Acceleration is not important in the twin paradox

[ACG52]

Far from gravitational masses electromagnetic accelerator accelerates a body.The accelerator feels the force

There is no electromagnetic accelerator. You can't just make things up.

[DimaMazin]

There is no electromagnetic accelerator. You can't just make things up.

You can just prove that there is no ACG52.

[Delta1212]

There are two events: the twins separating, and the twins coming back together to compare ages.

 

These two events are separated in spacetime. The longer the path through space you take from one to the other, the shorter your path through time. The shorter your path through space, the longer your path through time.

 

The separation and final comparison both take place at rest with respect to the twin who stays behind. In this frame, the twin never moved and therefore experienced the maximum amount of time.

 

According to this frame, the twin that leaves and returns traveled some distance, and this means he experienced proportionately less time.

 

While the twin is initially traveling, he can say that he is the one at rest and the other twin is moving. At this point, both twins will say that the other has aged less, but this isn't a problem because you can only compare clocks if they are co-located. However, once the twin turns around, he can no longer say that he spent the whole time at rest while the other twin traveled because there is no inertial path that goes from the twins separating through the turnaround point and then to the return where the clocks are compared.

 

To answer your question:

 

If 50 years pass on Earth during which one twin is traveling out and back at 99% of the speed of light, that twin will age 7 years.

 

If, during that time, the other twin takes a ten second trip out and back at 90% of the speed of light, that twin will be about 6 seconds younger than he would have been if he'd stayed on Earth the whole time.

 

So their ages would be, if they started when they were first born, 7 years old and 49 years, 364 days, 23 hours, 59 minutes and 54 seconds, respectively, after 50 years as measured on Earth.

[esbo]

There are two events: the twins separating, and the twins coming back together to compare ages.

 

These two events are separated in spacetime. The longer the path through space you take from one to the other, the shorter your path through time. The shorter your path through space, the longer your path through time.

 

The separation and final comparison both take place at rest with respect to the twin who stays behind. In this frame, the twin never moved and therefore experienced the maximum amount of time.

 

According to this frame, the twin that leaves and returns traveled some distance, and this means he experienced proportionately less time.

 

While the twin is initially traveling, he can say that he is the one at rest and the other twin is moving. At this point, both twins will say that the other has aged less, but this isn't a problem because you can only compare clocks if they are co-located. However, once the twin turns around, he can no longer say that he spent the whole time at rest while the other twin traveled because there is no inertial path that goes from the twins separating through the turnaround point and then to the return where the clocks are compared.

 

To answer your question:

 

If 50 years pass on Earth during which one twin is traveling out and back at 99% of the speed of light, that twin will age 7 years.

 

If, during that time, the other twin takes a ten second trip out and back at 90% of the speed of light, that twin will be about 6 seconds younger than he would have been if he'd stayed on Earth the whole time.

 

So their ages would be, if they started when they were first born, 7 years old and 49 years, 364 days, 23 hours, 59 minutes and 54 seconds, respectively, after 50 years as measured on Earth.

 

 

But given an object has no memory of previous accelerations, then the two twins are briefly together in space they are in the same

situation as in the the initial twin paradox proposition where both start on earth, one accelerates way for a period of time and then comes

back being much younger.

 

So in one analogous situation a twin jets away at high speed and stays very young but in another he remains at almost the same age.

[Delta1212]

 

 

But given an object has no memory of previous accelerations, then the two twins are briefly together in space they are in the same

situation as in the the initial twin paradox proposition where both start on earth, one accelerates way for a period of time and then comes

back being much younger.

 

So in one analogous situation a twin jets away at high speed and stays very young but in another he remains at almost the same age.

One experiences time dilation for 10 seconds (Earth time). The other experiences time dilation for 50 years (Earth time). The fact that they don't have the same difference in aging isn't particularly surprising.

[DimaMazin]

One experiences time dilation for 10 seconds (Earth time). The other experiences time dilation for 50 years (Earth time). The fact that they don't have the same difference in aging isn't particularly surprising.

How does gravitational force define kinematic traveler?Why size of gravitational force isn't important?

[esbo]

One experiences time dilation for 10 seconds (Earth time). The other experiences time dilation for 50 years (Earth time). The fact that they don't have the same difference in aging isn't particularly surprising.

 

 

But to me it depends on whose perspective you are looking from.

When they are together in space they are in the same inertial frame.

One accelerates off to earth thus he should appear to age slower than the one in space.

