## LDF as Gravity--Fringe stuff

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### LDF as Gravity--Fringe stuff

Okay, so for those of you who don't know about the London Dispersion Force, http://en.wikipedia.org/wiki/London_dispersion_force

To summarize: We know about an r-dependent always-attractive force between any two macroscopic objects composed of electric dipole-atoms, which should be roughly proportional to the number of charged particles in said objects. Maybe I'm missing something really key here, but it seems to me like if Newton had known about (kqq/r^2) and the charged structure of the atom before he tried to explain why things stick to the earth, he wouldn't have invented a separate "Stuff-dependent force" to account for it. Is it possible to construct a (perhaps temperature-dependent) model of a system that uses induced dipole-induced dipole attraction to explain gravitation?
Let's hash this out-- I want to hear your guys' retorts to the idea.
skolnick1

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### Re: LDF as Gravity--Fringe stuff

If Newton knew about these forces, then he probably would have known that the attraction of matter would not go down as 1/r^2, but rather 1/r^6. He would have seen that planetary orbits were unstable, and he would throw this explanation out. I'm not entirely sure exactly how this force is affected when rotation is added to the mix (the Earth and Sun are spinning while always being attracted to each other). The best-case scenario for attraction between molecules would be a permanent dipole which goes down as 1/r^3 iirc, and even then orbits would be unstable.
cyanyoshi

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### Re: LDF as Gravity--Fringe stuff

The r-dependence is based on the geometry of the field lines, yes? I think the best you can do is a cubed-dependence if.you model your dipole as a point particle. A dipole's field lines all curl around if the positive and negative charges are at the same location. If you account for the separation between the proton and electron's respective centers of charge, one little tiny field line is gonna pop off the end of that dipole.
It's gonna be really, really weak, like orders of magnitude weaker than anything else we see, but that force-vector arrow is going to point straight at a neighboring atom, and it's going to attract, whether you've got gravity in your system or not.http://www.chem.ufl.edu/~itl/2041_f97/matter/FG11_005.GIF is a particularly helpful visual for me
skolnick1

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### Re: LDF as Gravity--Fringe stuff

First, we know gravity has to be more-or-less 1/r in potential. Why? Closed, elliptic orbits. 1/r potentials are special, and have a hidden symmetry. That means that the Laplace-Runge-Lenz vectors for the elliptical orbits are nice and conserved, so the elliptical orbits are fixed. As soon as you go to a different central force, you destroy that symmetry in general major-axes of the elliptical orbits will precess.

So my question for you- can you get a 1/r force from induced multiple moments of a charged body? I claim its impossible (whats the monopole moment for an object thats neutral?)
SU3SU2U1

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### Re: LDF as Gravity--Fringe stuff

SU3SU2U1 wrote: (whats the monopole moment for an object thats neutral?)

Some of the folks here at IU are working on the neutron EDM experiment right now, actually.

But to address your point, I would argue that the r-dependence of the potential isn't some "special" intrinsic property that gives shape to the field lines. Your argument reduces to the fact that gravity appears to be a 1/r potential, which one typically associates with monopolar forces like electric charge, which are spherically symmetrical. However, an apparent symmetry is achieved if one considers a test dipole rather than a test + or - point charge.
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skolnick1

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### Re: LDF as Gravity--Fringe stuff

Some of the folks here at IU are working on the neutron EDM experiment right now, actually.

Yes, what neutrons MIGHT have are dipole moments. NOT monopole moments. The definition of neutral basically is no monopole moment.

Your argument reduces to the fact that gravity appears to be a 1/r potential, which one typically associates with monopolar forces like electric charge, which are spherically symmetrical.

No, that is not my argument. At long range, gravity IS 1/r. If it weren't 1/r, planetary orbits would be very different, as I described.

Now, electric MONOPOLES have 1/r potentials, so every higher moment will have a faster drop off at long range- that makes it impossible for gravity to be an induced multi-pole moment of the electric force.
SU3SU2U1

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### Re: LDF as Gravity--Fringe stuff

I'm not disagreeing that higher moments don't have a 1/r potential. I'm saying that in the proper orientation (the most energetically favorable), the electron's charge should effectively "shield" its proton's charge from the protons of neighboring atoms. Along the line that is the perpendicular bisector of the vector pointing from the positive to the negative charge, true, one sees no net "monopole moment". However, along the pointing vector itself (or any line with a component parallel to that vector) in either direction there is either a positive or negative apparent net charge, and a "monopole moment" is the sum of net charges in a system, yes?
skolnick1

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### Re: LDF as Gravity--Fringe stuff

However, along the pointing vector itself (or any line with a component parallel to that vector) in either direction there is either a positive or negative apparent net charge, and a "monopole moment" is the sum of net charges in a system, yes?

