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Twins paradox explained without forces

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Is this explanation ok that avoids using forces?  Thanks

Rather than twin flying it is best to consider synchronised clocks.  This avoids talking about acceleration at certain times is the flight.  Special relativity doesn’t require an understanding of acceleration – this is dealt with in general relativity.

It is not necessary for an actual twin to follow the out and back path or to experience an acceleration at the turning point.  Outgoing and incoming spaceships could simply exchange clock readings or videos when they pass each other.  Clock A stays on earth and is synchronised to clock B on a spaceship flying past (3/5c).  After 5 earth years the spaceship will be at 3 light years and synchronises the clock C on a spaceship flying in the opposite direction. This clock will arrive back at earth after another 5 earth years.  The moving clock will only show that a total of 8 years has elapsed.

As the twins move apart both will see the other age slower by the same amount (4/5).  This is after the time has been corrected for the video to travel between them.

When the clocks are synchronised at the turning point there is a change in the frame of reference.  For the spaceships to continue watching each other (videos and heart pings) the clocks need to be resynchronised.  

As they pass by the outbound rocket twin will relay the total number of pings counted so far to the inbound rocket and then the inbound rocket will continue counting all the way to Earth.  Outbound rocket twin will be getting low frequency pings and the inbound rocket will be getting high frequency pings.  The pings represent the heartbeats so the number of heartbeats by Earth twin will be greater and therefore Earth twin will be older.

On the inward journey, there is a new meaning of simultaneity.  There is a new clock synchronisation.  Remember there is no background absolute time.  The synchronisation of clocks must include the time for the clock reading to travel between them at the speed of light.

The outgoing spaceship (B) receives pings from the clock left on Earth (A) and from the clock on returning spaceship (C).  Clock C will be slow due to time dilation so that when it arrives at Earth it will show a total of 8 years has elapsed whereas the clock left on Earth will show 10 years has elapsed.  This is consistent with the twin paradox.  Note that clocks A and B will continue to run slow when compared to each other.  This is consistent with special relativity as there is no absolute time reference.

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You should do this with the relevant formulas. A simple description is insufficient. Once you sit down with the formulas you would then realize the twin paradox doesn't depend on the measurement device but upon the symmetry relations between the two events and the solution involves the stages where that symmetry is removed. Ie the acceleration stages (rapidity)

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51 minutes ago, scuddyx said:

Is this explanation ok that avoids using forces?  Thanks

Rather than twin flying it is best to consider synchronised clocks.  This avoids talking about acceleration at certain times is the flight.  Special relativity doesn’t require an understanding of acceleration – this is dealt with in general relativity.

SR can deal with accelerations, but the twins paradox is often presented such that the detail of the acceleration is irrelevant, so "an understanding of acceleration" isn't required.

 

Quote

When the clocks are synchronised at the turning point there is a change in the frame of reference.  

So you have an acceleration, you just don't call it that.

 

 

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2 hours ago, swansont said:

SR can deal with accelerations, but the twins paradox is often presented such that the detail of the acceleration is irrelevant, so "an understanding of acceleration" isn't required.

 

So you have an acceleration, you just don't call it that.

 

 

I am trying to explain the twin paradox without acceleration.  In my example all the spaceships are moving at constant speed.  No startup accelerations, decelerations or turning around.

The clock synchronisation is done instantaneously as they pass very close to each other.  Is there still an acceleration I am missing?

My explanation emphasises the fact that there is no absolute time reference.  Many twin paradox explanations flippantly claim its all about the spaceship turning around.

3 hours ago, Mordred said:

You should do this with the relevant formulas. A simple description is insufficient. Once you sit down with the formulas you would then realize the twin paradox doesn't depend on the measurement device but upon the symmetry relations between the two events and the solution involves the stages where that symmetry is removed. Ie the acceleration stages (rapidity)

Hi Mordred.  Thanks for your comments.  Do you agree with my reply to swansont?

Edited by scuddyx
typo fixed

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2 hours ago, swansont said:

So you have an acceleration, you just don't call it that.

What is accelerating? For how long? Is it an infinite acceleration? Where is the acceleration accounted for in formulas? Do you have acceleration every time you consider a new frame of reference?

3 hours ago, Mordred said:

You should do this with the relevant formulas. A simple description is insufficient.

This has been done countless times with the relevant formulas. This description is sufficient because it's a description of the result of those formulas. Here v = .6c and gamma = 1.25, anyone who properly applies the formulas will find that the description checks out. The proper time between when A and B meet, and B and C meet, plus the proper time between when B and C meet, and C and A meet, is 8 years.

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15 minutes ago, scuddyx said:

I am trying to explain the twin paradox without acceleration.  In my example all the spaceships are moving at constant speed.  No startup accelerations, decelerations or turning around.

How do they change from one frame to another?