 

You seem to be picking earth as an absolute 0 reference frame, but there is no such thing (apparently).

 

For example say we to go back to before any of these experiment sstarted and we discovered the twins were born on a different planet

which was moving near the speed of light (with respect to earth) and they then jetted off to earth to big the experiments.

 

That seems to turn everything on it's head if you see what I mean?

[Delta1212]

 

 

But to me it depends on whose perspective you are looking from.

When they are together in space they are in the same inertial frame.

One accelerates off to earth thus he should appear to age slower than the one in space.

 

You seem to be picking earth as an absolute 0 reference frame, but there is no such thing (apparently).

 

For example say we to go back to before any of these experiment sstarted and we discovered the twins were born on a different planet

which was moving near the speed of light (with respect to earth) and they then jetted off to earth to big the experiments.

 

That seems to turn everything on it's head if you see what I mean?

Earth's frame is being used because it's the one where they are brought back together for comparison. If, instead of turning around, the traveling twin continued on his path away from Earth, and, after some years, the twin on Earth blasted off and caught up with the traveling twin, the twin who had been on Earth initially would be younger. He'd also be the one who traveled away from the other twin (along with the Earth) and then turned around and came back.

[Markus Hanke]

Strictly speaking the escaping twin would have a gravitation effect on the other twin.

 

 

Of course it does, but that effect is far too small to cause any measurable time dilation effects - unless of course you can show us some numbers proving otherwise, but then again, you appear to reject any maths, so obviously you are unable to.

I should also remind you that the travelling twin's proper time dilation explicitly depends on the exact path it takes before returning to earth, including all maneuvers undertaken while far away from the stationary twin, i.e. far outside the range of any measurable gravitational interaction. Likewise, it is easy to show that the total proper time dilation experienced by the observer does not depend on their masses; if you were to substitute the twins with, say, muons, you would find that the particles' lifetime is dilated by the exact same amount as the much more massive spaceship would have been. Therefore it is obvious that the time dilation in this experiment is not the result of gravitational interactions between the observers.

[DimaMazin]

 

 

Of course it does, but that effect is far too small to cause any measurable time dilation effects - unless of course you can show us some numbers proving otherwise, but then again, you appear to reject any maths, so obviously you are unable to.

I should also remind you that the travelling twin's proper time dilation explicitly depends on the exact path it takes before returning to earth, including all maneuvers undertaken while far away from the stationary twin, i.e. far outside the range of any measurable gravitational interaction. Likewise, it is easy to show that the total proper time dilation experienced by the observer does not depend on their masses; if you were to substitute the twins with, say, muons, you would find that the particles' lifetime is dilated by the exact same amount as the much more massive spaceship would have been. Therefore it is obvious that the time dilation in this experiment is not the result of gravitational interactions between the observers.

Theoretically accelerating body can be more massive and can travel faster than its accelerator if the accelerator is cyclic.Then what is your math?

Edited June 14, 2013 by DimaMazin

[imatfaal]

!

Moderator Note

 

DimaMazin

 

This thread is a fairly well established argument on the importance of acceleration in the twin paradox - your introduction/discussion of electromagnetic accelerators, the gravitational attraction of one twin on the other, and cyclic acceleration is hijacking and not acceptable. Do not respond to this moderation within the thread - report this message if you wish to complain.

 

Get back to the OP

 

[lukemcleod]

Hold on forgive me if I'm wrong I'm not a genius like all you guys but I read about this in another blog and I believe the question its about says what would the triplets age be if triplet a set off at 99% of the speed of light for 50 years then returned to earth, twin b set of at 99% the speed of light from earth for 10 seconds before returning and twin c stayed on earth and observed them isn't the answer in the question they said 99% the speed of light that's not the same as the speed of light and does Einstein not say that you need to be traveling at least the speed of light before experiancing time travel so I would say all the twins remain the same age

Edited June 14, 2013 by lukemcleod

[Delta1212]

Hold on forgive me if I'm wrong I'm not a genius like all you guys but I read about this in another blog and I believe the question its about says what would the triplets age be if triplet a set off at 99% of the speed of light for 50 years then returned to earth, twin b set of at 99% the speed of light from earth for 10 seconds before returning and twin c stayed on earth and observed them isn't the answer in the question they said 99% the speed of light that's not the same as the speed of light and does Einstein not say that you need to be traveling at least the speed of light before experiancing time travel so I would say all the twins remain the same age

Nothing with mass can travel at the speed of light. You can only approach it. You also experience time dilation traveling at any speed, but it's too little to notice unless you're going at a significant fraction of light speed for a measurable period of time. The closer you get to the speed of light, the more time dilation you experience.

