# Time Dilation. I don't get it!

1. Dec 12, 2006

### eiapeteides

Suppose, you live on a planet in orbit around Alpha Centaury which is approximately 4 light years away from our own Sun. You board a spaceship bound for Earth, and fire up the engines at the exact moment you observe the Earth to be located at the position of the Vernal Equinox. Your observation of the equinox does of course trail its occurrence by 4 years due to your distance from the actual event. Your spaceship accelerates almost instantly from zero to about 0.999 c. For the most part of the journey you travel at constant speed.

Now, here is the question: How many Vernal Equinoxes are you going to observe during your transit to Earth?

According to Special Relativity, you, the passenger of the spaceship, are at rest since the reference system of the ship is a valid inertial system as long as the ship moves at a steady rate. You see surrounding space, Sun and Alpha Centaury with it, move past your position at a constant speed of 0.999 c.

The solar system is in steady linear motion in respect to the ship's frame of reference hence you observe terrestrial clocks running slower then your own. One Earth-hour is dilated to 22 of your Ship-hours. Simultaneously, the distance between Sun and Alpha Centaury appears contracted to 1/22 of the initial distance measuring now a bit less then 0.2 light-years in the ship's reference frame.

If special relativity were correct, then according to the ship's clock the trip would last less then 0.2 years. Since for the ship-bound observer 22 ship-years correspond to one Earth-year, the passenger won't see the Earth move much past the position of the Vernal Equinox during the journey.

But what happened to the 4 Vernal Equinoxes which did already occur by the time the start signal, the information of the first equinox, reached the ship at Alpha Centaury? How can the ship miss the visual information of those events?

If, on the other hand, the passenger does indeed observe those four Vernal Equinoxes, then he does so within 0.2 ship-years. That would mean that he witnessed the passage of at least 4 Earth-years within 0.2 ship-years. This observation contradicts the time-dilation effect predicted by special relativity. Remember! In the reference frame of the ship the Earth is in motion and must therefore "Age" slower, not faster, then the ship.

An Earth-bound observer sees things slightly different. In his frame of reference, 4 Vernal Equinoxes occur while the ship is in transit, a total of 8 occurred since the one (Vernal Equinox) that served as the start signal. The ship arrives 0.004 years after the information of its departure from Alpha Centaury. The Earth-bound observer expects the passenger to count 8 Vernal Equinoxes during his 4 Earth-years in transit.

There is a difference between counting 0, 4 or 8 occurrences of the Vernal Equinox. The observation of the passenger, assuming he can count and remains focused on the Earth during the journey, can impossibly confirm all those predictions.

What am I doing wrong?

2. Dec 12, 2006

### Janus

Staff Emeritus
What's throwing the monkey wrench in is the period in which you accelerate nearly instantly to .999c. It is during this period in which those "extra" equinoxes occur according to the passenger. Simple time dilation transforms do not apply to an accelerating observer.

3. Dec 12, 2006

### eiapeteides

How can that be? Lets say, at the end of my acceleration phase I'm at a distance of 0.001 light-years from Alpha Centauri how can I have encountered the information of the Vernal equinoxes at that point? They haven't propagated that far yet!

4. Dec 12, 2006

### JesseM

The time dilation equation doesn't tell you how fast you will see a clock ticking using light-signals, it tells you how fast you'll judge it to have been ticking after you correct for the time it took the signal from each successive tick to reach you. For example, suppose at t=10 seconds according to my clock I see the light from a ship at a distance of 10 light-seconds from me, with its clock reading "5 seconds", and then at t=14 seconds I see the same ship at a distance of 4 light-seconds, with its clock reading "13 seconds". If I then factor out the signal delays in each case, I'll conclude the ship was "really" 10 light seconds away at t=10-10=0 seconds in my frame, and that it was "really" 4 light-seconds away at t=14-4=10 seconds in my frame (meaning it travelled 6 light-seconds in 10 seconds, moving at 0.6c). Because the "real" time between those events is judged to have been 10 seconds, yet its clock only advanced by 13-5=8 seconds, I conclude the clock was slowed by a factor of 0.8.

