Twin Paradox: Understanding Who Ages Less

[matheinste]

Hello Al68

Quote:-

---- agree. I never said otherwise. I said: "Whichever twin measures the most distance between events will have the most elapsed time between those events."----

As Hurkyl pointed out to us both, distance is the wrong word. However your quote is still wrong. It should read " Whichever twin measures the most 'distance' between events will have the least elapsed time between those events"

Edit. Reading the latest post from Hurkyl #105 i t seems i have also been using the phrase elapsed time rather loosely but you will know wehat i mean

Matheinste.

 

[Hurkyl]

Here's a scenario: Alice and Bob are both sitting on Earth. They never leave each other's side. They consider the worldline of a rocket traveling from here to alpha centauri. I choose as my two events: "The rocket taking off" and "The rocket arriving". Alice decides to measure things relative to an Earth-centered inertial frame. Bob decides to measure things relative to an inertial frame where the rocket is (mostly) stationary. Bob measures the shorter coordinate distance between the two events (zero!), but also measures the longer coordinate time between those two events. And the aging of the twins seems entirely irrelevant to anything in the setup.

Same scenario, but this time Bob uses a rescaled version of the chart Alice uses: one that doubles both lengths and times. This time, Bob measures the longer coordinate distance between the two events, and the longer coordinate time between the events.


Here's another fun one. This time, Bob gets on the rocket, and uses the coordinate chart he originally used. This time, I will involve four events involved:
1. The rocket's departure
2. The rocket's arrival
3. Earth, at the time simultaneous with (2), as measured by Alice's chart
4. Earth, at the time simultaneous with (2), as measured by Bob's chart
Everyone can compute that:
(A) Alice's aging between (1) and (3) is more than Bob's aging between (1) and (2).
(B) Alice's aging between (1) and (4) is less than Bob's aging between (1) and (2).
This is a common one-way travel setup... how would you like to treat it?

 

[Al68]

The acceleration itself has absolutely no effect. What has an effect is choosing a turning point which is at rest relative to one of the twins and it this which causes the symmetry break - not the acceleration.

cheers,

neopolitan

neopolitan, you and I seem to be the only people around who see this obvious source of asymmetry in the twins paradox. The only asymmetry that cannot be eliminated simply by a minor change in the scenario. The only asymmetry that actually comes into play in the SR equations. The only asymmetry that can be used to show the ship's twin to age less just by doing the math and not making claims that are not even mentioned in SR.

Al

 

[Al68]

When people seek to resolve the twin paradox, they seek to point out a flaw in the logical argument in the twin paradox. If you are talking about something that isn't the twin paradox (e.g. any experiment involving twins that don't reunite), then that is something irrelevant.


In a one-way journey, it is impossible for both twins to be present at both events. And there is no intrinsic way to compare their ages.

Just have a third observer at the second event. Easy. And he's at rest with earth, so even clock synch is easy.

And an experiment involving twins who don't reunite is relevant to my point. Which is that a different experiment could yield a similar result even without acceleration involved.

Al

 

[Al68]

Events don't travel; they're events. This doesn't make sense.

I never mentioned "traveling events".

Similarly, you mentioned 'elapsed time'; elapsed time of what?

a clock.

You do realize that, in any experimental setup, by choosing the appropriate coordinate chart, I can make either twin (properly!) compute any value I want for any coordinate-dependent quantity I want, right?

Yes, but I have no interest in getting that far off track.

Al

 

[Hurkyl]

Just have a third observer at the second event. Easy. And he's at rest with earth, so even clock synch is easy.

Being at the second event doesn't mean he's at rest with Earth. To wit, the spacebound twin is at the second event, and he's not at rest with Earth.

I assume your plan is to synchronize the wristwatches of the Earthbound twin and the third observer according to the Einstein convention, and rather than compare the the aging of the two twins (which is not a well-defined question), you intend to compare the aging of the spacebound twin with the difference between the third observer's age at the second event with the Earthbound twin's age at the first event.

But why do it that way? Why not have the third observer at rest with the spacebound twin, and arrange things so that the third observer's clock when he meets the Earthbound twin reads the same as the spacebound twin's clock when he arrives at the second event? Or why not use a pair of observers at rest with respect to each other, but not relative to either of the twins?

And an experiment involving twins who don't reunite is relevant to my point.

But your point is not the twin paradox -- you have no business criticizing resolutions of the twin paradox on the grounds that they are not relevant to your alternative scenario.

 

[MeJennifer]

The "spacetime path" traveled by an object¹ is time-like; the notion of 'distance' doesn't make sense. The 'duration' of the path, however, is exactly what the observer's wristwatch measures.

I completely disagree. In spacetime there is a clear definition of the distance between two events. In spacetime there are no durations only distances between events.

