I Why do the signs alternate when using Gauss' law for gravity?

AI Thread Summary
Gauss's law for gravity explains how the signs of the normal forces change based on the position of a mass relative to a closed surface. When a mass is outside the surface, the cone of influence intersects the surface an even number of times, leading to alternating signs that ultimately cancel out, resulting in zero total normal force. Conversely, if the mass is inside the surface, the cone intersects an odd number of times, producing a net negative force. The direction of flux is determined by whether the vector from the mass to the surface points inward or outward, with inward flux being negative and outward flux being positive. This understanding clarifies the behavior of gravitational fields in relation to closed surfaces.
TimeRip496
Messages
249
Reaction score
5
"Gauss's theorem can be established as follows. consider an attracting particle of mass m at the point P, and let a cone of small solid angle ω be generated by radii through P. This cone cuts the surface at the points Q1, Q2, ... taken in order from P; the parts of the surface cut off by the cone at these points are S1, S2, ... and the outward-drawn normals are denoted n1, n2, ...

Now if P lies outside Σ (region of masses, each at points Q1, Q2, ...), the cone will cut S an even number of times, and the signs will be plus and minus alternately, so that the total normal force across S1, S2, ... will be zero. On the other hand, if P lies inside Σ, the cone will cut S an odd number of times, the sign being minus and plus alternately, so that the total normal forces across S1, S2, ... will be -mω."

Source: Spherical harmonic T.M.Macrobert

Help with this text! I don't understand the second paragraph as to why the cone will cut S an odd number of times, the sign being minus and plus alternately when P lies inside Σ vs when P lies outside Σ. I know the cone is just a mathematical object but I can't follow the author thought process here. How did he use the cone to cut the surface at Q1, Q2, ... ? Why will the normal force across S due to particle at P be zero when P is outside the mass but not inside the mass?
 
Physics news on Phys.org
flux.jpg


Above is a picture illustrating the point. You have a point mass on the left, and a surface drawn on the right. The flux through a section of a closed surface is positive if the vector from the mass to the section of the surface points outward. The flux is negative if it points inward.

If the point mass is outside the surface, the positive flux through one section is canceled by the negative flux through another section. So the total flux adds up to zero. If the point mass is inside the surface, then you just have outward flux (positive).

You can make the surface more complicated by putting in "folds" in the surface. But each fold will have cancellation of the positive and negative fluxes, so folds don't change the total flux.

So the even/odd business is:

If the point mass is inside the surface, you have, for each cone coming away from the mass, one section with outward flux, plus 2 sections (one outward and one inward) for each fold. So the total number of sections intersecting the cone is odd.

If the point mass is outside the surface, you have two sections (one with positive flux and one with negative flux) plus 2 more for each fold. So the total number is even.
 

Attachments

  • flux.jpg
    flux.jpg
    16.3 KB · Views: 904
stevendaryl said:
View attachment 231296

Above is a picture illustrating the point. You have a point mass on the left, and a surface drawn on the right. The flux through a section of a closed surface is positive if the vector from the mass to the section of the surface points outward. The flux is negative if it points inward.

If the point mass is outside the surface, the positive flux through one section is canceled by the negative flux through another section. So the total flux adds up to zero. If the point mass is inside the surface, then you just have outward flux (positive).

You can make the surface more complicated by putting in "folds" in the surface. But each fold will have cancellation of the positive and negative fluxes, so folds don't change the total flux.

So the even/odd business is:

If the point mass is inside the surface, you have, for each cone coming away from the mass, one section with outward flux, plus 2 sections (one outward and one inward) for each fold. So the total number of sections intersecting the cone is odd.

If the point mass is outside the surface, you have two sections (one with positive flux and one with negative flux) plus 2 more for each fold. So the total number is even.
I am sorry I don't quite get the direction of the flux part. For the "flux through a section of a closed surface is positive if the vector from the mass to the section of the surface points outward. The flux is negative if it points inward.", how are the direction for the flux determined? How do I know whether the vector is pointing outwards vs inwards? If we are looking at gravity, should't the direction of flux be just one as gravity only attracts and flux is just the amount of field lines passing through that surface?
 
TimeRip496 said:
I am sorry I don't quite get the direction of the flux part. For the "flux through a section of a closed surface is positive if the vector from the mass to the section of the surface points outward. The flux is negative if it points inward.", how are the direction for the flux determined? How do I know whether the vector is pointing outwards vs inwards? If we are looking at gravity, should't the direction of flux be just one as gravity only attracts and flux is just the amount of field lines passing through that surface?

A closed surface such as a balloon has an inside and an outside. The line from the point mass to a point on the surface has a direction. If the line is pointing inward, toward the inside of the surface, then the flux through that point on the surface is negative. If it is pointing outward, toward the outside of the surface, then the flux is positive.

.
flux.png
 

Attachments

  • flux.png
    flux.png
    10.8 KB · Views: 563
  • Like
Likes TimeRip496
In the figure below, I show an oddly-shaped surface. The line from the mass to the surface pierces the surface 6 times. 3 times the direction is inward--from outside the surface toward inside the surface. 3 times the direction is outward--from inside toward outside.
flux2.png
 

Attachments

  • flux2.png
    flux2.png
    11 KB · Views: 556
  • Like
Likes TimeRip496
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
Back
Top