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Shooting someone's past self using special relativity
In Special Relativity, is it allowed to ask 'How much time has elapsed in a second inertial frame at a particular moment in the first inertial frame'?Hypothetical effect of Solar System/Orion Spur traveling near speed of lightWhy don't two observers' clocks measure the same time between the same events?Synchronization of clocks and observations of time in special relativityIs time dilation an illusion? Variation on the twins paradoxTwin paradox, with a meeting at a halfway pointQuestion/Doubt about Time Dilation Symmetry in Special RelativityExplanation for a much simpler version of the twin paradox?Relativity of simultaneityUnable to Reconcile my Understanding of Length Contraction with the Lorentz TransformationHow do we explain this new take on the old Twins Paradox?In Special Relativity, is it allowed to ask 'How much time has elapsed in a second inertial frame at a particular moment in the first inertial frame'?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Suppose A and B are a long distance apart initially. B takes off in a spacecraft in A's direction at a really high speed. Both were aged 0 when B took off. When B about to cross paths with A, A observes him to be 30 years old (while A is 60). At this point, 60 year old A shoots the 30 year old B. B dies.
Now, from B's frame, the bullet has to be fired by A when B is 120. That's because the event of 'bullet firing' must happen after the event of 'A turning 60', and A turns 60 in B's frame only when B is 120. So B has to die at 120 as seen from his own frame of reference.
How can B die at both 30 years old and 120 years old?
special-relativity relativity
asked 11 hours ago
Ryder RudeRyder Rude
1616
$endgroup$
add a comment |
$begingroup$
Suppose A and B are a long distance apart initially. B takes off in a spacecraft in A's direction at a really high speed. Both were aged 0 when B took off. When B about to cross paths with A, A observes him to be 30 years old (while A is 60). At this point, 60 year old A shoots the 30 year old B. B dies.
Now, from B's frame, the bullet has to be fired by A when B is 120. That's because the event of 'bullet firing' must happen after the event of 'A turning 60', and A turns 60 in B's frame only when B is 120. So B has to die at 120 as seen from his own frame of reference.
How can B die at both 30 years old and 120 years old?
special-relativity relativity
asked 11 hours ago
Ryder RudeRyder Rude
1616
$endgroup$
1
$begingroup$
This is a follow-up question to physics.stackexchange.com/q/486482/123208
$endgroup$
– PM 2Ring
11 hours ago
$begingroup$
So will your next question involve the cousin C?
$endgroup$
– Solar Mike
11 hours ago
2
$begingroup$
This question has two different answers depending on what you mean by "Both were aged 0 when B took off". Do you mean to assume that this is true in A's frame or in B's frame?
$endgroup$
– WillO
10 hours ago
3
$begingroup$
The silver bullet for almost all of SR paradoxes: draw a spacetime diagram.
$endgroup$
– Feynmans Out for Grumpy Cat
9 hours ago
1
$begingroup$
@FeynmansOutforGrumpyCat : I have a long history (on this site and elsewhere) of telling people that the best approach to almost every problem in SR is to draw the spacetime diagram. But in the present case, I actually think that the step-by-step reasoning in my answer is likely to be more educational for someone who is confused at the level of this OP.
$endgroup$
– WillO
9 hours ago
add a comment |
$begingroup$
Suppose A and B are a long distance apart initially. B takes off in a spacecraft in A's direction at a really high speed. Both were aged 0 when B took off. When B about to cross paths with A, A observes him to be 30 years old (while A is 60). At this point, 60 year old A shoots the 30 year old B. B dies.
Now, from B's frame, the bullet has to be fired by A when B is 120. That's because the event of 'bullet firing' must happen after the event of 'A turning 60', and A turns 60 in B's frame only when B is 120. So B has to die at 120 as seen from his own frame of reference.
How can B die at both 30 years old and 120 years old?
special-relativity relativity
asked 11 hours ago
Ryder RudeRyder Rude
1616
$endgroup$
Suppose A and B are a long distance apart initially. B takes off in a spacecraft in A's direction at a really high speed. Both were aged 0 when B took off. When B about to cross paths with A, A observes him to be 30 years old (while A is 60). At this point, 60 year old A shoots the 30 year old B. B dies.