No. You are talking about a dipole-moment. Charge separation doesn't increase the net charge in the system, but it can increase the dipole moment. There is an apparent, induced dipole moment. The potentials you are talking about (induced dipoles) drop off like 1/r^6, and therefore cannot be gravity, which has to be 1/r, as discussed.
SU3SU2U1

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### Re: LDF as Gravity--Fringe stuff

SU3SU2U1 wrote:
However, along the pointing vector itself (or any line with a component parallel to that vector) in either direction there is either a positive or negative apparent net charge, and a "monopole moment" is the sum of net charges in a system, yes?

No. You are talking about a dipole-moment. Charge separation doesn't increase the net charge in the system, but it can increase the dipole moment. There is an apparent, induced dipole moment. The potentials you are talking about (induced dipoles) drop off like 1/r^6, and therefore cannot be gravity, which has to be 1/r, as discussed.

That makes sense. I'm not completely sure it's true, though. I have some handwaving that possible could be turned into an actual argument if it were to turn out there was nothing wrong with it.

First off, look at the ratio of electric to gravitational force. With 32 bits gravity is like ten billionth of the rounding error. It's tiny.

At first sight, a force that aligned dipoles wouldn't seem to be very effective as a stand-in for gravity. It's far stronger for close things, so distant charges wouldn't have much effect. Also speed of light would be important. If a distant force does manage to nudge dipole into the orientation that's best for attraction, by the time the attractive force from the nudged dipole reaches the original source it will be far too late for it to return the favor. The original dipole that affected it will have changed orientation many times by then.

But we're talking about tiny effects. (Here I wave my hands more vigorously.) If two dipoles line up together a tiny fraction more than they line up opposed, that could give you a force as strong as gravity. And with so many dipoles needed before you see an observable effect, that's a lot of room to allow some tiny correlations. It doesn't take a whole lot. And as skolnick1 pointed out, charges can sometimes screen other charges. Electric fields just add together with no interference, but electric charges don't. Maybe somehow (here I wave my arms very fast) all these dipoles could get a very slight correlation in orientations that is gravity. And maybe somehow it would be 1/r potential.
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J Thomas
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### Re: LDF as Gravity--Fringe stuff

J Thomas wrote: But we're talking about tiny effects. (Here I wave my hands more vigorously.) If two dipoles line up together a tiny fraction more than they line up opposed, that could give you a force as strong as gravity. And with so many dipoles needed before you see an observable effect, that's a lot of room to allow some tiny correlations. It doesn't take a whole lot. And as skolnick1 pointed out, charges can sometimes screen other charges. Electric fields just add together with no interference, but electric charges don't. Maybe somehow (here I wave my arms very fast) all these dipoles could get a very slight correlation in orientations that is gravity. And maybe somehow it would be 1/r potential.

But do not forget the most important point of the 1/r versus 1/r^6 arguments: While you can take two blobs of matter, rearrange the charges and get the exact same attraction between the two electrically as you would gravitationally, the measured force would only match for that one specific distance given that arrangement of charges. If the charges are held in place on the two objects then you would have the attraction between two dipoles which falls off with distance much faster then gravity does - any slight displacement from those relative positions would result in a force different then what would be expected from gravitational attraction. You move them a bit further from eachother and the force would be too weak, a bit closer and the force would be too strong. And if you changed their orientation, all of the sudden your faux-gravity would begin applying a torque!

Now, if you were able to freely manipulate the charge distribution of the electrically-neutral objects, you could artificially keep the electrical attraction at the same strength as you would expect from gravity: forcing the dipole moment to be smaller at close distances and larger at larger distances would indeed do the trick, as long as you kept the two dipole moments in the proper alignment. (This would, however, fall apart if you happened to have more then two objects: you could not change object A's dipole moment when you moved it in relation to object B without it also affecting some third object in a not-so-gravity-like manner.{except in a few very limited and cool cases})

But, it is important to remember, this is not how the London Dispersion Forces work - in fact it is quite the opposite. In general the closer two objects are, the larger the dipole moment created in each object is. That is why when you halve the distance between two neutral objects, the electrical attraction between the induced dipoles will increase by a factor much greater then a factor of four (which you would expect with gravitation or electric monopoles) or a factor of sixteen (which you would expect if you simply moved two dipoles closer to eachother). When you halve the distance between two neutral objects, charges redistribute on them which will create stronger dipoles, increasing the net attractive force by a factor of 64!

