Acceleration is not important in the twin paradox

[xyzt]

Not even an hour later, a great example of cherry picking quotes.

You have just reduced yourself to trolling my posts.

All I did was use the relativistic rocket equation to solve the easiest acceleration / deceleration time dilation problem known to man.

 

Is the number 2.853 correct? If not, please correctly calculate. If so, please say so. If you don't know, please say you don't know.

The number is correct, I showed you the formalism, you know better than to tell me that I don't know. Congratulations for applying the formula correctly, now can you do the same for a circular trajectory? The C twin goes in a circle, following the same profile, so how much longer do you plan to keep up this charade? You already admitted that acceleration is important, so where do you plan to go with this game?

 

 

 

 

 

[math]\tau_B = 4 \frac{c}{a} \mbox{ln} \left( \frac{at}{c} + \sqrt{1+ \left( \frac{at}{c} \right)^2} \right)[/math]

 

[math]\tau_B = 4 \frac{1}{2.06314646} \mbox{ln} \left( \frac{(2.06314646)(1)}{1} + \sqrt{1+ \left( \frac{(2.06314646)(1)}{1} \right)^2} \right)[/math]

 

[math]\tau_B = 2.853 \ \mbox{years}[/math]

 

Is this acceptable? I know we aren't finished, but I wanted to stop and see if we're good.

 

A is stationary and spends 4 years watching B accelerate away then back at 2g. The round trip takes B 2.853 years.

So, according to you, [math]\tau_B[/math] does not depend on acceleration, right? Or does it? Or you can't make up your mind?

Edited June 6, 2013 by xyzt

[Iggy]

The number is correct

 

Excellent. I shall use the same method to calculate the proper time of the pacing twin.

 

Person A notes that C accelerates away from him for 1 day, decelerates for 1 day, turns around, accelerates towards him for 1 day, and decelerates for 1 day (all at 2g). According to A, this round trip takes C 4 days. The time according to C is:

 

[math]\tau_C = 4 \frac{c}{a} \mbox{arcsinh} \left( \frac{at}{c} \right)[/math] (for one round trip)

 

c = 1 lightyear / year

a = 2.06314646 lightyears / year²

t = 0.002739726 years (i.e. one day @ 365 days/year -- A's proper time for one segment of the trip)

 

[math]\tau_C = 4 \frac{1}{2.06314646} \mbox{arcsinh} \left( \frac{(2.06314646)(0.002739726)}{1} \right)[/math]

 

[math]\tau_C = 0.010958846 \ \mbox{years}[/math]

 

A watches C make this round trip 365 times taking four days each time for a total of four years. At the end of four years A and C both land. The proper time for each trip for C was 0.010958846 years. He made the trip 365 times.

 

[math]\tau_C = 0.010958846 \ \mbox{years} \times 365 = 3.99997879 \ \mbox{years}[/math] (for 365 round trips)

 

Person A (accelerating linearly at 2g) aged 2.853 years while person B (accelerating linearly at 2g) aged 3.99997879 years.

 

 

 

Let me explain where you went wrong...

 

The total elapsed times for the twins will be :

 

[math]\Delta \tau1 = 2 T_c / \sqrt{ 1 + (a1 \ T_a/c)^2 } + 4 c / a1 \ \text{arsinh}( a1 \ T_a/c )[/math]

 

[math]\Delta \tau2 = 2 T_c / \sqrt{ 1 + (a2 \ T_a/c)^2 } + 4 c / a2 \ \text{arsinh}( a2 \ T_a/c )[/math]

 

Making [math]a1=a2=2g[/math] one gets:

 

[math]\Delta \tau1=\Delta \tau2[/math]

 

The 4 in each equation refers to the number of segments of acceleration or deceleration. That equation assumes that only one trip is being made (accelerate / decelerate / accelerate / decelerate). Four segments.

 

You assumed also that T is the same in both equations, but one trip (the pacing trip) is much shorter. One day versus one year undergoing constant acceleration makes a big difference as far as top speed.

