It would be much easier to use the sphere as nothing more than a frame and waterproof the lighting system. I doubt the lights take up much more than 5% of that sphere volume, so why introduce 95% air where you don't need it?
Think of it this way, for all the air volume you have, you need to carry that much water weight around with you to make this thing neutral buoyant. (ignoring the density of the lights and sphere material)
Steve C is off by at least a factor of four since he got the volume formula wrong.
If you need to weight it down, you need to add weight, no way around that.
sulphur hexafluoride ? strange idea. That is a gas and will add more buoyancy, the opposite of what you need.
And the more dense the weight, eg, lead, the better it is, as the volume of the added weight (assuming it is on the outside) will add little to the buoyancy.
edit, rereading, I suspect this is a hollow sphere, which you don't mention. Are you talking about adding the SF6 to the hollow portion? Again, you didn't hint at that. That will increase the overall density a bit. You could fill the inside with water for even more increase in density.
For the sphere to sink it has to have greater density than the water or be hauled down by anchor lines, etc. For it to have greater density that water is has to weigh at least 63 pounds per cubic foot of volume because fresh water is 62.42 lb/ft^3. A 3' sphere has a volume of 14.14 cubic feet and would thus have to weigh at least 890 pounds. No way a person can carry it. And no way around the physics.
volume of a sphere = (1/3)* pi*r^3
or (1/24) *pi*d^3
36 inch is is 0.889 m . so a sphere of that diameter would have a volume of about 0.092 [m^3]
1[m^3] of fresh water is approximately 1000kg so to counter the mass of displaced water for neutral buoyancy you'd need the ball+contents to have a mass of 92 kg
How is this for an idea... If the outer ball can't flood because you want to keep the lights dry, can you place a smaller ball inside the large ball, and allow the smaller ball to flood/empty using tubes going to the outside? That keeps a dry volume inside the ball. Say an inner ball was 34 inch diameter (1 inch gap between spheres to put the lights in to). that's 0.8636 m diameter and it would be flood with up to 84kg of water. If you do that the balls would only need to have an out of water mass of 8 kg to achieve neutral buoyancy.
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thanks TomG for pointing out the error in the above. yeah formula was out by factor of 4
V=(4/3) pi *r^3 = (4/24) pi*d^3
so would need mass of 368 kg to sink a completely unfilled 36" sphere in water
mass of water holdable by a 34" diameter sphere =336kg
difference =32 kg
now if a 1/2" gap between spheres was feasible...
mass of water holdable by a 35" diameter sphere= 367.88 kg
difference in mass water holdable by 36" and 35" spheres aprox =120g
it is sealed and needs to not take on water. it will have lights inside. It wants to be light enough that a person can carry it, it should be about 36" diameter. how about using sulphur hexafluoride to weight down the sphere but not add too much weight to it.
