Pressure with 1/d² forces

neufneufneuf

Hi,

It's a theorical study. Balls are free to move. Balls repuls themselves like a force function of 1/d². The black object is free to turn and it's neutral for balls. Balls are under fixed pressure. All volumes are constant. Sure, pressure must be aqual everywhere, but if I think about a ball at contact with black object and for know the pressure at each point I need to calculate all forces from all others balls, because even force decrease like d² there are a lot of balls and for me the pressure depend of all balls. Like the object is no symetrical I don't know how pressure can be compensate in each point. But if the pressure is not the same the torque on black object must be to 0. What's wrong with my thoughts ?

I give a solution of pressure if balls are particle point, with Gauss law it's possible to see the pressure is not the same everywhere.

Can you help to find my error ?

Thank you in advance

dlouth15

Is the [latex]1/d^2[/latex] force the only force between the particles? Or do the particles also bounce off each other in the way they might in a gas?

The object at the centre is neutral. What are the forces between the particles and this object? In a normal gas the pressure comes from the change in momentum of the particles as they bounce off the gas. Is this happening here or does it come from the [latex]1/d^2[/latex] force?

Do the particles have any velocity?

neufneufneuf

No temperature here, no velocity, just pressure but all volumes are constant

yes, force 1/d² is between the particles only, walls and object are neutral

I think like a theorical problem, I imagine particule with a charge and particle interact with all particles in the system, in a gas: pressure come from temperature and not from 1/d² forces and don't interact with all particles, for me it's not the same case

I think the presence of the black object itself change the local pressure because it changed the sum of forces, I don't know how to think about this problem

dlouth15

When you say neutral what do you mean? Do you mean that the [latex]1/d^2[/latex] force doesn't apply between particles and the object?

neufneufneuf

black object is not charged

dlouth15

Ok thanks. I've got a picture now of what is going on.

dlouth15

OK, here's my thinking.

In order for the particles to exert pressure on the object in the centre they must be free to move. If they are free to move, however, they will make their way to the inside surface of the containing sphere (in the same way that free electrons in a conductor migrate to the surface of the conductor). This means that there's now no field within the cavity of the sphere and no particles to exert pressure on the object within. The only pressure is the outward pressure of the particles on the inside wall of the containing sphere.

neufneufneuf

Do you open "im2" image in my first message ? The pressure in circle is not homogenous it's function of (x²,x). If a lot of particules are placed in the circle the pressure change with radius, no ? It's not possible to put infinite particles at the inside wall (only one layer ?) of the containing sphere (or part of cylinder: easier to resolve). For me, particules are placed with one layer, but with limited number, when this layer is full, a second layer is placed, etc.

dlouth15

Right, so the particles are all jammed up against one another filling the entire sphere? I was treating them as point charges. I will look again at your mathematics.

neufneufneuf

yes, the sphere is full of particules. Thanks

dlouth15

OK, assuming that all the pressure is due to the [latex]1/d^2[/latex] force, what I get is that the pressure is zero at the centre of the sphere but increases in proportion to [latex]r^2[/latex] as we move out towards the edge.

i.e. [latex]P\left(r\right)=\frac{2}{3}\,\rho\pi\gamma r^{2}[/latex]

neufneufneuf

yes, 0 in the center and function of r², and if an object is added (object is not charged), this object change the local pressure because it take place (volume where there are no particule) and I don't know how the torque can be to 0.

dlouth15

I don't think the torque would be zero. If we can assume that the charged particles can move around one another whilst maintaining contact in the way that the particles of a liquid might, then the neutral object will tend to migrate towards the centre as the charged particles try to move towards the edge of the sphere. So there will therefore be a force acting on the neutral object and this force will vary according to the way it is oriented in the container in relation to the centre of the sphere. Different parts of the object will experience a different force. Therefore there will be a non-zero torque. What do you think?

