Do antimatter and negative mass experience gravity differently?

[George Jones]

Isn't it a hole in a sea of negative-energy states, so the hole represents absence of negative-energy?

 

[ExactlySolved]

The sea of negative energy states is normally packed full of electrons, and this vacuum state has no net gravitational attraction, since the sea of electrons is homogeneous and isotropic. Now if we remove one of these electrons and send it some where far away, then all that is left is a hole where that electron used to be. Because of the hole the sea is no longer isotropic or homogeneous, and so if you are an ordinary matter particle looking at the hole then you have an infinite line of mass behind you, and an infinite - 1 line of electrons in front of you, therefore you will be pulled backwards, away from the hole.

Mathematically, I had to think carefully because normally a single point does not contribute to an integral, but in this case the mass of the hole, if we imagine it as being an empty site on a lattice with all other sites occupied, is less than the mass on neighboring sites by a factor of infinity, and so a delta function is appropriate which renders allows the integral to be effected in a finite way by the subtraction of a single point.

 

[Vanadium 50]

protons are uud, neutrons are udd, I believe that anti-protons are u-bar,u-bar,d-bar, anti-neutrons are u-bar,d-bar,d-bar.

That's a very superficial view. Those are the valence quarks. The proton is also composed of gluons, and sea quark-antiquark pairs. In fact, the parts other than the valence quarks carry the majority of the momentum of a proton.

I maintain my previous position - a different pull on matter and antimatter would yield a composition dependent force. Which is not observed.

 

[George Jones]

I know almost nothing about this, so I'll probably find myself in the sea over my head, but here goes ...

The sea of negative energy states is normally packed full of electrons,

So isn't it a sea of negative energy electrons?

and this vacuum state has no net gravitational attraction, since the sea of electrons is homogeneous and isotropic.

This is true if the sea were all the electrons were positive energy, or if the see were all negative energy electrons.

Now if we remove one of these electrons and send it some where far away, then all that is left is a hole where that electron used to be.

Where a negative energy electron used to be? Thus, if a normal positive energy electron drops into the hole, it falls to a lower energy (in this case, negative) and emits radiation; pait annihilation.

Because of the hole the sea is no longer isotropic or homogeneous, and so if you are an ordinary matter particle looking at the hole then you have an infinite line of mass behind you, and an infinite - 1 line of electrons in front of you, therefore you will be pulled backwards, away from the hole.

Shouldn't "infinite line of mass behind you" be "infinite line of negative energy behind you", etc., so a normal electron is pulled (gravitationally) towards the hole?

 

[ExactlySolved]

It is correct that the electrons in the dirac sea have negative energy, so this together with what GR says about the gravitational effect of negative energy would resolve the contradiction, thanks George.

It looks like what I said would be true for a large homogeneous mass of ordinary matter with a hole in it, the hole could be treated as a gravitational repeller, my mistake was extending this picture too far.