[lukemcleod]

Oh right I see makes sense really thanks for clearing that up for me

[victorqedu]

This initiative is very good.

This forum should have a section "Problem of the week" or something like this.

[JVNY]

Perhaps members can comment on the following diagram of triplets, whether the calculations are correct, and if it makes sense then how it relates to their explanation of the paradox. Two triplets are at rest at the origin of the inertial frame in the Minkowski diagram. The third triplet is traveling inertially with respect to the inertial frame at 1/3c. When the third triplet reaches the origin, all three agree on simultaneity because they are at the same point. They set their clocks at zero. Simultaneously, the first triplet begins constant acceleration at 1 light year per year^2, and the second triplet initiates a pattern of acceleration at 2 LY per Y^2 for 0.125 LY in the inertial reference frame, then deceleration at the same rate for the same distance.

 

The result is that all three meet again in the inertial frame at (0.25 LY, 0.75 Y). They agree on simultaneity because they are at the same point, and they compare clocks. What results? The first triplet's clock reads 0.693 LY. One can calculate this by the formulas for relativistic travel at a constant acceleration found at: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html.

 

The second triplet's clock reads 0.693 LY. One can calculate proper time for him by the same formulas first for the acceleration leg, then double it for the total (as the website states, you can use the same formula because you are just running the clock backward on the second leg).

 

The third triplet's clock reads 0.707 LY. This is the 0.75 LY in the inertial frame dilated by the Lorentz factor for 1/3c travel (that is, divided by 1.06).

 

Each triplet had the same average speed, 1/3c (0.25 LY / 0.75 Y). However, the accelerating triplets' clocks ran slower than the inertial triplet's clock. The first triplet underwent constant proper acceleration of 1, whereas the second triplet underwent (a) zero net acceleration and (b) constant absolute proper acceleration of 2. Yet their clocks ran at the same rate. I would appreciate it if anyone could confirm or correct the calculations.

 

If the calculations are correct, then average speed is not relevant. The accelerating triplets achieved higher velocity (0.6c according to the formulas at the website above). Time dilation is disproportionately higher at higher velocities, which should explain why the accelerating triplets' clocks ran slower than the inertial triplet's clock. Also, I don't think that the absolute amount of acceleration that one feels is relevant. Here one triplet felt absolute acceleration of 2 over the same distance that the other felt absolute acceleration of 1. Yet their clocks ran at the same rate. I suspect that this is because they traveled at the same speeds for the same times although in different patterns. This seems different from gravity. If a clock experienced gravity of 2 from the left and then 2 from the right, I think that the gravitational time dilation would be greater than that for a clock that experienced gravity of 1 from a single side for the same total time.

 

[xyzt]

Perhaps members can comment on the following diagram of triplets, whether the calculations are correct, and if it makes sense then how it relates to their explanation of the paradox. Two triplets are at rest at the origin of the inertial frame in the Minkowski diagram. The third triplet is traveling inertially with respect to the inertial frame at 1/3c. When the third triplet reaches the origin, all three agree on simultaneity because they are at the same point. They set their clocks at zero. Simultaneously, the first triplet begins constant acceleration at 1 light year per year^2, and the second triplet initiates a pattern of acceleration at 2 LY per Y^2 for 0.125 LY in the inertial reference frame, then deceleration at the same rate for the same distance.

 

The result is that all three meet again in the inertial frame at (0.25 LY, 0.75 Y). They agree on simultaneity because they are at the same point, and they compare clocks. What results? The first triplet's clock reads 0.693 LY. One can calculate this by the formulas for relativistic travel at a constant acceleration found at: http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html.

 

The second triplet's clock reads 0.693 LY. One can calculate proper time for him by the same formulas first for the acceleration leg, then double it for the total (as the website states, you can use the same formula because you are just running the clock backward on the second leg).

 

The third triplet's clock reads 0.707 LY. This is the 0.75 LY in the inertial frame dilated by the Lorentz factor for 1/3c travel (that is, divided by 1.06).

