Spacetime distance and ruler-measured length on an XT chart

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  • Thread starter KDP
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  • #1
KDP
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spacetime.png

With reference to the above diagram, we can see that if we used a ruler to measure the segments on the XT chart and added up all the segment lengths, that this ruler measured length is longer for the craft on the right which experiences the least elapsed proper time. Now unlike the regular spacetime distance (as correctly depicted in the picture) this XT chart ruler length is not invariant and different inertial observers will disagree on its length. However, from a quick check using a few examples in different reference frames, it seems that if one ruler measured chart path is longer than another in a given reference frame, then all observers will agree on which object has the longer XT chart ruler measured distance (Is there a formal name for this measurement?). Is it possible to show that this always the case or are there counter examples?

Just to be clear, I am talking about using the normal Pythagoras method of using $$ \sqrt{T^2 + X^2} $$ to calculate the segment lengths instead of the normal method of using$$\sqrt{T^2-X^2}$$ to calculate the spacetime distance segments.
 
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  • #2
KDP said:
However, from a quick check using a few examples in different reference frames, it seems that if one ruler measured chart path is longer than another in a given reference frame, then all observers will agree on which object has the longer XT chart ruler measured distance
Consider two paths starting at the origin and ending at ##t=t_1##, ##x=\pm x_1##. The interval along each is ##t_1^2-x_1^2##, but the Euclidean length of either line drawn on the Minkowski diagram can be made longer while the other one is made shorter by a boost opposite directions. So this is a counter example to the claim that all frames agree which worldline has a longer Euclidean representation.
 
  • #3
Lets call the two paths, A and B and take the case where the first observer sees the two paths as equal and the second observer sees path A as shorter than path B. This is the outward trip. Now if they turn around so that they both return to x=0 at time t1*2, then the return path of A will be longer than that of B and by symmetry the round trip Euclidean lengths of the trips will be equal for the second observer (or any other observer) if they both return to the origin at the same time.
 
  • #4
The implication of that is tthat you have a restriction not present in your OP, that you are only considering pairs of paths that travel from one common event to another. Is that what you intended?
 
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  • #5
Ibix said:
The implication of that is that you have a restriction not present in your OP, that you are only considering pairs of paths that travel from one common event to another. Is that what you intended?
You are right. That is what I intended. Sorry for not making that clear.
 
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  • #6
KDP said:
However, from a quick check using a few examples in different reference frames, it seems that if one ruler measured chart path is longer than another in a given reference frame, then all observers will agree on which object has the longer XT chart ruler measured distance (Is there a formal name for this measurement?). Is it possible to show that this always the case or are there counter examples?

Counter-example (see screenshot):
PF-Euclid.png

Alice moves ##E_0## -> ##E_1##-> ##E_3##.
Bob moves ##E_0## -> ##E_2##-> ##E_3##.
 
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  • #7
Hi Sagittarius. You are right. I checked your figures. There are counter examples. Thanks for clearing that up.
 
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