# Very Long Distance Phone Call

1. Jul 29, 2009

### PhanthomJay

This is probably a repeat question from somone somewhere, but here goes. Please fill in the blanks. I don't need a full explanation, because as much as I tried, I probably wouldn't understand it anyway:

Ann and Bob live on Earth. Bob is bored and decides to take a journey to a space station that is 1 light year away, as measured from his stationary location on earth. Ann stays home. Bob travels at very near lightspeed (say 0.99999c) to his destination 1 light year away. He departs on January 1, 2010, at noon.

Bob arrives at the space station, and immediatley calls Ann with a signal that travels at lightspeed. "Hi, Ann, this is Bob. My calendar watch says that it is _______________(please fill in month, day, time.year). What's is your date, year and time as you receive this message?".

Ann ultimately gets the message, and immediately responds "Hi, Bob, today is ____________ please fill in month, day, time,year). Thanks for calling."

Just fill in the blanks. I'm trying to se if time dilation is still in effect, even though Bob never physically returns to earth to physically compare that he has aged less than she. I'm guessing that the fill in for the first blank (Bob's call) is about January 2, 2010, and Ann's response (the fill in for the second blank) is January 2, 2012. More or less. Good guess, or wayyyyy off? Thanks.

Last edited: Jul 29, 2009
2. Jul 29, 2009

### Finbar

you implying bob can travel a light year in one day(as mesured by his clock)? And just over a year by ann's clock? I think this only makes sense only if bob is alawys traveling at 0.99999c. Really he has to accelerate first to that speed. This makes some difference...

3. Jul 29, 2009

### v2kkim

I agree. If we change the question like bob is moving all the time at the same speed passing the earth and another far away planet taking a straight course, then we can calculate exactly thanks to S. R. theory.

4. Jul 29, 2009

### PhanthomJay

Let's forget light speed. Let's forget acceleration. Let's put it in everyday speeds. Ann is standing stationary, in New York City, at 5th Ave and 42nd street. Bob jogs by her at 20mph, constant speed. At the instant he passes by her, slapping her hand with a high five as he does so, they synchronize their watches so they read exactly the same time. Bob keeps on running at 20 mph, constant speed, non stop to Los Angeles. When he arrives there, he calls Ann, and reports the reading of his elapsed time. Allowing for the speed of the phone call transmission, and the return call to Bob, does his elapsed time record as micro-micro seconds less than hers when she reports her time?

5. Jul 29, 2009

### solveforX

Sorry that no one has adequately answered your question, but I believe (and hope) that I am capable of answering. You say that you don't want a full explanation so I will perform estimate calculations, which are actually quite simple, and give you the results. Others, please don't scrutinize my results as they are estimates.

The difference between ann and bob's calendar would be about 223 and a half years. Which means that Ann is either six feet under after Bob only experienced one year of travel (I estimated one year of travel even though he is going slower than lightspeed, but either way she's dead).

for your second question, my calculator literally cannot calculate how small of a difference that is; it just rounds up to be no difference. But technically there is stil a difference, only it exists in nature not in my calculator or even our experience.

6. Jul 29, 2009

### sylas

You have given all the information needed to answer the question. Acceleration doesn't matter here; all that matters is the velocity at which Bob traveled to the station. The gamma factor is
$$\gamma = \frac{1}{\sqrt{1-0.99999^2}} = 223.6$$​

Bob's clock reads January 3, 2010, 03:12:06.5.

He's taken a bit over 39 hours for the trip, by his clock.

Ann gets the message about two years later. A year, in this case, is 365.242 or so days, and 2012 is a leap year; but we don't get up to February.

By Ann's clock, Bob took 1/0.99999 years, and the light trip back was 1 year. Bob took about 5 and a quarter minutes more than a year for the trip.

On the other hand, a light year is defined by a year of 365.2422 days or so, and so there will also be an extra half day tacked on to a calendar clock. But basically, Ann gets the message back after two years, plus 5 minutes, 15.57 seconds.

Hence she gets the message from Bob on January 1, 2012, at 23:42:47.5

Good guess. You've got it right.
Minor quibble added in edit. If we have a "short" leap year of 365 light-days, Bob indicates a time 39 hours, 12 minutes and 6.5 seconds after they part. Ann receives the message a time of 2 years, 5 minutes and 15.6 seconds after they part. Hence she gets the message after only a tiny fraction over two years, and Bob records a significant amount of time (as measured by someone listening to a lecture, for example) for the trip.​

Cheers -- sylas

Postscript for question two:
Yes.

Another way to do that is for Ann to have two synchronized clocks, and then carry one of them, slowly, to New York. Bob is synchronized with Ann's clock in LA. On reaching NY, he'll be out of synch with the other clock due to his time dilation.

