md65536

2 hours ago, KJW said:

In the case of gravitational redshift and blueshift, equivalent to accelerational redshift and blueshift, there is no relative velocity between the source and receiver. The redshift and blueshift is due to the acceleration itself, and depends on the displacement between source and receiver.

If the equivalence principle is used, then there should be an equivalent relative velocity between source and receiver.

Consider 2 sources falling into a black hole, and only one of them realizes it and accelerates outward so that it can remain stationary relative to the EH. Near the EH, it would need to have a velocity approaching c relative to the free-falling source, so there would be a Doppler shift between the two. The sources would have to appear differently, and external sources would have to appear different to them, depending if they're stationary or falling.

Consider a rocket accelerating upward, so that the bottom of the rocket is equivalent to being deeper in a gravitational well relative to the top. Light from the top is blue-shifted when seen at the bottom, but the top and bottom remain at relative rest. However, consider two sources at the top of the rocket, both emitting a single pulse of light. One of the sources is fixed to the rocket and accelerating, and the other is inertial but set up so that it is momentarily at rest with the other source at the moment the pulse is emitted. Both pulses should be blue-shifted the same amount when seen by the bottom of the rocket, even though the rocket will have a relative velocity with one of the sources when it is seen. Or to put it another way, since the rocket is accelerating and light takes time to cross the distance of the rocket, the velocity of the receiver at the moment of reception will be different than the velocity of the source at emission. The blueshift can be entirely attributed to this difference in velocity, based on reasoning when source and emitter are replaced with equivalent but inertial particles.

2 hours ago, DanMP said:

let's add another, bigger, "wheel" on the starship, with a different rotation speed, but with the same centripetal/centrifugal acceleration, g. The 4^th clock, on the rim of the bigger wheel, would record the same time as the clock #3, on the rim of the smaller wheel? If not, why not?

No. The centripetal force is Fc = mv^2/r, so greater r means greater v, which means a larger Lorentz factor relative to the centre of the wheel.

Would that give the same answer as a gravitational redshift analysis assuming negligible spaceship mass?

On 4/20/2024 at 11:39 AM, Halc said:

So it seems. Imagine how many posts I've made that are then wrong. My definition of proper velocity seems right, but that's different than rapidity.

I think we'd both have been helped a lot if it was more common on this site for people to discuss and accept corrections, but it always feels like a fight to try. Because of that I think this site is harmful for learning at least about relativity.

On 4/20/2024 at 3:57 PM, Mordred said:

 

 One of its useful relations is how it increases. Rapidity increases linearly when describing the four momentum and four force of the object. So if you have some object/craft/particle undergoing constant acceleration. You need some constant force as well. So as we all know the particle velocity will never reach it will always approach but reach c. However the rapidity of the the object under constant acceleration grows linearly so it can grow arbitrarily large. Velocity has c as a limiting factor but that isn't true for rapidity. That is very useful when you want to use the four force to maintain the constant acceleration of the spacecraft.

All of that makes sense now. The constant acceleration implies constant proper acceleration here, I just point that out because I've confused it with constant coordinate acceleration in a single reference frame before. Am I wrong to assume that someone who understands what you wrote here would see that previous statements from you (going back years) don't make sense?

10 minutes ago, Mordred said:

pray tell how am I not using it correctly in your uderstanding

Rapidity is not equal to proper velocity. The chart at the top of https://en.wikipedia.org/wiki/Proper_velocity shows how they (and velocity in natural units) relate.

Also,

On 4/18/2024 at 5:16 PM, Mordred said:

It's simply a type of boost called rapidity.

Rapidity is not a type of boost.

5 hours ago, Halc said:

The way I've heard 'rapidity' used is as a proper velocity.  It is proper acceleration integrated over proper time, and thus it adds the normal way.

You're referring to "celerity", which is different. https://en.wikipedia.org/wiki/Proper_velocity

https://en.wikipedia.org/wiki/Rapidity "For one-dimensional motion, rapidities are additive."

6 hours ago, Halc said:

Anyway, since the usage comes from Mordred, he may be using the term in a different way than I tend to see it.

I'm sorry to hear that, but yes he's not using the terms correctly.

1 hour ago, Mordred said:

Now that's the Lorentz boosts in terms of rapidity.

So that's the same Lorentz boost, just represented using a different expression for the constant velocity.

1 hour ago, Mordred said:

for changes in {x,t} we are boosting the velocity or alternately the rapidity .

Where's the acceleration? Everything you wrote seems to be for a constant velocity ie. a constant rapidity. If one changes, the other changes. I don't see that expressed anywhere.

