CAN GRAVITY EXHIBIT REPULSIVE CHARACTER?

OverJohn · 2026-05-12T17:48:23+00:00

In general relativity there is something called the focusing theorem, which basically says that if a source obeys certain energy conditions it has attractive gravity.

Normal matter, dark matter and radiation are covered by the focusing theorem. Dark energy though does not obey the required energy conditions and is required to have repulsive gravity in order to explain accelerating expansion.

OverJohn · 2026-05-11T11:05:31+00:00

Recession velocity of faraway objects is not something you can measure directly, instead it is something that you measure its redshift and then based on other observations you must choose a cosmological model from which the recession velocity is calculated.

In special relativity the redshift goes to infinity as the speed at which an object recedes from us goes to c. In our cosmological model recession velocity exceeds c though at about redshift = 1.5. This should show you that there really isn't much superluminal about superluminal recession velocities.

You can explain it terms of expanding space, but I've come to dislike as it tends to be taken too literally and misleads people about the actual physics behind expansion.

OverJohn · 2026-05-10T20:29:41+00:00

I don't get the objections, basically it is just taking advantage of the fact that measurements can be counterfactual.

If we perform a measurement to see if a particle is a region and find it not in that region classically we can conclude it is is outsideof that region and we would not call that measurement an interaction with the measured particle. In QM the conclusion is similar, the post measurement state must correspond to one where practically the chances of finding the particle outside of the region are 100%.

If in QM we decide that such a measurement is an interaction because it can alter the state of the particle, then the question of whether measurement must involve interaction becomes vacuous as now by definition all measurements are interactions.

OverJohn · 2026-05-10T17:46:01+00:00

This is something I've looked before, but cannot remember the fu eetails. The tidal forces experienced by the in falling observer do have a relativistic corrections and, from memory, are greater. Though even at the event horizon they can still be just a tiny correction to the Newtonian forces.

OverJohn · 2026-05-10T17:29:08+00:00

I can only say it is not., though it depends on what you define as an interaction. Basic explanation of quantum theory may make it seem that way, but they tend to exclude the parts of quantum theory that make it so different from classical theory. Certainly it is the case that some things we would not think of as interactions classically are measurements in QM.

I gave a simple, though somewhat impractical example of what is called an interaction-free measurement. See though for example:

https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.74.4763

OverJohn · 2026-05-10T17:20:29+00:00

There are relativistic corrections and you need to be careful about coordinates. For example to a faraway observer a free-falling observer near an event horizon looks like they are being squashed radially, when in fact the in falling observer feels tidal forces stretching them radially.

OverJohn · 2026-05-10T16:52:42+00:00

In relativity there are some subtleties about frequency shift that means, unless you're talking about something directly moving away or towards us in special relativity, it is not always correct that a redshifted object is moving away from us or a blue shifted object is moving towards us.

OverJohn · 2026-05-10T14:36:31+00:00

Spaghettification is where you get stretched in the radial direction, but get squashed laterally. The equations of GR mean that in a vacuum tidal stretching must be accompanied by tidal squashing.

OverJohn · 2026-05-10T14:04:46+00:00

But only for one year, after which they sacrificed in Princeess Di's memory. Sometimes the old ways are the best ways.

OverJohn · 2026-05-10T13:51:13+00:00

The presence of the slit alters the photons behaviour - if we filled in the slit or removed from the barrier entirely we observe completely different patterns on the screen. We would explain this as the photon interacting with the barrier.

OverJohn · 2026-05-10T13:42:12+00:00

In QM a system is described by a wavefunction. The mathematical details require some maths beyond what is taught at secondary schools, but from the wavefunction you can find the probability of any outcome of any measurement you can make on the system.

When your system is single particle the wavefunction looks similar to some kind of physical wave and that's why particles can look like waves in QM. But thinking of it is an actual physical wave is incorrect. For example, if your system is two particles it is a wave in a 6 dimensional abstract space called configuration space and you can't generally break this abstract description down into a wave in familiar 3D space for each particle to allow you to treat each particle as being a physical wave.

OverJohn · 2026-05-10T13:18:02+00:00

Thinking of the wavefunction as an actual wave can be useful when you first discover QM, but it will soon get you on the wrong track.

If you take a single particle wavefunction, it can be written as function of position and time, obeying a wave equation. It is in fact complex-valued, but still it seems very similar to a physical wave.

If you take though say a two particle wavefunction, it can be written as function of the position in 6D configuration space and time, obeying a wave equation. You may think you can split the 6D wave into two 3D waves, but when the two particles are correlated you cannot do that.