If the object is allowed to settle in equilibrium at the centre, then the torque will disappear. The system will have settled into it's lowest energy state. But in the diagram you have drawn I would expect a torque.

neufneufneuf

The object can only turn like a hand of a watch and yes, for me it's logical there is a torque.

I imagine another object: an asymetrical thread (like asymetrical screw) like image show. Here the axis is not the same, the thread can only turn around itself.

But it's not possible to have a torque because in this case the object can give energy, so where energy come from ?

dlouth15

Sorry, I misunderstood the first image, im1.png. I don't think in this case there will be a torque about the point indicated for the reason that this point coincides with the centre of the sphere. The reason being that due to symmetry, there's no change in the relationship between the object and the overall charge distribution no matter how you rotate the object about the central point.

Where you would get a torque would be if the pivot point is somewhere off centre. Then a rotation will change the relationship between the object and the charge distribution.

It is the same with the helical screw. It can rotate about the axis but there's no overall movement towards the centre so there's no torque.

It is really the field created by the charged particles that is important here; not the pressure. If you imagine a small neutral sphere of volume [latex]V[/latex]. Then this small sphere will displace [latex]\rho V[/latex] overall charge.

If the field is a function of the distance from the centre [latex]r[/latex] and is given by [latex]E\left(r\right)[/latex], then it will move towards the centre with a force of [latex]E\left(r\right)\rho V[/latex].

The field [latex]E\left(r\right)[/latex] is linearly proportional to the distance from the centre so [latex]E\left(r\right)=kr[/latex], where [latex]k[/latex] is a constant. So the force will be [latex]kr\rho V[/latex]

So we get a torque where we have some sort of extended object like a rod, with one end a bit closer to the centre than the other. Since the field is stronger away from the centre we will get a stronger force on the end further away from the centre than the end which is nearer and therefore we will get a torque. The rod will tend to align itself so that it is pointing radially away from the centre.

neufneufneuf

The charge displaced by small sphere, change all configuration of particles in the sphere not only at local volume. If all particles are stable I can calculate pressure by adding all forces from all others particles, no ? why it's not possible to calculate like that ? Because if I use this method, the black volume change the local pressure.

Take a big sphere (black object) with center at R/2 (see r2.png), for me it's logical the volume change the local pressure, the sum of forces are very different, sure no torque in this case but just for show the difference of pressure. If you're right, when you calculate the sum of forces what do you have ?

dlouth15

It would be very complicated to calculate the sum of forces on r2.png but I think it is fair to say that the pressure on the top of the black sphere will be greater than the pressure on the bottom and therefore the overall force will be towards the centre. We know that in the absence of any black sphere there's a pressure proportional the square of the distance from the centre. The presence of the sphere will change this relationship but there would still be greater pressure on the top than on the bottom and so an overall downward force.

If you wanted to calculate pressures around the black sphere I doubt if there's a neat analytical solution but you could do it numerically. You could do it numerically for the black object in im1.png also, but I think you would find that the sum of forces add up in such a way that there's no torque if the object is attached a the centre of the containing sphere.

Going back to r2.png, another way of explaining it is that the particles want to distribute themselves around the edge of the container and the way to do this is to have the black sphere migrate to the centre. Applying this same principle to the object in im1.png, can you see that regardless of which way this object rotates around the fixed point at the centre, there's no net movement of charged particles towards the edge which is what they want to do. Hence there's no torque. It is really just the exact same configuration of charge but viewed from a different angle.

Rotating the helix (or indeed any object) about an axis which runs through the centre of the containing sphere also does not change anything. Again, it merely results in essentially the same system but as viewed from a different angle. Therefore no torque.

So without any numerical calculation of net forces due to pressure, we already know that there will be no torque in either of these cases.

neufneufneuf

Tanks. Sure, if torque exist scientists saw it before. But I like calculations for watch what's happen with particules when there is an object. I would like to do a numerical solution for im1.png but the density of particules change with pressure and I don't know how to take this parameter for calculate, you know ?

dlouth15

Well I think we're dealing with a uniform charge density, therefore the density of particles itself will also be uniform since each carries the same charge. I was treating it as a sort of uniformly charged liquid of uniform density.

neufneufneuf

If the pressure increase with r², how the density can be the same everywhere ?

dlouth15

neufneufneuf wrote: »

If the pressure increase with r², how the density can be the same everywhere ?

