md65536

It's the constant speed of light that allows times to be expressed in terms of a distance that light travels, and distances to be expressed (and literally defined) by how far light travels in a given time. Consider a beam of light from one event to another event. Those two events will be generally different distances apart for different moving observers, meaning different frames will have the same beam of light travel different distances, meaning that the time between the events must be different in different frames. Draw a beam of light as a line. If you draw it in the ct dimension, that represents the time between two events at the same spatial location, like an abstract beam whose distance is measured only by time, or you could make it a real beam in the y direction. Now consider the same "beam" in a frame that is relatively moving in the x direction. It is orthogonal to ct, so if you draw this out, you end up with a right triangle. The hypotenuse can represent the same time as measured in another frame where the the start and end events are at different locations, ie. the distance between them is longer ie. the time is dilated. The relationship between the lengths of the edges of the triangle is given by the Pythagorean theorem. The meaning of it comes from the maths. Maybe you could say it's a geometrical representation of the invariance of the speed of light. For any timelike spacetime interval, there's always a rest frame where the spatial distance between the two events is zero, so you can always represent the interval with simple right triangles. I suspect if you want more meaning than that, it would be found in a more mathematical description.

To add to this, these are choices that are made in defining the interval, to make it useful and simple, not to make it somehow supremely meaningful. It's not even a distance, but a square (so that you can deal with negative squares, instead of imaginary times or distances). However since it's made up of distances and times, you can use maths to convert it into something with the meaning you want. The proper time along an arbitrary world line should be the same as an integral of infinitesimal times measured in momentary rest frames at each event on the world line, so you could integrate the square root of infinitesimal spacetime intervals.

Yes, a timelike spacetime interval is the square of the proper time measured by an inertial clock moving between the 2 events in flat spacetime. If the sword remained at rest the whole time, and gravity was neglected, the interval would be the square of how much the sword aged between the two events.

I thought there might be a paradox but I can't create one after all. Suppose the universe "wraps around" 1 light year in distance, and assume it behaves the same as it it was flat. Then you could see what appears to be an infinite row of Earths, each subsequent one looking one year older than the last. One that looks n years older is "old light" from Earth that has made n loops around the universe before reaching you. Lets say you can travel near enough the speed of light that it takes about a year Earth time to loop around the universe. If you leave Earth at the start of 2024, you'll return to Earth at the start of 2025 Earth time, even though the journey is almost instantaneous according to proper time of the traveler. Before you start, the clock on Earth as seen 1 LY away shows 2023, the one beyond it shows 2022. Thinking only of how things appear, there's no need to worry about relativity of simultaneity. As you travel one loop, you see 2 years pass on the "destination Earth" clock, so you see it showing 2023 when you start, and 2025 when you arrive. The next clock beyond it shows 2022 when you start, and also must have 2 years appear to pass during your journey, so it shows 2024 when you arrive. If you keep going, it shows 2026 when you get to it. There's no paradox there. From the perspective of Earth, the ship and the image of Earth in 2024 travel around the donut in opposite directions and meet at the far end after half a year, and the ship returns at the start of 2025. Now if you add another loop that's 2 light years long, it's the same thing, just double everything. You could have one ship travel the first loop twice, and meet a ship that travels the longer loop once, after 2 years Earth time. Negligible time would pass for both travelers. Or, you could have one traveler do the long loop in 2 years Earth time, and the other do the shorter loop in 2 years, ie. at a speed of c/2. One would see 4 years pass on their "destination Earth" and the other would see 3, where they would meet. Ie. they both start in 2024, and one sees Earth around the long loop looking like 2022, and arrive in 2026; the other sees Earth looking like 2023, and arrives in 2026. One would have aged a negligible time and the other would age 2x .866 years (according to Lorentz factor). In terms of distances everything should be similar. Traveling at near c, the lengths would be negligible. At half c, traveling a proper light year would be measured as .866 light years traveled distance. I can't see any paradoxes here. I think it would be equivalent to if you had a flat universe with a set of copies of Earth spaced a light year apart, all at relative rest and with synchronized clocks. Also add copies of the traveler so they could see "their distant selves". Creating that with copies wouldn't introduce any paradoxes.

Two objects falling directly towards a BH can diverge (as with spaghettification). Two objects falling indirectly and parallel can diverge, eg. if only one of them has escape velocity due to different distance from the BH. I think the analogy needs more details. On the other hand, if you have two side-by-side geodesics both directed toward a single point (like a CoM or barycenter), they shouldn't be parallel at any finite distance, in general?

Or use wheels that are the same size and curve the road intrinsically... like with a trampoline. How is your analogy "correct"? What do the wheels represent and are you saying that spacetime (the road) is not really curved??? This also shows that gravity is not needed to show curvature in the trampoline analogy. Pin a rubber sheet flat against a wall in zero-g. Stick a large ball representing a gravitational mass under the sheet, stretching it (or even a long pipe sticking out from the wall, to imagine it more extremely). Roll an axle with 2 wheels of the same size along it, and the path will curve, analogous to null geodesics.

