# Stupider Twins Question

Wait, there's still one asymmetry: time on Earth's clock in Earth frame is longer than time on spaceship clock in spaceship frame. But that creeps in because we say that spaceship(I assume you meant spacestation here) is at rest with respect to earth, so, proper length is 8 ly. If we considered, say, another spaceship at 0.8c, 8 light years behind our spaceship, and considered the time difference between events C and D which are "spaceship one passes earth" and "spaceship two passes earth" respectively, we'd get exactly the reverse results.
Well, you beat me to it. That was going to be my next point.

But I still say that this asymmetry is present in the "twins paradox". Not the only asymmetry there, but it's still there. My goal was not to "make the situation symmetrical". My goal was to have a scenario like the "twins paradox" without the asymmetry of accelerating vs. non accelerating frames, but still with the asymmetry of having the distance specified by a location at rest with earth.

Thanks,
Al

rahuldandekar
Well, you beat me to it. That was going to be my next point.

But I still say that this asymmetry is present in the "twins paradox". Not the only asymmetry there, but it's still there. My goal was not to "make the situation symmetrical". My goal was to have a scenario like the "twins paradox" without the asymmetry of accelerating vs. non accelerating frames, but still with the asymmetry of having the distance specified by a location at rest with earth.

Thanks,
Al

I see your point. But in the original twins paradox, the asymmetry stays even if you do the interchanging of proper distance i talked about. This is an additional asymmetry, and it doesn't matter.

The real difference between the comparing of events in the original twin paradox is that the two events occur at the same place, and at different times. Here, the two events occur at different places and different times.

Staff Emeritus
Gold Member
OK, I think I get what you mean by "asymmetrical" now. This scenario with two guys moving at different velocities is perfectly "symmetrical" in the sense that whatever the first twin says about the second, the second can say the same about the first. No such "symmetry" is present in the twin "paradox", because one twin turns around and that tilts his simultaneity planes the other way. (Before he turns around, sets of simultaneous events are parallel to the blue lines. After he turns around, it's the red lines instead. http://web.comhem.se/~u87325397/Twins.PNG [Broken]).

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I have no idea what you mean by "asymmetrical". I certainly wouldn't describe the fact that two different coordinate systems assign different coordinates to the same event as "asymmetrical". This is what coordinate systems do. It's how they are defined. If they would assign the same coordinates to every event they'd be the same coordinate system.

It also shouldn't come as any surprise to someone who understands simultaneity that event B has different time coordinates in the two frames. Relativity of simultaneity means that the two twins have completely different ideas about which slices of space-time are "space", but they would have to agree on simultaneity to be able to assign the same time coordinate to event B.
By asymmetrical, I mean the situation is different for the Earth twin than for the ship's twin.
This is asymmetrical:
Al68 said:
Everyone will agree that 10 yrs passed in Earth's frame between the two events, and 6 yrs passed in the ship's frame between the same two events.
And no, it's no surprise to me. What's surprising to me is that this asymmetry is never mentioned anywhere. Even though it is obviously present in the "twins paradox". Yes I know the "twins paradox" also has acceleration as an asymmetry. But my scenario does not, and we still have the above asymmetry.

Thanks,
Al

Staff Emeritus
Gold Member
It's not the fact that the time coordinates of B are different in the two frames that breaks the symmetry. What breaks the symmetry is that you chose two events on the world line of one of your twins. If you had replaced B with an event half way between their world lines, they would have agreed about its time coordinate.

Note that if you instead of considering only the pair of events (A,B), also consider the pair of events (A,C) you are also restoring the symmetry in a way. Now you can make your statement and the same statement with B replaced by C and the twin's roles reversed. I would say that the situation is perfectly symmetrical.

I see your point. But in the original twins paradox, the asymmetry stays even if you do the interchanging of proper distance i talked about. This is an additional asymmetry, and it doesn't matter.
I would disagree that the asymmetry stays. It doesn't stay, it switches sides.
The real difference between the comparing of events in the original twin paradox is that the two events occur at the same place, and at different times. Here, the two events occur at different places and different times.
The original twins paradox has three events. the first two are the same as the two in my scenario. The second two in the twins paradox are also equivalent to the two in my scenario (reversed). The only difference is in the twins paradox, my scenario is done once, then reversed and done again. This "reversal" between the two halves of the twins paradox is treated as crucial in the resolutions, even though the end result (elapsed time between events) is the same as in my scenario. Except that simultaneity is regained by the twins meeting again.

Thanks,
Al

Staff Emeritus
Gold Member
As I said, there's no asymmetry at all in your scenario unless you choose your events in an asymmetrical way. The same can not be said about the twin paradox.

It's not the fact that the time coordinates of B are different in the two frames that breaks the symmetry. What breaks the symmetry is that you chose two events on the world line of one of your twins. If you had replaced B with an event half way between their world lines, they would have agreed about its time coordinate.
You must be reading my mind now.
Note that your statement above is also true in (each half of) the twins paradox.

Thanks,
Al

As I said, there's no asymmetry at all in your scenario unless you choose your events in an asymmetrical way. The same can not be said about the twin paradox.
Yes, I chose my events in an asymmetrical way, like you say. That was on purpose. Events A and B in my scenario are identical to (each half of) the twins paradox, except for the acceleration in between.

matheinste
Hello all.

Surely, stripped down to basics you have a frame which you can regard as at rest and something moving relative to it. All we are doing is measuring elapsed time in the rest frame as, measured at start and finish events, with elapsed time in the frame moving relative to it and finding the latter elapsed time to be less. This is just normal time dilation so why introduce twins.

