#57805 Twin Paradox

[vorpal blade]

I appreciate the time you are spending in trying to explain this, Tao. I’ve read several times what you have written, and I’m trying to follow you. No offense meant, but so far it doesn’t look to me as though you have answered any of my questions or explained anything.

A few comments on your last post. I can’t access youtube from this computer, so I wasn’t able to watch the video. I’ll try to do that tonight.

I wouldn’t say that conceptualizing time dilation is any more difficult than space contraction for me. I don’t think the problem is my lacking of conceptualizing or my lack of understanding the theory. I think the problem is that the theory is wrong. But I’m giving it a try in case I have missed some important concept in the theory.

I have not chosen to anchor my perception at the first space station, or any one point at all.

I see no reason to believe that I would think that the space station and earth were not in synch or meshing at any time during the journey. I don't see why I would care, though.

I think I could rephrase the problem without acceleration, but then it becomes an academic exercise of what each reference frame perceives to be happening, without any necessity of coming to the reality of what is actually happening. Accelerating out of one frame and going back into that frame forces you to confront reality.

I realize that two time lines will only agree at a single point, or else they wouldn’t be different time lines. My point is not that they disagree, but that they disagree in an inconsistent manner. The theory of Relativity can be used to show that each one is shorter than the other, which is a logical impossibility.

You seem to confuse abstract notions of time and space with reality. The four axes of space and time do not really exist. They are mental constructs to assist in our description and model of the universe or reality. The number 2, for example, is not a real object. It is an abstraction. It is useful in seeing a pattern when we see two pencils, and two pieces of paper, and two fingers. The common thread of “twoness” is a figment of the mind. As a mathematician I easily manipulate 4 dimensional spaces, but I recognize that this is mathematics, with a possible correlation to things in reality, but mathematics is not reality. This may be just an aside.

As I understand it you are possibly saying that at the time of deceleration the disparity in what the space ship has been observing in passing space stations “catches up” as the space ship starts to land. I admit that I have never penetrated the mathematics of General Relativity, and I don’t know how to do the calculations. Never tried, really. Nevertheless it seems to me that you must be using two sets of rules on how to do the calculation. At the start of the trip we agree that the space ship and earth are only going to different in clock calculation by a day or two, due to the acceleration of the space ship and in accordance with the General Theory of Relativity. It seems to me that after the space ship has ceased to accelerate and is traveling at a constant velocity then we are done with the General Theory and we can use only the Special Theory, until we use it again when we decelerate. When we come to the point of deceleration the ship is just accelerating in the direction of earth. It will use its thrusters in exactly the same manner it used them to leave earth. We must use the equations just as we used them before on takeoff.

Therefore the time differential between the two frames cannot change at this time by more than a few days. There cannot be any “catching up” or “stacking up” of time differences. This is a new event and is logically and physically independent of what happened years ago near earth. The General Relativity calculation of what time differences occur during landing cannot depend on how long it has been since the space ship left the earth. True, clocks will have aged differently, but nothing significant to change that will occur just before landing. The idea that the discrepancies and inconsistencies all come out at this time seems to me to be an ad hoc explanation. It doesn’t come from the theory, but is useful to explain things this one time. Nevertheless it should be consistent with the way the calculation was done on takeoff.

Unfortunately a lot of the explanations of the time paradox sound like this to me: “Okay, first from one point of view, heads I win, and tails you lose. Now let’s take it from the other point of view. This time heads you lose, and tails I win.” In other words they aren’t really fair in giving each reference frame equal status.

[Tao]

I think I could rephrase the problem without acceleration, but then it becomes an academic exercise of what each reference frame perceives to be happening, without any necessity of coming to the reality of what is actually happening. Accelerating out of one frame and going back into that frame forces you to confront reality.

Okay. I can handle that. As long as we stick with realistic elements across the board, not picking and choosing, the math becomes more complicated, but the end result is the same.

My point is not that they disagree, but that they disagree in an inconsistent manner. The theory of Relativity can be used to show that each one is shorter than the other, which is a logical impossibility.

Simply being in different locations creates the same logical impossibility. Each of us standing in an end zone with a ruler at midfield would give measurements showing that each one of us is shorter than the other... It would be foolish to make a fuss over this, largely because we experience it so much, but if you think on it, the parallels to this situation seem to pan out.

You seem to confuse abstract notions of time and space with reality. The four axes of space and time do not really exist. They are mental constructs to assist in our description and model of the universe or reality. The number 2, for example, is not a real object. It is an abstraction. It is useful in seeing a pattern when we see two pencils, and two pieces of paper, and two fingers. The common thread of “twoness” is a figment of the mind. As a mathematician I easily manipulate 4 dimensional spaces, but I recognize that this is mathematics, with a possible correlation to things in reality, but mathematics is not reality. This may be just an aside.

