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Any finite & non empty set of masses has a computable center of gravity: $\vec{OG} = \frac{\sum_i m_i \vec{OM}_i}{\sum_i m_i}$ .

Does the contrapositive permits to conclude that a mass system with physical evidence that it doesn't have a gravity center is an infinite set of mass (i.e. of cardinal larger than $\aleph_0$) ?

On the other hand, an infinite set of masses may have a computable center of gravity. Ex. : within a 2D infinite plan, an infinite set of equal masses linearly distributed along the x and y axis will have a gravity center at its origin O. Unfortunately, this example doesn't have a center of gravity since the integral of masses in this particular topology doesn't converge.

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And why would there exist a system with no computable center of gravity? If it exists it will have coordinates and masses. –  anna v Apr 6 '13 at 14:39
    
Within a 1D infinite space, an infinite set of equal masses linearly distributed along the x axis doesn't have a gravity center. –  daniel Azuelos Apr 6 '13 at 14:58
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"Center of gravity" should probably be read "first moment of mass distribution" in a formal mathematical context. And looking for work on the moments of distributions may yield more results than using the physicists term. –  dmckee Apr 6 '13 at 17:02
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@danielAzuelos Implicitly your third paragraph assumes Cauchy principal values in the sums, but it is not a priori clear this is justified. See the answers and discussion here, here, and here –  Chris White Apr 6 '13 at 20:27
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@user12345: an infinite set of equal masses is a non-physical situation Not true. We frequently idealize a continuum by taking the limit of infinite particle number. Also, the universe may be infinite. Even then, why would it not be defined? If the masses are uniformally distributed then is the centre of gravity not everywhere? If it's "everywhere" then it's not defined. Also, you need to consider causality. What is the logical link here? –  Ben Crowell Sep 6 '13 at 14:51
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3 Answers 3

You are right that a (nonempty) finite set of positive masses must always have a center of mass. But so, too do continuous distributions of mass: in general, any positive Borel measure $\mu$ on $\mathbb R^3$ can be interpreted as a mass distribution, and it has a center of mass at $$\mathbf r_\text{CM}=\frac{\int\mathbf r\, \text d\mu(\mathbf r)}{\int\text d\mu(\mathbf r)}.$$ This will always exist provided that

  • The support of the measure is bounded. (That is, the system has a finite size.)
  • The distribution has finite total mass $M={\int\text d\mu(\mathbf r)}$.

These are sufficient conditions but the first one is not strictly necessary, as improper integrals can also converge. A necessary condition for $\mathbf r_\text{CM}$ to be well defined is absolute convergence of ${\int \|\mathbf r\|\,\text d\mu(\mathbf r)}$.

Your naïve assumption that non-existence of the center of mass implies an infinite set of masses is thus incorrect.

However, I'm puzzled at the phrasing of your question. All physical systems we deal with - with the possible exception of the Universe as a whole - have a finite size and a finite mass, and therefore have a center of gravity. There is no such thing as "evidence that a system doesn't have a gravity center" - we would just call it "new physics".

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I think the OP is asking about what happens if you have a continuous mass distribution, which, would be, in a sense, "an infinite set of masses", and you obviously answer this question, though for clarity, I might write your answer as $d\mu({\vec r}) = \rho({\vec r})d^{3}x$, as the latter notation might be more readable for beginning students. –  Jerry Schirmer Sep 5 '13 at 17:08
    
@JerrySchirmer yes, though then you have to jump through a number of hoops to describe point particles, infinitely thin wires, and so on. –  Emilio Pisanty Sep 5 '13 at 17:10
    
Sure. But, pedagogically, I"m not horribly opposed to saying "just use Gauss's law or Ampère's law" until the student's ready to deal with delta functions. –  Jerry Schirmer Sep 5 '13 at 17:50
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All physical systems we deal with - with the possible exception of the Universe as a whole - have a finite size and a finite mass, and therefore have a center of gravity. There is no such thing as "evidence that a system doesn't have a gravity center" - we would just call it "new physics". It wouldn't be new physics, it would just be general relativity. Centroids aren't well defined in a curved space, essentially because you can't add vectors that live in different tangent spaces. –  Ben Crowell Sep 6 '13 at 4:34
    
Your naïve assumption that non-existence of the center of mass implies an infinite set of masses is thus incorrect. I think it is correct, because he's not counting an arbitrary continuous mass distribution as one mass. –  Ben Crowell Sep 6 '13 at 14:56
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I think your analysis is basically right for Newtonian mechanics with countable sets of positive point masses. As an explicit example where the c.m. is undefined, let point masses $m>0$ be located at every point $(i,j,k)$, where $i$, $j$, and $k$ are integers. This is similar to Newton's cosmology in which the universe is homogeneous and isotropic, and he believed that gravitational collapse would be avoided because all gravitational forces from distant objects would cancel by symmetry. (In fact it would be an unstable equilibrium.)

