# B Can the fish move the ball?

#### A.T.

How can you prove when the fish hits the wall the water stops? Some glitter in the water should be able to test that proposal.
Glitter would not show the net linear momentum of the water, which can be zero even if water is swirling around.

#### metastable

If both of these encapsulated hollow glass spheres are floating on the space station, and filled with vacuum (empty), and a spring is released, the outer ball moves to the right, and the inner ball moves to the left as shown:

It is exactly the same momentum equation as a frictionless push off between 2 initially at rest ice skaters:

$m_1v_1=m_2v_2$

Now we fill the spheres with water so the push off between the skaters is no longer frictionless, and so a drag term is needed to describe the drag force counteracting the spring. By lowering the frontal area or streamlining the inner sphere without changing its mass, this drag term can be reduced via:

$F=(1/2)*C_d*rho*A_f*V^2$

I would like to see an actual experiment but my dollar says when the spheres are filled with water, the modification that the drag equation causes to momentum equation describing the frictionless push off scenario is the spring needs slightly more force to achieve the same acceleration of the 2 tanks, and the outer tank still moves to the right, and the inner tank to the left as shown.

#### jbriggs444

Homework Helper
my dollar says when the spheres are filled with water, the modification that the drag equation causes to momentum equation describing the frictionless push off scenario is the spring needs slightly more force to achieve the same acceleration of the 2 tanks, and the outer tank still moves to the right, and the inner tank to the left as shown.
Drag is not the problem. Momentum conservation is. You have persistently failed to recognize this.

#### metastable

Do you deny changing only the shape of the inner sphere without changing its mass will affect the force required from the spring for a given acceleration?

#### A.T.

..., this drag term can be reduced via:

$F=(1/2)*C_d*rho*A_f*V^2$
This is the drag equation for steady movement in a large body of a fluid. It doesn't apply if the body is accelerated. It doesn't apply if the body is occupying a non-negligible fraction of a filled enclosed container.

#### jbriggs444

Homework Helper
Do you deny changing only the shape of the inner sphere without changing its mass will affect the force required from the spring for a given acceleration?
You have not specified what scenario this question applies to.

I will guess that you are talking about an outer spherical container and an internal container whose shape is variable but whose volume is fixed. The internal container, the fluid it contains and the fluid between the containers is incompressible and all of the same uniform density.

You ask about the acceleration of the inner shape as it is propelled away from the wall of the outer container.

I do not deny that drag affects this acceleration.

You expect that the acceleration of the outer container as a result of the force of the spring is non-zero and that it is related to the acceleration of the inner shape in some way.

I do deny that drag affects this acceleration since it is identically zero. The force of the spring on the outer container is exactly and immediately countered by pressure and drag forces from the fluid on the outer container.

What you consistently fail to recognize is that the rigid container acts as a constraint on the behavior of the fluid within. When the inner container moves, the fluid between the two containers is constrained to move in a manner which is consistent with this. It has to stay inside the outer container and outside the inner. Pressure gradients will immediately form which are sufficient to enforce this constraint.

Given the uniform density, this means that the center of mass of the outer container's contents cannot move relative to the outer container. Assuming that the outer container is spherically symmetric, this means that the combined center of mass of the outer container plus contents will always be at the geometric center of the outer container. Since the spring force is internal to the system, the center of mass cannot move as a result. And the outer container cannot move [linearly] relative to its center of mass.

The pressure gradients which enforce flow consistency are also the pressure gradients which ensure that there is no unbalanced net force on the outer container.

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#### DaveC426913

Gold Member
Every time I read a new reply, this post from Van jumps out at me:
... the inventor of the overbalanced wheel who knows that he's just one additional complication from making it work.
Tweaks of the fish bowl setup have one of two results; either
- they violate one of the premises (such as no external input)
or
- they have no effect on movement.