Yeah, not really seeing what you're driving at here. I agree that notation is abstract, and that mathematical models remain just that, but that doesn't stop them from giving us an understanding of the world around us. The realization that time is unilateral does not negate the increased understanding we receive when treating it akin to a spatial dimension. Rejecting a theory because it has been made to correlate with more conceivable concepts would be akin to abandoning the teachings of Christ because the Kingdom of God simply cannot be both an Olive tree and a little child. If you'd like to start another thread considering the metaphysical meanings of what exists and what are simply mental constructs, I'd be interested, but doubt I've the time to do that topic any more justice than what I appear to be doing here.

As I understand it you are possibly saying that at the time of deceleration the disparity in what the space ship has been observing in passing space stations “catches up” as the space ship starts to land. I admit that I have never penetrated the mathematics of General Relativity, and I don’t know how to do the calculations. Never tried, really. Nevertheless it seems to me that you must be using two sets of rules on how to do the calculation. At the start of the trip we agree that the space ship and earth are only going to different in clock calculation by a day or two, due to the acceleration of the space ship and in accordance with the General Theory of Relativity. It seems to me that after the space ship has ceased to accelerate and is traveling at a constant velocity then we are done with the General Theory and we can use only the Special Theory, until we use it again when we decelerate. When we come to the point of deceleration the ship is just accelerating in the direction of earth. It will use its thrusters in exactly the same manner it used them to leave earth. We must use the equations just as we used them before on takeoff.

Therefore the time differential between the two frames cannot change at this time by more than a few days. There cannot be any “catching up” or “stacking up” of time differences. This is a new event and is logically and physically independent of what happened years ago near earth. The General Relativity calculation of what time differences occur during landing cannot depend on how long it has been since the space ship left the earth. True, clocks will have aged differently, but nothing significant to change that will occur just before landing. The idea that the discrepancies and inconsistencies all come out at this time seems to me to be an ad hoc explanation. It doesn’t come from the theory, but is useful to explain things this one time. Nevertheless it should be consistent with the way the calculation was done on takeoff.

I'll try to address this once I understand where you are coming from.

Unfortunately a lot of the explanations of the time paradox sound like this to me: “Okay, first from one point of view, heads I win, and tails you lose. Now let’s take it from the other point of view. This time heads you lose, and tails I win.” In other words they aren’t really fair in giving each reference frame equal status.

I find this final paragraph ironic, as it mirrors my thoughts exactly while trying to see where you are getting your ideas. By measuring acceleration from earth, and then abandoning that frame of reference when dealing with deceleration, declaring those equations independent of your original stance, you seem to be declaring a change of rules mid-equation in order to make deceleration a reciprocal to negate the effects of GA.

I really don't know what it is you want. If we run the equations without acceleration, taking in to account each of the three inertial reference frames, everything pans out, all reference frames agree. When you run them in GA, everything pans out, the numbers not only mesh with each other, but the end result is correlated to the results of SA. To say that we want to use acceleration to avoid the simultaneity issues with SA, but don't want to use the equations for GA, because we've made the most of the trip at constant velocity and like the simplicity of SA better makes it a little bit of a challenge to address.

From satellites 1-9 we've an acceleration-free field, and to calculate time appearances between them requires acknowledgment that simultaneous events in one perspective are not simultaneous in another. If we want to declare acceleration as having no effect, we still are faced with the fact that everything adds up between those two points. To say what can and cannot be in the extreme ends of the thought experiment while admitting that you've never studied the mechanics thereof seems strange to me. I have my own understandings of GA, and can crunch the numbers, yet I remain hesitant to make any claims as to what you can and can not use them for.

[vorpal blade]

I always wonder whether it is better to just try to explain myself again, or address the points where I don't seem to have made myself clear. Let's try both. In this post I'll point out where your interpretation of what I said is not what I meant. It is amazing how we can't seem to even begin to discuss this, and I'm trying hard.

My point is not that they disagree, but that they disagree in an inconsistent manner. The theory of Relativity can be used to show that each one is shorter than the other, which is a logical impossibility.

Simply being in different locations creates the same logical impossibility. Each of us standing in an end zone with a ruler at midfield would give measurements showing that each one of us is shorter than the other... It would be foolish to make a fuss over this, largely because we experience it so much, but if you think on it, the parallels to this situation seem to pan out.

I see no parallel here. There is an appearance that each is smaller than the other, but we are quite capable of doing the calculations to determine that the appearance is not reality. In the twin paradox it is not the appearance that one time frame is moving at a slower pace than the other, but this is supposedly the actual reality. Furthermore I have made pains to not deal with clocks or objects at a distance, but when objects are at the same position in space.