Then the radiation field of an accelerating charged particle drops off as 1/r because it propagates perpendicular to the acceleration of the charge, the field lines distributed over a circle for a given r rather than a sphere? An oscillating charge radiates EMR with a frequency equal to that of the oscillation. Apparently, Maxwell's equations imply that even a charge with a constant acceleration must also radiate. However, the frequency and energy would be zero, or at least approach zero as time approaches +/- infinity. So one could say that a charge at rest on the surface of Earth does not radiate energy, or that it radiates light with infinite wavelength, which is not physically detectable nor has an absolute meaning, but is consistent with all physical laws. I hope this is right instead of me just getting more confused.

I don't know! Is it all/only electromagnetic radiation, ie. photons? I see references to "radiation field", but it's described separately from the electromagnetic field? I assume that if electromagnetic radiation is detected, that means a photon is emitted at one event and absorbed at another event. That seems at odds with the quote from the paper, "the detection of radiation has no absolute meaning", so I'd already concluded my assumption of EMR was wrong. Then I (mis?)interpreted the quote as a description of what the radiation is, as something that remains when things like EMR are separated out. Doesn't EMR drop off as 1/R2? It's all very confusing...

I read https://arxiv.org/abs/physics/0506049 [1] from swansont's earlier link, and it clears up some of my misconceptions. The basic conclusion, in the case of a comoving observer and a uniformly accelerated charge, is that there is only a certain region of spacetime that "would allow us to detect unambiguously the radiation emitted by the charge," and that region is outside the light cone of any event on the particle's world line, meaning that no radiation is detectable at all. Within the light cones, "the detection of radiation has no absolute meaning because the detection depends both on the radiation field and the state of motion of the observer." So I guess if you wanted to argue that any radiation could be detectable, you'd have to be really creative with definitions in order support that conclusion. Also it is not electromagnetic radiation as I'd assumed. "The radiation content can be extracted by separating the components that drop off as 1/R from the usual Coulomb 1/R2 fields." 1. The radiation of a uniformly accelerated charge is beyond the horizon: a simple derivation Camila de Almeida, Alberto Saa

Oh, right. I see that was already resolved earlier. The frame where no light (of any wavelength) is radiated is an accelerating frame.

What does this mean? Wouldn't it radiate as light? If so, what frame wouldn't it radiate in? Isn't the problem that if there was radiation, the expected energy would be undetectably low?

Yes, what I described, at the limit where the BH can be removed leaving just the photon sphere, fits the definition of a geon. I can't imagine that such a spherically symmetric shape wouldn't work, and that a geon needs a different shape.* It would be unstable because if any photon deviated slightly outward, its orbit would be wider and it would escape, leaving less energy, reducing the photon sphere radius and letting other photons escape. I imagine a photon deviating inward, without a central BH, would cross the photon sphere again and escape, but I'm not sure (maybe it could collapse the geon?). * Actually, I see Wheeler's 1955 paper "Geons" describes this and interesting complications I hadn't thought of. PDF: https://blackholes.tecnico.ulisboa.pt/gritting/pdf/gravity_and_general_relativity/Wheeler_Geons.pdf To make it stable, it would have to be a quantum geon. Those are theoretical only and seem to require quantum gravity. I guess the basic idea is that if energy can only leak in specific amounts, one could be coherent enough to prevent that. There are papers on them but I haven't yet found anything I can make sense of.

I'm interested in any situation or metric, or any simplification (or complication) involving a system of trapped light and a minimum of anything else. (Now that I say that, I have a vague memory of well known physicists speculating on astronomical objects made of light itself, gravitationally bound to itself but not collapsed, but I can't remember what they're called and I think that might be harder to reason about.) It does seem like if you think of the system of a black hole and a photon at 1.5 rs with at least as much energy as the black hole, and consider it inside a sphere of size 2 rs, it should collapse, but that assumes all the energy is contained within that radius, but it is not spherically symmetric, and it should have angular momentum (unless you contrived it not to by giving the BH itself the right angular momentum, but that just further complicates things), and like you say Markus, the Schwarzschild metric can’t be used. It also seems like all these complications are just more "stress" than a Schwarzschild case, and more certain to collapse. But are there ways to remove stress from the system so you could increase energy without collapse. eg. a cosmological constant. What might happen if the original photon was moved farther away to avoid collapse, such as at the photon sphere of the new black hole you describe? Or to make it symmetric, many uniformly distributed photons in a photon sphere. It seems like in general, for a real black hole photon sphere of a given size, if you add more energy to the photon sphere, you could get away with a smaller black hole, to the point that you don't need the black hole at all (which sounds reasonable now that I remember the idea of objects made of gravitationally bound light).

I shouldn't have used the word "stable", what I meant was just "circular orbit" for some time (several orbits or so) because apparently circular already implies it's on the photon sphere. A bit of a digression on this: I see in https://en.wikipedia.org/wiki/Photon_sphere , "all circular orbits have the same radius". At the event horizon, light aimed directly outward will have a constant r, and at the photon sphere, light aimed tangentially will have a constant r. Is that correct? Then, everywhere in between, there is some direction that will let light have a constant r. These photons would circle the black hole, but they're not called circular orbits?

I figure they would because their orbits should have an effect on the electromagnetic field detectable at a distance?