How Does the Twins Paradox Apply to Clocks in a Closed Room?

[Stephanus]

Dear PF Forum,
I have a tought experiment here.
I'm asking about twins paradox, but instead of using twins, I'm using clocks to lock them up in a closed room. Sort of Einstein elevator. (unlike Schrödinger, even in tought experiment, I can't imagine locking human being -- or cat -- in a closed room).
So, here is the experiment.
Two clocks are sealed inside a closed room. Clock E on earth, clock R in a rocket.
Both clock are reset, and the rocket is fired away with acceleration 10 m/s².
Clock R practically 'feels' 1 g.
After 3 billions seconds (about 95 years) the rocket stops and turn around heading toward Earth and fired again with acceleration 10 m/s².
Of course the speed when it reaches Earth would be about zero. The rocket actually deccelerates.


A. How far away does the rocket travel right before it turns around? Will it reach ½at² = 45 thousand trillions KM? Actually after 1 year the rocket will travel 1 speed of light, after 10 years = 10 speeds of light?
I think the energy consumption is very big here. It's Newton's, right?


B.And this is my question.
How does the clocks run? Does clock E run faster, in twins paradox it would age faster, than clock R?
Both clocks are in a closed room. If they were twins, both twins would feel no different with the acceleration.
Clock E accelerates toward the center of the Earth in 1 g, the floor holds it up.
Clock R accelerates toward the floor of the room in a rocket, again 1 g.C. What about the twin who orbits the Earth in geostationary orbit.
Twin E accelerates 1 g, so it actually feels that it moves.
Twin O, in orbit, doesn't feel acceleration at all tough it travels 11 thousands KM per hour.
Which one ages faster.
Supposed both twins are in different rockets in space.
Twin E accelerates 1 g, and twin O's rocket's machine doesn't run. So twin O actually doesn't feel acceleration at all as in geostationary orbit. Will twin O ages slowlier?

Thanks for you enlightment.

Steven

 

[Orodruin]

Of course the speed when it reaches Earth would be about zero. The rocket actually deccelerates.

With your given setup, the rocket is always traveling away from Earth. It will not come back just when it stops. Furthermore, you say "95 years", this is ambiguous, 95 years as measured by whom?

A. How far away does the rocket travel right before it turns around? Will it reach ½at2 = 45 thousand trillions KM? Actually after 1 year the rocket will travel 1 speed of light, after 10 years = 10 speeds of light?

A relative velocity will never be faster than the speed of light. You also need to specify who measures the acceleration.

Time dilation is related to space time intervals, not to acceleration.

 

[Stephanus]

Thanks Orodruin for the answer.
95 years by clock R, I mean.
And I think i made some error condition. I have corrected below.

After 3 billions seconds (about 95 years) the rocket stops and turn around heading toward Earth and fired again with acceleration 10 m/s2.
Of course the speed when it reaches Earth would be about zero. The rocket actually deccelerates.

Correction

Time R A: _{(95 years)}
After 3 billions seconds (about 95 years) the rocket stops and turn around. Heading toward Earth but not speeding toward earth. Actually it still speeds away from the earth. And the rocket is fired again with acceleration 10 m/s². But the rocket actually decelerates.

Time R B: _{(190 years)}
By around this time the rocket does accelerate toward Earth with acceleration 10 m/s².

Time R C: _{(285 years)}
By around this time the engine stops and the rocket turn around again heading away from Earth but speeding toward earth. Engine starts again. Now the rocket actually decelerates.

Time R D: _{(380 years)}
By around this time the rocket should reach the Earth again. Of course the speed when it reaches Earth would be about zero. The rocket actually deccelerates.

 

[Janus]

You might want to check out this:

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

 

[yogi]

The two clocks won't feel a difference in the acceleration. But the clock which has been put in motion will gradually accumulate a large velocity relative to the Earth frame - the traveling clock that has been given the energy boost will have been found to run slower when the clocks are ultimately compared - whether they are brought together physically or simply brought to rest in the same Earth frame millions of miles apart. All such experiments are disguised versions of a lab experiments where a charged particle in the lab frame is given a boast and its decay time is extended. There is never any doubt which particle gets the long life - i.e., the real "slower clock" In SR Einstein hypothesized that both frames are inertial equivalents and that each clock can be considered at rest, but the reciprocal situation where the high speed clock views the Earth clock as running slow does not comport with what is measured when the clocks are brought to rest in the same frame. During any constant velocity phase of the clock experiment (aka the twin paradox), any measurement made by the boosted clock showing the Earth clock running slow, is illusory. When the clocks are reduced to the same speed, the boosted clock will have logged a greater time. There is one experimental result that seems to question whether the boosted clock always runs slow, but that teaches away from SR which is not permitted on these forums.

