Saturday, 5 December 2015

general relativity - Calculating the determinant of a metric tensor


Suppose the line element is $$ds^2 = -A(t,r)^2dt^2+B^2(t,r)dr^2+C^2(t,r)d\theta^2+C^2\sin^2\theta d\phi^2.$$

Since the metric is diagonal, to find the determinant I can multiply the diagonal entries, $$\det g_{ab} = g = -A^2B^2C^4 \sin^2\theta.$$ I have a few questions about this.



  1. First off, why do we call the metric determinant $g$?

  2. Why isn't it true that $g = g_{ab} g^{ab} = 4$? Isn't that how $g$ is defined?

  3. When will it be true that $g = 1$?




No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...