1. Dec 19, 2011

Let's say there are two synchronized clocks on Earth.

One leaves the Earth going at .8c and travels in uniform motion for one year. At the one year mark the clock is handed off to another spaceship coming towards earth at the same speed of .8c.

The clock eventually makes it back to Earth and two years have gone by for it. How much time has gone by on Earth or what does the other clock on Earth say?

2. Dec 19, 2011

### ghwellsjr

3.333 years of earth time.

3. Dec 19, 2011

Can you show me how you get that?

4. Dec 19, 2011

### mathman

te=ts/√(1-(v/c)2) = 2/.6 = 3.333...

te = earth time
ts = ship time

5. Dec 19, 2011

Where does the actual change take place?

I assume it takes place the moment the clock is switched to the space ship that is heading towards Earth. Does this count as an acceleration for the clock or is it just a change of reference frame?

I understand that after the first year of travel for the spaceship heading away from Earth he will see the clock on Earth as 1.667 years which I see is the same thing as the lorentz factor.

Now I don't understand how it actually works because I am under the impression that for the clock on Earth only one year has really passed for it in it's own FOR. It just appears to have aged 1.667 years from the FOR of the space ship. I guess it has to be the same for the return. Would this mean there is no acceleration at all in this twin paradox example?

Why is this so? Can someone prove to me why there clock on Earth will say have elapsed 3.33 years and the clock on the ship will only have passed 2 years when it returns.

There has to be something in here that doesn't necessarily have to do with relativity but with the fact that the clock on board is moving at .8c.

What happens if they call each other after a year out for the spaceship? Will they both say one year has passed?

The real problems I am having with SR is that I am the type of person that wants to understand what the equations actually mean and not just how to perform them.

6. Dec 19, 2011

### bobc2

I don't know if you are comfortable with space-time diagrams. If so, you can put the problem in the context of comparing paths traveled through 4-dimensional space. Here are a couple of sketches that tell the story. I've set this example up for a 10yr trip out and 10yr trip back. You can essentially disregard the effect of the acceleration, other than the fact that it puts the traveler on a different path (associated with change of velocity).

Basically, the traveling twin is taking a shorter path from the starting point to the ending point. It may look like the traveling twin takes a longer path, because the lines are longer on the computer screen. But, look carefully and you will see hyperbolic calibration curves that calibrate the screen distances properly for the Minkowski space.

7. Dec 19, 2011

Unfortunately I am having trouble with that graph.

I'm just trying to understand why 2 years only passes for him and 3.33 years passes for the clock on Earth. I want to understand this without just saying it's because of some equation which I may not even understand.

8. Dec 20, 2011

### ghwellsjr

The actual change is taking place during the entire trip. When we're talking about time dilation, we're talking about the rate at which a clock ticks not the time displayed on the clock.

Think about two clocks that you may have in your possession such as a wall clock and a windup clock. The wall clock keeps perfect time but the windup clock runs a little slow, losing about two seconds per day. You set the windup clock to the correct time that the wall clock displays. After a month, the windup clock is a minute behind the wall clock. Your windup clock has a little lever on it that allows you to change the rate at which it ticks. If you move that lever, the clock immediately starts ticking at a different rate but the hands on the clock don't make any change at all, do they? It takes time for the changed tick rate to show up on the clock as a difference in the transpired time, right? So if you make an adjustment in the tick rate, you will have to wait some time for it to show up on the hands of the clock.

