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## / Gravity |

krikoman - * * on 06 Jun 2018

Not had a long time to think about this, and I may have confused myself already.

But:-

As we move away from the surface of the Earth the acceleration due to gravity (AdG) decreases inversely in proportion to distance away.

Discounting the impossibility, if you were stood at the centre of the earth the (AdG) would be zero (leaving out the moon and the sun etc.).

With this in mind where is the maximum position of (AdG)?

I'm sure I've got something wrong here, and I've got loads of works to do, so you lazy lot can answer this

The Earth is **NOT **on a treadmill

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mal_meech on 06 Jun 2018

In reply to krikoman:

Quite a complex simple question, as if you are below the surface of the earth, you are surrounded by mass providing a gravitational effect. So simplistically, the maximum value of AdG is at the surface, ignoring variances due to changes in density... ;)

On you're centre of the earth point, are you familiar with lagrange point concepts for "zero" gravity effects?

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Rob Parsons on 06 Jun 2018

In reply to krikoman:https://en.wikipedia.org/wiki/Gravity_of_Earth

Analysis of that question in: https://en.wikipedia.org/wiki/Gravity_of_Earth

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elsewhere on 06 Jun 2018

In reply to krikoman:

> As we move away from the surface of the Earth the acceleration due to gravity (AdG) decreases inversely in proportion to **square **of distance away.

Slight type correction - inverse square law.

> Discounting the impossibility, if you were stood at the centre of the earth the (AdG) would be zero (leaving out the moon and the sun etc.).

> With this in mind where is the maximum position of (AdG)?

From memory, at the surface of the earth.

Again from memory of a derivation long ago...

acceleration due to gravity increases linearly with radios from centre then decreases with inverse square law from the surface outwards.

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GrahamD - * * on 06 Jun 2018

In reply to krikoman:

Surface of the earth. AdG increases linearly from 0 at the centre to about 9.8m/s/s at the surface and then falls away as inverse of the square of the distance above the surface

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Rob Parsons on 06 Jun 2018

In reply to GrahamD:

> Surface of the earth. AdG increases linearly from 0 at the centre to about 9.8m/s/s at the surface and then falls away as inverse of the square of the distance above the surface

That assumes uniform density, which is in practice not the case. See the graph in the link I posted above.

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elsewhere on 06 Jun 2018

In reply to Rob Parsons:

The earth was uniform and cows were spherical in my day ;-)

Interesting graph to see some more realistic.

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krikoman - * * on 06 Jun 2018

In reply to krikoman:

Great stuff, thanks everyone for the replies.

I love how UKC can at the same time complicate and supply answers to life's mysteries.

This all came about by estimating the depth of a hole at Coppermines, Coniston.

I'd like to pretend I asked the question based on the varying density, but it was in fact based on constant density

Thanks again.

Edit: I'm off to read about Lagrangian Points

Post edited at 15:42

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NottsRich on 07 Jun 2018

In reply to GrahamD:

Does it matter where on the surface you are? e.g. Chimborazo in Ecuador, or the Dead Sea?

GrahamD - * * on 07 Jun 2018

In reply to Rob Parsons:

> That assumes uniform density, which is in practice not the case. See the graph in the link I posted above.

True enough, but to a first order approximation ...

GrahamD - * * on 07 Jun 2018

In reply to NottsRich:

> Does it matter where on the surface you are? e.g. Chimborazo in Ecuador, or the Dead Sea?

Yes, it does, the answer I gave is an idealised approximation. Unfortunately I doubt my maths is up to anything other than the idealised approximation !

NottsRich on 07 Jun 2018

In reply to GrahamD:

And I'm sure I wouldn't understand that levels of maths either! How about this then, as idle speculation...

Gravity is at a maximum at the surface of the earth. It is influenced by the volume of stuff (mass) below your feet. Therefore on a point with a lot of mass below your feet (e.g. Chomborazo, furthest from the centre of the earth), gravity is stronger.

Just a guess!

cb294 - * * on 07 Jun 2018

In reply to NottsRich:

Not quite. If you are on a continental shield (70km of solid rock below you) there will be more mass than at the same level but with thin oceanic crust below you (both above a relatively lighte mantle). This accounts for minor variations in the local gravity.

A single mountain below you does not do much in this context. Here, the question of distance from the centre of the earth is more important (gravity falls with the square of the distance).

Chimborazo is always chosen as the example as it so close to the equator. Because earth is slightly flattened at the poles and expanded at the equator its summit is further from the centre of the earth than the summit of higher mountains further North or South.

CB

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NottsRich on 08 Jun 2018

In reply to cb294:

> Chimborazo is always chosen as the example as it so close to the equator. Because earth is slightly flattened at the poles and expanded at the equator its summit is further from the centre of the earth than the summit of higher mountains further North or South.

