should receive an infinite amount of energy). Jorrie pointed out that the traveller is not static but, instead, he falls in a speed very close to c. So he experiences an extreme retardation of the time flow of the sorrounding universe (red shift). The result of the two is that the traveller (finally) experiences a time retardation (a finite red shift) of the universe. If (in the case of the traveller) the "movement close to c" overcomes the "extreme gravity" I think that the same must happen in the case of a photon but in an "extreme way". I mean that (in the case of a photon) this time retardation must be infinite.

I just consider it this way intuitivelly. Jorrie could help me on this... ...

"but I'm still thinking about the case of the photon's fall inside a Black Hole".

I'm thinking in terms of the photon's energy , at C, as it strikes OUR EYE BALL . It is absorbed as energy and heat and mass, and then what happens may be more important to the nature of things than the black hole which seems kinda finite as C. As this photon travels along the optic nerve and the brain, well who knows, what kind of new energy appears as an altered photon outward into reality.

Maybe the dual nature of the photon , black hole and beyond on one end and the eye ball and beyond at the other, makes the energy "travel" both ways (except for heat and mass) at the same time. Even if the photon sees time slow way down and come to a stop, maybe as it enters the eye ball, it has its own speed which is unassociated with time, and is now in "brain" time, and the photons (vibrations) are coming from our sun and the stars in the universe, and man made photons, and reflected light etc. I think that "later civilizations" have fogged knowledge about our sun, and called it pagan,where as earlier ones enjoyed and understood the sun rise and sun set of the sun and the moon and stars.

Photons that have lost their energy would certainly explain the amount of Dark Matter in the universe...............

Hi Duckinthepond, you wrote: "Photons that have lost their energy would certainly explain the amount of Dark Matter in the universe..............."

How certain are you about this? Any reference to research in that direction?

Jorrie

It was Scherrer who (I think) proposed this idea alongside the theory of everything.

To prove would require a collider of astronomical proportions.

There is an intuitive elegance to the theory.

Relativistic time dilation when two events occur at differrent points with respect to the observers referance frame. Adding velocity becomes an approximation.

Any speed added to the speed of light gives the speed of light.

Hi Jorrie,

You've done a good job of explaining the paradox, and the graphs and pictures are well done.

"Do you agree?"

Well, I didn't see anything that was obviously wrong, and I trust your math (which was not shown here). However, some things seem strange. If I understood right, the outcome was the same in both parts (8 vs 12 years of aging). In the second part, Pam had a constant 0.8G of acceleration while Jim had 1.0 from Earth's gravity. Since gravity affects clock speed, it seems that Jim should be the younger one here.

The red loop doesn't match the blue peak. The reason is not obvious to me. You said the reversal of time was not real, so it's not much concern to me.

Regards,

S

Hi S, you wrote: "If I understood right, the outcome was the same in both parts (8 vs 12 years of aging). In the second part, Pam had a constant 0.8G of acceleration while Jim had 1.0 from Earth's gravity. "

That's true for both parts; Pam always accelerates away at 0.8g and back again, presumably from low Earth orbit, otherwise she wouldn't get off the ground!

"Since gravity affects clock speed, it seems that Jim should be the younger one here."

Earth gravity has ~1 part in 10⁹ (= 1/(1-√[1-2GM/(Rc²)]) effect on the rates of clocks, so it is totally negligible over the ~10 year timescale, compared to the size of the effect here. Remember that acceleration per se does not have an effect on atomic clocks...

"The red loop doesn't match the blue peak. The reason is not obvious to me."

Actually, they do match (4 years in Pam's frame and 6 years in Jim's). The cause is Pam's velocity time dilation; her speed increases (non-linearly) from 0 to 0.924c and back (twice), for an average of 0.667c (8 ly / 12 y).*

Jorrie

* Question: will it be the same average speed in Pam's frame?

"The fact that Pam accelerates and Jim stays inertial makes the situation asymmetrical and there is no valid argument for a paradox."

But you said that acceleration does not affect atomic clocks (therefore it doesn't affect time), so there is still argument for a paradox because speed is relative!

Hi S, you wrote: "But you said that acceleration does not affect atomic clocks (therefore it doesn't affect time), so there is still argument for a paradox because speed is relative!"

Yea, we've been through this argument before and it was obviously inconclusive...

OK, let's try this angle: acceleration (per se) does not affect clocks, but the combination of acceleration and the resultant velocity does. Acceleration changes the relationship between space and time for an object. Inertia may just be the resistance that objects have to this change.*

Jorrie

* See e.g.: Origin of Inertia on my website.

"Acceleration changes the relationship between space and time for an object. Inertia may just be the resistance that objects have to this change.*"

An interesting idea. I enjoyed your writeup in your e-book. Understanding inertia would go a long way here. You have given me food for thought. 'Current wisdom' says that there is no center to the universe, and also says there are 3 dimensions at our size. An object with 3 dimensions + time has a center! We don't even know how many dimensions we are living in - speed relative to a center would explain the time difference for Pam and Jim. I am 'forced' to be skeptical of current wisdom.

