twin paradox

[xyzt]

According to the internet, the twin paradox can be resolved because "only one twin underwent acceleration." But, from the frame of reference of someone traveling away from Earth near the speed of light, they would have observed Earth moving away from themselves at nearly the speed of light instead. To both frames of reference, both observed the other undergoing near-light acceleration, so why wouldn't it be symmetric? Let's say there wasn't turn-around acceleration involved and Earth and the ship are in a hyper-doughnut where by traveling in a straight line away from Earth, one would end up back at Earth eventually, after let's say, 10 light years of distance.

Even if there was turn-around acceleration, couldn't the near-light traveler just say they saw Earth traveling in a semi-circle from their view instead?

Accelerated motion, unlike uniform rectilinear motion, is absolute. You can tell which twin is accelerating from the one who isn't by using accelerometers attached to each twin. So, your scenario is flawed.

[md65536]

So if youre floating in deep space when suddenly a clock flies past. You see whether it's ticking slow from your perspective... Then catch up to it and look at it again.

 

Are you suggesting that you can definitively detect which one accelerated more during their trip by noting which clock jumps forward in time when the inertial frames meet back up?

[...]

If SR is a matter of relative velocity, it can't be asymmetrical.

You can definitely calculate that. Not everything is symmetrical.

 

Remember that these things are relative. The clock that flies by is inertial in this example. If you look at if from its perspective, it remains at rest while you fly by and then turn around and return. The usual calculations work out correctly.

According to the internet, the twin paradox can be resolved because "only one twin underwent acceleration."

Who undergoes acceleration and when doesn't matter except in that it affects the spacetime path of the observer. You can have any number of observers, each taking different paths between the same two points, and each enduring their own particular proper time (aka. aging). Any of the observers' relative aging could be calculated in SR. If two twins accelerate symmetrically, their paths will be symmetrical, and they'll age the same. I've been told that the longer the path is spatially, the shorter it will be in time. An inertial path is the spatially shortest path between 2 points in flat spacetime, so if you include one inertial observer it simplifies the situation because that one will always age the most out of any possible paths taken.

 

It matters that their spacetime paths are different lengths, not that one must not accelerate.

 

As a quick example: Say you have a typical twin setup where an inertial observer ages 4 years, and a traveling twin ages 2, and add another faster traveling twin which ages 1. If you remove the inertial twin, then you still have 2 twins that both travel and return, at different speeds, and the first ages twice as much as the other.

Edited April 4, 2014 by md65536

[Janus]

So if youre floating in deep space when suddenly a clock flies past. You see whether it's ticking slow from your perspective... Then catch up to it and look at it again.

 

Are you suggesting that you can definitively detect which one accelerated more during their trip by noting which clock jumps forward in time when the inertial frames meet back up?

As the clock flies by you, you will note that it is running slow. you note the time on both it and your own clock as it passes. If you then accelerate so as to catch up with the clock, you will note that when you do, more time will have passed on it than on your clock since it originally passed you. However, if you do nothing and the other clock instead undergoes a acceleration which brings it back to you, less time will have passed on it than on your clock. Any past accelerations that either of you have undergone before you first meet will have no bearing on the result.

If both clocks experienced equal acceleration, it's been suggested that both clocks would jump forward and synch. It seems that one could compare the ratio of how each object had accelerated during it's trip by noting how much each clock adjusts when they come together.

 

 

Al goes left at .8c and Bert goes right at .7c, but turns around early so they still meet back at the same time. They still had a relative velocity of nearly the speed of light, so the whole trip, each would see the other's clock run slower. Right before they decelerate, they compare how far behind the other clock is. Even if they don't see the clockove forward during deceleration... Once they stop, they can then compare which clock jumped forward more than the other.

 

Is that a fair interpretation of your suggestion? If not, where do you disagree?

By "meet back at the same time" I assume you mean return to the origin point at the moment. so for example, Bert travels after one year as measured by a clock that remains at the origin and Al does the same. In this case, Al will have aged 1.2 years when they meet up and Bert will have aged 1.43 yrs.