Now, from B's frame, the bullet has to be fired by A when B is 120. That's because the event of 'bullet firing' must happen after the event of 'A turning 60', and A turns 60 in B's frame only when B is 120. So B has to die at 120 as seen from his own frame of reference.
How can B die at both 30 years old and 120 years old?
special-relativity relativity
special-relativity relativity
asked 11 hours ago
Ryder RudeRyder Rude
1616
asked 11 hours ago
Ryder RudeRyder Rude
1616
asked 11 hours ago
Ryder RudeRyder Rude
1616
asked 11 hours ago
Ryder RudeRyder Rude
1616
asked 11 hours ago
Ryder RudeRyder Rude
1616
1616
1
$begingroup$
This is a follow-up question to physics.stackexchange.com/q/486482/123208
$endgroup$
– PM 2Ring
11 hours ago
$begingroup$
So will your next question involve the cousin C?
$endgroup$
– Solar Mike
11 hours ago
2
$begingroup$
This question has two different answers depending on what you mean by "Both were aged 0 when B took off". Do you mean to assume that this is true in A's frame or in B's frame?
$endgroup$
– WillO
10 hours ago
3
$begingroup$
The silver bullet for almost all of SR paradoxes: draw a spacetime diagram.
$endgroup$
– Feynmans Out for Grumpy Cat
9 hours ago
1
$begingroup$
@FeynmansOutforGrumpyCat : I have a long history (on this site and elsewhere) of telling people that the best approach to almost every problem in SR is to draw the spacetime diagram. But in the present case, I actually think that the step-by-step reasoning in my answer is likely to be more educational for someone who is confused at the level of this OP.
$endgroup$
– WillO
9 hours ago
add a comment |
1
$begingroup$
This is a follow-up question to physics.stackexchange.com/q/486482/123208
$endgroup$
– PM 2Ring
11 hours ago
$begingroup$
So will your next question involve the cousin C?
$endgroup$
– Solar Mike
11 hours ago
2
$begingroup$
This question has two different answers depending on what you mean by "Both were aged 0 when B took off". Do you mean to assume that this is true in A's frame or in B's frame?
$endgroup$
– WillO
10 hours ago
3
$begingroup$
The silver bullet for almost all of SR paradoxes: draw a spacetime diagram.
$endgroup$
– Feynmans Out for Grumpy Cat
9 hours ago
1
$begingroup$
@FeynmansOutforGrumpyCat : I have a long history (on this site and elsewhere) of telling people that the best approach to almost every problem in SR is to draw the spacetime diagram. But in the present case, I actually think that the step-by-step reasoning in my answer is likely to be more educational for someone who is confused at the level of this OP.
$endgroup$
– WillO
9 hours ago
1
1
$begingroup$
This is a follow-up question to physics.stackexchange.com/q/486482/123208
$endgroup$
– PM 2Ring
11 hours ago
$begingroup$
This is a follow-up question to physics.stackexchange.com/q/486482/123208
$endgroup$
– PM 2Ring
11 hours ago
$begingroup$
So will your next question involve the cousin C?
$endgroup$
– Solar Mike
11 hours ago
$begingroup$
So will your next question involve the cousin C?
$endgroup$
– Solar Mike
11 hours ago
2
2
$begingroup$
This question has two different answers depending on what you mean by "Both were aged 0 when B took off". Do you mean to assume that this is true in A's frame or in B's frame?
$endgroup$
– WillO
10 hours ago
$begingroup$
This question has two different answers depending on what you mean by "Both were aged 0 when B took off". Do you mean to assume that this is true in A's frame or in B's frame?
$endgroup$
– WillO
10 hours ago
3
3
$begingroup$
The silver bullet for almost all of SR paradoxes: draw a spacetime diagram.
$endgroup$
– Feynmans Out for Grumpy Cat
9 hours ago
$begingroup$
The silver bullet for almost all of SR paradoxes: draw a spacetime diagram.
$endgroup$
– Feynmans Out for Grumpy Cat
9 hours ago
1
1
$begingroup$
@FeynmansOutforGrumpyCat : I have a long history (on this site and elsewhere) of telling people that the best approach to almost every problem in SR is to draw the spacetime diagram. But in the present case, I actually think that the step-by-step reasoning in my answer is likely to be more educational for someone who is confused at the level of this OP.
