# I Twin paradox in GR - negative time?

1. Jul 10, 2017

### MrBlank

The mechanism for the advancing of the stay-at-home twin's clock is gravitational time dilation. When an observer finds that inertially moving objects are being accelerated with respect to themselves, those objects are in a gravitational field insofar as relativity is concerned. For the traveling twin at turnaround, this gravitational field fills the universe. In a weak field approximation, clocks tick at a rate of t' = t (1 + Φ / c^2) where Φ is the difference in gravitational potential. In this case, Φ = gh where g is the acceleration of the traveling observer during turnaround and h is the distance to the stay-at-home twin.

My question is, if the traveling twin were to accelerate away from the stay-at-home twin, would Φ be negative? If yes, then if Φ < c^2 that would mean as time goes forward for the traveling twin, time would go backward for the stay-at-home twin (negative time).

2. Jul 10, 2017

### Staff: Mentor

3. Jul 11, 2017

### MrBlank

I read the paper. I didn't entirely understand it. A specific example would help.

There are two rockets R1 and R2 that are at rest in empty space in the same inertial frame. They are one billion light-years apart. There are no sources of natural gravity.

Senario 1: For 1 second of proper time for R2 it has proper acceeration of 1 m/s^2 toward R1. During this acceleration how much proper time passes for R1?

Senario 2: For 1 second of proper time for R2 it has proper acceeration of 1 m/s^2 away from R1. During this acceleration how much proper time passes for R1?

The solutions should use general relativity, not special relativity.

4. Jul 12, 2017

### Ibix

Time dilation will never be negative. The formula you quoted is a weak-field approximation, the definition of "weak" being $|\Phi|<<c^2$. So in any situation where this formula would give a negative number it would be wrong to apply it.

The answer to either of your scenarios depends on the simultaneity convention in use. Different simultaneity conventions are implied for the two scenarios by taking the approach you appear to want to take, so the answers would be different. This is more to do with the question boiling down to "how long is a piece of string" than any actual physics.

Last edited: Jul 12, 2017
5. Jul 12, 2017

### Ibix

Just to add - Dolby and Gull's figure 6 is closest to matching your scenarios. As drawn it shows Barbara accelerating towards Alex. If you erase Alex's worldline (the dotted vertical line) and redraw it on the left of the diagram you get Barbara accelerating away from Alex. According to Barbara the part of Alex's worldline that is simultaneous with her acceleration phase is approximately the bit between the middle three lines of simultaneity (the middle three of the seven solid lines that cross the diagram from side to side - and approximately because none of the lines drawn actually pass through the ends of the acceleration phase). Depending on which way she is accelerating (whether Alex's worldline is as drawn, or on the left as I described) the part of Alex's worldline that Barbara considers to be "during her acceleration phase" is different - hence my "how long is a piece of string" comment.

Edit: Dolby and Gull's definition of "simultaneous" need not be the one you use. I think it is the one implied by the gravitational time dilation approach, but haven't checked this. But Dolby and Gull is certainly relevant in a qualitative sense, even if it turns out that their simultaneity convention isn't quite the one you are planning to use.

Last edited: Jul 12, 2017
6. Jul 12, 2017

### Chris Miller

It was my understanding from answers provided here that acceleration, unlike gravity, has no effect on time.

7. Jul 12, 2017

### Ibix

Your acceleration doesn't appear directly in my calculations of your elapsed time - all I need is your velocity as a function of time. That was what was being discussed on the thread to which you refer.

Here we are discussing your notion of "now, over there" as you accelerate. That isn't a physical invariant, so whether or not your acceleration appears explicitly in the result is a matter of choice. It does appear in the OP's formula, but that's a result of his choice to use a particular simultaneity criterion. He could choose otherwise and acceleration might not appear.

This is a different topic from the thread about the clock postulate.

8. Jul 12, 2017

### jartsa

"Negative time dilation" is the effect where now-moment shifts towards the past, at the home planet, according to the person that is accelerating away.

I think a good way to think for an accelerating person is: During one second of my time clocks at the home planet tick towards the future one second, and in addition to that the clocks move towards the past n seconds, where n is proportional to my acceleration and distance, so n is proportional to "gravitational potential difference".

The clock mechanism causes the clock to tick forwards, the change of frame causes the clock's reading to change towards the past or towards the future, depending on the direction of the acceleration.

By the way, the above is general relativity, to be more specific, the part of general relativity that is special relativity. You see, general relativity includes special relativity. Clocks in the pseudo gravity field seem to behave like clocks in a real gravity field, as long as the pseudo gravity field behaves like a real gravity field, which it does if the acceleration stays constant forever.

