# Invariance of timelike Killing vector of Schwarzschild sol.

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• TomServo
In summary, under isotropic coordinates, the norm squared of a vector is timelike everywhere except at the point where it is null.
TomServo
I use the ##(-,+,+,+)## signature.

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?

TomServo said:
he norm squared of a vector ##g_{ab}K^aK^b## is invariant under coordinate transformations, correct?

More precisely, the norm squared of a vector at a given spacetime point (i.e., at a given value for all four coordinates) is invariant under coordinate transformation. Yes, this is true.

TomServo said:
why does the spacelike/null/timelike norm squared of this vector, a scalar, vary when it's supposed to be invariant?

You have confused yourself about the meaning of the word "invariant". Try re-thinking your question in the light of my more precise phrasing above.

TomServo said:
Is this because the Schwarzschild coordinates don't properly cover the entire manifold

Not in this case (although it is true that Schwarzschild coordinates are singular at ##r = 2m##--see further comments below); it's because isotropic coordinates cover the manifold a different way. Try computing the area of a 2-sphere at isotropic radial coordinate ##\rho## for values of ##\rho## from zero to infinity and comparing those areas with the area of a 2-sphere at Schwarzschild radial coordinate ##r## for values of ##r## from zero to infinity.

TomServo said:
I should disregard anything the Schwarzschild metric tells me for ##r<2m##

No, you shouldn't. See above. The coordinate singularity at ##r = 2m## actually doesn't affect your ability to calculate the area of a 2-sphere at ##r = 2m##, because the metric coefficient ##g_{rr}##, which is the only one that becomes singular at ##r = 2m##, does not come into play in that calculation.

TomServo said:
Is that what they mean when they say the Schwarzschild coordinates "break down" here, that the "invariantness" of scalars breaks down?

No.

TomServo said:
Or does this indicate something else, that KaK^a and K′aK'^a are different vectors somehow?

No.

TomServo
I wasn't exact enough with my question about invariance. I know that invariance of a scalar field refers to the invariance of the value of that scalar at that point in spacetime.

What I see by plotting r vs. ##\rho## is that for ##0<\rho<\infty## we never go below ##2m##. It dips to the point where ##r=2m## as ##\rho\rightarrow \frac{m}{2}## but for ##\rho<\frac{m}{2}## it goes to infinity. IOW it seems like the portion of the manifold covered by ##0<r<2m## isn't present in the isotropic coordinates, so points on the inside of the event horizon are not covered by ##\rho<\frac{m}{2}## in isotropic coordinates like I thought they were, but rather those points (##\rho<\frac{m}{2}##) correspond to...something else?

TomServo said:
t seems like the portion of the manifold covered by ##0<r<2m##

That's right.

TomServo said:
those points (##\rho < \frac{m}{2}##) correspond to...something else?

Yes. They correspond to the region ##2m < r < \infty## in a parallel universe that can't be reached from the "normal" one. More precisely, that's what those points are in the maximal analytic extension of the mathematical solution. Physically, a real object would either be something like a planet or star, which stops being vacuum at some radius well above ##\rho = \frac{m}{2}##, or it would be a black hole that formed by gravitational collapse of something like a star, which would still have a non-vacuum region in isotropic coordinates starting somewhere above ##\rho = \frac{m}{2}##, for reasons that are more complicated. So in any real case, the vacuum region covered by isotropic coordinates would never reach ##\rho = \frac{m}{2}## so the issue would never arise.

TomServo

## 1. What is the Schwarzschild solution?

The Schwarzschild solution is a mathematical solution to Einstein's field equations in general relativity. It describes the spacetime curvature around a spherically symmetric, non-rotating mass.

## 2. What is a timelike Killing vector?

A timelike Killing vector is a vector field that describes a symmetry in a spacetime, meaning that it remains unchanged under certain transformations. In the case of the Schwarzschild solution, the timelike Killing vector represents the symmetry of time translation.

## 3. Why is the invariance of the timelike Killing vector important in the Schwarzschild solution?

The invariance of the timelike Killing vector is important because it is a conserved quantity in the Schwarzschild solution. This means that it remains constant along the worldline of a particle, which is crucial for understanding the motion of objects in the gravitational field of a massive object.

## 4. How does the invariance of the timelike Killing vector affect the behavior of light in the Schwarzschild solution?

The invariance of the timelike Killing vector plays a role in determining the behavior of light in the Schwarzschild solution. It affects the paths of light rays, causing them to bend towards the massive object due to the curvature of spacetime. This is known as gravitational lensing.

## 5. Can the invariance of the timelike Killing vector be generalized to other solutions in general relativity?

Yes, the concept of a timelike Killing vector and its invariance can be generalized to other solutions in general relativity. It is a fundamental concept in understanding the symmetries and conserved quantities in different spacetimes and is applicable to a wide range of solutions beyond the Schwarzschild solution.

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