The effect will be very small.

Last edited: Jul 29, 2009
7. Jul 29, 2009

### solveforX

Oh wow whoops I made a lightyear a year on Bob's clock. That's embarassing. Don't be confused, sylas has the correct answer.

8. Jul 30, 2009

### vin300

I copied sylas' post This is a landmark in PF!

9. Jul 30, 2009

### sylas

Honoured, I am; but in fact I made an error in the last bit, with reference to the second question. Carrying a clock slowly to New York is not an adequate way to get synchronized clocks for measuring the dilation of the jogger. Apart from that error, I think its okay. Oops.

Cheers -- sylas

10. Jul 30, 2009

### Finbar

sylas,

How do you justify ignoring acceleration in the first question?

If bob and ann begin in the same reference frame and both measure the distance to the space station as one light year, can it really be true that bob only takes 39 hours to ge there?

By this logic if bob can go even faster 0.9999999999999999999999999999999c he could do the jorney in secounds?

Sure if bob is passing the earth at this speed and never slow down we can use the SR result. Maybe your right and we can just ignore any accerleration. But we know from the twin paradox that we have to worry about acceleration in these matters.

11. Jul 30, 2009

### michelcolman

Accelleration does not matter for these results since we can calculate the whole thing using Ann's reference frame, which does not move.

I will give you an example of a different question which WOULD depend on accelleration/decelleration: what time would Ann have at the moment of Bob's arrival, according to Bob? Assume an extremely quick accelleration and decelleration.

While traveling at cruising speed, Bob would say that the distance between Ann and the space station is only around 39 light hours (due to length contraction), which explains the short trip time. If, just before decelerating at the space station, he looks at Ann's watch using very, very good binoculars, he will see an indicated time of just over 5 minutes after his departure time. Since Ann is traveling away from Bob at a constant speed of almost the speed of light, and the image of her watch traveled towards Bob at the speed of light relative to Bob as well, the image must have been emitted when she was only about half as far as she is now, so during the entire trip about twice as much time must have passed on her watch. This means her watch is "now" indicating around 10 and a half minutes after Bob's departure, which is quite logical since she is moving at almost the speed of light so her time is passing more slowly by a factor of 223.6. 39 hours divided by 223.6 indeed gives about 10 and a half minutes. Of course Ann will not agree, but she has a different definition of "now" in her reference frame.

Now let's see what happens when Bob decellerates to zero. Ann is all of a sudden no longer 39 light hours away, but a full light year, and not moving. Bob now redoes his calculation, taking into account the new fact that Ann and the space station are not really moving after all. If you assume he is still seeing the same image (watch indicating just over 5 minutes past departure time), he would come to the (again, correct) conclusion that, since the light must have traveled one year to reach the space station, Ann's watch is "now" indicating a year and 5 minutes after departure time. This time, Ann agrees.

Now try to figure out what happens on the return trip (not as easy as it seems) and you've got a complete description of the Twin Paradox!

So you see, accellerations only matter if the observer is the one who is accellerating. The observed objects are free to do whatever they like (in special relativity, at least, because gravitational fields do complicate things a little bit). This is what solves the Twin Paradox: the twin that stays behind does not have to take accellerations into account, and comes to the correct conclusion that less time has passed on the traveling twin's watch. The traveling twin will be convinced that less time passed on the home twin's watch during the two journeys, but during the turnaround the watch at home suddenly jumped into the future! This is only because he switched to a new reference system.

12. Jul 30, 2009

### PhanthomJay

Thanks for the replies. But now I have another question. Concerning the NY to LA trip. Since Ann is stationary and Bob is moving at constant speed, both are in inertial reference frames. SR I believe claims that it makes no difference whether Ann is stationary and Bob is moving away from her at constant speed, or whether Bob is considered stationary and an is moving away from him at constant speed. So why would the elapsed time of the 'moving' traveller (Bob) be micro micro seconds less than her recording of the elapsed time? She can very well be considered as the 'moving' traveller, per SR, so it his her elapsed time that could be recorded as slightly less than Bob's. Since this makes no sense, I must conclude that the elapsed time recorded by each must be exactly the same. Is this a flawed statement?

13. Jul 30, 2009

### Staff: Mentor

Yes, it is a flawed statement. Realize that for Ann to measure the elapsed time between Bob's passing NY and LA requires the use of two clocks (or the equivalent)--one in NY and one in LA. According to Bob's frame, these clocks are not synchronized. When Bob adjusts for that lack of synchronization, he sees that--according to him--each of Ann's clocks registers a shorter elapsed time than his clock does. Each frame sees the other frame's clocks run slow.