Thanks, yeh I know that it's summit is further from the centre than any other place on earth - that's why I chose it. And as a result of this, I assumed it also had the greatest mass below it (i.e. thicker crust). Might have got that bit wrong, as like you said, the rock below it may actually be quite thin (who knows!).

subtle on 08 Jun 2018

In reply to krikoman:

> As we move away from the surface of the Earth the acceleration due to gravity (AdG) decreases inversely in proportion to distance away.

> Discounting the impossibility, if you were stood at the centre of the earth the (AdG) would be zero (leaving out the moon and the sun etc.).

> With this in mind where is the maximum position of (AdG)?

As the Earth is flat it doesn't really matter where you stood, the (AdG) will be the same

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kamala - * * on 08 Jun 2018

In reply to cb294:

> Not quite. If you are on a continental shield (70km of solid rock below you) there will be more mass than at the same level but with thin oceanic crust below you (both above a relatively lighte mantle). This accounts for minor variations in the local gravity.

Sorry, this is niggling at me...The mantle is relatively light(er) compared to what?

cb294 - * * on 08 Jun 2018

In reply to NottsRich:

There are several phenomena that account for differences in gravity between different places on the Earth surface.

In first approximation you can treat the mass distribution within the planet as spherically symmetric, in which case it is the same as if it were concentrated in one point only, and gravity should be the same everywhere.

This is obviously not quite true, though: As you say, Earth is flattened a bit. You can still approximate mass distribution with rotational symmetry. Under this model, moving further out from the centre (i.e. closer to the equator, or on a mountain top vs. a valley) will give you lower gravity, which is exactly what is observed..

Now if you can measure gravity even more precisely you will find that the assumption of a symmetric mass distribution fails as well. Gravity is not only affected by geometry, but also by the local composition of the planet: In some places the material underfoot will be more dense in some places than others, resulting in a locally stronger gravity. The strongest such effect is seen at large scales, e.g. when comparing continents (with a thick, reasonably cool and hence dense crust) vs. expanding oceanic floor (a thin crust covering warm and hence less dense material closely below).

If you can measure gravity even better you can find even weaker and more local variations. This is e.g. used for measuring from the air how far an underground deposit of dense ore (higher gravity) extends, and where the lighter rock around begins.

CB

cb294 - * * on 08 Jun 2018

cander - * * on 08 Jun 2018

In reply to cb294:

It’s been a while since I last looked but I recall the crust has an average density of 3.0 g/cm3 (oceanic) and 2.7 g/cm3 (continental), whilst the mantle has an average density of 4.5 g/cm3 increasing with depth due to increasing pressure. Unless of course my undergraduate knowledge (which to be fair was a bit clouded with beer) has been superseded I'm thinking you’ve got some wrong info here.

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cb294 - * * on 08 Jun 2018

In reply to cander:

Quite likely, so I stand corrected. Maybe it was continental vs. oceanic crust I had in mind. I only did some geology as a secondary subject as an undergrad ages ago, so I will bow to the experts!

CB

kamala - * * on 08 Jun 2018

In reply to cb294:

All the gravity modelling I've ever seen (or done) uses denser mantle values. Cander's numbers look about right to me, though upper mantle is less dense than lower mantle.

Saw your expanded post and realise where you're coming from now. It's not just about density variation with temperature, there are big compositional variations between continental crust, oceanic crust, and mantle.

Post edited at 14:40

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Robert Durran - * * on 08 Jun 2018

In reply to krikoman:

Assuming a spherically symmetrical earth (same density anywhere at equal radius) the maximum will be at the surface. It is a bit technical to prove this.

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Pero - * * on 08 Jun 2018

In reply to cb294:

To add to this the gravity we experience is reduced by the Earths rotation. The required centripetal force reduces with latitude and is greatest at the Equator and zero at the Poles. The resulting nett force is, therefore, least at the Equator and greatest at the Poles.

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wintertree - * * on 08 Jun 2018

In reply to krikoman:

> As we move away from the surface of the Earth the acceleration due to gravity (AdG) decreases inversely in proportion to distance away.

This is not stated precisely enough. Distance away from what? The answer is the centre of the Earth.

All mass within a sphere of maximum radius *R*, with only radial density dependence, acts on a test mass at distance *r>R* as if it was a point mass at the centre of the sphere - ie at *r=0*. It acts with a force that is inverse **square** in nature with respect to *r*.

Moving in to the sphere as above, at a radius rr exerts a net force on the test mass, again as if it was a point mass located at r=0.

With uniform radial density of material it’s thus trivial to show that force and acceleration scale linearly from 0 at the centre of the earth to little g at the surface (*r^3* mass scaling and *r^-2* distance scaling) before decreasing of the form *r^-2* after this point.

*S*o acceleration is unambiguously greatest at the planets surface.

However, the Earth’s core is denser than its crust. This means mass no longer scales as a simple power of radius, making it possible that the acceleration is greater closer to the higher density core. My gut feeling is that this isn’t the case, but I don’t have the means to check right now.

Edit: The interesting one to consider is acceleration vs r inside a gas planet made from an ideal gas...

