Re-creation of the Cavendish Experiment

Quote from: JackBlack on April 27, 2017, 02:47:54 AM

Quote from: ScintillaOfStars on April 26, 2017, 04:22:20 PM

Unfortunately not. Because there are diagonal forces imposed by objects on the theoretical plane, you cannot use Gauss's Law to model it. That isn't to say there aren't other ways of modelling it, but Gauss' Law is not it. You cannot 'merely add' the variations because they don't add up to a Gaussian model.

Again, that's not to say an infinite plane wouldn't be possible to model, merely not with the current Gaussian derivation.

You don't use the same surface for each one. You use multiple "models" to create a unified model. It wont have a simple relation like the normal ones though. Just like we can do with multiple spheres

That's what I'm saying; you can't use the same surface for each one. You have to create another function than a derived version of Gauss' Law of Gravitation. Gauss' Law doesn't work for it. I just don't know what that new function would look like.

My bad, just noticed I missed the simple model.

Yes, you can't use the same surface for each one, but you can get a function which takes the x, y and z position (potentially as a piece-wise/step function) and use that to calculate the force.
You can do that for each part, and then add it all together.

For an infinite plane, located at the origin (i.e. 0,0,0) parallel to the x,y plane, with a thickness of t, and density of p, then the acceleration due to gravity will be:
<0,0,-2*pi*G*p*t> if z>t/2;
<0,0,2*pi*G*p*t> if z<-t/2;
<0,0,-4*pi*G*p*z> otherwise

For a point mass, of mass m, located at x0,y0,z0, the acceleration due to gravity will be:
[Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5]*<x0-x,y0-y,z0-z>

You can expand that out to give you:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5>

And then you can add the 2 together.

It becomes a lot more complex because you can no longer just set the object to be at the origin and just measure the distance from it.
Instead you need to note the position of the object.

Quote from: ScintillaOfStars on April 27, 2017, 03:06:14 AM

Quote from: JackBlack on April 27, 2017, 02:47:54 AM

Quote from: ScintillaOfStars on April 26, 2017, 04:22:20 PM

Unfortunately not. Because there are diagonal forces imposed by objects on the theoretical plane, you cannot use Gauss's Law to model it. That isn't to say there aren't other ways of modelling it, but Gauss' Law is not it. You cannot 'merely add' the variations because they don't add up to a Gaussian model.

Again, that's not to say an infinite plane wouldn't be possible to model, merely not with the current Gaussian derivation.

You don't use the same surface for each one. You use multiple "models" to create a unified model. It wont have a simple relation like the normal ones though. Just like we can do with multiple spheres

That's what I'm saying; you can't use the same surface for each one. You have to create another function than a derived version of Gauss' Law of Gravitation. Gauss' Law doesn't work for it. I just don't know what that new function would look like.

My bad, just noticed I missed the simple model.

Yes, you can't use the same surface for each one, but you can get a function which takes the x, y and z position (potentially as a piece-wise/step function) and use that to calculate the force.
You can do that for each part, and then add it all together.

For an infinite plane, located at the origin (i.e. 0,0,0) parallel to the x,y plane, with a thickness of t, and density of p, then the acceleration due to gravity will be:
<0,0,-2*pi*G*p*t> if z>t/2;
<0,0,2*pi*G*p*t> if z<-t/2;
<0,0,-4*pi*G*p*z> otherwise

For a point mass, of mass m, located at x0,y0,z0, the acceleration due to gravity will be:
[Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5]*<x0-x,y0-y,z0-z>

You can expand that out to give you:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5>

And then you can add the 2 together.

It becomes a lot more complex because you can no longer just set the object to be at the origin and just measure the distance from it.
Instead you need to note the position of the object.

This is all well and good, but what happens if those forces don't cancel out?

And, even in this simulation, you're only accounting for the force due to gravity with a vector directly down. But a mass will exert forces diagonally on objects around it as well, as just a fact of having mass. And the Gaussian model for an infinite flat earth cannot, cannot be modeled properly if diagonal forces are involved. There has to be some other model for an infinite flat earth to work. Simply Gauss simply doesn't cut it.