But note that before I correct for the time for light-signals to reach me, the clock actually looked sped-up rather than slowed down, because it only took 4 seconds for me to see the clock advance from 5 to 13 seconds. The reason for this difference is just that the light from the second time reading had a shorter distance to travel to reach me than the light from the first time reading, owing to the fact that the ship had moved closer to me during that time. This effect is known as the "Doppler shift", which changes the observed frequency of any regular signal like a ticking clock when the emitter is moving relative to me (you can see the equations for the relativistic Doppler effect here). Basically, because of this any clock which is moving towards me will appear to be ticking faster than my own clock when I observe it using light, although when I factor out the signal delays I'll conclude it was actually ticking slower. If you go to the Doppler shift explanation from the Twin Paradox FAQ you can see more explanation of how the Doppler effect explains what each twin sees, and there's also a nice spacetime diagram of this in the Too Many Explanations section.

Last edited: Dec 12, 2006
5. Dec 12, 2006

### eiapeteides

So how many Vernal equinoxes are witnessed by the passenger. I can factor in the effect of acceleration and construct two possible scenarios:

1. The ship accelerates, travels at 0.999c and decelerates at arrival at Earth:
a) Acceleration phase: passenger counts 4 Equinoxes in quick succession
b) Cruise phase: Earth appears frozen in space passenger counts 0 Equinoxes
c) Deceleration phase: passenger counts 4 Equinoxes in quick succession

makes a total of 8 Vernal Equinoxes counted which agrees with the prediction of the Earth-bound observer.

2. The ship accelerates, travels at 0.999c and flies by the Earth:
a) Acceleration phase: passenger counts 4 Equinoxes in quick succession
b) Cruise phase: Earth appears frozen in space passenger counts 0 Equinoxes ship passes by earth.

Makes a total of 4 Vernal Equinoxes counted. During fly-by the number is communicated with Earth-bound observer and we got a contradiction!

Any Ideas why or if indeed that contradiction appears?

6. Dec 12, 2006

### JesseM

Forget acceleration, it isn't important to your question. Just suppose the traveler was coming in from infinity at constant velocity, and we just want to know how many equinoxes he saw between passing Alpha Centauri and passing Earth.
Either you didn't understand what I just said about the Doppler effect, or you're ignoring it. The answer is that because of the Doppler shift, throughout the journey from Alpha Centauri to Earth the ship will see the Earth orbiting faster than normal, not slower than normal. If you want to have the velocity be 0.999c, then using the Doppler equation here, we can calculate that the orbits of the Earth will appear to be sped up by a factor of $$\sqrt{\frac{1 + 0.999}{1 - 0.999}}$$, or around 44.7. So, the ship-observer will see one vernal equinox every 0.0224 years according to his own clock. Due to length contraction he measures the distance between Earth and Alpha Centauri as $$4*\sqrt{1 - 0.999^2}$$ = around 0.17884 light-years, so he sees the time to cross that distance as 0.17884/0.999 = 0.1790 years. So, he sees a total of (0.1790 years)*(44.7 vernal equinoxes/year) = 8 vernal equinoxes in that time.

Last edited: Dec 12, 2006
7. Dec 12, 2006

### eiapeteides

You are right that simplifies matters a lot! And based on your calculation, the passenger will count 8 Vernal equinoxes. That does indeed confirm with the prediction of the earth bound observer and sounds absolutely reasonable.

But here is the problem: If the passenger counts 8 Vernal equinoxes, he can right fully expect the earth-bound observer to have aged 8 Earth-Years during the transit time of 0.2 Ship-Years. Now in the reference frame of the ship the Earth is moving at 0.999c. So should the passenger not expect the Earth-bound observer to be subject to time-dilation and age therefore slower?

At such a high velocity, should not a signal coming from earth appear red-shifted to the passenger due to the Time dilation effect on terrestrial emitter?

However I'm not sure i really understand the concept of relativistic Doppler shift. Clarifications are very welcome.

8. Dec 12, 2006

### JesseM

He should expect to see the Earth-observer as aging 8 years, but when he takes into account the light signal delays, he will conclude the Earth-observer was "really" aging slower in his frame. See my earlier post for an explanation of the difference between how fast you see a clock ticking and how fast you calculate it was "really" ticking after you take into account the different distances that the light-signals from each tick had to travel to reach you. In your example, suppose that when the ship-observer passes Alpha Centauri, his own clock reads t=200 years, and he sees the Earth-observer celebrating his 40th birthday. Now he wants to know at what time this event "really" happened in his coordinate system. Well, he knows that the distance between Earth and Alpha Centauri is 0.17884 light-years in his frame, and that both are moving towards him at 0.999c. So if the Earth is at a distance of 0.17884 light-years from him at t=200 years, then at some time T years prior to t=200 years (i.e. t=200-T), the Earth would have been at a distance of 0.17884 + 0.999T light-years from his position. And since light travels at 1 light-year per year, if the signal was emitted T years ago, it must have taken 0.17884 + 0.999T years to reach him. We want to figure out a T such that if the signal was emitted T years ago, it would take T years to reach him (so it would be arriving right now), which we can find by solving:

0.17884 + 0.999T = T
0.17884 = 0.001T
T = 178.84

So, he concludes that the image of the Earth-observer celebrating his 40th birthday really happened 178.84 years ago, when the Earth was at a distance of (0.17884 + 0.999*178.84) = 178.84 light-years from his current position. Since his current time is t=200 years, this must have happened at t=200-178.84=21.16 years.

Then after he passes Alpha Centauri, it takes him 0.1790 years to reach Earth, and when he reaches Earth he finds the Earth-observer to be 48 years old. Since they are now in the same location there is no signal delay, in his frame the Earth-observer is "really" 48 years old at a time of t=200+0.1790 = 200.1790 years. So although he saw the Earth-observer age a little over 8 years between the time he reached Alpha Centauri and the time he passed Earth, when he corrects for signal delays, he concludes it "really" took the Earth-observer 200.1790 - 21.16 = 179.02 years to age a mere 8 years! So in his frame the Earth-observer was aging at about 0.0447 the normal rate, which is exactly what you get if you plug 0.999c into the time dilation factor of $$\sqrt{1 - v^2/c^2}$$.

This example involves some hairy algebra though, I think the example I gave in my first post is a lot easier to follow--did you look it over, and if so was there anything you didn't understand? For example, do you understand why, if I look through my telescope at t=14 seconds and see a ship 4-light seconds away whose own clock reads 13 seconds, then because of the 4-second light signal delay, I would conclude that the other ship's clock "really" read 13 seconds at t=14-4=10 seconds in my own frame?
Blue-shifted, actually. If you work it out, a clock moving towards you always appears to tick faster than your own, not slower, due to the fact that the light from each successive signal has a significantly shorter distance to travel to reach me than the previous one, and this outweighs the fact that rate the signals are emitted goes down as the speed increases due to time dilation.
Again, the basic concept is very simple--if you have an emitter transmitting signals at some regular frequency, then if it is moving towards you, each signal has a shorter distance to travel to reach you than the previous one, and this causes you to see the frequency as faster than it actually is in your frame. This is true of the nonrelativistic Doppler shift as well, the only difference with the relativistic Doppler shift is that you also have to take into account that the "real" frequency that the signals are being emitted (as opposed to the observed frequency that you are receiving them, which is affected by the aforementioned difference in signal delays) changes due to time dilation.

Last edited: Dec 12, 2006
9. Dec 12, 2006

### eiapeteides

OK. At least I get the concept of relativistic Doppler shift now. When you approach you get always a blue shift. The time dilation of the emitter is compensated by the shorter signal path ways. And you are right this is indeed evident from the equation. Just had to play around with it!

But the Aging issue I still don't get. When the passenger passes by the Earth, the image of the Earth-bound observer doesn't have to travel that far! So if the passenger takes a picture what is he going to see?

A 200 year old skeleton or a guy who age by 8 years compared to the image of the same guy the passenger took from alpha centauri?

10. Dec 12, 2006

### JesseM

He'll see the Earth-observer as being aged 48 years when he passes Earth. But again, although he saw the Earth-observer aged 40 years when he passed Alpha Centauri, he'll conclude this image was "really" emitted 178.84 years earlier when Earth was 178.84 light-years away, and the light is just reaching him now as he passes Alpha Centauri. Since it then takes him 0.1790 more years to reach Earth, he'll conclude the Earth-observer was aging verrrrry sloooowly, growing only 8 years older in 178.84+0.1790=179.02 years. And this is exactly what he'd expect based on the time dilation equation, which tells him that if someone is moving at 0.999c, their clocks are slowed down by a factor of $$\sqrt{1 - 0.999^2}$$ = 0.0447, so in 179.02 years it makes sense the Earth-observer only ages 179.02*0.0447 = 8 years.

Last edited: Dec 12, 2006
11. Dec 12, 2006

### eiapeteides

Why should he think that? In the inertial frame of the ship the distance between Sun and Alpha Centauri is only about 0.2 light-years (I've rounded it up).

Unless of course I apply the Lorentz-contraction erroneously!