In a similar way, when a clock accelerates to a new speed, its tick rate immediately changes but there is no immediate change in the time on the clock. You have to wait some time for the change to show up and the longer you wait, the more the difference in time between the accelerated clock and the unaccelerated one.
If the clock was transferred from a ship going away from the earth to a ship going toward the earth, then it will experience a huge acceleration which could immediately change its tick rate (but not the time displayed on the clock). There is no reason to consider a change of reference frame. A clock can be accelerated within a single reference frame and this will account for a change in its tick rate. If you wanted to consider the clock to be at rest in a non-inertial frame, then it's tick rate will never experience time dilation. You will get very confused if you think in terms of a clock changing frames just because it changed the direction of its motion.
Your understanding is mixed up. Each twin sees the other ones clock as running slower than their own, not faster. But it's not going to be helpful to think about what each twin sees of the other twin if you want to use a FOR for your analysis and your numbers are correct for a FOR in which the Earth is at rest, that is, after one year for the traveling clock, 1.667 years will have passed for the Earth clock.
Where did you get the idea that only one year has passed for the Earth in its FOR? Even if you consider the FOR for the clock in the spaceship as it is traveling away from the Earth, after one year for the traveling clock, only 0.6 years will have passed for the Earth clock, not 1.667. Remember, in a clock's rest frame, meaning that it is stationary, all other clocks in relative motion will experience time dilation. But if you want to use this FOR, you must also use it during the inbound portion of the trip where the clock will no longer be at rest but will be traveling at an even much higher rate and experiencing much more time dilation so that it ends up with less time on it than the Earth clock.

Again, if you use just one inertial frame of reference, it will be much easier to understand. The Earth frame is the easiest to use. So during the outbound portion of the trip, the traveling twin has aged 1 year but the Earth-bound twin has aged 1.667 years. The same thing applies for the inbound portion of the trip so the traveling twin has aged 2 years while the Earth-bound twin has aged 3.333 years.

Do you understand that time dilation occurs for any clock in motion with respect to a give Frame of Reference?
Yes, moving at .8c relative to a FOR will cause that clock to experience a tick rate that is 0.6 of normal.
This is a tricky question because, although we can identify when one year has passed for the traveler, we cannot associate that same instant with the Earth Twin.
Me too. I'm sure you'll get to the level of understanding you desire if you keep at it.

9. Dec 20, 2011

### harrylin

Special relativity only gives the equations that follow from the postulates which were based on observation; our explanations depend on how we think about it. For starters, the first explanation of why a change of velocity (and thus acceleration) breaks the symmetry has been given in the sections of p.47-53 of:
http://en.wikisource.org/wiki/The_Evolution_of_Space_and_Time

Basically the explanation there is that this problem reveals the existence of physical space (3D Space or "stationary ether"), wrt which a change of velocity can be detected.
A different way of thinking about the same problem is already described in post#6: that alternative explanation is that the traveling twin is taking a shorter path through Spacetime (4D Space or "block universe").
And there are other points of view such as here:
http://ndpr.nd.edu/news/25025-physical-relativity-space-time-structure-from-a-dynamical-perspective/
Whatever our opinion, obviously there must be something wrt which the change of velocity or the shorter path takes place.

As you see this gets you into philosophy; for you to figure out what makes most sense to you, depending on how your brain is wired.

Last edited: Dec 20, 2011
10. Dec 20, 2011

### nitsuj

Specifically, imagine that the Earth clock and ship clock also emit a beep every second with a radio signal that ideally travels at exactly c from start to reception.

Imagine how, as the ship speeds away from earth, the earth beeps will be received less frequently then 1 second.

When the ship is on the return, the beeps recieved from the earth clock are more frequent then 1 second.

Last edited: Dec 20, 2011
11. Dec 20, 2011

### nitsuj

Ohh those graphs are dandy for a non mathimatical approach. Maybe look into a right angle triangle, pathagoreons theorem and c. From there it translates well into those spacetime diagrams. Well at least the less decorated ones, unlike those above

To understand "why" it's 3.33 years for Earth clock and 2 years for the ship look into intervals.

it brings a time component into distances.

The interval between the ship starting its journey to the clock "handoff" (return trip) is the same for both the Earth FoR and ship FoR. However the time taken to travel this distance and the distance travelled are both calculated to be different values then what the other FoR calculates.

I hope someone here with the skills can let you know what figure each FoR comes up with for distance travelled. Seems like it is important in answering the question how a different amount time passed for each FoR. (the ship claims it travelled less distance then what Earth FoR claims, this is the exact same relationship as the different amounts of time passed issue)

Last edited: Dec 20, 2011
12. Dec 20, 2011

### harrylin

I'm afraid that that is a bit too vague: it's even true in classical physics with absolute time. :tongue2:
Thus it's necessary to be a little more quantitative about relativistic Doppler - like Langevin did in his description on p.51 to which I already referred (his description was still very approximate, just enough to sketch the main effects).