Quote from: JackBlack on April 27, 2017, 04:35:30 AM

Quote from: ScintillaOfStars on April 27, 2017, 03:06:14 AM

Quote from: JackBlack on April 27, 2017, 02:47:54 AM

Quote from: ScintillaOfStars on April 26, 2017, 04:22:20 PM

Unfortunately not. Because there are diagonal forces imposed by objects on the theoretical plane, you cannot use Gauss's Law to model it. That isn't to say there aren't other ways of modelling it, but Gauss' Law is not it. You cannot 'merely add' the variations because they don't add up to a Gaussian model.

Again, that's not to say an infinite plane wouldn't be possible to model, merely not with the current Gaussian derivation.

You don't use the same surface for each one. You use multiple "models" to create a unified model. It wont have a simple relation like the normal ones though. Just like we can do with multiple spheres

That's what I'm saying; you can't use the same surface for each one. You have to create another function than a derived version of Gauss' Law of Gravitation. Gauss' Law doesn't work for it. I just don't know what that new function would look like.

My bad, just noticed I missed the simple model.

Yes, you can't use the same surface for each one, but you can get a function which takes the x, y and z position (potentially as a piece-wise/step function) and use that to calculate the force.
You can do that for each part, and then add it all together.

For an infinite plane, located at the origin (i.e. 0,0,0) parallel to the x,y plane, with a thickness of t, and density of p, then the acceleration due to gravity will be:
<0,0,-2*pi*G*p*t> if z>t/2;
<0,0,2*pi*G*p*t> if z<-t/2;
<0,0,-4*pi*G*p*z> otherwise

For a point mass, of mass m, located at x0,y0,z0, the acceleration due to gravity will be:
[Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5]*<x0-x,y0-y,z0-z>

You can expand that out to give you:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5>

And then you can add the 2 together.

It becomes a lot more complex because you can no longer just set the object to be at the origin and just measure the distance from it.
Instead you need to note the position of the object.

This is all well and good, but what happens if those forces don't cancel out?

And, even in this simulation, you're only accounting for the force due to gravity with a vector directly down. But a mass will exert forces diagonally on objects around it as well, as just a fact of having mass. And the Gaussian model for an infinite flat earth cannot, cannot be modeled properly if diagonal forces are involved. There has to be some other model for an infinite flat earth to work. Simply Gauss simply doesn't cut it.

They don't cancel they add.

I'm not only accounting for a force with a vector directly down. That is only for the infinite plane.
For other things, like a point source, it is accounted for with a more complex vector, like this one:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5>
with an x, y and z component.

You can then add these 2 together, such that if z>t/2, the total will be:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5-2*pi*G*p*t>

So no, Gauss can cut it, you just need to apply it once to each source, and then add them up.

Quote from: ScintillaOfStars on April 27, 2017, 01:54:35 PM

Quote from: JackBlack on April 27, 2017, 04:35:30 AM

Quote from: ScintillaOfStars on April 27, 2017, 03:06:14 AM

Quote from: JackBlack on April 27, 2017, 02:47:54 AM

Quote from: ScintillaOfStars on April 26, 2017, 04:22:20 PM

Unfortunately not. Because there are diagonal forces imposed by objects on the theoretical plane, you cannot use Gauss's Law to model it. That isn't to say there aren't other ways of modelling it, but Gauss' Law is not it. You cannot 'merely add' the variations because they don't add up to a Gaussian model.

Again, that's not to say an infinite plane wouldn't be possible to model, merely not with the current Gaussian derivation.

You don't use the same surface for each one. You use multiple "models" to create a unified model. It wont have a simple relation like the normal ones though. Just like we can do with multiple spheres

That's what I'm saying; you can't use the same surface for each one. You have to create another function than a derived version of Gauss' Law of Gravitation. Gauss' Law doesn't work for it. I just don't know what that new function would look like.

