Special relativity question

[Jagube]

Let's say there are twin brothers A and B standing next to each other. Then B gets on a spaceship and goes on a 10 year space journey at some speed V close to C. Special relativity says when B comes back, he'll be younger than A.

However, Newtonian relativity tells us that when B is moving at velocity V in relation to A, A is moving at V in relation to B. So we could as well say it was A who went on a space journey in spaceship Earth, and B stood still. Which means A is younger than B.

It all depends on the frame of reference. From A's point of view, B is younger than A. From B's point of view, A is younger than B. But that can't be at the same time, right? It's in different points in spacetime or something like that.

Let's say I'm A and I'm meeting with B for the first time after his epic journey. I can clearly see B is younger than me and if I say to him "Hey B, you look so much younger than me!" he will confirm "Yes, indeed! Lucky me!" So the guy I'm talking to is not real B, because in real B's point of view *I* am the one who has traveled for 10 years at crazy speeds and who is therefore younger, so B would never confirm my statement that he is younger. So the guy I'm talking to can't be B, it must be some version of B that only exists in my frame of reference, my own private universe. Real B exists in his own universe (at its center) and in that universe I'm younger than him and he's probably jealous of my younger age just as I am of his in mine.

Both real me (A) and real B have reasons to be jealous. The only guys that have reasons to be happy that they're younger than their twin brother are the fake versions: fake B in my universe, and fake me in my brother's universe. However, since they're not real, the fact that they're happy doesn't matter.

Is my understanding correct?

 

[Ulim]

However, Newtonian relativity tells us that when B is moving at velocity V in relation to A, A is moving at V in relation to B. So we could as well say it was A who went on a space journey in spaceship Earth, and B stood still. Which means A is younger than B.

It all depends on the frame of reference. From A's point of view, B is younger than A. From B's point of view, A is younger than B. But that can't be at the same time, right? It's in different points in spacetime or something like that.

You are wrong because in special relativity the speed c is fixed.
And I quote

In order for the speed of light to be the same to all observers, the moving light clock should experience time slower than an observer attached to the first ,non moving, clock

The fact that you move can be easily checked for. Its never unknown who the moving observer is in special realtivity. But it is unkown in newtonian.

If you say its impossible to tell who is the moving observer you also say c isnt constant.

Also every observer experiences time the same. So neither observer could tell who was being slowed/sped up. But from the fact that if you reunite the observers or exchange info they will know immeadiatly. The guy in the spaceship has aged less thus he is the one to be slowed. And the one to be slowed is the one who was moving.

The frame of the observer also doesnt know if he is moving at speed of light.

Also if im in a train going 0.9c and i shine a flashlight down the train.
The light will go 1c in the train.
And 1c viewed from the outside not 1.9c because c is constant.
Because the time in the train will get slowed and the train will get smushed in the direction he is moving by the lorentz factor keep the light also 1c viewed from the outside.

Thus it can be concluded that the slowed and smushed observer is the moving observer.

 

[Ulim]

Also the effects are reversed. So the guy in the train will see the world outside speed up and stretch to like 2.3 its original lenght in the direction of movement.

 

[pitubo]

The fact that you move can be easily checked for. Its never unknown who the moving observer is in special realtivity. But it is unkown in newtonian.

Not being a physicist, but a curious layperson, I wonder if you could explain to me how this fact (who is moving and who is not) is easily checked in the special relativity universe, but not in the newtonian universe?

Also if im in a train going 0.9c and i shine a flashlight down the train.
The light will go 1c in the train.
And 1c viewed from the outside not 1.9c because c is constant.
Because the time in the train will get slowed and the train will get smushed in the direction he is moving by the lorentz factor keep the light also 1c viewed from the outside.

Hold on there.

If I stand next to the railway track and shine a flashlight in the direction of the approaching train, the light would not be seen inside the train as moving at 1.9c either. The ensuing logic appears to apply just the same in this case. Time outside the train appears "smushed" to the observers inside the train.

Thus it can be concluded that the slowed and smushed observer is the moving observer.

