The Davis Model
« on: May 03, 2019, 09:12:56 PM »
Today I found this absolute masterpiece of misuse on the wiki, trying to "prove" that an infinite plane could have finite gravity:



Let's start with statement (1). It appears to be invoking Gauss's law, but there are a lot of problems with it:

  • S is a surface, you can't do a single integral over it
  • Surface integrals are over dS, not dA
  • You don't want flux, you want a triple vector integral

Just these problems are enough to reject this proof.
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Re: The Davis Model
« Reply #1 on: May 03, 2019, 09:21:42 PM »
Welcome to part of why we have tfes.org and theflatearthsociety.org as far as I understand it. Deep rooted differences in how to approach the problem. From what I recall of my time on the other site Davis was favoring some form of Relativity based flat Earth, where (to put it extremely simply and inelegantly) space is curved rather than the Earth.

I'm kind of curious however where you dug that bit up, I don't recall seeing it in there before and I thought I had perused most of the wiki at one time or another.

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Re: The Davis Model
« Reply #2 on: May 03, 2019, 09:38:05 PM »
I'm kind of curious however where you dug that bit up, I don't recall seeing it in there before and I thought I had perused most of the wiki at one time or another.

It is on the UA page in the wiki under the heading "Alternatives to Universal Acceleration"
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Re: The Davis Model
« Reply #3 on: May 04, 2019, 01:03:38 AM »
Today I found this absolute masterpiece of misuse on the wiki, trying to "prove" that an infinite plane could have finite gravity:



Let's start with statement (1). It appears to be invoking Gauss's law, but there are a lot of problems with it:

  • S is a surface, you can't do a single integral over it
  • Surface integrals are over dS, not dA
  • You don't want flux, you want a triple vector integral

Just these problems are enough to reject this proof.

I’m afraid you are mistaken, it is possible to have finite gravity from infinite plane. The FES has depicted an accurate calculation.

Yes, I am surprised as well.

Your issues:

1. Surface integrals are often done this way in physics.

/integral{2*pi*r*dr}=pi*r^2.

I just computed the area of a circle with a single integral.

2. dS or dA are used interchangeably to denote a surface integral, depending on the textbook.

3. No. A triple vector integral makes no sense here. You want to compute a the flux of gravitational field lines through the closed surface, exactly how you would for an electric field.

The computation is correct and this is not a surprise. You can also have a finite electric field from an infinite plane of charge.

Many, many things on the wiki are incorrect physics or baseless unphysical claims, but this is not one of them.
The fact.that it's an old equation without good.demonstration of the underlying mechamism behind it makes.it more invalid, not more valid!

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Re: The Davis Model
« Reply #4 on: May 04, 2019, 01:22:59 AM »
It is an infinite plan. But does it have infinite mass?

Re: The Davis Model
« Reply #5 on: May 04, 2019, 01:46:31 AM »
Today I found this absolute masterpiece of misuse on the wiki, trying to "prove" that an infinite plane could have finite gravity:

https://wiki.tfes.org/images/e/ec/Infinite_Plane.gif

Let's start with statement (1). It appears to be invoking Gauss's law, but there are a lot of problems with it:

  • S is a surface, you can't do a single integral over it
  • Surface integrals are over dS, not dA
  • You don't want flux, you want a triple vector integral

Just these problems are enough to reject this proof.

I’m afraid you are mistaken, it is possible to have finite gravity from infinite plane. The FES has depicted an accurate calculation.

Yes, I am surprised as well.

Your issues:

1. Surface integrals are often done this way in physics.

/integral{2*pi*r*dr}=pi*r^2.

I just computed the area of a circle with a single integral.

2. dS or dA are used interchangeably to denote a surface integral, depending on the textbook.

3. No. A triple vector integral makes no sense here. You want to compute a the flux of gravitational field lines through the closed surface, exactly how you would for an electric field.

The computation is correct and this is not a surprise. You can also have a finite electric field from an infinite plane of charge.

Many, many things on the wiki are incorrect physics or baseless unphysical claims, but this is not one of them.
  • Sure, but most of those are shorthands just to skip single integrals. The use of dA indicates that this isn't so.
  • Yes, I am aware of that.
  • I'm not really following. The goal is the gravitational force on an arbitrary point, no?
You'll notice that I never rejected the claim, only the proof.
Recommended reading: We Have No Idea by Jorge Cham and Daniel Whiteson

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Re: The Davis Model
« Reply #6 on: May 04, 2019, 02:28:29 AM »
Today I found this absolute masterpiece of misuse on the wiki, trying to "prove" that an infinite plane could have finite gravity:

https://wiki.tfes.org/images/e/ec/Infinite_Plane.gif

Let's start with statement (1). It appears to be invoking Gauss's law, but there are a lot of problems with it:

  • S is a surface, you can't do a single integral over it
  • Surface integrals are over dS, not dA
  • You don't want flux, you want a triple vector integral

Just these problems are enough to reject this proof.

I’m afraid you are mistaken, it is possible to have finite gravity from infinite plane. The FES has depicted an accurate calculation.

Yes, I am surprised as well.

Your issues:

1. Surface integrals are often done this way in physics.

/integral{2*pi*r*dr}=pi*r^2.

I just computed the area of a circle with a single integral.

2. dS or dA are used interchangeably to denote a surface integral, depending on the textbook.

3. No. A triple vector integral makes no sense here. You want to compute a the flux of gravitational field lines through the closed surface, exactly how you would for an electric field.

The computation is correct and this is not a surprise. You can also have a finite electric field from an infinite plane of charge.