My bad, just noticed I missed the simple model.

Yes, you can't use the same surface for each one, but you can get a function which takes the x, y and z position (potentially as a piece-wise/step function) and use that to calculate the force.
You can do that for each part, and then add it all together.

For an infinite plane, located at the origin (i.e. 0,0,0) parallel to the x,y plane, with a thickness of t, and density of p, then the acceleration due to gravity will be:
<0,0,-2*pi*G*p*t> if z>t/2;
<0,0,2*pi*G*p*t> if z<-t/2;
<0,0,-4*pi*G*p*z> otherwise

For a point mass, of mass m, located at x0,y0,z0, the acceleration due to gravity will be:
[Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5]*<x0-x,y0-y,z0-z>

You can expand that out to give you:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5>

And then you can add the 2 together.

It becomes a lot more complex because you can no longer just set the object to be at the origin and just measure the distance from it.
Instead you need to note the position of the object.

This is all well and good, but what happens if those forces don't cancel out?

And, even in this simulation, you're only accounting for the force due to gravity with a vector directly down. But a mass will exert forces diagonally on objects around it as well, as just a fact of having mass. And the Gaussian model for an infinite flat earth cannot, cannot be modeled properly if diagonal forces are involved. There has to be some other model for an infinite flat earth to work. Simply Gauss simply doesn't cut it.

They don't cancel they add.

I'm not only accounting for a force with a vector directly down. That is only for the infinite plane.
For other things, like a point source, it is accounted for with a more complex vector, like this one:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5>
with an x, y and z component.

You can then add these 2 together, such that if z>t/2, the total will be:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5-2*pi*G*p*t>

So no, Gauss can cut it, you just need to apply it once to each source, and then add them up.

Yes, you are only accounting for a force with a vector directly down. Because any other vector direction breaks the Gaussian model (please excuse me for not copy+pasting all the working out from the two links I posted much earlier, but you can check them out if you like).

I'm not arguing that the Gaussian-derived formula for an infinite flat plane operates just fine for the vector direction you describe. It does, quite adequately as far as I'm concerned. But it breaks once you introduce vectors of differing directions, not least because they aren't contained within the plane, but because they aren't parallel or perpendicular to the plane.

(Just as a funny side-note: Perhaps ironically, the infinite-plane-Gauss-function would work fine if your infinite plane were in fact an infinite sphere, as that would make all vectors perpendicular.)

Yes, you are only accounting for a force with a vector directly down. Because any other vector direction breaks the Gaussian model (please excuse me for not copy+pasting all the working out from the two links I posted much earlier, but you can check them out if you like).

I'm not arguing that the Gaussian-derived formula for an infinite flat plane operates just fine for the vector direction you describe. It does, quite adequately as far as I'm concerned. But it breaks once you introduce vectors of differing directions, not least because they aren't contained within the plane, but because they aren't parallel or perpendicular to the plane.

(Just as a funny side-note: Perhaps ironically, the infinite-plane-Gauss-function would work fine if your infinite plane were in fact an infinite sphere, as that would make all vectors perpendicular.)

Like I said, it isn't a single Gaussian model with a single surface.
You use the Gaussian modelling to get each part, convert that "down" into a vector in 3D space, and then add them together.
What I did was take a Gaussian model for the plane and a Gaussian model for a point mass, convert the acceleration to a 3D vector, and add them together.

You guys are awesome.

Quote from: ScintillaOfStars on April 27, 2017, 02:08:09 PM

Yes, you are only accounting for a force with a vector directly down. Because any other vector direction breaks the Gaussian model (please excuse me for not copy+pasting all the working out from the two links I posted much earlier, but you can check them out if you like).

I'm not arguing that the Gaussian-derived formula for an infinite flat plane operates just fine for the vector direction you describe. It does, quite adequately as far as I'm concerned. But it breaks once you introduce vectors of differing directions, not least because they aren't contained within the plane, but because they aren't parallel or perpendicular to the plane.