Unfortunately a lot of the explanations of the time paradox sound like this to me: “Okay, first from one point of view, heads I win, and tails you lose. Now let’s take it from the other point of view. This time heads you lose, and tails I win.” In other words they aren’t really fair in giving each reference frame equal status.

I find this final paragraph ironic, as it mirrors my thoughts exactly while trying to see where you are getting your ideas. By measuring acceleration from earth, and then abandoning that frame of reference when dealing with deceleration, declaring those equations independent of your original stance, you seem to be declaring a change of rules mid-equation in order to make deceleration a reciprocal to negate the effects of GA.

I really don't care whether the acceleration is measured from earth or not. It is consistent, however, to measure it from earth at takeoff, and then measure it from the 10th space station, which is in the same inertial frame as the earth. The point is that the acceleration can be made inconsequential. I was able to watch the youtube video you referenced, and I noticed he ignored acceleration. If we completely ignore acceleration than everything is relative, and just as you can show that the girl (if you like) twin ages less than the boy twin, you can turn the problem around and show the boy twin has aged less than the girl twin when they are brought back together. Without acceleraton it is a perfectly symmetrical problem, and each is younger than the other when reunited, which is the true paradox in this problem. The video never discussed the true paradox, only the weird thing that the theory predicts when interpreted from a particular point of view.

It is not uncommon to take a problem and use the approximations or formulas that apply to each part of the problem. I'm not trying to negate the effects of GA, merely show they are so small they can be neglected in this problem. Which is why I don't think it is necessary to go through the calculations.

I really don't know what it is you want. If we run the equations without acceleration, taking in to account each of the three inertial reference frames, everything pans out, all reference frames agree. When you run them in GA, everything pans out, the numbers not only mesh with each other, but the end result is correlated to the results of SA. To say that we want to use acceleration to avoid the simultaneity issues with SA, but don't want to use the equations for GA, because we've made the most of the trip at constant velocity and like the simplicity of SA better makes it a little bit of a challenge to address.

But everything does not pan out when you take three inertial reference frames, they disagree if you do it right, although they can be made to agree if you do it in a certain way. I do not want to use acceleration to avoid the simultaneity issues with SA. I think acceleration is often brought in to save the paradox, but I would like to show that the effect can be minimized and there is still a paradox. I'd be happy to dispense with acceleration as long as you dispense with a preferred reference frame. But if you do it in GA you will be doing it in a preferred reference frame for a reason.

What do I want? I've been thinking of that. I don't need or want an explanation of how the equations can be manipulated to show a given result. I can do that as well. What the paradox is all about is that, yes, one can manipulate the equations to give a given result, but one can also manipulate correctly the same equations and get a different result. So, obviously, showing me alternate ways to get the same result is not going to help. The only thing that will help will be to show how the alternate way that I propose is wrong. Not just that it gives a different result, but that my use of the equations goes against the principles and theory.

From satellites 1-9 we've an acceleration-free field, and to calculate time appearances between them requires acknowledgment that simultaneous events in one perspective are not simultaneous in another. If we want to declare acceleration as having no effect, we still are faced with the fact that everything adds up between those two points. To say what can and cannot be in the extreme ends of the thought experiment while admitting that you've never studied the mechanics thereof seems strange to me. I have my own understandings of GA, and can crunch the numbers, yet I remain hesitant to make any claims as to what you can and can not use them for.

Actually, I've read what Einstein had to say about the General Theory of Relativity. I don't see the need to study the mechanics of it. But I can tell you what they claim it can and cannot be used for. Though it has been a while since I studied it, and I could be mistaken.

I submit that when the space ship passes the space stations this is a simultaneous event in both time frames. The clocks will not of course read the same, but the event is simultaneous even though the clocks read different times. What I have been trying to do with the space stations along the way is to nail down the two rubber rulers, or variable clocks. What does the clock say on space station 1 when the space ship goes through it. What does the clock on the space ship say at that same event? This nails down the two clocks at that event. If you mean by simultaneous what happens on the space ship after 1.25 years as read on the space ship and compare that to what happens on the space station after the clock on the space station reads 1.25 years is not important to me. The simultaneous event of the space ship traveling through the space station occurs at different times, according to their clocks, but in another sense at the same time. The event of the space ship traveling through the space station does occur simultaneously with the space station passing the space ship, though their funny clocks don't read the same.

