# B Can the fish move the ball?

#### metastable

Isn’t it necessary to drop the assumption the water is incompressible? Can’t the water store and release internal stress energy in the form of differential pressure in different areas of the ball? This CFD animation clearly shows what appears to my untrained eye to be pressure waves in the wake of a simulated fish...

#### A.T.

Isn’t it necessary to drop the assumption the water is incompressible?
If you do, moving the CoM relative to the ball is possible and we can also use the hamster method.

#### metastable

Its center of mass starts at x=50 cm and ends at x=51 cm. It moves at one centimeter per second.
1 Nanosecond later, based on the work done in one whole second by the drag equation $W=(1/2)*C_d*rho*A_f*V^3$ the fish has not transferred all the kinetic energy it acquired relative to the ground to the water or the glass
This equation is for a steady movement through the fluid
The drag equation you are using is invalid during the push-off. See post #115.
^ @A.T. I don't think the equation is invalid in the context I used it. @jbriggs444 specified constant motion in the scenario I was responding to.

#### jbriggs444

Homework Helper
@jbriggs444 doesn’t your proof that a swimmer can’t push off the side of a covered pool ignore the fact that pressure and sound waves have a finite speed in water, so if the pool is big enough it would take more time for any pressure waves in the water to reach the covered surface, reflect and have some effect on the diver than the time it takes for the diver to push off from the side and travel an extremely short distance?
Yes indeed. It assumes incompressible flow and a rigid container. If this is the rock on which you were planning to rest your claim that the ball can be made to move then go ahead and claim victory.

you are claiming the only factor that determines whether a neutrally buoyant diver can push off from the side is whether or not the pool has a rigid cover. I am not satisfied that the validity of this assertion is assured because I can keep the fish the same mass, same drag coefficient, same initial velocity, but change the frontal area and change the amout of energy the fish transfers to the water in one second. @Vanadium 50 @jbriggs444 ’s proof does not allow for such a variation in energy transfer from fish to water per second based on changes to frontal area only.
But then you go back to your same old tired screed. @Vanadium 50 hit the nail on the head. You are always just one more complexity away from getting your potential energy machine to work.

#### A.T.

^ @A.T. I don't think the equation is invalid in the context I used it. @jbriggs444 specified constant motion in the scenario I was responding to.
Okay, but your method was involving accelerating the fish by pushing of the wall. You cannot use the steady state drag equation to analyse that.

However, once you drop the incompressibilty of the water or the perfect rigidy of the ball, it's trivial that propulsion is possible.

#### sophiecentaur

Gold Member
There are (not quite directly) analogous situations in a space station or satellite and on a kids’ playground roundabout. The water will have some friction force on the walls of the ball and that can have the same effect (but less easy to exert) as the force / torque that the rider can exert on the rails of the roundabout or the torque wheel of a satellite. Momentum will be conserved. When the fish stops swimming there will be an opposite torque due to water against the fish’s nose (returning the ball to where it started) but a net forward displacement due to energy loss.

#### metastable

Okay, but your method was involving accelerating the fish by pushing of the wall. You cannot use the steady state drag equation to analyse that.
I thought the steady state drag equation for the work done on the water by the constant speed fish in one second:

$W=(1/2)*C_d*rho*A_f*V^3$

is derived from the drag equation describing the instantaneous force in newtons acting on the fish relative to the glass:

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

Suppose the mass of the fin that pushes off the side is insignificant compared to the mass of the overall fish, and the mass of the overall fish equals the mass of the water and the glass, and the point of contact between the ball and the ground has no friction.

If the nose of the massive front part of the fish initially gains 0.01m/s relative to the tip of the pushing fin and glass while pushing off the wall (& the pushing fin has insignificant mass), and we compare the force acting on the fish as a whole from the water immediately after fin tip separation from the wall in 3 different scenarios-- where the fish has the same initial velocity, mass, and drag coefficient in all cases but different frontal areas in each case.

Assuming the drag coefficient remains 1 in all cases, fluid density of water 1000kg/m^3, initial velocity fish nose is 0.01m/s, overall fish mass is 0.1kg, push fin mass is insignificant compared to fish, and the frontal area is either 0.1cm^2, 1cm^2 or 10cm^2:

Results:

0.1cm^2 Frontal Area: 5*10^(-7) Newtons Force on Fish From Water
1cm^2 Frontal Area: 5*10^(-6) Newtons Force on Fish From Water
10cm^2 Frontal Area: 5*10^(-5) Newtons Force on Fish From Water

@jbriggs444 does your proof allow for a different amount of force to act on the fish with the same initial "nose velocity," based on changes only to the fish's frontal area?

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#### jbriggs444

Homework Helper
@jbriggs444 does your proof allow for a different amount of force to act on the fish with the same initial "nose velocity," based on changes only to the fish's frontal area?
It is completely general regardless of fish geometry, water viscosity and phase of the moon.

The drag between fish and water is an internal force pair. It is obvious that it can have no effect on the combined momentum of fish plus water. This obvious fact makes it puzzling why you persist in focusing on an equation for a quantity that has no relevant effect.

#### metastable

I thought of a new one... an octopus creates a volume of vacuum on one side of the ball with the muscles of its suction cup against the glass, increasing the pressure and density of the water slightly, which makes the ball heavier on one side. If the ball has friction with the ground, the ball rolls.

