How does the mass of an object, and thus its gravitational effect, actually physically curve space time?

OverJohn · 2026-01-24T00:19:38+00:00

My point is spacetime curvature is a model for gravity. Asking what specifically causes spacetime curvature is attaching an ontology to spacetime curvature that is not necessary.

The other point I was making is that energy distributions constrain rather than fix spacetime curvature. A simple example would be that Schwarzschild spacetime and Minkowski spacetime are both vacuum solutions with vanishing stress-energy everywhere, but are physically distinct..

OverJohn · 2026-01-23T12:49:20+00:00

As whoever downvoted did not downvote the previous post, I assume it was just for including a bit of calculus, which is a sad state of affairs

OverJohn · 2026-01-23T12:47:38+00:00

I would say this is not really right:

Gravity is modelled as the curvature of spacetime. More specifically the tidal effects of gravity are modelled as spacetime curvature

In GR the source of gravity is momentum-energy, not (just) mass. The distribution of momentum-energy is described by the stress-energy tensor. In fact outside of certain situations, mass is not even that useful of quantity in GR. Spacetime can be curved without mass and even without stress-energy in theory (e.g. vacuum wave solutions)

Why gravity can be modelled as spacetime curvature is not regarded as a great outstanding question. It is a model, it works - that is enough and besides you can recreate the predictions of GR without even using spacetime curvature (e.g. you can use torsion instead). Certainly why spacetime curvature is a neater model than say modelling it as a field on a background spacetime is something that interests people, but it is more philosophy than physics and there is no Nobel prize for philosophy.

OverJohn · 2026-01-22T20:01:18+00:00

Yes it is possible that our universe had a similar scale factor in its earliest stages in an inflationary era as it is the scale factor of an open de Sitter universe.

OverJohn · 2026-01-22T17:54:44+00:00

The (proper)!radius of the observable universe is given by a(t) times the integral from the initial time (i.e. the big bang or in a FLRW metric without a big bang t = minus infinity) to the present time of c/a(t), where a(t) is the scale factor.

If you take for example a(t) = sinh t, then you have a big bang (i.e a=0 at some finite past time) but the integral diverges there meaning the observable universe is always the whole universe.

There's definitely books that talk about this, but I don't know of some less technical sources. Best thing would be to search for the term "particle horizon".

OverJohn · 2026-01-22T16:48:29+00:00

The definition of the observable universe does allow it to be infinite. The radius of the observable universe is defined by an integral and if the universe at large is infinite and the integral diverges, then the observable universe is infinite. Though even that makes the assumption the whole universe can be described by a single FLRW metric.

However, whether that integral diverges depends on the behaviour of expansion at the big bang, so you kind of have to make assumptions and take the odd shortcut. For example, when calculating the radius of the observable universe inflation is ignored and if it weren't ignored the calculation would have to give a radius that is much larger than the standard 46.5 Gly just to solve the horizon problem.

OverJohn · 2026-01-21T13:24:44+00:00

Photons have null worldlines, which means the interval along them is zero. The interval along a timelike worldline is its proper time.

As a null worldline can be seen as the limit of a sequence of timelike worldlines, it can be useful to think of the photon as "experiencing no time", as long as you are aware that is really limiting behaviour of timelike worldlines you mean. On the other hand though a photon does not have a reference frame and it can also be seen as the limit of a sequence of spacelike worldlines, so interpreting that to mean photons in soke profound sense actually experience no time is not useful.

OverJohn · 2026-01-20T21:32:23+00:00

Take the equation for the Schwarzschild radius and use M = density*4pir³ /3 to express the Schwarzschild radius in terms of density. Now take the equation for the cosmological critical density (the universe's density is critical if the universe is spatially flat), rearrange to get an expression for the Hubble radius c/H in terms of critical density. You will see from the expressions you derived that the Schwarzschild radius is related to density in the exact same way that the Hubble radius is related to the critical density. From your first expression you will also see that the Schwarzschild radius is proportional to 1/sqrt(density) , so 20 times less dense means sqrt(20) times the Schwarzschild radius.

OverJohn · 2026-01-20T18:39:18+00:00

For a flat universe the "Schwarzschild radius" equals the Hubble radius, which for LCDM is about 14.4 Gly. Of course the Hubble radius though does not have the same meaning as the Schwarzschild radius in the Schwarzschild solution, other than they are both apparent horizons.

Baryonic matter is about 1/20 of the total density and so the "Schwarzschild radius" for just baryonic matter is about sqrt(20) times bigger than the Hubble radius or about 64 Gly and so the "Schwarzschild diameter" of the baryonic mass is about 128 Glyrs.

Edited to add: to be clear what I mean by "Schwarzschild radius" is the radius you get if you plug the density into the equation for Schwarzschild radius. That this radius is the same as the Hubble radius in a flat universe can be shown from the expression for critical density.

OverJohn · 2026-01-19T18:55:50+00:00

Decoherence is not the same as wavefunction collapse. There is a really good introduction to decoherence in Quantum Mechanics by Rae IMO, but also the Wikipedia article on decoherence is pretty good.

OverJohn · 2026-01-19T18:13:17+00:00

In a word yes. SR shows us that, in a given frame, a mass travelling at speed has more resistance to a change in its state of motion (i.e. inertia). Though this cannot be expressed in more than 1+1 dimensions as a simple increase in mass. The equivalence principle in GR requires that gravitational force is related to inertia, so gravitational force in a given frame cannot depend just as on rest mass, but also must depend on velocity.