 

Each triplet had the same average speed, 1/3c (0.25 LY / 0.75 Y). However, the accelerating triplets' clocks ran slower than the inertial triplet's clock. The first triplet underwent constant proper acceleration of 1, whereas the second triplet underwent (a) zero net acceleration and (b) constant absolute proper acceleration of 2. Yet their clocks ran at the same rate. I would appreciate it if anyone could confirm or correct the calculations.

 

If the calculations are correct, then average speed is not relevant. The accelerating triplets achieved higher velocity (0.6c according to the formulas at the website above). Time dilation is disproportionately higher at higher velocities, which should explain why the accelerating triplets' clocks ran slower than the inertial triplet's clock. Also, I don't think that the absolute amount of acceleration that one feels is relevant. Here one triplet felt absolute acceleration of 2 over the same distance that the other felt absolute acceleration of 1. Yet their clocks ran at the same rate. I suspect that this is because they traveled at the same speeds for the same times although in different patterns. This seems different from gravity. If a clock experienced gravity of 2 from the left and then 2 from the right, I think that the gravitational time dilation would be greater than that for a clock that experienced gravity of 1 from a single side for the same total time.

triplets.png

 

The total elapsed proper time is given by the integral [math]\tau=\int_C{\sqrt{1-(v(t)/c)^2}dt}[/math]. The only thing you need to know is how does v vary as a function of the coordinate time t and the path C followed by the twin. Plug in these two and you will get the answer. Depending on how you make v to vary over time , you can get any result you want. While the result does not depend explicitly on the acceleration (t depends explicitly on v), it surely depends implicitly since v depends on the acceleration, as you found out in your exercise above.

Now, if we remember that [math]v(t)=\frac{at}{\sqrt{1+(at/c)^2}}[/math] it follows immediately that

 

 

[math]\tau=\frac{c}{a} arcsinh \frac{at}{c}[/math]

 

The takeaway is that the total elapsed proper time, depends on the acceleration (profile).

Edited November 27, 2013 by xyzt

[md65536]

The total elapsed proper time is given by the integral [math]\tau=\int_C{\sqrt{1-(v(t)/c)^2}dt}[/math]. The only thing you need to know is how does v vary as a function of the coordinate time t and the path C followed by the twin. Plug in these two and you will get the answer. Depending on how you make v to vary over time , you can get any result you want. While the result does not depend explicitly on the acceleration (t depends explicitly on v), it surely depends implicitly since v depends on the acceleration, as you found out in your exercise above.

Now, if we remember that [math]v(t)=\frac{at}{\sqrt{1+(at/c)^2}}[/math] it follows immediately that

 

 

[math]\tau=\frac{c}{a} arcsinh \frac{at}{c}[/math]

 

The takeaway is that the total elapsed proper time, depends on the acceleration (profile).

We're in total agreement.

 

In addition, and the point of this thread, is that the proper time measured over the path traveled by a single clock, is the same as the sum of the proper times measured by multiple clocks each traveling a section of the same path (where every subsection of the path is traveled by exactly one clock).

 

If the path can be split into inertial sections, as in post #1, then the proper time along the path can be measured by clocks that don't accelerate.

[xyzt]

We're in total agreement.

 

It is good to see that I managed to convince you.

[md65536]

Perhaps members can comment on the following diagram of triplets, whether the calculations are correct, and if it makes sense then how it relates to their explanation of the paradox.

It makes sense to me. I'm not an expert and am also interested in how others would comment on it.

[decraig]

I'll try this again!

 

 

What's important in the twin paradox, which results in the asymmetry between the twins, is that one of the twins remains in an inertial frame while the other uses primarily two different inertial frames. It is tempting then to think that a mechanical switch between the frames somehow "causes" the relativistic effects---and further that the only way to switch frames is to accelerate---but this is not true.

 

This can be shown by running the experiment with 3 moving clocks, none of which needs to accelerate during the experiment.

 

Start with 2 passing clocks, A and B, which are each set to zero at passing. Let B travel some distance at velocity v, where it passes clock C traveling in the opposite direction at the same speed, and have C set its clock to match B's as they pass. When C and A meet they'll record the same difference in proper time as if the experiment was run with clock B instantly turning around as it passes clock C.

 

What this shows is that the effect of time dilation does not depend on physically turning around, it only depends on what happens while in the various inertial frames.

 

 

If B and C pass at a distance of 0, their meeting can be considered a single event that is simultaneous according to all observers. However at that moment, their different frames are important (not any mechanical effect of switching anything between their frames) as they have very different measures of simultaneity relative to A. C will have measured A's aging as greater than B has measured A's aging, even though they are at the same place at the moment of their passing.