 

 

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3 hours ago, scuddyx said:

It is not necessary for an actual twin to follow the out and back path or to experience an acceleration at the turning point.

This all looks okay to me. This has been discussed in the past and someone referenced a paper written in the 1900s that describes the same experiment, but I can't find it quickly. Wikipedia has references, see https://en.wikipedia.org/wiki/Twin_paradox#Role_of_acceleration

I don't like using the word "synchronized", but it's used in wikipedia. You must be careful because none of the clocks remain synchronized for any length of time, which I think does not satisfy the definition of synchronized. I'd just say the clocks are set the same.

The path length that you describe, between meetings of AB, BC, and CA, has an invariant length of 8 years. This is true whether or not a real object follows it. If you compare path lengths, you'll find that what you're describing is true; one path length between AB and CA is 8 years, and another is 10 years. Describing things in terms of path length is incontrovertible.

However even if you do this, people will disagree.

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13 minutes ago, md65536 said:

What is accelerating? For how long? Is it an infinite acceleration? Where is the acceleration accounted for in formulas? Do you have acceleration every time you consider a new frame of reference?

Whichever one is changing frames. As long as that happens quickly, it won't matter — any effect will be small compared to the constant-velocity part of the problem, as it is normally presented. 

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17 minutes ago, scuddyx said:

I am trying to explain the twin paradox without acceleration.  In my example all the spaceships are moving at constant speed.  No startup accelerations, decelerations or turning around.

The clock synchronisation is done instantaneously as they pass very close to each other.  Is there still an acceleration I am missing?

My explanation emphasises the fact that there is no absolute time reference.  Many twin paradox explanations flippantly claim its all about the spaceship turning around.

No I think it is hiding the fact that there is no absolute time.

Your problem is that you blythely say 'the clocks are synchronised', but do not define what this means.

As I recall Einstein realised this difficulty and spent a couple of pages going through it in meticulous detail in his papers.

He also realised that you can't have a time reference (absolute or otherwise) without a means to refer to it. Note I said absolute time not absolute time reference.

It is these small apparently innocuous, things that make such a big difference.

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1 minute ago, swansont said:

Whichever one is changing frames. As long as that happens quickly, it won't matter — any effect will be small compared to the constant-velocity part of the problem, as it is normally presented. 

3 clocks pass by each other at 3 events. What changes frames?

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7 minutes ago, md65536 said:

3 clocks pass by each other at 3 events. What changes frames?

I am going by the OP's description of changing frames at a turning point. So don't ask me, ask the OP.

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3 minutes ago, swansont said:

That's not the scenario described in the OP. Don't move the goalposts.

I read it again and I think it's exactly what OP posted. What difference do you see in our descriptions?

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Just now, md65536 said:

I read it again and I think it's exactly what OP posted. What difference do you see in our descriptions?

Sorry. I edited my comment. 

The OP describes "changing frames" and mentions a turning point...and then says there is no turning point. Also 3 clocks, but this is the twins paradox...but then, it isn't the twins paradox if there are 3 clocks.

I was thrown by the mass of confusion in the description.

 

41 minutes ago, scuddyx said:

My explanation emphasises the fact that there is no absolute time reference.  Many twin paradox explanations flippantly claim its all about the spaceship turning around.

The issue of "synchronizing*" B and C is taking the place of the acceleration in the twins paradox. Trading one issue for another. The whole point of the acceleration is that it represents a change in reference frames. There's nothing flippant about it.

 

(*strictly speaking, you can't synchronize clocks in two different frames of reference. Two clocks are synchronized when they have the same frequency (rate) and phase (time). No two clocks will run at the same rate in this scenario.)

 

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4 hours ago, scuddyx said:

When the clocks are synchronised at the turning point there is a change in the frame of reference.

This statement seems to be causing confusion and is unnecessary. Wikipedia instead calls it 'at the point corresponding to "turnaround" of a single traveller,' which emphasizes that nothing turns around.

"there is a change in the frame of reference" sounds like a description of something physical happening. Nothing physical changes frame of reference. There's only a change in the frame of reference that we're considering, and it doesn't have to be described as a change. Instead maybe something like, "for the inbound leg, we're considering a different reference frame."

That doesn't change the experiment that you described. What changes reference frame? Nothing physical that you've described.

 

Another way to emphasize that nothing changes at the BC meeting point, is to not set clock C there at all. One way to do this is to set C beforehand so that it will already have the time you want (ie. same time as B) when it passes B. Another way is to just add up the proper times of clocks B and C clock between the events, without even mentioning or caring what they were set to.

34 minutes ago, swansont said:

Also 3 clocks, but this is the twins paradox...but then, it isn't the twins paradox if there are 3 clocks.

I see it as using 3 clocks to describe the differential aging along two different world lines, which is the heart of the twin paradox.