 

The thing with this example is that since both Al and Bert undergo accelerations during their turn around according both the other's clock will "jump forward" at when they turnaround. When Bert turns around Al's clock jump's forward and when Al turns around Bert's clock jumps forward. How much the other clock jumps forward depends on both the acceleration and distance separating them at the moment they turn around. The thing is that in this situation, according to both Al and Bert, the other will not have reached his turn around point before he himself reaches his, but after he has completed the turn around, the other will have already completed his.

 

As far as what he actually "sees": Up until he reaches his turn around point he sees the other clock run slow at a rate of 1/14 of his. Immediately after turn around he will see the other clock run at 1.27 the rate of his. (The small amount is due to the fact the light he is receiving doesn't yet reflect the other ship's change in speed, as it will take some time for the light carrying that information to reach them.) He will continue to see the other clock running at this rate for most of the return trip until the light from the other ship's turnaround gets to him. Then he will see the other clock run 14 times faster. The actual durations for these segments will vary for Bert and Al due to the difference in turnaround point for the two.

 

If SR is a matter of relative velocity, it can't be asymmetrical.

Again: Time dilation is the relative time rate between two relatively moving inertial frames ( each will measure the others as running slower.)

 

Accumulated time difference between two observers that separate and meet again is the result of other Relativistic effects in addition to time dilation. You cannot understand the Twin Paradox by only considering time dilation part.

[SamBridge]

You can definitely calculate that. Not everything is symmetrical.

 

Remember that these things are relative. The clock that flies by is inertial in this example. If you look at if from its perspective, it remains at rest while you fly by and then turn around and return. The usual calculations work out correctly.

Who undergoes acceleration and when doesn't matter except in that it affects the spacetime path of the observer. You can have any number of observers, each taking different paths between the same two points, and each enduring their own particular proper time (aka. aging). Any of the observers' relative aging could be calculated in SR. If two twins accelerate symmetrically, their paths will be symmetrical, and they'll age the same. I've been told that the longer the path is spatially, the shorter it will be in time. An inertial path is the spatially shortest path between 2 points in flat spacetime, so if you include one inertial observer it simplifies the situation because that one will always age the most out of any possible paths taken.

 

It matters that their spacetime paths are different lengths, not that one must not accelerate.

 

As a quick example: Say you have a typical twin setup where an inertial observer ages 4 years, and a traveling twin ages 2, and add another faster traveling twin which ages 1. If you remove the inertial twin, then you still have 2 twins that both travel and return, at different speeds, and the first ages twice as much as the other.

But their space time paths could only be different to a 3rd observer, not to each other. If people on Earth observe me leaving, then that means I observe Earth leaving or moving away. How could it ever be anti-symmetric with only two frames? That's like saying when you draw a line, it's shorter when you go one of the two directions.

Edited April 4, 2014 by SamBridge

[Schneibster]

Here's a simple case that will help define where the poles/zeroes are for you, md65536:

 

Suppose one twin goes to point A, five light years away, at 0.8c and gets there in seven and a half years by his clock. Now suppose he stays there, and the other twin comes to him.

 

When the other twin arrives they'll be the same age again, and they'll both be fifteen years older.

 

The fact that, simply having the first twin turn around and come back, as opposed to the scenario I have just explained, makes such a huge difference in their histories, suggests that the act of rotating makes all the difference. Einstein referred to this as "Mach's Hypothesis."

 

Another pedagogical trick to make this clear to you is to say that, in my case, neither twin sees the entire universe rotate; however, in the original case, where the first twin returns, from the first twin's point of view the entire universe rotates 180 degrees. And this is, in fact, apparently a huge difference; look at the results.

Edited April 4, 2014 by Schneibster

[ACG52]

 

 

Are you suggesting that you can definitively detect which one accelerated more during their trip by noting which clock jumps forward in time when the inertial frames meet back up?

Neither clock jumps forward in time. Synching up when they meet means they both now tick at the same rate, but the difference in the time elapsed between the two remains.