 

 

Because there is no difference in relativistic effects whether we abstractly switch from B to C, or physically cause B to turn around, it cannot be said that the mechanics of turning around has any physical bearing on the paradox. If we use a single traveling observer, then the resulting frame switch is what's important, but any acceleration used to cause the frame switch should not be confused as being the cause of the relativistic effects.

 

That which you have written, that I have place in bold text, is exactly correct. And well written. (Better than I could do it.) That which I did not bold I did not understand.

 

In fact, this construction better captures the asymmetry of the twins "paradox" better than most. Recall that the objections to relativity theory were due to an incorrect comparison of symmetry of clocks implied by relativity and symmetry of velocity.

 

You have illucidated that the velocites are not symmetrical in the twins 'paradox' scenario.

 

This is a very long thread, so I haven't read the ensuing attacks, and have no desire to wade through them.

Edited December 2, 2013 by decraig

[md65536]

That which you have written, that I have place in bold text, is exactly correct. And well written. (Better than I could do it.) That which I did not bold I did not understand.

Thanks. I think my point in the non-bold text was something like...

 

Assume B and C pass at an event (a definite location and time). At that moment, with B and C both remaining inertial, they predict/calculate different relative simultaneity with A (even though A appears the same to both at the time). With the example values (gamma=2, B and C are 1 travel year away from A at their meeting), B calculates that A is 3 years younger than C does.

 

Now, if you imagine B turning around to travel with C, there is a change in relative simultaneity that corresponds to B now agreeing with C's determination of simultaneity. If one tries to assert that the turn-around "causes" a differential aging in A corresponding to that change in simultaneity, it requires cumbersome interpretation I think, because that aging at A has already occurred according to C, whether B turns around or not. Nobody's turn-around or acceleration has caused what C measures.

 

The importance of acceleration in this example, is that it determines whether B agrees with C by switching to its inertial frame.

 

 

 

 

 

(On the other hand, if B doesn't turn around then B doesn't measure any twin paradox effect itself. So while I think it's okay to say B's acceleration doesn't cause any physical advanced aging at A, its own determination of what happens depends on its own path and thus on its own acceleration.)

[phyti]

I thought this was settled long ago!

There are references online for experiments proving time dilation is not affected by acceleration. This is why the lorentz factor only contains a term for speed, v/c.

The integral expression involving acceleration, used for a continuously changing or easily defined path, is only a mathematical convenience.

Sometimes it's easier to integrate a volume of discrete masses by assuming a continuous medium, than summing a large number of particles.

[decraig]

"Well, Her Einstein, you think you know better than Newton. If all this stuff is relative as you say, then if clock A is faster than B, B should also be faster than A, you pretentious fool."---or something like that.

 

Folks, it occurs to me that there is a way to the bring symmetry back into the mix.

 

md2¹⁶--

Two clocks, A and B, pass each other in relative motion. Instead of using your "C" clock, I will call it " B' ".

Now let us include an A' clock, as well, in a symmetrical way. A' rushes back and meets-up with B. We can do the the comparison of A and B', which is what all the fuss was about, but also compare A' and B.

 

Now everybody is happy. B' is slower than A, and A' is slower than B. We have both the symmetry demanded by the critics, as well as special relativity.

 

phyti--

This discussion of the twins paradox is perpetual. There seem to be two major camps. The critics who are still skeptical and the freshmen who have an epiphany they credit to acceleration while still harboring notions of universal time.

Edited December 2, 2013 by decraig

[JVNY]

Decraig, this example is quite good, but I don't "feel" the release from the paradox using it. Others might.

 

md65536's explanation based on simultaneity seems the better way. The most fundamental difficulty seems to me to be reconciling symmetry with the outcome of the inertial triplet and the a=1 accelerating triplet in post 391 above. Each is in motion relative to the other, and they meet up again after their first crossing. So a naive application of special relativity would say that each's clock runs slow to the other, and thus at their second meeting each must observe the other's clock to be behind his own -- which is plainly false. It cannot be that each clock runs slower to the other. The accelerating triplet's clock runs slower to the inertial triplet, and the inertial triplet's clock runs faster to the accelerating twin. The two are simply not symmetrical to each other and cannot be -- and it is the supposed symmetry of the two that drives the paradox. It may be better to give up symmetry here and focus on the relativity of simultaneity. Although it would be great if the example resonates with other readers.