Post edited at 20:54

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Dave Garnett - * * on 08 Jun 2018

In reply to Robert Durran:

> Assuming a spherically symmetrical earth (same density anywhere at equal radius) the maximum will be at the surface. It is a bit technical to prove this.

Although intuitively fairly obvious.

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Dave Garnett - * * on 08 Jun 2018

In reply to wintertree:

> Edit: The interesting one to consider is acceleration vs r inside a gas planet made from an ideal gas...

... which is the key plot device of Iain M Banks' The Algebraist!

wintertree - * * on 08 Jun 2018

In reply to Robert Durran:

> Assuming a spherically symmetrical earth (same density anywhere at equal radius) the maximum will be at the surface. It is a bit technical to prove this.

I don’t think so.

Consider a very small, very dense core inside a very large, very low density outer. Gravity would be strongest at the surface of the core. That is the case for earth where you consider the surface of the crust as the “core” and the top of our atmosphere as the “surface of the outer”.

Consider that a less artificial form of my example - smooth density increases with depth below the surface - is a natural consequence of gravity in gas planets (compression) and liquid core planets (light stuff floats, heavy stuff sinks).

There is a critical rate at which density has to increase with depth to tip the scales and make gravity stronger below the surface, and I don’t think earth will be anywhere near it, excepting the crust/atmosphere interface.

Edit: My hunch was wrong - the PREM model in Rob Parson’s wiki link above has a step change in radial density giving a higher acceleration at some depth well below the surface of Earth, about half way to the center!

Post edited at 21:43

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Robert Durran - * * on 08 Jun 2018

In reply to Dave Garnett:

Why? Mass outside your radius cancels out - ok seems reasonable. Mass inside your radius smaller but closer - not really intuitive that decreases continuously with radius (?).

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Robert Durran - * * on 08 Jun 2018

In reply to wintertree:

> I don’t think so.

Sorry, yes, you are right. I made a stupid mistake when I wrote down an integral.

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Dave Garnett - * * on 08 Jun 2018

In reply to Robert Durran:

> Why? Mass outside your radius cancels out - ok seems reasonable. Mass inside your radius smaller but closer - not really intuitive that decreases continuously with radius (?).

The way I visualise it, if you are standing on the surface of the sphere, all of its mass is pulling down on you (at least that's the resultant force). At any point within the sphere, at least some of its mass is pulling you towards the surface and, at the centre, exactly half of it is.

Oh, and of course once you leave the surface the force must decrease according to the inverse square rule if you imagine yourself standing on a larger virtual sphere the surface area of which increases as a square of its radius.

Post edited at 22:38

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Robert Durran - * * on 08 Jun 2018

In reply to Dave Garnett:But, as Wintertree has pointed out, you are wrong!

> The way I visualise it, if you are standing on the surface of the sphere, all of its mass is pulling down on you (at least that's the resultant force). At any point within the sphere, at least some of its mass is pulling you towards the surface and, at the centre, exactly half of it is.

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Dave Garnett - * * on 08 Jun 2018

In reply to Robert Durran:

OK, I was just describing the simplest situation. Once you have discontinuous zones of greatly different density then, as Wintertree explains, it all depends on the relative densities and what proportion of the volume they occupy.

Robert Durran - * * on 08 Jun 2018

In reply to Dave Garnett:

> OK, I was just describing the simplest situation. Once you have discontinuous zones of greatly different density then, as Wintertree explains, it all depends on the relative densities and what proportion of the volume they occupy.

I still don't think it's really obvious for uniform density - your argument doesn't seem to take account of being closer to the smaller mass inside your radius.

krikoman - * * on 09 Jun 2018

In reply to subtle:

> As the Earth is flat it doesn't really matter where you stood, the (AdG) will be the same

What happens when you fall off the edge though?

krikoman - * * on 09 Jun 2018

In reply to wintertree:

> Edit: The interesting one to consider is acceleration vs r inside a gas planet made from an ideal gas...

That's where I was hoping this discussion was going to lead

krikoman - * * on 09 Jun 2018

In reply to krikoman:

OK so, and I could work it out but I'm really busy with work .

Is the point for linear density, the point where the volume above, equals the volume below?

Robert Durran - * * on 09 Jun 2018

In reply to Robert Durran:

> I still don't think it's really obvious for uniform density - your argument doesn't seem to take account of being closer to the smaller mass inside your radius.

For uniform density, gravity is proportional to your distance from the centre of the planet. Proving this requires the "shell theorem" (that the gravity due to a uniform shell is the same of that of an equal point mass at the centre if you are outside the shell, but zero if you are inside it). The proof only works out "nicely" because gravity follows an inverse square law. It is not at all obvious that it would give a simple relationship if the law was otherwise (I got stuck with the integration for a general power!), so I think this confirms it is by no means intuitive that gravity is greatest at the surface.

wintertree - * * on 09 Jun 2018

In reply to Robert Durran:

I found it interesting to stop and ponder why things may be intuitive...

