Time and right angle traveling twins paradox

[KojoMott]

TL;DR Summary

Twin paradox with twist

So in scenario in which two ships(A and B) are moving relative to each other, and towards each other; each ship will measure the other to have a velocity. Within special relativity no observer himself moves. Do you agree?

If you agree, assume each of the above two ships sends out shuttle along a trajectory perpendicular to the axis of motion of the first ships(shuttle C from ship A and D from B). Ship A would see Ship B moving towards it at some velocity and shuttle D departing Ship B on a perpendicular trajectory while also moving towards Ship A at the same velocity of Ship B. Ship B would see A moving towards B, and C from A departing on a perpendicular trajectory while also moving towards B at the same velocity of A. Do you understand the situation I have presented, and do you agree that I have accurately described what each frame for A and B would observe?

If you agree, can you please explain how the velocity in the A/B direction for C and D are handled with regards to the time dilation C and D experience when they turn around to reunite with A and B respectively?

My concerns: if C and D are said to experience less time than A and B, then they are only subject to the velocity of the AB axis(the momentum they inherit from their mother) for the time recorded on their clocks, which must be less the time recorded on A and B’s clocks. So I think A will not see D reunite with B, and B will not see C reunite with A, but both will say their shuttle returned without issue.

If we say shuttles C and D perceive A and B as going slower due to their time dilation then we have simply pushed the discrepancy to the distance between A and B. A and B will disagree about distance between themselves.

Is it clear?

 

[Dale]

TL;DR Summary: Twin paradox with twist

Within special relativity no observer himself moves. Do you agree?

Any observer can use a frame where they are moving. However, by convention in describing scenarios it is traditional for each observer to choose to use a frame where they are stationary. We can do that here.

TL;DR Summary: Twin paradox with twist

If you agree, assume each of the above two ships sends out shuttle along a trajectory perpendicular to the axis of motion of the first ships(shuttle C from ship A and D from B). Ship A would see Ship B moving towards it at some velocity and shuttle D departing Ship B on a perpendicular trajectory while also moving towards Ship A at the same velocity of Ship B.

If ship C is on a perpendicular trajectory in ship A’s frame then the trajectory will not be perpendicular in ship B’s frame. Similarly with ship D.

TL;DR Summary: Twin paradox with twist

If you agree, can you please explain how the velocity in the A/B direction for C and D are handled with regards to the time dilation C and D experience when they turn around to reunite with A and B respectively?

These are each just standard twins scenarios.

TL;DR Summary: Twin paradox with twist

So I think A will not see D reunite with B, and B will not see C reunite with A

Why not. I mean, please show your actual calculations that led you to this conclusion.

 

[Nugatory]

TL;DR Summary: Twin paradox with twist

Ship A would see Ship B moving towards it at some velocity and shuttle D departing Ship B on a perpendicular trajectory while also moving towards Ship A at the same velocity of Ship B.

Ship A would not see that. There are several ways of recognizing this, but the easiest is to remember that angles are frame-dependent: if shuttle D is moving perpendicular to the axis along which A is moving when using the frame in which B is at rest, then when we use the frame in which A is at rest shuttle D will not be moving perpendicular to the axis along which B is moving.

There's no relativity required here, the same thing happens in classical physics. Consider a child bouncing a ball up and down on the deck of a ship. Using the frame in which the ship is at rest, the ball's trajectory is up-and-down vertical and perpendicular to the deck, but when we use a frame in which the ship is moving (say we're watching from the shore) the ball's trajectory makes an angle relative to the deck.

At the risk of pushing this thread beyond the B level that you chose, you might want to take a look at the wikipedia article on relativistic velocity addition - you've presented a classic example of the non-parallel case.

 

[sdkfz]

... the ball's trajectory makes an angle relative to the deck....

Relative to the deck, or the ocean?

 

[Nugatory]

Relative to the deck, or the ocean?

Both, and whether the angle relative to the ocean is the same as the angle relative to the deck will depend on whether the deck is parallel to the surface of the ocean - it will not be if the ship is down at the bow or the stern.

The important point here (and if this is not clear, you will want to google for "Galilean relativity" and be very sure that you understand that concept before you take on Einstein's relativity) is that.... no matter how the ship lies relative to the water, if the ship is moving the angle the ball makes relative to the deck and the ocean will be different in the ship frame and the shore frame.

