1

I'm midway through a problem at the beginning of a GR course, my question is simply this:

If
$$ x=x'\cos\Omega t-y'\sin\Omega t $$ where $x'$ and $y'$ indicate the rotated frame of reference. What does that make $dx^2$?

I need this so I can make substitutions into the equation: $$ ds^2=c^2dt^2-dx^2-dy^2-dz^2 $$

Kyle Kanos
  • 28,229
  • 41
  • 68
  • 131
  • Closely related/possible duplicates: https://physics.stackexchange.com/q/53769/226902 ("Metric coefficients in rotating coordinates"), see also https://physics.stackexchange.com/q/427492/226902 and https://physics.stackexchange.com/q/651120/226902. Also related: https://physics.stackexchange.com/q/613281/226902 – Quillo Jan 30 '23 at 12:41

1 Answers1

1

$$ dx = \cos (\Omega t) dx' -x' \Omega \sin (\Omega t) dt - \sin (\Omega t) dy' -y' \Omega \cos (\Omega t) dt$$

It's basically just the product rule and the chain rule.

DanielSank
  • 24,439
Jold
  • 3,023
  • 1
  • 16
  • 22