You could have a single physical object follow the same path, and it would be a true twin paradox experiment. If you run OP's experiment alongside that, OP's is a way to measure a twin paradox experiment. It *is* a measure of the ageing in a twin paradox experiment.

Edited by md65536

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1 hour ago, md65536 said:

What is accelerating? For how long? Is it an infinite acceleration? Where is the acceleration accounted for in formulas? Do you have acceleration every time you consider a new frame of reference?

This has been done countless times with the relevant formulas. This description is sufficient because it's a description of the result of those formulas. Here v = .6c and gamma = 1.25, anyone who properly applies the formulas will find that the description checks out. The proper time between when A and B meet, and B and C meet, plus the proper time between when B and C meet, and C and A meet, is 8 years.

Sure now apply the turnaround acceleration. However your missing the point behind the request for the OP to look directly at the formulas.

 It isn't a shut up and calculate reason but to have the OP look directly at the symmetry relations between emitter and observer reference frames 

Axiom there is no frame preference.

Edited by Mordred

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51 minutes ago, md65536 said:

 That doesn't change the experiment that you described. What changes reference frame? Nothing physical that you've described.

Nothing physical changes reference frame in this description (contradicting the statement in the OP), but there is a physical change, since you are using a different clock. It is functionally the same as turning the clock around, which requires an acceleration.

 

51 minutes ago, md65536 said:

Another way to emphasize that nothing changes at the BC meeting point, is to not set clock C there at all. One way to do this is to set C beforehand so that it will already have the time you want (ie. same time as B) when it passes B. Another way is to just add up the proper times of clocks B and C clock between the events, without even mentioning or caring what they were set to.

That's cheating a bit, though, since you are admitting to set up the problem to get a particular answer, and it only works for one scenario, instead of in general. That's not a good approach.

 

51 minutes ago, md65536 said:

I see it as using 3 clocks to describe the differential aging along two different world lines, which is the heart of the twin paradox.

You could have a single physical object follow the same path, and it would be a true twin paradox experiment. If you run OP's experiment alongside that, OP's is a way to measure a twin paradox experiment. It *is* a measure of the ageing in a twin paradox experiment.

It gets the same answer because the crux of the matter is changing from one frame to another. One of the errors of the OP is failing to recognize this, and another is insistence on the subjective assessment that one is better than the other, simply because they like it better, and presenting it as objective.

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29 minutes ago, swansont said:

Nothing physical changes reference frame in this description (contradicting the statement in the OP), but there is a physical change, since you are using a different clock. It is functionally the same as turning the clock around, which requires an acceleration.

 

That's cheating a bit, though, since you are admitting to set up the problem to get a particular answer, and it only works for one scenario, instead of in general. That's not a good approach.

OP's description still has 3 clocks passing by each other. What physically changes?

It's not cheating to set initial conditions to make measurements easier. Only one scenario is described, ie. 4 years of proper time measured by C at a speed of .6 c relative to A. Whether clock C reads 0 or 4 years or 500 when it meets B, it will read 4 years later when it meets A. You get the same answer either way; the path from AB to BC to CA is 8 years long, regardless of how the clocks are set. Also regardless of how many clocks are used to measure sections of that path.

49 minutes ago, Mordred said:

Sure now apply the turnaround acceleration.

Can you show how to apply turnaround acceleration using the experiment described by OP?

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1 hour ago, md65536 said:

 

Can you show how to apply turnaround acceleration using the experiment described by OP?

I will post the mathematics when I get time tonight though I will not complete the scenario for the third reference. It should be trivial enough for the OP to implement.

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4 hours ago, scuddyx said:

I am trying to explain the twin paradox without acceleration.  In my example all the spaceships are moving at constant speed.  No startup accelerations, decelerations or turning around.

The twin paradox based purely in Special Relativity is truly a paradox, meaning it cannot be solved this way.  In this case, both observers would actually measure the others clock to run slower.  Then it is impossible to imagine how this can take place.  

The problem is that it assumes that two observers could have always existed in the universe with a constant relative velocity, but this would actually be impossible.  Everything accelerated from the point of the Big Bang, so every two objects would have had to had acceleration relative to each other in their history.  Even if you assumed that they were at a constant velocity relative to each other and their acceleration was due to the acceleration of spacetime itself, it would still have relative Doppler shifts.  Then Doppler shifts solve the paradox in terms of acceleration.

Edited by Conjurer

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6 minutes ago, Conjurer said:

The twin paradox based purely in Special Relativity is truly a paradox, meaning it cannot be solved this way.

!

Moderator Note

This is nonsense. I think you need to take some off to consider whether this is the right place for you. 

 

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16 hours ago, Mordred said:

I will post the mathematics when I get time tonight though I will not complete the scenario for the third reference. It should be trivial enough for the OP to implement.

