matheinste
- 1,068
- 0
Hello granpa.
This is the explanation i was referring to.
Quote originally posted by robphy in "Twin Paradox"
-----------Here is the spacetime-geometric interpretation of the situation... which hopefully amplifies the key points.
The elapsed-proper-time of an observer (as read off from his wristwatch) is the arc-length of his worldline in spacetime.
Fix once and for all event A (the separation event) and event B (the reunion event),
as if one marked two points on a piece of paper.
There are many ways to experience both A and B.
That is, there are many worldlines that pass through both A and B.
In Minkowski spacetime,...
- there is one inertial observer that can experience both A and B, and that observer records the longest elapsed proper-time (as read off from his wristwatch). That is, there is one geodesic (one straight worldline) that meets both A and B, and this worldline has the longest arc-length (using the Minkowski metric)
- every other observer experiencing both A and B will not be an inertial observer, and such an observer will record a shorter elapsed proper-time (as read off from his own wristwatch). That is, every other worldline meeting both A and B will not be a geodesic and will not be a straight worldline... and thus must have at least a portion that is curved (a smooth portion representing a finite acceleration, or a kink representing an impulsive acceleration).
--Note that a kink itself will not contribute any additional arc-length to the worldline.
(That is, the impulsive acceleration at an event (say, a turnaround event) will not change (i.e., "will not cause a jump in") the reading of that observer's wristwatch experiencing that event. [Slightly off topic... what will change are the sets of distant events that this observer will regard as simultaneous. Possibly interesting... but secondary... and often a distraction from the key point.])
--[He will be a non-inertial observer and will be able to detect this because an ice cube placed over a spot on his frictionless table will be displaced at some time during the trip from A to B.]
The supposed paradox is due to ignoring the distinction between the inertial and noninertial observer... and falsely thinking that it is sufficient to consider "being at rest (in your own reference frame)". Said another way... "being at rest (in your own reference frame)" does not make you an inertial observer... especially when an ice cube on your table suddenly gets displaced.
The symmetry break (between inertial and noninertial) is the "presence of an acceleration (worldline curvature) somewhere during the trip" for the noninertial observer. Neither are causes of the shorter-elapsed-proper-time from A to B... they are correlated with the shorter-elapsed-time because they indicate that a noninertial (i.e. nongeodesic) worldline was used to experience both A and B.--------------------
Many thanks to robphy for this enlightening explanation
Matheinste.
This is the explanation i was referring to.
Quote originally posted by robphy in "Twin Paradox"
-----------Here is the spacetime-geometric interpretation of the situation... which hopefully amplifies the key points.
The elapsed-proper-time of an observer (as read off from his wristwatch) is the arc-length of his worldline in spacetime.
Fix once and for all event A (the separation event) and event B (the reunion event),
as if one marked two points on a piece of paper.
There are many ways to experience both A and B.
That is, there are many worldlines that pass through both A and B.
In Minkowski spacetime,...
- there is one inertial observer that can experience both A and B, and that observer records the longest elapsed proper-time (as read off from his wristwatch). That is, there is one geodesic (one straight worldline) that meets both A and B, and this worldline has the longest arc-length (using the Minkowski metric)
- every other observer experiencing both A and B will not be an inertial observer, and such an observer will record a shorter elapsed proper-time (as read off from his own wristwatch). That is, every other worldline meeting both A and B will not be a geodesic and will not be a straight worldline... and thus must have at least a portion that is curved (a smooth portion representing a finite acceleration, or a kink representing an impulsive acceleration).
--Note that a kink itself will not contribute any additional arc-length to the worldline.
(That is, the impulsive acceleration at an event (say, a turnaround event) will not change (i.e., "will not cause a jump in") the reading of that observer's wristwatch experiencing that event. [Slightly off topic... what will change are the sets of distant events that this observer will regard as simultaneous. Possibly interesting... but secondary... and often a distraction from the key point.])
--[He will be a non-inertial observer and will be able to detect this because an ice cube placed over a spot on his frictionless table will be displaced at some time during the trip from A to B.]
The supposed paradox is due to ignoring the distinction between the inertial and noninertial observer... and falsely thinking that it is sufficient to consider "being at rest (in your own reference frame)". Said another way... "being at rest (in your own reference frame)" does not make you an inertial observer... especially when an ice cube on your table suddenly gets displaced.
The symmetry break (between inertial and noninertial) is the "presence of an acceleration (worldline curvature) somewhere during the trip" for the noninertial observer. Neither are causes of the shorter-elapsed-proper-time from A to B... they are correlated with the shorter-elapsed-time because they indicate that a noninertial (i.e. nongeodesic) worldline was used to experience both A and B.--------------------
Many thanks to robphy for this enlightening explanation
Matheinste.