 

[yogi]

Correction - second to last line of post 5 - the boosted clock will log less time

 

[1977ub]

When the ship turns around, it is in a new frame of reference in which the Earth clock "instantly" gains a lot of time, as the FAQs point out. Just looking at coordinates makes it look as a completely reciprocal situation, but actually only one of the clocks will be in a single inertial frame the whole time. That's the only one where the single set of SR calculations makes sense.

 

[yogi]

Response to post 7:

My take on the proverbial twin thing, aka clock paradox, is to forget about the round trip initially. Ask the question, if a spaceship is accelerated from the Earth to a high velocity in a short amount of time and then coast the rest of the way and ether slows to stop on an faraway planet "alpha, or simply reads the time on a clock located on "alpha" then assuming there is no relative motion between the Earth and alpha, the time logged by the traveling clock will be less (measured by reference to the Earth clock or the alpha clock). Since the Earth clock and the alpha clock have been in the same frame during the entire voyage, they will measure same time, and the traveling clock will have measured less as predicted by SR. To figure the round trip time difference, simply double the result for the one way trip

Einstein actually got into a bit of a problem when he was pushed by a number of critics about the clock paradox. He wrote a paper in 1918 solving the twin paradox using GR - reasoning similar to your resolution ...that the Earth clock gains a lot of time when the flying clock turns around - Einstein imposed an artificial G field - Max Born conjured a similar result by reasoning that acceleration field at turn around corresponded to a lot of time passing on the Earth clock because the formula for dilation takes into account the distance between the clocks. These solutions give the right numerical value at the expense of fracturing the real physics as to why the passage of time in relatively moving frames is different.

 

[harrylin]

[..] Einstein actually got into a bit of a problem when he was pushed by a number of critics about the clock paradox. He wrote a paper in 1918 solving the twin paradox using GR - reasoning similar to your resolution ...that the Earth clock gains a lot of time when the flying clock turns around - Einstein imposed an artificial G field [..]. These solutions give the right numerical value at the expense of fracturing the real physics as to why the passage of time in relatively moving frames is different.

You will likely appreciate the Usenet Physics FAQ (they certainly do not mean inertial reaction force but the magical force that appears to pull the children outward):
"You may be bothered by the Big Coincidence: how come the uniform pseudo-gravitational field happens to spring up just as Stella engages her thrusters? You might as well ask children on a merry-go-round why centrifugal force suddenly appears when the carnival operator cranks up the engine. There's a reason why such forces carry the prefix "pseudo"."
- http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html

 

[harrylin]

[..] both twins would feel no different with the acceleration.[..]

The "feeling of acceleration" should not be understood as physical explanation; texts that use that merely mean (or should mean) that in cases far from gravity it's easy to detect that one is accelerating.
In the original full "twin paradox" the traveler doesn't feel any acceleration as he is in a slingshot around a star. However, for a sufficiently long travel at sufficiently high velocity the clock slowdown from the Earth's gravity as well as the short-time clock slowdown from the star's gravity become relatively small so that they can be neglected (and perhaps a case can be construed in which those effects exactly compensate each other).

 

[m4r35n357]

I'm asking about twins paradox, but instead of using twins, I'm using clocks to lock them up in a closed room. Sort of Einstein elevator.

This video shows a first-person journey for the traveler leaving and returning home. For the purposes of your question, the clock at the upper left shows three simultaneous values: the red one is coordinate time (the time shown on the clock currently just outside a porthole), the green one is the traveler's proper time (the clock in his ship) and the yellow one shows the time actually seen on the home twin's clock by the traveler through a powerful telescope.
It is easy to see what all three clocks show for the traveler throughout the journey. You will need to view in full HD to see the clock clearly.
BTW more description is available in the channel notes here.

 

[Stephanus]

- http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_gr.html

Thank you very much, harrylin for you answer.
But one thing is borthering me. Again.
At one of those link
http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_intro.html#standard

In short, Stella felt the acceleration, while Terence felt nothing.

Stella, star, is the one who travel to the star.
Terence, terra, is the one who stays terresterial on earth.
"Terence feels nothing", but how can that be?
Surely Terence "travels" 0 km per hour, but doesn't he feel the Earth gravity?

My questions are:
A. Does Terence feel acceleration?
B. Is there a difference between accelerated by a rocket or "feel" accelerated by the gravity?

Thanks for anyone's attentionsSteven.

 

[m4r35n357]

Stella, star, is the one who travel to the star.
Terence, terra, is the one who stays terresterial on earth.
"Terence feels nothing", but how can that be?
Surely Terence "travels" 0 km per hour, but doesn't he feel the Earth gravity?

