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To be clear, this example can't apply to the Solar System, since the barycenter is within the Sun, similarly the Earth/Moon system's barycenter is within the Earth.

But, given a system of gravitationally attacting masses revolving about a barycenter that is not contained within any of the masses, i.e. a barycenter in open space, would a photon passing near that point behave in the same manner as a photon passing a point mass of the same mass as the system itself?

anna v
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user1003469
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  • Actually, the barycenter of the Sun-Jupiter system is above the Sun's surface, at 1.068 solar radii from the Sun's center, according to Wikipedia. (It might dip below the surface when Jupiter and Saturn are at opposition; I haven't crunched the numbers.) – Keith Thompson Nov 18 '11 at 08:37
  • I meant apply to the Solar System as a whole, which Wikipedia told me was inside the sun. – user1003469 Nov 21 '11 at 19:39

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The barycenter has no mass and therefore no forces emanating. This is evident by your example of the moon earth barycenter which continually moves in the mantle 1700 km down or so. If it had any effect it would be working as a whip in cream, generating from quakes to volcanoes!

It is just a geometrical point whose use is to give an observer outside the system a reference point for calculations at large distances from the system.

The photon, or any other particle, will feel the gravity forces of the individual masses according to the laws of gravitation, the distances it has from the individual masses. BTW the forces felt by the photon will be very short and transient.

anna v
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  • Yes, I certainly understand that. I'm not sure that answers the question about the modeling, however. A photon passing relatively near the earth can be accurately modeled as if the earth was a point source. A photon passing relatively near the outer solar system should, at some sufficient distance, be successfully modeled as if it were passing a point source at the solar system's barycenter. My question was whether the collective influences of the multiple bodies in a system having an open space barycenter would have the same effect as passing a point source (neglecting motion of the system). – user1003469 Nov 17 '11 at 18:22
  • So just to be clear, your actual answer to the question would be 'No', correct? It would not behave the same? – user1003469 Nov 17 '11 at 18:24
  • It is no for the case of of the photon passing by the barycenter through the solar system, and yes if you are at a great distance so that individual constituents of the conglomerate mass are not discernible. In the intermediate range, i.e. outside the solar system but not extremely far, it could be used as a first order approximation, the closer, the worse. – anna v Nov 17 '11 at 20:02
  • I should clarify that "solar system" above means a hypothetical one, when the barycenter is in the vacuum as the question asks. – anna v Nov 18 '11 at 04:44
  • @user1003469: The Earth is a special case. Any spherically uniform body is gravitationally equivalent, for any object outside the surface of the sphere, to a point source at the sphere's center. – Keith Thompson Nov 18 '11 at 08:27
  • Sure. This extends to a system of masses, outside their orbits. The large distance is necessary for accurate solutions using the total mass in the barycenter, because if individual masses are "seen" by a test particle (the photon in your case) the two body forces will change the solution by pulling it differently than if all the mass were at the barycenter, depending on their orbits and their individual distances from the test particle.( a bit like tides) – anna v Nov 18 '11 at 10:26
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The answer is no, but for the weak-field, slow moving, Newtonian limit, the far away light can be thought of as deflected by the mass as concentrated in two dimensions to the Barycenter, when the mass distribution has a spherically symmetric 2d projection.

You can see that The deflection cannot only depend on the Barycenter using two orbiting two black holes. Then the light which is off-center can swing by the black holes and get deflected an enormous amount, while light going through the center mostly goes straight through.

But in the weak field limit, the deflection of light as it passes through a gravitating region is small, and the angular deflection is given by the sum of

$$ \Delta\theta = \sum_i {4GM_i\over c^2R_i} $$

Where $M_i$ is the i-th mass, and R_i is the impact parameter--- the distance of closest approach of the light to the mass. All but the factor of 2 in the formula is simple to derive, and a quick run through is contained in this answer, which has a detailed description of the weak-field bending of light by gravity: How does gravitational lensing account for Einstein's Cross? .

The upshot of the weak field limit is that light is deflected by a collection of nearby compact masses as if all the mass were squashed in a two dimensional sheet perpendicular to the direction of motion, and the light felt an impulsive force equal to the two-dimensional gravitational potential that would be produced by this source.

This means that if light is going in the z-direction, and it passes by a blob of matter, the deflection is according to the projection of that blob of matter to the x-y plane. If this blob of matter is spherically symmetric, a light ray passing through the center is undeflected. The precise deflection at any distance goes as the solution to the 2d Laplace equation, and it can be solved exactly in many circumstances. The linked answer works out many cases.

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Ron Maimon
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