 

[Ibix]

Another way to see what's going on is to imagine each shuttle reeling out a cable as it travels. Each cable will be perpendicular to the line of motion of the ship. But in frames where the ship is moving, so is the cable - it must be if it is to remain beside the ship. So the shuttle that is deploying it must be moving diagonally in this frame, so it has a component of velocity parallel to the ship (so it can stay alongside the ship) and a component away from the ship (so it can move further away).

Here's a quick diagram. The ship moves left-to-right along the red line. The cable is deployed upwards and I have sketched it at several different times in orange, getting longer as we move to the right. Note that the cable is always perpendicular to the path of the ship. The vehicle deploying the cable must always be at the end of the cable, so we can determine its path simply by linking the ends of the cable sketches. That's the blue line.

You can see the (blue) shuttle is moving on a diagonal but is keeping pace with the ship. So it is always level with the ship. So you can also see that in the ship's rest frame it is moving perpendicular to the line joining the two ships.

 

[KojoMott]

Another way to see what's going on is to imagine each shuttle reeling out a cable as it travels. Each cable will be perpendicular to the line of motion of the ship. But in frames where the ship is moving, so is the cable - it must be if it is to remain beside the ship. So the shuttle that is deploying it must be moving diagonally in this frame, so it has a component of velocity parallel to the ship (so it can stay alongside the ship) and a component away from the ship (so it can move further away).

Here's a quick diagram. The ship moves left-to-right along the red line. The cable is deployed upwards and I have sketched it at several different times in orange, getting longer as we move to the right. Note that the cable is always perpendicular to the path of the ship. The vehicle deploying the cable must always be at the end of the cable, so we can determine its path simply by linking the ends of the cable sketches. That's the blue line.
View attachment 343153
You can see the (blue) shuttle is moving on a diagonal but is keeping pace with the ship. So it is always level with the ship. So you can also see that in the ship's rest frame it is moving perpendicular to the line joining the two ships.

Yes, this is what I see in mind, but the issue to me is the shuttle is exposed to the horizontal velocity for less time. Do you see? How can cable remain perpendicular if red gets horizontal velocity for longer? Blue experience less time, not just less ticks on clock. So it gets horizontal velocity component only for duration shown on its clock. Less than red. I think all have thought I do not understand the velocity of shuttle is at angle, I see that, but velocity has two components like you say, and horizontal component become time dependent.

 

[Mister T]

So I think A will not see D reunite with B, and B will not see C reunite with A, but both will say their shuttle returned without issue.

That's not possible. Each reunification is an event. All observers will agree that the events took place. They may not assign the same position and time coordinates to those events, but there will not be a dispute that the events occurred.

This is rather like a lot of thought experiments where something like a bomb explodes when an event occurs. If the bomb goes off in one frame of reference, it goes off in all frames of reference.

 

[KojoMott]

That's not possible. Each reunification is an event. All observers will agree that the events took place. They may not assign the same position and time coordinates to those events, but there will not be a dispute that the events occurred.

This is rather like a lot of thought experiments where something like a bomb explodes when an event occurs. If the bomb goes off in one frame of reference, it goes off in all frames of reference.

Okay but if shuttle experience less time how can it not experience less exposure to inherited velocity component from its mother? A will see D have a velocity component it inherits from B, and will also see it experiences less time, if it didn't D clock would not be less than B.

 

[PeroK]

Okay but if shuttle experience less time how can it not experience less exposure to inherited velocity component from its mother? A will see D have a velocity component it inherits from B, and will also see it experiences less time, if it didn't D clock would not be less than B.

Perhaps something has been lost in translation, but that isn't physics. Physics is about coordinates and calculations.

 

[Ibix]

Yes, this is what I see in mind, but the issue to me is the shuttle is exposed to the horizontal velocity for less time. Do you see? How can cable remain perpendicular if red gets horizontal velocity for longer?

Red doesn't "get horizontal velocity for longer", in short. You are mixing up measurements in different frames, which will always give you silly answers.

If you want to analyse things in the frame where the blue shuttle is at rest you need to transform all the quantities into that frame, not just the time (which is what you are implicitly doing by considering multiplying velocity by the elapsed time measured by the shuttle). That way you will also take into account the relativity of simultaneity - I rather suspect that not doing this is part of how you are confusing yourself.

 

[Dale]

exposed to the horizontal velocity for less time

What does this mean? I understand each of those words, but not what you are trying to say.

