Thread: Twin Paradox, Doppler Effect and the Symmetry

Some claim that the Twin Paradox is explained by the fact that that the Symmetry is broken when taking the relative Doppler Shift into a consideration. Comparing the Earth and the rocket that made a round trip, the rocket travels along the signal from Earth and when it made a turn, it will experience an immediate shift from red to blue, while the Earth will experience a delay factor by t(del) = d/c because it takes time for the signal from the rocket to reach the Earth after the rocket has made a turn. Therefore, the rocket will have traveled fast forward in time with respect to Earth.
However, consider one scenario. Since the rocket cannot use booster all along for its acceleration, the rocket uses nearby celestial bodies to accelerate away from Earth but manages to stay relatively stationary to the center of the Galaxy, while the Earth, along the Sun, travels around the Galaxy. Both the Earth’s and rocket’s relative travel is due to a free fall towards the center of Galaxy and to the nearby celestial bodies, it is Gravitation. The Earth made a round trip and the rocket now maneuvers to land on the Earth.
In this case, which one made a round trip, and which went through an acceleration? Also, how does the Doppler Shift work for both parties?
My conclusion is that the Symmetry never breaks, or the Symmetry breaks in a symmetrical way to both the Earth and the rocket, and the Doppler Shift delay factor by t(del) = d/c to both parties because both stayed in their own “inertial” frame of reference. While the Earth will observe rocket’s time elapsed less, if you had been on the rocket, you will observe the Earth’s time elapsed less. Then, where is the agreement? See the Loaf Model of Space-Time, this explains well.
While Space-Time can have any vector mathematically, since we are locked in our own perceptual Space-Time, one can only meet the other for an infinitesimal period of moment and never again. The original Earth and the rocket never meet up again, each will experience the other newly constructed within each one’s frame of reference.
The Symmetry has never broken, and this is not a way to fast forward time travel.
Next topic will be on the Proper Acceleration, I call it “Inertial Acceleration”, and related phenomena we can observe, including the fast forward time travel.

Originally Posted by tachyon1

Some claim that the etc.

Whut?
Shouldn't this be in New Hypotheses? (At best).

My conclusion is that the Symmetry never breaks

Your conclusion is wrong. Let's say (1) labels the event when both observers start off at rest, and (2) labels the event when, at the end of the experiment, they come back together at rest. The proper times recorded on the clocks then are ( earth's gravity disregarded for simplicity ) :



and



The metric tensor for the traveller is isomorphic, but not identical, to the Minkowski metric, and will be of the general form



This is straightforward during the outbound and inbound segments of the journey, but potentially very non-trivial at the turn-around point. The exact form of the acceleration function there will explicitly depend on the geometry of the world line at that region, i.e. on how exactly the turn-around is executed. However, at no point during the turn-around phase does the acceleration function vanish.

It is easy to see that these metric tensors are not the same for the two observers, hence the two frames are not symmetric, contrary to your conclusion. The only times when these frames are symmetric is when they are both in relative inertial motion ( in which case the metric tensor becomes the Minkowski tensor in both frames ).

While the Earth will observe rocket’s time elapsed less, if you had been on the rocket, you will observe the Earth’s time elapsed less.

This is incorrect. For this to be the case, the two proper time integrals above would need to be the same - which they obviously aren't.

The Symmetry has never broken

Yes it was. See above.

Originally Posted by tachyon1

However, consider one scenario. Since the rocket cannot use booster all along for its acceleration, the rocket uses nearby celestial bodies to accelerate away from Earth but manages to stay relatively stationary to the center of the Galaxy, while the Earth, along the Sun, travels around the Galaxy. Both the Earth’s and rocket’s relative travel is due to a free fall towards the center of Galaxy and to the nearby celestial bodies, it is Gravitation. The Earth made a round trip and the rocket now maneuvers to land on the Earth.
In this case, which one made a round trip, and which went through an acceleration? Also, how does the Doppler Shift work for both parties?

This is a different scenario to the one of the standard twin-clock "paradox". The standard twin-clock "paradox" is in a Minkowskian spacetime and the explanation given for that is valid. Your scenario involves complicated motion in curved spacetime and the explanation given for Minkowskian spacetime does not apply, nor does it invalidate the explanation given for Minkowskian spacetime. General relativity will provide the correct answer for your scenario, but there is no reason to expect that it will be as simple as the standard twin-clock "paradox".

Just as a further exercise, and one I think is more direct for the theory, how would GR handle the "twin paradox" if the situation were instead two spaceships that both synchronized their total accelerations to be identical over the course of the journey, but not their total time spent in linear motion?

Eg. Take 2 identical spaceships A and B, who are both positioned next to each other and who's stopwatches are both showing zero. A and B then both start their stopwatches and accelerate to .99c at the same rate, but then upon reaching that velocity, A immediately decelerates back to rest at the same rate, leaving B to travel constantly at .99c for some time, t. Then, B decelerates to rest, then immediately accelerates back up to .99c in the opposite direction, as does A. A once again decelerates to rest upon reaching .99c, leaving B to travel back towards it for the same time, t, then B decelerates back to rest, to meet back up with A and compare their stopwatches.

Considering the total accelerations were the same (both A and B underwent the same rates of accelerations over the same duration), the stopwatch on ship B should theoretically be further behind that of ship A, since it spent more time traveling at a higher velocity than A. The accelerations of the ships would seemingly be unable to be used as a reason, due to them being identical for the round trip. This, in turn, would seem to imply absolute motion (acceleration in itself kind of already does anyway).

Originally Posted by Waveman28

Considering the total accelerations were the same (both A and B underwent the same rates of accelerations over the same duration), the stopwatch on ship B should theoretically be further behind that of ship A, since it spent more time traveling at a higher velocity than A

I think you are confusing proper time and coordinate time. What the stop watch measures is proper time, and that is dilated only by the presence of acceleration, but not relative motion. In the specific scenario you have presented ( as in the standard twin paradox ), the outcome ( difference in clock readings ) will depend only on the exact acceleration profiles of both observers, but not on the phases of inertial motion, since those do not contribute to the time dilation experienced.

OK, is g a function of x?, Then, u should also be a function of x', when viewed from the rocket, then the equation is symmetric. Your equation states that g is a function of x but u is not a function of x’, correct?

(I have been away from academia for decades, so my math is rusty, anyway)

More precisly, the symmetry is broken mutually to both observers, respectively.


It is important to define the acceleration precisely.
1. Gravitational Acceleration = Free Falling = Free Drifting = Local Inertial Reference Frame
2. Geometric Acceleration = (d^2)x/d(t^2) of one point with respect to another point
3. Inertial Acceleration, a(i) = a(geo) - g, where the pseudo gravity takes in place (they call it "proper acceleration")


On the surface of the Earth, a(i) = -g

The correct definition of the Inertial Frame is where a(i) = 0, or a(geo) = g(local value).