[vorpal blade]

Let's consider just a small part of the example I gave. The space ship sees the clock on earth to read Jan 1, 3001. The clock on the space ship also reads Jan 1, 3001. This is just before takeoff. Now let's ignore acceleration, or we could assume that the space ship is merely passing earth at a constant velocity and sets the clock on the space ship to read the same as the clock on earth at the moment of passing earth. When the space ship passes the first space station those on the space ship see the clock on the space station and read that 0.45 years have passed since the space ship passed earth. However, the event of the space ship passing the space station is the same event as the space station passing the space ship. The people on the space station read their clock when this event occurs and read that their clock says that 1.25 years have passed since the space ship passed earth. My question is, how do you explain the fact that the people on the space ship read the clock on the space station when this event occurs and see that that clock reads 0.45 years, while the people on the space station read their clock during that same event and see that their clock reads 1.25 years? How can they see different readings on the same clock, located in a particular inertial time frame, and see different things at the same time event? I don't know how to make that any clearer.

Explaining how they both see the same thing doesn't help. What I need to know is what is wrong with my analysis, if anything.

[Tao]

Let's consider just a small part of the example I gave. The space ship sees the clock on earth to read Jan 1, 3001. The clock on the space ship also reads Jan 1, 3001. This is just before takeoff. Now let's ignore acceleration, or we could assume that the space ship is merely passing earth at a constant velocity and sets the clock on the space ship to read the same as the clock on earth at the moment of passing earth. When the space ship passes the first space station those on the space ship see the clock on the space station and read that 0.45 years have passed since the space ship passed earth. However, the event of the space ship passing the space station is the same event as the space station passing the space ship. The people on the space station read their clock when this event occurs and read that their clock says that 1.25 years have passed since the space ship passed earth. My question is, how do you explain the fact that the people on the space ship read the clock on the space station when this event occurs and see that that clock reads 0.45 years, while the people on the space station read their clock during that same event and see that their clock reads 1.25 years? How can they see different readings on the same clock, located in a particular inertial time frame, and see different things at the same time event? I don't know how to make that any clearer.

Explaining how they both see the same thing doesn't help. What I need to know is what is wrong with my analysis, if anything.

Awesome. This one is clear to me. I’ll use the following notations for the standard Lorentz equations: stationary observer records time t, velocity v, and distance away on the x axis x. He records his observations of objects in motion as seeing time t’, velocity v, and distance x’.

t’=γ(t-vx/c²)
x’= γ(x-vt)

where γ = 1/√1-(v²/c²)

What the people on Earth see: at t₀, the ship syncs with them, so t₀’=0 ship flies by at speed (v).8c, in 1.25 years (∆x/v=t₁) it passes the space station 1 ly out (x=1), noticing that the trip took .75 years for the ship. (t₁’=.75)

What the space station sees: at t₀=0 the clock on the spaceship reads t₀’=1.33
The ship is at earth, 1 ly away,(x=-1) headed towards them (x=0) at .8c. In 1.25 years (∆x/v=t₁) the spaceship passes them and they note it has taken the ship 2.08333 years to arrive.

What the spaceship sees: at t₀=0 clocks on board and on earth sync as it passes them (x=0) at .8c. The space station is .6 ly away (x= 0.6) is approaching at the same velocity. However they notice that the time on the space station is t₀’=.8 As they reach the space station, their own clock reads that 0.75 years have passed (∆x/v=t₁), they show the space station’s reference frame as being dilated, leaving only 0.45 years gone by (t₁’)

Thus, according to an earth observer, setting t₀=0=Jan 1 3001 their calendar and that of the space station reads t₀+ t₁=April 3002 when the ship passes. But the ship reads t₀’+ t₁’ Oct 3001.

To the observers on the space station at the time of passing t₀+ t₁=April 3002 but the spaceship reads t₀’+ t₁’ Oct 3001.

To the observers on the spaceship, t₀+ t₁= Oct 3001, whereas they would perceive the space station at t₀’+ t₁’ April 3001.

One nice thing with this application of relativity is that every equation here predates Einstein. He’s just the one who cobbled it all together.

[FauxRaiden]

*Brain esploded*

[vorpal blade]

Excellent. We agree on the conclusions. There is nothing wrong with my analysis.

Actually there is a slight difference. We both figured that the spaceship would see the station's clock to have advanced 0.45 years. I calculate this as mid-June 3001, while you gave it as April 3001.

Now we are ready to address my question. Your (and my) calculations show that those on the space station see the clock on the space station to read April 3002 during the event in which the space ship passed by. The calculations also show that those on the space ship read the clock on the space station as June 3001 (April 3001 in your post). How is this possible? According to this theory it really is April 3002 on the space station, and it really is June 3001 on the space station as the space ship passes. It seems to me that it can't be both April 3002 and June 3001 at the same time in the same place. And we are talking about the same place and same time--the event of the spaceship passing the space station.

[Tao]

Lo siento, that's a typo on my part. Should read:"To the observers on the spaceship, t₀+ t₁= Oct 3001, whereas they would perceive the space station at t₀’+ t₁’ April 3002."

I think you're using t₁’ =.45 just as I am, without factoring in t₀’=.8 That would put a date near enough June 3001.

