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Gravity decreases as 1/r

  1. Mar 26, 2003 #1
    Ok the above statement is not STRICTLY true. It should read...

    "The gravity of a light emmitting star will appear to decrease as 1/r, at large distances"

    I think this is what MOND theory holds but has never actually said why. This is a proof I discovered to possibly explain this phenomenon.
    God I hope these scripts work.


    OK my bold assumption is as follows. Light, or photons, exerts a gravitational force on the universe around it. A photon with energy E has associated with it a gravitational acceleration

    ap= Gm/d2

    where m = E/c2 because E=mc2

    I'll sight gravitational lensing and redshift as examples of this phenomenon. In both cases the momentum of the light photons has been changed. Direction has changed in the former, and magnitude has changed in the latter.

    In both phenomena the momentum of the photons has been changed and so by Newtons second law the mass that shifted them, usually a star, galaxy, or black hole, will experience an equal and opposite change in momentum.
    This is my justification, flimsy as it may be for setting light to have gravity.

    Imagine a star,alone in space, no planets or space dust about it. Now imagine a sphere of radius a surrounding the star and a mass just hanging on the edge of this sphere, as badly shown.




    We know that the Force of Gravity on the mass m due to the star of mass Ms is

    Fs = GMsm/a2

    Imagine the sphere, of radius a. What is contained within this sphere?(Apart from the star at its center of course.

    Our first responce, nothing! Empty space. But of course if we look more closely we realise that there is indeed something in this void. The light from the star!

    If light has gravity as we have supposed then what is the gravity on the object on the edge of the sphere, due to the "mass" of the light within this sphere.

    Let us call this supposed force FL

    N.B. From here on in the maths gets really hairy so if you hate calculus the just skip to the end.

    It is known that

    M = &int;&int;&int; &rho; dV , where &rho; is the density at a point.

    Thus ML = &int;&int;&int; &rho;L dV

    but what is &rho;L , the density of the light at a point.
    We can regard &rho;L as either the mass or enegy density at a point. But the energy density is the same as the Intensity of the light at that point.

    So &rho;L= I
    Where I is the Intensity of the light at that point. Convince yourself of this, or convince me otherwise! :E

    Note : I tried to find a better symbol for proportionality but all I came up with was &alpha;
    So take it that
    y &alpha; x means
    y = kx , for some constant k.

    We know that I &alpha; 1/r2
    where r is the distance from the light source.
    thus &rho;L&alpha; 1/r2

    &rho;L = k/r2

    k is an arbitrary constant


    ML= &int;&int;&int; k/r2 dV

    ML = &int;&int;&int; k/r2 dxdydz

    it is more convinient for us to use spherical co-ordinates in this integral so we use the fact that

    dxdydz = r2Sin&Theta; drd&Theta;d&phi;
    r - radius from origin
    &Theta; - co-latitude
    &phi; - latitude
    Note to the Side:

    I'm unable to find a good script for the limits of integration so I'm using this convention. The integral from x=a and x=b is
    &int;x=a..b dx


    and so

    ML = &int;&int;&int; kr-2r2Sin&Theta; drd&Theta;d&phi;

    ML= &int;&int;&int; k*Sin&Theta; drd&Theta;d&phi;

    Subing in the usual limits for &phi; and &Theta; and r ranges from 0 to a

    ML = &int;&phi;=0..2[pi] &int;&Theta;=0..[pi]&int;r=0..a k*Sin&Theta; drd&Theta;d&phi;

    ML = &int;&phi;=0..2[pi] &int;&Theta;=0..[pi] r*k*Sin&Theta; |r=0..a d&Theta;d&phi;

    ML= &int;&phi;=0..2[pi] &int;&Theta;=0..[pi] ak*Sin&Theta; d&Theta;d&phi;

    ML = &int;&phi;=0..2[pi] -ak*Cos&Theta;|&Theta;=0..[pi] d&phi;

    ML = &int;&phi;=0..2[pi] -ak*Cos([pi]) + ak*Cos(0) d&phi;

    ML = &int;&phi;=0..2[pi] 2ak d&phi;

    ML = 2ak&phi; |&phi;=0..2[pi]

    ML = 4ak[pi] - 2ak(0)

    ML = 4ak[pi]

    So the mass of the sphere of light is 4ak[pi]
    where a is the radius, and k is the coefficient of E.M. intensity

    So the gravity due to this collection of shells is the gravity due to a point mass of the same mass as the sphere, at its centre. This is due to Newton's shell theorem.

