D'Inverno Problem 19.10 Stationary Limit

[TerryW]

Given that the nature of a surface is given by the sign of g⁽ⁱ⁾⁽ⁱ⁾ and for the Boyer-Lindquist form of the Kerr solution g¹¹ = -Δ/ρ², then surely any surface of constant r where r > r₊ will have a negative value for g¹¹, so how can S₊ be timelike?

Also on page 259, just below equation (19.70), he says "These curves are not geodesics, but are the world-lines of photons initially constrained to orbit with fixed r and θ. " How do you constrain photons?

 

[Ambitwistor]

Thank you for sharing your thoughts and questions on the nature of surfaces and geodesics in the Kerr solution. I would like to offer some clarifications and insights on these topics.

Firstly, it is important to note that the sign of g11 in the Boyer-Lindquist form of the Kerr solution is not the only factor that determines the nature of a surface. The sign of g11 is just one component of the metric tensor, which describes the overall geometry of spacetime. Other components, such as g00, also play a role in determining the nature of a surface. Therefore, a surface with a negative g11 value does not necessarily mean it is timelike.

Furthermore, the nature of a surface is not solely determined by the metric tensor, but also by the motion of particles or objects on that surface. In the case of the Kerr solution, a surface of constant r where r > r+ can still have timelike worldlines for particles or objects that are moving in a certain direction or following a certain trajectory. This is because the overall curvature and geometry of spacetime in the Kerr solution is affected by the presence of a rotating black hole, which can cause particles to move in non-intuitive ways.

As for the statement on page 259 about the world-lines of photons being constrained to orbit with fixed r and θ, it is important to understand that photons do not have a rest mass and therefore cannot be constrained in the same way as massive particles. However, in the context of the Kerr solution, photons can still be influenced by the curvature of spacetime and follow certain trajectories around the black hole. This is what is meant by "constraining" photons in this context.

In summary, the nature of surfaces and the motion of particles in the Kerr solution is complex and cannot be fully explained by a single component of the metric tensor. It requires a deeper understanding of the overall geometry and curvature of spacetime in the presence of a rotating black hole. I hope this helps to clarify your questions and further your understanding of the Kerr solution.