[Tao]

*Brain esploded*

Heh. Too true. It has been so long since I've done any of the advanced calculations that previous nights' attempts to run General Relativity equations in the wee hours of the morning seemed nigh on to triggering an aneurysm. It was a bit relieving to get back to simple equations, with minimal errors (found thus far...).

[vorpal blade]

I found I was really rusty on the Lorentz coordinate transformation equations as well. I think I have it straight in my mind now. I believe you are incorrect in some of the intermediate steps, and in the interpretation of the meaning of the Lorentz transforms, but amazingly you got to the correct conclusions. Since a correct derivation of the results might be useful later, I offer my solution. Feel free to locate any errors and tell me about them.

Awesome. This one is clear to me. I’ll use the following notations for the standard Lorentz equations: stationary observer records time t, velocity v, and distance away on the x axis x. He records his observations of objects in motion as seeing time t’, velocity v, and distance x’.

t’=γ(t-vx/c²)
x’= γ(x-vt)

where γ = 1/√1-(v²/c²)

What the people on Earth see: at t₀, the ship syncs with them, so t₀’=0 ship flies by at speed (v).8c, in 1.25 years (∆x/v=t₁) it passes the space station 1 ly out (x=1), noticing that the trip took .75 years for the ship. (t₁’=.75)

Agreed. I assume that you mean that the people on earth notice that the trip took 0.75 years according to the clocks on the spaceship, as according to earth’s own clocks the trip took 1.25 years, as you pointed out. I’m just clarifying the language.

What the space station sees: at t₀=0 the clock on the spaceship reads t₀’=1.33
The ship is at earth, 1 ly away,(x=-1) headed towards them (x=0) at .8c. In 1.25 years (∆x/v=t₁) the spaceship passes them and they note it has taken the ship 2.08333 years to arrive.

This really makes no sense. The ship takes 2.08333 years to arrive? And leaves at t₀’=1.33 years? What you seem to have done is shifted the x-axis origin of the space time coordinates systems, without shifting the time axis. I think it is simpler not to shift the origin. (But if you do, so time zero occurs when the spaceship passes the space station, you find the time on the spaceship as it passes the earth is -3/4 years (as seen from the space station), and it all works out. I think.)

At t₀= 0 yr on the space station, t₀= 0 yr on earth, and t₀’= 0 on the spaceship, by design. At t₁ = 5/4 yr on the space station it is 5/4 yr on the earth, since they are in synch, as seen from the space station (or earth).

However, on the spaceship’s clock, at t₁ = 5/4 yr, x = 1 ly (at the space station and spaceship, when ship passes the space station):
t₁’ = (5/3)*(5/4 – (4/5)*1) = ¾ yr.

Elapsed on various clocks as seen from the space station:
ship’s clock is t₁’ - t₀’ = ¾ - 0 = ¾ yr.
Earth: t₁ - t₀ = 5/4 yr
Space station : t₁ - t₀ = 5/4 yr

What the spaceship sees: at t₀=0 clocks on board and on earth sync as it passes them (x=0) at .8c. The space station is .6 ly away (x= 0.6) is approaching at the same velocity. However they notice that the time on the space station is t₀’=.8 As they reach the space station, their own clock reads that 0.75 years have passed (∆x/v=t₁), they show the space station’s reference frame as being dilated, leaving only 0.45 years gone by (t₁’)

The spaceship is assumed stationary, and we use the x and t coordinates for the inertial frame of the spaceship, while x’ and t’ refer now to the moving frame of space station and earth. If the space station is approaching from positive x values then the velocity is negative, or v = -4/5 c.

At t₀= 0 yr, x = 0 ly (earth and the ship at the time the earth passes the spaceship)
t₀’= (5/3)*(0 - 0) = 0 (time on earth’s clock as seen from spaceship)

At t₀= = 0 yr, x = 3/5 ly (at the space station, when earth passes spaceship)
t₀’= (5/3)*(0 – (-4/5)(3/5)) = 4/5 yr (time on space station’s clock as seen from spaceship)

At t₁ = 3/4 yr, x = -3/5 ly (at the earth, when space station passes spaceship)
t₁’ = (5/3)*(3/4 – (-4/5)*(-3/5)) = 9/20 yr. (time on earth’s clock as seen from spaceship)

At t₁ = 3/4 yr, x = 0 ly (at the space station and ship, when space station passes spaceship)
t₁’ = (5/3)*(3/4 – (4/5)*0) = 5/4 yr. (time on space station’s clock, as seen from spaceship)

Elapsed time on the various clocks as seen from the spaceship:
Earth: 9/20 - 0 = 9/20 yr
Space station: t₁’ - t₀’ = 5/4 – 4/5 = 9/20 yr
Spaceship: t₁ - t₀ = ¾ yr

Thus, according to an earth observer, setting t₀=0=Jan 1 3001 their calendar and that of the space station reads t₀+ t₁=April 3002 when the ship passes. But the ship reads t₀’+ t₁’ Oct 3001.

