Graduate Asymmetry parameter while relating proper time with distance

[Damodar Rajbhandari]

In special relativity, we know, (proper time)^{2} = - (proper distance)^{2}. But, in Causal Dynamical Triangulations (CDT), they introduce an asymmetry parameter \alpha as, (proper time)^{2} = - \alpha (proper distance)^{2}

[Q. 1] Can you please explain me about, why we need to introduce \alpha ? And, Is there is any useful resources to learn more about the role of \alpha in Quantum Gravity? Or, Any derivation relating to asymmetry parameter with proper time and proper distance?

[Q. 2] In most of the research in CDT, why they prefer to choose \alpha to be 1? Concrete reason needed!

[Q. 3] Does CDT prefer Time-reversal symmetry?

With thanks,
Damodar

P.S: This question was primarily asked in https://www.researchgate.net/post/Question_relating_to_Quantum_asymmetry_between_proper_distance_and_proper_time.

 

[haushofer]

In special relativity, we know, (proper time)^{2} = - (proper distance)^{2}.

I hope I'm not disappointing you, but if you mean by "proper distance" the spacetime interval, then this is not true in general. It is only true if the proper time is that of an inertial observer. The Euclidean analog is that the length of a path between two points equals the distance if the path is straight.

 

[Demystifier]

I hope I'm not disappointing you, but if you mean by "proper distance" the spacetime interval, then this is not true in general. It is only true if the proper time is that of an inertial observer. The Euclidean analog is that the length of a path between two points equals the distance if the path is straight.

I hope I'm not disappointing you, but proper time is equal to (minus) proper distance along any time-like curve $C$, i.e.
$$\int_C d\tau =-\int_C ds $$
not only along trajectory of an inertial observer. Indeed, many people tried to explain that to you in another thread, but you still seem not to get it.

Anyway, it doesn't help to answer the OP's question.

 

[haushofer]

Yes, I wrote that post right before I started my topic on grav.time dilation. Infinitesimally, as you write it down, you're right of course.