> so I think this confirms it is by no means intuitive that gravity is greatest at the surface.

I think that all hinges on how intuitive one finds it that (a) gravity is proportional to *r^-2 *and volume to *r^3 *and (b) the radial distribution of mass inside a virtual sphere has no effect on the gravitational field at its surface.

**(a)** I think someone happy with basic geometry will see this as intuitive, otherwise not. I say this as inverse square laws are very intuitive if you imagine forces, fields, power etc being transmitted by particles, and particles obeying conservation of stuff laws.

**(b)** I don’t find the “shell theorem” approach to proving this intuitive. In fact it’s a mystery to me why people talk about shell theorem for gravity at all. Gauss’ law is much more general (and therefore useful to know), and with the additional constraint of no radial density dependence you arrive promptly at the same conclusion as shell theorem. It’s a much clearer and more intuitive way of proving it. Of all of Physics, Gauss’ law has to be one of the nicest, simplest things to intuitively understand without even needing any maths.

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Robert Durran - * * on 09 Jun 2018

In reply to wintertree:

> I found it interesting to stop and ponder why things may be intuitive...

I think that once you have a geometrical picture of Gauss' law as lines of flux then the inverse square law, the shell theorem (and therefore that radial distribution of mass makes no difference) do become intuitive. However, from a starting point of Newton's inverse square law, the shell theorem certainly isn't (unless you think the flux picture of Gauss' law follows intuitively from Newton's law - I don't think it does, otherwise Newton would have come up with it rather than the step requiring he insight of another genius!). So yes, if you have the flux picture as your starting point, then everything is intuitive, but most laymen will be starting from Newton.

In general, if you have to stop and think about whether something is obvious or intuitive then it probably isn't.

wintertree - * * on 09 Jun 2018

In reply to Robert Durran:

> So yes, if you have the flux picture as your starting point, then everything is intuitive, but most laymen will be starting from Newton

Which is why this is so interesting - why is teaching at A-Level and undergraduate level so Newton centric when a Gaussian flux approach is so much more intuitive? Other than several centuries of precedent I don’t really know.

> In general, if you have to stop and think about whether something is obvious or intuitive then it probably isn't.

Obvious, I ageee. Intuitive - often something is intuitive if only one looks at it the right way. Sometimes that way is not obvious.

Paz - * * on 09 Jun 2018

In reply to wintertree:

> Which is why this is so interesting - why is teaching at A-Level and undergraduate level so Newton centric when a Gaussian flux approach is so much more intuitive?

You've clearly never taught at A-Level or undergraduate level. Gaussian Flux needs too much grounding in vector calculus, more than the vast majority of students posess, to be truly called "so much more intuitive" I'd say.

Post edited at 23:45

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wintertree - * * on 10 Jun 2018

In reply to Paz:

> You've clearly never taught at A-Level or undergraduate level.

Clearly.

> Gaussian Flux needs too much grounding in vector calculus, more than the vast majority of students posess, to be truly called "so much more intuitive" I'd say.

Funny, but I teach it by drawing some circles and splatting some lines through them.

That covers flux and surface integrals, and is a far more intuitive way of understanding how the radial distribution of mass within a sphere has no effect on surface gravity than the reams of tedious and uninsightful calculus in shell theorem.

If you read my post I didn’t call for only teaching a flux based approach - as you say to fully use it instead of the Newtonian metohd you need vector calculus beyond A-level.

But I suggest than an hour spent thinking about the Gaussian flux approach with words and pictures helps develop far more intuition than learning the Newtonian maths does. Which is why I am surprised it’s not more common to see such a mixture, with a qualitative teaching of flux methods to help put pictures in people’s heads before going on to teach Newton’s stuff.

Post edited at 05:45

Robert Durran - * * on 10 Jun 2018

In reply to wintertree:

> Which is why this is so interesting - why is teaching at A-Level and undergraduate level so Newton centric when a Gaussian flux approach is so much more intuitive? Other than several centuries of precedent I don’t really know.

At A-level standard it will be normal to think of everything as point masses, so Newton gives a simple formula from which another simple formula for potential energy is easily derived, and that is all that is needed. A picture of Gaussian lines of flux might be nice, but a proper treatment involves vector calculus way beyond A-Level to get far with it, whereas Newton with point passes allows quite easy calculations about orbits and so on.

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wintertree - * * on 10 Jun 2018

In reply to Robert Durran:

> At A-level standard it will be normal to think of everything as point masses, so Newton gives a simple formula from which another simple formula for potential energy is easily derived, and that is all that is needed. A picture of Gaussian lines of flux might be nice, but a proper treatment involves vector calculus way beyond A-Level to get far with it, whereas Newton with point passes allows quite easy calculations about orbits and so on.

Sure I get all that. I’d hoped that was clear from my last post.

I’m not advocating the Gaussian approach for calculations at A-level.