Until you can point out the flaw in my argument above, it is not clear at all to me.

Jagube, I don't understand from where you conjure up the many worlds. The difference that I can point out between the twins is that the traveling twin is accelerating relative to his original frame of reference, whereas the other twin isn't.

Again, I am not a physicist. Any seasoned physicists who can lucidly wield occam's razor to cut through the paradoxes is very welcome to point out my errors.

 

[Loveall]

It all depends on the frame of reference. From A's point of view, B is younger than A. From B's point of view, A is younger than B. But that can't be at the same time, right? It's in different points in spacetime or something like that.

This great question is called the twin paradox. It was used by very smart people trying to constructively find holes in Einstein's theories when they came out. There does seem to be a contradiction in special relativity, but it goes away when you account for acceleration. The traveling Twin needs to accelerate and decelerate during the journey to come back to the meeting. It is during these acceleration stages that he sees the twin back on earth age very quickly, so when they meet everything makes sense. At other times during the trip (constant speed), you are correct: both twins see the other twin aging more slowly.

If you are interested in the subject I recommend a beautiful book: "It's about time".

 

[Loveall]

The fact that you move can be easily checked for. Its never unknown who the moving observer is in special realtivity. But it is unkown in newtonian.

Not being a physicist, but a curious layperson, I wonder if you could explain to me how this fact (who is moving and who is not) is easily checked in the special relativity universe, but not in the newtonian universe?

Also if im in a train going 0.9c and i shine a flashlight down the train.
The light will go 1c in the train.
And 1c viewed from the outside not 1.9c because c is constant.
Because the time in the train will get slowed and the train will get smushed in the direction he is moving by the lorentz factor keep the light also 1c viewed from the outside.

Hold on there.

If I stand next to the railway track and shine a flashlight in the direction of the approaching train, the light would not be seen inside the train as moving at 1.9c either. The ensuing logic appears to apply just the same in this case. Time outside the train appears "smushed" to the observers inside the train.

Thus it can be concluded that the slowed and smushed observer is the moving observer.

Until you can point out the flaw in my argument above, it is not clear at all to me.

Jagube, I don't understand from where you conjure up the many worlds. The difference that I can point out between the twins is that the traveling twin is accelerating relative to his original frame of reference, whereas the other twin isn't.

Again, I am not a physicist. Any seasoned physicists who can lucidly wield occam's razor to cut through the paradoxes is very welcome to point out my errors.

Looks like you posted as I was writing my answer. Your intuition is correct, acceleration (or changing inertial frames during the trip) is the key here. Without it, there is a paradox when the twins meet, hence the OP's motivation to generate new universes :d

 

[Jagube]

Thanks guys. I'll read it again when I have more time to digest it.

So are these statements correct?:
When two objects are moving in relation to each other, Newtonian relativity doesn't distinguish between them and treats them symmetrically. However, special relativity breaks this symmetry and always tells us which of the two is moving / moving faster?

 

[muladharma]

I enjoyed these videos about space-time diagrams, maybe you will find them enlightening or someone can share their opinion about them?

 

[Loveall]

Thanks guys. I'll read it again when I have more time to digest it.

So are these statements correct?:
When two objects are moving in relation to each other, Newtonian relativity doesn't distinguish between them and treats them symmetrically. However, special relativity breaks this symmetry and always tells us which of the two is moving / moving faster?

No, there is no symmetry breaking in special relativity. Both travelers see each other aging more slowly in a perfectly symmetrical situation (as long as they don't accelerate). This is not a contradiction and when the travelers are brought together their realities are matched.

 

[Ulim]

If I stand next to the railway track and shine a flashlight in the direction of the approaching train, the light would not be seen inside the train as moving at 1.9c either. The ensuing logic appears to apply just the same in this case. Time outside the train appears "smushed" to the observers inside the train.

You are going in between the two closed systems. Thats another thing.
The light outside is not connected to the light inside thats why the light outside will hit the train at 1c from the outside. Thats because the ship and the space between you and the ship will get bent too.