(Just as a funny side-note: Perhaps ironically, the infinite-plane-Gauss-function would work fine if your infinite plane were in fact an infinite sphere, as that would make all vectors perpendicular.)

Like I said, it isn't a single Gaussian model with a single surface.
You use the Gaussian modelling to get each part, convert that "down" into a vector in 3D space, and then add them together.
What I did was take a Gaussian model for the plane and a Gaussian model for a point mass, convert the acceleration to a 3D vector, and add them together.

Oh I see what you are getting at, my apologies. I've been misinterpreting what it is you're saying. But modelling a point mass that is hovering above the plane can't be done with Gauss' Law for Gravity. The point mass has to be inside the plane.

Quote from: JackBlack on April 27, 2017, 03:36:55 PM

Quote from: ScintillaOfStars on April 27, 2017, 02:08:09 PM

Yes, you are only accounting for a force with a vector directly down. Because any other vector direction breaks the Gaussian model (please excuse me for not copy+pasting all the working out from the two links I posted much earlier, but you can check them out if you like).

I'm not arguing that the Gaussian-derived formula for an infinite flat plane operates just fine for the vector direction you describe. It does, quite adequately as far as I'm concerned. But it breaks once you introduce vectors of differing directions, not least because they aren't contained within the plane, but because they aren't parallel or perpendicular to the plane.

(Just as a funny side-note: Perhaps ironically, the infinite-plane-Gauss-function would work fine if your infinite plane were in fact an infinite sphere, as that would make all vectors perpendicular.)

Like I said, it isn't a single Gaussian model with a single surface.
You use the Gaussian modelling to get each part, convert that "down" into a vector in 3D space, and then add them together.
What I did was take a Gaussian model for the plane and a Gaussian model for a point mass, convert the acceleration to a 3D vector, and add them together.

Oh I see what you are getting at, my apologies. I've been misinterpreting what it is you're saying. But modelling a point mass that is hovering above the plane can't be done with Gauss' Law for Gravity. The point mass has to be inside the plane.

Have a look at The Physics Hypertextbook, Gravity of Extended Bodies, practice problem 2.

Quote from: ScintillaOfStars on April 27, 2017, 04:42:47 PM

Oh I see what you are getting at, my apologies. I've been misinterpreting what it is you're saying. But modelling a point mass that is hovering above the plane can't be done with Gauss' Law for Gravity. The point mass has to be inside the plane.

Have a look at The Physics Hypertextbook, Gravity of Extended Bodies, practice problem 2.

Many apologies, but I don't see the point you're getting at. This agrees with everything I have looked at so far. My point refers to this statement (which in this case I've pulled from the source you have given):

Given the geometric simplicity of the situation, the field must strike either side of the slab at a right angle. The two faces of the imaginary box outside the slab are parallel to the sides of the slab. These two faces are the only ones that can capture any gravitational field.

If the surface of the slab does not hit at right angles to your field at every point along the x-y plane (surface of the slab), then the field can't be applied and Gauss' Law can't be used to model gravity. It only works if your infinite plain is an infinite plane.

Sorry if this isn't what you are trying to say, but I couldn't really understand your point from what you wrote.

If the surface of the slab does not hit at right angles to your field at every point along the x-y plane (surface of the slab), then the field can't be applied and Gauss' Law can't be used to model gravity. It only works if your infinite plain is an infinite plane.

Sorry if this isn't what you are trying to say, but I couldn't really understand your point from what you wrote.

Agreed, I thought that you were claiming that Gauss' Law couldn't be applied and part of what I was pointing out was that there were also other ways of analysing it.

But have you read some of the paradoxes that arise when you bring GR into it?
An infinite plain (or plane) can be also looked on as the limiting case of a spherical shell, as the radius approaches infinity.

But, you are probably much better at this than I.

Quote from: JackBlack on April 27, 2017, 03:36:55 PM

Quote from: ScintillaOfStars on April 27, 2017, 02:08:09 PM

Yes, you are only accounting for a force with a vector directly down. Because any other vector direction breaks the Gaussian model (please excuse me for not copy+pasting all the working out from the two links I posted much earlier, but you can check them out if you like).

