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I thought so, thank you! The trouble we were having on that other forum was that the OP rejected the validity of the Penrose diagram for Vaidya spacetime (saying it doesn’t accurately reflect the physics of this scenario), which is why I was attempting to explicitly work through the maths. But...

Thanks for replying - and I am sorry, I should have given more background. Essentially the question arose in a discussion we were having on another forum. The thread became very convoluted, but in the end boiled down to two questions - given an observer stationary very far away, will he ever be...

Consider an observer starting a purely radial free fall from rest at infinity in outgoing Vaidya spacetime - this being a simple model for a radiating black hole. Does anyone have an explicit expression for the coordinate in-fall time (assuming purely radial motion) from infinity to event...

Very nicely, clearly, and succinctly presented - thank you for this :)

How does the Lanczos tensor fit into all of this ? That's something I have always wondered about.

Perhaps it would help if you turn things around a bit, and ask yourself - what are the fundamental conserved quantities in a region of Minkowski spacetime ? The answer is given by Noether's theorem - to every differentiable symmetry generated by a local action, there corresponds a conserved...

Empirically speaking - because every experiment we perform in a small enough local region matches the predictions of Special Relativity. Since that model is based on flat Minkowski spacetime, we can deduce that spacetime is locally flat. There are of course also more theoretical reasons, but I...

The Einstein tensor is a linear machine with two slots to pass an "input" to. If you choose a time-like unit vector ##t^{\mu}## and insert it into each of these slots, you find that \displaystyle{G_{\mu \nu}t^{\mu}t^{\nu}=-\frac{1}{2}R_{(D-1)}} So, if you choose a "time direction" on your...

Because the laws of physics - including your wave equation - are the same in all inertial frames of reference. Hence, the line element must be preserved when you go from one frame into another; this is possible only if you treat time and space equally.

Think of it this way - the Schrödinger equation is of first order with respect to time, but second order with respect to spatial derivatives. This is a problem once you go into the realm of relativistic physics, quite simply because in relativity time and space are on equal footing, so the wave...

Yes, it can be taken to be the same everywhere, in the sense that it manifests in the exact same manner no matter where you are.

The world lines of photons are null geodesics, so you cannot parametrise them with proper time. The separation between any two neighbouring events on a null geodesic is zero, hence the name. As such, the concept of 4-velocity is not defined for photons. No, because for photons you have ##ds^2=0##.

The Higgs field is a quantum field, not a mechanical entity, so concepts such as "thickness" do not make any sense in this context. All I can really say is that the vector bosons of the weak interaction have the same rest mass everywhere in spacetime, so in that sense the effect of the Higgs...

When you go to work/school tomorrow morning, how will you move relative to time ? What does that even mean ? Do you mean how much time will it take you to get to where you are going ? Or something else ? You need to make your question mathematically and physically precise, then we'll try to...

Would it make sense to argue that it is due to the principle of least action ? Without that principle, I do not see any mechanism by which free fall world lines would be guaranteed to be spacetime geodesics.

What a metric does, in physical terms and in the context of General Relativity, is tell you how neighbouring events are related to one another. Since events are points in space at given instances in time, this means that a metric enables you to calculate ( among other things ) everything there...

Ah - I misunderstood the OP. Sorry.

I would say the curvature invariants formed from the Riemann tensor do the trick here, for example the Kretschmann scalar. If at least one of those invariants diverges, you are dealing with a curvature singularity, otherwise it is just an issue with your coordinate system. I am not sure though...

Ricci can be taken as the trace of the Riemann tensor, hence it is of lower rank, and has fewer components. If you have a small geodesic ball in free fall, then ( ignoring shear and vorticity ) the Ricci tensor tells you the rate at which the volume of that ball begins to change, whereas the...

You see, you are thinking of position and momentum as two completely separated, isolated entities. But the thing is, they are not - you can describe any given system in terms of position, or in terms of momentum, and you will find that these two descriptions are related via an operation called a...

Allow me to quote one sentence in particular from the link in post #23 which captures perfectly how I feel at the moment ( bottom page 1 ) : "Often, one is left with the impression that there is some blind faith required on the side of the physicists or at least that some black magic is helping...

Absolutely not, and to be honest I am quite surprised that this is how you interpreted my post. QFT is manifestly a very successful theory, there can be no doubt or argument over that. It is just that there are elements to it that feel ad-hoc and "messy" to me; it's a matter of intuition, I...

This makes me wonder just how fundamental and valid QFT as a framework really is, in general terms. I have only just begun studying QFT in recent weeks, so I'm still largely ignorant of the finer points and issues, but even in the most simplistic of textbook cases ( e.g. quantisation of...

Actually, the orbit of a free fall test particle is a geodesic of spacetime, and if you parametrise these geodesics by proper time, then their defining characteristic is that they are world lines which extremise total proper time between given events ( = the principle of extremal ageing ). As...