$endgroup$
– WillO
9 hours ago
$begingroup$
@FeynmansOutforGrumpyCat : I have a long history (on this site and elsewhere) of telling people that the best approach to almost every problem in SR is to draw the spacetime diagram. But in the present case, I actually think that the step-by-step reasoning in my answer is likely to be more educational for someone who is confused at the level of this OP.
$endgroup$
– WillO
9 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Case I: When B takes off, A and B are both aged 0 in A's frame.
1) When B passes A, B is 30 (given in your setup).
2) But B's clocks run at half-speed according to A, so A says that B has been traveling 60 years.
3) Therefore A is 60.
4) But B says A's clocks run at half speed, so B says A was born 120 years before the shooting --- that is, 90 years before B was born.
5) So B's story is this: "90 years before I was born, A was born. He aged at halftime, so on my birthday, he was 45. At that time I started my journey to earth, which took 30 years. During that time, A aged another 15 years, so he was 60 when we met. Then he shot me. I died at 30."
Your mistake: You said that "from B's frame, the bullet has to be fired by A when B is 120". That's not correct. The correct statement is "from B's frame, the bullet has to be fired 120 years after A was born". Since B is 30 at the time of the shooting, A must have been born 90 years before B.
Your bigger mistake: You assumed that two different problems (namely this one and the one you asked in your last post) have to have exactly the same answer. In the other problem, A and B were in the same place at the same time when both were born. In this problem they weren't.
Case II: When B takes off, A and B are both aged 0 in B's frame.
1) When B reaches earth, he is aged 30 (given in the problem).
2) According to B, A ages at half-speed. Therefore A is 15, not 60 as you supposed. The 15-year-old A shoots the 30 year old B. Game over for B.
answered 10 hours ago
WillOWillO
7,18622334
$endgroup$
$begingroup$
Another mistake is also the OP saying "Both were aged 0 when B took off" without specifying which frame this is in. You have implicitly addressed this through having your two cases.
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
Also you have B saying "At that time I started my journey to earth", but doesn't the entire analysis depend on B moving at some speed relative to A the entire time?
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
@AaronStevens: If you assume B has been moving forever, then "started my journey at that time" means something like "I was born on a rocket ship that passed Alpha Centauri at that time". But I think it's more consistent with the OP's vision to assume B is born on Alpha Centauri, which is stationary with respect to earth, and then immediately begins moving toward earth at some large fraction of the speed of light. The same analysis applies either way.
$endgroup$
– WillO
9 hours ago
$begingroup$
If B's frame (before B is born, sounds odd haha) is stationary with respect to Earth then how can you assert that A is aging at half-time before B leaves though?
$endgroup$
– Aaron Stevens
9 hours ago
1
$begingroup$
@AaronStevens: Thanks for confirming that we agree instead of just disappearing. I hate when that happens!
$endgroup$
– WillO
8 hours ago
|
show 2 more comments
$begingroup$
You suppose that A and B are far apart "initially" and that they were "both aged 0 when B took off". It means that you assume that simultaneity is an absolute notion, independent of the frame of reference, which is why you come to a paradox. It is exactly the symetric case of two people starting at the same point, diverging fast, and observing each other: each one will see the other one aging slower than him, because they compare moments in non parallel time sections. This is best shown by a drawing, this is the one I have suggested in this thread: link to a video at the second displaying the graph.
answered 11 hours ago
MattMatt
39216
$endgroup$
add a comment |
$begingroup$
In addition to Willo's answer
As it was being said by @WillO and @PM 2Ring and others, your problem here is that you have assumed:
'Both were aged 0 when B took off'
Well that's simply wrong, because like this problem, you have not specified in which frame?! Of course you could have asked
1.When B takes off, A and B are both aged 0 in A's frame.
or
2.When B takes off, A and B are both aged 0 in B's frame.
or
3.before B takes off (where he is at the rest relative to A, but in different position), A and B are both aged 0 in both frame. Indeed, their frame would be same, because they are at rest at that moment. Then instantly, B accelerate toward A and make his velocity constant after a moment in his frame.
But they are different questions. In case 3 You might think that if B zero y/o (after taking off) sends a signal to A, in his frame, he'd assume that A is zero y/o too but that's 100% wrong, in his frame A will be a lot older than B.