It would take some ridiculously thorough control over charges just to simulate gravity's attraction between two neutral objects with electrical forces alone, and would be quite impossible should you happen to bring a third object into play.

J Thomas wrote:Also speed of light would be important. If a distant force does manage to nudge dipole into the orientation that's best for attraction, by the time the attractive force from the nudged dipole reaches the original source it will be far too late for it to return the favor. The original dipole that affected it will have changed orientation many times by then.

I do not quite follow you here, the LDF's that we can measure do rely on the electric fields which travel at the speed of light. Are you talking about molecular distances (a fair number of molecules have rotational frequencies in the gigahertz, and so distances up to a few centimeters would easily allow for light to travel from one molecule to another to relay information) or are you talking something much larger?
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Jakell

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### Re: LDF as Gravity--Fringe stuff

Jakell wrote:
J Thomas wrote: But we're talking about tiny effects. (Here I wave my hands more vigorously.) If two dipoles line up together a tiny fraction more than they line up opposed, that could give you a force as strong as gravity. And with so many dipoles needed before you see an observable effect, that's a lot of room to allow some tiny correlations. It doesn't take a whole lot. And as skolnick1 pointed out, charges can sometimes screen other charges. Electric fields just add together with no interference, but electric charges don't. Maybe somehow (here I wave my arms very fast) all these dipoles could get a very slight correlation in orientations that is gravity. And maybe somehow it would be 1/r potential.

But do not forget the most important point of the 1/r versus 1/r^6 arguments: While you can take two blobs of matter, rearrange the charges and get the exact same attraction between the two electrically as you would gravitationally, the measured force would only match for that one specific distance given that arrangement of charges. If the charges are held in place on the two objects then you would have the attraction between two dipoles which falls off with distance much faster then gravity does - any slight displacement from those relative positions would result in a force different then what would be expected from gravitational attraction. You move them a bit further from eachother and the force would be too weak, a bit closer and the force would be too strong. And if you changed their orientation, all of the sudden your faux-gravity would begin applying a torque!

I have no idea how to handle that. I have to wave my arms and suppose there's a way that averaged over many trillions of dipoles which are typically unaligned but which randomly align a little more than expected, it could work out.

Now, if you were able to freely manipulate the charge distribution of the electrically-neutral objects, you could artificially keep the electrical attraction at the same strength as you would expect from gravity: forcing the dipole moment to be smaller at close distances and larger at larger distances would indeed do the trick, as long as you kept the two dipole moments in the proper alignment. (This would, however, fall apart if you happened to have more then two objects: you could not change object A's dipole moment when you moved it in relation to object B without it also affecting some third object in a not-so-gravity-like manner.{except in a few very limited and cool cases})

But, it is important to remember, this is not how the London Dispersion Forces work - in fact it is quite the opposite. In general the closer two objects are, the larger the dipole moment created in each object is. That is why when you halve the distance between two neutral objects, the electrical attraction between the induced dipoles will increase by a factor much greater then a factor of four (which you would expect with gravitation or electric monopoles) or a factor of sixteen (which you would expect if you simply moved two dipoles closer to eachother). When you halve the distance between two neutral objects, charges redistribute on them which will create stronger dipoles, increasing the net attractive force by a factor of 64!

So London dispersion would be part of what's left over after you take out the part that fits gravity.

It would take some ridiculously thorough control over charges just to simulate gravity's attraction between two neutral objects with electrical forces alone, and would be quite impossible should you happen to bring a third object into play.

J Thomas wrote:Also speed of light would be important. If a distant force does manage to nudge dipole into the orientation that's best for attraction, by the time the attractive force from the nudged dipole reaches the original source it will be far too late for it to return the favor. The original dipole that affected it will have changed orientation many times by then.

I do not quite follow you here, the LDF's that we can measure do rely on the electric fields which travel at the speed of light. Are you talking about molecular distances (a fair number of molecules have rotational frequencies in the gigahertz, and so distances up to a few centimeters would easily allow for light to travel from one molecule to another to relay information) or are you talking something much larger?