 

---

So, according to you, [math]\tau_B[/math] does not depend on acceleration, right? Or does it? Or you can't make up your mind?

 

It does. Acceleration is 2.06314646 light-years / year². It is in the equation you quoted.

Edited June 6, 2013 by Iggy

[xyzt]

 

Excellent. I shall use the same method to calculate the proper time of the pacing twin.

Do you have a reading difficulty ? I have explained to you (repeatedly) that the situation is the one described in your post 297, whereby observer C goes in a circle. I have also explained to you (repeatedly) that the speed profiles are the same for both B and C. Last time I explained that to you was right before yours.

You assumed also that T is the same in both equations, but one trip (the pacing trip) is much shorter. One day versus one year undergoing constant acceleration makes a big difference as far as top speed.

Yep, I explained this to you about 4 times as well, observer A measures B and C start together and come back together after elapsed time T on A's clock. Do you have so much difficulty following simple explanations?

 

Besides, the guy "pacing" ( C ) must pace REPEATEDLY, until B returns, so your calculation, under your own scenario, is dead wrong.

 

Finally, with all your mistakes piling up, what are you trying to prove? The results (if you eventually learn how to do the calculations correctly) depend on acceleration.

Edited June 6, 2013 by xyzt

[Iggy]

Do you have a reading difficulty ?

 

 

You said this:

 

 

 

Do you mean that someone accelerating and decelerating back and forth at 2g, over a distance of 100 meters (back and forth, back and forth, over and over) ages the same as someone who accelerates and decelerates back and forth once at 2g over a much larger distance? When they reunite their clocks would be equal?

 

I'm curious if I understand you correctly.

 

Yep

 

 

I asked very explicitly because I knew you would try to take it back.

 

 

I have explained to you (repeatedly) that the situation is the one described in your post 297, whereby observer C goes in a circle.

 

The velocity of the rotating twin is constant in post 297.

 

The point is that two different twins can experience 2g acceleration throughout the duration of a thought experiment (matching accelerometers) and their clocks will disagree upon reuniting.

 

You claimed that this was a result of centripetal acceleration as opposed to linear acceleration. I have now given an example with linear acceleration.

[xyzt]

The velocity of the rotating twin is constant in post 297.

We've been over this already, you agreed to stop the sleigh of hand and apply the acceleration tangentially , remember?

 

The point is that two different twins can experience 2g acceleration throughout the duration of a thought experiment (matching accelerometers) and their clocks will disagree upon reuniting

...because you are using a sleigh of hand that I have already exposed as a cheap cheat. Several times. You are running around in circles, excuse the pun

 

You claimed that this was a result of centripetal acceleration as opposed to linear acceleration. I have now given an example with linear acceleration.

You did it obviously incorrectly , as already pointed out, you "forgot" that the time, as measured by A, needs to be the same for both B and C and that you needed to have the same acceleration profile. Using a different acceleration profile gives (predictably) different answers. I already anticipated that you'll try this sleight of hand in a prior post. Even with all this cheap cheats, the answers still depend on acceleration, so you fail to make any valid point.

Now, try to do this honestly, by applying the acceleration profile of B to C's circular motion. What's the answer?

Edited June 6, 2013 by xyzt

[Iggy]

Besides, the guy "pacing" ( C ) must pace REPEATEDLY, until B returns, so your calculation, under your own scenario, is dead wrong.

He made 365 trips. I multiplied the proper time of the round trip by 365. You didn't notice?

We've been over this already, you agreed to stop the sleigh of hand and apply the acceleration tangentially , remember?

I don't know what purpose that would serve. You're welcome to calculate it if you like. If you think it proves some point. The results certainly wouldn't disagree with anything I've done. Time dilation won't depend on acceleration. Whatever the velocity of the rotating twin -- that will give the time dilation.

You did it obviously incorrectly , as already pointed out, you "forgot" that the time, as measured by A, needs to be the same for both B and C and that you needed to have the same acceleration profile. Using a different acceleration profile gives (predictably) different answers. I already anticipated that you'll try this sleight of hand in a prior post.

You're talking nonsense. All of my calculations have been correct and they have shown exactly what I've claimed they've shown.

 

You're just throwing rocks now. I'm going to honestly stop feeding the... uh... yeah...

Edited June 6, 2013 by Iggy

[xyzt]

You're welcome to calculate it if you like.

I already did, twice. I showed you that [math]\tau_B=\tau_C[/math]

Time dilation won't depend on acceleration.

"Time dilation" is the incorrect term, we are discussing "elapsed proper time", please try to learn the correct terminology. Your own calculations contradict your claim:

The time according to C is:

 

[math]\tau_C = 4 \frac{c}{a} \mbox{arcsinh} \left( \frac{at}{c} \right)[/math] (for one round trip)

So, according to you, the elapsed time does not depend on acceleration, right? What does [math]"a"[/math] represent in the above formula? This is the third time I am asking you this question. Please answer.

 

Excellent. I shall use the same method to calculate the proper time of the pacing twin.

 

Person A notes that C accelerates away from him for 1 day, decelerates for 1 day, turns around, accelerates towards him for 1 day, and decelerates for 1 day (all at 2g). According to A, this round trip takes C 4 days. The time according to C is:

 

[math]\tau_C = 4 \frac{c}{a} \mbox{arcsinh} \left( \frac{at}{c} \right)[/math] (for one round trip)

 

c = 1 lightyear / year

a = 2.06314646 lightyears / year²

t = 0.002739726 years (i.e. one day @ 365 days/year -- A's proper time for one segment of the trip)

 

[math]\tau_C = 4 \frac{1}{2.06314646} \mbox{arcsinh} \left( \frac{(2.06314646)(0.002739726)}{1} \right)[/math]

 

[math]\tau_C = 0.010958846 \ \mbox{years}[/math]

 

A watches C make this round trip 365 times taking four days each time for a total of four years. At the end of four years A and C both land. The proper time for each trip for C was 0.010958846 years. He made the trip 365 times.

 

[math]\tau_C = 0.010958846 \ \mbox{years} \times 365 = 3.99997879 \ \mbox{years}[/math] (for 365 round trips)

 

Person A (accelerating linearly at 2g) aged 2.853 years while person B (accelerating linearly at 2g) aged 3.99997879 years.

 

 

 

Excellent, so according to your own scenario: [math]\tau_A \ne \tau_B \ne \tau_C[/math]. Also according to you, [math]\tau_B,\tau_C[/math] depend clearly on the proper acceleration and on the fact that the acceleration profiles are different. So, congratulations! you just proved yourself wrong.

Edited June 6, 2013 by xyzt

[md65536]

 

 

Do you mean that someone accelerating and decelerating back and forth at 2g, over a distance of 100 meters (back and forth, back and forth, over and over) ages the same as someone who accelerates and decelerates back and forth once at 2g over a much larger distance? When they reunite their clocks would be equal?

 

I'm curious if I understand you correctly.

Yep

 

 

Do you see now your mistake? The observers have the same total acceleration (but split up differently with different timing) with different velocities, and the difference in elapsed time depends only on the velocity and time. If you assume that the difference in timing is a direct effect of acceleration, without considering the actual velocity involved, it's easy to make mistakes like this. Don't worry, remember by your own words there is honor in admitting mistakes. There's no problem using calculations based on acceleration if they correctly account for velocity.

Yet, Minguzzi's answer is a direct function of proper acceleration. Fancy that! The explanation is that, in SR, "accelerated observer" and "acceleration" are not one and the same thing: "accelerated observers" are prohibited in SR, by definition, whereas SR deal with acceleration just fine.

As your wikipedia link demonstrates, it is velocity that matters, and when velocity can be expressed as a function of acceleration then time dilation can be expressed as a function of acceleration. That doesn't contradict anyone's statements in this thread.

 

Still, as the relevant quote that I cherry-picked from the source you recommended says, "The acceleration, however, is not the ultimate source of

differential aging". That's been demonstrated in several different ways in this thread. Another example, if you still don't see it, is to start the twins off with a relative velocity. Then it is possible for the traveling twin to use the exact same acceleration profile with the exact same timing in each case, but come up with different results depending on the initial velocity, because the "velocity profile" is not the same.

 

I don't understand your explanation of why "The acceleration, however, is not the ultimate source of differential aging". What is accelerating here, if not observers?

[xyzt]

I don't understand your explanation of why "The acceleration, however, is not the ultimate source of differential aging". What is accelerating here, if not observers?

I already answered that, you need to go back and read the answer. Once again, SR forbids "accelerating observers", this is basic special relativity, only inertial frames (observers) are allowed. As to mistakes, I am quite sure Iggy realized by now the mistake in his two misguided attempts at showing that "equal" acceleration produces different effects on elapsed proper time (he is using the same scalar for acceleration but different profiles or different vectors, depending on the scenario). I have debunked both attempts and I will continue to debunk any further attempt because such attempts simply violate mainstream physics, as explained long ago by Markus, at the beginning of your thread. One day, you might understand it as well.

Another example, if you still don't see it, is to start the twins off with a relative velocity. Then it is possible for the traveling twin to use the exact same acceleration profile with the exact same timing in each case, but come up with different results depending on the initial velocity, because the "velocity profile" is not the same

I'd love to see your calculations for the above. Like I told you several times, physics isn't English composition, you need to be able to express your thoughts in a mathematical formalism. So, far,we have yet to see that from you, rest assured that I will take less than 30 minutes from your post to point out your mistakes. You can also stop the cowardly act of negging my posts trying to help you learn physics.

Edited June 6, 2013 by xyzt

[Phi for All]

You can also stop the cowardly act of negging my posts trying to help you learn physics.

 

!

Moderator Note

I checked into this, for all concerned. The vast majority of negative rep points for everyone in this thread seems to be for condescending personal remarks rather than simple disagreement. Yet another piece of evidence that sniping and uncivil behavior don't help anyone or anything, and often cause further problems. EVERYONE needs to stop it.

[esbo]

Hi all!!

 

Glad I found this thread because I was going to start a thread on this subject, however I see there is already one going on

so I will use this one!!

 

Forgive me for not reading all 18 pages now I do not have time but I will read them later.

I will just throw my tuppence ha'penny worth in though!!

 

1. Firstly the basic proposition goes one twin (B) remains young because he is travelling very fast and the other (A) is at rest (on earth or whatever).

 

2. Then someone says "but from the perspective of the the fast twin (B) the twin (A) back on earth is moving fast so twin (A) should be younger.

 

3. So the problem is that contradiction both cannot be the youngest

 

4. I have seen acceleration put forward as the explanation (with out and explanation of exactly how that works). Other sources it is not acceleration but hen don't seem to give an explanation of the solution (many just into a pile of dense maths at this point). I don't think we need maths here, we are not looking for a numerical answer rather a logical one.

 

5. If acceleration is the solution then it seems to me there is a limit to the age difference that can arise because you can only accelerate to the (near) speed of light and back, so that would seem to put a cap on the age difference.

 

6. Once you get to near the speed of light you can effectively spend millions of years flying about at that speed and that would seem to make any

acceleration effects very small indeed effectively zero in comparison.

 

7.So you seem to have the situation where an age difference of millions of years should be able to arise according to standard theory of fast moving objects. But then that leads to the contradiction mentioned in point 2.

 

So can someone explain this to me, in simple language, I do not believe maths is needed for the explanation and will pretty much regard mathematical answers as a failure to give the required answer.

 

Thanks a bunch, look forward to being educated but kind of doubt I will be.

 

I will now have a wade through the thread to see if I can find anything helpful (I'm not too optimistic given there are so many mathematical posts at this stage!!)

[ACG52]

Acceleration itself does not have an effect on relative time dilation, that's solely due to the relative velocities. What it does do is break the symmetry between the inertial frames.