1 hour ago, Mordred said:

you can also boost the rapidity

Are you still using "boost" to refer to a Lorentz transformation or are switching between meanings of the word "boost" here? If it's "a type of boost called rapidity" you're saying you can boost the boost? What does that mean?

It looks to me like you're just showing the rapidity form of the (constant velocity) Lorentz boost, which still describes constant rapidity. However since it's easier to use changes in rapidity to describe relativistic acceleration because rapidity is additive, you're assuming that rapidity describes the change in velocity itself (ie. change in rapidity) rather than just representing the velocity alone.

Your problem example was, "Lets have a constant acceleration for however many years." What's the result?

13 hours ago, Mordred said:

Both velocity and accelerations are boosts in the the Lorentz transforms. Rapidity is just a particular type of boost. I know you and I had tried discussing this in the past. Later on when I'm not at work I will try to get you far better detail on the difference of a boost due to velocity as opposed to acceleration. Part of the confusion  is that both velocity and accelerations are also described by rapidity. However the transforms for each slightly  different .

 

Well I look forward to the details because nothing you wrote here makes sense to me. You are an expert on this? I was hoping someone else would chime in because the only confusing parts of it are things you wrote.

A boost is a Lorentz transformation, do you agree? Rapidity is a measure for relativistic velocity, do you agree?

9 hours ago, Mordred said:

Acceleration is easily handled in both SR and GR. It's simply a type of boost called rapidity.

I've seen this posted on this site many times over the years and I think it's wrong but never saw a correction or explanation. It's repeated often in posts labelled "expert" but I don't understand what it means. As a Lorentz transformation doesn't a boost imply constant velocity? How can a measure of velocity be called an acceleration? How is a measure of velocity a type of Lorentz transformation?

Is there some sensible meaning to what I quoted that I'm just not comprehending?

On 4/8/2024 at 7:51 PM, geordief said:

Was it Minkowski who arranged the axes so that they all had the same units of spatial distance?

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.

8 hours ago, MigL said:

It is multiplied by a 'conversion' factor c , and because it must be orthogonal to the three spatial dimensions, we make it imaginary.

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.

1 hour ago, geordief said:

Are they related?Can they be the same under any  particular conditions?

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.

31 minutes ago, Mordred said:

The reason being that two falling objects as they approach either CoM of either BH would converge.

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?

1 hour ago, KJW said:

A correct analogy that I discovered recently is two wheels of unequal radius joined by an axle. As this rolls along a flat road, the trajectory will curve towards the smaller wheel, and the larger the difference between the radius of the two wheels, representing time dilation, the larger the curvature of the trajectory, representing the acceleration we feel as gravity.

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.

1 hour ago, swansont said:

The Coulomb field is static; that’s the 1/r^2 field. EMR intensity drops off as 1/r^2 from a point source, but intensity is the square of the field strength.

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.

16 hours ago, swansont said:

How does your conclusion follow from the quote? What kind of radiation is it?

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

11 hours ago, Markus Hanke said:

For a stationary charge supported in a gravitational field, the result I am familiar with from the literature (see link further up in the thread) would indicate that a comoving detector would not detect any radiation, but another detector freely falling past the charge, would.

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

On 2/26/2024 at 10:44 PM, Markus Hanke said:

Sure - isn’t that already a suitable model for the situation at hand? The charge is seen to radiate in some frames but not in others.

What does this mean? Wouldn't it radiate as light? If so, what frame wouldn't it radiate in?

On 2/26/2024 at 10:44 PM, Markus Hanke said:

Ok - this doesn’t sound like too hard of a test to perform, I wonder if this has been done?

Isn't the problem that if there was radiation, the expected energy would be undetectably low?

On 2/20/2024 at 10:16 PM, Markus Hanke said:

the non-collapsed version would be some form of gravitational  geon.

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 r_s with at least as much energy as the black hole, and consider it inside a sphere of size 2 r_s, 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.

On 2/19/2024 at 1:19 AM, Genady said:

If the light orbiting the BH is not a test particle but rather has the energy as described in the OP, then I think it is not on a photon sphere anymore. The "photon sphere" of the original black hole would be inside the new black hole.

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?

Is it possible trap light in a stable circular orbit around a tiny Schwarzschild black hole such that the energy of the light is greater than that of the black hole?

5 hours ago, swansont said:

E=mc^2 is not directly predicated on the existence of the electromagnetic interaction; c is the speed of any massless particle.

However, physics is inter-related. You can’t just arbitrarily change part of it and think that change would be isolated.

You should still be able to model a universe with an absence of light without removing the rules for it, such as a universe made up only of dark matter, or maybe including uncharged black holes. You'd have to make assumptions, but in accepted models, energy/mass equivalence holds for dark matter on its own.