[md65536]

But their space time paths could only be different to a 3rd observer, not to each other. If people on Earth observe me leaving, then that means I observe Earth leaving or moving away. How could it ever be anti-symmetric with only two frames? That's like saying when you draw a line, it's shorter when you go one of the two directions.

It's more like saying "If you draw a straight line between two points, and then draw a curved or bent line between the same two points, the straight line will be shorter." I could try to explain that analogy but it might not help.

 

If you look at your example from any single inertial frame, it is not symmetric. "Your" frame of reference where you see the Earth leaving and then returning, isn't an inertial reference frame.

 

The Doppler shift analysis of the situation http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#doppler clearly shows how there is no symmetry here. If you turn around, you see the Earth change its relative velocity immediately. From Earth, your turn around isn't seen for some time due to delay of light. If you want to claim symmetry, you'll have to face it that the two observers aren't observing the situation symmetrically.

[phyti]

But their space time paths could only be different to a 3rd observer, not to each other. If people on Earth observe me leaving, then that means I observe Earth leaving or moving away. How could it ever be anti-symmetric with only two frames? That's like saying when you draw a line, it's shorter when you go one of the two directions.

The motion is symmetrical, but the physics is not. When you reverse your course to return, you experince inertial forces. Via the equivalence principle, you can assume a pseudo rest frame in a gravitational field for the duration of your reversal. The earth reverses and moves toward you, as a result of that g-field (your perception by choice). If you ask them if they felt any forces, they will say no. No symmetry there.

Even without acceleration, the case is asymmetrical, since the speed profiles are different, which results in differential aging.

[SamBridge]

It's more like saying "If you draw a straight line between two points, and then draw a curved or bent line between the same two points, the straight line will be shorter." I could try to explain that analogy but it might not help.

 

If you look at your example from any single inertial frame, it is not symmetric. "Your" frame of reference where you see the Earth leaving and then returning, isn't an inertial reference frame.

 

The Doppler shift analysis of the situation http://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_vase.html#doppler clearly shows how there is no symmetry here. If you turn around, you see the Earth change its relative velocity immediately. From Earth, your turn around isn't seen for some time due to delay of light. If you want to claim symmetry, you'll have to face it that the two observers aren't observing the situation symmetrically.

But if Earth observes me accelerating, then I must observe Earth accelerating, I still don't see how there could possibly be any room for difference. If Earth observes me turning in a semi circle to turn around, then I instead observer Earth traveling in a semi circle around the star and then heading towards me. Also, how can you model it completely like the Doppler shift because you have the same relativistic effect in both directions? And can you address the hyperdoughnut situation? Would this situation ever occur if there was no turn around?

Edited April 4, 2014 by SamBridge

[Schneibster]

But if Earth observes me accelerating, then I must observe Earth accelerating, I still don't see how there could possibly be any room for difference.

 

Earth doesn't undergo unbalanced acceleration forces. Only you do.

[swansont]

But if Earth observes me accelerating, then I must observe Earth accelerating

Acceleration is not relative. You can tell if you are accelerating (and in an accelerating frame the laws of physics aren't the same)

[SamBridge]

Acceleration is not relative. You can tell if you are accelerating (and in an accelerating frame the laws of physics aren't the same)

So you're telling me if I get in a car right now and accelerate away from a house, I won't observe the house accelerating away from me? What else could I possibly observe in that scenario? If acceleration isn't relative, then that means every frame of reference can agree on the same value of acceleration of an object just as it use to be thought that everyone in the universe measures time as passing the time, but I know you know that's not true.

Edited April 4, 2014 by SamBridge

[md65536]

But if Earth observes me accelerating, then I must observe Earth accelerating, I still don't see how there could possibly be any room for difference.

I'd recommend working through the details of a Doppler shift analysis example with a Minkowski diagram or with numbers. This analysis doesn't rely on anyone feeling acceleration, and it will work with the free fall gravitational turnaround idea. If you work through the details you'll see where the difference is. Earth and the traveling twin really do observe a different situation. Without seeing those details it will be difficult to understand them.

[Strange]

So you're telling me if I get in a car right now and accelerate away from a house, I won't observe the house accelerating away from me? What else could I possibly observe in that scenario?