[esbo]

Acceleration itself does not have an effect on relative time dilation, that's solely due to the relative velocities. What it does do is break the symmetry between the inertial frames.

 

 

That is kind of typical of a lot of answer I see.

It says something which may well be true, it might even be untrue, but as it does not actually answer the question it is neither here not

there really.

Do you seriously believe you answer the question?

 

Precisely what effect does breaking the symmetry have?

 

Unfortunately is seems just like so many answers I have seen, ie it does not really answer anything in a meaningful way, but thanks for trying.

Sorry if that sounds slightly rude, but I am just being honest.

[ACG52]

Do you seriously believe you answer the question?

Yes. Because you didn't understand the answer doesn't mean the answer wasn't given. It means you didn't understand it.

 

Precisely what effect does breaking the symmetry have?

When you have two inertial frames in relative motion, each will see the other's clock as running slow. When the two frames are brought back together, and the clocks compared, the clock which will show the effect of time dilation (i.e. the slower clock) will be the one which underwent acceleration (to bring the two frames back together, at least one of them must be accelerated).

 

That's as simple as it can be explained. If you've seen that answer a lot, and don't understand it, the problem is not with the answer.

[md65536]

Precisely what effect does breaking the symmetry have?

Wherever the twins are at a place where they can agree on their relative aging, if they've followed a symmetrical path---ie. the twins are interchangeable---they'd have to age the same amount. Symmetry must be broken to have an agreed-upon different aging.

 

Breaking symmetry alone doesn't lead to different aging. You can set it up so that the twins follow very different non-inertial paths configured so that they end up aging the same amount when they reunite.

 

Two ways to think about how the broken symmetry or acceleration makes a difference:

1. It forces different path lengths for the twins. On spacetime diagrams, the inertial twin will be a straight line and the shortest spatial distance, corresponding to the largest proper time (greatest aging) of any path between the events in flat spacetime.

2. The accelerating twin does not use just one inertial reference frame. A naive application of time dilation works for an inertial observer. Switching frames means dealing with relative simultaneity (with standard simultaneity defined with Einstein clock synchronization).

[Markus Hanke]

Precisely what effect does breaking the symmetry have?

 

It changes the arc length of the worldline connecting the same two events in space-time.

In other words - the proper times as measured by the two twins will no longer agree, which is precisely what we wanted to explain in the first place.

 

1. Firstly the basic proposition goes one twin (B) remains young because he is travelling very fast and the other (A) is at rest (on earth or whatever).

 

Too imprecise. The basic proposition is that both twins start off at rest in the same frame of reference, with synchronized clocks; one twin remains at rest, whereas the other twin makes a round trip of arbitrary distance and duration, and then returns to the stationary twin. In the end both twins are once again together at rest within the same frame. They then compare their clocks which each of them has been carrying, and find that the readings don't agree.

 

3. So the problem is that contradiction both cannot be the youngest

 

There is no contradiction. The stationary twin feels no forces acting on him, so will always consider himself at rest. The travelling twin feels the forces of his own acceleration and deceleration during his round-trip journey, so he will always consider himself moving. These two frames are physically distinct, and not interchangeable, hence there is no contradiction, and the fact that their clocks don't agree at the end of the experiment is not a paradox because they know precisely who was moving and who was stationary at the end of the experiment.

 

So can someone explain this to me, in simple language, I do not believe maths is needed for the explanation and will pretty much regard mathematical answers as a failure to give the required answer.

 

Mathematics is just simply a language to express physical ideas in concise form. You learn maths just as you learn any other foreign language. Dealing with physics and refusing to use maths is like living in China and refusing to use Mandarin; not only are you going to make your own life extremely difficult ( trust me, I've been there ), but you will also forego any chance of ever really understanding or appreciating what it really is you are dealing with. That is a simple truth of life. In this example, everything that has been so verbously elaborated on over the course of 18 pages can be mathematically expressed in just a handful of lines - if everyone here spoke the language of maths to the same standard, we would not have had to engage in this lengthy discussion. It is very easy to show mathematically that the proper times of the two twins cannot agree in this scenario, and more crucially also precisely why that is. Doing the same verbally on the other hand takes literally a wall of text to do it right; everyone needs to decide for him/herself which one they prefer.