Breaking the speed of light relative to a moving object

You can't go faster than light, and light can't be additive (if you shine a light from a spaceship, the light is not going $c$+"speed of spaceship", it's just going like it always does). But what

physics.stackexchange.com

Thats the thing about special realitvity if you go 0.9c in the same direction and flash a light at another ship that goes 0.9c it the light goes still at c speed.
Because the spaceship will get smushed and slowed.

 

[Ulim]

Thanks guys. I'll read it again when I have more time to digest it.

So are these statements correct?:
When two objects are moving in relation to each other, Newtonian relativity doesn't distinguish between them and treats them symmetrically. However, special relativity breaks this symmetry and always tells us which of the two is moving / moving faster?

Special relativity tells us that velocity is relative because time and space is relative.
In newtonian those are absolutes and dont change.
The one true thing thats always correct is only c. The speed of light.

 

[Ulim]

Not being a physicist, but a curious layperson, I wonder if you could explain to me how this fact (who is moving and who is not) is easily checked in the special relativity universe, but not in the newtonian universe?

Jagube, I don't understand from where you conjure up the many worlds. The difference that I can point out between the twins is that the traveling twin is accelerating relative to his original frame of reference, whereas the other twin isn't.

It can be checked for because c is a constant. Because even if you move at 0.9c its still c.
You cannot break this law. This makes its so that due to all the effects it carries.
(Extreme slowdown of time for the moving thing, Extreme smushing in movement direction, extreme increase in weight to outside observers)
These effects tell you everything.
The effect will always be that the moving partner gets multiplied by 0.xx

Also a third observer actually gets "created" if you compare the clocks on the outside of the ship.
Breaking down any ambiguitiy for people who dont want to spend time to crunsh the formulas.

A third observer can always tell.

 

[Ulim]

As long as you all wont accepct the fact that c stays the same for every point of view and for every interaction between systems you will never understand it.

 

[Ulim]

You cant tell if you are moving in a closed system.
But if you have two systems interacting its clear.
But a third observer is needed.

Also regarding the twin paradox.
Its true that both twins will see the time from eachother going slow.
But as i already said. Compare clocks and that will make the time collapse showing one is slower which was the moving one.

[YOUTUBE]

 

[pitubo]

Not being a physicist, but a curious layperson, I wonder if you could explain to me how this fact (who is moving and who is not) is easily checked in the special relativity universe, but not in the newtonian universe?

It can be checked for because c is a constant. Because even if you move at 0.9c its still c.
You cannot break this law. This makes its so that due to all the effects it carries.
(Extreme slowdown of time for the moving thing, Extreme smushing in movement direction, extreme increase in weight to outside observers)
These effects tell you everything.
The effect will always be that the moving partner gets multiplied by 0.xx

Thank you for answering (so many times) Ulim, but you don't really answer my question in any of your answers. I try to point out that all that can be said about the observer outside of the train can also be said about the observer inside the train. Nowhere do I claim that c is not a constant or not a limit, that is irrelevant to the argument of symmetry.

Also a third observer actually gets "created" if you compare the clocks on the outside of the ship.

How is your seeming creation of a third observer not actually a doubling of the first observer in the same frame of the first, implying it to be a preferential (absolute?) frame of observation. What if the third observer is moving at 0.9c respective to both the other observers?

A third observer can always tell.

But isn't this third observer's experience tied to his own frame of reference?

This nice animation does not address my question either. However I do note that one of the two bodies in motion is undergoing a continuous acceleration.

The difference that I can point out between the twins is that the traveling twin is accelerating relative to his original frame of reference, whereas the other twin isn't.

Perhaps our debate needs clarification by a third observer, since we both seem to be stuck in a preferential frame of mind.

 

[Ulim]

This nice animation does not address my question either. However I do note that one of the two bodies in motion is undergoing a continuous acceleration.

There is no acceleration in this animation.
Both have a constant speed and a fixed distance to eachother so none of them are accelerating in respect to eachother.