In this scenario, except the rocket briefly underwent an Inertial Acceleration using booster, which we neglect here, both the Earth and the rocket stayed in their own "Inertial" frames throughout the trip, though they did not stay in the same “Local” Inertial Frames in terms of (d^2)x/d(t^2).

The definition of the proper time essentially falls back to “my perceptual time”. Both the Earth and the rocket have their own “my perceptual time”

I think we all have fundamental understanding, except, what is "acceleration"?

If we define the Acceleration in terms of the Inertial Acceleration, it is the realm of General Relativity, and yes, this results in a fast-forward movement in time, and there is no confusion regarding Twin Paradox. KJW is correct in saying that Minkowskian relation does not apply in this case, or it applys symmetrically to both observers. If I remember correctly, Minkowskian relation is for a Euclidean view of the Space-Time.

The time dilation of the astronaut should only result from the Inertial Acceleration that the astronaut went through, as Markus said.

It seems like our precise understanding of the "accelerations" is the stepping stone in this issue. Depending on the volume of understanding audience, my next session will be on some phenomena relating to the Inertial Acceleration.

By the way, if you are moving backwards in time, would you remember the past or the future?

I think what Waveman is describing involves a third observer, which is not the case I raised a question on.

Originally Posted by tachyon1

It is important to define the acceleration precisely.
1. Gravitational Acceleration = Free Falling = Free Drifting = Local Inertial Reference Frame
2. Geometric Acceleration = (d^2)x/d(t^2) of one point with respect to another point
3. Inertial Acceleration, a(i) = a(geo) - g, where the pseudo gravity takes in place (they call it "proper acceleration")

For the twin clock "paradox", acceleration has to be the general relativistic definition. Newtonian theory and general relativity has the opposite viewpoint of acceleration with regards to gravity. For example, if I throw a clock upwards and catch it when it returns, it is me who has experienced less time than the clock because it is me who has accelerated while the clock is in inertial motion.

Hey, at last, some body has a correct understanding. Do you know how long it took me to find a person like you? ... 30 years, not that others have zero understanding.
Like I said, my math is rusty and I lack time due to my profession in a different field, while it takes devoted hours to put out precise mathematical expression, not a copy-paste version form the internet that makes you look professional.

I think you confirmed my question, and let me know what you think of the Loaf Space-Time Model. I think this makes good sense in a lot of ways, and as a result, in a more realistic situation of the acceleration, the rocket makes same conclusion on the Earth's time delayed as the Earth made on the rocket, if we consider the two frames separately and respectively. I just needed that one more person other than myself who has the same understanding vector, though you may not agree on the Loaf Model.

Once this topic is settled, I have other topics to share, from the definitions of the accelerations. The Inertial Acceleration (proper, they say) seems to squeeze a lot of juice that we can use for the interpretation of our Space and Time.

By the way, throwing and catching of the clock is an Inertial Acceleration (non-Gravitational), so maybe this is not an accurate example? .... Anyway, it is enough that you have the right view point. The mathematical detail is not simple to achieve but a correct concept is a good stepping stone.

Originally Posted by Markus Hanke

My conclusion is that the Symmetry never breaks

Your conclusion is wrong. Let's say (1) labels the event when both observers start off at rest, and (2) labels the event when, at the end of the experiment, they come back together at rest. The proper times recorded on the clocks then are ( earth's gravity disregarded for simplicity ) :



and



The metric tensor for the traveller is isomorphic, but not identical, to the Minkowski metric, and will be of the general form



This is straightforward during the outbound and inbound segments of the journey, but potentially very non-trivial at the turn-around point. The exact form of the acceleration function there will explicitly depend on the geometry of the world line at that region, i.e. on how exactly the turn-around is executed. However, at no point during the turn-around phase does the acceleration function vanish.

It is easy to see that these metric tensors are not the same for the two observers, hence the two frames are not symmetric, contrary to your conclusion. The only times when these frames are symmetric is when they are both in relative inertial motion ( in which case the metric tensor becomes the Minkowski tensor in both frames ).

While the Earth will observe rocket’s time elapsed less, if you had been on the rocket, you will observe the Earth’s time elapsed less.

This is incorrect. For this to be the case, the two proper time integrals above would need to be the same - which they obviously aren't.

The Symmetry has never broken

Yes it was. See above.

To proceed, from Markus' excellent post:








But, (by definition, it is the gravitational potential), so, meaning that independent of the integration path from 1 to 2.

The above is equally true in accelerated frames (Rindler coordinates) as well as in the presence of a gravitational field (through the equivalence principle). This simple math answers Waveman question as well.

This equation is with respect to the Earth, and we can construct the same with respect to the rocket. I think you first need to understand the three definitions of accelerations before you allpy this equation. Also, this is Euclidean expression, and the Twin Paradox only gets resolved when we apply the General Theory of Relativity, with Inertial Acceleration (proper acceleration). Before you say "wrong", you need to understand the equation you wrote more precisely.

Once again, in my scenario, both the rocket and Earth maintains a free fall, where the difference in the gravitational potential applies to both because they are both in their own Inertial Reference Frames.

Originally Posted by tachyon1

This equation is with respect to the Earth, and we can construct the same with respect to the rocket. I think you first need to understand the three definitions of accelerations before you allpy this equation. Also, this is Euclidean expression, and the Twin Paradox only gets resolved when we apply the General Theory of Relativity, with Inertial Acceleration (proper acceleration). Before you say "wrong", you need to understand the equation you wrote more precisely.

Another crackpot just joined in....

I should remark that my previous post required nothing more than a proper understanding of the equivalence principle and the standard twin-clock "paradox". Over the small distance scale involved (so that tidal effects can be ignored), the clock's inertial frame of reference can be considered to be Minkowskian, making me the "travelling twin" relative to the inertial clock.

I often read statements where the notion of a straight line in general relativity is different to the normally understood notion of a straight line. I consider it one of my missions to eradicate this viewpoint by pointing out that what general relativity regards as a straight line is truly a straight line even in the normally understood sense. The equivalence principle is key to this mission.

Originally Posted by tachyon1

OK, is g a function of x?

Yes, because it is not uniform along the world line.

Inertial Acceleration

This is a not a term I am familiar with. By definition, an inertial frame is one in which an accelerometer would read zero ( i.e. space-time is homogeneous, isotropic and independent of time ), so the term "inertial acceleration" makes little sense.

More precisly, the symmetry is broken mutually to both observers, respectively.

Of course. The frames simply aren't symmetric, as demonstrated earlier.

It is important to define the acceleration precisely.

It seems like our precise understanding of the "accelerations" is the stepping stone in this issue.

The definition is quite simply the covariant derivative of the 4-velocity with respect to proper time :



This is the only definition which is valid for all observers.

The definition of the proper time essentially falls back to “my perceptual time”.

Proper time is what clocks physically record, so it is an invariant quantity. Mathematically, it can be understood to be the length of the ( appropriately parametrised ) world line through space-time.

xyzt - I don't like arguing but it seems like you just cut and paste the equations, not knowing how to truly interpret those equations in different situations.

