Which observer measures proper time in the twins paradox?
$begingroup$
I know that proper time is defined as the time which the clock moving relative to that observe shows. That is, a clock attached to observer A will always be As proper time. I also understand that this time will be the shortest time which passes as seen by A. However in the twins paradox I am very confused. What is the proper time. From Earths perspective, its clock is measuring proper time and from the travellers perspective its clock is measuring proper time. But when the travellers clock returns it shows the shortest time and therefore is proper time. Pretty much, my question is which observer measures proper time and why. Is this even an valid question? Is there an absolute proper time? My understanding of relativity is really basic, pretty much high school level.
special-relativity reference-frames coordinate-systems inertial-frames observers
$endgroup$
add a comment |
$begingroup$
I know that proper time is defined as the time which the clock moving relative to that observe shows. That is, a clock attached to observer A will always be As proper time. I also understand that this time will be the shortest time which passes as seen by A. However in the twins paradox I am very confused. What is the proper time. From Earths perspective, its clock is measuring proper time and from the travellers perspective its clock is measuring proper time. But when the travellers clock returns it shows the shortest time and therefore is proper time. Pretty much, my question is which observer measures proper time and why. Is this even an valid question? Is there an absolute proper time? My understanding of relativity is really basic, pretty much high school level.
special-relativity reference-frames coordinate-systems inertial-frames observers
$endgroup$
add a comment |
$begingroup$
I know that proper time is defined as the time which the clock moving relative to that observe shows. That is, a clock attached to observer A will always be As proper time. I also understand that this time will be the shortest time which passes as seen by A. However in the twins paradox I am very confused. What is the proper time. From Earths perspective, its clock is measuring proper time and from the travellers perspective its clock is measuring proper time. But when the travellers clock returns it shows the shortest time and therefore is proper time. Pretty much, my question is which observer measures proper time and why. Is this even an valid question? Is there an absolute proper time? My understanding of relativity is really basic, pretty much high school level.
special-relativity reference-frames coordinate-systems inertial-frames observers
$endgroup$
I know that proper time is defined as the time which the clock moving relative to that observe shows. That is, a clock attached to observer A will always be As proper time. I also understand that this time will be the shortest time which passes as seen by A. However in the twins paradox I am very confused. What is the proper time. From Earths perspective, its clock is measuring proper time and from the travellers perspective its clock is measuring proper time. But when the travellers clock returns it shows the shortest time and therefore is proper time. Pretty much, my question is which observer measures proper time and why. Is this even an valid question? Is there an absolute proper time? My understanding of relativity is really basic, pretty much high school level.
special-relativity reference-frames coordinate-systems inertial-frames observers
special-relativity reference-frames coordinate-systems inertial-frames observers
edited Nov 12 '18 at 3:01
Qmechanic♦
105k121881202
105k121881202
asked Nov 12 '18 at 2:18
Max WILLIAMSMax WILLIAMS
4612
4612
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
The objects may age different amounts between meetings.
There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.
One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.
Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.
$endgroup$
$begingroup$
Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
$endgroup$
– WillO
Nov 12 '18 at 15:31
$begingroup$
(PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
$endgroup$
– WillO
Nov 12 '18 at 17:03
add a comment |
$begingroup$
I know that proper time is defined as the time which the clock moving
relative to that observe shows.
This isn't quite correct so let's start here.
First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.
Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:
In relativity, proper time along a timelike world line is defined as
the time as measured by a clock following that line.
In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.
The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:
An accelerated clock will measure a smaller elapsed time between two
events than that measured by a non-accelerated (inertial) clock
between the same two events. The twin paradox is an example of this
effect.
$endgroup$
add a comment |
$begingroup$
In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.
There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation
$$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$
The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.
$endgroup$
1
$begingroup$
P.S. the metric equation is in units of $c=1$.
$endgroup$
– Zack Hutchens
Nov 12 '18 at 2:37
$begingroup$
Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
$endgroup$
– Alfred Centauri
Nov 12 '18 at 11:46
add a comment |
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3 Answers
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3 Answers
3
active
oldest
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active
oldest
votes
$begingroup$
Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
The objects may age different amounts between meetings.