S

"...skeptical of current wisdom..."

In my experience, this is almost ALWAYS a wise course! Ideas are subject to change given additional data, or at least they should be. Ideological inertia may just be the resistance that brains have to this change. (Thanks, Jorrie!)

Hi Jorie,

You said:

"In fact the apparent speed eventually becomes superluminal, with a slope "flatter" than that of light (in this case, with a slope of less than -1). This is not a "real speed", but simply as things appear from an accelerating frame of reference. One must remember that even the speed of light is not the same in all directions when measured in an accelerating frame of reference."

I'd like to understand more about how to analyze accelerating frames of reference in General Relativity. In GR, take the example of a dust filled homogeneous universe with Lambda=0. Every individual local frame is decelerating due to gravity (the frames are moving closer together as a function of time), but (again assuming complete homogeneity) every individual local frame is decelerating at the same rate. So a frame isn't accelerating relative to a nearby local frame, both are accelerating in unison. The deceleration action is omnidirectional.

In that scenario, is it valid for superluminal speeds of light to occur in a frame relative to a nearby (or distant) local frame? I think so. After all, as a photon emitted in the early universe passes by a succession of galaxies receding in the Hubble flow, each succeeding galaxy it passes has a faster recession speed than the previous galaxy passed. (relative to the photon's origin point) So the photon must continually speed up as it moves, relative to the origin, to keep its speed exactly at c in each local frame it passes. Eventually it passes galaxies which themselves are receding superluminally relative to the origin, and so the photon is travelling multiple times the speed of light relative to the distant origin frame.

I'm just wondering if there is an accepted mathematical approach to calculating superluminal motion for this GR issue, given that decelerating frames of reference are involved?

Jon

Oops, I said:

"Every individual local frame is decelerating due to gravity (the frames are moving closer together as a function of time),"

What I meant the parenthetical to say is "(the frames are moving apart ever more slowly as a function of time)".

Jon

Hi Jon.

Yes, I think you're right with the "superluminal photons" from the perspective of a local observer.

The only GR formalism that I know of is the "inverse Shapiro time delay". To an observer held static near a massive object, light will seem to travel faster the farther it goes from the mass. A return signal will takes less time than what it would have taken in flat space.

The Schwarzschild radial coordinate speed of light is given by c' = c(1-2GM/(rc²)), where r is the radial distance parameter. If the observer sits at r, the effective radial speed of light at infinity will be the inverse of that expression.

Hope it helps.

Jorrie

Hi Jorrie. I need some help in post #41. Thanks.

Hi Jorrie, thanks for the reference to inverse shapiro time delay.

While perusing that subject, I ran across the interesting website of Curt Renshaw, who offers an alternative to special relativity, RCM, which is simpler than SR and seems to be internally consistent. I imagine you might have run across this fellow's writings before, so I wonder what you think of it. He seems to have received zero acceptance in the mainstream, and most of his material was written over a decade ago. http://renshaw.teleinc.com/index.asp

Even if his alternative theory is wrong, his book chapter excerpts provide a lot of useful intuitive description about how traditional SR and some aspects of GR theory work, including the twins paradox. He explains several aspects normally attributed to relativity as more properly arising from energy conservation and the equivalence principle.

Jon

Why it is so hard this twin para, and why still try to make up the paradox.It is not the amount of different illustration who we will change some thing.delay, and cheating the time or worst beleiving wrong math.Too much graph, and not enought correction will keep the same wrong direction.small and too small or fast and too fast= ratio scale.difference into the position, and diference into other point into multiple possition in time to me philippe martin mean huge problem of trying puting all into the same dimention,avoiding matter,mass, and stick with .88.and what about the simple triangle who alaways zoom into a delay?

This is not a regular spacetime diagram.

There are two transformations: a spacetime and a space transformations.

The first is a Lorentz transformation, than is the x coordinate streching by sqr(1-v²/c²).

Therefore You get speeds above the ligth speed.

The loops lock like a singularity in spacetime, wich is not.

The left one is a Rindler spacetime diagram for constant acceleration, but with a reversal at 3 years, as viewed in Jim's inertial frame.

The right hand frame is as "stationary" objects are viewed in the spacetime of an accelerating observer with a thrust reversal at 2 years own time.

There are obviously no singularities, but it shows that changing relative simultaneity under acceleration can case loops in some worldlines.

I consider this a much more intuitive view than the "instantaneous" jump in the usual discussion from the "away-twin's" perspective.

The downside is that it is much more difficult to calculate.

-J

Thans you werry much for your replay.

I just want to draw your diagrams in autocad.