(An asside to my main point - like I said before, one can prove the shell theory result using t without any calculus or maths so I would argue that you can in some cases get further with it with less maths)

But why is it an inverse square law for a point mass? Here is where a bit of time talking about flux theory and drawing some circles and spheres pays off. Bang - there’s an “aha” moment where the students understand why a physical law is as it is, rather than regurgitating equations and putting numbers through them.

One of those two options is learning physics, the other is memorising some equations. From my experience the A-level approach is excellent at producing students who can regurgitate reams of equations - so long as they recognise which one to use by the format of the question - but is very poor at developing much deep understanding or intuition as to what is going on.

There’s no vocational need to learn the Newtonian - or any - approach to gravity. Universities don’t need an intake with a large number of equations rote learned beyond the level of decent understanding. So why are we so equation led?

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Robert Durran - * * on 10 Jun 2018

In reply to wintertree

> .........the reams of tedious and uninsightful calculus in shell theorem.

Having spent an hour or two playing around with it yesterday, I wouldn't call it tecious. Indeed, hitting on an efficient approach which actually offered some insight into why everything simplifies (and indeed disappears when inside the shell) with some nice symmetry under an inverse square law was rather gratifying.

Robert Durran - * * on 10 Jun 2018

In reply to wintertree:

> But why is it an inverse square law for a point mass? Here is where a bit of time talking about flux theory and drawing some circles and spheres pays off. Bang - there’s an “aha” moment where the students understand why a physical law is as it is, rather than regurgitating equations and putting numbers through them.

Yes, I agree up to a point, but I don't think you need to introduce lines of flux to make it intuitive why nature has chosen an inverse square law. I think I would talk in general terms about sound and light spreading out from a point over an ever larger sphere so that the intensity follows an inverse square law to establish the general principle.

Anyway, I don't know of any fundamental reason why nature shouldn't have chosen, say, an inverse cube law for gravity, except for appealing to "beauty", but that is getting a bit philosophical.

> One of those two options is learning physics, the other is memorising some equations. From my experience the A-level approach is excellent at producing students who can regurgitate reams of equations - so long as they recognise which one to use by the format of the question - but is very poor at developing much deep understanding or intuition as to what is going on.

As maths (rather than a physics) teacher, I agree entirely!

> There’s no vocational need to learn the Newtonian - or any - approach to gravity. Universities don’t need an intake with a large number of equations rote learned beyond the level of decent understanding. So why are we so equation led?

Manipulating equations can also be fun and instructive in the power of mathematics - it is one of my favourites to combine Newton's gravity and the equations of circular motion to calculate periods of orbits, and, at a higher level, derive Kepler's laws from Newtonian gravity which, of course, was Newton's own triumph.

wintertree - * * on 10 Jun 2018

In reply to Robert Durran:

> Anyway, I don't know of any fundamental reason why nature shouldn't have chosen, say, an inverse cube law for gravity, except for appealing to "beauty", but that is getting a bit philosophical.

If your force carrier is radiated, and is conserved (because it has energy, and energy is conserved), the scaling is determined by the number of spatial dimensions alone.

There are other scalings out there but they rely on near field effects or on decay of the force mediator to something else that doesn’t mediate the force (eg the Yukawa potential).

Conservation laws have a solid basis in geometry and mathematics (Noether’s theorem).

Of course you could rephrase your line of thought as to why nature “chose” 3 spatial dimensions. That’s rather intangible to a dullard like me, apart from the cop out anthropic answer that stable orbits (and hence stuff) only exist with inverse square forces.

Post edited at 11:38

Robert Durran - * * on 11 Jun 2018

In reply to wintertree:

> > Anyway, I don't know of any fundamental reason why nature shouldn't have chosen, say, an inverse cube law for gravity, except for appealing to "beauty", but that is getting a bit philosophical.

So you are saying that anything other than an inverse square law would necessarily violate conservation of energy? I didn't know that and it sounds really interesting! Is it easily proved?

wintertree - * * on 11 Jun 2018

In reply to Robert Durran:

> So you are saying that anything other than an inverse square law would necessarily violate conservation of energy? I didn't know that and it sounds really interesting! Is it easily proved?

I’m hoping someone smarter than me chimes in.

It’s obvious as you’ve said before that light from a point must be inverse square due to conservation of energy. Light is radiation in the EM field aka photons. Call it flux or photons this is a conserved quantity.

Under quantum field theory, electric forces are mediated by “virtual” (a confusing term) photons in the EM field. These are subject to the same conservation laws and behaviours as “real” photons even if they only have “borrowed” energy under uncertainty.

It’s a massive liberty on my behalf to generalise the same behaviours to gravity given the lack of a complete quantum field theory of gravity - but it shows similar behaviour of virtual (fields) and real (waves) gravitons.

Good enough for me - but then my general relativity is crap bordering on non existent.

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Robert Durran - * * on 11 Jun 2018

In reply to wintertree:

Thanks. So it's pretty technical then........ To put it extremely mildly, I don't think my very rusty GR or quantum field theory would (or ever would have been) up to the details!