Last edited: Dec 12, 2006
12. Dec 12, 2006

### JesseM

Yes, but both the Earth and Alpha Centauri are moving in his frame, at a speed of 0.999c. So, if you use the rounded figure for the distance of 0.2 light-years, he should predict that T years before Alpha Centauri whizzed by him, Alpha Centauri was 0.999*T light-years away and Earth was 0.999*T + 0.2 light-years away. I used a more exact figure of 0.17884 light-years for the distance bt. Earth and Alpha Centauri in my previous post, but this is where I got the equation in bold when I said:

13. Dec 12, 2006

### eiapeteides

Your math is wrong:

Ooops No it is not! My bet. Have to think about that!

Last edited: Dec 12, 2006
14. Dec 12, 2006

### JesseM

Sure, and that equation is true of the numbers I have, since the light he sees as he passes Alpha Centauri was emitted 178.84 years earlier, and the Earth was 178.84 light-years away at that point (just plug T=178.84 into the equation 0.1784 + 0.999T which gives the Earth's distance as a function of time T before the moment the traveler passes Alpha Centauri).
The Earth's distance now is irrelevant, since the light the traveler sees when passing Earth was not emitted now, it was emitted T years in the past. In your equation, are you assuming T represents how long ago the light that the traveler is seeing now was originally emitted? If so, T*0.999c would represent the distance that the Earth has moved between the time the light was emitted and the current time, it wouldn't represent the current distance of the Earth from the traveler. If you assume T=178.84 years, and that the Earth was 178.84 light years away at that time, then T*0.999 = 177.05, meaning that the Earth has moved 177.05 light-years closer since that time, so it would now be at a distance of 178.84 - 177.05 = 1.79 light-years from the traveler, which is just the distance between Earth and Alpha Centauri.
No, but it moves in relation to the traveler, so in the past it would have been at a much greater distance--do you disagree?
T is not 0.2. Maybe you're ignoring the traveler and thinking that since the distance between Earth and Alpha Centauri is 0.17884 light-years, then any signal from Earth must take 0.17884 years to reach Alpha Centauri? If so, what you're forgetting is that in this frame Earth is moving away from the point the signal was emitted at 0.999c, so it takes a lot longer for the signal to catch up with the Earth than if the Earth were at rest. If we introduce a new symbol t to represent the time after the signal is emitted from Earth, then at t = 0, Alpha Centauri is 0.17884 light-years from the point in space where the light was emitted, and since it's moving at 0.999c, Alpha Centauri's distance from the point in space the signal was originally emitted (not from the Earth, which is moving too) as a function of t is given by 0.999ct + 0.17884 ly. Meanwhile, since the signal itself moves at c in this frame, its distance from the point it was originally emitted is just ct. So, you can figure out the value of t at which the signal will finally catch up to Alpha Centauri by solving:

ct = 0.999ct + 0.17884 ly

which you can solve for t to get t = 178.84 ly/c = 178.84 years.

Last edited: Dec 12, 2006
15. Dec 12, 2006

### eiapeteides

Your math is right but your logic is not. You are sliding along the time line. Lets apply time dilation on DeltaT instead on T:

The earth bound observer is borne on a vernal equinox. Let the passenger start counting at the moment he witnesses the birth of the earth bound observer.

At the time he reaches Alpha centauri he witnesses the 40th birthday on earth. By the time the passenger reaches earth the 48th birthday party is going on:

So from the birth to fly-by the the passenger counts 48 vernal equinoxes.
Since he counts 44 of them within one ship year, the time elapsed according to the ships clock is 1.09 years!

The passenger Ages 1.09 Years while he sees the Birthday boy age 48 earth years!

16. Dec 12, 2006

### JesseM

Can you define what you mean by DeltaT and T? For me, T is the time (as measured in the traveler's frame) between the event of the Earth-observer turning 40 and of the light from this event finally catching up with Alpha Centauri, at the same moment the traveler passes it.
No, since he already saw the Earth-observer as aged 40 when he passed Alpha Centauri, that means he sees him get older by 48 - 40 = 8 years, not 48 years.
Where do you get 1.09 years? In the traveler's frame the Earth is 0.17884 light-years away when he passes Alpha Centauri (although he sees it much further due to light-signal delays), and the Earth is coming at him at 0.999c, so it must take 0.17884/0.999 = 0.1790 years for him to reach Earth after passing Alpha Centauri, according to his own clock.
No, the passenger ages 0.1790 years while he sees the Earth-observer age 8 years. But he concludes that the apparent faster aging of the Earth-observer is actually due to Doppler shift, and that the Earth-observer was really aging at 0.0447 the normal rate.