Someone might wondering how this asymmetry possible even though their initial state were same, well in short in accelerated frames like B, time can elapse faster or slower depending on position. i.e in B frame, A's clock will elapse a lot faster than his as long as B is accelerated. And if you assume that B will accelerate instantly, then you will see that in B's frame A ages instantly as well! If you are intested in last case i suggest that you read
Explorations in Mathematical Physics from Don Koks
Accelerated frame section.
answered 7 hours ago
paradoxyparadoxy
16416
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Case I: When B takes off, A and B are both aged 0 in A's frame.
1) When B passes A, B is 30 (given in your setup).
2) But B's clocks run at half-speed according to A, so A says that B has been traveling 60 years.
3) Therefore A is 60.
4) But B says A's clocks run at half speed, so B says A was born 120 years before the shooting --- that is, 90 years before B was born.
5) So B's story is this: "90 years before I was born, A was born. He aged at halftime, so on my birthday, he was 45. At that time I started my journey to earth, which took 30 years. During that time, A aged another 15 years, so he was 60 when we met. Then he shot me. I died at 30."
Your mistake: You said that "from B's frame, the bullet has to be fired by A when B is 120". That's not correct. The correct statement is "from B's frame, the bullet has to be fired 120 years after A was born". Since B is 30 at the time of the shooting, A must have been born 90 years before B.
Your bigger mistake: You assumed that two different problems (namely this one and the one you asked in your last post) have to have exactly the same answer. In the other problem, A and B were in the same place at the same time when both were born. In this problem they weren't.
Case II: When B takes off, A and B are both aged 0 in B's frame.
1) When B reaches earth, he is aged 30 (given in the problem).
2) According to B, A ages at half-speed. Therefore A is 15, not 60 as you supposed. The 15-year-old A shoots the 30 year old B. Game over for B.
answered 10 hours ago
WillOWillO
7,18622334
$endgroup$
$begingroup$
Another mistake is also the OP saying "Both were aged 0 when B took off" without specifying which frame this is in. You have implicitly addressed this through having your two cases.
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
Also you have B saying "At that time I started my journey to earth", but doesn't the entire analysis depend on B moving at some speed relative to A the entire time?
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
@AaronStevens: If you assume B has been moving forever, then "started my journey at that time" means something like "I was born on a rocket ship that passed Alpha Centauri at that time". But I think it's more consistent with the OP's vision to assume B is born on Alpha Centauri, which is stationary with respect to earth, and then immediately begins moving toward earth at some large fraction of the speed of light. The same analysis applies either way.
$endgroup$
– WillO
9 hours ago
$begingroup$
If B's frame (before B is born, sounds odd haha) is stationary with respect to Earth then how can you assert that A is aging at half-time before B leaves though?
$endgroup$
– Aaron Stevens
9 hours ago
1
$begingroup$
@AaronStevens: Thanks for confirming that we agree instead of just disappearing. I hate when that happens!
$endgroup$
– WillO
8 hours ago
|
show 2 more comments
$begingroup$
Case I: When B takes off, A and B are both aged 0 in A's frame.
1) When B passes A, B is 30 (given in your setup).
2) But B's clocks run at half-speed according to A, so A says that B has been traveling 60 years.
3) Therefore A is 60.
4) But B says A's clocks run at half speed, so B says A was born 120 years before the shooting --- that is, 90 years before B was born.
5) So B's story is this: "90 years before I was born, A was born. He aged at halftime, so on my birthday, he was 45. At that time I started my journey to earth, which took 30 years. During that time, A aged another 15 years, so he was 60 when we met. Then he shot me. I died at 30."
Your mistake: You said that "from B's frame, the bullet has to be fired by A when B is 120". That's not correct. The correct statement is "from B's frame, the bullet has to be fired 120 years after A was born". Since B is 30 at the time of the shooting, A must have been born 90 years before B.
Your bigger mistake: You assumed that two different problems (namely this one and the one you asked in your last post) have to have exactly the same answer. In the other problem, A and B were in the same place at the same time when both were born. In this problem they weren't.
Case II: When B takes off, A and B are both aged 0 in B's frame.
1) When B reaches earth, he is aged 30 (given in the problem).
2) According to B, A ages at half-speed. Therefore A is 15, not 60 as you supposed. The 15-year-old A shoots the 30 year old B. Game over for B.
answered 10 hours ago
WillOWillO
7,18622334
$endgroup$
$begingroup$
Another mistake is also the OP saying "Both were aged 0 when B took off" without specifying which frame this is in. You have implicitly addressed this through having your two cases.