Edited June 13, 2013 by Markus Hanke

[DimaMazin]

It changes the arc length of the worldline connecting the same two events in space-time.

In other words - the proper times as measured by the two twins will no longer agree, which is precisely what we wanted to explain in the first place.

 

 

Too imprecise. The basic proposition is that both twins start off at rest in the same frame of reference, with synchronized clocks; one twin remains at rest, whereas the other twin makes a round trip of arbitrary distance and duration, and then returns to the stationary twin. In the end both twins are once again together at rest within the same frame. They then compare their clocks which each of them has been carrying, and find that the readings don't agree.

 

 

There is no contradiction. The stationary twin feels no forces acting on him, so will always consider himself at rest. The travelling twin feels the forces of his own acceleration and deceleration during his round-trip journey, so he will always consider himself moving. These two frames are physically distinct, and not interchangeable, hence there is no contradiction, and the fact that their clocks don't agree at the end of the experiment is not a paradox because they know precisely who was moving and who was stationary at the end of the experiment.

Wrong. Stationary twin feels forces action on him from escaping twin.You do not define a quantitative disagreement of these feelings which are necessary that you would be right.

[ACG52]

Stationary twin feels forces action on him from escaping twin.

What forces?For example, one twin is on earth, the other twin is moving at a constant .9 c at a distance of 10 au. So what force does the twin on earth feel?

Edited June 13, 2013 by ACG52

[md65536]

Wrong. Stationary twin feels forces action on him from escaping twin.You do not define a quantitative disagreement of these feelings which are necessary that you would be right.

No, the twins' individual clock readings do not depend on any physical connection between them. They age how they age even if they don't even know the other exists, even if they measure nothing from each other. Edited June 13, 2013 by md65536

[esbo]

It changes the arc length of the worldline connecting the same two events in space-time.

In other words - the proper times as measured by the two twins will no longer agree, which is precisely what we wanted to explain in the first place.

 

 

Too imprecise. The basic proposition is that both twins start off at rest in the same frame of reference, with synchronized clocks; one twin remains at rest, whereas the other twin makes a round trip of arbitrary distance and duration, and then returns to the stationary twin. In the end both twins are once again together at rest within the same frame. They then compare their clocks which each of them has been carrying, and find that the readings don't agree.

I was just trying to be brief, we all under stand that bit we don;t really need it spelt out.

What we need spelt out is the bit people do not understand but of course this is the bit which is glossed over with an inadequate explanation.

 

 

 

Yes. Because you didn't understand the answer doesn't mean the answer wasn't given. It means you didn't understand it.

 

When you have two inertial frames in relative motion, each will see the other's clock as running slow. When the two frames are brought back together, and the clocks compared, the clock which will show the effect of time dilation (i.e. the slower clock) will be the one which underwent acceleration (to bring the two frames back together, at least one of them must be accelerated).

 

That's as simple as it can be explained. If you've seen that answer a lot, and don't understand it, the problem is not with the answer.

 

 

OK so lets say both twins experience acceleration.

 

So this is what happens.

 

Initial both at rest, twin A is accelerated to 99% the speed of light for 50 years and then returns back to where he started.

The same is done to twin B be he only spends 10 seconds at 90% of the speed of light.

 

What are their respective ages (approximately)

 

They have both experienced the same acceleration thus that effect will cancel out.

 

Wherever the twins are at a place where they can agree on their relative aging, if they've followed a symmetrical path---ie. the twins are interchangeable---they'd have to age the same amount. Symmetry must be broken to have an agreed-upon different aging.

 

Breaking symmetry alone doesn't lead to different aging. You can set it up so that the twins follow very different non-inertial paths configured so that they end up aging the same amount when they reunite.