That's exactly what I was looking for, thank you :) Very interesting, I wasn't aware of this. So basically you'd need a rocket with thrusters, instead of just a free fall particle. Do you happen to know precisely where and how that acceleration would have to take place ? I'm not trying to get...

Thank you @tionis, I understand what you are saying, and agree on the instability issue. However, I don't think I can agree with your statement that the inner ergo surface is the same as the ring singularity - mathematically these seem to be quite distinct, unless I am missing something...

I have two questions regarding Kerr black holes, which I am hoping some of you might be able to shed some light on for me. 1. What is the physical significance / meaning of the inner ergosurface, the one beyond the inner horizon ? If considered as a boundary surface, what would it separate from...

Time dilation is a real, physical effect with real, physical consequences - it is not just some trick of perception. With that being said, it is best to regard time dilation as a relationship between observers, rather than something that "happens" to a clock; as such, the amount of time dilation...

Why ?

@Simon Phoenix : I like your post, very well explained :)

I think ( though not 100% sure ) that the semicolon is used in these examples to separate variables from parameters; for example, in G(r1,r2;z), the z is a complex parameter which you choose to be of a specific value ( and which doesn't vary within the function ), whereas r1 and r2 are actual...

In what context is this ? Can you link to the original source ?

Maybe I'm wrong, but my first impulse would be to model this as a special case of a Tipler cylinder, with small radius and zero angular momentum, in which case your ansatz looks pretty good. Let's wait and see what the experts say :smile:

Yup indeed :) Given the correct expansion of this bracket, I actually got the rest of the exercise done with no problems, it was really just this one upper index that tripped me up. But all is good now :smile:

My goodness, of course ! I had come across that already in my studies of GR, but it had completely slipped my mind in this context. I knew I was missing something elementary. Thank you for taking the time to reply in such detail, I understand it now :woot:

How can a tensor being the same for all observers be an incorrect statement ? The fact that tensors are invariant under changes of coordinate basis is the entire point of using them !

I have begun teaching myself Lagrangian field theory in preparation for taking the plunge into quantum field theory ( it's just a hobby, not any kind of formal course ). When working through exercises, I have run across the following issue which I don't quite understand. I am being given a...

Should that not read "massless" instead of "massive" ?

Is the diverging coordinate in-fall time for an external observer also true in R-N spacetime, which is different in many respects from the Schwarzschild case ? Also, if you let a charged particle fall towards the black hole, then that charge should be affected by its very own electromagnetic...

Good point @PeterDonis, I overlooked that as well o_O That's what happens when an amateur's enthusiasm for the subject outpaces his limited knowledge !

I suspected that, thank you ! So the (3+1) case really does seem to have a privileged character...

I think detecting the gravitational effect of a single particle ( massless or not ) is monumentally difficult, since the effect is so small. It also makes me wonder just what - given quantum effects such as position/momentum uncertainty, the superposition principle etc - that effect would really...

You are right, I overlooked that :oops: However, it seems that the theorem can be extended to cover pseudo-Riemannian manifolds as well, so long as they are orientable : https://duetosymmetry.com/files/An.%20Acad.%20Brasil.%20Ci.%201963%20Chern.pdf

I am swimming in deep water here, but isn't it possible to define some notion of "gravity conservation" by considering the topological invariants of the spacetime in question ? In particular I am thinking of the ( generalised ) Gauss-Bonnet theorem here, which connects a measure of total...

Lol, just as I thought I had the whole EPR-entanglement thing hammered down, someone comes along and throws a spanner in my mental works o0) I am going to have to digest your last three posts, and definitely do a bit more reading on this subject, as I wasn't aware of the phenomena you have...

The question refers to this paper : http://arxiv.org/abs/1606.00672 I am having difficulty understanding the idea behind this. How can we possibly be able to unambiguously detect any left-over traces of entanglement here ? If I understand this right, the thought process is that an entanglement...

I don't think there is any physical relevance to either Alice or Bob in isolation. Their reality is the same, regardless of what Charlie - who has information about both particles - concludes ( entanglement or no entanglement ). That is why I think there should be some notion of...

Well, in that case it comes down again to the issue of interpretations. To me, I do not see how you can meaningfully speak of an entanglement relationship, if all you have is two eternally separated observers. The outcome of all measurements is random for Alice, just as it is for Bob; neither of...

From a purely mathematical point of view, having an additional time dimension would fundamentally change the form of the general wave equation, turning it into what I believe is called an "ultra-hyperbolic equation" ( correct me on this if I'm wrong ). The question then becomes how the solutions...

If the two particles are eternally separated ( i.e. in the past and in the future ), then I don't see how you can have entanglement - not only will it be impossible to verify the correlation, but the entanglement cannot exist in the first place, since its creation requires an initial interaction...