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
Also you have B saying "At that time I started my journey to earth", but doesn't the entire analysis depend on B moving at some speed relative to A the entire time?
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
@AaronStevens: If you assume B has been moving forever, then "started my journey at that time" means something like "I was born on a rocket ship that passed Alpha Centauri at that time". But I think it's more consistent with the OP's vision to assume B is born on Alpha Centauri, which is stationary with respect to earth, and then immediately begins moving toward earth at some large fraction of the speed of light. The same analysis applies either way.
$endgroup$
– WillO
9 hours ago
$begingroup$
If B's frame (before B is born, sounds odd haha) is stationary with respect to Earth then how can you assert that A is aging at half-time before B leaves though?
$endgroup$
– Aaron Stevens
9 hours ago
1
$begingroup$
@AaronStevens: Thanks for confirming that we agree instead of just disappearing. I hate when that happens!
$endgroup$
– WillO
8 hours ago
|
show 2 more comments
$begingroup$
Case I: When B takes off, A and B are both aged 0 in A's frame.
1) When B passes A, B is 30 (given in your setup).
2) But B's clocks run at half-speed according to A, so A says that B has been traveling 60 years.
3) Therefore A is 60.
4) But B says A's clocks run at half speed, so B says A was born 120 years before the shooting --- that is, 90 years before B was born.
5) So B's story is this: "90 years before I was born, A was born. He aged at halftime, so on my birthday, he was 45. At that time I started my journey to earth, which took 30 years. During that time, A aged another 15 years, so he was 60 when we met. Then he shot me. I died at 30."
Your mistake: You said that "from B's frame, the bullet has to be fired by A when B is 120". That's not correct. The correct statement is "from B's frame, the bullet has to be fired 120 years after A was born". Since B is 30 at the time of the shooting, A must have been born 90 years before B.
Your bigger mistake: You assumed that two different problems (namely this one and the one you asked in your last post) have to have exactly the same answer. In the other problem, A and B were in the same place at the same time when both were born. In this problem they weren't.
Case II: When B takes off, A and B are both aged 0 in B's frame.
1) When B reaches earth, he is aged 30 (given in the problem).
2) According to B, A ages at half-speed. Therefore A is 15, not 60 as you supposed. The 15-year-old A shoots the 30 year old B. Game over for B.
answered 10 hours ago
WillOWillO
7,18622334
$endgroup$
Case I: When B takes off, A and B are both aged 0 in A's frame.
1) When B passes A, B is 30 (given in your setup).
2) But B's clocks run at half-speed according to A, so A says that B has been traveling 60 years.
3) Therefore A is 60.
4) But B says A's clocks run at half speed, so B says A was born 120 years before the shooting --- that is, 90 years before B was born.
5) So B's story is this: "90 years before I was born, A was born. He aged at halftime, so on my birthday, he was 45. At that time I started my journey to earth, which took 30 years. During that time, A aged another 15 years, so he was 60 when we met. Then he shot me. I died at 30."
Your mistake: You said that "from B's frame, the bullet has to be fired by A when B is 120". That's not correct. The correct statement is "from B's frame, the bullet has to be fired 120 years after A was born". Since B is 30 at the time of the shooting, A must have been born 90 years before B.
Your bigger mistake: You assumed that two different problems (namely this one and the one you asked in your last post) have to have exactly the same answer. In the other problem, A and B were in the same place at the same time when both were born. In this problem they weren't.
Case II: When B takes off, A and B are both aged 0 in B's frame.
1) When B reaches earth, he is aged 30 (given in the problem).
2) According to B, A ages at half-speed. Therefore A is 15, not 60 as you supposed. The 15-year-old A shoots the 30 year old B. Game over for B.
answered 10 hours ago
WillOWillO
7,18622334
answered 10 hours ago
WillOWillO
7,18622334
answered 10 hours ago
WillOWillO
7,18622334
answered 10 hours ago
WillOWillO
7,18622334
7,18622334
$begingroup$
Another mistake is also the OP saying "Both were aged 0 when B took off" without specifying which frame this is in. You have implicitly addressed this through having your two cases.
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
Also you have B saying "At that time I started my journey to earth", but doesn't the entire analysis depend on B moving at some speed relative to A the entire time?