I think we agreed we're not talking about a gas here but charge carrying balls that are pressed up against one another filling the container. The density is therefore uniform throughout and therefore also the charge density is also uniform.

You don't need to have the density varying for the pressure to vary. The outermost layer of balls are under the most pressure because not only is that layer pressing against the wall of the container but every layer further in is also contributing to that pressure.

It is like in water the pressure increases as you go down but the density remains more or less constant. This is because each higher layer of water contributes to the pressure below. The further down you go the more layers above exist to contribute.

In fact the [latex]r^2[/latex] pressure function depends on the density being constant.

neufneufneuf

We're agreed, it's not a gas, this is particles with a charge. The pressure in water is not the same because water is not compressible, here particles are compressible, no ? And even in water, big pressure change very very few the density because in practise water is a little compressible, no ? I imagine space between particle, not you ?

Maybe it's because I'm thinking with sum of forces in each particle like image r3.png, if I do the sum of force at external point I have more force in one direction. If I do the sum of forces near center, a big part of forces cancel themselves, so the pressure is near 0 at the center. What do you think about that ?

I'm thinking of the sum of forces in all the system (particles+object+circle), sum must be to 0. If pressure is not changed by object, this would say the sum of forces of all the system is not 0.

Lbeard

dlouth15 wrote: »

OK, here's my thinking.

If they are free to move, however, they will make their way to the inside surface of the containing sphere (in the same way that free electrons in a conductor migrate to the surface of the conductor).

This means that there's now no field within the cavity of the sphere and no particles to exert pressure on the object within. The only pressure is the outward pressure of the particles on the inside wall of the containing sphere.

I think you're thinking of electrons on a charged sphere. They are not going to be in the volume if they are electrons. They will be on the surface.

Lbeard

neufneufneuf wrote: »

We're agreed, it's not a gas, this is particles with a charge.

Okay, and the pressure is the force of repulsion the charges are experiencing?

Maybe it's because I'm thinking with sum of forces in each particle like image r3.png, if I do the sum of force at external point I have more force in one direction. If I do the sum of forces near center, a big part of forces cancel themselves, so the pressure is near 0 at the center. What do you think about that ?

From the way I'm looking at it, if the sphere has no charge, and is neutral, the charged particles have no momentum, then they will be arranged like atoms in a crystal. They will all be equidistant from each other (you can also think of them as point particles)


I'm assuming an initial condition that when you place the balls in the sphere they will move into position.

The force between all the point particles should be constant, therefore the distance should be constant. - I'm assuming this, I can't tell why it would be any different.

I'm thinking of the sum of forces in all the system (particles+object+circle), sum must be to 0. If pressure is not changed by object, this would say the sum of forces of all the system is not 0.

If nothing is moving, the sum of forces is zero.

If the object has no charge and has the same permitivity as the emptiness of the volume, it shouldn't interact will the balls unless it collides directly with them.

I'm a little confused about charge density. More I'm a bit fuzzy on it, if there is a gradient between particles, or if there's another way of looking at it, where the charge density is considered to be uniform. If there were an infinite number of points in the sphere the distance between each would be infinitesimally small, so there would be a uniform charge distribution, but I think that's a cheat. I'm not sure.

dlouth15

neufneufneuf wrote: »

We're agreed, it's not a gas, this is particles with a charge. The pressure in water is not the same because water is not compressible, here particles are compressible, no ?

The pressure function of [latex]P(r)=r^2[/latex] depends on density being constant. Remember that you came up with this function in the first place. In your initial analysis you have a constant [latex]\rho[/latex] standing for charge density throughout the system. This means that the density of particles must also be constant.

And even in water, big pressure change very very few the density because in practise water is a little compressible, no ?