What is exposure to a velocity? Horizontal in which frame? Which time is less than which other time?

velocity has two components like you say, and horizontal component become time dependent

The horizontal component of the velocity is not time dependent during the inertial motion. Inertial means that the velocity is constant.

experience less exposure to inherited velocity component

What does this mean? Now we have not only the previous confusion but we are adding "experience" and "inherited". What does it mean to experience an exposure? Which exposure is less than what other exposure? What is an inhereted velocity component?

I get the impression that you are trying to understand relativity through some sort of story rather than through any math. I understand the part of your story where you introduce the motion of the shuttles and the ships, but this part about experiencing an amount of exposure and inheriting a velocity component is completely opaque to me.

Here is how I would approach this problem. In ##A##'s frame (unprimed) define ##A##'s worldline as ##A(\tau_A)=(t,x,y,z)=(\tau_A,0,0,0)## and define ##C##'s worldline as $$C(\tau_C)=
\begin{cases}
(\tau_C \gamma, 0,L + u \tau_C \gamma,0) & 0<\tau_C \\
(\tau_C \gamma, 0,L -u \tau_C \gamma,0) & t<0
\end{cases}$$

You can set up similar equations for ##B## and ##D## in ##B##'s frame. And you can use the Lorentz transform to express everything in any frame you like.

Now, based on those equations, what is your concern? Can you please write it down mathematically and not with all of these strange terms you are introducing.

 

[Ibix]

What does this mean? I understand each of those words, but not what you are trying to say.

What is exposure to a velocity? Horizontal in which frame? Which time is less than which other time?

That's a good question.

My interpretation of the phrase was that @KojoMott was assuming that the distance travelled by an object as measured in a frame S ought to be its velocity in that frame multiplied by its proper time, which is wrong (except in the trivial case where it's stationary). It should, of course, be the distance travelled as measured in S divided by the time taken in S, and to do otherwise is mixing quantities from different frames, which will just give you meaningless numbers.

If that is not what @KojoMott meant, my answer may be off-track.

 

[KojoMott]

It's not about what A frame sees for C, it call itself at rest. What does A see for D? It has motion.

Inherited motion I thought well known... You stand on train moving and play catch, you and ball has inherited velocity of train. In space with ships, shuttle inherit velocity component of its mother on A/B axis. I ask why if shuttle have less time it is not subject to that inherited velocity for less time. For instance does planet earth move through solar system more in ten seconds or five? Same concern here. Shuttle have less time under inherited component of velocity compared to mother when viewed by ship that say itself at rest.

Put other way you race friend and both run same speed but you only allowed to run for 8 seconds and him 10.

I don't understand why time dilation not include this motion, but affect all other motion of shuttle, like biology systems and clock.

Do you see it?

 

[Ibix]

I ask why if shuttle have less time

It doesn't. Not in the way you need it to for your argument to make sense.

 

[KojoMott]

It doesn't. Not in the way you need it to for your argument to make sense.

This seems odd if all other motion slows for shuttle such as clocks and life. Clocks and life too have same inherited motion contained within them so it must slow in some cases but not others?

 

[PeterDonis]

In space with ships, shuttle inherit velocity component of its mother on A/B axis. I ask why if shuttle have less time it is not subject to that inherited velocity for less time.

The "inherited motion" is in the rest frame of the mother ship. The "less time" is in the rest frame of the shuttle. You are mixing up quantities in two different frames, which, as has already been commented, will give you nonsensical answers.

In the rest frame of the mother ship, the shuttle has its "inherited velocity" for the time according to that frame, which is not "less time".

As has already been commented, throwing words around like you are doing is not physics. Physics is done with math.

 

[Ibix]

This seems odd if all other motion slows for shuttle such as clocks and life.

You are using "motion" in at least two different senses here.

The sense you are using it for the shuttle is just measured distance divided by measured time. In this sense, my heart walls currently travel at a few centimetres per second. But if I were in a train doing 60mph (about 2700cm/s) my heart walls would be doing 2700cm/s plus or minus a few. Their motion is much faster! And if I were in a ship doing two thirds of the speed of light my heart walls would be doing twenty billion centimetres per second plus or minus a few. In that sense, absolutely nothing in the ship has slowed motion.

The other sense you are using the word in is as a synonym for the period of some cyclic process, such as a heartbeat. But that has nothing to do with the speed of the shuttle. Why eould that care what heart beats are?