#### metastable

-There are 2 rigidly enclosed containers, not 1– the outer container and the inner container

-There are 2 centers of mass, not 1– the outer glass and its water, and the inner glass and its water, and these 2 centers of mass can move in relation to one another via

-the outside force which is the energy stored in the spring which is equivalent to the energy stored in the fish’s muscles

-it hasn’t been shown conclusively by your proof @jbriggs444 that the very same water atoms which were in front of the fish end up behind the fish

-further, even if it can be shown the same water atoms which were in front were somehow transported all the way behind the sphere as it moves from the spring... it can be shown certain modifications to the shape will cause the mass of the water molecules flowing past the inner container as it moves to travel a longer distance in the same amount of time which takes greater force. I struggle to see where this difference in force has been accounted for in your proof that the 2 enclosed bodies can’t move in opposite directions with a spring force or fish muscle between them.

#### sophiecentaur

Gold Member
You are falling into the same trap as @metastable. The ball does not move, even if the fish hits the side.

The gotcha is that if the fish comes to an abrupt stop then so does the water. Any momentum transferred from fish to wall is matched by an opposite momentum transferred from water to wall.

Subject to background assumptions of rigid wall, incompressible fluid, etc.
I get that now. It's easy to forget the incompressable and rigid constraints.

#### A.T.

-There are 2 centers of mass, not 1
All that matters is the combined CoM of the outer container and all its contents, and whether that CoM can move relative to the outer container.

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#### jbriggs444

Homework Helper
-it hasn’t been shown conclusively by your proof @jbriggs444 that the very same water atoms which were in front of the fish end up behind the fish
Irrelevant. All that matters is that it is a flow that does not result in a shift of the center of mass. There are many possible flow fields with that property. Almost none of them move the molecules in front to a position behind. Most put a different set behind.

The constraints imposed by the rigid container, the incompressible fluid and its uniform density provide a guarantee that all permitted flows will preserve the position of the center of mass of the contents relative to the container.

#### metastable

In asymmetrical mass interactions (Such as a push off between two different mass skaters), The lighter body obtains more of the kinetic energy relative to the ground. In the diagram above The fish could weigh 1000 kg, with the outer tank just slightly larger so the outer tank and water only weighs 100 kg. In a push off between 1000 kg fish and 100 kg water/tank, according to the momentum equation, which gets more kinetic energy relative to ground initially during the internal push off?

#### jbriggs444

Homework Helper
In asymmetrical mass interactions (Such as a push off between two different mass skaters), The lighter body obtains more of the kinetic energy relative to the ground. In the diagram above The fish could weigh 1000 kg, with the outer tank just slightly larger so the outer tank and water only weighs 100 kg. In a push off between 1000 kg fish and 100 kg water/tank, according to the momentum equation, which gets more kinetic energy relative to ground initially during the internal push off?
The fish gets more than the tank. Because the tank gets none. But the water gets more than the fish, because it has to move much more rapidly.

As usual, subject to the assumptions about rigid tanks, completely filled, uniform density, incompressible, yadda yadda yadda. There is no inner tank this time, right? Just a fish inside an outer tank against which it pushes off?

#### metastable

Are you able to prove a fish can swim normally in “incompressible water?”

#### DaveC426913

Gold Member
Are you able to prove a fish can swim normally in “incompressible water?”
I'm not sure that infinitely incompressible water is a realistic scenario worth discussion, unless we're leaving the land of reality altogether (akin to the infinitely rigid rod in relativity).

Are we?

Water is essentially incompressible, yet provides no barrier to swimming.

#### metastable

So are we 100% sure the fish’s normal swimming ability doesn’t rely entirely on water’s slight compressibility?

#### DaveC426913

Gold Member
So are we 100% sure the fish’s normal swimming ability doesn’t rely entirely on water’s slight compressibility?
What if we're not 100% sure? Does it make sense to talk about infinitely incompressible water and infinitely rigid tanks?

BTW, note that water for fish has dissolved gases in it, which are compressible, so the water cannot be infinitely compressible.

#### jbriggs444

Homework Helper
So are we 100% sure the fish’s normal swimming ability doesn’t rely entirely on water’s slight compressibility?
We can be 100% sure that in the limit as compressibility decreases toward zero, fish's normal swimming ability is unimpaired.