13. Dec 20, 2011

### nitsuj

Ah okay, thanks.

I see now, of course it's true even in pre SR classical physics but, in post SR we know that c is the maximum, as I noted in the post. This is what makes it significant. I see my comment and your reference as simular.

However I agree with you that my comment (and your reference) are too vague. I don't know what time the travelling observer would measure in between the beeps emitted from Earth, the fact the traveller is moving away from the pulse + the travelling observers dilated time.

For all I know the traveller measures the time between beeps as one second still. Since I don't know, I shouldn't have posted it. I thought the concept would be helpful.

Last edited: Dec 20, 2011
14. Dec 20, 2011

### mathman

One way to look at is by thinking about the distance travelled. Special relativity (Lorentz transformation) tells us that the distance is forshortened (for the guy in the rocket ship) in the direction of travel so he has to cover less distance.

There is no way to understand it without the Lorentz transformation, which affects the distance and the clock.

15. Dec 20, 2011

What does wrt stand for?

I guess could someone show me how the clock example would work with the lorentz transformation? Thanks.

16. Dec 20, 2011

Is this right?

One year passes on the clock on Earth and from the FOR of the earth the traveling clock at .8c is .6 of a year, correct?

Isn't this relationship symmetrical?

Doesn't this mean that when the traveler sees one year on his clock he sees .6 on the clock on earth?

17. Dec 20, 2011

### Fredrik

Staff Emeritus
wrt = with respect to

It does.

The reason why this isn't an obvious contradiction is that statements about what things are like in Earth's frame of reference are (usually) statements about the coordinates assigned by the inertial coordinate system associated with the curve in spacetime that describes Earth's motion, while statements about what things are like in the traveler's frame of reference are statements about the coordinates assigned by the inertial coordinate system associated with the curve in spacetime that describes the traveler's motion. So we are talking about coordinate assignments made by two different coordinate systems.

To fully understand why it's not a contradiction, you must understand Minkowski spacetime and the standard procedure to associate inertial coordinate systems with timelike curves. In particular, you must understand relativity of simultaneity. What I said in the previous paragraph only explains why you can't immediately, without any further thought, conclude that the two statements are contradictory.

18. Dec 21, 2011

### Snip3r

if i m in space ship wouldnt i think earth's tick rate immediately changes?(though i have noticed explanations for twin paradox completely without acceleration)i m completely blind of where the symmetry breaks

19. Dec 21, 2011

### Fredrik

Staff Emeritus
Yes, if you define the ship's "frame of reference" as its comoving inertial coordinate system, then the ticking rate of the Earth clock immediately changes in the ship's frame. However, it is also pushed forward by several years when the rocket turns around, because of relativity of simultaneity. An event on the ship's world line just before the turnaround is simultaneous in the ship's frame with a much earlier event on Earth, than an event immediately after the turnaround. This is a result of the choice to define their "experiences" using the comoving inertial coordinate systems.

Let's consider some specific numbers. c=1 (choice of units). v=0.8. This implies $\gamma=1/\sqrt{1-v^2}=1/0.6=5/3\approx 1.667$, and $1/\gamma=3/5=0.6$. Let's say that the ship is gone for 20 years in Earth's frame. Then the journey will take 20*0.6=12 years according to clocks and people on the ship. In the ship's frame, Earth's clock is slow by a factor of 0.6, so just before the turnaround, it's only at 6*0.6=3.6 years. Similarly, on the way home the time it displays will only increase by 3.6 years. But at the turnaround event, the number it displays will change from 3.6 to 16.4 in no more time than it takes to turn the ship around. If we idealize the scenario by taking the turnaround to be instantaneous, then the clock and everything on Earth instantly ages 12.8 years.

20. Dec 21, 2011

### MikeLizzi

Nicely stated Fredrik. Several times I have posted statements to the effect that the astronaut must consider the earth clock to be running faster than his own while accelerating thru the turnaround. All I got was negative comments.