I'm not arguing that the Gaussian-derived formula for an infinite flat plane operates just fine for the vector direction you describe. It does, quite adequately as far as I'm concerned. But it breaks once you introduce vectors of differing directions, not least because they aren't contained within the plane, but because they aren't parallel or perpendicular to the plane.

(Just as a funny side-note: Perhaps ironically, the infinite-plane-Gauss-function would work fine if your infinite plane were in fact an infinite sphere, as that would make all vectors perpendicular.)

Like I said, it isn't a single Gaussian model with a single surface.
You use the Gaussian modelling to get each part, convert that "down" into a vector in 3D space, and then add them together.
What I did was take a Gaussian model for the plane and a Gaussian model for a point mass, convert the acceleration to a 3D vector, and add them together.

Oh I see what you are getting at, my apologies. I've been misinterpreting what it is you're saying. But modelling a point mass that is hovering above the plane can't be done with Gauss' Law for Gravity. The point mass has to be inside the plane.

Like I said, you model the 2 masses separately, using Gauss' law, and then combine them.

Quote from: ScintillaOfStars on April 27, 2017, 04:42:47 PM

Quote from: JackBlack on April 27, 2017, 03:36:55 PM

Quote from: ScintillaOfStars on April 27, 2017, 02:08:09 PM

Yes, you are only accounting for a force with a vector directly down. Because any other vector direction breaks the Gaussian model (please excuse me for not copy+pasting all the working out from the two links I posted much earlier, but you can check them out if you like).

I'm not arguing that the Gaussian-derived formula for an infinite flat plane operates just fine for the vector direction you describe. It does, quite adequately as far as I'm concerned. But it breaks once you introduce vectors of differing directions, not least because they aren't contained within the plane, but because they aren't parallel or perpendicular to the plane.

(Just as a funny side-note: Perhaps ironically, the infinite-plane-Gauss-function would work fine if your infinite plane were in fact an infinite sphere, as that would make all vectors perpendicular.)

Like I said, it isn't a single Gaussian model with a single surface.
You use the Gaussian modelling to get each part, convert that "down" into a vector in 3D space, and then add them together.
What I did was take a Gaussian model for the plane and a Gaussian model for a point mass, convert the acceleration to a 3D vector, and add them together.

Oh I see what you are getting at, my apologies. I've been misinterpreting what it is you're saying. But modelling a point mass that is hovering above the plane can't be done with Gauss' Law for Gravity. The point mass has to be inside the plane.

Like I said, you model the 2 masses separately, using Gauss' law, and then combine them.

I'll essentially rephrase what I said to rabinoz.

Even using one mass, if the field you are using doesn't strike at right angles at every point, then Gauss's Law doesn't work for an infinite plane.

Agreed, I thought that you were claiming that Gauss' Law couldn't be applied and part of what I was pointing out was that there were also other ways of analysing it.

Ah I see. Thanks for clarifying for me.

But have you read some of the paradoxes that arise when you bring GR into it?
An infinite plain (or plane) can be also looked on as the limiting case of a spherical shell, as the radius approaches infinity.

No I haven't. That sounds very interesting; I'll give a look into it. Thanks!

But, you are probably much better at this than I.

This part is false

I'll essentially rephrase what I said to rabinoz.

Even using one mass, if the field you are using doesn't strike at right angles at every point, then Gauss's Law doesn't work for an infinite plane.

Not that I have done it myself, but I think that Gauss's Law has been used on the infinite plane by

• evaluating the field at the of a finite disk.
• taking the limit as the radius approaches infinity,
• then reasoning that the field is uniform over the whole surface.

A nice little point is that if the plane is "not quite infinite" the plane will collapse into a "near infinite sphere", but in "near infinite" time.

Quote from: rabinoz on April 28, 2017, 02:01:39 AM

Agreed, I thought that you were claiming that Gauss' Law couldn't be applied and part of what I was pointing out was that there were also other ways of analysing it.