We need not bother with what happens for perfect incompressibility as long as we can reason about what happens for anything short of that. We can hand-wave and take the limiting behavior as an apt description of the behavior at the limit.

#### metastable

Well if there’s some compressibility then It stands to reason the water in front of the fish get squeezed a bit first before it moves behind the Fish…

#### jbriggs444

Homework Helper
Well if there’s some compressibility then It stands to reason the water in front of the fish get squeezed a bit first before it moves behind the Fish…
We've been over this before.

#### metastable

If it can’t be proved that a fish can swim in 100% incompressible water, then it can’t be proved the other methods of locomotion involving the fish swimming around will work either... Can it?

#### jbriggs444

Homework Helper
If it can’t be proved that a fish can swim in 100% incompressible water, then it can’t be proved the other methods of locomotion involving the fish swimming around will work either... Can it?
What point are you trying to make here?

It seems perfectly clear that for anything short of 100% incompressible fluid, everything works normally. It also seems perfectly clear that the time interval over which any momentary propulsion can be achieved with springs on containers and fish noses touching glass reduces toward zero as incompressibility approaches 100%.

As I've already suggested, we can consider the limiting behavior rather than the behavior at the limit if the latter bothers you so much.

#### A.T.

So are we 100% sure...
If you want 100% certainty in anything, try religion instead of physics.

... the fish’s normal swimming ability doesn’t rely entirely on water’s slight compressibility?
Compressibility reduces the propulsion efficiency in a fluid, so it doesn't help. We use the incompressibility limit, just like we use the rigid body limit, where appropriate.

#### metastable

Even if someone can show evidence that increasing the compressibility of water could make the fish’s swimming less efficient, it does not necessarily follow that a fish is still capable of swimming in 100% incompressible fluid.

If it can’t be proven a fish can move its nose forward through 100% incompressible fluid, the realism of any scenario modeling a fish swimming in such a fluid is called into question.

For example for a fish to move forward through a fluid (as in swimming) some of the water in front of the fish must move behind the fish as it translates.

What force (if not differential pressure) causes the water molecules in front of the fish to move behind the fish as it swims in 100% incompressible fluid?

If the water molecules can’t move behind the fish because it can’t create differential pressure because the fluid is 100% incompressible, how can the fish swim at all?

Is there any realism in modeling a scenario about the locomotion of a fish that relies on a fluid which entirely prevents the fish from swimming?

@jbriggs444 would you infer that if the water has even slight compressibility (as it does, see: http://sites.bsyse.wsu.edu/joan/teaching/bsyse558/W2/Lec5Stu.pdf ), it can at least temporarily shift the outer glass ball slightly by pushing directly off the glass, regardless of whether the ball is floating in space, water or resting on a ground with or without friction between the ball and the contact point with ground?

#### jbriggs444

Homework Helper
If it can’t be proven a fish can move its nose forward through 100% incompressible fluid, the realism of any scenario modeling a fish swimming in such a fluid is called into question.
The physics of incompressible flow are well accepted. If you want to call that approximation into question, please start another thread.

Please try to find a mainstream reference however. Personal speculation that incompressible flow is impossible may run afoul of the forum guidlines.

@jbriggs444 would you infer that if the water has even slight compressibility (as it does, see: http://sites.bsyse.wsu.edu/joan/teaching/bsyse558/W2/Lec5Stu.pdf ), it can at least temporarily shift the outer glass ball slightly by pushing directly off the glass, regardless of whether the ball is floating in space, water or resting on a ground with or without friction between the ball and the contact point with ground?
I along with others have already agreed several times in this thread that if you want to discard the assumptions of incompressibility and rigidity then the problem of propulsion by the fish becomes trivial.

Yes, making the fluid compressible de-couples the motion of the fish from the motion of the fluid sufficiently to allow brief mismatches between their respective momenta.

"Can the fish move the ball?"

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