As in almost all of these puzzles, it's the relativity of simultaneity that is most often forgotten.

14. Jul 30, 2009

### sylas

Because time dilation is a result of velocity, not acceleration. Acceleration is relevant only as a way of changing velocity.

Yes; but Bob then moves into a new reference frame when he starts to move towards the station. THIS is where acceleration matters; it is how Bob changes to a new frame. But in the problem statement, we simply have the velocity at which Bob makes the trip, which is all you need. If Bob has a non-trivial period of acceleration, then you would need to integrate proper time over the trip; but it is still the velocity at each point which you are using.

In the frame where Bob is moving from Ann to the station, Bob will see the station approaching him at 0.99999c; but in this frame the distance to the state is about 39 light-hours.

Yes, certainly. Elapsed time for Bob approaches zero as Bob approaches light speed. The integration of proper time along a light path is exactly zero.

I've explained in several threads how you can express and resolve the twin paradox without accelerations. In this specific problem, there's no role for acceleration; we are simply told Bob makes the trip at 0.99999c. Finis.

If you want to describe a different problem where the the speed is attained more gradually, you can solve that also; but always you integrate proper time along the path Bob takes. In the case where Bob simply moves at 0.99999c, the proper time for the trip is as I have calculated. If you want to imagine a very short period of very large acceleration to get started, or if you want to imagine Bob simply moving past Ann at constant velocity and synchronizing with Ann as he passes, you get the same answer.

Cheers -- sylas

15. Jul 30, 2009

### Finbar

Cool cool. This makes sense. So if I insist on an acceleration period this will change the result slightly but if i take this acceleration period to zero(and hence the acceleration to infinity) I get 39 hours still?

how much energy would this take though?

16. Jul 30, 2009

### solveforX

There is nothing practical about special relativity. I want to clarify this because it seems important to me at least. Basic special relativity ignores things like acceleration, gravity, and motion in 3 dimensions so that it CAN simply and elegantly illustrate time dilation, length contraction, etc. Hence, general relativity.

The energy I assume that you are looking for would require that one knows the mass of Bob's spaceship, The gravitational field that he leaves and enters, and other conditions which make the problem too practical for special relativity.

17. Jul 30, 2009

### solveforX

You have dug into what I think is the most fruitful part of the special theory, and I encourage asking questions that challenge even such a well known theory.

The answer is that the speed of light needs to be constant for both Ann and Bob. If, at a certain point a light beam is sent, equidistant of both Ann and Bob, it has to hit them at the same time; even though Bob is moving away from the beam. This means that something else has to change about the situation. As it turns out, time for moving observers (Bob) slows in order to allow the speed of light to be constant for everyone.

Now, why does the speed of light need to be constant? That was found to be true experimentally, which you could find more details on by researching Michelson and Morley.

18. Jul 30, 2009

### Finbar

Well we ignored the gravitational field in all earlier questions. So i think the energy is just E=mC^2 * gamma

gamma = 223.6

m is of order 10^3 kg say

so its roughly 10^22 joules

several million atomic bombs?

19. Jul 30, 2009

### sylas

Yes.

As you note... the energy is also independent of the acceleration; it is determined exclusively by the final velocity, and the mass. If you get to that velocity quickly, or slowly, the energy is the same. You need more force to accelerate quickly; but you need it for less time.

Your order of magnitude for the one ton mass is good. (For 99.999% light speed, the energy is about 2*1019 joules per kilogram.)

Comparing with nuclear bombs: one Megaton is 4.184*1015 Joules. So yes; a one ton mass at 99.999% light speed has the energy of about 5 million Megatons.

And another comment...

Special relativity handles accelerations and three dimensions just fine. The only thing it omits is gravity. It's very practical indeed for problems where you don't need to worry about gravity. For example, if Bob launches from the surface of the Earth, the gravity contribution is completely negligible. Hence it would be very impractical to use full general relativity. Special relativity is the most practical way to get the correct answer. This applies also if Bob travels in various different directions and accelerations on his way to the station. If you've got his trajectory, and it does not go into strong gravitational fields, then you apply special relativity; not general.

Cheers -- sylas

20. Jul 30, 2009

### Finbar

Yeah using SR for accelerating observers is ok as long as your careful. SR applies to all inertial observers therefore the easiest way to calculate things is to do the calculation using the inertial observer.

Lets suppose bob is traveling at a constant speed v=0.9c(relative to ann) and he travels past ann at which point they synchronizes watches . Then if bob continues on for some time and then ann accelerates to v=0.99c and catches bob up. Then it will be ann who's watch is slower because it is bob who has remained in an inertial frame and ann who has accelerated.

My point is that you do need to know who is accelerating to resolve "twin paradox" type problems.