You don’t want to add the clock readings at two different times. When you compare what the clocks say when the spaceship passes the space station you compare t₁ with t₁’ . In this case t₁ = 5/4 and t₁’ = ¾ . So setting t₀=0=Jan 1 3001 gives the correct result.

To the observers on the space station at the time of passing t₀+ t₁=April 3002 but the spaceship reads t₀’+ t₁’ Oct 3001.

t₁= 5/4 yr = April 3002 and
t₁’ = ¾ yr = Oct 3001. (I don’t know how you got ¾ from t₀’+ t₁’ when you had t₀’ = 1.33)

To the observers on the spaceship, t₀+ t₁= Oct 3001, whereas they would perceive the space station at t₀’+ t₁’ April 3001.

t₁= ¾ = Oct 3001
t₁’ = 5/4 = April 3002.

So, it looks like I was wrong that two people in different frames were looking at the same clock at the same time and seeing different readings. At least, when you do it the preferred way of using Lorentz transformations. But I still have the question of whether or not you get a different answer using a different approach to the problem. But thanks for making me work through the math and figure this out.

[vorpal blade]

The next question is this, what happens if you add deceleration to the problem? I’d like to see what happens if you factor in the spaceship accelerating toward the speeding away earth in such a way that it lands on the space station. Let’s assume that the spaceship accelerates at a constant force of 5 gs. That’s unrealistic over a long period of time, but this is just a thought exercise anyway. For convenience we could assume that

g = 10 m/s^2
c = 3*10^8 m/s
v = 0.8*c = 2.4*10^8 m/s
a = 5*g

If I remember correctly

v = a*t

So the time it takes for the space ship to be up to the speed of the earth and space station is

t = v/a = 240,000,000/50 = 4,800,000 seconds = 1333.33 hours = 55.56 days.

I’m sure the answer is going to depend on whether we consider the acceleration in the inertia frame of the spaceship or the earth. Whichever is easier. I’m interested in knowing what the clock on the spaceship reads when it lands on the space station, and what the clock on the space station reads.

So, if this will not cause an aneurysm, could Tao (or someone else) please calculate it, showing your work so that I can verify the calculations?

[Tao]

You don’t want to add the clock readings at two different times. When you compare what the clocks say when the spaceship passes the space station you compare t₁ with t₁’ . In this case t₁ = 5/4 and t₁’ = ¾ . So setting t₀=0=Jan 1 3001 gives the correct result.

You are correct. I think I used a negative x value (or vice verse) at some point and ended up with -1.33 in one run of the equations and so jumped to summation, where I should have been showing a difference.

The next question is this, what happens if you add deceleration to the problem? I’d like to see what happens if you factor in the spaceship accelerating toward the speeding away earth in such a way that it lands on the space station. Let’s assume that the spaceship accelerates at a constant force of 5 gs. That’s unrealistic over a long period of time, but this is just a thought exercise anyway. For convenience we could assume that

g = 10 m/s^2
c = 3*10^8 m/s
v = 0.8*c = 2.4*10^8 m/s
a = 5*g

If I remember correctly

v = a*t

So the time it takes for the space ship to be up to the speed of the earth and space station is

t = v/a = 240,000,000/50 = 4,800,000 seconds = 1333.33 hours = 55.56 days.

I’m sure the answer is going to depend on whether we consider the acceleration in the inertia frame of the spaceship or the earth. Whichever is easier. I’m interested in knowing what the clock on the spaceship reads when it lands on the space station, and what the clock on the space station reads.

So, if this will not cause an aneurysm, could Tao (or someone else) please calculate it, showing your work so that I can verify the calculations?

Are we just wanting to focus on deceleration, so start with the ship already at .8c and in a different inertial reference frame, or doing the acceleration and deceleration? To be honest, I'm not sure if one would be more difficult than the other.

[vorpal blade]

As the equations show, a time period of 0.75 yrs seems to have passed on the spaceship, as seen from earth, the space station, and the spaceship during the interval between the ship passing the earth and passing the space station. Funny how that all works out to be the same. However, the spaceship sees time passing on earth and the space station of only 0.45 years, while the earth and space station see a passage of time of 1.25 years. So the Twin Paradox wishes to explore the question, what is the reality? What happens to the clocks when you bring them together in the same place and at rest with one another?

The full problem would have the twins at rest in one frame, one of them accelerates to a high velocity. After a time the traveling one declerates and then accelerates back to the first twin, and finally decelerates to land on earth. In between these periods of accleration would be long periods of constant velocity and no accelerations.