Last edited: Jul 12, 2017
9. Jul 12, 2017

### Ibix

I have to say I don't like this approach; it seems to me to be taking sonething fairly straightforward (different lines can have different lengths, especially if we're flexible about where those lines start and stop) and making it complex for the sake of it. But each to their own, I guess. Can I recommend that you don't use n for your ratio? It's usually used for an integer but, as noted above, in this case $|n|=|\Phi|/c^2<<1$.

10. Jul 12, 2017

### Staff: Mentor

I will not just work it out for you. If you want, I can help you work through it but I won't just do it for you. However, your numbers are very inconvenient. I would strongly recommend using units where c=1 and picking more reasonable distances, times, and acceleration.

To work this problem you would write down the expression for the accelerating clock and the inertial clock in some inertial reference frame. Then, find the starting and ending events on the accelerating clock's worldline. Then, from each of those events calculate the worldline of an incoming and an outgoing pulse of light and find where those pulses intersect with the inertial clock's worldline. Finally, apply the definition of radar coordinates to determine the radar time for each event.

FYI, this is all special relativity since you specified that the spacetime is flat. There is no need to solve the EFE.

11. Jul 12, 2017

### Staff: Mentor

That is altogether impossible, as special relativity is a special case of general relativity (which is why it's called "special"), and this particular problem happens to fall into that special case of GR.

12. Jul 12, 2017

### Staff: Mentor

13. Jul 13, 2017

### MrBlank

It's now clear to me that the equation t' = t (1 + Φ / c^2) is only valid when |Φ | << c^2. What equation is used when it is not the case that |Φ| << c^2?

14. Jul 13, 2017

### Ibix

Don't know. The Wikipedia expression comes from approximating the result from the Schwarzschild metric, but I don't think that simply reversing that is correct for the twin paradox case outside the weak field approximation. Happy to be corrected on that, though.

Depending on the details of the experiment you want to describe there may not be a solution in the form you want. It's easy enough to figure out a set of coordinates appropriate for a constantly accelerating observer. You can then use them to construct coordinates for the case of an observer who undergoes a finite period of constant acceleration. But in the former case you get a Rindler horizon, and it's not possible to define time dilation between all pairs of observers because some of them are on opposite sides of the horizon. And I don't think the latter case can be treated as a stationary gravitational field, so I expect that a potential cannot be defined.

It's easy enough to work through this by the procedure Dale outlined in #10. You marked this thread as I level, and the maths is straightforward. If you can find (or derive, if you're comfortable with calculus) the equations of motion of an object undergoing constant proper acceleration then the rest is easy. Happy to help you through it and see how far we get. What are those equations of motion? (Hint: the Latex tutorial linked below the reply box will show you how to enter maths here).

Edit: or we could cheat and use the fact that Dolby and Gull already did a lot of the work in their paper. Points about there not necessarily being a solution of the type you want still apply.

Last edited: Jul 13, 2017
15. Jul 13, 2017

### jartsa

First check that the formula above works when we are dealing with a normal twin paradox.

(The home clock is moving during the acceleration according to the traveling twin, which slows down the clock, don't forget to take that into account)

(Or maybe just check that the above formula agrees with special relativity's simultaneity convention)

Then consider what happens if the twin on his return journey, after having accelerated, decides to undo the acceleration.

Last edited: Jul 13, 2017
16. Jul 13, 2017

### MrBlank

I've noticed something odd. In the limit as $\phi$ goes to negative infinity, presumably for each second of proper time that passes for the accelerating twin, zero time passes for the non-accelerating twin (time stands still). This implies that in the limit as $\phi$ goes to infinity, for each second of proper time that passes for the accelerating twin, 2 seconds pass for the non-accelerating twin. This would mean that in the twin paradox it could never be the case that more than twice as much time would pass for the non-accelerating twin compared to the accelerating twin.

17. Jul 13, 2017

### jartsa

if c=1 and t=1 and $\phi$=100 then
t'= t (1 + Φ / c^2) = 101
so t'=101 t

if c=1 and t=1 and $\phi$= -100 then
t'=t (1 + Φ / c^2) = -99
so t'= -99 t

Right?

Ah I see. You think $\phi$ can not be larger than 1. Is there some good reason for that?

18. Jul 14, 2017

### Ibix

See post #4, if you want to use this approximate time dilation formula.

19. Jul 14, 2017

### Ibix

This makes absolutely no sense to me.

