# I Momentum not conserved for Contracting Concentric Shells

1. Dec 5, 2016

### FallenApple

So I have two concentric spherical shells of masses. Say outer is M(out) and inner is M(in) with radii R(out) and R(in) respectively. Let's assume that they are the same magnitude for mass and also that there is nothing but empty space between them.

Assume that both are flexible and can stretch like balloons at the same time, the particles that comprise them are solid like billiard balls. So stretching/contracting simply means that the spacing of these particles get smaller/bigger with time.

Well, initially, they are both M(in) and M(out) at rest.

But then they attract each other via the inverse square law of gravity. The inner shell doesn't stretch outward at all since we know that the gravitational field due to the outer shell is 0 everywhere inside of it. But from the outer spherical shell's perspective, the inner spherical shell can be considered a point mass. So every piece of the outer shell is being pulled on by M(in). This makes M(out) contract towards M(in) with increasing velocity while M(in) doesn't move.

Here is what happens: M(out) contracts inward towards M(in). M(in) will stay fixed since there is no influence from M(out).

Intially, as the outer sphere contracts, due to the symmetry, the vector momentum of each piece of M(out) will cancel out with another piece of M(out), specifically the opposite side. So momentum is conserved if the system is M(out) only.

However, if we consider M(out) as a whole, it would have a radial momentum inward. But M(in) has no radial momentum outward. So momentum is not conserved.

Can this be fixed if we consider momentum of gravitational field?

If we create a mathematical spherical volume V ecapsulating both shells at time t=0, then we can analyze how the momentum changes with time.

As the outer sphere contracts, the change in volume delta V becomes filled with the field lines from the outer shell. Since the field due to the M(in) is fixed for any region outside R(in), the displacement in volume corresponds to an increase field density. We see that the field density would increase with time within V.

So there are two contributions to momentum. The momentum of p=m*v and the momentum of the gravitational field. Both are from the outer shell, M(out) only.

So the flux momentum of this field is increasing inwards through time and the radial momentum of M(out) mechanical motion is increasing inwards as well and are both in the same direction.

So does this mean momentum is not conserved for this situation? Perhaps its a time delayed effect? That is the situation changes after the collision. But I thought momentum of a isolated system doesn't depend on time.

2. Dec 5, 2016

### A.T.

This is not how conservation of a vector quantity is defined.

And why make stuff so complicated? Just take two equal masses, initially together at rest, which fly apart so they now have "outward momentum". The actual total momentum is still zero.

3. Dec 5, 2016

### Staff: Mentor

This "radial momentum" doesn't exist. Or rather, it is a personal concept that you invented which has nothing to do with standard momentum concepts. Specifically, your "radial momentum" is not conserved, but the statement that "momentum is not conserved" is false and does not follow from the fact that your invented "radial momentum" is not conserved.

Please stick to standard terminology and concepts.

4. Dec 5, 2016

### albertrichardf

Momentum would still be conserved in this situation, even without looking at the momentum of the g-field. It is true that the M(in) mass is a point mass for all intents and purposes, so it can be ignored. Instead, it is M(out) we should concentrate on. Each point on M(out) moves with the same speed due to symmetry, and they each move towards the centre. And as you said, the momentum of pairs of opposite pieces is zero, so p is conserved.

However, you mention M(out) "as a whole" and from that stems the idea of a net radial momentum. But considering each bit of M(out), as was done in the analysis above is considering M(out) as a whole. There is no net radial momentum, because even though an individual piece of M(out) has a net radial momentum, its opposite would have radial momentum directed in precisely the opposite direction, so that overall there is no net momentum.

If you still wish to look at it "as a whole", then that would analysing the motion of the centre of mass of the system M(out) and M(in). M(out) is symmetrical, and thus the centre of mass is at the centre of M(out), and since M(in) is considered as a point mass, it too is at the centre of M(out). Thus, we can ignore M(in) since it is not moving and look at the motion of M(out)'s CM instead. For a sphere, the centre of mass is always at its centre, and since the whole of M(out) is moving with the same speed, M(out) never deforms. It hence remains a sphere throughout, so the centre of mass does not move at all. Since the CM does nor move at all, the momentum, which the mass of the system times the velocity of the CM, is zero throughout. While it is true that each component of M(out) moves, they always move in such a way that their positions form a sphere