You don't need any compression in a liquid whatsoever for there to be a big pressure change.

I imagine space between particle, not you ?

In the system you describe, if there's any space between the particles initially, that will disappear as the particles will pile up against the inside wall of the container leaving a void in the middle.

The only thing that makes sense is a sort of incompressible liquid of charged particles. The density remains constant but particles can still move to allow for motion of the neutral object inside.

dlouth15

Lbeard wrote: »

I think you're thinking of electrons on a charged sphere. They are not going to be in the volume if they are electrons. They will be on the surface.

Yes but that was because I was not clear on the system being described at that point. I'm still not totally clear but see my later posts where I discuss the alternative which is incompressible balls of charge which fill the volume but can move about one another behaving a bit like an incompressible liquid. This is the only scenario I think which is consistent with a pressure function which is proportional to [latex]r^2[/latex].

neufneufneuf

Thanks for your help

Okay, and the pressure is the force of repulsion the charges are experiencing?

Yes, imagine a lot of particule in the sphere. No temperature, no velocity, no gravity here.

A/ With hypothesis:

Particle is like point. Object is neutral (no charge) like walls. Particle are not compressible. I put a lot of particles inside the volume. If I resume:

1/ Pressure change like r² ?
2/ Particles are at equidistant from each others everywhere, density is constant ?
3/ If I add an object, this don't change the pressure at sphere wall ?

Lbeard

neufneufneuf wrote: »

Yes, imagine a lot of particule in the sphere. No temperature, no velocity, no gravity here.

Yeah, anything you take out, you don't have to worry about. But remember anything you remove is removed. The sphere is neutral, and has no charge - it is just a physical barrier. The object you've place in there, also has no charge, it shouldn't have any effect on the particles with charge unless it comes into direct contact with them. If the object had a permitivity, - the vacuum actually has a permitivity, everything has - just off the top of my head, someone should correct me if I'm wrong, an object with a different permitivity if you move it around inside the sphere, the balls will move, because the path of the field lines will be longer. And this will change the displacement of the balls.

Particle is like point. Object is neutral (no charge) like walls. Particle are not compressible. I put a lot of particles inside the volume. If I resume:

I think the word you're looking for there is presume, instead of resume. Sorry, I wouldn't normally correct anyone, it's just presume and resume have two very different meanings. Resume means to start, or start again. To assume, means you have a bunch of ideas you believe to be true. Presume is a lot like assume, but it kind of means before you start, you have a bunch of ideas you believe to be true.

1/ Pressure change like r² ?

If the pressure is the force between the particles, yes. But, I'll explain something important. Don't use the term 'like' . The pressure change is proportional to r². The word proportional is very important. It means there is a constant relationship between the two values. But proportional just by itself does not tell you whether one gets bigger as the other gets smaller, or if as one gets bigger the other gets bigger (and vice versa). For this we have to qualify the proportionality as either direct, or inverse.

If I know there's a proportionality, I can relate one variable to another using a constant of proportionality - For the purpose of demonstration, I'll call my constant of proportionality k.

If the pressure gets bigger as the square of the distance; r, gets bigger, and smaller when it gets smaller, then we can say there is a direct proportionality between the values. And the equation would be:

p = kr²

If the pressure gets smaller as the the square of the distance; r, gets bigger, there we can say there is an inverse proportionality between the values. And the equation would be:

p=k(1/r²) or p = k/r²

Without knowing the precise details, if I know there is a force between two bodies, and the bodies are in three dimensional space, and I know it's to do with their size (volume, mass, some property), and I know it's inversely proportional to their distance.

Then I can write the formula Force = k (X1.X2)/r² That's Newton's law of gravitation, and a few more formula.