For Lorentz transformation, I am using the following formulas, to calculate (t,x) evens of Pam's wordline agains τ - proper time of Jim's frame.

x= 1/0.8 *cosh(0.8τ)- 1.25

t= 1/0.8 * sinh(0.8τ)

where c=1 and g=1.

In this way, with acceleration of 0.8g of Pam's ship, Pam has to start braking at speed of th(κ)=3/2, than she will stop nearby the start in 4 year proper time from the begining of travel from the Eart. Wich means 6 year in Jim proper time. The start is 4ly far from Jim. If Y construct the simmultanity lines for the proper time events of Pam, I can calculate the Minkowski distance from the Eart of each this evenst, by measuring the intersection points of the sim.lines with axel t.

To draw you left hand diagram- Jims's frame, I have to multiply each of x coordinates by 3/4?

What exactly have I do, to draw the right hand diagram?

x= -1/0.8 *cosh(-0.8τ)- 1.25

t= -1/0.8 * sinh(-0.8τ)-2

These formulas are correct for draw the diagram, where τ is the proper time of Pam?

t= -1/0.8 * sinh(-0.8τ)-2

wrong.

t= -1/0.8 * sinh(-0.8(τ-2))

correct

th(τ)=3/2 wrong th(τ)=2/3 correct

To draw you left hand diagram- Jims's frame, I have to multiply each of x coordinates by 3/4? wrong, I do not have to do anything with my diagram. The reversal time is at 3year in Jim's proper time, as You have written, and becouse the accelerating and the braking has the same time in proper time of Pam, we have is two simmetrical section of the same hiperbola. 2 by 3 is 6.

Yes, I think your last equations are correct. If you need more help, feel free to ask.

I made a drawing, it is your left hand diagram, this is a spacetime diagram, and it is geometrically correct.

The braking point is after 3 year in Jim's frame, and the max. speed of Pam is 2/3c.

This diagram was made with the following formular:

x= 1/0.8*cosh(τ*0.8)-1/0.8

t=1/0.8*sinh(τ*0.8)

c=1, g=1, the uniform acceleration of Pam is 0.8g, and -0.8g.

The blue lines are the simultaneity plans (line in 2D) of Pam. You can measure directly the distance beetwen Pam and Jim on each line, at the braking point of Pam, this is 2.71 ly.

I tray to make a drawing, using Pam frame as standing (inercial), and Jim is accelerating, with the following formulas:

x= -1/0.8*cosh(τ*0.8)+1/0.8

t=1/0.8*sinh(τ*0.8)

so, just changing the acceleration from 0.8g to -0.8g. The curve (which is not a hiperbola), at Pam's time τ =2.4ly is turning spacelike. But Jim has a proper time of 12ly.

What I am doing wrong?

I have made also the same diagrams for constant velocity of 3/5c, in that case at the turning point, a Jim's frame has to turn a left, otherway he cannot meet Pam.

Sorry for my bad english.

Your Rindler diagram for the inertial frame (Jim) looks perfect.

The right-hand diagram is not Rindler, but it keeps Pam static as if she is sitting in a uniform (but reversing) gravitational field and as you know her simultaneity definition is then different from Jim's and the star's, which are free-falling to the left at 0.8g.

I have just used Pam's lines of simultaneity to plot their wordlines from the same data that I calculated for the left-hand diagram. Again the top half is just a mirror image of the bottom.

Hi Jorrie

Finalli I draw the accelerated twins from Pam's wiew. It is not a Minkowski spacetime diagram, is just a use a tricky geometrical issue to succed.

From the Jim IR frame we have for

x=1/0.8*ch(0.8τ); t=1/0.8*sh(0.8τ).

But the leng of I0 line is 1.25 fy, which is the same for the IA' line, due the fact that the orange line is a hiperbola.

The intersection of line IA' line with the wordline of Jim (magenta), is the point C.

α is the angle between IA and IA'. We know, the cos(α)=1.25*ch(0.8τ)/1.25=1.25/IA.

So, IA= 1.25/(1.25*ch(0.8τ)), and CA'=1.25-IA=1.25- 1.25/(1.25*ch(0.8τ)).

So the distance between the Jim and Pam at any τ, is x=-1.25+1.25/(1.25*ch(0.8τ))

For the star we have a similar relation: IB=5.25, cos(α)=1.25*ch(0.8τ)/1.25=5.25/IB'. Then IB'=5.25*1.25/(1.25**ch(0.8τ)), and B'A'=1.25-5.25*1.25/(1.25**ch(0.8τ))

the distance between the start and Pam at any τ, is x=-1.25+5.25/(1.25*ch(0.8τ))

For the rest of the curve I use

x=-1.25+1.25/(1.25*ch(0.8(τ-2))); x=-1.25+5.25/(1.25*ch(0.8(τ-2)))

So I got the following diagram:

This is not a Minkowski one, ewen the distance between the star end Jim IR frame is depend on τ. But they are in the same IR only the distance between them is 4ly.