Dave Garnett - * * on 11 Jun 2018

In reply to Robert Durran:

> To put it extremely mildly, I don't think my very rusty GR or quantum field theory would (or ever would have been) up to the details!

Let alone my entirely fallible 'intuitively obvious' test!

wintertree - * * on 11 Jun 2018

In reply to Dave Garnett:

> Let alone my entirely fallible 'intuitively obvious' test!

Sure the full theoretical rabbit hole isn’t obvious, but the idea that a force is transmitted by some sort of Stuff, and that Stuff is concerned seems pretty intuitive? Not necessarily obvious for sure.

Dave Garnett - * * on 11 Jun 2018

In reply to wintertree:

> Sure the full theoretical rabbit hole isn’t obvious, but the idea that a force is transmitted by some sort of Stuff, and that Stuff is concerned seems pretty intuitive?

Yes, but my sense of the obvious has been somewhat dented by the revelation that gravity isn't a force.

I'm no longer even sure how quickly it travels. I heard someone say recently that the distortion of space-time caused by mass was instantaneous, so that if, say, black hole evaporated suddenly, we'd know about it immediately...

wintertree - * * on 11 Jun 2018

In reply to Dave Garnett:

> Yes, but my sense of the obvious has been somewhat dented by the revelation that gravity isn't a force.

I should say that the jury is still out on that question.

Is gravity a fundamental thing or does it say emerge from quantum mechanics below the plank scale?

If it is a fundamental thing, is object interaction described by a quantum field theory? If so I should say semantically “force” is accurate.

But we don’t know. From a GR view there are no gravitational forces, but GR is not reconciled with quantum mechanics yet.

Its to soon to proclaim if gravity is actually a force or not.

> I'm no longer even sure how quickly it travels. I heard someone say recently that the distortion of space-time caused by mass was instantaneous, so that if, say, black hole evaporated suddenly, we'd know about it immediately...

I should say they were wrong. Both flux (gravitational force) and radiation (gravity waves) travel at the speed of light.

Even if they were right, they’d be wrong. The mass/energy (particles and radiation) emitted by the evaporating black hole would expand in a shell around its centre at a maximum speed of that of light. Assuming we are far from said black hole, that mass acts on us as if located at the centre of that shell. Conservation of momentum tells us that must be where the black hole otherwise would have been.

Gravity travelling at a finite speed is related to how “gravitomagnetic” effects arise in the same way magnetism arises as a result of motion with respect to electric charges. Fascinatingly whilst this is derived from relativity, Oliver Heaviside largely discovered it before relativity was published.

I think Oliver Heaviside is the physicist from that era I would most like to meet.

Post edited at 12:41

Dave Garnett - * * on 11 Jun 2018

In reply to wintertree:

> I think Oliver Heaviside is the physicist from that era I would most like to meet.

Certainly he was in the right field with a name like that!

wintertree - * * on 11 Jun 2018

In reply to Dave Garnett:

> Certainly he was in the right field with a name like that!

He invented the Heaviside step function. People tend to forget his name when talking about the step function, although they remember Dirac’s name with the Dirac delta. A shame really as the later is purely derivative of Oliver’s work.

Pero - * * on 11 Jun 2018

In reply to Dave Garnett:

> Yes, but my sense of the obvious has been somewhat dented by the revelation that gravity isn't a force.

When you are standing on the surface of the Earth, there is only one measurable force acting on you: the ground pushing you up. There is no measurable force of gravity pushing you down.

Instead, the curvature of spacetime around a massive body like the Earth creates a compulsion to move towards the centre of the Earth almost exactly like the Newtonian inverse square force.

It's not unlike turning sharply in a car. You imagine that a (centrifugal) force is pushing you against the car door, but in fact there is only the car door pushing you in the opposite (centripetal) direction.

And, in fact, if you fall off a climb, then while you are falling there are no measurable forces acting on you (except air resistance). As you are falling you feel no force pulling you down. It's simply that the force that was holding you onto the rock has been released.

However, when we take the spacetime theory of gravity regarding the Earth, we are left with - to all practical intents and purposes - the same gravitational effects as in the Newtonian theory. There are fundamental differences in terms of, most notably, the curvature of light around a massive body and the extreme gravitation around a black hole. But, almost all practical applications still model gravity as the Newtonian force.

Post edited at 17:27

wintertree - * * on 11 Jun 2018

In reply to Pero:

I generally agree with your post, but to me the bit below isn’t indicative of the rest in the way you infer.

> When you are standing on the surface of the Earth, there is only one measurable force acting on you: the ground pushing you up. There is no measurable force of gravity pushing you down.

Consider two equal sign test charges/masses connected by an infinitely light spring seperated in the direction of a uniform electric field - the masses are accelerated equally by the field and the spring shows no compression - “it feels no force”.

Unopposed, uniformly applied acceleration from any known force can not be “felt” - gravity is no different in that sense. We tend to perceive it differently because (a) being apparently a monopole force it isn’t screened over large distances and (b) everything is obliged to interact with it.

