# Talk:Gauss's law for gravity

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## Can a point mass be a source of gravity?

How is it assumed that a point mass will have a gravity field? Example, in order to measure the gravity field another mass must be used and therefor disturb the experiment. I cannot agree that a point mass will have a gravity field in the absence of any other mass. Basically, I would say that gravity does not contain a monopole. Two mass objects must be present for gravity to exist. —Preceding unsigned comment added by 66.125.82.34 (talk) 19:45, 29 June 2009 (UTC)

1. A gravitational field does not require observable effects to exist, it is just a mathematical tool. Regardless, we can calculate the gravity field of a point mass by finding that of a system containing the mass and the same system without that mass, and calculating the gravitational contribution of the point mass.
2. Gravity most certainly does have monopoles, indeed it is an always-attractive force, and so all poles are monopoles.
3. How do you know gravity does not exist if there are not two "mass objects" to test its existence? The existence of gravity in this instance is trivial and a matter of semantics. For convenience gravity may be assumed to exist in this case.
All of this is irrelevant for discussion here as the purpose of this encyclopaedia is to explain theorems, not to question their correctness! But hopefully I have helped explain things a little. Remember this theorem was only intended to deal with classical gravity, and not the far more complex relativistic/quantum gravity used today. Take care – Ikara talk → 23:29, 29 June 2009 (UTC)

## Teaching the theory

Is there anyone who has been using this wiki page to help their pedagogy in teaching classical mechanics involving gravity? I always found the emphasis on forces a la Newton's more cumbersome techniques to unnecessarily cloud the fundamentals that arise from solving the force equations. The main problem, it seems, is that many textbooks take all of a page or even less to briefly define the gravitational field and then never mention it again. Gauss's Law of gravitation is equally treated as a curiosity in E&M texts (kind of an "ooh look, see what you can do!"); I think it is a real shame that the pedagogy seems to demand a forces-based treatment instead of using fields to shortcut the process. --142.90.99.60 (talk) 17:40, 15 December 2009 (UTC)

## Indeterminacy

There is an ambiguity in the equation

$\nabla \cdot \mathbf{g} = -4 \pi G \rho \,,$

that is, it does not uniquely determine the value of g, even if we assume that some boundary conditions are added. Specifically one could add the curl of any vector field to g without changing the value of the left-hand side of the equation. At the very least, we should also add another equation

$\nabla \times \mathbf{g} = 0 \,$

to make this equivalent to the form mentioned in the section Gauss' law for gravity#Poisson's equation and gravitational potential. The article should also mention the Lagrangian of the field as described at Lagrangian#Newtonian gravity. JRSpriggs (talk) 15:51, 26 April 2012 (UTC)

Curl g = 0 was already there but not obvious. I think you want it to be part of "Gauss's law for gravity" and I'm putting it as a separate law that supplements "Gauss's law for gravity". Just terminology ... and I could be wrong. Anyway, I tried to make the equation more obvious. I also added a short section on the Lagrangian, nice suggestion. --Steve (talk) 04:11, 27 April 2012 (UTC)

## Derivation from Lagrangian?

I don't think that the derivation from a Lagrangian, as presented here, is appropriate and complete. (I also note that it is presented without appropriate citations, so it may be a piece of original research.)

The problem: as introduced, ρ is also a degree of freedom that must be varied. When you do that, you get a second field equation, which is simply φ = 0. This is of course not very useful. The Lagrangian, therefore, must also include some dynamical term for ρ, so the second field equation would capture how ρ responds to the presence of gravity. For a more complete (and I think, more valid) derivation, see, e.g., [1]. vttoth (talk) 02:19, 24 January 2014 (UTC)

## Does this show the speed of gravity?

Newton's Law does not take that into account, does this? — Preceding unsigned comment added by 88.104.111.5 (talk) 14:14, 21 June 2014 (UTC)

Effectively, it assumes that the speed of gravity is infinite.
If you attempt to replace the Laplace operator with the D'Alembert operator to get a wave equation which includes a speed of gravity equal to the speed of light and thus make the theory consistent with special relativity, you run into problems. If at long last, you manage to fix all the problems, you end up with general relativity's Einstein field equations. JRSpriggs (talk) 10:23, 22 June 2014 (UTC)