Recent content by Antarres

I apologize for the long pause in replies, I was kept away from the computer recently. So to answer @PAllen and @PeterDonis, terminology 'null orthonormal frame' can surely be called an abuse of standard terminology. However, I've seen it used in papers and textbooks where this formalism is...

My question wasn't about how to extract the right component. Of course this is done using a dual basis. It was more about the interpretation. Like for example, if I contract with a null vector ##k## that's part of a null orthonormal frame, then this component would be a component along...

I'm having a somewhat silly confusion, about interpretation of components of tensors in null basis(such as Newman-Penrose or similar formalisms). For example, let's say that we're working in a 4d spacetime, and we have a null real orthonormal tetrad, consisting of null vectors ##k## and...

I was busy these days, so it took me a while to write up the re-derivation of the algebra from the paper. I've used the following steps: Let ##p^\mu##, ##q^\mu## be the spacelike vectors on the horizon. Then the decomposition of the metric reads: $$g_{\mu\nu} = l_\mu n_\nu + n_\mu l_\nu - p_\mu...

I was replying to post by @PeterDonis, who said that the vectors with respect to which I should transport surface gravity are not spacelike, but the post isn't appearing now, maybe it was deleted. And yes, it is in the paper, I'm just rederiving it to understand the proof well, that's what I...

@PeterDonis Aren't vectors tangent to the horizon(apart from the generators) spacelike? Those would be the vectors constructed from the angular part of the metric in usual models, in this case, the tetrad vectors that are orthonormal with respect to the induced metric.

Yes, I was just about to post that I saw that. I guess using the covariant derivative of the defining relation then makes it circular. I'm gonna look to correct it. @Edit:I think I'm going to solve this soon, so I won't post anything until then, but if there is an obvious hint, feel free to post.

I was looking at the proof of zeroth law of thermodynamics from the original paper by Bardeen, Carter, Hawking, which can be found here. Now, we have the Killing vector which is the generator of the horizon, we call it ##l^\mu##, and auxiliary null vector field ##n^\mu##, which we define to be...

Indeed, I was a bit sloppy when writing the last post about interpretation of those theorems, thanks for correcting me. It looked at first that this proves that event horizon must be generated by null geodesics, but not further than that. However, this corollary also implies that there's no...

Consulting Hawking&Ellis proved to be illuminating. Proposition 6.3.1 proves exactly what I stated in the last post from definition. Thank you very much for the reference, I refrained from using it thinking that there are less contrived proofs but this works just fine.

Yes, it definitely seems plausible, what @Dale said. I looked into Wald, for the theorem you mentioned, but it doesn't exactly clarify what I want to prove(at least not yet). For example, I was able to prove that horizon is definitely swept by null geodesics, however not every surface swept by...

I want to show that it is impossible to construct a timelike curve between two points on the event horizon of a black hole. This should be an obvious fact, for example, by considering any particular model of a black hole, one can go to coordinates that extend over the horizon and by observing...

Yeah, I'm aware of that, quadratic equation was just the simplest example I could think of. For example, if ##f(x) = x^2 - ax + b##, and ##f(x)=0## is a surface defined through this equation, then for particular ##a## and ##b## this function will have a double root. In that case ##\sqrt{f(x)}...

@martinbn In case of Kerr solution, the horizon is defined in the coordinates in the link by ##\Delta=0##, and this is a quadratic function in the radial coordinate. In the extremal case, it has a double root. Taking the gradent(or just exterior derivative, to make it simpler), you'll find it's...

I'm not sure about your example of a bump function, that is, I don't understand what you mean by piecing it together to put up the function ##g##. I understand that principally, you don't have to define the normal vector with respect to a function, but this function defines a surface, and it's...