Firstly, my advice is ignore all talk of acceleration until you understand the basics. It is not irrelevant, but it is unnecessary to describe the effect and will just confuse the hell out of you and everyone else (search past threads if you don't believe me!). Secondly, the younger twin is the one that goes somewhere and returns, the older one is the one who goes nowhere. They each know perfectly well which one they are!
The easiest source to learn from IMO is the Baez site.

 

[Stephanus]

the red one is coordinate time (the time shown on the clock currently just outside a porthole), the green one is the traveler's proper time (the clock in his ship).

Thanks m4r35n357 for your answer.
But I would like to ask for clarity here.
What's the different between the red clock and the green clock.
The green clock is in his ship, what about the red clock? Is it located out side his window, 2 meters away? Where is the red clock located?

the yellow clock shows the time actually seen on the home twin's clock by the traveler through a powerful telescope.

Through a powerful telescope:
The ship is moving away from the earth, so the yellow clock runs slowler, because the light from the clock must takes time to reach the ship.
What if the ship is heading toward earth? Does the yellow clock will run faster?
Supposed, the ship is 10 light year away.
If the ship compares its clock vs Earth clock, there will be 10 year different right? The clock on the Earth is 10 years lates. What if the ship is heading toward Earth with say... a very slow speed about 1000 km / seconds so it is affected very little relativity effect.. Altough it takes 300 000 years to reach the earth, the yellow clock seems run faster doesn't it such as shown in the video? Because the clocks is rather synchronized.

Thanks for anyone's attentions.Steven

 

[m4r35n357]

Thanks m4r35n357 for your answer.
But I would like to ask for clarity here.
What's the different between the red clock and the green clock.

Have you read the YouTube channel notes that I linked to? To clarify, the red clock dot is the instantaneously nearest one of an "imaginary" field of clocks in the coordinate frame (the one the traveler is currently passing). All the rotating stuff (including the home station) is showing coordinate time where it is, but there is a visible lag caused by the distance between it and the traveler. You have to look at a clock exactly where you are to read coordinate time.
The green clock is the traveler's wristwatch.
As to your other questions, please see the Baez link; he explains it better than I could.

 

[Stephanus]

To clarify, the red clock dot is the instantaneously nearest one of an "imaginary" field of clocks in the coordinate frame (the one the traveler is currently passing).

Ahh, yes, mea culpa. It's the "red clocks", plural.
Sorry.

And THANK YOU VERY MUCH.
This really GIVES ME ENLIGHTMENT!
The stationary clocks outside the ship!

 

[m4r35n357]

Yes. That is what I was struggling to say (I chose to avoid using the word stationary, perhaps wrongly), but it sounds like you have understood.

 

[Janus]

Thanks m4r35n357 for your answer.
But I would like to ask for clarity here.
What's the different between the red clock and the green clock.
The green clock is in his ship, what about the red clock? Is it located out side his window, 2 meters away? Where is the red clock located?

The red clock represents what a clock just outside the ship's window and stationary with respect to the Earth would read at the instant you pass it. It isn't really a physical clock.

Through a powerful telescope:
The ship is moving away from the earth, so the yellow clock runs slowler, because the light from the clock must takes time to reach the ship.
What if the ship is heading toward earth? Does the yellow clock will run faster?
Supposed, the ship is 10 light year away.
If the ship compares its clock vs Earth clock, there will be 10 year different right? The clock on the Earth is 10 years lates. What if the ship is heading toward Earth with say... a very slow speed about 1000 km / seconds so it is affected very little relativity effect.. Altough it takes 300 000 years to reach the earth, the yellow clock seems run faster doesn't it such as shown in the video? Because the clocks is rather synchronized.

Thanks for anyone's attentions.Steven

What you see on the yellow clock is a combination of classic Doppler shift due to the changing distance and Relativity or Relativistic Doppler shift.

If you are separating you see the yellow clock run slow, if you are coming together, you see it running fast.

The factor follows the equation

\sqrt{\frac{1-\Beta}{1+\Beta}}

Where beta = v/c and is positive when you are separating.

So in your example of returning at 1000 km/sec from 10 light years, you get a factor of 1.0033389.

It actually takes 3000 yrs (not 300,000) to travel ten light years at 1000 km/sec, so this works out to you seeing the Earth age by 3010.0167 years during the trip, 10.0167 years more than you do, 10 years of that is due to the decreasing time lag of the light, and the 0.0167 yr (6+ days) will be due to relativistic effects.

If we beef up the speed to 30,000 km/sec and check the progress of the yellow clock through the telescope for both the outbound and inbound leg we can get the following results.

Assuming that our ship travels for 10 years by its own clock, it will by its reckoning traveled 1 light year in that ten years. During that time, it will see the yellow clock on Earth run at a rate of ~0.904534 and advance 9.04534 years. On the return leg which takes another 10 years by its clock, it sees the yellow clock run at a rate of 1.105542, and advance another 11.05542 years for a total of 20.1008 years or just a tad or 1/10 of year more than the 20 years it measured by its own clock.

