Proper Time Interval: Explaining Δs Relationship

Click For Summary
Δs is directly related to the proper time interval (ΔTau) through the equation Δs = c*ΔTau. In Minkowski space, a positive Δs² indicates a real proper time interval, corresponding to a clock moving inertially between two events. When Δs² equals zero, it signifies a light-like interval where ΔTau is also zero. A negative Δs² indicates an imaginary proper time interval, meaning no real particle can travel between those events, categorizing the interval as spacelike. This relationship highlights the connection between proper time and the geometry of spacetime.
xma123
Messages
9
Reaction score
0
Could anyone explain how Δs is related to the proper time interval?
 
Physics news on Phys.org
There's no real difference between proper time and ds except for (maybe) a factor of \pm c to make sure that a real trajectory has a positive proper time and that the units are right. Either way, proper time is the analogue of arc length in Euclidean spaces, and for a curved trajectory, one integrates to get the right result (the same way you would in 3D).
 
xma123 said:
Could anyone explain how Δs is related to the proper time interval?

Δs is really c*ΔTau where ΔTau is the proper time interval. In Minkowski space and using the (+---) sign convention, when Δs2 is positive, then the proper time interval is real and represent the proper time of a clock that moves inertially between the two events. If Δs2 is zero then it represents a light like interval. (i.e. ΔTau is zero). If Δs2 is negative, the proper time interval is imaginary and in that case, no real particle or physical clock can physically travel between those two events and the interval is said to be spacelike and after reversing the signature to (-+++) represents the proper distance (ruler) measurement between the two events.
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 37 ·
2
Replies
37
Views
4K
  • · Replies 35 ·
2
Replies
35
Views
6K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
795
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 51 ·
2
Replies
51
Views
4K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 31 ·
2
Replies
31
Views
2K
  • · Replies 53 ·
2
Replies
53
Views
5K