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
@AaronStevens: If you assume B has been moving forever, then "started my journey at that time" means something like "I was born on a rocket ship that passed Alpha Centauri at that time". But I think it's more consistent with the OP's vision to assume B is born on Alpha Centauri, which is stationary with respect to earth, and then immediately begins moving toward earth at some large fraction of the speed of light. The same analysis applies either way.
$endgroup$
– WillO
9 hours ago
$begingroup$
If B's frame (before B is born, sounds odd haha) is stationary with respect to Earth then how can you assert that A is aging at half-time before B leaves though?
$endgroup$
– Aaron Stevens
9 hours ago
1
$begingroup$
@AaronStevens: Thanks for confirming that we agree instead of just disappearing. I hate when that happens!
$endgroup$
– WillO
8 hours ago
|
show 2 more comments
$begingroup$
Another mistake is also the OP saying "Both were aged 0 when B took off" without specifying which frame this is in. You have implicitly addressed this through having your two cases.
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
Also you have B saying "At that time I started my journey to earth", but doesn't the entire analysis depend on B moving at some speed relative to A the entire time?
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
@AaronStevens: If you assume B has been moving forever, then "started my journey at that time" means something like "I was born on a rocket ship that passed Alpha Centauri at that time". But I think it's more consistent with the OP's vision to assume B is born on Alpha Centauri, which is stationary with respect to earth, and then immediately begins moving toward earth at some large fraction of the speed of light. The same analysis applies either way.
$endgroup$
– WillO
9 hours ago
$begingroup$
If B's frame (before B is born, sounds odd haha) is stationary with respect to Earth then how can you assert that A is aging at half-time before B leaves though?
$endgroup$
– Aaron Stevens
9 hours ago
1
$begingroup$
@AaronStevens: Thanks for confirming that we agree instead of just disappearing. I hate when that happens!
$endgroup$
– WillO
8 hours ago
$begingroup$
Another mistake is also the OP saying "Both were aged 0 when B took off" without specifying which frame this is in. You have implicitly addressed this through having your two cases.
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
Another mistake is also the OP saying "Both were aged 0 when B took off" without specifying which frame this is in. You have implicitly addressed this through having your two cases.
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
Also you have B saying "At that time I started my journey to earth", but doesn't the entire analysis depend on B moving at some speed relative to A the entire time?
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
Also you have B saying "At that time I started my journey to earth", but doesn't the entire analysis depend on B moving at some speed relative to A the entire time?
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
@AaronStevens: If you assume B has been moving forever, then "started my journey at that time" means something like "I was born on a rocket ship that passed Alpha Centauri at that time". But I think it's more consistent with the OP's vision to assume B is born on Alpha Centauri, which is stationary with respect to earth, and then immediately begins moving toward earth at some large fraction of the speed of light. The same analysis applies either way.
$endgroup$
– WillO
9 hours ago
$begingroup$
@AaronStevens: If you assume B has been moving forever, then "started my journey at that time" means something like "I was born on a rocket ship that passed Alpha Centauri at that time". But I think it's more consistent with the OP's vision to assume B is born on Alpha Centauri, which is stationary with respect to earth, and then immediately begins moving toward earth at some large fraction of the speed of light. The same analysis applies either way.
$endgroup$
– WillO
9 hours ago
$begingroup$
If B's frame (before B is born, sounds odd haha) is stationary with respect to Earth then how can you assert that A is aging at half-time before B leaves though?
$endgroup$
– Aaron Stevens
9 hours ago
$begingroup$
If B's frame (before B is born, sounds odd haha) is stationary with respect to Earth then how can you assert that A is aging at half-time before B leaves though?
$endgroup$
– Aaron Stevens
9 hours ago
1
1
$begingroup$
@AaronStevens: Thanks for confirming that we agree instead of just disappearing. I hate when that happens!
$endgroup$
– WillO
8 hours ago
$begingroup$
@AaronStevens: Thanks for confirming that we agree instead of just disappearing. I hate when that happens!