 

Two ways to think about how the broken symmetry or acceleration makes a difference:

1. It forces different path lengths for the twins. On spacetime diagrams, the inertial twin will be a straight line and the shortest spatial distance, corresponding to the largest proper time (greatest aging) of any path between the events in flat spacetime.

2. The accelerating twin does not use just one inertial reference frame. A naive application of time dilation works for an inertial observer. Switching frames means dealing with relative simultaneity (with standard simultaneity defined with Einstein clock synchronization).

 

 

Again you gloss over the tricky bit.

You have not given a space time diagram or explained what it is .

 

I don't believe the accelerating twin remain in the same inertial frame in the first place so restating the obvious is not much help.

 

I gave an example where both twin accelerate but remain at a high speed for different lengths of time.

Perhaps you could use that example to describe the ageing process of each twin.

 

Now that would be helpful.

[Markus Hanke]

Wrong. Stationary twin feels forces action on him from escaping twin.

 

News to me. Can you specify exactly what those "forces acting on him from escaping twin" would be, both qualitatively and quantitively.

I was once present at the launch of a rocket carrying a satellite - what forces from the escaping rocket was I supposed to be feeling ? Please be very exact here, and provide a textbook reference to the force you are proposing.

Edited June 13, 2013 by Markus Hanke

[esbo]

 

 

 

Mathematics is just simply a language to express physical ideas in concise form. You learn maths just as you learn any other foreign language. Dealing with physics and refusing to use maths is like living in China and refusing to use Mandarin; not only are you going to make your own life extremely difficult ( trust me, I've been there ), but you will also forego any chance of ever really understanding or appreciating what it really is you are dealing with. That is a simple truth of life. In this example, everything that has been so verbously elaborated on over the course of 18 pages can be mathematically expressed in just a handful of lines - if everyone here spoke the language of maths to the same standard, we would not have had to engage in this lengthy discussion. It is very easy to show mathematically that the proper times of the two twins cannot agree in this scenario, and more crucially also precisely why that is. Doing the same verbally on the other hand takes literally a wall of text to do it right; everyone needs to decide for him/herself which one they prefer.

I know what maths is, it is just a code for expressing ideas which can be expressed in English.

 

 

Maths is a description of the real world in a coded form, I just want the description not the code thanks.

 

I don;'t want to learn a new language, most of us here speak ENGLISH so no need to put it in code we can cut the middleman out.

 

If you were in china woudl you talk to you English friends in Chinese?

 

I expect you would!!!!!!

 

News to me. Can you specify exactly what those "forces acting on him from escaping twin" would be, both qualitatively and quantitively.

I was once present at the launch of a rocket carrying a satellite - what forces from the escaping rocket was I supposed to be feeling ? Please be very exact here, and provide a textbook reference to the force you are proposing.

 

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

 

IN your example as the rocket moved away it's gravitation pull on you woudl have progressively decreased

however you probably would not have noticed.

[ACG52]

I know what maths is, it is just a code for expressing ideas which can be expressed in English.

Except that English, or any other language is imprecise and unwieldy.

 

Math is the language of physics, and is precise. .

[esbo]

Except that English, or any other language is imprecise and unwieldy.

 

Math is the language of physics, and is precise. .

 

 

 

It is precise enough to answer my question(s)

As you see above there is an unanswered question, maybe you can answer that? Use maths if you must

but it really should not be necessary,

[DimaMazin]

What forces?For example, one twin is on earth, the other twin is moving at a constant .9 c at a distance of 10 au. So what force does the twin on earth feel?

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

 

Very fast free planet can create gravitational forces to the accelerating body,though the body can have insignificant acceleration in space.

 

News to me. Can you specify exactly what those "forces acting on him from escaping twin" would be, both qualitatively and quantitively.

I was once present at the launch of a rocket carrying a satellite - what forces from the escaping rocket was I supposed to be feeling ? Please be very exact here, and provide a textbook reference to the force you are proposing.

You have forgotten electromagnetic accelerator.I am exacter here.

Edited June 13, 2013 by DimaMazin