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What about gravity's relativity?

  1. Apr 2, 2012 #1
    Why is not gravity relative like the space-time continuum, since it operates on and is modifying it?
    Why is gravity exactly inversely equal to the square root of the distance between objects and not some number like pi or the speed of light?
  2. jcsd
  3. Apr 3, 2012 #2
    I don't have time for a proper response right now but I can quickly address one thing.

    Why isn't gravity's force inversely equal to the speed of light?
    I don't even know what this means. C is a velocity, not a measure of distance. I think you need to be more specific with what units you're using and exactly what you mean.
  4. Apr 3, 2012 #3
    Thank you, Technoir, for your help.
    What i meant was: 1) Is gravity absolute or relative like space-time?

    2) Why is gravity EXACTLY relative to the square root of the distance, how come it is not a fraction of it, why precisely the square root? It seems too neat to me.
    Why not .9999 or 1.0001 or some infinite number like pi?
  5. Apr 3, 2012 #4
    Gravity is just as relative as space-time is. Keep in mind you need to use general relativity to describe relativistic gravity, instead of newtonian gravity.

    That is an excellent question. I don't think there is an entirely definitive answer as of yet. An appealing idea to explain exactly integer exponents (like '2') is by relating them to the exactly integer number of dimensions. For example, the inverse square law would have to do with 2 being 1-less than the number of large spatial dimensions (i.e. because the surface of a 3D region is a 2D surface)....
    Last edited: Apr 3, 2012
  6. Apr 3, 2012 #5
    That is incorrect. The space-time is very much non-relative. Only space and time as separate concepts are relative.

    Do you mean square instead of square root?

    You have to understand here that in general relativity, this is not true. In general relativity, gravity is not a force at all. Particles just move as straight as they can, but as space-time is curved, the particle world-lines appear to be bent.

    Then the question becomes: why in the limit of weak fields, the apparent force goes down like 1/r². The reason is that gravitational force carriers are conserved. Force of gravity is proportional to the flux of these force carriers, and as they are conserved, the flux at a given radius is proportional to the area of a sphere of that radius.
  7. Apr 4, 2012 #6


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    Newton formulated the 1/r^2 relationship of gravity because it explained Kepler's laws. It's only approximately true. Note that this is an inverse square relationship, not an inverse square root relationship.
  8. Apr 4, 2012 #7
    To Zhermes, Clamtrox and Khashishi: thanks for your help.

    I take note:square not square root.
    Still, is it exactly square or approximately square , and if the latter what?
    How come magnetic attraction is inversely proportional to the cube of the distance?
    If there is a relativistic gravity, is it subject to the same effect as time (such as the so called twins paradox, which could as well be called the two clocks paradox or the two anythings paradox)? In other words does gravity depend on the speed something is moving?
    As for the two different explanations received, i am affraid i don't quite understand them; the link between the dimensions of space-time or the volume of a sphere and the force of attaction of gravity on one side and the square of the distance between two objects on the other is not easy to explain in plain english i suppose.
    It seems there is no consensus on this matter.
  9. Apr 4, 2012 #8


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    There is, but remember that most people here are not physics professors. We all have varying amounts of knowledge about a subject, and sometimes one person can simply be incorrect and not know it.

    I myself have had difficulty in understanding what you've been asking, as the terminology you used isn't quite correct. Of course this is simply due to lack of knowledge about the subject on both your end and my end. If I understood more I would probably realize exactly what your asking and be able to explain it the correct way.
  10. Apr 4, 2012 #9
    C. Bernard, there is a definite reason why Newton used the inverse square law for gravity. The inverse square law can be used to describe any force that radiates outward, like gravity. Since the surface area of a sphere is proportional to the square of the radius, as the emitted force gets farther from the source, it is spread out over an area that is increasing in proportion to the square of the distance from the source. Hence, the strength of the force at any unit area is inversely proportional to the square of the distance from the source.

    Obviously, we now know that gravity is not a force. But Newton's law provides a very accurate result, and is used for all space missions. Einstein's equations only become relevant in enormous gravitational fields.

    There is a consensus on the matter, it is not trivial in any way.

    Also, gravity is not relative. Remember Einstein's first postulate of special relativity, the laws of physics are the same for all inertial frames of reference. So, some object approaching the speed of light will not become a black hole, as some people tend to think.
    Last edited: Apr 4, 2012
  11. Apr 4, 2012 #10


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    many have provided good answers, but I'dd add 2c.

    Einsteinian gravity is a field. It is everywhere. It may have different values here and there, but every point in space has a value.

    Anything that radiates from a point (such as Newtonian gravity) will have an inverse square relationship. Its value is essentially that of the surface of an expanding sphere. Light does the same thing. If 100 photons leave a point source, the will spread out such that, at twice the distance, they cover 4x the area so their density will be quartered; at 3x the distance, their density will be 1/9th. 4x the distance, their density is 1/16th, etc.
  12. Apr 4, 2012 #11
    It is exact (as far as anyone can tell), and if it wasn't exact---to my knowledge---the universe wouldn't work anywhere close to the way it does. Kepler's laws, for example, are dependent on it being exactly an inverse square law. Similarly, extended massive objects wouldn't be able to be treated as point masses... etc etc.

