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...
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...
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.