I'm just wanting to study the problem one step at a time. What I am trying to avoid is losing the relativity of the problem. So, you could begin with the initial accleration if you wish, as long as you didn't claim that thereafter, when in constant velocity, you can't use the Lorentz transformations or can't consider it solely as a Special Relativity problem.

Incidentally, I don't see the Special Relativity Theory as a special case of the General Theory, but rather the General Theory is a special case of considering acceleration as equivalent to gravity. What do you think?

[Tao]

As the equations show, a time period of 0.75 yrs seems to have passed on the spaceship, as seen from earth, the space station, and the spaceship during the interval between the ship passing the earth and passing the space station. Funny how that all works out to be the same. However, the spaceship sees time passing on earth and the space station of only 0.45 years, while the earth and space station see a passage of time of 1.25 years. So the Twin Paradox wishes to explore the question, what is the reality? What happens to the clocks when you bring them together in the same place and at rest with one another?

They show the same thing:1.25 years later than sync. At sync the space ship sees earth and their own clocks set at Jan 1 3001, but they see the space station's clock at late October 3001 (.8 of a year). While they see .45 years pass, their view of the station's clock agrees at passing with it's inhabitants, 1.25 years after Jan 3001.

The full problem would have the twins at rest in one frame, one of them accelerates to a high velocity. After a time the traveling one declerates and then accelerates back to the first twin, and finally decelerates to land on earth. In between these periods of accleration would be long periods of constant velocity and no accelerations.

I'm just wanting to study the problem one step at a time. What I am trying to avoid is losing the relativity of the problem. So, you could begin with the initial accleration if you wish, as long as you didn't claim that thereafter, when in constant velocity, you can't use the Lorentz transformations or can't consider it solely as a Special Relativity problem.

Incidentally, I don't see the Special Relativity Theory as a special case of the General Theory, but rather the General Theory is a special case of considering acceleration as equivalent to gravity. What do you think?

Hard to say, really. The General Theory is fairly expansive in it's scope, but you can fairly easily see the development from SR, so I could see arguments either way.

[vorpal blade]

Let me show you why I think this problem cannot be solved with Special Relativity Theory alone. Let us now add in the 3rd inertial reference frame (x’’, y’’, z’’, t’’) containing the incoming return ship for the twin. This frame of reference is moving at velocity v = -4/5 c, as seen from the earth-space stations frame of reference. We set the origin such that
At t=t’=t’’=0 then x=x’=x’’ =0.

I. From the earth-space station reference frame,

A. Earth and the outgoing space ship are at x=0, but the return ship is at x=2 ly (the location of space station 2). One twin is moved from reference frame (x,y,z,t) to reference frame (x’,y’,z’,t’) at (0,0,0,0) in both reference frames.

B. At t=5/4 y the outgoing space ship is at x=(4/5](5/4}=1 ly, or at space station 1. In the earth-space station reference frame the time on the outgoing spaceship’s clock when passing space station 1 is

t1’ (t=5/4, x=1, v=4/5c) = (5/3)(5/4 – (4/5)1) = ¾ y

On the incoming ship, approaching space station 1 from space station 2,

t1’’ (t=5/4, x=1, v=-4/5c) = (5/3)(5/4 –(-4/5)1) = 41/12

when the outgoing ship and incoming ship meet at space station 1 at t =5/4 y as see in the earth-space station reference frame. One twin is now passed from reference frame (x’,y’,z’,t’) to (x’’,y’’,z’’,t’’) at (1,0,0,5/4) as seen in the earth-space station reference frame.


C. At t=5/2 y the incoming ship is now at the earth (0,0,0,5/2). The clock on the incoming space ship reads,

t2’’ (t=5/2, x=0, v=-4/5c) = (5/3)(5/2 - (-4/5)0) = 25/6 y

So, from the perspective of the earth-space station reference frame a total of 5/2 y has passed on earth, but when the returning twin lands the clock in the incoming space ship reads 25/6 y. The total elapsed time, as seen from the earth reference frame on the clocks in the moving space ships is

(t1’-t0’) + (t2’’-t1’’) = ¾ + (25/6)-(41/12) = ¾ + ¾ = 3/2.


II. Now let’s look at it from the reference frame of the incoming space ship.

A. At t=0 earth appears to be at x=0 and the return space ship at x=6/5. Observers in the return ship’s reference frame see the clock on earth to be

t0’’(x=0, t=0,v=4/5c) = (5/3)(0 – (4/5)(0) = 0

The earth and space station 1 are approaching the return ship at velocity v=4/5c. Space station 2 is at the same location as the return ship, and the clock on earth reads t0” = 0. Space station 1 appears to be at x=3/5 ly in the return ship’s reference frame.