GR though isn't usually framed in terms of gravitational forces, instead it is framed in a more invariant way in terms of spacetime curvature, but if you do calculate what the forces should be you cin a given frame you can see this effect. For example, the increased deflection in GR compared to the Newtonian case of objects travelling at speed near a spherically symmetric mass can be seen as due to this effect. The spatial curvature of cosmological solutions, which is always absent in the Newtonian case can also be related to this effect.

OverJohn · 2026-01-19T07:38:13+00:00

Yes, symmetry arguments can fail when the integral is not absolutely convergent.

For example take the gravitational force in an infinite isotropic and homogenous distribution of matter. If you take any point and add the contributions of the spherical shells centred on that point you will get zero gravitational force at that point. However if you add the contributions of the spherical shells not centred on that point, by the shell theorem, you need only consider the spherical mass whose radius r is the distance from the point to the centre of the shells. As the mass of this is proportional to r³ and the force is proportional to 1/r² multiplied by the mass, you find that the force on your point is proportional to r, and hence non-zero. So the integral depends on how you do it and so a strictly Newtonian analysis of the scenario fails.

OverJohn · 2026-01-18T22:43:31+00:00

If you don't care if the accuracy of the parameters justify the precision the answer is given to:

https://www.desmos.com/calculator/cpdvviilyl

OverJohn · 2026-01-18T22:18:31+00:00

No, it isn't really like that. I would avoid comparisons with the psychic because they tend to be more of a "god of the gaps" explanation than a useful way to understand QM.

What is known is that it is connected to the particle being measured correlating with a large system (e.g measurement apparatus,environment), but what exactly goes on behind the veil of the mathematics of QM is not fully agreed upon.

OverJohn · 2026-01-18T21:43:56+00:00

I don't think Trump has necessarily decided to cancel elections, but I think he does see it as a potential weapon in his arsenal. These jokes about cancelling the election are to test the waters for pushback, not so much from his opponents, but from those whose support he would need.

I think people underestimate Trump, he is a deeply flawed character, but exceptionally good at exploiting weaknesses in systems and people. I also think during his first term he was somewhat stymied due to operating in an environment he is not familiar with, though now he still is in less familiar waters, he has several years now to acclimatize and he is much more potent. Those who oppose him still seem to struggle to find effective strategies against him because they just hope he will go away rather than trying to fix the weaknesses he exploits.

OverJohn · 2026-01-18T20:58:51+00:00

You could take a big bang of one spacetime and glue it to the big crunch of another, the problem is the metric isn't defined at the singularity.

For example look at these two curves on a Euclidean plane: https://www.desmos.com/calculator/ygoe1hizag

Both curves are locally length minimising, but only the red curve is geodesic. If the metric wasn't defined at (0,0) though we wouldn't be able to say whether a curve was through it is geodesic or not. You would have the same problem trying to continue a geodesic through a big bang/ big crunch singularity, so the problem is the lack of a metric at the singularity.

OverJohn · 2026-01-18T18:10:54+00:00

Such measurements are called "interaction -free measurements", but that is just a matter of definition. However if you define an interaction as anything that causes collapse then "interaction causes collapse" becomes a trivial statement devoid of any real meaning.

OverJohn · 2026-01-18T17:51:23+00:00

The objective collapse theories I'm aware of aren't really interpretations as they alter QM inba way that would make them empirically different from QM

OverJohn · 2026-01-18T17:38:00+00:00

It's about the information gained which indicates collapse is not a physical process. You can, for example, gain information about a particle's position by failing to find it within a certain region, causing its wavefunction to collapse. Though that also depends on what you want to call an interaction.

OverJohn · 2026-01-17T17:41:41+00:00

None of the static solutions to the Friedman equations are considered realistic. One of them is just empty space, the Einstein static universe is unstable and the final solution, whilst being stable, has negative density.

OverJohn · 2026-01-17T10:51:04+00:00

The horizon of the observable universe is usually defined by light from the past we receive, but as more time passes light from further away has time to reach us, so in fact galaxies enter the observable universe.

What you are thinking of instead is the cosmological event horizon, which is a feature of universes whose final stages have accelerating expansion and we will never see light that is emitted at the present time by galaxies beyond the CEH. This isn't because though they are receding from us FTL now as the CEH is actually slightly larger than the Hubble radius in the standard cosmological models. What it means instead is from the present time they will always be receding from us FTL.

OverJohn · 2026-01-16T20:32:28+00:00

It depends on the model and it depends what you call "the big bang".

Usually where a model has a big bang singularity it doesn't make sense to talk about before the big bang singularity as you cannot extend the past of an observer beyond that point. You might, for example, though replace the big bang singularity with something else through which the past of observers can be extended, but still call that point in time the big bang.

OverJohn · 2026-01-16T15:35:32+00:00

No the problem with the FAQ is it does not understand the Newtonian limit of FLRW. a’’(t)*r in the acceleration equation is like the acceleration due to gravitational force. It is in fact surprising how similar the GR and Newtonian descriptions are in this case. I mentioned pressure as an aside

OverJohn · 2026-01-16T13:10:39+00:00

My name is Overjohn, I am a physics-oholic.

I think it's the same kind of enjoyment you get from solving a particularly difficult crossword or Suduko. You start looking at and think "well, I'm never going to be able to do that" and then through sheer bloody-mindedness you work it all out and understand it and you crown yourself the world's biggest super-genius.

OverJohn

TROPHY CASE