    FL = GMLm/a2

    FL = 4ak[pi]m/a2

    FL = 4k[pi]m/a

    And since 4,k, and m are constant

    FL &alpha; 1/a

    but Fs &alpha; 1/a2

    Thus FL although weaker initally will eventually become greater than Fm

    For our own sun this occurs at about 7.537 * 1014m from the sun, but my calculations are VERY rough so take that with a pinch of salt.
    Or maybe you should take this whole post with a pinch of salt, a very large one at that! :E
    Last edited: Mar 26, 2003
  2. jcsd
  3. Mar 26, 2003 #2
    Hmm, interesting theory. I'm suprised that this idea isn`t analogous to any spherical object, including planets. But of courses it can`t be, because planets exert 1/r^2 gravity. That's one of the differences between light and matter eh?, or just the effect of being inside the perimeter of the object ( e.g. gravity varys at 1/r for inside the Earth, while 1/r^2 outside ) ? OMF, your math doesn`t look too radical, surely others will have already deduced it (* implied when you refer to newtons shell theorm ) ?

    I'd be interested to know if this idea can explain the freaky slowing down of satellites as they travel out of the solar system.
    One important aspect of an explaination for this would be for your idea to exclude planets, as they don`t seem to be affected by the fraky slowing down force, i.e. your idea must only affect smaller objects. Of course, you should also check to see if it can explain the galaxy rotation problem.

    Yeah, I'm glad someone else accepts 1/r gravity, and thinks it isn`t hard to explain. It shouldn't be such a herectical thing to believe. But there's no way you'll get the majority of scienctists to go with it.
  4. Mar 26, 2003 #3
    OMF, would the math work for vacuum fluctuations in the same way you've done it for light? It should add to the overall strength of the 1/r effect.
  5. Mar 26, 2003 #4
    But hey, MOND is aimed at intergalactic distances, many magnitutes greater than 10^14, so really the 1/r effect should have have no problem dominating, even if we take a critical view of error bounds in your maths.
  6. Mar 27, 2003 #5
    Well at least someone is interested this time.

    This thought actually first came to me when I read about the voyager satellites expieriencing an abnormal slowdown. Then I read about those MOND people saying gravity decreased as 1/r

    I suppose if this is right the gravity, for a light emmitting object, would decrease as

    A/r2 + B/r

    so the 1/r term would dominate as r->infinity

    This is like the gravity inside planets, which is proportional to r, until you get to the surface, then it is proportional to 1/r2

    But unlike planets the "density" of light is not constant, but is itself proportional to 1/r2

    I'm not sure about vaccuum fluctuations though. Mostly because I havn't a clue what they are. :E
  7. Mar 27, 2003 #6
    I've just been checking the numbers myself basesd on the my approximation for light radiation from the sun being 5000000000kg per sec, and they don`t add up to a figure anywhere near big enough. Don`t know how you manage to get 10E14m.
    Light-spheres alone are not enough to explain MOND, galactic rotation or satelitte trajectorys.
    I still like the idea of intergrating a 1/r^2 field though, but which field if not light?
    Looks like Vaccuum fluctuations might be our only hope!
  8. Mar 28, 2003 #7
    My approximations are probobly wrong.

    The only unlnown is the value of k in the expression. I think I used the solar constant to figure that out but the solar constant only takes into account, infered radiation from the sun, whereas for the complete picture you need the whole spectrum.
  9. Apr 1, 2003 #8


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    BTW, there's a shortcut for your computation of the total energy of the light:

    The total energy of light contained in any spherical shell is constant (since any particular smaller shell will expand to any particular larger shell and total energy is conserved), so you simply need to integrate a constant over the permitted range of radii and you get a total energy proportional to distance!

    And P.S.: you mixed up &theta and &phi in your formulae!

    Interestingly, you might apply the same idea to gravitational energy! (GR shows, incidentally, that gravity doesn't just propagate but it generates more gravity)

    Assuming that initial gravitational energy density is 1/r^2, you get K/r for the gravity of the gravitational energy... then k^2/2 for the gravity of the gravity of the gravitational energy... then k^3 * r / 3!... then ...

    adding it all up, you get:

    F = (1/r^2) * e^(kr)

    for gravitational force. When kr << 1, gravitational force would look like 1/r^2, but when kr gets bigger gravitational attraction would quickly blow up towards infinity... this all assumes instantaneous gravitational transmission, though; the need for gravity to propagate would profoundly alter this calculation, so take this argument with fewer grains of salt than OMF's.