Why the distance is squared.........You'll have to look for a formal proof elsewhere but it is because of a square. It's for geometric reasons. The geometry of three dimensional space. Take a point particle with some kind of charge that creates a force. Draw a sphere around the point particle, with a distance (or radius r). Then in the surface of the sphere, cut a square. Fill the square with evenly spaced dots, equidistant from each other - these will represent the density of the charge - and these are all the dots you're getting. Then draw another sphere around the point particle with twice the radius. You cut another square in this sphere, the vertices (corners) of this square will be from straight lines, drawn from the point particle through the corners of the first square. Take your dots from the first square and place them equidistantly from each other - the distance between them will be greater.....The density will be less. If you look at all the geometry, you will see the density is inversely proportional to the square of the radius.

2/ Particles are at equidistant from each others everywhere, density is constant ?

Density in distribution through out the sphere, as far as I am aware, is constant.

Between the point particles the charge density will not be constant. The charges are repelling each other. To bring them together requires energy, if they're not moving it means that all the energy is stored as potential energy. By the minimum potential energy principle, the points will arrange themselves in the sphere to minimise their potential energy - the barrier of the sphere wall is there, if it weren't the point particles would try to get as far away from each other as possible. Instead they will get as far away from each other as they can. To mimimise the potential energy the point particles must minimise the charge density between each other, since all point particles have the same charge, they will be evenly distributed from each other. The potential difference between all points will be zero. (with electricity the potential difference is also known as the voltage)

Electrons on the surface of a charged sphere will follow the same principle - the charge will evenly distribute itself, there will be no potential difference between any two points on the sphere.

3/ If I add an object, this don't change the pressure at sphere wall ?

If the object has permittivity and that is a factor, then moving the object around inside the sphere will change the pressure at the sphere wall. If however you discount permittivity, giving the object the same permittivity as the vacuum, then there should be no pressure change at the sphere walls.

Lbeard

Now a question I'd like to know the answer to.

We can say if the charges particles have reached the minimum potential, and they're not moving, then we can say they've condensed.

Now, I want to just ignore the polarity of the particles. So, we're not necessarily talking about the real world.

If a charge wave enters the sphere. Forgetting scattering and reflections, the ripple passed through the charge densities of the particles, they move, but they have mass. What is the formula that relates the mass to the wave speed? I have a rough idea but I'd like to get a better perceptual and intuitive purchase on it.

neufneufneuf

Thanks for your help, really

I think the word you're looking for there is presume, instead of resume.

I would like to be sure if I understood all previous messages

means you have a bunch of ideas you believe to be true

it's only my intuition (the problem for me is very complex), I don't want to believe wrong results, I would like the true solution

giving the object the same permittivity as the vacuum, then there should be no pressure change at the sphere walls.

Take this example: I consider permittivity of black sphere like vaccum and I fixed black sphere like r2.png (the object can't move). If pressure at outer cirle (sphere in 3d) is constant, the sum of forces on outer wall is 0 but the black sphere receive a force in one direction because pressure is inverse proportionnal at r². So, the sum of forces of all the system is not 0, it's for that I presume the pressure is not the same everywhere on circle wall. I forget something ?

then we can say they've condensed.

If a charge wave enters the sphere. Forgetting scattering and reflections, the ripple passed through the charge densities of the particles, they move, but they have mass. What is the formula that relates the mass to the wave speed? I have a rough idea but I'd like to get a better perceptual and intuitive purchase on it.

Sorry, I don't understand

Lbeard

neufneufneuf wrote: »

Take this example: I consider permittivity of black sphere like vaccum and I fixed black sphere like r2.png (the object can't move). If pressure at outer cirle (sphere in 3d) is constant, the sum of forces on outer wall is 0 but the black sphere receive a force in one direction because pressure is inverse proportionnal at r². So, the sum of forces of all the system is not 0, it's for that I presume the pressure is not the same everywhere on circle wall. I forget something ?

Okay, if in r2.png you haven't drawn in the charged particles. If they're there I know what is going to happen, next.

the black sphere receive a force in one direction because pressure is inverse proportionnal at r²

No, it receives a net force in one direction. This is the particles pressing against it - it is a physical barrier.........inverse proportionality just means the force from an individual charge decreases with distance.