 

[Stephanus]

It actually takes 3000 yrs (not 300,000) to travel ten light years at 1000 km/sec,

Again, my mistake. I hastily calculated. You're right! 3000 years, not 300 000 years.

Thanks for the correction.

 

[m4r35n357]

Again, my mistake. I hastily calculated. You're right! 3000 years, not 300 000 years.

Thanks for the correction.

Many thanks to Janus for explaining what I had no appetite for!

I should also point out the trigger for my original post was the part of your original question where you describe sealed clocks. I felt compelled to point out that there is no real paradox here, and no secret concealed magic about time. All the clocks can be made visible to all observers throughout the entire process.

 

[Stephanus]

Many thanks to Janus for explaining what I had no appetite for!

... to point out that there is no real paradox here, and no secret concealed magic about time. All the clocks can be made visible to all observers ...

Thanks for enlighten me m4r35357, and to you too Janus.
So, what I mean about concealed clock there, has nothing todo with magic. What I'm trying to say is "concealed twins". So they don't know about the outside world. They don't know whether they are in a gravitational object, or they are in a rocket. But can't imagine human beings locked. I use clocks instead.

Thanks.

 

[Stephanus]

Dear PF Forum,
Thanks for members taking effort answering my endless questions.
But one thing still borthering me. Still on twins paradox.
Perhaps someone can explain me, more.
I have read Jane and Joe, Stellar and Terrence, Prime and Umprime, even Alphonse and Gaston if I must.
Supposed, the twins stay on earth, synchronize their watches and twin B zaps into space 0.6c, assuming not destroying his ships or himself for that matter, for one or two years and come back.
Both twins see their counterpart moving, right?
So, how does the paradox assymetric occur?
A. Because twin A stays on earth, experiencing gravity ALL THE TIME?
B. Because twin B zaps, experiencing tremendous acceleration, although for only 1 second? (two for the run around trip)
C. Because twin B turns around, CHANGINGS inertial frame?
D. or something else?

Thanks for any helpSteven

 

[1977ub]

Dear PF Forum,
Thanks for members taking effort answering my endless questions.
But one thing still borthering me. Still on twins paradox.
Perhaps someone can explain me, more.
I have read Jane and Joe, Stellar and Terrence, Prime and Umprime, even Alphonse and Gaston if I must.
Supposed, the twins stay on earth, synchronize their watches and twin B zaps into space 0.6c, assuming not destroying his ships or himself for that matter, for one or two years and come back.
Both twins see their counterpart moving, right?
So, how does the paradox assymetric occur?
A. Because twin A stays on earth, experiencing gravity ALL THE TIME?
B. Because twin B zaps, experiencing tremendous acceleration, although for only 1 second? (two for the run around trip)
C. Because twin B turns around, CHANGINGS inertial frame?
D. or something else?

Thanks for any helpSteven

It is not about coordinates that the twins use to measure the other's motion. It is about the # of inertial frames they occupy before they meet again. "Seeing the other twin's clock move slowly" turns out to be a red herring. The twin which was in one inertial frame all along is the one who aged more.

 

[Janus]

Dear PF Forum,
Thanks for members taking effort answering my endless questions.
But one thing still borthering me. Still on twins paradox.
Perhaps someone can explain me, more.
I have read Jane and Joe, Stellar and Terrence, Prime and Umprime, even Alphonse and Gaston if I must.
Supposed, the twins stay on earth, synchronize their watches and twin B zaps into space 0.6c, assuming not destroying his ships or himself for that matter, for one or two years and come back.
Both twins see their counterpart moving, right?
So, how does the paradox assymetric occur?
A. Because twin A stays on earth, experiencing gravity ALL THE TIME?
B. Because twin B zaps, experiencing tremendous acceleration, although for only 1 second? (two for the run around trip)
C. Because twin B turns around, CHANGINGS inertial frame?
D. or something else?

Thanks for any helpSteven

C is the best answer.

You keep bringing up gravity and acceleration forces, so I feel that I need to bring up the fact that a difference in local gravity or acceleration has no direct effect on time dilation.

For example, Identical clocks sitting at the North pole and on the equator will weigh differently, with the Equator clock weighing less, However these clocks will urn at the same rate.

You can also imagine a situation where two clocks at rest with respect to each other and experiencing the exact same g force run at different rates.

One is to put these clocks in a uniform gravity field (one that dos not change in strength with altitude) at different altitudes in this field. The higher clock will run faster even though both feel the same g force. Or you could place two clocks in a rocket, one in the tail and one in the nose. Have the rocket accelerating so that both clocks feel the exact same g force. The clock in the nose will run faster than the one in the tail.

