I How does gravity vary in deep mines and interstellar gas?

[KurtLudwig]

The equations for gravity vary differently depending on the distances involved: Inside a body, such as a planet, the weight varies directly with the distance from the center - as r. Outside of the planet or sun, gravity varies as the inverse square law - as 1/r^2. At distances at the edges of a galaxy and presumably further, gravity decrease less gradually - as 1/r, as calculated by an equation such as MOND.
Question 1: In the vast reaches of the universe, between galaxies, is the strength of gravity calculated by 1/r? I think that gravity decreases more slowly as 1/r.
Question 2: How is gravity calculated for interstellar gas? How can the inverse square law apply to a point mass, such as a hydrogen atom?
Question 3: Has gravity been actually measured in deep mines or bore holes? It cannot be directly proportional to r, since mass is more dense at the center of a planet or sun.

 

[PeterDonis]

The equations for gravity vary differently depending on the distances involved

No, the particular solution of the equations that describes how gravity behaves varies depending on the distribution of masses (more generally, stress-energy) that is present.

Inside a body, such as a planet, the weight varies directly with the distance from the center - as r.

More correctly, this is the variation inside a planet that is assumed to be a perfect fluid of uniform density and negligible pressure.

Outside of the planet or sun, gravity varies as the inverse square law

Yes, because now you're outside the planet or star, and you are assuming that there is vacuum there (zero stress-energy).

At distances at the edges of a galaxy and presumably further, gravity decrease less gradually - as 1/r,

A galaxy is not vacuum; on the scale of the galaxy, the behavior of gravity depends on the distribution of stars and the average density that distribution gives rise to. To say that it is a simple 1/r law in the region further out towards the edges is a serious oversimplification. (And if you are implying that it's an inverse square law closer in, that's not just an oversimplification, it's wrong; closer into the center, gravity looks more like it does inside a planet or star, the "acceleration due to gravity" increases with distance from the center.)

(Also, once you're completely outside the galaxy, gravity due to that galaxy will indeed be an inverse square law, just like it is in the vacuum region outside any mass distribution, assuming that distribution is roughly spherically symmetric.)

 

[PeterDonis]

Question 1: In the vast reaches of the universe, between galaxies, is the strength of gravity calculated by 1/r? I think that gravity decreases more slowly as 1/r.

Neither of these is right. As noted in my previous post, in the vacuum region outside a galaxy, gravity due to that galaxy goes like 1/r^2, just like it does in the vacuum region outside any massive object. The only complication is that there will probably be multiple galaxies close enough to you to have gravity effects that are comparable, so you have to figure the 1/r^2 effect due to each of them and add them all up (similar to in a system like the solar system where you can be affected by gravity from multiple bodies).

Question 2: How is gravity calculated for interstellar gas? How can the inverse square law apply to a point mass, such as a hydrogen atom?

This is two questions. Interstellar gas is typically modeled as a fluid with a density. If you're inside the gas, gravity works the way it does inside any fluid with that density.

I'm not sure why you think there's a problem with the inverse square law applying to a point mass. Strictly speaking, in Newtonian gravity, that is the fundamental case that everything else is derived from. Newton had to use calculus, integrating over the effects of a continuum of point masses, to show that the gravity due to a large mass, like the Earth, in the vacuum region outside it goes like 1/r^2 where r is distance from the center of the Earth--i.e., as if the Earth were a point mass with the same position as its center.

Question 3: Has gravity been actually measured in deep mines or bore holes? It cannot be directly proportional to r, since mass is more dense at the center of a planet or sun.

I don't know if there have been any direct measurements of gravity inside the Earth, but we have fairly detailed indirect measurements from seismic data, as shown in this Wikipedia article:

https://en.wikipedia.org/wiki/Gravity_of_Earth#Depth

You're right that inside a mass with non-uniform density, gravity does not increase like 1/r from the center out to the surface. The data in the Wikipedia article linked to above shows that gravity actually increases as you go in from the surface (rather than decreases) down to a depth of about 2500 km; only further in than that does it start to decrease with depth, down to zero at the center.

 

[strangerep]

[...] Outside of the planet or sun, gravity varies as the inverse square law - as 1/r^2. At distances at the edges of a galaxy and presumably further, gravity decrease less gradually - as 1/r, as calculated by an equation such as MOND.

Note that MOND is only an empirical formula, without a theoretical foundation to account for it. As such, it's a bit like Kepler's empirical laws before Newton worked out his foundational framework in which the Kepler laws could be derived from first principles.

In the vast reaches of the universe, between galaxies, is the strength of gravity calculated by 1/r? I think that gravity decreases more slowly as 1/r.

Certainly, in the outer regions of galaxies, many rotational velocity profiles can be well-fitted to a MOND formula, which is dominated by the 1/r term at those scales. But the MOND formula doesn't do as well for even larger structures such galaxy clusters, etc.

