- #1
whatif
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- 5
With regard to special relativity…
Whenever, I come across the spacetime interval, written like this, say, (Δs)2 = (Δt)2 – (Δx)2 – (Δy)2 – (Δz)2 , it is as if it has to be that way. However, it seems to me it is this way by definition and does not have to be so. Sometimes, it seems to be referred to as the modified Pythagoras theorem.
What if we apply the usual Pythagoras theorem so that (Δs)2 = (Δt)2 + (Δx)2 + (Δy)2 + (Δz)2 and apply the following analysis. For simplicity, say the coordinates are arranged such that Δy = 0, and Δz = 0, so that (Δs)2 = (Δt)2 + (Δx) 2. Then, for timelike events:Consider 3 events, where E1 and E2 are timelike, and E1 and E3 are lightlike:
(Δs12)2 = (Δt12)2 + (Δx12)2
(Δs13)2 = (Δt13)2 + (Δx13)2)
(Δs12)2 - (Δs132)
= {(Δt12)2 + (Δx12)2} - {(Δt13)2 + (Δx13)2}
= {(Δt12)2 + (Δx12)2} - {(Δx13)2 + (Δx13)2}
= {(Δt12)2 + (Δx12)2} - {2(Δx13)2}
= {(Δt12)2 + (Δx12)2} - {2(Δx12)2}
= (Δt12)2 + (Δx12)2 - 2(Δx12)2
= (Δt12)2 - (Δx12)2
= the same invariant as when the “modified Pythagoras theorem” is used
In words, for timelike events E1 and E2 ...
... the square of the invariant, is equal to the square of the spacetime interval between events E1 and E2, minus the square of the spacetime interval for light to get from E1 to the spatial location of E2.
A similar case can be made for spacelike and lightlike events.
Lightlike events in different inertial frames would have different spacetime intervals between the same 2 events, but would not have the oddity of no spacetime interval between separate events.
?
Whenever, I come across the spacetime interval, written like this, say, (Δs)2 = (Δt)2 – (Δx)2 – (Δy)2 – (Δz)2 , it is as if it has to be that way. However, it seems to me it is this way by definition and does not have to be so. Sometimes, it seems to be referred to as the modified Pythagoras theorem.
What if we apply the usual Pythagoras theorem so that (Δs)2 = (Δt)2 + (Δx)2 + (Δy)2 + (Δz)2 and apply the following analysis. For simplicity, say the coordinates are arranged such that Δy = 0, and Δz = 0, so that (Δs)2 = (Δt)2 + (Δx) 2. Then, for timelike events:Consider 3 events, where E1 and E2 are timelike, and E1 and E3 are lightlike:
(Δs12)2 = (Δt12)2 + (Δx12)2
(Δs13)2 = (Δt13)2 + (Δx13)2)
(Δs12)2 - (Δs132)
= {(Δt12)2 + (Δx12)2} - {(Δt13)2 + (Δx13)2}
= {(Δt12)2 + (Δx12)2} - {(Δx13)2 + (Δx13)2}
= {(Δt12)2 + (Δx12)2} - {2(Δx13)2}
= {(Δt12)2 + (Δx12)2} - {2(Δx12)2}
= (Δt12)2 + (Δx12)2 - 2(Δx12)2
= (Δt12)2 - (Δx12)2
= the same invariant as when the “modified Pythagoras theorem” is used
In words, for timelike events E1 and E2 ...
... the square of the invariant, is equal to the square of the spacetime interval between events E1 and E2, minus the square of the spacetime interval for light to get from E1 to the spatial location of E2.
A similar case can be made for spacelike and lightlike events.
Lightlike events in different inertial frames would have different spacetime intervals between the same 2 events, but would not have the oddity of no spacetime interval between separate events.
?