A First Course in String Theory/Invariant Interval/Metric

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
Cosmology2015
Messages
31
Reaction score
1
Hello,
Before starting, I would like to apologize for any errors in the use of symbols. This is my first time :sorry:.
I am studying the wonderful book of Barton Zwiebach, "A First Course in StringTheory".
In chapter 02, I am experiencing for the first time with the mathematics of special relativity (Minkowski Spacetime).
My question is on the definition of invariant interval ds[2]. By definition, the invariant interval is given by -ds[2]=η[μν]dx[μ]dx[ν]
I am not able to understand the minus sign on ds[2]. Is there any relationship with the idea of positive-definite condition? Others books use only ds[2] for the invariant interval. Is there any advantage in using this convention?
Another question would be about the invariant interval -ds[2.]. The definition of the invariant interval is very similar to the definition of Riemannian metric (metric tensor) g[ij].
(a) invariant interval → -ds[2]=η[μν]dx[μ]dx[ν]
(b) Riemannian metric → g=∑g[ij]dx⊗dx[j]
Is there any direct relationship? What is the difference between them?
I sincerely thank any reply :smile:.
 
Last edited:
Physics news on Phys.org
Hello,
I would like to apologize for the errors in the use of symbols. As I told before, it was my first time, and I am still learning how to use the resources of this forum.
I sincerely thanks any reply :smile:.
 
In your case you have not the Riemannian metric, but the Minkovski metric. In writing any metric as a sum [itex]\sum_{ij}\,g_{ij}\ dx^i\otimes dx^j[/itex] the differentials [itex]dx^i[/itex] and [itex]dx^j[/itex] are formal symbols. The interval notation [itex]ds[/itex] or [itex]ds^2[/itex] is used if you want to calculate the length of a parametric curve [itex]x^i=x^i(\theta)[/itex], [itex]\theta\in [0,1][/itex], in the Minkovski space. In this case you write [itex]dx^i=(x^i)'_\theta\,d\theta[/itex] and then integrate [tex]s=\int^1_0 ds.[/tex]