 

If you get in your car and accelerate away from your house, you will be pressed into the seat. If you are at home watching someone accelerate away in their car, you will not be pressed into your seat. Therefore, you can always tell who is accelerating.

[Schneibster]

So you're telling me if I get in a car right now and accelerate away from a house, I won't observe the house accelerating away from me?

 

No. We're telling you if you get in a car right now and accelerate away from a house, you will feel acceleration forces and the house won't.

swansont and Strange have both said the same thing I did, in different ways. It's all the same thing.

[Didymus]

So then if clocks are said to catch up in time based on the difference in symmetry, would the duration of each trip be irrelevant? Since it's been claimed that time will catch back up when two symmetrical objects stop.

[ACG52]

 

 

Since it's been claimed that time will catch back up when two symmetrical objects stop.

The rate time passes catches up when the two objects stop (relative to each other). The time displayed on their respective clocks does NOT change.

[Schneibster]

So then if clocks are said to catch up in time based on the difference in symmetry, would the duration of each trip be irrelevant? Since it's been claimed that time will catch back up when two symmetrical objects stop.

 

No. Different durations would result in different amounts of dilated time on each clock.

[SamBridge]

I'd recommend working through the details of a Doppler shift analysis example with a Minkowski diagram or with numbers. This analysis doesn't rely on anyone feeling acceleration, and it will work with the free fall gravitational turnaround idea. If you work through the details you'll see where the difference is. Earth and the traveling twin really do observe a different situation. Without seeing those details it will be difficult to understand them.

 

If you get in your car and accelerate away from your house, you will be pressed into the seat. If you are at home watching someone accelerate away in their car, you will not be pressed into your seat. Therefore, you can always tell who is accelerating.

There wouldn't happen to be an opposite to length contraction that results in simple relativity scenarios would there? Like length expansion? If a photon source accelerates towards and away from something, there will be either a blueshift or a red shift from the FOR of whatever is observing the source, but, as far as I know, time dilation and length contraction work the same no matter what direction someone is going, so I still don't see how a fictitious force breaks this symmetry with simultaneity.

And in the space scenario, people on Earth might not get pushed into their seat, but what actual force explains the difference in the scenarios between the space ship and the Earth if both observers observe each other traveling in a semi circle around the star?

And still no one has addressed the hyperdonought situation.

Edited April 5, 2014 by SamBridge

[Schneibster]

There wouldn't happen to be an opposite to length contraction that results in simple relativity scenarios would there? Like length expansion? If a photon source accelerates towards and away from something, there will be either a blueshift or a red shift, but, as far as I know, time dilation and length contraction work the same no matter what direction someone is going, so I still don't see how a fictitious force breaks this symmetry with simultaneity.

 

GR specifically deals with acceleration; technically to discuss what happens when you accelerate we have to discuss gravity, because the Equivalence Principle says gravity is locally indistinguishable from acceleration; this is where the "enclosed elevator" scenarios come from, where one observer is in a closed elevator in Earth's gravity, and the other is on top of a rocket accelerating at 1g. GR says neither observer can perform an experiment inside their respective elevator that will tell them whether they're on Earth or on the rocket. That's the Equivalence Principle.

 

Unlike movement and velocity, acceleration is not relative. And, of course, neither is gravity. You're thinking of the Special Relativity Principle, which says there is no special frame of reference against which motion can be measured. There is no such principle for acceleration/gravity.

 

Also, are you sure "fictitious" or "pseudo" force means what you think it means?

Edited April 5, 2014 by Schneibster

[SamBridge]

 

GR specifically deals with acceleration; technically to discuss what happens when you accelerate we have to discuss gravity, because the Equivalence Principle says gravity is locally indistinguishable from acceleration; this is where the "enclosed elevator" scenarios come from, where one observer is in a closed elevator in Earth's gravity, and the other is on top of a rocket accelerating at 1g. GR says neither observer can perform an experiment inside their respective elevator that will tell them whether they're on Earth or on the rocket. That's the Equivalence Principle.