    It isn't. In general the 'magnetic attraction' depends on the particular geometry, but is based on the same inverse square law as electricity... for two bar magnetic the force is approximately an inverse-square law (if they're end to end); for a general dipole I think its an inverse fourth-power. It all stems from the inverse square law of the electrostatic force.

    Mark M gives a good explanation of the surface-area---inverse-square-law connection.
  13. Apr 4, 2012 #12


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    Take a square n units on a side. Its area is exactly n^2.

    This does not seem unusual does it? You don't wonder if it's approximately ^2, or whether there's some strange coincidence at work?

    When you relate linear dimensions to area dimensions, you get ^2, exactly , whether it's squares or circles or surfaces of spheres.
  14. Apr 6, 2012 #13
    Many thanks to everybody who has intervened in this discussion, i think i get it.
    As for the magnetic attraction being the inverse of the cube of the distance, i found on wikipedia article "magnet", paragraph "fields of a magnet", so it is not entirely reliable.
    Still it is amazing than a natural phenomena conforms to a geometrical figure which is a virtual reality more or less invented by man, after all a perfect sphere does not exist in nature.
  15. Apr 6, 2012 #14
    Yes, the field, in that case, goes like one over the distance-cubed---not the force.

    It doesn't matter if the shape is perfect or not---that just changes the constant of proportionality, as long as you're talking about the surface area - it scales like [itex]r^2[/itex], where the exponent is exactly 2. If general relativity teaches us anything, its that the underlying geometry of nature is incredibly decisive in its behavior.
  16. Apr 7, 2012 #15


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    I feel compelled to contribute here and say that 1/r^2 is NOT the whole story. In Einstein's general relativity, for one, gravity isn't even thought of in the traditional sense of a force, so this kind of thing isn't really well defined. BUT what you can do is in the limit of slow moving objects that aren't massive compared to black holes, you can see what gravity looks like. It turns out that when you reduce to the case of Newtonian mechanics, you do get 1/r^2, but it actually looks something like
    [tex] F(r)=\frac{\alpha_2}{r^2}+ \frac{\alpha_3}{r^3} + \frac{\alpha_4}{r^4} + \cdots [/tex]

    Where the [itex]\alpha[/itex]s are some proportionality constants (They're related to v/c, the velocity divided by the speed of light. So when this is small, each next term in the series is much smaller than the last, so is less important). What's important is that the deviation from Newton DOESN'T appear in a difference from two in the lowest order term, but rather the existence of these higher order terms like 1/r^3, 1/r^4 etc.
  17. Apr 7, 2012 #16


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    If you prefer to picture gravity as a force that permeates space, the answer is intuitively simple - its declines with distance squared because the volume of space it encompasses increases by a factor of distance squared.
  18. Apr 7, 2012 #17


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    You're looking at it wrong.

    When you examine a single factor influencing a phenomenon, the formula describing the effect will be correspondingly simple.

    Perfect spheres do exist in nature: the surface where all points are equidistant from a central point is a perfect sphere. So the surface where all points are exactly 1 light second from Earth's centre-of-gravity is a perfect sphere 1 light second in diameter. (This it trivial).

    That formula is quite exact - and simple. And it is that principle that we are talking about in this thread.

    However, when you examine the reality - you take all other factors into account - for example, there are more and less dense areas on Earth's surface that will cause local gravity over that point to vary. So that means that the surface where all points are under the same force of gravity is not a perfect sphere - it is quite lumpy.
  19. Apr 8, 2012 #18
    Well, i may have been overly optimistic when i wrote i got it.I sure don't get it anymore.
    Magnetic fields are function of a volume, but magnetic forces of a surface. But both gravity forces and field are a function of a surface.
    It seems gravity is a function of "v"/c. What is v: the speed of gravity? Or is gravity instantaneous? So gravity was relative after all.
    Or is v the speed two objects are moving relative to each other?
    If r=1, does F=1? If r=0,does F=∞?Does the sum of all the ∂ tend to 1? If so, is F≤1/r^2?

    Please somebody, help me.
  20. Apr 8, 2012 #19


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    I apologize if I confused the issue any with my response! v is intended to be the relative velocity of the bodies in question, yes. Keep in mind though, that the expansion I wrote is just that: an expansion. To say that gravity is a function if v/c is a little misleading, then. The ACTUAL equations of gravity are those of general relativity, in which you cannot even write F(r)=??? So again, keeping in mind that we're in this expansion framework, not full general relativity, the equation to first order is this:

    As you can see, very complicated! My schematic description was merely to illustrate that it's not the full story.

    If r=1, F= whatever, nothing special. In general, though, F(0) is undefined, as you'll be dividing by zero.
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