I’m not saying that gravity is, or isn’t, fundamentally different to other forces, but I don’t like the approach of saying it’s different because you don’t feel it.

Post edited at 18:35

1

Dave Garnett - * * on 11 Jun 2018

In reply to Pero:

> the Earth creates a compulsion to move

Sounds a lot like a force to me! And why does the curvature of space-time create this compulsion and what is the frame of reference that gives it direction?

I get the centripetal/centrifugal thing but this is just sophistry isn't it? I thought the whole point of relativity was to point out that it all depends on your frame of reference...

Pero - * * on 11 Jun 2018

In reply to Dave Garnett:

> Sounds a lot like a force to me! And why does the curvature of space-time create this compulsion and what is the frame of reference that gives it direction?

> I get the centripetal/centrifugal thing but this is just sophistry isn't it? I thought the whole point of relativity was to point out that it all depends on your frame of reference...

The compulsion to move is the Lagrangian principle. Most, if not all, popular science authors and TV programs shy away from this - for reasons I don't really understand.

Lagrange was an 18th Century French-Italian mathematician. He reformulated Newtonian mechanics using the Lagrangian principle that nature acts as to minimise or maximise certain quantities. Newtonian mechanics, including his law of gravity, can be reformulated without the direct use of forces into something completely equivalent but based on the Lagrangian principle.

When you move to General Relativity, you have no force, but you still have the Lagrangian principle. Applying this to curved spacetime gives you not only the motion of particles but the paths of light rays through curved spacetime as well. Light rays are massless, so any attempt to model the motion of light rays using F = ma is impossible. But, as they do in classical optics, they obey the Lagrangian principle.

On your second point, some things depend on your frame of reference. But, if you were in a sealed laboratory orbiting the Earth, you would not be able to distinguish that from a laboratory floating in gravity-free space. There would be no local experiment to determine the "force" of gravity keeping you in orbit. Whereas, if you were being spun round on an ice-rink, say, your arms would feel the real force of the rope pulling you into circular motion. There would be a clear experiment to measure the force you were experiencing. The astronauts in the space-station feel no such force. They are, in fact, in an inertial reference frame and experience no external force that keeps them in their circular orbit.

PS on a more technical note, the external force experienced by something leads to what is called "proper" acceleration. This is an "invariant", which means it's the same in all reference frames: all reference frames agree on the proper acceleration of the space-station and someone standing on the surface of the Earth. The proper acceleration of the space-station is zero and the proper acceleration of someone on the Earth is g (upwards). This is different from the "co-ordinate" acceleration in the Earth's reference frame. But, coordinate acceleration is relative to your reference frame and not invariant.

Post edited at 20:34

Pero - * * on 11 Jun 2018

In reply to wintertree:

> I’m not saying that gravity is, or isn’t, fundamentally different to other forces, but I don’t like the approach of saying it’s different because you don’t feel it.

The theory of GR is clear: gravity is spacetime geometry and not a force. You don't have to like it, but there it is!

wintertree - * * on 11 Jun 2018

In reply to Pero:

> But, if you were in a sealed laboratory orbiting the Earth, you would not be able to distinguish that from a laboratory floating in gravity-free space. There would be no local experiment to determine the "force" of gravity keeping you in orbit.

Releasing 6 balls initially stationary to each other at equal distances either side of the origin along three orthogonal axes would distinguish that as well as telling you which way Earth was.

> Whereas, if you were being spun round on an ice-rink, say, your arms would feel the real force of the rope pulling you into circular motion. There would be a clear experiment to measure the force you were

Only because the rope only excerpts a force on a small part of the person. This isn’t an intrinsic property of the electrostatic forces behind the macroscopic perceived force, but of macroscopic systems build out of electrostatics in the presence of macroscopically balances positive and negative charges. If the rope was attached to and pulling on every atom in the ice skater, they wouldn’t feel it. The reasons this can’t happen are not intrinsic to the forces involved but to the systems we build from them.

Pero - * * on 11 Jun 2018

In reply to wintertree:

> Releasing 6 balls initially stationary to each other at equal distances either side of the origin along three orthogonal axes would distinguish that as well as telling you which way Earth was.

That's a measure of "tidal" gravity or spacetime curvature. Such an experiment, over a long enough period, would show that there is a local variation in spacetime curvature. But, it doesn't give you the classical gravitational force that keeps you in orbit. In classical physics the gravitational force of a circular orbit is F = GMm/r^2 etc. That force is not measurable. An orbit is entirely a coordinate acceleration. It's not a real (proper) acceleration. That's the fundamental difference between Newton and GR.

None of the balls feels any force or any proper acceleration. Each is on a separate (locally inertial) geodesic. Their gradual separation is not a result of any measurable force.

See also my comments about about the difference between coordinate and proper acceleration.

Post edited at 20:51

wintertree - * * on 11 Jun 2018

In reply to Pero:

> The theory of GR is clear: gravity is spacetime geometry and not a force. You don't have to like it, but there it is!