B. At t=3/4 y, space station 1 has moved from x=3/5 to 3/5 + (4/5)(3/4) = 6/5 ly, the position of the return ship. So space station 1 and the return ship are collocated at t=3/4 y. The time on earth’s clock, as seen by observers in the inertial frame of reference of the return ship, is

t1’’ (x=3/5, t=3/4, v=4/5c) = (5/3)(3/4 - (3/5)(4/5)) = 9/20 y.


Meanwhile the ship passing earth going toward space station 1 is seen by the return ship as having the combined velocities of the earth moving toward the return ship (v’’=4/5c), plus the velocity of the outgoing ship relative to the earth (v=4/5c). So, using the relativistic addition of velocities formula,

v’ = [v’’+v]/[1+(v’’v/cc)] = c[(4/5) + 4/5)]/[1 + (4/5)(4/5)] = 40/41 c

From this we can calculate that

γ = 1/√1-(v²/c²) = 1/√1-(40²/41²) = 41/√(1681- 1600) = 41/9.

The time on the outgoing spaceship clock, as seen from the return spaceship frame of reference, when the return spaceship is just passing space station 1, is

t1’(x=3/5, t=3/4, v=40/41c) = (41/9)(3/4 – (3/5)(40/41)) = ¾ y.

As the twin in the outgoing space ship passes the return space ship the twin transfers reference frames and begins to head home, and the readings on the outgoing and incoming clocks match.

C. At t=3/2 y earth has reached the return ship, and the twin in the spaceship has returned home. At this time the clock on earth reads, according to the observers in the return spaceship,

t2’(x=6/5, t=3/2, v=4/5c) = 5/3(3/2 – (4/5)(6/5)) = 9/10

So, here is the paradox. The twin in the space ship saw an elapsed time of 9/20 y on the clocks of the earth-space station reference frame going out, and an elapsed time of 9/20 y coming home. So, according to his observations the twin on earth cannot have aged more than 9/10 y while he was away. Furthermore, the observers in the reference frame of the return ship, who do not see themselves as having moved, read the clock on earth as having gone from 0 to 9/10 y, so the twin on earth has to be just 9/10 years old. In the reference frame of the return ship a total of 1.5 years has passed from the time the space ship passed earth and when earth arrived at the return ship, so the inhabitants of the return ship have aged more than the inhabitants on earth, from their point of view.

The point of view of the twin who got on the space ship is that he aged ¾ years getting to space station 1, and ¾ years in returning, so he is 1.5 years older than when he left earth. Since his observation of earth’s clock shows that earth has aged 9/10 years he is older than the twin left on earth.

However, the twin left on earth measures 5/4 years for the twin to reach space station 1, and another 5/4 years to return, so according to his clock he is 2.5 years older than he was. In his observations he agrees with the traveling twin that the traveling twin is just 1.5 years older.

How is it possible for the return ship to read the clock on earth to be 9/10 y, still the year 3001, while those on earth see the same clock, in the same location, at the same time to be 2.5 years, or mid-year 3003? How can each twin see that the other twin has aged less?

[NerdGirl]

I'm dragging this back up because I want to finish reading it and have something to say about it!

[Tao]

sigh.

If I ever get the time set aside to dive back into this, I'd like to run a breakdown of GR's equations on Vorpal's numbers. I could always copy and paste the equations wholesale, but don't feel that does it justice.

I've not forgotten it, Life just seems to be in the way, at the moment.

[NerdGirl]

I've not forgotten it, Life just seems to be in the way, at the moment.

Yeah, that's my problem. My internet was mostly broken when this started, so I haven't read most of it, but I really want to because I love to talk about relativity. We'll see if it happens any time soon. I think there was also a thread about quantum mechanics sometime around Christmas that I still want to say something about and haven't yet.

[vorpal blade]

sigh.

If I ever get the time set aside to dive back into this, I'd like to run a breakdown of GR's equations on Vorpal's numbers. I could always copy and paste the equations wholesale, but don't feel that does it justice.

I've not forgotten it, Life just seems to be in the way, at the moment.

So, do you now agree with me that Special Relativity alone can't resolve the Twin Paradox, and you have to go to General Relativity? If it can be answered with the Special Relativity theory I don't much care what the GR equations say. But I think that I have proved the point that Special Relativity theory isn't adequate. If you don't think so, what is wrong with my use of the equations?

[Tao]

I apologize, Vorpal. I'd not seen your previous post and was still under the assumption that we'd worked out the SR equations and you'd wanted to run it again with acceleration. Without crunching all the numbers, I'd guess where it seems the numbers fall out of sync is the assumption that t"=0, t'=0, and t=0 are simultaneous. Remember, different inertial reference frames will disagree on what is simultaneous.