Many, many things on the wiki are incorrect physics or baseless unphysical claims, but this is not one of them.
  • Sure, but most of those are shorthands just to skip single integrals. The use of dA indicates that this isn't so.
  • Yes, I am aware of that.
  • I'm not really following. The goal is the gravitational force on an arbitrary point, no?
You'll notice that I never rejected the claim, only the proof.

Ahh, quite true. Still though, the calculation seems fine to me. Gauss law for planar geometry with electric fields uses this same approach. The integral returns 2A, where A is the area or the Gaussian surface. You need the factor of 2 to close the surface.

The result indicates that the gravity is independent of 3D location. It is constant everywhere for an infinite plane. Which is exactly what you find for the electric field from an infinite plane of charge.
The fact.that it's an old equation without good.demonstration of the underlying mechamism behind it makes.it more invalid, not more valid!

- Tom Bishop

We try to represent FET in a model-agnostic way

- Pete Svarrior

Re: The Davis Model
« Reply #7 on: May 04, 2019, 06:06:13 AM »
Today I found this absolute masterpiece of misuse on the wiki, trying to "prove" that an infinite plane could have finite gravity:

https://wiki.tfes.org/images/e/ec/Infinite_Plane.gif

Let's start with statement (1). It appears to be invoking Gauss's law, but there are a lot of problems with it:

  • S is a surface, you can't do a single integral over it
  • Surface integrals are over dS, not dA
  • You don't want flux, you want a triple vector integral

Just these problems are enough to reject this proof.

I’m afraid you are mistaken, it is possible to have finite gravity from infinite plane. The FES has depicted an accurate calculation.

Yes, I am surprised as well.

Your issues:

1. Surface integrals are often done this way in physics.

/integral{2*pi*r*dr}=pi*r^2.

I just computed the area of a circle with a single integral.

2. dS or dA are used interchangeably to denote a surface integral, depending on the textbook.

3. No. A triple vector integral makes no sense here. You want to compute a the flux of gravitational field lines through the closed surface, exactly how you would for an electric field.

The computation is correct and this is not a surprise. You can also have a finite electric field from an infinite plane of charge.

Many, many things on the wiki are incorrect physics or baseless unphysical claims, but this is not one of them.
  • Sure, but most of those are shorthands just to skip single integrals. The use of dA indicates that this isn't so.
  • Yes, I am aware of that.
  • I'm not really following. The goal is the gravitational force on an arbitrary point, no?
You'll notice that I never rejected the claim, only the proof.

Ahh, quite true. Still though, the calculation seems fine to me. Gauss law for planar geometry with electric fields uses this same approach. The integral returns 2A, where A is the area or the Gaussian surface. You need the factor of 2 to close the surface.

The result indicates that the gravity is independent of 3D location. It is constant everywhere for an infinite plane. Which is exactly what you find for the electric field from an infinite plane of charge.

To address the OP's concerns;

1. This is a surface integral - the S beneath the integral denotes the integration over the surface, regardless of whether two integrals are visibly shown.
2. Note that the normal vector is dotted with the infinitesimal surface element. Hence n*dA = dS
3. Anyone who has studied multivariate calculus ought to be familiar with Gauss' law. Though not explicitly enumerated in the steps, the flux integral (over a Gaussian pillbox) is converted into a volume integral by taking the divergence of the vector field. The surface integral taken over a closed surface is then the volume integral of an associated finite measure.

Now, while the mathematics itself may be sound, it is erroneously applied to the given situation. The very "inverse-square" nature was ascertained through the point-light-source analog, wherein flux density (intensity) decays according to the aforementioned inverse square relation. Gauss' law can be used to derive Newton's Law of Gravitation (and vice versa) by representing the uniform gravitational field in terms of a volume integral (the divergence of the 1/r^2 term results in a Dirac delta function centered about the chosen point of reference). Spherical symmetry is present.

Now, about the "infinite sheet" argument. While the gravitation field is constant given this assumption (due to the obvious fact that more flux lines appear as one venture further from the sheet), people would be absolutely stupid to argue that the earth is infinite. In a realistic sense, the gravitational field would only by constant near the center of the flat earth disc.
« Last Edit: May 04, 2019, 03:10:41 PM by Poiseuille04 »

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Re: The Davis Model
« Reply #8 on: May 04, 2019, 11:36:54 AM »
/integral{2*pi*r*dr}=pi*r^2
Just a heads-up, we do provide an instance of mathtex, so you don't have to suffer with bad pseudo-latex

[tex]\int2\pi r\,dr=\pi r^2[/tex]

becomes

%5Cint2%5Cpi%20r%5C%2Cdr%3D%5Cpi%20r%5E2
« Last Edit: May 04, 2019, 11:41:10 AM by Pete Svarrior »
Read the FAQ before asking your question - chances are we've already addressed it.
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Re: The Davis Model
« Reply #9 on: May 04, 2019, 11:55:32 AM »
/integral{2*pi*r*dr}=pi*r^2
Just a heads-up, we do provide an instance of mathtex, so you don't have to suffer with bad pseudo-latex

[tex]\int2\pi r\,dr=\pi r^2[/tex]

becomes

%5Cint2%5Cpi%20r%5C%2Cdr%3D%5Cpi%20r%5E2

Bad-ass! Thanks Pete!
The fact.that it's an old equation without good.demonstration of the underlying mechamism behind it makes.it more invalid, not more valid!

- Tom Bishop

We try to represent FET in a model-agnostic way

- Pete Svarrior