There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.
One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.
Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.
$endgroup$
$begingroup$
Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
$endgroup$
– WillO
Nov 12 '18 at 15:31
$begingroup$
(PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
$endgroup$
– WillO
Nov 12 '18 at 17:03
add a comment |
$begingroup$
Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
The objects may age different amounts between meetings.
There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.
One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.
Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.
$endgroup$
$begingroup$
Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
$endgroup$
– WillO
Nov 12 '18 at 15:31
$begingroup$
(PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
$endgroup$
– WillO
Nov 12 '18 at 17:03
add a comment |
$begingroup$
Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
The objects may age different amounts between meetings.
There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.
One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.
Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.
$endgroup$
Each object has its own proper time. If two objects meet each other twice, taking different journeys in between, then each one will experience its own amount of elapsed proper time between meetings, and they may be different.
object $A$'s elapsed proper time describes how much object $A$ aged between meetings, and
object $B$'s elapsed proper time describes how much object $B$ aged between meetings.
The objects may age different amounts between meetings.
There is no absolute time, in the sense of a time that is the same for all objects. However, whenever two objects meet twice, they both agree about which meeting occurred first. In other words, they both agree about the sequence from past to future. But they can only make such direct comparisons when they meet. If we try to compare the timing of things that are far away from each other, things become more complicated, because then we need to account for the light (or other signal) that must travel from one location to the other so that we can actually make comparisons.
One of the keys to thinking about this is to think in terms of meetings between objects. Each object has its own proper time, which depends on how it behaved between meetings. When two objects meet, they can directly compare their (usually different) proper times.
Here's a related post, which is fresh in my mind because I just wrote it yesterday: https://physics.stackexchange.com/a/440209/206691.
edited Nov 12 '18 at 17:04
WillO
6,88322132
6,88322132
answered Nov 12 '18 at 2:36
Dan YandDan Yand
11.1k21540
11.1k21540
$begingroup$
Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
$endgroup$
– WillO
Nov 12 '18 at 15:31
$begingroup$
(PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
$endgroup$
– WillO
Nov 12 '18 at 17:03
add a comment |
$begingroup$
Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
$endgroup$
– WillO
Nov 12 '18 at 15:31
$begingroup$
(PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
$endgroup$
– WillO
Nov 12 '18 at 17:03
$begingroup$
Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
$endgroup$
– WillO
Nov 12 '18 at 15:31
$begingroup$
Dan: You are absolutely right; my brain appears to have been temporarily broken. I'm deeleting my silly comment.
$endgroup$
– WillO
Nov 12 '18 at 15:31
$begingroup$
(PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
$endgroup$
– WillO
Nov 12 '18 at 17:03
$begingroup$
(PS: While in my brainfog, I had downvoted your answer, and was not allowed to retract my downvote until the answer was edited. I therefore edited in a brief "...", cancelled the downvote, and am now going back to undo my edit.)
$endgroup$
– WillO
Nov 12 '18 at 17:03
add a comment |
$begingroup$
I know that proper time is defined as the time which the clock moving
relative to that observe shows.
This isn't quite correct so let's start here.
First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.
Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:
In relativity, proper time along a timelike world line is defined as
the time as measured by a clock following that line.
In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.
The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:
An accelerated clock will measure a smaller elapsed time between two
events than that measured by a non-accelerated (inertial) clock
between the same two events. The twin paradox is an example of this
effect.
$endgroup$
add a comment |
$begingroup$
I know that proper time is defined as the time which the clock moving
relative to that observe shows.
This isn't quite correct so let's start here.
First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.
Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:
In relativity, proper time along a timelike world line is defined as
the time as measured by a clock following that line.
In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.
The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:
An accelerated clock will measure a smaller elapsed time between two
events than that measured by a non-accelerated (inertial) clock
between the same two events. The twin paradox is an example of this
effect.
$endgroup$
add a comment |
$begingroup$
I know that proper time is defined as the time which the clock moving
relative to that observe shows.
This isn't quite correct so let's start here.
First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.
Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:
In relativity, proper time along a timelike world line is defined as
the time as measured by a clock following that line.