I'm talking about much larger, like planetary or galactic distances. But it would need to work down to centimeters if gravitation gets measured accurately down to centimeter distances, and the answers come out right for gravity.

LDFs would result in changed correlation for dipoles that are close together, which I'd expect would effectively reduce the population size for statistical effects at larger distances.

I couldn't begin to make a real argument for this vague idea. But I think it might take a whole lot of work to demonstrate conclusively that it could not happen. Or maybe it would be easy.
The Law of Fives is true. I see it everywhere I look for it.
J Thomas
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### Re: LDF as Gravity--Fringe stuff

Here's the thing; using k-Q-Q/r^2, if you just model the attractive and repulsive forces between four electric monopoles (2 of each charge) in a plus-minus/plus-minus orientation, (not counting the repulsion an electron should feel from its "own" proton) the leading term in the taylor expansion of the resultant equation looks exactly like a 1/r^2 force when the distance between the two dipoles is 0 (relative to charge separation) and looks exactly like 1/r^4 when the distance between the two is infinite (again relative to charge separation). What do you think of the idea that gravity is a 1/r^2 force at comparatively small distances that scales up as we go outward?
skolnick1

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### Re: LDF as Gravity--Fringe stuff

if you just model the attractive and repulsive forces between four electric monopoles (2 of each charge) in a plus-minus/plus-minus orientation, (not counting the repulsion an electron should feel from its "own" proton) the leading term in the taylor expansion of the resultant equation looks exactly like a 1/r^2 force when the distance between the two dipoles is 0 (relative to charge separation)

In this case, you really don't want to use a dipole expansion- when the distance between the dipoles is 0, you have two negative charges on top of each other, and two positive on top of each other. Thats just two monopoles repelling, and not at all the situation you want to be describing.

What do you think of the idea that gravity is a 1/r^2 force at comparatively small distances that scales up as we go outward?

Its not, its 1/r^2 out to very large scales. Otherwise planets would not orbit in closed ellipses.

I have no idea how to handle that. I have to wave my arms and suppose there's a way that averaged over many trillions of dipoles which are typically unaligned but which randomly align a little more than expected, it could work out.

It can't. Electric fields are linear- so averaging a bunch of dipoles can get you either a dipole or a higher multipole.
SU3SU2U1

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### Re: LDF as Gravity--Fringe stuff

SU3SU2U1 wrote:
I have no idea how to handle that. I have to wave my arms and suppose there's a way that averaged over many trillions of dipoles which are typically unaligned but which randomly align a little more than expected, it could work out.

It can't. Electric fields are linear- so averaging a bunch of dipoles can get you either a dipole or a higher multipole.

That sounds plausible to me. But I don't know very much about this stuff, would you mind helping me with a beginner's question?

Say you have a 2 cm nylon ball and a 2 cm polyurethane ball. You put a unit positive charge on the surface of the nylon ball, and a four-unit negative charge on the surface of the polyurethane ball.
You arrange them so their centers are 5 cm apart, and the charges are not strong enough to spark across the gap.
How strong will be the electric field 5 cm from the opposite side of the nylon ball? 10 cm? 15 cm?

I have the impression this should be easy. The charges on each ball behave exactly like they are a point charge at the center of the ball. The superposition principle says the charges and the balls each have no effect on the electric fields produced by the other ball. So the result should be

5 cm: 1 - 1 = 0
10 cm: 1/4 - 4/9 = -7/36
15 cm: 1/9 - 1/4 = -5/36
20 cm: 1/16 - 4/25 = -39/400

From the intro textbook that looks just right, except it seems like they did it with metal balls and they talked about the charges quickly distributing on all surfaces. Is this right? I know the superposition principle works fine through empty space. Can I depend on it to work through charged plastic balls too?
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J Thomas
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### Re: LDF as Gravity--Fringe stuff

SU3SU2U1 wrote:In this case, you really don't want to use a dipole expansion- when the distance between the dipoles is 0, you have two negative charges on top of each other, and two positive on top of each other. Thats just two monopoles repelling, and not at all the situation you want to be describing.

I think the breakdown in understanding here happens because you keep trying to approximate your dipole as a point halfway between your charges. I'm modelling it as four monopoles; I was referring to the distance between the two middle charges being 0.