Matheinste

yuiop
The traveling twin is a bit rusty on his relativity and he can't quite rememebr if the Earth twin is supposed to be ageing faster or slower so he decides to work it out. As he passes the space station his clock is showing 6 years and the station clock is showing 10 years and through his telescope he sees an image of the Earth clock showing 2 years. Initially he assumes the Eath twin is ageing faster so he figures that when the Earth clock was reading 2 years his own clock must have been reading 1.2 years. The furthest the Earth twin could have moved away in that time is 1.2*0.8= 0.96 lght years so that is how long the image of the clock would take to reach him so the signal should have seen the Earth read 2 years when his own clock read 1.2+0.96= 2.16 years. Clearly that is not correct and he must have made a wrong assumption. He tries again assuming the Earth twin is ageing slower so when the Earth clock read 2 years his own clock must have read 3.33 years. In that time the Earth twin must have moved a distance of 3.33*0.8= 2.666 light years. So he adds the travel time of the Eart light signal to the time his own clock must have been reading when it was emitted and sees that signal should have arrived at 3.333+2.666 = 6 years which agrees with what he observes. Therefore the traveling twin concludes that the Earth twin is ageing slower and that the Earth twin must have been 3.6 years old when he passes the space station.

There is no asymmetry or "breaking of the symmetry" in this situation until one of the twins turns around. In fact claiming there is asymmetry is pretty close to claiming that one of the twins has absolute motion or that there is some sort of detectable absolute reference frame.

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Staff Emeritus
Gold Member
It's more like claiming that these are asymmetries:

Right now there's a plastic plant on my left. If I turn $\pi/2$ clockwise, it's going to be behind me.

Right now it's 2 meters away from me. If I move one meter to the right, it's going to be 3 meters away.

It's not the fact that the time coordinates of B are different in the two frames that breaks the symmetry. What breaks the symmetry is that you chose two events on the world line of one of your twins.
This is really my point. I chose (a priori) event B to be at 6 yrs in the ship's frame and at 10 yrs in the Earth frame, by specifying that the space station is 8 ly from, and at rest with earth. In the twins paradox, the turnaround event is chosen (a priori) to be at 6 yrs in the ship's frame and at 10 yrs in Earth's frame, the same way.

By defining the turnaround event (and the departure and return events) this way, the outcome is determined a priori. The fact that we have to consider the ship to change frames (instead of the earth) is determined a priori. All the textbook resolution does is show how to do the math and draw the spacetime drawings for a pre-chosen outcome.

I'm not saying the resolution is wrong, just that it's only a resolution for the specific situation in which we assume ahead of time that we must consider the ship's (proper) acceleration, and not just the (relative) coordinate acceleration between the ship and earth.

Thanks,
Al

Therefore the traveling twin concludes that the Earth twin is ageing slower and that the Earth twin must have been 3.6 years old when he passes the space station.
Of course, but there is no event on the Earth's twin's worldline at t = 3.6 yrs. With the Earth's calculation of 6 yrs for the ship's clock at event B, that event is actually on the ship's worldline at t' = 6 yrs. And that same event is actually on the Earth's worldline at 10 yrs.

Of course we could define an event corresponding to t = 3.6 yrs on Earth's worldline and the same event at t' = 6 yrs on the ship's worldline.

But look at what happens if we say this event is a "turnaround". We would have to say that the Earth twin accelerated, not the ship. That would also result in concluding that the space station and the Earth did not "turnaround" simultaneously in the ship's frame. Which means that they did not stay "at rest" with each other according to the ship's frame. OMG, that would also mean that the Earth twin would be younger than the ship's twin when they reunite. All because we defined the turnaround event differently and didn't take into account who "felt" the acceleration. We could even draw a spacetime diagram of this. But the prohibition on ignoring proper acceleration prevents us from creating such a scenario, right?

This is what I mean by saying that the presentation of the twins paradox assumes we have to take acceleration into account to begin with. Then "resolves" it by concluding that we have to take acceleration into account. Sure it's a good exercise in math, spacetime diagrams, the Doppler effect, etc. But it just shows how to do the math for an a priori conclusion. Again, that doesn't mean the conclusion is wrong, just that it's not "resolved" in the sense that Einstein wanted to "resolve" it.

matheinste
Hello Al68.

But what you have described is not the twin paradox without acceleration. For the apparent paradox to be present they must be reunited. Also in your scenario they cannot have started life as twins and pass each other at the start as twins.

I do understand what you are trying to do but twins are irrelevant in your scenario as you are only showing that an accumulated time difference appears without acceleration which we already knew. The lack of need of acceleration is removed by the twins never starting/passing as twins and not being reunited.

Matheinste.

Hello Al68.

But what you have described is not the twin paradox without acceleration. For the apparent paradox to be present they must be reunited.
Why does the twins reunion cause an apparent paradox? To anyone familiar with SR.
Also in your scenario they cannot have started life as twins and pass each other at the start as twins.
Yeah, you're right, I wondered when someone would notice that. Of course it doesn't really matter here.
I do understand what you are trying to do but twins are irrelevant in your scenario as you are only showing that an accumulated time difference appears without acceleration which we already knew. The lack of need of acceleration is removed by the twins never starting/passing as twins and not being reunited.

Matheinste.
Well, for something we already knew, it sure seems controversial.

Thanks,
Al

matheinste
Hello Al168

I said apparent paradox because it is not a real paradox. I have no problem with it. Others do have a problem. I see nothing controversial.

Matheinste