At galactic scales and above, cosmological effects start to have an effect. Instead of taking the usual $1/r^2$ rule, one can solve the Einstein equations with cosmo constant $\Lambda$ and obtain the Schwarzschild-deSitter solution, in which $$g_{00} ~=~ 1 ~-~ \frac{2GM}{r} ~-~ \frac{\Lambda}{3}\, r^2 ~.$$This implies an equation of motion for the radial component of acceleration: $$a_r ~=~ - \partial_r g_{00} ~=~ -~ \frac{2GM}{r^2} ~+~ \frac{2\Lambda}{3}\, r ~.$$Due to the sign change in the 2nd term from taking the derivative, we find the whole expression goes to 0 at $$r_\lim^3 ~:=~ \frac{3GM}{\Lambda} ~.$$For a stable circular orbit where r = const we have $$ a_r ~=~ -\,v^2_{\mbox{tan}}/r ~.$$ But $v^2_{\mbox{tan}}$ cannot go negative, so we conclude that stable circular orbits are no longer possible beyond $r_\lim$. For the Milky Way, with radius $\sim 24 Kpc$, I calculate $r_\lim \approx 1.4 Mpc$.

However, this is rather more complicated in reality by the presence of other galaxies nearby.

 

[KurtLudwig]

First, I want to thank you both for your explanations. Your answers show me that I have a long way to go. I need to get acquainted with cosmological equations and their derivations, which I am doing.

Being careful not to come up with hypotheses and using only physics from textbooks and Wikipedia, please explain the errors of the logic in the paragraph below.

When there will be a quantum gravity theory, it must involve a boson, such as the hypothesized graviton, to interact with protons and neutrons. Protons and neutrons do have cross sections, 10^-15m. To interact, I assume that a graviton will have a cross section too. I do not see how the 1/r^2 law can reduce the cross section of a proton while it reacts with a graviton. It will be a quantum reaction. Then the frequency of the gravitational reactions will only depend on the time it takes the gravitons to reach these protons of the hydrogen gas. Since this is time related, the frequency of interactions will be proportional to 1/r from the source of the gravitons. The source being a nearby star.

 

[PeterDonis]

When there will be a quantum gravity theory, it must involve a boson, such as the hypothesized graviton

This is not necessarily true as a statement about the fundamental theory. We think it's true that any quantum gravity theory will have to have some approximation in which it can be viewed as a quantum field theory of a massless spin-2 field. But that doesn't mean that QFT has to be the fundamental quantum gravity theory.

Further discussion of this topic should be in a new thread since this is not the right forum for it. See further comments below.

to interact with protons and neutrons.

Why are you picking out protons and neutrons particularly? Gravity interacts with everything.

Protons and neutrons do have cross sections, 10^-15m.

The term "cross section" has a particular technical meaning in physics, and it's not what you appear to be thinking.

Protons and neutrons can, for many purposes, be modeled as having fairly sharp boundaries at around the distance scale you give. However, this is just a model and doesn't work for all purposes, nor do we have any way AFAIK of deriving such a result from first principles.

However, the cross section of protons and neutrons for interacting with each other or with other particles can vary widely, depending on the type of interaction. It is certainly not always the equivalent of a distance scale of $10^{-15}$ meters.

To interact, I assume that a graviton will have a cross section too.

In the technical sense I just described, we would expect to be able to calculate interaction cross sections for the graviton with various other particles, yes. But in the sense you were using the term, as some kind of "size" of a boundary, a graviton, or indeed any massless particle, can't have a well-defined size. (The technical reason for this is that massless particles cannot be localized the way massive particles can; more precisely, Newton-Wigner localization doesn't work for massless particles.)

I do not see how the 1/r^2 law can reduce the cross section of a proton while it reacts with a graviton.

The 1/r^2 law has nothing whatever to do with cross sections, either in your sense of the size of a "boundary" for a particle, or in the more correct sense of interaction cross sections. It's a classical approximation that's only valid in certain regimes, and that views the interaction as a force between objects that are considered point particles for purposes of the approximation. If you're trying to view the interaction as an exchange of virtual particles, you're not using the classical model of force at all, and trying to mix them together won't work, it will just confuse you.

Then the frequency of the gravitational reactions will only depend on the time it takes the gravitons to reach these protons of the hydrogen gas. Since this is time related, the frequency of interactions will be proportional to 1/r from the source of the gravitons. The source being a nearby star.

This is all personal theory and is out of bounds for discussion here.

If you want to discuss the classical models of gravity that are used to study galaxies, galaxy clusters, and the like in cosmology, that is what this thread would be appropriate for. Trying to investigate quantum gravity models has nothing whatever to do with that.

If you want to discuss particular quantum gravity models that appear in the literature, you can start a new thread in the appropriate forum, which would probably be the Beyond the Standard Model forum.

If you want to speculate about your own home-brewed model of quantum gravity, you'll have to go somewhere else; that's not what PF is for.

 

[KurtLudwig]

Understood
I do not want to get kicked off Physics Forums since I have questions reading textbooks, such as Modern Cosmology by Andrew Liddle.