For a sufficiently brutal acceleration you can make the elapsed time for the travelling twin tend to zero, giving an arbitrarily large age ratio.

20. Jul 14, 2017

### jartsa

This formula:

t'=t
(1 + Φ / c^2)

Well, that formula is the formula of gravitational time dilation, and also a formula in the special relativity that tells us how many seconds now-moment shifts at some distance from here, when we accelerate one second with some constant acceleration.

Just derive that special relativity thing , then you'll see yourself. Or wait a few days and I derive it.

Last edited: Jul 14, 2017
21. Jul 14, 2017

### Ibix

Good luck. I can't immediately see where the hyperbolic functions drop out if you're claiming it's an exact formula, but I haven't completely worked through it. They do surprise me like that sometimes.

22. Jul 14, 2017

### MrBlank

I'll give an example to illustrate what I'm talking about.

There are three rockets R1, R2 and R3. They are initially at rest in the same inertial reference frame. Initially R1 and R2 are together and R3 is 1 billion light-years away. R2 accelerates at 1 m/s^2 for 1 second toward R3. Then R2 accelerates at 1 m/s^2 away from R3 for 2 seconds. Finally, R2 accelerates at 1 m/s^2 for 1 second towards R3. R2 is back where it started at rest next to R1.

Four seconds pass for R2. R1 and R2 experience the twin paradox in "miniture" but the distance and acceleration (and velocity) are so small that the effect can be ignored. So, 4 seconds pass for R1. Since R1 and R3 were at rest relative to each other, from R1's point of view, 4 seconds passed fo R3. R2 and R3 are a large distance appart, so $\phi$ will be large. I will arbitrarily say that for each second that passes for R2 while it is accelerating towards R3, 100 seconds pass for R3. When R2 is accelerating away from R3 assume that for each second that passes for R2, zero second pass for R3. From R2's point of view, 200 seconds have passed for R3.

This creates the non-self consistent situation where at the begining and end R1 and R2 are at rest beside each other and they agree that 4 seconds have passed for each of them, but R1 thinks that 4 seconds have passed for R3, and R2 thinks that 200 seconds have passed for R3.

The only way to make it self consistent is if when R2 is accelerating towards R3, for each second that passes for R2, 2 seconds pass for R3.

Let f($\phi$) be the time dilation factor. It appears to be the case that (f($\phi$) + f(-$\phi$)) / 2 = 1.

23. Jul 14, 2017

### Staff: Mentor

That is one advantage of the Dolby and Gull method. It is always self consistent for any region it covers.

24. Jul 14, 2017

### jartsa

Well don't assume that "for each second that passes for R2, zero second pass for R3", if that leads to any problems.

Assume that for each second that passes for R2, x seconds pass for R3, where x is a number that leads to no problems.

A negative number is needed.

Last edited: Jul 14, 2017
25. Jul 15, 2017

### Ibix

@MrBlank - the problem here is that the distant rocket is on the other side of the Rindler horizon I mentioned earlier, and it's difficult to even define "during the turnaround phase" using gravitational time dilation in the naive way that you want.

The underlying problem is that the acccelerating rocket has three different acceleration phases. During each one it is at rest in a different reference frame. But one of Einstein's key insights was to realise that different reference frames have different ideas of what "at the same time" means for things that aren't in the same place. This means that if you talk about "at the same time as the end of the acceleration phase" and "at the same time as the beginning of the deceleration phase" then you aren't talking about the same moment in general. So you can't just add up the time "during the acceleration phase" "during the turnaround phase" and "during the deceleration phase" without being careful about what those reference frames are calling "the beginning and end of the phase a long way from the rocket".

I have a street atlas of my town. Each page shows an area of the town, with a small overlapping border between pages. Imagine photocopying two pages and butting them up against each other. You can measure distances with a ruler, but it will go wrong if you try to measure distances across the join. And there are two copies of some streets so I can have two distances to them. This is because the two pages had different notions of "the position at the edge of the map" and I made a silly mistake by putting them together side by side.

I can handle this in two ways. I can remember that for any distance across the map join I need to measure it, then measure the distance across the map join and subtract that off, and I can come up with some rule about which distance of the two possible ones I should use for things that appear twice. Or I could just stick the two maps together overlapping them so that the streets that appear on both maps are overlapped.

The same options are available when sticking together reference frames (a complete set of all possible frames covering a spacetime is even called an atlas). @jartsa's negative time dilation is the first approach. Dolby and Gull's approach is the second. Your approach (and, I suspect, Wikipedia's) is not noticing that the overlaps are a problem and coming up with contradictory answers as a result.