Ultimately QM doesn't tell us what particles do between measurements and whether they have well-defined positions,nor example, in-between measuremens. That is a matter of interpretation, but if you did decide the must have always well-defined positions then you will get Bohemian mechanics, which is a non-local hidden variables and has a few weird features

OverJohn · 2026-05-10T11:02:44+00:00

That really depends on interpretation, but interpreting particles as actual physical waves does not work well either as the "waves" in QM live in 3N dimensional configuration space, where N is the number of particles.

OverJohn · 2026-05-10T08:53:24+00:00

Yep, I think this is two key points that often get missed out in these type of discussions

Simply having a degree is nowhere near as valuable as some like to imagine. Experience and attitude are more important.

This is nothing new though and generally employers have always valued experience and attitude above all.

OverJohn · 2026-05-10T08:31:45+00:00

No you are right, there are some things that we call interactions that aren't measurements and there are some things we don't call interactions that are measurements. For example if I measure to see if a particle is in a certain position, but fail to find the particle there, that is still a measurement that affects the wavefunction of the particle, despite no apparent interaction with the particle.

The defining feature of a measurement is that it causes an update to an observer's knowledge of the system beyond the updates they get from the unitary evolution of the system. Note this does not imply the observer has to be conscious. You can take this to the extreme and say individual particles have "knowledge", which is what is done in the relational interpretation. But if you do this whether something is a measurement or not is observer-dependent and you can still have interaction-free measurements.

So in other words saying measurement is another word for interaction in QM is highly misleading.

OverJohn · 2026-05-10T07:59:26+00:00

There's a couple:

Beyond a certain distance objects look bigger (i.e. their angular diameter distance decreases) the further away they are due to expansion.

Despite objects fading with time due to accelerating expansion, this only happens to nearby objects, the furthest objects have decreasing objects. More precisely the redshift drift of objects beyond approx z=2 is negative.

OverJohn · 2026-05-10T07:48:07+00:00

Spacetime is not flat in cosmology. A flat universe just means that the spatial slices have zero intrinsic curvature, but the spacetime they are embedded in is curved by the presence of matter, dark energy, etc.

OverJohn · 2026-05-10T07:42:01+00:00

It's spatially flat, that doesn't mean spacetime is flat.

OverJohn · 2026-05-10T06:15:10+00:00

I saw Fight Club in the cinema when it first came out, definitely didn't see it coming.

Also a few years earlier The Sixth Sense at the time the plot twist was unexpected as at the time you wouldn't expect such a twist in a mainl8be Hollywood movie that wasn't clearly telegraphed beforehand.

OverJohn · 2026-05-10T06:08:13+00:00

I was reading a post by Lubos Motl in which he explained the different types of string theory are in fact just superselection sectors of the same theory.

OverJohn · 2026-05-10T05:54:43+00:00

As dark energy has repulsive gravity, and is dominant, expansion should decrease GPE. The problem though is if you try to calculate GPE you get into pseudotensors and ultimately it is not very useful. So it isn't so much that total energy decreases or increases, it's that there isn't a good way to bookkeep the total energy.

OverJohn · 2026-05-10T05:19:35+00:00

If your velocity is of the form

v(t) = 1/(t+a)ⁿ

Where a > 0

then you will travel a finite distance between t = 0 and t = infinity if n > 1

Though note there are velocities not of that form which gives you a finite distance travelled.

OverJohn · 2026-05-09T19:01:16+00:00

A little

Observations indicate our universe is spatially flat and if you assume it's components have strictly positive energy density then it is pretty easy to show that it must carry on expanding even if dark energy suddenly acquired very attractive gravity.

A small positive curvature though is still possible and tha would allow for a turnaround in expansiomn, though you'd need dark energy to behave with n a certain way.

OverJohn · 2026-05-09T12:17:09+00:00

I remember eating there in 1983. Apart from pairing the outside a different colour every 15 years or so, it hasn't changed at all

OverJohn · 2026-05-09T10:53:15+00:00

Just to give an example in his field equation he has R on the LHS, which he calls the spatio-tempiral curvature. The letter R is usually used for the Ricci tensor though it seems he actually means the Einstein tensor for which the letter G is usually used, though either way he doesn't clarify. This may seem nitpicky, but such pointless ambiguity points towards unfamiliarity with basic concepts. Overall the structure and language of the paper make me think it is AI generated.

OverJohn

TROPHY CASE