 

Unlike movement and velocity, acceleration is not relative. And, of course, neither is gravity. You're thinking of the Special Relativity Principle, which says there is no special frame of reference against which motion can be measured. There is no such principle for acceleration/gravity.

I was thinking about the equivalence myself and that gravitational red-shifts and blue shifts aren't symmetric, but still don't see how the gaps are filled in. A system with multiple objects accelerating has the potential to be symmetric, but gravity doesn't, they have some differences, they aren't 100% the same. it's just that Einstein used the mathematical extrapolations and the elevator scenario you mentioned to say gravity is merely an acceleration in the curvature of space time, and that's why there's similarities between going through space and space itself curving.

 

 

Also, are you sure "fictitious" or "pseudo" force means what you think it means?

When you get "pushed back into the seat of your car", it is a fictitious force, there isn't suppose to be anything pushing you back, the car is suppose to be accelerating into your body and that's it.

Edited April 5, 2014 by SamBridge

[Schneibster]

 

I was thinking about the equivalence myself and that gravitational red-shifts and blue shifts aren't symmetric, but still don't see how the gaps are filled in. A system with multiple objects accelerating has the potential to be symmetric, but gravity doesn't, they have some differences, they aren't 100% the same. it's just that Einstein used the mathematical extrapolations to say gravity is merely an acceleration in the curvature of space time, and that's why you get similar results because in both scenarios, whether your in a field or just flying in a rocket, you still have to measure the speed of light as constant in a local frame, so there will be similar dilation effects in both instances.

 

There is a place where this gets hinky, but it's not the place you're thinking. It mostly has to do with the shape of a gravity field. A gravity field created by a mass is spherical. In real terms, this means, there is in fact an experiment that can tell you if you're accelerating or in a spherical gravity field. It's quite simple: as two objects at the same altitude fall in a real gravity field they move closer together. And this does not happen in an accelerating rocket. You will hear "local" used a lot; this is an acknowledgement of this real difference in the behavior of gravity and acceleration. The key point is, however, if you can imagine a planar instead of spherical gravity field, that planar field would, in fact, be completely indistinguishable from acceleration within the elevator. When relativists talk about a "local experiment," they mean one that doesn't cover enough volume to allow this difference to be measured.

 

It is, BTW, more correct to say "acceleration due to the curvature of spacetime."

 

However, that is a shot in the dark; I'm not entirely clear what you're objecting to so I'm quoting some doctrine that's similar to what I perceive you as talking about, but I'm not confident I'm really answering your questions. However, if I'm not, the above background is necessary to understand the correct answer anyway, so it's not wasted time.

 

When you get "pushed back into the seat of your car", it is a fictitious force, there isn't suppose to be anything pushing you back, the car is suppose to be accelerating into your body and that's it.

 

 

The car pushes you forward. This is a real force. You can also define matters (by using the frame in which the car is motionless and accelerates the Earth to its rear) in the opposite direction, and it is this that is called a "pseudo" or "fictitious" force; however, it is always (or, rather, it's components always add up to) Newton's Second Law, as modified by relativity.

 

Later: One of the things that's always gotten my rear end beat by relativists is forgetting that motion and speed/velocity are relative, but acceleration is absolute. You can always perform a local experiment that will tell you if you are accelerating; you can never perform one that will tell you if you are moving.

Edited April 5, 2014 by Schneibster

[swansont]

 

And still no one has addressed the hyperdonought situation.

 

You didn't give any details about how such a geometry would exist, but space would have to be curved in that situation, so special relativity would not apply. SR works only in flat space. (another way of saying no accelerations)

[Schneibster]

 

You didn't give any details about how such a geometry would exist, but space would have to be curved in that situation, so special relativity would not apply. SR works only in flat space. (another way of saying no accelerations)

 

You can sneak them in, but they always make weirdness, like the twin paradox and so forth.

[uncool]

 

You can sneak them in, but they always make weirdness, like the twin paradox and so forth.

To be exact, it means no acceleration of frames. Objects can accelerate, but in SR inertial frames are in constant rectilinear motion with respect to each other.