 

[harrylin]

[..]
Supposed, the twins stay on earth, synchronize their watches and twin B zaps into space 0.6c, assuming not destroying his ships or himself for that matter, for one or two years and come back.
Both twins see their counterpart moving, right?
So, how does the paradox assymetric occur? [..]

A. Because twin A stays on earth, experiencing gravity ALL THE TIME?
Due to gravity alone, twin A would be behind twin B (but as I already clarified, experiencing gravity is irrelevant). That is also an asymmetry, but it is neglected in simple examples and it works against the one that you ask about.

B. Because twin B zaps, experiencing tremendous acceleration, although for only 1 second? (two for the run around trip)
It is assumed that the force of acceleration has no effect. If the clock (bioclock) is sensitive to that, the additional delay needs to be added.

C. Because twin B turns around, CHANGINGS inertial frame?
Yes, a constant velocity of A with a change of velocity of B breaks the symmetry. A is all the time co-moving with an inertial frame so that it may be considered "at rest" all the time by designating that inertial frame "rest frame". Not so with B.

 

[1977ub]

Or you could place two clocks in a rocket, one in the tail and one in the nose. Have the rocket accelerating so that both clocks feel the exact same g force. The clock in the nose will run faster than the one in the tail.

I've been thinking about this case a lot lately (and comparing it to the Bell's Spaceship Case). It appears that to keep the spaceship (close to) rigid, the force you apply at the top of the rocket to minimize stresses is less than the force applied at the bottom - and so doesn't this suggest that the g forces experienced at top and bottom are different? From the original frame S, the front clock is also seen to run faster, but for S, this is because the front of the rocket has lower acceleration and thus the back end clock with the greater acceleration is gaining on the front, thus causing the Lorentz contraction? Alternately, if both clocks experience the same g force, is this not equivalent to the Bell case, and the ship must therefore be stretching?

 

[PeterDonis]

It appears that to keep the spaceship (close to) rigid, the force you apply at the top of the rocket to minimize stresses is less than the force applied at the bottom - and so doesn't this suggest that the g forces experienced at top and bottom are different?

This is correct; for rigid motion, the rear must have greater proper acceleration than the front.

From the original frame S, the front clock is also seen to run faster, but for S, this is because the front of the rocket has lower acceleration and thus the back end clock with the greater acceleration is gaining on the front, thus causing the Lorentz contraction?

In the original frame S, at any time after the spaceship starts moving, the front clock is moving slower than the rear clock. That accounts for the contraction of the spaceship relative to frame S, and also accounts for the front clock running faster relative to frame S.

However, the front clock is also running faster relative to the rear clock, and the rear clock is running slower relative to the front clock. Observers at the two clocks can verify this by exchanging repeated round-trip light signals and observing that (a) the round-trip light travel time is constant, indicating that the spaceship is indeed undergoing rigid motion; and (b) more time elapses on the front clock between two successive round-trip light signals than on the rear clock, indicating that the front clock is running faster. This is an invariant sense in which the front clock runs faster; the observations I have just described are independent of any choice of coordinates.

if both clocks experience the same g force, is this not equivalent to the Bell case, and the ship must therefore be stretching?

Yes. In this case (the Bell case), relative to frame S, at any instant of time, the front and rear of the ship are moving at the same speed, and the front and rear clocks are ticking at the same rate. So in frame S, the length of the ship remains constant. But in the momentarily comoving inertial frame of either the rear clock or the front clock, the other clock is moving away (i.e., the ship is stretching). This means that the round-trip light travel time between the clocks is not constant, so there is no invariant way to tell which one is running faster or slower.

 

[Stephanus]

Dear forum,
Thanks for taking any efforts to answer me!

And, 1977ub, please don't counter Janus about head and tail clocks, it has already blown my mind (away).
These relativity stuffs are really confusing me, much less head and tail clocks. But, it seems that "changing intertia frame" that matters most.
Again, thank you, thank you.

Steven.

 

[yogi]

You will likely appreciate the Usenet Physics FAQ (they certainly do not mean inertial reaction force but the magical force that appears to pull the children outward):
"You may be bothered by the Big Coincidence: how come the uniform pseudo-gravitational field happens to spring up just as Stella engages her thrusters? You might as well ask children on a merry-go-round why centrifugal force suddenly appears when the carnival operator cranks up the engine. There's a reason why such forces carry the prefix "pseudo"."

Richard Feynman referred to inertial reactions as "pseudo forces" He also mused that since pseudo forces, like gravity, are always proportional to the masses, perhaps gravity is a pseudo that we do not recognize as such because we don't have the right reference frame.

 

[PeterDonis]

perhaps gravity is a pseudo

According to GR, it is.

that we do not recognize as such because we don't have the right reference frame.