Originally Posted by tachyon1

xyzt - I don't like arguing but it seems like you just cut and paste the equations,[

I did not cut and paste, I used the same notation as Markus for reasons of logical continuity.

not knowing how to truly interpret those equations in different situations.

Let me assure you that people who know physics understood my post perfectly. Doesn't apply to you. Based on your posts, you are in no position to judge.

Originally Posted by tachyon1

Once again, in my scenario, both the rocket and Earth maintains a free falls

In order for us to meaningfully compare proper times, the two observers must start off at rest in the same frame of reference ( earth's surface ), and end up the same way again; how do you separate them ( i.e. change their state of relative motion ) without applying acceleration to one or both of them ? Surely you are not proposing that that rocket free-falls away from the earth, whereas the earth itself free-falls away from the rocket...

Originally Posted by Markus Hanke

Originally Posted by tachyon1

OK, is g a function of x?

Yes, because it is not uniform along the world line.

Inertial Acceleration

This is a not a term I am familiar with. By definition, an inertial frame is one in which an accelerometer would read zero ( i.e. space-time is homogeneous, isotropic and independent of time ), so the term "inertial acceleration" makes little sense.

More precisly, the symmetry is broken mutually to both observers, respectively.

Of course. The frames simply aren't symmetric, as demonstrated earlier.

It is important to define the acceleration precisely.

It seems like our precise understanding of the "accelerations" is the stepping stone in this issue.

The definition is quite simply the covariant derivative of the 4-velocity with respect to proper time :



This is the only definition which is valid for all observers.

Excellent post, Markus. In flat spacetime the Christoffel coefficients are all null () so the proper acceleration coincides with the derivative of the speed wrt proper time ().
In a uniform gravitational field, the Christoffel coefficients are not all null, so, the proper acceleration no longer coincides with the derivative of the speed wrt proper time.

KJW - I agree, I am going to spend more time to articulate this topic. People who are good in non-Euclidean geometry are still very much fixed in a Euclidean picture of the interpretation of the equations. Unless this topic is clear enough to more people, maybe not all, I cannot post the next topic, as this topic is a prerequisite to the next step. You are correct in saying that a clear understanding of the Equivalence Principle is the key to the understanding of this topic, not just a copy-paste of equations that are specific to one scenario.
I still want to hear from you regarding the Loaf Model. I may have to re-join next week for more specifics.

Originally Posted by tachyon1

not just a copy-paste of equations that are specific to one scenario.

Not sure who you are referring to, but what I posted is pretty much as general as it gets.

If you are accelerating towards a very far, very large mass, how do you define a flat Space-Time? There is no such a thing as flat Space-Time in reality. If you are free falling, you are in a local frame of reference and therefore you are in a flat Spacetime with respect to you but not necessarily with respec to another in a different inertial frame of reference. The Christoffel, I think, depends on the reference fram you are in and it is also relative.

Originally Posted by Markus Hanke

Originally Posted by tachyon1

not just a copy-paste of equations that are specific to one scenario.

Not sure who you are referring to, but what I posted is pretty much as general as it gets.

He's trolling you, there is no meaning to his posts, he's just another pretender.

Originally Posted by tachyon1

The Christoffel, I think, depends on the reference fram you are in and it is also relative.

What gives you this bright idea? You obviously have no clue, yet you make all kinds of pronouncements.

I understand your post, but I was trying to add my understanding to the conventional understanding. I think KJW understands.

With respect to O(0), the proper time is calculated t = t(0)[sqrt(1-v^2/c^2)]
At the same token, if you had been O(1), the proper time is also t = t(1)[sqrt(1-v^2/c^2)], meaning that the entire reference point is still with respect to you, except that you are now O(1).

You see the symmetry right here.

Once again, the accelration defined by a = (d^2)x/d(t^2) - g(local) "only" contributes to the breaking of the Symmetry, and as KJW mentioned, this requires a clear understanding of the Equivalence Principle, not just using a century old equation.

I took Tensor Analysis 20 years ago, just not my second nature any more but I understand with time spent. I am just rusty, not clueless.

Originally Posted by tachyon1

I understand your post, but I was trying to add my understanding to the conventional understanding. I think KJW understands.

With respect to O(0), the proper time is calculated t = t(0)[sqrt(1-v^2/c^2)]
At the same token, if you had been O(1), the proper time is also t = t(1)[sqrt(1-v^2/c^2)], meaning that the entire reference point is still with respect to you, except that you are now O(1).

You stumble right out of the gate, with the basic stuff. Why do you keep pretending to understand the more advanced topics when you clearly don't know the basics?

Originally Posted by tachyon1

The Christoffel, I think, depends on the reference fram you are in and it is also relative.

The connection coefficients are a function of the derivatives of the metric, so they will depend on the choice of coordinates. The Christoffel symbol in itself is thus not covariant, however, the covariant derivative obviously is ( as the name implies ).

Stop giving me BS, I don't know everything but I have a clear understanding of the topic I am dealing with. OK, you have better conventional math skill, maybe it is the blockade for you to understand new concepts. This is not new to me, this was my question not answered since highschool time, and I can't make the long story short. You are not applying your math in the right way. Also, I can't waste imte on illogical slurs and condemnation.

Originally Posted by tachyon1

Stop giving me BS, I don't know everything but I have a clear understanding of the topic I am dealing with.

Based on your posts, you clearly DON'T. You just pretend to do.

Originally Posted by tachyon1

Some claim that the Twin Paradox is explained by the fact that that the Symmetry is broken when taking the relative Doppler Shift into a consideration.

The symmetry is broken due to the fact that one twin accelerates, the other one doesn't. Nothing to do with any Doppler effect.

OK, I will resume next week, I must go for now. At least there is some heat generated. What you guys are talking about is just the geometry itself. You are not considering the effect of the Equivalence Principle, and the three categories of acceleration as I mentioned. The effect is different when the acceleration is by Gravity and when it is by External Force such as e-m force, or by propulsion. The time dilation is only influenced by the a = d(x^2)/dt(^2) - g(local) factor, not by d(x^2)/dt(^2).

There is a delay in Doppler Shift from red to blue to both parties.

Originally Posted by tachyon1

OK, I will resume next week, I must go for now.

Try to stay gone, refrain from spamming more nonsense for a while.

At least there is some heat generated.

More like noise.

What you guys are talking about is just the geometry itself. You are not considering the effect of the Equivalence Principle,

What gives you this bright idea?

and the three categories of acceleration as I mentioned.

Tough, we refuse to "learn" from crackpots.

The effect is different when the acceleration is by Gravity and when it is by External Force such as e-m force, or by propulsion. The time dilation is only influenced by the a = d(x^2)/dt(^2) - g(local) factor, not by d(x^2)/dt(^2).

Excellent, you went full out crank.

Your very statement "start off at rest in the same frame of reference (earth's surface)" = "start off at rest in the same frame of reference (rocket's surface)". Why are you so math smart and yet so physics stupid? Once again, "at rest" can only mean free falling or freely drifting frame of reference, which is equivalent to a local inertial frame of reference, which inevitably was the original condition for both.

The rocket is using the nearby celestial bodies for its acceleratin away from Earth since the booster is very very very very limted. Isn't this how all our space crafts accelerate away from Earth into the deep space? This is equivalent to a free fall.

In my scenario, both the rocket and the Earth are free falling away from each other, then merging each other yet again via a free fall. I made this very clear to begin with, and I don't know how else to make it more clear - revisit the Principle of Equivalence, and just disregard the tidal effect for our practice.

As a result, as much as there is no absolute velocity, there is no absolute acceleration.

It looks like you are book smart but not too intellectual in thinking.

Three categories of acceleration is not even my bright idea, if you understand the Equivalence Principle, it is a common sense that not too many people looked into.

xyzt - I don't think you belong here. Stay away from my post.

I got this bright idea 30 years ago in high school but I was very disappointed becasue it was the very fundamental of the Equivalence Principle, and was not my idea at all.

Originally Posted by tachyon1

Isn't this how all our space crafts accelerate away from Earth into the deep space? This is equivalent to a free fall.

No. They all start off at rest on the earth's surface, and are accelerated into the starting point of their trajectories by rockets, which use thrusters. All our space-craft undergo an initial period of acceleration. Of course you can say also that the Earth starts off on the rocket's surface, but that doesn't change the fact that the rocket observer will measure acceleration on his trajectory through space, whereas the Earth observer won't. The two frames are not symmetric, and the rocket thrusting off the surface is certainly not equivalent to a free fall.

As a result, as much as there is no absolute velocity, there is no absolute acceleration.

Proper acceleration is a physically measurable quantity on which all observers agree, because no outside point of reference is required to define it. Velocity on the other hand is relative and thus observer-dependent; in the absence of another point of reference, velocity cannot be determined. Again - the rocket observer feels acceleration when he thrusts off the Earth's surface, whereas the Earth-bound observers left behind do not.

It looks like you are book smart but not too intellectual in thinking.

In that case it seems that our frames of reference are symmetrical, because to me you are thinking too much, but lack the prerequisite knowledge of the basics.

The effect is different when the acceleration is by Gravity (...), or by propulsion.

You are not considering the effect of the Equivalence Principle

Huh ?

The time dilation is only influenced by the a = d(x^2)/dt(^2) - g(local) factor, not by d(x^2)/dt(^2).

Originally Posted by tachyon1

I got this bright idea

What idea, exactly ? Can you formulate it in a general and concise manner ?

Originally Posted by Markus Hanke

Originally Posted by tachyon1

I got this bright idea

What idea, exactly ? Can you formulate it in a general and concise manner ?

Don't feed the troll, let him starve.

Originally Posted by tachyon1

I still want to hear from you regarding the Loaf Model.

I don't understand the model enough to comment at this point in time. However, at the outset, I will say the following:

If it disagrees with general relativity (or a standard extension of it such as Einstein-Cartan theory), you will find it extremely difficult to convince me.

If it agrees with general relativity, then I'll need to be convinced of what it adds to the understanding.

Originally Posted by KJW

Originally Posted by tachyon1

I still want to hear from you regarding the Loaf Model.

I don't understand the model enough to comment at this point in time. However, at the outset, I will say the following:

If it disagrees with general relativity (or a standard extension of it such as Einstein-Cartan theory), you will find it extremely difficult to convince me.

If it agrees with general relativity, then I'll need to be convinced of what it adds to the understanding.

Don't feed the troll, let him starve.

KJW - OK, I will spend time to articulate the details of thei model. Let's discuss.

I put down the acceleration equation - I don't think you belong here at all. The fact that you put out slur is the fact that you feel defeated, though my purpose is not to defeat anyone. Go away.

I think you need to have better understanding of the Equivalence Principle - the rocket acceleration by thruster is minmal compared to the entire trip. There is no absolute acceleration "according to the Equivalence Principle". If you insist that the Earth is the standard point, then you are saying that the Earth is the center of the Universe. There is no standard velocity, there is no standard acceleration.

Originally Posted by tachyon1

I put down the acceleration equation -

....which is utter rubbish.

I don't think you belong here at all. The fact that you put out slur is the fact that you feel defeated, though my purpose is not to defeat anyone. Go away.

I agree , the purpose is to sweep away trolling cranks. You don't belong here, you belong in Pseudoscience.

It is so amazing to discover supid individuals like you everyday.

Originally Posted by tachyon1

It is so amazing to discover supid individuals like you everyday.

Markus - I think at least you have some understanding of where I come from by now, except you are having a hard time deviating from your conventional thinking.
Your statement, “Again - the rocket observer feels acceleration when he thrusts off the Earth's surface, whereas the Earth-bound observers left behind do not.” Let me explain to you in a concise manner.

1. According to the Equivalence Principle, the booster acceleration is equivalent to standing on surface of the Earth, though the values may differ – let’s even disregard this factor for simplicity
2. The rocket can, then, only experience the time delay factor during its acceleration “by the booster”, which makes the long distance travel completely meaningless – this is in agreement with KJW’s statement

Can you give me a logical answer to this?
I am going to put names in order to avoid confusion.
KJW – you do understand my point but cautious in admitting because you don’t know where I come from. This is understandable and I don’t expect a full agreement from a stranger.
“xyzt” – you are completely out of whack – stop your slur, it is only an indication of your inferiority. A dog barks when it is in fear, a confidence makes you silent. OK, let’s say I am an idiot, then explain to me in a logical way rather than slur, so I understand. Otherwise, stay away permanently. I did not start this slur, you did.
I understand all the Tensor, Minkowski, … you guys use. I am only cautious in using those equations because I have been away from academia for so long and I don’t want to put out an inaccurate math. Soon, I am going to put together math for everyone.

Sorry, I am new to the layout of this forum, I did not mean to tackle PhDemon, I meant to say to xyzt

Markus Hanke - are you the owner of this forum?

Originally Posted by tachyon1

Markus Hanke - are you the owner of this forum?

Member Markus Hanke is one of the moderators. He oversees sub-fora and he has the ability to edit and delete posts, move and close threads, etc.
If you are looking for the owner (member Admin​) of the Science Forum, then scroll to the bottom of the page and click "Contact Us".

Originally Posted by tachyon1

KJW – you do understand my point but cautious in admitting because you don’t know where I come from.

You have fully agreed with me on a point I made involving an understanding of general relativity, so in spite of your unconventional approach, I am willing for the time being to give you the benefit of the doubt.

Cogito, thank you for your response. Different opinion, different belief, ... it's all alright, but a meaningless attack via slurs, I don't think it should be allowed in this forum, just because I have a different idea. I was a physicist and I understand the conventional thinking clearly, and this topic is my 30 year old question that none of the then known professors was able to answer "logically". OK, I respect Markus Hanke's knowledge level, but we ended in a corner now that the booster based acceleration is the only acceptable portion of the acceleration (this was my original statement), and you can see this from Markus's later statement. I would like to see Markus's answer to my latest question, and I am eager to see his logical answer. Would you not agree? Just because of the difference in understanding, you can't just repell the idea. If you are only accepting the conventional ideas, then they are all over the internet and this forum has absolute zero value, except to boast your conventional knowledge. I don't think I need to go to the owner at this point, I just want to hear Markus's logical answer to my latest question regarding the booster based acceleration and the free fall(=free drift) based acceleration. Lets try hard to remove politics to the best we can, and get to the bottom of pure physics. I am a Wall Street worker (not too low a position at all) and I know the politics I have to challenge everyday.

Besides, I have a lot of topics to put out but if the audience is not there, I am only willing to put them in trash, I am getting old and things are meaningless to me to fight for. Why bother? I was truly surprised that KJW was the only fellow( not the first, however) who grasped the principle, not that he had to agree with all my ideas.

Thank you,

KJW, we are living in a tough world. Yes, I will try my best in the midst of this bread-n-butter world, which is what I do at work. Give me some time, I will try hard until my heart stops. Thank you for not dumping my idea to trash (do I really have to say this much?). At this point, you just have to know that we both understand the Principle of Equivalence at an acceptable level. Talk to you soon.

Originally Posted by tachyon1

According to the Equivalence Principle, the booster acceleration is equivalent to standing on surface of the Earth, though the values may differ

I highlighted in the important bit. The two frames are not symmetric, because accelerometers in both will show different values. They can thus be physically distinguished, unlike inertial frames.

The rocket can, then, only experience the time delay factor during its acceleration “by the booster”, which makes the long distance travel completely meaningless

Correct, provided of course that there is no other acceleration happening along the world line. Also, if you control the rocket's trajectory by gravitational slingshot ( as you seem to imply ), then the rocket will spend at least some time under the influence of a gravitating body - even though it is in free fall, its clock will still get gravitationally dilated during that time. What I am trying to say to you is that the long distance travel is meaningless only if there is no acceleration and no gravity present at all ( i.e. if the journey is purely inertial all the way ). If there is either, as seems to be the case here, then dilation accumulates along the world line.

OK, I respect Markus Hanke's knowledge level, but we ended in a corner now that the booster based acceleration is the only acceptable portion of the acceleration

See above. Gravitational influence due to the presence of massive bodies will also dilate the on-board clock.

Markus Hanke - are you the owner of this forum?

No. I am just a member of the moderator team, and I moderate some other platforms also.

Markus, you are still disregarding the fact that Earth is gravitationally accelerating with respect to the center of the Galaxy, and the very fact that there is no measure as to which is accelerating at a higher rate. You are in an inertial reference frame (a local inertial reference fram to be more precise) as long as you are gravitationally pulled or freely drifting, and this is the very fundamental of the Equivalence Principle. Besides, an accelerometer can only register d(x^2)/(d^2)/t - g(local value). Accelerometer does NOT register Gravitational pull - basic physics

"Gravitational influence due to the presence of massive bodies" is a realm of General Relativity, and we will discuss this later. Besides, name on celestial body that is truly not under the influence of gravity. If you observe a free falling object, the curvature is not flat, but if you fall along with the free fallling object, then the curvature flattens (work out your Tensor Analysis), so you do not experience the difference between free fall and free drift.

Originally Posted by tachyon1

If you observe a free falling object, the curvature is not flat, but if you fall along with the free fallling object, then the curvature flattens (work out your Tensor Analysis), so you do not experience the difference between free fall and free drift.

A point of clarification is in order here:

The Riemann curvature tensor is a covariant quantity such that whether or not it is zero is independent of the frame of reference (coordinate system). Because an accelerated frame of reference is just a coordinate transformation of an inertial frame of reference, whether or not an observer is accelerating does not affect whether or not the spacetime is curved. Spacetime curvature manifests itself as tidal effects that are present even in inertial frames of reference. However, for an arbitrary spacetime, it is possible to establish a coordinate system such that at an arbitrary single point, the metric tensor is Minkowskian and the Christoffel symbols (connection) are zero. It is this that is the equivalence principle. Indeed, there is even the stronger statement that it is possible to establish a coordinate system such that the metric tensor is Minkowskian and the connection is zero along an arbitrary curve. What is not possible in an arbitrary spacetime is to establish a coordinate system such that the metric tensor is Minkowskian and the connection is zero over an extended region of the spacetime, as the Riemann curvature tensor defines an obstruction to the existence of such a coordinate system.

Originally Posted by tachyon1

"Gravitational influence due to the presence of massive bodies" is a realm of General Relativity

Well, isn't that what we are discussing all along ? This is about gravitational time dilation of proper times.

Besides, name on celestial body that is truly not under the influence of gravity.

There isn't any, nor did I claim there was.

If you observe a free falling object, the curvature is not flat, but if you fall along with the free fallling object, then the curvature flattens (work out your Tensor Analysis), so you do not experience the difference between free fall and free drift.

Yes, but that is true only locally, in a region on the order of 1/g, not for the entire trajectory of the rocket in your example. As KJW rightly points out :

What is not possible in an arbitrary spacetime is to establish a coordinate system such that the metric tensor is Minkowskian and the connection is zero over an extended region of the spacetime, as the Riemann curvature tensor defines an obstruction to the existence of such a coordinate system.

Thus you need to account for the effects of curvature, which, in the general case, will have an impact on the proper time recorded in the rocket. A much simpler example is a satellite with a clock on board ( e.g. a GPS satellite ) - even though the satellite is technically in free fall, its clock does not tick at the same rate as a reference clock on earth ( or in another orbit ), because these clocks are located in different gravitational potentials. This needs to be accounted for.

Generally speaking, my point is just this - even if it were possible for the rocket to be in free fall along its entire trajectory, we would still potentially record a different proper time than a stationary reference clock travelling through the same two fixed events. The proper time recorded between fixed events depends on the geometry of the clock's world line through space-time, which is affected by both acceleration and gravity. That is the meaning behind the first line integral I gave in post 3.

KJW – you know something. OK, correct me if I am wrong, KJW’s statement is simple. When a = g(local), “a” meaning (d^2)x/d(t^2) = g(local) to a nearby center of mass, you are in a “local” inertial frame of reference (zero curvature). When a <> g(local), you are not in a local inertial frame of reference (non-zero curvature). Once again, any acceleration under Gravity is considered zero acceleration (local inertial frame of reference) by the principle of Equivalence. Minkowski diagram applies to each of those “local” inertial reference frames, not an inch deviated from that point. As long as you are either sling shot or free fall or free drift under Gravity, less the tidal force, you are in a local inertial frame of reference. Therefore, both the rocket and the Earth are in their own inertial frame of reference, in which case under Einstein’s principle, they are both NOT accelerating.
The delayed clock of the Gemini astronaut, is it a one time factor under booster acceleration or is in a continual cumulative delay factor that builds up over the Earth time? Likewise, is the GPS time delay factor a one- time delay factor while booster accelerating to the orbit, or a continual, cumulative delay factor?
Please re-visit the Equivalence Principle and understand the definition of acceleration. Then we resume. Somehow, I don’t want to let go of this discussion. I am not putting out a new idea at all, just a re-visit of the old theory that not many had clear understanding of.
The reason I am not putting out math is 1) copy paste is easy 2) I have to add to existing math 3) In order to do this, I need to establish the principle first (I already have). Math without the correct principle is correct in itself, but not in it’s resultant.

My point is this, the equation for a proper time with respect to you on the other applies to the other on you in a symmetric way. Therefore, if one observer observes a delayed time of the other, the other also observes a delayed time of the one.

Please re-visit the Principle of Equivalence, and understand what acceleration really means.

Markus, your (actually not yours) equation is correct, however, it is correct bothways.

By the way, I am mourning the fact that I had to depict the center of our Galaxy as the center for acceleration. There is no way to define the center of anything in fact, this was my point all along.

If you are standing on the surface of the event horizon, the space-time is very curved, but if you are free-falling into the black hole (less the tidal effect), your relative curvature is zero - again, the Principle of Equivalence.

Suppose you and I were initially together, then we go our separate ways, and eventually meet again at some point in the future. But we are somehow able to directly observe each other's clocks the entire time we are separate. Relative to an inertial frame of reference local to you, whenever you accelerate, you are changing your velocity and hence the Doppler effect of the light emitted by your clock. This change in the Doppler effect propagates to me at the speed of light and I see this change delayed. But when you accelerate, because you are changing your velocity relative to the inertial frame of reference local to you, you see a change in Doppler effect of the light that is local to you at the time you changed velocity. This light is also delayed from when I emitted it, but it is your change in velocity that resulted in the immediate change in Doppler effect you observe.

To summarise, an observer observes the effect of their own change in velocity immediately, but the effect of a distant object's change in velocity only after the light from that change has reached them.

Note that in the above, acceleration is always the general relativistic definition.

Originally Posted by tachyon1

My point is this, the equation for a proper time with respect to you on the other applies to the other on you in a symmetric way

Not it doesn't, because we are in different gravitational potentials, even if and when both observers are in free fall. In other words - both observers use the same equation, but they don't use the same metric tensor in it.

If you are standing on the surface of the event horizon, the space-time is very curved, but if you are free-falling into the black hole (less the tidal effect), your relative curvature is zero - again, the Principle of Equivalence.

And again, the in-falling observer's clock will be gravitationally dilated along his world line as compared to a reference clock far away outside the black hole ( as much as this even makes sense ), even though he is in free fall. Zero proper acceleration ( free fall ) in the vicinity of a massive body means that the observer falls along a geodesic, which is a straight line ( parallel-transport law ! ) in globally curved space-time.

Please re-visit the Principle of Equivalence, and understand what acceleration really means.

I do understand quite well what it means - it is what an accelerometer at rest in a frame will physically measure, just like proper time is what a clock at rest in a frame will measure. It is very simple.
I have a feeling though that you don't seem to understand the difference between local and global in a curved space-time; I urge you to revisit that ( start with KJW's excellent post ), it is crucial for this scenario where gravity is intrinsically involved. Just because your frame of reference appears locally Minkowskian does not mean you can extend that trivial coordinate system to span your entire in-fall trajectory.

Originally Posted by Markus Hanke

Just because your frame of reference appears locally Minkowskian does not mean you can extend that trivial coordinate system to span your entire in-fall trajectory.

Actually, you can all the way along your worldline, just not away from it. This is why our local frame of reference appears constantly Minkowkian and not just at some instant in time.

Originally Posted by KJW

Actually, you can all the way along your worldline, just not away from it. This is why our local frame of reference appears constantly Minkowkian and not just at some instant in time.

Sorry for having been imprecise, and yes, you are right of course - what I really meant to say was that in the presence of energy-momentum you can't extrapolate your locally flat coordinate system to global space-time, not just the trajectory alone.
I really need to be more careful how I word things

One further point I should mention that is relevant is that the ability to locally transform to the Minkowskian metric and zero connection also implies the ability to locally transform to any other flat spacetime metric and connection. Thus, the frame of reference of an observer standing on the earth can be considered to be locally Rindler, thus directly expressing the equivalence between gravity and acceleration.

KJW, eacept the fact you like jargon too much, you are very close. I am beat from long hours of work but I am off tomorrow. Tomorrow, I will use a very simple example and simple question. I want to test you both to see if you are off the course and to see if it is worth my further trial. I have no time to waste! - I am disappointed that you put my post into the pseudo science before even catching the key point. You can use complex math, I understand all, but my point is that a correct math based on a wrong reference point will only result in a wrong result. I always first get the principle idea settled, then add math one step at a time - the application of math has to be precise. I shared this with my colleagues - PhD, engineer, math professional, of course, a few of whom were my actual colleagues from school, and they were surprised that the audience of this forum had no understanding of the very basic concept with so much jargno they use, I really must say.

This is not too precise a statement, I will talk to you tomorrow.

"local frame of reference appears constantly Minkowskian" = Free Fall or free Drift, less tidal effect (tidal effect can always be mathematically removed, so this is really not an issue ). Booster based acceleration <> Minkowskian (= the realm of General Relativity) - how many more times do you want me to repeat?

Thanks greatly to your inputs, I have decided not to get into the Delay Effect in Doppler Shift in this forum, I am done at the most elementary level.

The key points were.

In Special Relativity, the Lorentz Transformation for Length and Time Contraction and Mass Increase was "Symmetric" both the the ground observer and the moving observer (the equation is not inverse) - Twin Paradox does not get resolved in Special Relativity.

General Relativity resolves the Twin Paradox, but it is complex.

You imply all the Gravitational pulls to the rocket only, when Earth is pulled towards the center of Galaxy, it is also Gravilationally pulled and accelerated, therefore the clock of Earth will also be Gravitationally dilated by the Grand Center of Galaxy, far larger than the celestail bodies the rocket uses (actually the Gravitational time dilation is not precisely understood by people - it will take me a long way to explain, but I decided to use this analogy to meet your level. Also, the rocket's total acceleration by Gravity and the Earth's total acceleratin by Center of Galaxy actually cancel each other in net value) - therefore the symmetry being maintained.

Originally Posted by tachyon1

I am disappointed that you put my post into the pseudo science before even catching the key point.

Personally, I would've placed it in the "New Hypotheses and Ideas" section of the forum, but it was not my decision to make.

Originally Posted by tachyon1

General Relativity resolves the Twin Paradox, but it is complex.

It is not complex at all, it is very straightforward, and really just a one-liner - see earlier posts.

You imply all the Gravitational pulls to the rocket only, when Earth is pulled towards the center of Galaxy

Why are you adding a third frame of reference to a very simple scenario ? There is no need, it will only obfuscate the main point. But have it your way - as earth is in free fall wrt to the sun ( which is what it orbits, not the galactic centre ), so the rocket feels the sun's gravitational influence as well. So what have you gained ?

therefore the clock of Earth will also be Gravitationally dilated by the Grand Center of Galaxy

Same for the rocket.

therefore the symmetry being maintained.

The gravitational potentials are still different, hence no symmetry.

Booster based acceleration <> Minkowskian (= the realm of General Relativity)

GR is precisely what is not needed in Minkowski space-time.

it will take me a long way to explain, but I decided to use this analogy to meet your level.

You are an incredibly arrogant and snobby individual, do you know that ? And you wonder why people react the way they do to your presence - go figure.

Originally Posted by tachyon1

"local frame of reference appears constantly Minkowskian" = Free Fall or free Drift

There is a subtle issue concerning what I said that is a potential source of confusion. What I was referring to was not an inertial frame of reference. It can be shown that for an arbitrary curve in an arbitrary spacetime, there exists a coordinate system in which the connection is zero along that curve (implying that the metric can also be Minkowskian along that curve). The important point is that this coordinate system is not the frame of reference of the observer whose worldline is that curve in that the observer is accelerating in the traditional sense with respect to the coordinate system (because the connection is zero along the curve).

Originally Posted by tachyon1

tidal effect can always be mathematically removed, so this is really not an issue

It cannot be removed by any coordinate transformation. The tidal effect is a manifestation of the spacetime curvature via the geodesic deviation equation. The non-zero Riemann curvature tensor is the mathematical obstruction to coordinate transforming an arbitrary connection to zero

Originally Posted by tachyon1

Booster based acceleration <> Minkowskian (= the realm of General Relativity)

A coordinate system need not represent the frame of reference of any observer. Thus, while the frame of reference of an accelerated observer in flat spacetime is Rindler, not Minkowskian, the underlying spacetime is still Minkowskian in that there exists a Minkowskian coordinate system for this spacetime.

Originally Posted by tachyon1

In Special Relativity, the Lorentz Transformation for Length and Time Contraction and Mass Increase was "Symmetric" both the the ground observer and the moving observer (the equation is not inverse) - Twin Paradox does not get resolved in Special Relativity.

The standard twin clock "paradox" is perfectly resolved in Special Relativity. One only needs to consider the path-lengths in spacetime of the worldlines of the two clocks within an inertial frame of reference. This inertial frame of reference need not be the frame of reference of any of the clocks. While it is interesting to consider the twin clock "paradox" from the perspective of each of the clocks, it is not necessary to do so.

KJW - "While it is interesting to consider the twin clock "paradox" from the perspective of each of the clocks, it is not necessary to do so." You would say this because humans currently do not have the means to have a space travel of this scenario. If humans are advanced enough to have this type of space travel, both perspectives will certainly play an important role. Your very statement "it is not necessary to do so" means that your perspective is fixed to Earth, which is equivalent to saying, "Earth is the center of the Universe", and one only makes this type of statement when he has run out of logical reason.

Markus - regarding the Gravitational Potential, it is my understanding that when you are on the ground floor of the Empire State Building, your clock moves slower than the one at the top of the Building. In this case, the rocket should have gained more time elapsed - your example is wrong, and you are constantly confusing between the realms of Special Relativity and General Relativity.

You guys are confusing the following, or stubbornly holding onto your conventional knowledge.
1) the definition of inertial frame and inertial motion - when you free fall, free drift, sling-shot under Gravity, you are in an inertial frame/motion and the Minkowskian diagram works with respect to you as long as you are in this condition
2) this is why I always differentiate the Gravitational and Non-Gravitational accelerations - countless results get generated by using this simple principle, but I have no willingness to share in this forum, not mature enough
3) Lorentz contraction of length, time, and the Lorentz transformation of increased mass are "symmetric" to both observers – for simplicity, we need to keep this within Special Relativity
4) In essence, all bodies in the Universe are accelerating towards the center of the gravity, somewhere very far and very large, and the tidal force is then experienced by all bodies. A long stretched or a large, non-Gravitationally accelerating rocket will also experience a tidal force because the surrounding Space-Time curvatures are different in each point on the rocket – a true local inertial frame of reference can only apply to a point mass and for a point mass, there is no tidal issue regardless of the case, but since there is no end to this type of argument, I don’t put too much value in this type of discussion
I can easily paste the Minkowskian diagram but again, math upon a wrong principle generates wrong result. This is why I always get to the bottom of the principle first.
If you claim that the Earth must be the only reference point and completely disregard the rocket itself, you people are correct. However, what if you were the astronaut? The key to Relativity is the proper use of reference frames in various scenarios.
People are strange, I am not even bringing up a brand new issue, this is just a re-visit to the century old confusion that once you understand, it is much simpler. Maybe your knowledge is your own blockade in understanding simple stuff.
20 years ago, I spent much time on this topic with my real Physics colleagues, and at least half of them did not have trouble clarifying the principles I have mentioned in their minds.
Once again, when you watch a fast moving bicycle, the Lorentz transformation applies to the bicycle. Likewise, if you were on the fast moving bicycle, the same applies to the background – this simple.

The GPS Satellite issue – I don’t know how long the satellite has been orbiting but any way, it is too much to get into all the details. Earth is observing contracted time of the satellite's clock, I surely agree but we are not on the satellite observing the Earth's clock (symmetry). If you are purely applying the Gravitational potential, then the satellite should gain more time elapsed (we are closer to the center of gravity of the earth) – I am sure there are (+) and (-) factors that all add up to an observed time contraction of the satellite with respect to Earth. Once again, a dynamic thinking is the key to Relativity, and I can't cover all scenarios especially when even cut-pasting existing equations is a chore. Did you ever come up withyour own equation? I have.

Markus, "therefore the clock of Earth will also be Gravitationally dilated by the Grand Center of Galaxy Same for the rocket" - you said "same for the rocket" - "Correct!"

Therefore, the departure of the rocket from Earth using Gravitational Slingshot, except for the booster portion, will be towards two different centers of net-gravity for the Earth and for the rocket, and except to say the space is curved and both the Earth and the rocket are "geometrically" accelerating away from each other, both are still within their inertial frames of reference [free fall, free drift, slingshot (thou I feel a strong tidal force in slingshot-hahaha)].

Originally Posted by tachyon1

KJW - "While it is interesting to consider the twin clock "paradox" from the perspective of each of the clocks, it is not necessary to do so." You would say this because humans currently do not have the means to have a space travel of this scenario.

No. I say this because proper time is an invariant. It is the same in all coordinate systems. Therefore, it doesn't matter which coordinate system I choose to consider the twin clock "paradox" from, the result will be the same regardless. I only mentioned choosing to consider the standard twin clock "paradox" from an inertial frame of reference so that only Special Relativity is required. Otherwise, any coordinate system could be chosen. We do not need to choose the frame of reference of either of the clocks, because once we have the result for any coordinate system, we have the result for every coordinate system, including the frames of reference of both the clocks.

I don't know why you think we're confused about the twin clock "paradox". To me, it is very straightforward. Confusion only results if you misunderstand relativity. Look at my signature below.

Originally Posted by KJW

Originally Posted by tachyon1

KJW - "While it is interesting to consider the twin clock "paradox" from the perspective of each of the clocks, it is not necessary to do so." You would say this because humans currently do not have the means to have a space travel of this scenario.

No. I say this because proper time is an invariant. It is the same in all coordinate systems. Therefore, it doesn't matter which coordinate system I choose to consider the twin clock "paradox" from, the result will be the same regardless. I only mentioned choosing to consider the standard twin clock "paradox" from an inertial frame of reference so that only Special Relativity is required. Otherwise, any coordinate system could be chosen. We do not need to choose the frame of reference of either of the clocks, because once we have the result for any coordinate system, we have the result for every coordinate system, including the frames of reference of both the clocks.

How do you define the Inertial Frame of Reference? Inertial Reference Frame is, e.g., when between two masses, (d^2)x/(dt^2) = g(local-to barycenter), not necessarily when (d^2)x/(dt^2) = 0 with respect to each other, correct?

The Proper Time,



This equation is an integral of Lorentz equation for varying velocity which is just the Relativistic Time, I know you know. However, this is still with respect to you without considering “with respect to rocket”
The definition of a curved space-time is that we have an infinite array of the “Local Inertial Frames of References”, where each Local Inertial Frame of Reference is defined as (d^2)x/(dt^2) = g(local-the net value to barycenter).
While the proper time is global, the symmetry still holds because the rocket will have same proper time with respect to itself and will observe the contraction of Earth’s time.
When you are falling to Sun and I am falling to Earth, we are both in an Inertial Frame of Reference, not when we have (d^2)x/(dt^2) = 0 to each other.

The Proper Time,

This equation is an integral of Lorentz equation for varying velocity which is just the Relativistic Time, I know you know. However, this is still with respect to you without considering “with respect to rocket”
The definition of a curved space-time is that we have an infinite array of the “Local Inertial Frames of References”, where each Local Inertial Frame of Reference is defined as (d^2)x/(dt^2) = g(local-the net value to barycenter).
While the proper time is global, the symmetry still holds because the rocket will have same proper time with respect to itself and will observe the contraction of Earth’s time.
When you are falling to Sun and I am falling to Earth, we are both in an Inertial Frame of Reference, not when we have (d^2)x/(dt^2) = 0 to each other.

Yes, proper time is same and global, therefore it is symmetric - this simple.

Originally Posted by tachyon1

How do you define the Inertial Frame of Reference? Inertial Reference Frame is, e.g., when between two masses, (d^2)x/(dt^2) = g(local-to barycenter), not necessarily when (d^2)x/(dt^2) = 0 with respect to each other, correct?

I prefer the notation of tensor calculus. Given a metric tensor field , the connection can be obtained:



Then an inertial trajectory satisfies the equation:



I think I follow your notation sufficiently to say that it is essentially correct.

Originally Posted by tachyon1

The definition of a curved space-time is that we have an infinite array of the “Local Inertial Frames of References”

As in the vierbein formalism?

Originally Posted by tachyon1

Yes, proper time is same and global, therefore it is symmetric - this simple.

In this particular context, what do you mean by "it is symmetric"? Symmetric with respect to what?

KJW – I am so surprised that you both understand a lot and not understand a thing at the same time. You are a very interesting person next to myself. My simple statement for you is that you understand but you never was able to deviate from your own perspective. Think in two distinct terms, one you are on Earth and the other you are on rocket, respectively and exclusively. When you are on rocket, you will certainly forget about the reference point of the Earth. The same proper time works when you are on the rocket and you will only conclude in terms of the rocket. The term, “invariant” is not too accurate, it is not wrong though.

To yourself, all that is "you" is the ultimate measure. Relativity tries the hardest to break this concept. I am getting way too old for this type of rudimentary quarrel, and so far, I have not brought up any new idea at all, just a repetition of the age-old quarrel.

Forget all about my real theory that were to come, you people are all stuck in the very first step.

What more do you want to know about the symmetry?

I am rusty in Tensor Analysis though it is all a converted term from Sin-Cos-Tan to directional cosine, so I am hesitant - math is sensitive, so I avoid flamboyant math expression. Forgive me, I am not in physics now - the very time factor.

It all depends on how comprehensive you want to be, for example, Fractal Geometry is visualized with computer, the equation itself gives an educated person the principle idea, but not enough to plot the graph for everyone.

Yeah, "i" with respect to "j" and "j" with respect to "i", I am liking Tenson now becasue it trys to generalize the view points.

Originally Posted by tachyon1

What more do you want to know about the symmetry?

I only wanted to know if we were on the same page with regards to your use of the term "symmetry". Because we are talking about the twin clock paradox, "symmetry" was used in a different way and I wanted to make sure you weren't using the word "symmetry" in this particular way. Symmetry with respect to coordinate transformations I can accept. And yes, "invariant" is the correct term.

You appear to be assuming that my knowledge is much less than yours. Please don't do that. You have no basis for that assumption.

KJW - I took a day off today, and yes I am tired, but it seems like we are on a similar (not yet the same) page, so I am glad. I am no stupid knucklehead, so stay in this spseudo forum. I have to have arrogant, less intuitive Markus put this topic back to Physics from Pseudo first. I only put people down because I first was put down. I will resume this evening, I am tired.

Originally Posted by KJW

I don't know why you think we're confused about the twin clock "paradox". To me, it is very straightforward. Confusion only results if you misunderstand relativity. Look at my signature below.

I completely agree, and have said so several times; in fact the twin scenario is one of the simplest demonstrations of relativity I can possibly think of.

How do you define the Inertial Frame of Reference?

A frame is inertial if an accelerometer at rest in that frame measure exactly zero.

While the proper time is global, the symmetry still holds because the rocket will have same proper time with respect to itself and will observe the contraction of Earth’s time.

No, this is where you are wrong. Only one of the frames ( the rocket ) experiences acceleration, and the two frames are not in the same gravitational potential; hence their clocks are dilated differently with respect to one another. To illustrate the point, consider two purely inertial frames ( no acceleration, no gravity ) - if frame A sees frame B dilated by some amount, then frame B will see frame A dilated by exactly the same amount ( relative time dilation ). That is the meaning of symmetry - that all frames experience the exact same laws of physics, and that these frames are indistinguishable. Not so in the rocket example - only one of the frames experiences acceleration ( no matter how short ), which makes these frames physically distinct, and immediately destroys all symmetry. Furthermore, during the slingshot phase, the rocket will first approach and then recede from a massive body, so there is a varying degree of gravitational time dilation. Again, this is not the same as the other frame.