In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.
The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:
An accelerated clock will measure a smaller elapsed time between two
events than that measured by a non-accelerated (inertial) clock
between the same two events. The twin paradox is an example of this
effect.
$endgroup$
I know that proper time is defined as the time which the clock moving
relative to that observe shows.
This isn't quite correct so let's start here.
First, the proper time associated with two events A and B, with timelike interval, is just the elapsed time recorded by an unaccelerated clock with world line through both events. It is the longest elapsed time given by any clock along any other world line through both events.
Second, the proper time along any world line is just the time according to an ideal clock with that world line. For example, from the Wikipedia article Proper time:
In relativity, proper time along a timelike world line is defined as
the time as measured by a clock following that line.
In the twin 'paradox', the clock of the 'stay-at-home' (unaccelerated) twin records the proper time associated with the events A (other twin leaves) and B (other twin returns) and thus, records the longest elapsed time.
The clock of the traveling twin records the elapsed proper time along that twin's world line between A and B and this is less than the other twin because the world line is not unaccelerated. Again, from the same Wikipedia article:
An accelerated clock will measure a smaller elapsed time between two
events than that measured by a non-accelerated (inertial) clock
between the same two events. The twin paradox is an example of this
effect.
edited Nov 12 '18 at 3:24
answered Nov 12 '18 at 3:08
Alfred CentauriAlfred Centauri
48.2k350150
48.2k350150
add a comment |
add a comment |
$begingroup$
In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.
There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation
$$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$
The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.
$endgroup$
1
$begingroup$
P.S. the metric equation is in units of $c=1$.
$endgroup$
– Zack Hutchens
Nov 12 '18 at 2:37
$begingroup$
Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
$endgroup$
– Alfred Centauri
Nov 12 '18 at 11:46
add a comment |
$begingroup$
In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.
There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation
$$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$
The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.
$endgroup$
1
$begingroup$
P.S. the metric equation is in units of $c=1$.
$endgroup$
– Zack Hutchens
Nov 12 '18 at 2:37
$begingroup$
Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
$endgroup$
– Alfred Centauri
Nov 12 '18 at 11:46
add a comment |
$begingroup$
In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.
There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation
$$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$
The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.
$endgroup$
In the twin paradox, both observers measure the proper time, because both observers are present at both events - which are the departure and arrival from Earth, respectively.
There is an "absolute" proper time, at least of sorts. The "spacetime interval" $Delta s$ is the longest possible proper time between two events, and it is given by the Minkowski spacetime metric equation
$$ Delta s^2 = Delta t^2 - Delta x^2 - Delta y^2 - Delta z^2. $$
The Earth and the traveler, despite being different observers, can still measure proper times as long as they are present at both events, because the two paths between those events can differ, which includes acceleration, of course. But there is only spacetime interval for the pair of events.
answered Nov 12 '18 at 2:36
Zack HutchensZack Hutchens
1,8841716
1,8841716
1
$begingroup$
P.S. the metric equation is in units of $c=1$.
$endgroup$
– Zack Hutchens
Nov 12 '18 at 2:37
$begingroup$
Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
$endgroup$
– Alfred Centauri
Nov 12 '18 at 11:46
add a comment |
1
$begingroup$
P.S. the metric equation is in units of $c=1$.
$endgroup$
– Zack Hutchens
Nov 12 '18 at 2:37
$begingroup$
Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
$endgroup$
– Alfred Centauri
Nov 12 '18 at 11:46
1
1
$begingroup$
P.S. the metric equation is in units of $c=1$.
$endgroup$
– Zack Hutchens
Nov 12 '18 at 2:37
$begingroup$
P.S. the metric equation is in units of $c=1$.
$endgroup$
– Zack Hutchens
Nov 12 '18 at 2:37
$begingroup$
Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
$endgroup$
– Alfred Centauri
Nov 12 '18 at 11:46
$begingroup$
Somewhat pedantically, the interval is $Delta s^2$ according to many in SR and not $Delta s$
$endgroup$
– Alfred Centauri
Nov 12 '18 at 11:46
add a comment |
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Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown