When I studied Physics, we viewed that in the center (x/0) the value of gravity was infinite. But now I see that you say it's 0 (null), because all the force vectors cancel each other, in the exact center. Is near the center near infinite?
your question is quite interesting. If you are referring to gravitational force at the center of a "sphere", only then it's value is said to be zero '0' that too it is '0/0' form. Gravitational force for a UNIFORMLY DENSE sphere is proportional to (mass enclosed with its center coinciding with center of the spherical object/distance from the center^2), where mass is proportional to (distance from the center^3). Hence finally it comes out to be proportional to (distance from the center), hence at the center, gravitational force is '0'. But still there might be a question that what if we substitute '0' before cancellation of (distance), it gives us 0/0? It is solved by using the method of limits; to be simple, the numerator decreases at a faster rate than that of denominator and hence reaches zero faster.
In fact to be very specific, the net gravitational force at the center of the sphere is zero even for the nonuniform spherically symmetric mass distribution! Since, the density of the sphere could only be a function like D = r^i + r^j + r^k + ...... where r = distance from the center and i, j, k, ... are non negative integers or else the negative integer powers mean infinite mass at the center ( which is impossible). Using this kind of density function, we can verify that mass of the sphere always comes out to be proportional to distance raised to the powers greater than or equal to that of 3.
With a notion that infinite force is not possible, we may conclude the same for all kinds of geometrical shapes!
No Zafar, the vector indicating net force decreases gradually from the surface of the sphere to its center. Consider a simple case of a spherical symmetric mass distribution sphere, now as we move from the surface to the center, the net force vector gets smaller and smaller since the force vectors start getting cancelling each other.
On the surface, forces act only from the side of sphere whereas as when we consider a point inside the sphere, with some mathematical support we can conclude that it is the spherical mass enclosed that effectively exerts g-force. So, as we approach the center, the mass enclosed has to decrease, hence the vector of net force decreases continuously until it approaches zero at the center!
No, the mass enclosed decreases continuously as we move from the surface of the sphere to the its center and in no case the force is increasing as we go inside in our model. The net force vector is decreasing continuously as we approach the center.
Yes, indeed the vectors do have a value near the center but near to zero.
No good physical quantity should have two values at the same time (for example zero and infinity), right? Of course, apart from singular points, where its value is simply undefined at all, like at phase transition temperature. Did you think about the following idealized situation: Take two point masses, not necessarily equal; what happens at their center of mass? The algebra is simple and the result is funny: neither the gravitational potential nor the gravitational force at the center of mass is equal to zero, both are finite an well defined. Since the force there appears proportional to (m2/m1-m1/m2), and inversely proportional to the distance between masses, then the earlier claim that it is 'near to zero' is clearly unsupported.
Sorry, lost some second powers. The force is proportional to:
(m1+m2)^2 * [m2/m1^2-m1/m2^2] / (x2-x1)^2.
The expression in square brackets is only equal to zero when m1=m2 and is by no means "small" otherwise.
No, not always, as my counterexample shows. If you don't like calculations think this way: to find center of mass you have to sum expressions like m*r. To calculate the gravitational force you have to sum m/r^2. The dependence of both expressions on distance is quite different, so what makes you think that location of center of mass and the point at which |F|=0 coincide?
Yes, it may happen, but only in very special circumstances, for example for a single spherical body, with finite radius, and with density depending only on distance from its center. So the answer is: gravitational force at center of mass is finite, sometimes equal to zero. The same goes for gravitational potential.
You must specify the geometry of the mass distribution and how its density varies at each point. If, say, the distribution is spherical symmetric then you can use Gauss' law for gravitation to calculate g inside and outside the distribution
There is always a center, although it varies every moment.
I am not sure if I understand the nature of your problem, but if your distribution is homogeneous (the density is constant), isotopic and infinite in extension, yes, the gravitational field is zero everywhere. This is Newton's model of the universe.
But, if I restrict myself to the original question: no physical meaningful quantity can have two different values at the same point. It is either zero or infinity. Which one is valid depends on the nature of your distribution.
@Andrej, I didn't get what is so specific about understanding zero or infinity. I think you are saying about application of limits i.e 0/0 form. Anything else sir?
Some experiments seems to prove the force of gravity is really the sum of all forces exerted on a body; for example, on earth, the foce above is bigger because planet earth is blocking some of the force coming from the other side.If gravity is really a pushing force then the formula suggesting that at distance zero, the force would be inifnite, that formula has to be expanded for small distances because the pushing force will never be infinite.
Pierre Amiot, retired, Université Laval
There is no infinite field in Newton's gravity. If your body is spherical and uniform (in angles), the gravity at the center of mass is zero. Gauss' theorem tells you this in this special case. Now, the center of mass has a geometrical definition, while the zero gravity point has a physical definition and the two will not coincide in general, but there is no way to have an infinite gravitational force from a finite mass.
Disregarding any analytical solutions, including Shell theorem, Gauss' law for gravity, and especially Birkhoff's theorem (relativity), it seems there should be a critical distinction between spherically symmetric gravitatonal solutions for the attraction vectors of classical physics and vacuum field equations for a kinetic force: the centerpoint of attraction vectors should cancel out, producing a zero result, while a pushing force effect should produce incremental (rather than infinite) maximal results...
Physically, either a zero or infinite solution would seem to indicate that no spherical mass should produce sufficient pressure to induce particle fusion at its center point, or that any spherical mass should produce sufficient pressure to induce fusion. Neither implication is satisfactory.
I'm not capable of reasonably evaluating, but a recent paper might suggest potential errors and alternative analytical approaches...
Please see
http://en.wikipedia.org/wiki/Shell_theorem
http://en.wikipedia.org/wiki/Gauss_law_for_gravity
http://en.wikipedia.org/wiki/Birkhoff_theorem_(relativity)
also see:
SHUANG-NAN ZHANG and SHUXU YI, (2012), "ON A COMMON MISUNDERSTANDING OF THE BIRKHOFF THEOREM AND LIGHT DEFLECTION CALCULATION: GENERALIZED SHAPIRO DELAY AND ITS POSSIBLE LABORATORY TEST", http://www.worldscientific.com/doi/abs/10.1142/S2010194512006642
Dear Sai Krishna, you stated:
"No, the mass enclosed decreases continuously as we move from the surface of the sphere to the its center and in no case the force is increasing as we go inside in our model. The net force vector is decreasing continuously as we approach the center."
"Yes, indeed the vectors do have a value near the center but near to zero."
Most interestingly, these statements seem entirely consistent with Shell theorem, etc., analyses, but inconsistent with observations as, for example, pressure seems to increase within stars so that fusion occurs generally in their central cores. Also, within the Earth it is the heaviest elements (iron) that are most commonly found near the center.
The discrepancy seem to lie in the idea that only mass within any radial location relative to the center applies an attractive effect to a test particle. Surely I'm simply overlooking some quite basic consideration, as it seems the idea that more external points within peripheral shells of mass contribute nothing to a specific test particle's gravitational effects must be false, perhaps falsifying the idea that gravity can be represented as an attractive force... What am I missing?
Dear Ana Maria,
your question about the gravity in the centre of an infinite spherical matter distribution is very important for modern cosmology and actually leads to some paradoxies in Friedmann model: what is the reason of the Friedmann force (and acceleration) between two galaxies in the homogeneous infinite matter distribution which determine the dynamics of the whole Universe?
In the Newtonian gravity theory there is Holtzmark theorem stated that in infinite homogeneous Poisson matter distribution the average gravity force equals to zero, and the fluctuations of the force is determined by the nearest neighbor particles. Hence there is no reason for global dynamics of such homogeneous galaxy distribution (global expansion or contraction) and only local strutures will be developed.
Another important note is that in NGT there are two different quantities - gravitational potential Fi and gravutational force - grad(Fi). For homogeneous infinite matter distribution the gravitational potential Fi = - infinity, while the gravity force grad(Fi) equals zero which follows from spherical symmetry and also confirmes for Poisson random distribution by Holtzmark theorem.
Another interesting paradox was already mension in this discussion by James Dwyer about the behavior of a clock in the centre of the matter ball, see reference
SHUANG-NAN ZHANG and SHUXU YI (2012) in his notes.
Actually this effect opens possibility for interpretation of cosmological redshift as gravitational effect without expansion (see our book Baryshev & Teerikorpi (2012), Springer).
Yurij Baryshev,
I presume you're referring to http://en.wikipedia.org/wiki/Holtsmark_distribution...
I can only guess, but the cosmological implications do seem intriguing but, as you mention, applicable only to an attractive gravitational force effect, by extension to the EM attraction of (unlike) charged particles.
However, in any expanding spacetime in which contracting (pushing) gravitational effects diminish as separation distances increase (as for any radial dispersal effect), the cosmological influence of gravitation also temporally diminishes, which should introduce temporal inhomogeneity into the distribution of matter. Of course, derived expansion need not have originated from a maximally interpolated singularity - more than one factor may be contributing to the apparent conditions of the universe...
To discuss the gravity physics one needs to use a known gravity theory, e.g. like Newtonian gravity theory, relativistic tensor field gravity theory or geometric general relativity.
Any questions on the properties of gravitational interaction have definite sense only within fixed gravitation theory.
In the case of infinite random Poisson distribution of massive particles interacting via Newtonian force the theorem is named Holtsmark-Chandrasekhar which state that the net average force acting on any particle from the whole infinite Universe equals to zero, and flactuations of the force equal infinity. But this infinity occur due to close collision between assumed pointness of particles and may be easyly removed.
@James (Dec 30, 2012): What you consider 'inconsistent with observations' actually completely agrees with a gravity force field that decreases towards 0 when we approach the center of gravity.
The scholary explanation is that pressure does not equal the force field (a scalar can't equal a vector anyway) whereas the divergence of the pressure does.
The natural explanation is: if you are near the center of gravity all the mass above you tends to squeeze you wheras gravitational forces pull in all directions and compensate each other as to become hardly noticible.
Of course, there is a rich theory of the interior of stars which studies the interaction of pressure (including radiation pressure) gravity and mass density in detail. There is nothing mysterious with this matter.
Dear Ulrich, in the case of finite radius body you are right, there is nothing misterious,
besides the case of black hole which has singularity in the center of mass and infinite force at the Schwarzschild radius.
Another problem exists for the cosmological case of infinite homogeneous mass distribution (or finite without a border) where Newtonian potential is infinit in each point, while the gravity force should be zero from symmetry condition.
The strange thing is that in general relativity the Friedmann equation which determined the dynamics of the whole Universe is exactly Newtonian force equation
where the gravitating mass is the mass of the ball having center in one galaxy and
radius equal distance between two considered galaxies. Though the force in this homogeneous case should be zero (according to the Holtsmark distribution).
Dear Yurij:
I had in mind Ana Maria's original question which reflects a wide-spread misunderstanding resulting from (unfortunately also wide-spread) bad physics teaching that mixes facts on point masses and on extended bodies. (I was told that one finds in some physics text-book the problem: Consider a tunnel that extends till the center of the Earth's body. Pump all air out of it and let a stone fall down the tube. Show that it will fall to the center and will be reflected from there and return to the starting point! Of course, if all the Earth's mass would (and could!) be concentrated in their center point, this would happen.)
The problematic (in my view ridiculous) idealization of on infinitely extended homogeneous mass distribution can hardly add to resolving Ana Maria's puzzle.
I ask this many years ago because the formula: g= G . M / r2
if radius is 0, result x/o. But we think that this be only by the surface of a mass.
Now I look, that is only for surface. But if the formula inside the mass needs r in the denominator, we have the same problem.
About this image of a tunnel, I think that because pressure and temperature, this is false and impossible on Earth.
Stars, planets, etc.:
First of all, the mass of a "body" in general relativity or in relativistic gravity can be well defined provided the spacetime it generates is asymptotically flat and sometimes of other classes (e.g. asymptotically de Sitter). So assuming this is the case, then for many bodies in the Universe, like planets, stars or other extended bodies that are no singular at the origin, then certainly one can define a center or origin and gravity is finite there. Now, it depends what do you mean by "gravity" here if you want to asign a value. If you mean the scalar curvature (ie. the Ricci scalar) then the value of gravity at the center is very simple: R_0= -8pi G T_0, where T_0 is the trace of the EMT evaluated at the center. So if the body is described by a perfect fluid then T= -rho + 3p. Therefore, if you know both the density of the body and its pressure at the center then you can easily evaluate its Ricci curvature. Notice that IT IS NOT ZERO NOR INFINITE, in this case. For "ordinary" (non compact) bodies, you can neglect the pressure.
If by gravity you mean the Riemann curvature, then, it is also finite, and you can also compute it at the center. It is a bit more complicated but not so much.
If by gravity you mean the acceleration "it feels" a static observer placed at r=0, then it is zero (the "grav. potential" is flat at the center and so its gradient is zero).
What about other bodies like black holes which are singular at the origin or "systems" like the whole Universe, that if homogeneous and isotropic, does not have a center or origin ?
Black Holes:
In the case of BH you can evaluate its surface gravity (it is finite ~1/M for a Sch. BH), but it does not make too much sense to compute gravity at "the center" because at the interior of BH strange things can happen. But even forgetting about these nuances, for instance, the Ricci scalar can be zero everywhere, while the Riemann tensor or its Kretschmann can be infinite at the coordinate r=0. So in this case you find an answer similar to what was in your question.
Universe:
since the Universe is not asymptotically flat it is difficult to asign a mass to it, even if intuitively you might think you can. If it closed you might though, but all the current evidence shows that it is open. In any case, there is not a center here. Gravity is finite everywhere and at all times except at the bigbang (assuming classical GR ).
Concerning the formula g=GM/r^2, you must have written g=GM(r)/r^2.
Outside the body M= const., but inside the body it is not. In fact, near the origin
M (r)~ r^3. Therefore g~ r . So at the origin r=0, you got g=0. This coincides with my previous answer concerning the acceleration felt by a static observer at r=0, since the potential phi~ const. + r^2 and its gradient at the origin vanishes. This analysis is more Newtonian, but you got the same conclusion if you do it within a relativistic description.
No singularity unless the mass is a point mass (a bit more generally, if the mass density has a singularity there).
Ulrich Mutze - I agree, in terms of analytical evaluations, but to elaborate...
I'm not even a proper student of physics, but as I understand, center of mass and point mass and singularity all represent mathematical abstractions used to simplify approximations of various effects. The principal difference is that it is commonly suggested that a singularity somehow exists as a physical entity - an enormous amount of mass physically contained within a dimensionless point - supposedly leading to physical infinities.
In gravitational evaluations, isn't the center of mass often used to represent the net of a distribution of dynamical effects occurring throughout a dimensional object containing a distribution of mass? Likewise, a point mass in classical dynamics can represent the gravitational effects produced by a spherically symmetrical distribution of mass in accordance with shell theorem...
From this perspective, a center of mass is a much more general term than either a singularity, which represents a dimensionless mass, or a point mass, which can properly represent a spherically symmetrical distribution of mass or at sufficient distance, any discrete distribution of mass (at sufficient distance the distribution of gravitational effects produced by even a binary star or an edge-on spiral galaxy effectively resolves to a single point).
Dear James,
science is not about having many notions available for free mixing, but about using well-defined notions that allow us to express thoughts with a precision that exceeds the precision of discussions in daily live.
The center of mass is a point-valued descriptor of a mass distribution. If the mass distribution is that of a mass point, the center of mass of this distribution equals the position of the mass point.
If the mass distribution is fully contained in some finite volume of space, as you said, the gravitational field outside that volume approximately equals the gravitational field of a point mass which collects all mass of the mass distribution at the center of mass of that distribution. (All my statements refer to non-relativistic (Newtonian) gravity). Actually this point-mass field is the first contribution in the so-called multipole expansion, all higher contributions fall off with distance much faster than the point-mass contribution. If the mass distribution is spherically symmetric, all these higher contributions vanish.
A field is said to be singular at a point if it grows over all bounds when approaching this point. So a point-mass (i.e. an idealized peaked mass distribution) creates a gravitational field which is singular at the location of the point-mass (which is the center of mass).
A continuous mass distribution creates a gravitation field that is not singular at any point (not even at the center of mass).
Dear Ulrich,
Thanks for elaborating. The reason I interjected my comment/question is that I think that your technically correct terse answer was an inadequate response to someone who was considering whether a center of mass might be a singularity. In simple, imprecise terms, they are apples and oranges and should not commonly be subject to confusion - I think then that a more complete explanation was necessary.
Let me be perfectly clear: I am not a scientist and have no education or background in physics, mathematics or engineering - I am merely a retired information systems analyst. Thanks for patiently providing the more precise elaboration I was hoping for.
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