So, the sum of forces of all the system is not 0

Okay. If nothing is moving in a system then it means all forces are canceling out. If any forces do not cancel out, then you will have movement, the system will try to rearrange itself until all forces cancel out. When all forces sum to zero, the movement stops. Potential energy is released when a potential barrier is removed - that is an equal and opposing force - in 3d space the forces are in 3 dimensions.

If the sphere is empty apart from the particles with charge, they will equally distribute themselves through out the sphere.

If you add a neutral sphere, like in r2.png, in the position it is in in, the net forces on it will not be zero. ........It will move.....I'm guesstimating but I would say it's pushed to the edge of the sphere. I might be able to be stable in the certain of the sphere, but if the net force is anything greater than zero, it will move. So the most likely place it will be will be a the edge of the sphere. That would seem to me to be the most stable point.

Have you got any maths equations?

I'd like to see them.

Sorry, I don't understand

Don't worry, it's classical electro magnetics and wave mechanics. I don't have a fantastic grasp on it. An argument for refractive index off a transparent media is the light wave has to lift and drop the atoms in the material, the particles cannot move at the speed of light, so the wave moves slower.

neufneufneuf

Yes, I imagine the system stable, nothing move in it (particles). The black sphere is fixed to the support (and there is nothing in it for simplify the problem), so if the sphere is fixed it can't move to outer "circle". But the sphere give a net force to the system. And if pressure at outer circle is identical everywhere the sum of forces on all system is not 0. What's wrong ?

Have you got any maths equations?

No, too difficult for me, I don't find the solution, I'm waiting a simulation on Ansys software, but I would like to understand with equations too...

Lbeard

neufneufneuf wrote: »

Yes, I imagine the system stable, nothing move in it (particles).

Okay, this is a fundamental rule of physics. If nothing is moving, all forces are being canceled out. It's just a case of finding where they are.

The black sphere is fixed to the support (and there is nothing in it for simplify the problem), so if the sphere is fixed it can't move to outer "circle".

But the sphere give a net force to the system. And if pressure at outer circle is identical everywhere the sum of forces on all system is not 0. What's wrong ?

If the black sphere is fixed to a support, that is where the other forces have gone. The strain is in the support. It's a potential barrier.

In the real world all solid materials are being held together by forces. If you stretch an elastic band, it will keep together until you apply more force than the forces holding it together. Then it will snap. The potential energy will be released in the snap, as kinetic energy.

Even if you're using a non real world barrier - even if the forces holding it together are infinite, the strain the barrier will experience will be infinitely small, but it's still there. In the real world, all barrier materials strain when an object is resting on them. In you put stones in a paper bag, one at a time, the bag will hold the stones and they will not move, but once you reach a point where the force of the stones exceeds the forces holding the paper together, the bag will tear and the stones will fall true.

If the black sphere had no support, there would be no barrier to stop it moving to the edge of the sphere.

neufneufneuf

If the black sphere is fixed to a support, that is where the other forces have gone. The strain is in the support. It's a potential barrier.

in this case the pressure at outer circle is no the same everywhere ? In the contrary, outer circle have forces like I drawn and another force is added to the support from black object, look s2.png please

neufneufneuf

Hi,

I changed the problem with gravity forces, it's easier to understand I think. The problem:

A disk full of helium liquid is put on Earth. This disk is fixed. Gravity is perpendicular to your screen. Friction of helium is near 0 but even friction or not the torque must be to 0. Inside the disk there is an object composed of one volume full of helium liquid and one volume with gas at low pressure (pressure lower as possible). These volumes are attached together and can only turn around black disk. It's not possible to deform the object.

The presence of the volume of gas change the density of helium inside disk. There is more molecules at d1 than at d3 and more molecules at d2 than at d4. Molecules attract each others like third image show. The sum of forces on object give a torque if there is more molecules at left than at right.

Like liquid has no friction, this could say the object turn and give energy, like it's not possible what are forces I forget ?