The right reference frame is any local inertial frame; in any such frame, there is no "force of gravity", just as there are no other pseudo forces.

 

[harrylin]

Richard Feynman referred to inertial reactions as "pseudo forces" [..].

I strongly disagree with that, but often such things are just a matter of definition. Action and reaction forces form a force pair, it would be nonsense to consider a real force to be balanced by an unreal force.

 

[yogi]

The right reference frame is any local inertial frame; in any such frame, there is no "force of gravity", just as there are no other pseudo forces

Obviously, Feynman has something else in mind - Einstein eliminated the local "g' force with a hypothesis - the idea that inert mass could curve static space - he was actually rather apologetic about that, later calling it "a house of straw." He never offered any physics to suggest how inert matter could bend space. It was a good solution for the time - but as you know or should know, Einstein's gravitational equations can be derived straightway from fundamental energy considerations and Newtonian mechanics as shown by Milne and McCrea - as can Freidman's equations. So whether there is a gravitational force or not, Feynman was, up to the time of his death, still asking questions about gravity, as are others. When it comes to gravity, pontification is risky.

 

[PeterDonis]

Einstein eliminated the local "g' force with a hypothesis - the idea that inert mass could curve static space - he was actually rather apologetic about that, later calling it "a house of straw."

Reference, please?

He never offered any physics to suggest how inert matter could bend space.

Um, what? That's exactly what the Einstein Field Equation is.

Einstein's gravitational equations can be derived straightway from fundamental energy considerations and Newtonian mechanics as shown by Milne and McCrea - as can Freidman's equations.

Reference, please?

 

[yogi]

General response to post 33. I am on vacation and do not have access to the these things except by memory. The "house of straw" was a quote from a book entitled "Out of my later Years" partially written by Einstein - at least edited by Einstein - if you are really curious I will give you exact citation when I return in about 3 weeks. With regard to the derivation of the equations from Newtonian principles ...This is a statement straight out of Harrison's 2002 edition - but you can find it about 3/4 of the way through - I think it was in the chapter called General Relativity ...earlier he derived the Friedmann equations from Newtonian principles.

As you know, the General Theory has only been resolved for a few cases and then by making simplifying assumptions - as did Schwarzschild. The fact that these that relationships are on some level founded upon energy, is a very useful thing to keep in mind.

This got me to thinking about our previous discussion where I proposed using the Earth in a train experiment - I tried to recall if I remembered reading about any time dilation experiment that were ever conducted at any time by anyone, that did not use the Earth as the inertial frame.. This led me to question whether acceleration is important.. that is, unless it changes the energy - a train traveling a curved path will experience the same time dilation as a train on a straight path - (remember my example of the centrifuge). What is curious is that both GR and SR reduce to equations that define the passage of time in terms of the terms that look like KE energy - for SR its KE of the frame taken to be moving and for GR its the potential energy change - not the acceleration - but even this reduces to KE because time dilation in GR is given by the escape velocity.

Obviously, we have gone beyond what the op had in mind - sorry about that.

Yogi

 

[PeterDonis]

the General Theory has only been resolved for a few cases and then by making simplifying assumptions

If by "resolved" you mean that exact solutions have only been found by making simplifying assumptions, that's true. (But it's more than "a few cases".) However, numerical solutions cover a lot more cases, including ones like highly non-symmetric gravitational collapse where no exact solutions are known.

these that relationships are on some level founded upon energy

How so? The Schwarzschild solution, for example, is a vacuum solution--no stress-energy anywhere.

a train traveling a curved path will experience the same time dilation as a train on a straight path - (remember my example of the centrifuge).

I'll have to look up your example, but your claim here is simply wrong as a general claim.

both GR and SR reduce to equations that define the passage of time in terms of the terms that look like KE energy - for SR its KE of the frame taken to be moving and for GR its the potential energy change - not the acceleration - but even this reduces to KE because time dilation in GR is given by the escape velocity.

What equations are you talking about? And are they equations that apply in all cases, like the Einstein Field Equation? Or are they, as I suspect, equations that only apply in one particular solution (which I suspect is the Schwarzschild solution)? You can't make general claims about "GR" based on one solution.

 

[m4r35n357]

Yogi

Serious answer this time. Non-geodesic paths encounter more proper time than geodesic ones, even in GR. GR just allows more than one geodesic path between two events. What more must GR provide? I don't see the problem . . .

 

[PeterDonis]

Non-geodesic paths encounter more proper time than geodesic ones, even in GR.

This is wrong in two ways. First, in SR, non-geodesic paths between the same pair of events have less elapsed proper time, not more. Second, in GR, both cases are possible: non-geodesic paths that have less proper time than geodesic ones between the same two events, and non-geodesic paths that have more.