Last edited: Dec 12, 2006
17. Dec 12, 2006

### eiapeteides

I apologize! i removed allready that argument.

18. Dec 12, 2006

### eiapeteides

DeltaT is the time interval elapsed between the observation of the birth and the 48 birthday of the Earth-bound observer from the reference frame of the ship.

The ship-bound observer witnesses the 40th birthday when he passes by alpha centauri. Since he counts 8 vernal equinoxes (earth-years) while he travels from alpha centauri to earth. He arrives at earth just in time for the 48th birthday party.

This is a consequence of the relativistic Doppler-shift you kindly explained to me. The passenger observes 44 equinoxes (passage of as many earth years) within one ship-year. So from the observation of the Birth of the earth-bound observer to the moment the ship passes by the earth (48th birthday) the passenger ages 1.09 ship-years.

The earth bound observer is 48 by the time the ship passe by Earth! The passenger at that point recollects that he witnessed the birth of the earth bound observer 1.09 Years earlier according to the ships clock and log!

19. Dec 12, 2006

### RandallB

Eiapeteides You're making this part much to hard.
Lets assume the travelers speed is actually “c” for simplicity even though we know it cannot be, and stick with a Planet you first described as being 4 light years away.

Both Earth & Planet are at t = 0 when the traveler starts off at an instant speed of “c”.
Only the traveler and Planet know the trip has started and based on distance Planet knows there is 4 V-Equinoxes already in route starting from t = -4 yr on earth. and 4 more to occur before the ‘light speed’ trip is complete; total 8.
The traveler of course does see 8 V-E’s but also knows that time on earth is not moving because it is moving at them at a speed of “c” therefore the earth at the travelers start must already be eight years in the future as he expects and does count a total of 8 V-E’s. (At least that is the time on earth from the travelers perspective at the start of the trip.)
Now on Earth some on happens to spot a signal beam from the traveler start from Planet and streak to earth just as the traveler at the same time as the traveler arrives. See that the traveler a been moving so fast it he knows the clock on the travelers ship would not have changed but would have started out 8 years ago in Earths past.
Pots right as everybody counts 8 V-E’s

---- But wait didn’t the traveler “KNOW” the time on earth was t = 0 not + 8yrs because they could see the time at Planet as they departed? After all Earth and Planet are in about the same frame, right ????

NO, because here is the #1 most important thing you must get about SR to understand it. Such measures or perspectives of what time it is “now” is not simultaneous but relative to just how fast the observer is moving.
Thus for the traveler to complete the trip in what is effectively “0” time in that traveling frame the distance covered is also effectively “0” in that frame.
So it is up to wildly different “start” times between Earth and Planet for this to work. This simultaneity issue is the most important point Einstein was making in SR.
Without it none of the rest of it works.
Simultaneity is the hard part, once you have that the rest you’re having so much trouble with will fall into place as time dilation will finally “fit”.

20. Dec 12, 2006

### JesseM

OK, so since DeltaT is stated in terms of when the traveler sees things rather than when they really happen in his coordinate system, you'd agree it's influenced by Doppler shift rather than being purely a matter of time dilation, right?
Sorry, I missed the fact that you had said "So from the birth to fly-by the the passenger counts 48 vernal equinoxes." I thought you were talking about how many vernal equinoxes the traveler would see between passing Alpha Centauri and passing Earth, which, as you say, would be 8.
Again, I missed that you were talking about the moment he observed the birth (although if you calculate it exactly it should be closer to 1.07 years, not 1.09 years). But wouldn't it be easier to stick to analyzing the two observations at Alpha Centauri and at Earth as we have been doing up until now, rather than introduce this third observation? If you find it easier to think of the Earth-observer's total age rather than just the amount he ages between two observations, then we could always just say the traveler sees the Earth-observer being born as he passes Alpha Centauri, and then finds that the Earth-observer is 8 years old when he passes Earth.
Yes, but again, the traveler should realize that the image he saw was speeded-up as compared to how fast the Earth-observer was "really" aging in his frame, due to the fact that the light from each successive equinox had a shorter distance to travel to reach him (the Doppler effect). Do you agree that when he factors out the signal delay, he concludes the Earth-observer was aging slower than him rather than faster?

Last edited: Dec 12, 2006