Newtonian mechanics assumes (1) that space is flat, and (2) that the topology of space is that of 3-dimensional Euclidean space.

You can relax assumption 2 while maintaining assumption 1. For example, you can make a space with the topology of a cylinder, but with zero intrinsic curvature. In such a space, there is no sensible way to define a center of mass, even for two point masses in general position (assuming, e.g., that you want the c.m. to be a continous function of their positions).

The 1st line of math. is surely false if we consider GR. Which formula should be used to compute the center of gravity of an infinite set of masses with (hypothetical) infinite distances and taking into account the GR?

In general relativity, 1 is violated, and 2 may be as well. When there's curvature, you get an ambiguity in the definition of the center of mass because parallel transport is path-dependent. The ambiguity lessens on smaller scales, so you get a well-defined c.m. only locally. This leads to some interesting possibilities (Wisdom 2004, Gueron 2005, Gueron 2007).

GR is also compatible with the existence of negative mass, and there are strong reasons for believing that negative mass exists, in a certain technical sense, although not as bulk matter in the ordinary sense (Barcelo 2002). If we have negative mass, then the c.m. is not really a meaningful thing.

Twilight for the energy conditions?, Barcelo and Visser, http://arxiv.org/abs/gr-qc/0205066

"Swimming in Spacetime: Motion in Space by Cyclic Changes in Body Shape" Jack Wisdom 2003, Science , 299 , 1865. http://groups.csail.mit.edu/mac/users/wisdom/

"The relativistic glider," Eduardo Gueron and Ricardo A. Mosna, Phys.Rev.D75:081501,2007. http://arxiv.org/abs/gr-qc/0612131

"'Swimming' versus 'swinging' in spacetime", Gueron, Maia, and Matsas, http://arxiv.org/abs/gr-qc/0510054

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You asked

Does the contrapositive permits to conclude ...

so I consider your question from the point of view of logic.

Consider the following propositions:

$P_1=$ The set of masses under consideration is finite.

$P_2=$ The set of masses under consideration is non-empty.

$Q=$ The system has a gravity center.

You claim that $P1 \wedge P2\to Q$.

This is not necessarily true, as shown by Ben Crowell. You may add some assumptions, especially if you want the center of mass be calculated by the formula you wrote. For instance that the space is Euclidean, so we are talking about Newtonian mechanics, or that we are talking about special relativity, provided that the masses are considered at rest w.r.t. a given inertial reference frame. Let the proposition containing these assumptions be $P_3$.

Now, your conditional statement is

$$P1 \wedge P2 \wedge P3\to Q.$$

The contraposition is

$$\neg Q \to \neg (P1 \wedge P2 \wedge P3).$$

Equivalently,

$$\neg Q \to \neg P1 \vee \neg P2 \vee \neg P3.$$

Hence, if the system doesn't have a gravity center, the contraposition only says that at least one of the propositions $P_1$, $P_2$, and $P_3$ is false.

It can be that $P_1$ is false, hence the set of masses is infinite. For instance, the space is Euclidean, and there are equal masses at the vertices of a cubic lattice extending in all the space. This set is countable. Or, there is a countable set of masses, which increase in a divergent way, such as, $m_i=i\cdot m_0$, where $m_0>0$.

Another example when a countable set of masses don't have a center of mass is the one you gave

Ex. : within a 2D infinite plan, an infinite set of equal masses linearly distributed along the x and y axis will have a gravity center at its origin O.

This set doesn't actually have a gravity center. You say it is at the origin, but there is no absolute origin in the Euclidean plane. If you change the origin, you will find the gravity center at another point.

But merely the negation of $P_1$ doesn't imply that there is no center of gravity, and Emilio Pisanty gave such an example from continuous mechanics.

Another way for $P1 \wedge P2 \wedge P3$ to be false is that $P_2$ is false, hence there are no masses. It follows equally from contraposition.

Or, if both $P_1$ and $P_2$ are true, then $P_3$ is false, for instance the theory is not Newtonian mechanics or special relativity. Ben Crowell exemplified with general relativity.

The main point is that, if your conditional statement is true, contraposition only implies that at least one of the conditions in your statement is false, and it doesn't allow you to pick which one you want to invalidate.

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