$endgroup$
– WillO
8 hours ago
|
show 2 more comments
$begingroup$
You suppose that A and B are far apart "initially" and that they were "both aged 0 when B took off". It means that you assume that simultaneity is an absolute notion, independent of the frame of reference, which is why you come to a paradox. It is exactly the symetric case of two people starting at the same point, diverging fast, and observing each other: each one will see the other one aging slower than him, because they compare moments in non parallel time sections. This is best shown by a drawing, this is the one I have suggested in this thread: link to a video at the second displaying the graph.
answered 11 hours ago
MattMatt
39216
$endgroup$
add a comment |
$begingroup$
You suppose that A and B are far apart "initially" and that they were "both aged 0 when B took off". It means that you assume that simultaneity is an absolute notion, independent of the frame of reference, which is why you come to a paradox. It is exactly the symetric case of two people starting at the same point, diverging fast, and observing each other: each one will see the other one aging slower than him, because they compare moments in non parallel time sections. This is best shown by a drawing, this is the one I have suggested in this thread: link to a video at the second displaying the graph.
answered 11 hours ago
MattMatt
39216
$endgroup$
add a comment |
$begingroup$
You suppose that A and B are far apart "initially" and that they were "both aged 0 when B took off". It means that you assume that simultaneity is an absolute notion, independent of the frame of reference, which is why you come to a paradox. It is exactly the symetric case of two people starting at the same point, diverging fast, and observing each other: each one will see the other one aging slower than him, because they compare moments in non parallel time sections. This is best shown by a drawing, this is the one I have suggested in this thread: link to a video at the second displaying the graph.
answered 11 hours ago
MattMatt
39216
$endgroup$
You suppose that A and B are far apart "initially" and that they were "both aged 0 when B took off". It means that you assume that simultaneity is an absolute notion, independent of the frame of reference, which is why you come to a paradox. It is exactly the symetric case of two people starting at the same point, diverging fast, and observing each other: each one will see the other one aging slower than him, because they compare moments in non parallel time sections. This is best shown by a drawing, this is the one I have suggested in this thread: link to a video at the second displaying the graph.
answered 11 hours ago
MattMatt
39216
answered 11 hours ago
MattMatt
39216
answered 11 hours ago
MattMatt
39216
answered 11 hours ago
MattMatt
39216
39216
add a comment |
add a comment |
$begingroup$
In addition to Willo's answer
As it was being said by @WillO and @PM 2Ring and others, your problem here is that you have assumed:
'Both were aged 0 when B took off'
Well that's simply wrong, because like this problem, you have not specified in which frame?! Of course you could have asked
1.When B takes off, A and B are both aged 0 in A's frame.
or
2.When B takes off, A and B are both aged 0 in B's frame.
or
3.before B takes off (where he is at the rest relative to A, but in different position), A and B are both aged 0 in both frame. Indeed, their frame would be same, because they are at rest at that moment. Then instantly, B accelerate toward A and make his velocity constant after a moment in his frame.
But they are different questions. In case 3 You might think that if B zero y/o (after taking off) sends a signal to A, in his frame, he'd assume that A is zero y/o too but that's 100% wrong, in his frame A will be a lot older than B.
Someone might wondering how this asymmetry possible even though their initial state were same, well in short in accelerated frames like B, time can elapse faster or slower depending on position. i.e in B frame, A's clock will elapse a lot faster than his as long as B is accelerated. And if you assume that B will accelerate instantly, then you will see that in B's frame A ages instantly as well! If you are intested in last case i suggest that you read
Explorations in Mathematical Physics from Don Koks
Accelerated frame section.
answered 7 hours ago
paradoxyparadoxy
16416
$endgroup$
add a comment |
$begingroup$
In addition to Willo's answer
As it was being said by @WillO and @PM 2Ring and others, your problem here is that you have assumed:
'Both were aged 0 when B took off'
Well that's simply wrong, because like this problem, you have not specified in which frame?! Of course you could have asked
1.When B takes off, A and B are both aged 0 in A's frame.
or
2.When B takes off, A and B are both aged 0 in B's frame.
or
3.before B takes off (where he is at the rest relative to A, but in different position), A and B are both aged 0 in both frame. Indeed, their frame would be same, because they are at rest at that moment. Then instantly, B accelerate toward A and make his velocity constant after a moment in his frame.
But they are different questions. In case 3 You might think that if B zero y/o (after taking off) sends a signal to A, in his frame, he'd assume that A is zero y/o too but that's 100% wrong, in his frame A will be a lot older than B.