 

[Hurkyl]

In spacetime there is a clear definition of the distance between two events.

Only for space-like separated events.

In spacetime there are no durations only distances between events.

There are durations for time-like separated events.

When $(c \Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$ is positive, the events are time-like separated, and the (proper) duration between the two events is $(1/c) \sqrt{(c \Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2}$. When it's negative, the events are space-like separated, and the (proper) distance between the two events is $\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2 - (c \Delta t)^2}$

 

[Al68]

Hello Al68

Quote:-

---- agree. I never said otherwise. I said: "Whichever twin measures the most distance between events will have the most elapsed time between those events."----

As Hurkyl pointed out to us both, distance is the wrong word.

I used the word distance so that someone wouldn't mistakingly think I meant spacetime path. I mean spatial distance.

However your quote is still wrong. It should read " Whichever twin measures the most 'distance' between events will have the least elapsed time between those events"

No, I'm pretty sure you've got it backward. In the twins paradox, the ship's twin measures less distance traveled than the Earth twin, hence less time elapsed.

Al

 

[matheinste]

Hello Al68.

If i have got it back to front ( i am sure i have not ) no-one else has picked up on it.

Matheinste

 

[Al68]

Being at the second event doesn't mean he's at rest with Earth. To wit, the spacebound twin is at the second event, and he's not at rest with Earth.

OK. I meant an observer at rest with earth.

But your point is not the twin paradox -- you have no business criticizing resolutions of the twin paradox on the grounds that they are not relevant to your alternative scenario.

Huh? I think you missed my point completely. It is easy to construct a scenario like the twins paradox, but where there is no acceleration, yet with the same result. My point is not that the answer given in the common resolutions is wrong, but that the claim of acceleration being important is unsubstantiated. Because you could have the same result if there were no acceleration involved.

Al

 

[MeJennifer]

Only for space-like separated events.


There are durations for time-like separated events.

When $(c \Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2$ is positive, the events are time-like separated, and the (proper) duration between the two events is $(1/c) \sqrt{(c \Delta t)^2 - \Delta x^2 - \Delta y^2 - \Delta z^2}$. When it's negative, the events are space-like separated, and the (proper) distance between the two events is $\sqrt{\Delta x^2 + \Delta y^2 + \Delta z^2 - (c \Delta t)^2}$

The metric of spacetime defines the distance between two events. This metric is well defined (ignoring for the moment if this space is actually Hausdorff) for both flat (Minkowski) and curved (Lorentzian) spacetimes.

http://en.wikipedia.org/wiki/Metric_(mathematics )

 

[Al68]

Hello Al68.

If i have got it back to front ( i am sure i have not ) no-one else has picked up on it.

Matheinste

Well, your statement would have the Earth twin younger than the ship's twin when they reunite.

Al

 

[Hurkyl]

but that the claim of acceleration being important is unsubstantiated.

Acceleration is important because it precisely demonstrates the flawed reasoning -- one twin accelerates, is not stationary in any coordinate chart, and thus the time dilation argument that the Earthbound twin should age less is invalid. If you do not bring up acceleration (or something equivalent to it), then you cannot invalidate the twin paradox.

 

[matheinste]

Hello Al68.

I think you are wrong. However i would rather let someone decide this for us rather than continue disagreeing.

It's 4AM here so i will be disappearg soon.

Matheinste.

 

[Hurkyl]

The metric of spacetime defines the distance between two events. This metric is well defined (ignoring for the moment if this space is actually Hausdorff) for both flat (Minkowski) and curved (Lorentzian) spacetimes.

http://en.wikipedia.org/wiki/Metric_(mathematics )

First off, you've cited the definition of the wrong word: see Metric tensor. Secondly, see proper length and proper time.

 

[MeJennifer]

Every serious book on relativity writes about distances between events in spacetime.

What I write seems to fall on deaf ears, so I won't respond anymore to this.

 

[Al68]

Hello Al68.

I think you are wrong. However i would rather let someone decide this for us rather than continue disagreeing.

It's 4AM here so i will be disappearg soon.

Matheinste.

I don't understand why there's a disagreement. In the twins paradox, the ship's twin measures the distance traveled to be less than that measured by the Earth twin. If I'm right, that would mean that the ship's twin would age less. I thought we all agreed on which twin aged less. I thought the only disagreement was why.

Al

 

[matheinste]

Hello Al68.

I have no disagreement about which twin ages less. I only disagree about the spactime distance. Tomorrow i will consult my textbooks and then can hopefully quote some relevant passages.

Matheinste.

 

[Al68]

Hello Al68.

I have no disagreement about which twin ages less. I only disagree about the spactime distance. Tomorrow i will consult my textbooks and then can hopefully quote some relevant passages.

Matheinste.

Well, there's the problem, I was referring to spatial distance, not spacetime length.

Al