Ah I see. Thanks for clarifying for me.

Quote from: rabinoz on April 28, 2017, 02:01:39 AM

But have you read some of the paradoxes that arise when you bring GR into it?
An infinite plain (or plane) can be also looked on as the limiting case of a spherical shell, as the radius approaches infinity.

No I haven't. That sounds very interesting; I'll give a look into it. Thanks!

Quote from: rabinoz on April 28, 2017, 02:01:39 AM

But, you are probably much better at this than I.

This part is false

Thanks, but we'll see.

There are two sorts of "expert",
     the "generalist" who knows "nothing about everything" and  the "specialist" who knows  "everything about nothing".
I'll let you put the members here into the appropriate slots  .

Quote from: ScintillaOfStars on April 28, 2017, 02:20:38 AM

I'll essentially rephrase what I said to rabinoz.

Even using one mass, if the field you are using doesn't strike at right angles at every point, then Gauss's Law doesn't work for an infinite plane.

Not that I have done it myself, but I think that Gauss's Law has been used on the infinite plane by

• evaluating the field at the of a finite disk.
• taking the limit as the radius approaches infinity,
• then reasoning that the field is uniform over the whole surface.

A nice little point is that if the plane is "not quite infinite" the plane will collapse into a "near infinite sphere", but in "near infinite" time.

That is a rather jarring counter-point against taking a limit.

And if the disk field has a height dimension, then its sides will still strike perpendicular to a perfectly flat plane, and not against a contoured one, so the same thing I describe applies, right?

EDIT: Congrats on 7000 comments, by the way!

I'll essentially rephrase what I said to rabinoz.

Even using one mass, if the field you are using doesn't strike at right angles at every point, then Gauss's Law doesn't work for an infinite plane.

Not necessarily right angles. Integer multiples of 90 degrees.

And each sub-model used will have that, and then you add the results together, after first converting them into 3D vectors.

Quote from: ScintillaOfStars on April 28, 2017, 02:20:38 AM

I'll essentially rephrase what I said to rabinoz.

Even using one mass, if the field you are using doesn't strike at right angles at every point, then Gauss's Law doesn't work for an infinite plane.

Not necessarily right angles. Integer multiples of 90 degrees.

And each sub-model used will have that, and then you add the results together, after first converting them into 3D vectors.

Yeah absolutely, I agree. Only integer multiples of 90 degrees.

So, if this is the case, say I were to build a wall (as a hypothetical example given by one of my sources that I posted ages ago). This wall will exert a vector down, but also vectors radiating outwards and upwards, as a result of gravity. These vectors will not be integer multiples of 90 degrees, and therefore will not perfectly line up. Or, say I were to have a mountain. Here, the entire side of the mountain is on an angle, and this angle changes at all points.

A possible way it could work was to have infinitely small sub-models, so that each point mass can add up, and those naughty little diagonal vectors don't exist anymore. The problem with that is, now you can't add up anything and you're left with a model for an infinite plane that has incalculable gravity.

Is this what you are trying to get at, or am I missing the point?

Not that I have done it myself, but I think that Gauss's Law has been used on the infinite plane by

• evaluating the field at the of a finite disk.
• taking the limit as the radius approaches infinity,
• then reasoning that the field is uniform over the whole surface.

That would be so horribly complicated it isn't funny.

The 2 easier options would be just going straight to the infinite plane, or starting with a large spherical shell and expanding it to infinity.
Start with a spherical shell of radius r, thickness t, density p, with you a distance d from the surface.
The mass of this shell will be (4/3)*pi*((r+t)^3-r^3)*p=(4/3)*pi*(r^3+3*r^2*t+3*r*t^2+t^2-r^3)*p=(4/3)*pi*(3*r^2*t+3*r*t^2+t^2)*p.
As it is spherically symmetric, if you are outside this spherical shell it can be modelled as a point mass.
Thus the value of g obtained will follow that for a point mass, and thus be:
g=Gm/(r+d)^2=Gm/(r^2+2*r*d+d^2)
Subbing in the mass you get:
g=G*[(4/3)*pi*(3*r^2*t+3*r*t^2+t^3)*p]/[(r^2+2*r*d+d^2)]