  10. Apr 1, 2003 #9


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    Re: Re: Gravity decreases as 1/r

    I think i've covered this with someone else before, but:

    Keppler's Law only applies to a single planet in a given orbit. It does not apply to different orbits.

    If you have two planet's in circular orbits, one at R1 and one at R2, Keppler's law does not say that each planet must sweep out an equal area in same time. Each planet will sweep out a different area in any given time period.

    What Keppler's law does say that a planet in orbit will sweep out equal areas at all points of its orbit. This is obvious for a circular orbit. But Keppler says it is true even for none circular orbits.

    For instance, If a planet is in a elipitical orbit, it will sweep out the same area per sec while at aphelion, perihelion, or any point in between.

    This is not the same as saying that objects in different orbits will sweep out equal areas in equal time periods.
  11. Apr 2, 2003 #10
    This isn't a shortcut. Although you are correct, the energy in each shell is constant, however the energy density is not, and it is upon the density that the equation depends.

    I know that &Theta; and &Phi; are mixed up to some, if not most.
    I'm using the convention where &Theta; is the co-latitude(longditude) and &phi; is the latitiude.

    Sorry About any confusion. The variables dissappear in the integrel in any case.
  12. Apr 2, 2003 #11


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    Yah... latitude only traces out a half circle (it goes 0 .. pi) but you used &theta as your variable for that.

    The shortcut does work... if you remember the section on solids of revolution from your calc class. The ordinary triple integral chops up all of space into little cubes and you add up the energy of each cube to get total energy; the shortcut only chops up space into spherical shells and adds up the energy of the shells to get total energy.

    Algebraically, in this case it's just skipping using &theta and &phi becuase spherical symmetry implies that integrating those variables will just result in a 4 pi constant factor.

  13. Apr 6, 2003 #12
    Yeah, Hurkyl's method is the best, it's the one I used.
    Right, so if light shells aren`t enough to explain the freaky diminishing gravity effect, what other possibilitys does that leave us with?
  14. Apr 8, 2003 #13
    The light shells explain an INCREASE in predicted gravitational effects.
  15. Apr 12, 2003 #14
    very interesting idea. I have always thought that light must possess some gravitational effects just for the same reasons you say... they BEND, so whatever source is bending them should feel this.

    I have a theory that I will be posting soon on the PhysicsPost, where the abnormal slow-down of satellites would be a natural effect. There are issues I have with Mond, but I haven't quite put together a logical proof to rule it out yet though.
  16. Apr 15, 2003 #15
    Excellent contribution, Hurkyl, but what about the gravity of the light? Is it enough to contribute to the apparent anomalous acceleration? as OMF says,
    acceleration = Gm/r2=GE/c2r2=Ghf/c2r2
    at one point.
  17. Apr 15, 2003 #16
    It all depends on the constants.

    The gravity of the light shells would be proportional to 1/r


    F = k/r

    depending on k it may explain the decelleration of the voyager satellites
  18. Apr 15, 2003 #17
    Very interesting,OMF. Could it be possible that the sun's gravitational force starts overiding the earth's gravitational pull on the satellite when the satellite is at a specific distance from the earth?

  19. Apr 15, 2003 #18


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    That doesn't make any sense...

    The voyager probes were not satellites, they did not orbit the earth. And I don't see how the sun can overide earth's field...
  20. Apr 15, 2003 #19
    Sorry, probes, satellites, whatever.

    The point of this post was that the sun's gravitational field will be surpassed by the gravitational effect of the sphere of light surrounding it at very large distances.
  21. Apr 15, 2003 #20
    Aha! I knew there was something fishy here, and I figured it out. :)

    OMF, your derivation is valid... for a "freeze frame" in time. But, it won't have any actual effect on motion. The basic reason you get the 1/r effect is because as r increases, more and more light is subsumed into the sphere of radius r behind you, which acts like all its mass was at the origin.

    *But*, in actuality, the energy of this light is coming from the star itself, and moving outward at c. So, if you were moving away from the star at c, you would be just keeping up with your current shell of light. More energy and hence gravitational pull would show up in the sphere of light -- but this would be exactly cancelled by the energy&gravitational pull loss from the star's mass decrease. So you get the usual 1/r^2 effect.

    If you're moving away from the star slower than c, gravitational pull will actually decrease faster than 1/r^2 , as more of the mass-energy of the star passes you. In the case where you are standing still, the effective mass will drop at a rate of Luminosity/c^2 and so gravitation force by a rate G*L/(c^2*r^2).

    For v<<c, this will dominate, and the outgoing object will see a force of G*(M/r^2 - L/(c^2*r*v))
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