 

I've thought about how to go about this so I decided to take a few posts to the four momentum form. (I will need that for the four acceleration)

 First let's get the symmetry relations behind the (supposed paradox). We will obviously be applying the Lorentz transformations

[math]\acute{\tau}=\gamma(\tau-\frac{vx}{c^2})[/math]

[math]\acute{x}=\gamma (x-vt)[/math]

Y and Z coordinates are equivalent respectively.

[math] \gamma=(\sqrt{1-\frac{v^2}{c^2}})[/math]

The inverse of each is simply switching the observer frame ie

[math]x= \gamma(\acute{x}-vt)[/math]

Now you can see under math the symmetry between frames. This being under constant velocity. Using coordinates [math]x^\mu={x^o,x^1,x^2,x^3}={ct,x,y,z}[/math]

Now let's see how acceleration gets involved under the Minkowskii metric. We will need the four momentum and four acceleration.

Four velocity [math]\mu^\mu=\frac{dx^\mu}{d\tau}=(c\frac{dt}{d\gamma},\frac{dx}{d\gamma},\frac{dy}{d\gamma},\frac{d}{dz\gamma})[/math]

The invariant distance or seperation between two events in Cartesian coordinates. [math]ds^2=-c^2dt^2+dx^2+dy^2+dz^2[/math]

[math]ds^2=\eta_{\mu\nu}dx^\mu dx^\nu[/math]

[math]\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/math] for the Minkowskii tensor.

Proper time defined as [math]ds^2=-c^2dt^2[/math]

Four acceleration [math]a^\mu=\dot{\mu}^{\mu}[/math]

[math]\eta_{\mu\nu}a^\mu\mu^\nu=a^\mu\mu_\nu=0[/math]

Now consider a ship in x direction with constant acceleration g the velocity and acceleration four vectors are

[math]c\frac{dt}{d\tau}=\mu^0, \frac{dx^1}{d\tau}=\mu^1[/math]

[math]\frac{d\mu^0}{d\tau}=a^0,\frac{d\mu^1}{d\tau}=a^1[/math]

[math]-(\mu^0)^2+(\mu^1)^2=-c^2[/math]

[math]a^\mu a_\mu=-(a^0)^2+(a^1)^2=g^2[/math]

As it's getting late will finish this tomorrow to arrive at the hyperbolic turnaround

[math]x2-ct^2=\frac{c^4}{g^2}[/math]

 

 

 

 

 

 

 

Edited by Mordred

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13 hours ago, scuddyx said:

I am trying to explain the twin paradox without acceleration.

The easiest and most straightforward way to do this would be to simply compare the geometric lengths of the two observers’ world lines in spacetime. This way the explanation is a purely geometric one, and makes no direct reference to forces, accelerations, or frames. 

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19 hours ago, md65536 said:

OP's description still has 3 clocks passing by each other. What physically changes?

The OP said turnaround. That's a physical change, requiring an acceleration.

But then, the OP also gave a different description of the situation. I don't think referring to the OP is going to be very illuminating in this regard.

What physically changes is the clock you are using. It is a different clock.

 

Quote

It's not cheating to set initial conditions to make measurements easier. Only one scenario is described, ie. 4 years of proper time measured by C at a speed of .6 c relative to A. Whether clock C reads 0 or 4 years or 500 when it meets B, it will read 4 years later when it meets A. You get the same answer either way; the path from AB to BC to CA is 8 years long, regardless of how the clocks are set. Also regardless of how many clocks are used to measure sections of that path.

If I move an arbitrary distance, the clocks don't read the same time.

The criticism of a setup like this is that it doesn't work, in general, and was contrived to get a certain result.

 

 

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The three clocks in this scenario can only be synchronised whilst they are all at rest relative to the same inertial frame, anything else is meaningless. From this point on it's just a more complicated version of the "Twins Paradox"

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Anyways I will finish the proof behind the hyperbolic angle tonight for completeness. The reason I wanted the OP to study the symmetry vs antisymmetric relations is as follows.

By changing the observer emitter from twin A to B B to A respectively we have mathematically defined The symmetric and identical relations 

[math]\acute{\tau}=\gamma(\tau-\frac{vx}{c^2})[/math]

[math]\acute{x}=\gamma (x-vt)[/math]

Hence we cannot from those equations show a priveleged frame nor prove which twin is the inertial twin to identify which twin will age slower.

During the acceleration the travelling twin will be in a non inertial frame flowing the spacetime coordinates of the hyperbolic angle which is also antisymmetric.

[math]x2-ct^2=\frac{c^4}{g^2}[/math]

The above is the result of change from  v+ to v-. 

If the turnaround is sufficiently long enough to model as a rotationary body however brief then we can incorporate the Sagnac effect which I will show that tensor tonight.

In essence we can mathematically show where the aging becomes antisymmetric.

 

Edited by Mordred

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