# Is Simultaneity "Real"?

Discussion in 'Physics & Math' started by Mike_Fontenot, Jul 7, 2019.

1. ### Neddy BateValued Senior Member

Messages:
1,838
Thanks for taking the time to answer as best as you could. I think that if GR can handle stitching together charts near black holes, then surely it should be able to handle stitching together some charts in flat space. I would argue that the traveling twin's experience (with assumed instantaneous accelerations) could be replicated without any acceleration at all, by simply considering the separate inertial frames. Surely a few inertial frames should not break GR.

Of course people are going to find it strange that the traveling twin could say, "I was riding a train from birth until I was 20 years old at which time I reckoned my sister was 10. I jumped off the train right then, and my sister was 40, so I jumped back on the train and she was 10 again. I rode that train until I was 80 years old, and by then my twin sister was 40 again."

It does not surprise me that people would reject the idea, and give various scientific-sounding reasons for doing so. I remember someone on this forum insisted it broke causality just on its face, without ever being able to point to any contradiction such as someone knowing a lottery number before it was drawn, or anything like that.

On the plus side, this idea of the stay-home twin getting even younger than she was before made me realize that I had never actually done the maths for that case. Using the above case, first let me do the maths for her being 10 years old after he jumps back on the train:
v = 0.866025c
γ = 1 / √(1 - (v²/c²)) = 2.000000
x = 34.641016
t = 40.000000
t' = γ(t - (vx / c²)) = 20.000000
T = t' / γ = 10.000000
There he consider her age to be 10 in that case.

Now let's imagine that after the traveling twin jumps off the train, he jumps on a slightly faster train. This should make his sister get even younger than 10, I would think. Let's see:
v = 0.900000c
γ = 1 / √(1 - (v²/c²)) = 2.294157
x = 34.641016
t = 40.000000
t' = γ(t - (vx / c²)) = 20.241546
T = t' / γ = 8.823085
Yep, there he considers her age goes down to about 9 in that case.

Now let's imagine that after the traveling twin jumps off the train, he jumps on an even faster train:
v = 0.990000c
γ = 1 / √(1 - (v²/c²)) = 7.088812
x = 34.641016
t = 40.000000
t' = γ(t - (vx / c²)) = 40.444466
T = t' / γ = 5.705394
Yup, there he considers her age goes down to about 6 in that case.

Now let's imagine that after the traveling twin jumps off the train, he jumps on an even faster train:
v = 0.999999c
γ = 1 / √(1 - (v²/c²)) = 707.106958
x = 34.641016
t = 40.000000
t' = γ(t - (vx / c²)) = 3789.399263
t = t' / γ = 5.359018
Yep, there he considers her age goes down to about 5 in that case.

I tried putting as many nines as I could into my spreadsheet, and the lowest I could get her age to be was 5.35898385000004 so she never gets so young that she and her brother are being born, which would prove this idea false. So that is a good sign.

Last edited: Jul 11, 2019

3. ### Mike_FontenotRegistered Senior Member

Messages:
126
Good job, as usual.

You can also see fairly easily from the Minkowski diagram that no matter what he does, he can't quite make her age decrease all the way to zero. I think he can get arbitrarily close to that, but not quite to zero age (if he starts out arbitrarily fast). The Minkowski diagram always shows a world line of a light pulse to have slope + or - 1, so the fastest possible worldline leaving the origin of the diagram (where the twins were born at time zero) is a straight line sloping upwards at exactly 45 degrees. And his worldline slopes upward at less that 45 degrees, but can be arbitrarily close to 45 degrees. And for every point on his worldline, his line of simultaneity passing through that point has the slope 1/v, which means that that line of simultaneity makes an angle with respect to the vertical (X) axis that is the SAME as his worldline makes with respect to the horizontal (T) axis. Her age, according to HIM, is given by the intersection of that line of simultaneity with the horizontal (T) axis. From that construction, you can see that the youngest age that she can be, AFTER he has left her at birth, can't quite be zero, but it can be arbitrarily close to zero.

(It's hard to describe the Minkowski diagram just using words, so the above description may be hard to follow. But the Minkowski diagram is SO useful that it is worth the effort to get a good mental image of what I described above.)

5. ### Neddy BateValued Senior Member

Messages:
1,838
Thank you. Yes, Minkowski diagrams can be very useful. Some folks don't like them, and prefer to use the Lorentz transformation (LT) equations to see what is going on. I prefer the LT's but even I admit the usefulness of Minkowski diagrams.

Even without drawing the Minkowski diagram, I knew intuitively that there would be no way for the traveling twin to calculate his sister was being born, because he would have had to have been there being born also!

7. ### DaveC426913Valued Senior Member

Messages:
12,303
OK, so all you're saying is that he can slow her age down arbitrarily close to zero rate of aging - but he can never make her younger than she was the day he left.

This is pretty a straightforward consequence of time dilation, either via LT or MD.

8. ### Neddy BateValued Senior Member

Messages:
1,838
Basically, yes, but I think you missed out on some of the details. This is partly because you were trying to work out what the twins might see with their eyes, and also because you were trying to include periods of finite acceleration and deceleration.

In the simplest twin arrangement, one twin (she) stays home and the other twin (he) travels out at a constant velocity, so that we don't really have to worry about the acceleration profile. Note that, even though we do not consider the traveling twin's actual acceleration profile, we do stipulate that he is the one who accelerates/decelerates, and she never does.

On the outbound journey (with constant velocity), each twin is free consider themselves at rest, and consider the other twin to be moving relatively. So during the outbound journey, the traveling twin concludes that the stay-home twin is aging at a fraction of his own aging rate. Let's call that fraction 1/γ where gamma (γ) is greater than 1. But when he reaches the turnaround point, he can momentarily stop, and if he does he has to conclude that her age is γ times his own age. You never did say, "Oh yes, that is because when he stops her age changes" or anything like that. You were saying he would have to turn around and accelerate toward her so that he could start to 'see' her aging faster (with his eyes). That is not required, because all he has to do is stop, and she must be older than he is, because she is the stay-home twin.

Furthermore, if the traveling twin makes the inward journey at constant velocity back to his twin, he is once again free to consider himself at rest, and consider her to be moving relatively. So during the in-bound journey, the traveling twin again concludes that the stay-home twin is aging at a fraction of his own aging rate, with that fraction being 1/γ again. But when he reaches his twin, he has to conclude that her age is γ times his own age again. This time her extra aging can't happen when he arrives, because he would be co-located with her then, so it would be nonsensical. So this time her extra aging has to happen right when he heads back toward her. That is what Mike was saying about her aging a large amount during the turnaround point. Note that in both cases her extra aging happens when they are far apart from each other, and that this has nothing to do with what anyone sees with their eyes.

You also seemed to be saying that none of that was correct.

Last edited: Jul 11, 2019
9. ### Mike_FontenotRegistered Senior Member

Messages:
126
You really ARE very good at applying the Lorentz equations correctly. But they are easy to make mistakes with. The Minkowski diagram is less error prone, I think, because some of it's characteristics are always the same, and you also just keep that visual image of a typical diagram in your head.

I should add a few other things that are important to know about Minkowski diagrams:

At the instant the twins are born (at T = 0), both twins have the same line of simultaneity ... it is the vertical (X) axis of the diagram. But immediately, he changes his velocity from 0 to some positive velocity v ly/y (say 0.866, for example). So then his line of simultaneity is no longer vertical, it is rotated clockwise (CW) so that it's slope is 1/v. The angle that line of simultaneity makes with the X axis is arctan(v), which is about 41 degrees for v = 0.866. And that is the SAME angle that his worldline makes with respect to the horizontal (T) axis. That particular line of simultaneity (which passes through the origin of the diagram) is HIS x axis (analogous to HER X axis). Note that his x axis and his world line are symmetrically oriented with respect to the 45 degree line coming out from the origin (representing the worldline of a light pulse leaving the location of their birth). They each have the same small angle (beta, say) with respect to that light pulse's worldline, but on opposite sides of it, one above, one below. For example, beta is 45 - 41 = 4 degrees for v = 0.866. SO, those two lines always look like a pair of scissors, symmetrically closing on the light pulse's worldline as we choose to make the velocity v closer and closer to one. That's a great image to keep in your mind.

Yes, you knew that that NEEDED to be true, in order for special relativity to NOT have an inconsistency. But the Minkowski diagram shows that that is INDEED true: since his velocity must be less than 1, the angle beta can't be zero, and so the intersection of his lines of simultaneity with the T axis, at any time after he has left home, MUST be to the right of the T=0 point, i.e., AFTER they were born.

10. ### Mike_FontenotRegistered Senior Member

Messages:
126
That's right.

I'd like to add one comment here. The CADO equation says that whenever v = 0, regardless of their separation then, the two twins will AGREE about their respective ages. So when he does his turnaround, and is momentarily at zero velocity with respect to her, he agrees with her about their respective ages. Since she says he is half of her age then (for v = 0.866 on the outbound leg), he must conclude exactly the same thing.

Also, the CADO equation says that whenever they are c0-located, regardless of their relative velocity, they must agree about their respective ages. So the CADO equation says they must agree with each other about their respective ages when they are reunited (which of course MUST be true, because they are standing there looking at each other then).

11. ### DaveC426913Valued Senior Member

Messages:
12,303
OK, I left out some subtleties in my setup. Yeah, he doesn't have to start his journey back before she ends up being older than him.

Yeah, I jumped on a specific point Mike made, taking it out of context.

I think maybe it's time for me to refresh my knowledge of relativity of simultaneity, so I don't trip over my own feet.

Hat-tip to Mike for being patient while I pointed at his shoelaces instead of my own.

Neddy Bate likes this.