For example, consider the following scenario: one astronaut hovers at a constant altitude over a non-rotating planet. A second astronaut is in a circular orbit about the planet at the same altitude. A third astronaut launches himself upward from the first's position at the same instant that the second one passes, in such a way that he free-falls upward and then free-falls back down so that he arrives back at the first astronaut at the same instant that the second one passes again in his orbit.

The three astronauts follow three different paths between the same pair of events. Astronaut #1's path is non-geodesic; the other two are geodesic. Astronaut #1 has more elapsed proper time than #2 between the pair of shared events, and less than #3.

 

[m4r35n357]

This is wrong in two ways. First, in SR, non-geodesic paths between the same pair of events have less elapsed proper time, not more. Second, in GR, both cases are possible: non-geodesic paths that have less proper time than geodesic ones between the same two events, and non-geodesic paths that have more.

Yeah, got that the wrong way round, oops! My head hurts a bit thinking about the scenarios, I'll need a bit of time to think about them. Still, my point was that there is no misunderstanding or mystery about "twin paradoxes" in GR; they can all be calculated and resolved.

 

[PeterDonis]

my point was that there is no misunderstanding or mystery about "twin paradoxes" in GR; they can all be calculated and resolved.

Yes, definitely.

 

[yogi]

How so? The Schwarzschild solution, for example, is a vacuum solution--no stress-energy anywhere

The Schwarzschild solution is a solution for the space and time outside a mass based upon the mass ...look at the coefficient of the time increment

ds^2 - (1-2M/r)dt^2 ...

the fact there is no mass embedded in the solution for the exterior space is not the same as saying "no stress energy anywhere"

While reasonable minds may differ upon interpretation, the formula for time dilation in GR is totally based upon energy - the escape velocity that determines the gravitational potential - If you believe there are any experiments validating the fact that an object traveling a curved path at constant velocity (as described by Einstein in his 1905 paper, will incur a time dilation component greater than that predicted by SR, I would like to see the proof of such.

Look at the elements that make up the time dilation in GR The time dilation is given by dt* = dt(1-2GM/rc^2)^1/2 = dt(1- v^2/c^2)^1/2 where v is the escape velocity

There is no change in time dilation unless acceleration results in a change in height (PE) or a change in velocity KE) and therefore I stand on my statement. A clock traveling a curved path at constant velocity keeps the same time as a clock on straight track traveling at constant velocity

Regards

Yogi

 

[PeterDonis]

the fact there is no mass embedded in the solution for the exterior space is not the same as saying "no stress energy anywhere"

It does if the solution is describing a black hole.

the formula for time dilation in GR is totally based upon energy - the escape velocity that determines the gravitational potential

First of all, you have this backwards; the gravitational potential determines the escape velocity, not the other way around.

Second, how is any of this "based upon energy"? The gravitational potential is a geometric feature of the spacetime (more precisely, it is directly definable in terms of a geometric feature, namely the spacetime's timelike Killing vector field).

If you believe there are any experiments validating the fact that an object traveling a curved path at constant velocity (as described by Einstein in his 1905 paper, will incur a time dilation component greater than that predicted by SR, I would like to see the proof of such.

You're going to have to be more specific about what you mean by "an object traveling a curved path at constant velocity". Do you mean an object traveling around in a circle?

You're also going to have to be more specific about what you mean by "time dilation component". Do you just mean the "tick rate" of the object's clock as compared to coordinate time in a fixed inertial frame?

There is no change in time dilation unless acceleration results in a change in height (PE) or a change in velocity KE) and therefore I stand on my statement. A clock traveling a curved path at constant velocity keeps the same time as a clock on straight track traveling at constant velocity

How does the comparison between two objects at different heights in a gravitational field have anything to do with "a clock traveling a curved path at constant velocity"?

 

[yogi]

It does if the solution is describing a black hole.
First of all, you have this backwards; the gravitational potential determines the escape velocity, not the other way around.

Second, how is any of this "based upon energy"? The gravitational potential is a geometric feature of the spacetime (more precisely, it is directly definable in terms of a geometric feature, namely the spacetime's timelike Killing vector field).
You're going to have to be more specific about what you mean by "an object traveling a curved path at constant velocity". Do you mean an object traveling around in a circle?

You're also going to have to be more specific about what you mean by "time dilation component". Do you just mean the "tick rate" of the object's clock as compared to coordinate time in a fixed inertial frame?
How does the comparison between two objects at different heights in a gravitational field have anything to do with "a clock traveling a curved path at constant velocity"?

