- #1

TomServo

- 281

- 9

In the Schwarzschild solution $$ds^2=-\left(1-\frac{2m}{r}\right)dt^2+\left(1-\frac{2m}{r}\right)^{-1}dr^2+r^2d\Omega^2$$ with coordinates $$(t,r,\theta,\phi)$$ the timelike Killing vector $$K^a=\delta^a_0=\partial_0=(1,0,0,0)$$ has a norm squared of $$K^2=\frac{2m}{r}-1$$ which is timelike when ##r>2m##, null when ##r=2m## and spacelike when ##r<2m##. And the norm squared of a vector ##g_{ab}K^aK^b## is invariant under coordinate transformations, correct?

Well my confusion arises when I transform to, say, the isotropic coordinates $$ds^2=-\left(\frac{1-\frac{m}{2\rho}}{1+\frac{m}{2\rho}}\right)^2dt^2+\left(1+\frac{m}{2\rho}\right)^4\left(d\rho^2+\rho^2d\Omega^2 \right)$$ with coordinates $$(t,\rho,\theta,\phi)$$ where $$r=\rho\left(1+\frac{m}{2\rho} \right)^2$$ and I look at how the timelike Killing vector from before, ##K^a## changes.

Under the transformation $$K'^0=\frac{\partial t}{\partial t}\delta^0_0=1$$ $$K'^1=\frac{\partial \rho}{\partial t}\delta^0_0=0$$ so that in the isotropic coordinates the timelike Killing vector is $$K'^a=\delta^a_0$$ as before, which makes sense to me since the isotropic metric is also independent of time. However, this time the norm squared $$K^2=-\left(\frac{1-\frac{m}{2\rho}}{1+\frac{m}{2\rho}}\right)^2$$ is timelike everywhere except at ##\rho=\frac{m}{2}## where it is null (when ##\rho=\frac{m}{2}## then ##r=2m##).

So why does the spacelike/null/timelike norm squared of this vector, a scalar, vary when it's supposed to be invariant? Why isn't there a portion of the range of ##\rho## where ##K^2>0## corresponding to the range ##r<2m##?

Is this because the Schwarzschild coordinates don't properly cover the entire manifold, and thus I should disregard anything the Schwarzschild metric tells me for ##r<2m##? Is that what they mean when they say the Schwarzschild coordinates "break down" here, that the "invariantness" of scalars breaks down? Or does this indicate something else, that ##K^a## and ##K'^a## are different vectors somehow?