 

[pitubo]

There is no acceleration in this animation.

Circular motion - Wikipedia
Acceleration - Wikipedia

Both have a constant speed and a fixed distance to eachother so none of them are accelerating in respect to eachother.

But the body in circular motion is accelerating relatively to its own frame, which is what I was referring to (read the part that you responded to but forgot to quote.) Also, while the absolute speed may be constant, the speed vector clearly isn't.

Two other tidbits about your animation that just spring to mind:
1) good luck finding any actual examples of orbital speeds approaching c.
2) even purely theoretically your animation is not realistic, because both bodies would orbit around a common center of mass (or something equivalent, depending on the type of bodies and forces involved.) Both bodies would be undergoing continuous acceleration. The animation takes one of the bodies variable position and turns it into an absolute reference. This only recreates a(n illusion of) circular in the case of two bodies. With three or more, this is generally not possible.

 

[Jagube]

The traveling Twin needs to accelerate and decelerate during the journey to come back to the meeting. It is during these acceleration stages that he sees the twin back on earth age very quickly, so when they meet everything makes sense. At other times during the trip (constant speed), you are correct: both twins see the other twin aging more slowly.[/url]

OK, that makes sense - unlike his brother, he expends energy to accelerate / decelerate, and that sets him apart. In a sense he's 'rewarded' for expending all that energy: it 'buys' him a slowdown of time. But traveling per se doesn't.

I always thought the time he spent traveling at v (close to c) would affect how much younger he'd be upon reuniting with his brother. But from what you say, that time irrelevant. The only thing that affects the difference in age between them is how close to c he got. Is that right?

 

[dragonrider]

The traveling Twin needs to accelerate and decelerate during the journey to come back to the meeting. It is during these acceleration stages that he sees the twin back on earth age very quickly, so when they meet everything makes sense. At other times during the trip (constant speed), you are correct: both twins see the other twin aging more slowly.[/url]

OK, that makes sense - unlike his brother, he expends energy to accelerate / decelerate, and that sets him apart. In a sense he's 'rewarded' for expending all that energy: it 'buys' him a slowdown of time. But traveling per se doesn't.

I always thought the time he spent traveling at v (close to c) would affect how much younger he'd be upon reuniting with his brother. But from what you say, that time irrelevant. The only thing that affects the difference in age between them is how close to c he got. Is that right?

Wouldn't that be true only, in case he travels away from his twin first, and then back?
In case of the circular motion, time would go slower for the moving object. So wouldn't that meant that for the moving twin, the not moving twin ages faster? I mean, if they had Skype or something, and he's getting N frames per second sent towards him, to him that would be more frames per second, because a second for him lasts longer. If you move away from the twin, or towards him, you get a sort of doppler effect ofcourse. But not in the case of circular motion (so in that sense, acceleration wouldn't matter either, or would it, because of it's gravitational effects on time?).

I mean, normally you have two things going on here: the doppler effect, AND the slowing down of time itself. But would that still be the case with circular motion?

 

[Ulim]

There is no acceleration in this animation.

Circular motion - Wikipedia
Acceleration - Wikipedia

Both have a constant speed and a fixed distance to eachother so none of them are accelerating in respect to eachother.

But the body in circular motion is accelerating relatively to its own frame, which is what I was referring to (read the part that you responded to but forgot to quote.) Also, while the absolute speed may be constant, the speed vector clearly isn't.

Two other tidbits about your animation that just spring to mind:
1) good luck finding any actual examples of orbital speeds approaching c.
2) even purely theoretically your animation is not realistic, because both bodies would orbit around a common center of mass (or something equivalent, depending on the type of bodies and forces involved.) Both bodies would be undergoing continuous acceleration. The animation takes one of the bodies variable position and turns it into an absolute reference. This only recreates a(n illusion of) circular in the case of two bodies. With three or more, this is generally not possible.

:? dude its just an animation to show dialation.
No newtonian physics here and also generally its only there to show that red (moves faster because way longer path) has time dialation.
What you are doin is just useless nitpicking