If electric fields are linear, then adding monopole forces along a straight line shouldn't give you anything other than a weak, weak monopole-looking force, yes? Obviously the electrons aren't always in a straight line with the protons--kinetic energy in the system effectively randomizes their positions above a threshold temperature. But below that, they have some r-dependent predisposition to appear in that attractive orientation rather than its repulsive counterpart. And if this system of particles is relatively free to swivel and orient as it pleases, it forms a force whose most apparent rate of change at this scale is 1/r^2.
skolnick1

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### Re: LDF as Gravity--Fringe stuff

I think the breakdown in understanding here happens because you keep trying to approximate your dipole as a point halfway between your charges. I'm modelling it as four monopoles; I was referring to the distance between the two middle charges being 0.

Yes, I'm using standard terms, and you are making yours up as you go. In this new orientation you now have a negative charge sitting right on top of a positive charge, so you still only have a monopole interaction. This is NOT the situation you want to model with gravity (the distance between the Sun and the Earth is significantly larger than there radii).

If electric fields are linear, then adding monopole forces along a straight line shouldn't give you anything other than a weak, weak monopole-looking force, yes?

No. Please look at the standard monopole expansion! The dominant 1/r^2 term can cancel out (if there is no monopole moment) and you are left with 1/r^3 or higher.

. And if this system of particles is relatively free to swivel and orient as it pleases, it forms a force whose most apparent rate of change at this scale is 1/r^2.

No! if there is no net charge on a system the apparent rate of change is not 1/r^2. Seriously, you need to sit down with the general multipole expansion and really try and understand it.
SU3SU2U1

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### Re: LDF as Gravity--Fringe stuff

I've figured out what it is I'm trying to say--Think of the dipole as a coin. It's got a plus and a minus to it, representing the positive and negative charge in a Hydrogen atom.
The way you're modelling it, as a 1/r^3 force, assumes no charge separation. It's a coin of zero thickness, and so your charges are directly "On top" of one another. In this model, you're right, there's no way you can orient the coin to get a better result than 1/r^3 on a test-charge or 1/r^6 if another dipole is involved.
Give some thickness to your coin, though. Space out your positive and negative charge, then zoom way in. Shrink down and stand on top of that atom; on top of the electron. Opposite you, there's a proton, right? But its charge is completely shielded by the electron. Equivalently, most of the electric field from that electron gets kinda "Used up" or "Sucked in" in the process of shielding you. However, since you're closer to that electron than to the proton you still see a net negative charge, an apparent "Monopole Moment". Which means that it'll look like a monopole (1/r) potential if r is 0 (or "Small" relative to your charge separation) and it'll look like a dipole potential (1/r^3) if r is infinity (or "Arbitrarily large").
The distance between the earth and the sun is only about a hundred times the sun's radius. Is that arbitrarily large? 100 times the radius SOUNDS like a lot, but you have to remember that 1/100 is 10^-2, and gravity is on the order of 10^-11 relative to your other forces.
I'm still not convinced.
skolnick1

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### Re: LDF as Gravity--Fringe stuff

The way you're modelling it, as a 1/r^3 force, assumes no charge separation. It's a coin of zero thickness, and so your charges are directly "On top" of one another. In this model, you're right, there's no way you can orient the coin to get a better result than 1/r^3 on a test-charge or 1/r^6 if another dipole is involved.

No, it assumes the separation between charges is small compared to the lengths you care about. In practice an order of magnitude or so is fine. Its also a controlled approximation, you can easily take higher terms in the expansion. Your non-zero thickness coin will have many non-zero moments, but each of them is higher than monopole, so its a sum of pieces, but the higher order terms shrink faster than 1/r^2.

Give some thickness to your coin, though. Space out your positive and negative charge, then zoom way in. Shrink down and stand on top of that atom; on top of the electron. Opposite you, there's a proton, right? But its charge is completely shielded by the electron.

I don't think you understand what shielded means. Fields are linear, they just add. You'll never see a pure monopole field.

The distance between the earth and the sun is only about a hundred times the sun's radius. Is that arbitrarily large? 100 times the radius SOUNDS like a lot, but you have to remember that 1/100 is 10^-2, and gravity is on the order of 10^-11 relative to your other forces.

Yes, gravity is weak, but its a weak 1/r^2 squared force. Any explanation of gravity has to explain its 1/r^2 nature, first and foremost.

I'm still not convinced.

Thats because, near as I can tell, you've made roughly 0 effort to actually work through the details yourself. I've outlined a general proof that your proposal is impossible- prove me wrong. Show me a charge configuration for a system where two neutral objects of size R separated by a distance 100R have a force has the right properties.
SU3SU2U1

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