Now lets try to simplify things a bit:
g=(G*r^2)*[4*pi*(t+t^2/r+t^3/3r^2)*p]/[r^2*(1+2*d/r+d^2/r^2)]
Then a bit more:
g=4*pi*G*(t+t^2/r+t^3/3r^2)*p/(1+2*d/r+d^2/r^2)

Then taking the limit as r->infinity means anything/r->0, giving us:
g=4*pi*G*t*p, where t*p is the area density, sigma, so:
g=4*pi*G*sigma.

This is almost guass' law for an infinite plane.

That is because this sphere isn't just an infinite plane.
There are 2 infinite planes, spaced infinitely far apart.
Basically you have the part of spherical shell near you, and the part which is all the way on the other side. So to find the final result, you need to divide this by 2, 2 get the force for a single infinite plane (which still requires you to be "above" the plane, and thus:
g=2*pi*G*sigma.

A possible way it could work was to have infinitely small sub-models, so that each point mass can add up, and those naughty little diagonal vectors don't exist anymore. The problem with that is, now you can't add up anything and you're left with a model for an infinite plane that has incalculable gravity.

Is this what you are trying to get at, or am I missing the point?

That is akin to what I am getting at.
Except treating the objects as simply as you can.
So you treat the plane as an infinite plane, and then you add other objects to it, similar to how you can treat the entire infinite plane as a series of point masses and add them all up.

Objects too far away you can deem to be insignificant and neglect.
Close by objects you can approximate in various ways.

Yes, it will be a simplification and not perfectly accurate, but pretty much everything is.
There is a wonderful quote about science (at least when you get to things like this) that I love:
Everything is either too simple to be real, or too complex to be useful.

Sure, you can treat everything as a point mass, but if you just wanted to treat them as atoms, then with just 60g of sand, you would have 6.022*10^23 points to deal with.
That would be far too complex for any sane person to try to deal with, and produce an accuracy far greater than that needed (especially as you would never get the distance accurate enough to care).
As such, it makes more sense to break the object down into small, but still quite large, simple objects, which you can use Gauss' law for, like spheres.

Assuming the density doesn't vary too much, cut out the biggest spheres you can, until you have a decent amount potentially overlapping some of the smaller ones a bit to make sure you account for all the mass, and then use that.
It wont be perfect, but it will likely be as good as you need.

Quote from: ScintillaOfStars on April 28, 2017, 04:33:45 AM

A possible way it could work was to have infinitely small sub-models, so that each point mass can add up, and those naughty little diagonal vectors don't exist anymore. The problem with that is, now you can't add up anything and you're left with a model for an infinite plane that has incalculable gravity.

Is this what you are trying to get at, or am I missing the point?

That is akin to what I am getting at.
Except treating the objects as simply as you can.
So you treat the plane as an infinite plane, and then you add other objects to it, similar to how you can treat the entire infinite plane as a series of point masses and add them all up.

Objects too far away you can deem to be insignificant and neglect.
Close by objects you can approximate in various ways.

Yes, it will be a simplification and not perfectly accurate, but pretty much everything is.
There is a wonderful quote about science (at least when you get to things like this) that I love:
Everything is either too simple to be real, or too complex to be useful.

Sure, you can treat everything as a point mass, but if you just wanted to treat them as atoms, then with just 60g of sand, you would have 6.022*10^23 points to deal with.
That would be far too complex for any sane person to try to deal with, and produce an accuracy far greater than that needed (especially as you would never get the distance accurate enough to care).
As such, it makes more sense to break the object down into small, but still quite large, simple objects, which you can use Gauss' law for, like spheres.

Assuming the density doesn't vary too much, cut out the biggest spheres you can, until you have a decent amount potentially overlapping some of the smaller ones a bit to make sure you account for all the mass, and then use that.
It wont be perfect, but it will likely be as good as you need.