 

[Ibix]

Another thing to consider is this: what is speed? It's distance travelled divided by time taken. In relativity both of those quantities are defined to be the ones measured by rulers and clocks at rest in your chosen inertial frame. Therefore multiplying a velocity by a time measured by a clock not at rest in that frame is meaningless. It would be like multiplying your velocity by the time since your last meal. It gives you a distance, just not one that's relevant to anything in particular.

You can define a speed-like quantity by dividing the distance travelled as measured in some frame by the proper time measured by the moving clock. This quantity is called celerity. Determining this for the shuttle can be done, and then you could multiply it by the proper time elapsed for the shuttle to get the distance travelled, as you appear to want to do. But celerity is not speed - it will always be higher by a factor of ##\gamma##, so you would get the correct distance out if you used this. Celerity is a bit pointless - it isn't invariant and the celerity of light is undefined due to a division by zero. But you can use it if you want - just don't mix it up with velocity.

 

[KojoMott]

So justification for time dilation not changing exposure to inherited motion because you always pull inherited motion from frame which is at rest on A/B axis? This is like pretending there is no motion between the mother ships.

 

[Ibix]

So justification for time dilation not changing exposure to inherited motion because you always pull inherited motion from frame which is at rest on A/B axis?

No. Again, you are confusing two different senses of the word "motion". The relevant sense is distance travelled and time taken, and since we defined the measure of that to be the distance travelled as measured in a frame divided by the time taken in that same frame you need to multiply by the time taken in that frame to get the distance travelled in that frame. You could have defined the measure of rate of change of position otherwise (using celerity), but you didn't. Using velocity as if it were celerity is simply self-contradiction - that the result is silly should be no surprise.

That is completely different from the other sense you are using "motion", where you refer to the time taken for cyclic processes to complete. Motion (in this sense) slows down when motion (in the other sense) speeds up. Things that are moving slowly (in this sense) are moving extremely quickly (in the other sense). I'd strongly suggest stopping using the same word for two very different things. That might help to reduce your confusion.

 

[Dale]

You stand on train moving and play catch, you and ball has inherited velocity of train

That is weird terminology. You don’t inherit the velocity of the train. You are moving relative to the train. In another frame your velocity is the composition of the velocity of the train relative to that frame and the velocity of you relative to the train. The correct term is “velocity composition” or even “velocity addition”. You do not inherit the velocity of the train but you can use your velocity relative to the train and the trains velocity relative to something else to figure out your velocity relative to that something else.

Put other way you race friend and both run same speed but you only allowed to run for 8 seconds and him 10.

This doesn’t make any sense. In which frame is their speed the same and in which frame are their times different?

So justification for time dilation not changing exposure to inherited motion because you always pull inherited motion from frame which is at rest on A/B axis? This is like pretending there is no motion between the mother ships.

It is like pretending that you are saying intelligible things.

Again, which speeds in which frames are the same and what times in what frames are different? Be clear

 

[Vanadium 50]

OP, are you using an auto-translator? If so, you might try a different one.

The whole premise of thread is suspect. Is the problem with the twin paradox really that it's not complicated enough? Do we really need quadruplets?

It might be better to try and be much more specific about events and their coordinates. As well as paring down the number of bodies you are talking about. Simplify, simplify, simplify.

 

[Dale]

Simplify, simplify, simplify.

@KojoMott pay attention to this.

One way to tell if you have simplified the problem enough is to actually write it down mathematically. You don’t have to solve it, but if you cannot even write it down then it is too complicated.

I gave you a start above.

 

[PeterDonis]

So justification for time dilation not changing exposure to inherited motion because you always pull inherited motion from frame which is at rest on A/B axis?

No. It's that the "motion" in any frame in which the shuttle is not at rest, is the motion in that frame--but what you call "less time" is not in that frame, it's in the frame in which the shuttle is at rest. It doesn't matter what frame you use: in any frame, you have to use the motion of the shuttle in that frame and the coordinate time in that frame. You can't use the motion of the shuttle in one frame, in which the shuttle is not at rest, and also use the time in a different frame, the frame in which the shuttle is at rest. You can't mix frames.

This is like pretending there is no motion between the mother ships.

It is nothing of the sort. See above.

 

[Mister T]

Okay but if shuttle experience less time how can it not experience less exposure to inherited velocity component from its mother? A will see D have a velocity component it inherits from B, and will also see it experiences less time, if it didn't D clock would not be less than B.

The difference in elapsed time in the twin paradox is due to, in this case, A and C (and also B and D) taking different paths through spacetime. It's more than just the fact that C moves relative to A, because A moves relative to C. (Likewise for B and D).