Someone might wondering how this asymmetry possible even though their initial state were same, well in short in accelerated frames like B, time can elapse faster or slower depending on position. i.e in B frame, A's clock will elapse a lot faster than his as long as B is accelerated. And if you assume that B will accelerate instantly, then you will see that in B's frame A ages instantly as well! If you are intested in last case i suggest that you read
Explorations in Mathematical Physics from Don Koks
Accelerated frame section.
answered 7 hours ago
paradoxyparadoxy
16416
$endgroup$
add a comment |
$begingroup$
In addition to Willo's answer
As it was being said by @WillO and @PM 2Ring and others, your problem here is that you have assumed:
'Both were aged 0 when B took off'
Well that's simply wrong, because like this problem, you have not specified in which frame?! Of course you could have asked
1.When B takes off, A and B are both aged 0 in A's frame.
or
2.When B takes off, A and B are both aged 0 in B's frame.
or
3.before B takes off (where he is at the rest relative to A, but in different position), A and B are both aged 0 in both frame. Indeed, their frame would be same, because they are at rest at that moment. Then instantly, B accelerate toward A and make his velocity constant after a moment in his frame.
But they are different questions. In case 3 You might think that if B zero y/o (after taking off) sends a signal to A, in his frame, he'd assume that A is zero y/o too but that's 100% wrong, in his frame A will be a lot older than B.
Someone might wondering how this asymmetry possible even though their initial state were same, well in short in accelerated frames like B, time can elapse faster or slower depending on position. i.e in B frame, A's clock will elapse a lot faster than his as long as B is accelerated. And if you assume that B will accelerate instantly, then you will see that in B's frame A ages instantly as well! If you are intested in last case i suggest that you read
Explorations in Mathematical Physics from Don Koks
Accelerated frame section.
answered 7 hours ago
paradoxyparadoxy
16416
$endgroup$
In addition to Willo's answer
As it was being said by @WillO and @PM 2Ring and others, your problem here is that you have assumed:
'Both were aged 0 when B took off'
Well that's simply wrong, because like this problem, you have not specified in which frame?! Of course you could have asked
1.When B takes off, A and B are both aged 0 in A's frame.
or
2.When B takes off, A and B are both aged 0 in B's frame.
or
3.before B takes off (where he is at the rest relative to A, but in different position), A and B are both aged 0 in both frame. Indeed, their frame would be same, because they are at rest at that moment. Then instantly, B accelerate toward A and make his velocity constant after a moment in his frame.
But they are different questions. In case 3 You might think that if B zero y/o (after taking off) sends a signal to A, in his frame, he'd assume that A is zero y/o too but that's 100% wrong, in his frame A will be a lot older than B.
Someone might wondering how this asymmetry possible even though their initial state were same, well in short in accelerated frames like B, time can elapse faster or slower depending on position. i.e in B frame, A's clock will elapse a lot faster than his as long as B is accelerated. And if you assume that B will accelerate instantly, then you will see that in B's frame A ages instantly as well! If you are intested in last case i suggest that you read
Explorations in Mathematical Physics from Don Koks
Accelerated frame section.
answered 7 hours ago
paradoxyparadoxy
16416
answered 7 hours ago
paradoxyparadoxy
16416
answered 7 hours ago
paradoxyparadoxy
16416
answered 7 hours ago
paradoxyparadoxy
16416
16416
add a comment |
add a comment |
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1
$begingroup$
This is a follow-up question to physics.stackexchange.com/q/486482/123208
$endgroup$
– PM 2Ring
11 hours ago
$begingroup$
So will your next question involve the cousin C?
$endgroup$
– Solar Mike
11 hours ago
2
$begingroup$
This question has two different answers depending on what you mean by "Both were aged 0 when B took off". Do you mean to assume that this is true in A's frame or in B's frame?
$endgroup$
– WillO
10 hours ago
3
$begingroup$
The silver bullet for almost all of SR paradoxes: draw a spacetime diagram.
$endgroup$
– Feynmans Out for Grumpy Cat
9 hours ago
1
$begingroup$
@FeynmansOutforGrumpyCat : I have a long history (on this site and elsewhere) of telling people that the best approach to almost every problem in SR is to draw the spacetime diagram. But in the present case, I actually think that the step-by-step reasoning in my answer is likely to be more educational for someone who is confused at the level of this OP.
$endgroup$
– WillO
9 hours ago