I don’t like it or dislike it. It’s clear that GR is a simplified limit of a quantum gravity theory, itself yet unknown to us. It emerges from something that demonstrably has to be quantum in nature at smaller scales. Small things have quantum behaviour/properties and they have gravity, and so their gravity must reflect their nature, be it uncertain or superposed.

So to me it is premature to accept our interpretation of GR as being “real”. Once this is understood then the largely semantic debate about calling gravity a force becomes worthwhile.

My post to which you replied and my 20:30 post are not challenging the interpretation of GR you give but are challenging what I see as your incorrect presentation of gravity as “not a force” because one can’t feel it. I challenge this not on grounds of GR, but on electrostatics and the observation that fundamentally electrostatics can’t be felt, but that real compound systems built out of actual stuff have more complex macroscopic behaviour emerge. But fundamentally a test electrostatic system also doesn’t “feel” the effect of a uniform electric field, only the reaction force from something resisting its acceleration.

wintertree - * * on 11 Jun 2018

In reply to Pero:

> That's a measure of "tidal" gravity or spacetime curvature.

I know, I agree and I don’t beleive I disagreed with anything you wrote in the message I am now replying to. It was a tangential point, but I wouldn’t want any UKC reader who ends up stranded in a box in zero g to think they couldn’t tell if they were in an orbit or not.

Then again perhaps I do disagree... Take 6 electrostatic charges of the same magnitude in the same configuration held apart from their origin by insulating springs. Place in orbit about a fixed electrostatic charge of the opposite sign. The resultant compression and extension of the springs will tell you the same thing as the differential motion of the balls, because they experience different forces due to the divergence of the field.

This is not exclusively a GR property, it’s the property of any non uniform force or field and behaves the same in all inverse square laws.

Sure - in GR there is no force, but using it as an example to justify the no force line does not ring true when again the same effect emerges from other forces.

Sorry if this seems like an incessant bug bear but using effects explicable by a classical field theory or by application of calculus to the inverse square law to justify GR as different and “not a force” seems wrong.

There are behaviours described only by GR and not Newtonian mechanics, such as gravitomagnetic effects like frame dragging. I don’t know if the limited subset of those applicable to linear motion emerges from considering SR and Newtonian mechanics, but clearly rotational effects can not.

Edit: I suspect you are well knowledgable in GR and you know that I am not (yet. It’s on my list. I may be to old to ever really become fluid with the time I can devote to it). My intention really is not to argue against GR - that would be a fools errand - but against the examples you are using to try and intuit that there is no force. I understand how they naturally confirm that when looking through the accepted interpretation of GR, but I suggest they confirm it but don’t require it in the way other Gr effects do.

Post edited at 21:07

Pero - * * on 11 Jun 2018

In reply to wintertree:

One fundamental problem with seeing gravity as an inverse square law is that this is a coordinate dependent description. Who is measuring the distance? There is no invariant distance in SR, let alone GR, on which to postulate an inverse square law.

In the rest frame of the Earth, in Schwartzschild coordinates, you have something close to the Newtonian energy equation, but it has an extra term, which causes the precession of the planetary orbits, among other things. And something close to an inverse square law.

But, in more general cases you cannot find a reference frame where everything is a superposition of inverse square forces, like you can in Newtonian physics.

What is invariant is the measurable spacetime curvature, which is the same in all coordinate systems. Or, you can describe the metric tensor, which transforms (as a tensor) between coordinate systems.

You will have trouble near and inside the event horizon of a black hole. If gravity were a force, then any force can be overcome and you could escape from inside the event horizon. But, the reason you cannot escape is that all valid paths through spacetime lead towards the centre of the black hole. You can blast yourself off in any direction with any force you like, but it only hastens your demise! Gravity is not a force to be overcome, it's the geometry inside the event horizon that leads inexorably to the singularity, in all directions.

Once inside the event horizon, there is simply no physical direction that leads away from the singularity. Not that there is an irresistible force.

Post edited at 22:24

1

wintertree - * * on 11 Jun 2018

In reply to Pero:

Excellent post. I agree with all of those effects as being beyond Newtonian theory - and some of them measurably so to high precision - in so far as I understand them (some more than others).

My point was simply that the examples you had been giving before do not differentiate GR from inverse square laws as they apply equally to classical field theories. The whole “you don’t feel gravity” argument does nothing to me to prove the GR view, any more than tidal forces do.

You’ve almost sold me on the side of “not a force” with the black hole view. Wormhole as hypothetical as they currently are challenge a “force” view even more. Then again the inside of a black hole is also quite hypothetical for hopefully the forseeable future.

Edit: Things have moved on since I last though about this. A framework has been built applying relativity principles to Newtonian mechanics and produces results compatible with GR for many of the areas you list - https://en.m.wikipedia.org/wiki/Relativistic_Newtonian_dynamics

Post edited at 22:57

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