The Black Hole is what generally comes to mind in the Schwarzschild solution - the mass is concentrated in a black hole and the exterior space is curved. This is a different subject than the curvature experienced by a spaceship tethered to a pole so that it travels at constant velocity and therefore constant energy with constant centripetal acceleration, ergo, there is no change in energy, time dilation relative to the Earth is uniform through-out the entire trip. That was my original analogy - in such a case, a round a trip voyage is the same as a one way voyage - there is no special turnaround acceleration involved since the curvature of the path is constant during the entire journey - that was Einstein's example in part IV of his "Electrodynamics of Moving bodies. As far as time dilation, we are talking about comparing the total time of a circular path which begins on Earth and ends at the same point - that was Einstein's example - it was perfectly correct as originally presented - there is NO correction imposed by GR for a circular path in free space - contrary to many misstatements on these and other forums where a curved path is immediately relegated to a problem requiring GR - it doesn't require GR. GR is not involved.

Finally to clarify the height energy - a clock traveling on Earth or in any other "g" field, if subjected to different heights, would to that extent, experience time changes due to GR since the PE is a function of the height in determining the time dilation in a gravitational field - but that potential energy is immediately seen as the KE that corresponds to the velocity acquired to leave the Earth and never return (7 mi/sec) which in this sense, points in the direction of an absolute reference frame rather than a relative one. My point in saying that time differences depend upon energy differences follows from the fact, that in all experiments to date (except one), are based upon adding energy and measuring a slower clock rate for the object put into motion wrt the earth.

 

[PeterDonis]

This is a different subject than the curvature experienced by a spaceship tethered to a pole so that it travels at constant velocity

Do you mean in flat spacetime? If so, there is no "curvature"; spacetime is flat. The spaceship's worldline is curved, but spacetime itself is not. Or...

so that it travels at constant velocity and therefore constant energy with constant centripetal acceleration, ergo, there is no change in energy, time dilation relative to the Earth is uniform through-out the entire trip.

...do you mean, for example, a spaceship tethered on a pole attached to the surface of the Earth, so it goes around in a circle at a constant altitude? If so, yes, spacetime is curved, but it's curved the same for this spaceship as it is for a spaceship "hovering" at the same altitude but at rest relative to the Earth. The only difference is in the curvature of the worldlines of the two ships.

in such a case, a round a trip voyage is the same as a one way voyage - there is no special turnaround acceleration involved since the curvature of the path is constant during the entire journey

I'm still confused as to whether you mean the flat spacetime or the curved spacetime case; but taking the flat spacetime case for discussion, since that's the one Einstein was talking about, yes, an inertial observer at rest at one point on the circular path of the observer going around in a circle will have more elapsed time between two successive meetings of the two. I'm still not sure what this has to do with comparing two observers at rest at different altitudes in a gravitational field, though.

As far as time dilation, we are talking about comparing the total time of a circular path which begins on Earth and ends at the same point - that was Einstein's example - it was perfectly correct as originally presented - there is NO correction imposed by GR for a circular path in free space

Assuming that by "free space" you mean "flat spacetime", yes, this is correct; SR is sufficient to analyze any scenario in flat spacetime, even if some of the worldlines involved are curved (i.e., accelerated). GR is only necessary if spacetime is curved.

However, if the scenario is set on Earth, then it is not in flat spacetime. If everything happens at exactly the same altitude, you can finesse that, which is basically what Einstein did. But why do that when you can just as easily set the scenario in flat spacetime to begin with?

a clock traveling on Earth or in any other "g" field, if subjected to different heights, would to that extent, experience time changes due to GR since the PE is a function of the height in determining the time dilation in a gravitational field

The PE does not determine the time dilation; the position of the observer with respect to the timelike Killing vector field of the spacetime determines both the PE and the time dilation.

but that potential energy is immediately seen as the KE that corresponds to the velocity acquired to leave the Earth and never return (7 mi/sec)

Again, this is determined by position relative to the timelike Killing vector field; it is that position which is the fundamental quantity, and it is a geometric quantity, not an "energy" quantity.

which in this sense, points in the direction of an absolute reference frame rather than a relative one.

There is a unique reference frame picked out by the timelike Killing vector field of the spacetime, yes. (In the case of Schwarzschild spacetime, this is Schwarzschild coordinates.) Put another way, the spacetime we are talking about has a particular symmetry, and any spacetime with a particular symmetry will have a particular reference frame picked out that matches up with that symmetry. Again, the fundamental fact is the symmetry, and that is a geometric fact.

all experiments to date (except one), are based upon adding energy and measuring a slower clock rate for the object put into motion wrt the earth.

What is the one exception?

 

[georgir]

http://www.convertalot.com/relativistic_star_ship_calculator.html
It can't take distances over 15 digits normally, but if you know your way around a browser's HTML inspector you can change that.
Then, assuming the formula it uses is correct, you can see that accelerating 1g for 95 on-board years can take you almost 1885540714000000000000 light years away :) Yes, that is ten orders of magnitude more than the observable universe.