Okay, I can see now where you're coming from. But explain to me (because I still haven't quite wrapped my head around this whole thing), how do you deal with the diagonal vectors these hypothetical spheres would create?

Okay, I can see now where you're coming from. But explain to me (because I still haven't quite wrapped my head around this whole thing), how do you deal with the diagonal vectors these hypothetical spheres would create?

You consider each part, e.g. the sphere, as a Gaussian model.
This means for this sub-model, the vectors would be perpindicular.
You then convert this sub-model centred model to a full model centred one, taking note that now, instead of just r, you have a position (x,y,z) of both the mass and the accelerated object, and use this to find the acceleration as a function of position (instead of r) with a direction, i.e. it is a vector.

For example, the one above for a point mass located at (x0,y0,z0), with a mass of m is:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5>

Then, as you have a bunch of vectors from each sub model, you simply add the vectors together.

Quote from: ScintillaOfStars on April 28, 2017, 04:54:26 AM

Okay, I can see now where you're coming from. But explain to me (because I still haven't quite wrapped my head around this whole thing), how do you deal with the diagonal vectors these hypothetical spheres would create?

You consider each part, e.g. the sphere, as a Gaussian model.
This means for this sub-model, the vectors would be perpindicular.
You then convert this sub-model centred model to a full model centred one, taking note that now, instead of just r, you have a position (x,y,z) of both the mass and the accelerated object, and use this to find the acceleration as a function of position (instead of r) with a direction, i.e. it is a vector.

For example, the one above for a point mass located at (x0,y0,z0), with a mass of m is:
<(x0-x)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(y0-y)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5,(z0-z)*Gm/((x-x0)^2+(y-y0)^2+(z-z0)^2)^1.5>

Then, as you have a bunch of vectors from each sub model, you simply add the vectors together.

What? The vectors are still diagonal. Are you saying you take the x, z, y force components and then these are all at 90 degree increments so it all adds up? Because that doesn't work. You can do it in theory, but in practice I still have a diagonal vector which conflicts with an infinite plane Gaussian model.

What your working shows is you finding the direct downward vector for your point mass, to which I say: I concede that. You can model for a downward vector. But in reality, vectors aren't nearly so simplistic.

What? The vectors are still diagonal. Are you saying you take the x, z, y force components and then these are all at 90 degree increments so it all adds up? Because that doesn't work. You can do it in theory, but in practice I still have a diagonal vector which conflicts with an infinite plane Gaussian model.

What your working shows is you finding the direct downward vector for your point mass, to which I say: I concede that. You can model for a downward vector. But in reality, vectors aren't nearly so simplistic.

Yes, the vectors are diagonal, but for each vector for each of the sub-models (the Gaussian model of each part of the overall model), this diagonal vector is at 90 degrees to the surface used for that sub-model, and thus Gauss' law would hold.

Yes, it would conflict with the infinite plane model, because this considers more than just the infinite plane. It is an infinite plane with other bits.

Quote from: ScintillaOfStars on April 28, 2017, 05:15:11 AM

What? The vectors are still diagonal. Are you saying you take the x, z, y force components and then these are all at 90 degree increments so it all adds up? Because that doesn't work. You can do it in theory, but in practice I still have a diagonal vector which conflicts with an infinite plane Gaussian model.

What your working shows is you finding the direct downward vector for your point mass, to which I say: I concede that. You can model for a downward vector. But in reality, vectors aren't nearly so simplistic.

Yes, the vectors are diagonal, but for each vector for each of the sub-models (the Gaussian model of each part of the overall model), this diagonal vector is at 90 degrees to the surface used for that sub-model, and thus Gauss' law would hold.

Yes, it would conflict with the infinite plane model, because this considers more than just the infinite plane. It is an infinite plane with other bits.

I think I see. Interesting. I'll see if I can fully wrap my head around it with some calculations. I'll get back to you if I find anything. Thanks so much!