#### pinball1970

Gold Member
If you do, moving the CoM relative to the ball is possible and we can also use the hamster method.
I thought of a new one... an octopus creates a volume of vacuum on one side of the ball with the muscles of its suction cup against the glass, increasing the pressure and density of the water slightly, which makes the ball heavier on one side. If the ball has friction with the ground, the ball rolls.
The op stated a fish not an octopus and I think it's possible all avenues have been explored on this one.
Just my opinion.

#### metastable

So what if the fish pushes off the wall in such a way that its fins temporarily form a suction cup so the water in front is compressed, and can't get behind the fish, but the fish moves forward relative to the ground, can the glass translate?

#### jbriggs444

Homework Helper
So what if the fish pushes off the wall in such a way that its fins temporarily form a suction cup so the water in front is compressed, and can't get behind the fish, but the fish moves forward relative to the ground, can the glass translate?

Staff Emeritus
Metatsable, you are 100% right. If fish are octopuses and momentum is energy is force and incompressible fluids are compressible (or is it the other way round?) then you're right.

#### metastable

incompressible fluids are compressible (or is it the other way round?) then you're right.

I thought referring to water as incompressible is only an approximation relative to the compressibility of gases. Water actually can be compressed, can it not? Obviously we are talking about potentially constructible robotic fishes— because real fish use swim bladders filled with air to maintain neutral buoyancy, and so do not meet the criteria of the current discussion— moving the glass while having uniform density with the water.

#### jbriggs444

Homework Helper
I thought referring to water as incompressible is only an approximation relative to the compressibility of gases. Water actually can be compressed, can it not? Obviously we are talking about potentially constructible robotic fishes— because real fish use swim bladders filled with air to maintain neutral buoyancy, and so do not meet the criteria of the current discussion— moving the glass while having uniform density with the water.
Re-read #130 for comprehension this time.

If we treat water as compressible then we can simply put a vacuum-breathing super-hamster in a water-filled ball, have him drink all the water and excrete it under enormous pressure into his bladder. Then he can trot around the room just like a regular hamster.

If we treat the ball as non-rigid then we can simply have the super hamster extrude legs from the ball and walk around on all fours.

Treating the ball as non-rigid and the water as compressible reduces the problem to a triviality.

Treating the ball as rigid and the water as incompressible makes for an interesting toy problem that the rest of us were trying to discuss.

#### .Scott

Homework Helper
You say the swimmer can push off. (which gives the swimmer momentum relative to the ground)
Under normal conditions, when the swimmer is pushing off, he is also pushing a quantity of water with him - although, for efficiency, as little as possible. The result is a wake that shows up at the water/air interface just as the swimmer pushes off - but it takes the right lighting to catch it. Here is a screen shot from the 2:39 mark in video Swimmer pushing off.

The wake can be seen in the areas circled - and their formation and movements can be seen in the video.

But when he is in a closed tank with no compressible fluids, he is simultaneously pushing himself forward and pushing an equal volume (and therefore mass) backwards. What exactly is happening is that he is pushing against a spreading column of water that reaches across the bowl to the opposite side. He is pushing the bowl in two opposite directions with equal force. Since we take the "water" to be incompressible and Newtonian, the speed of sound in that water would be infinite and that force would be applied simultaneous to the push-off.

#### metastable

What if the fish has an extremely long, extremely thin “fin” that unfolds like an arm with many elbows, and as the arm straightens out, it pushes the much more massive body of the fish all the way against the glass on the other side of the tank. When the nose of the fish reaches the other side, it’s arm can’t do any more work, and the fish’s body stops, but the water has additional energy that keeps it swirling in vortices for a time. If the main body of the fish translates across the ball, hits the other side and stops, but the water keeps moving, what momentum is equal and opposite to the continued swirling of the water after the fish stops, if translation of the entire ball doesn’t occur?

#### jbriggs444

Homework Helper
but the water has additional energy that keeps it swirling in vortices for a time. If the main body of the fish translates across the ball, hits the other side and stops, but the water keeps moving, what momentum is equal and opposite to the continued swirling of the water after the fish stops, if translation of the entire ball doesn’t occur?
Repeat after me: "Energy and momentum are two different things"

It is perfectly possible to have swirling water with zero total momentum. In the case at hand, the momentum of the water after the fish stops totals zero regardless of any swirling that may continue.

#### metastable

Repeat after me: "Energy and momentum are two different things"
If there are 2 masses that can move independently relative to the ground, & both masses are initially at rest to the frictionless ground, I can calculate the momentum of 1 mass directly from the kinetic energy in joules the other mass acquired relative to the ground:

$m_1v_1=\sqrt{2}*\sqrt{m_2}*\sqrt{W_{kinetic_m2}}$

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#### jbriggs444

Homework Helper
If there are 2 masses that can move independently relative to the ground, & both masses are initially at rest to the frictionless ground, I can calculate the momentum of 1 mass directly from the kinetic energy in joules the other mass acquired relative to the ground:

$m_1v_1=\sqrt{2}*\sqrt{m2}*\sqrt{W_{kinetic_m2}}$
In this case we have many many little bits of water swirling this way and that. The fact that their aggregate energy is non-zero says nothing about their aggregate momentum.

[An astute observer could note that a known total energy together with a known total mass does place an upper bound on the magnitude of the total momentum. However, it does not impose a lower bound].

What is the total momentum of a pair of counter-rotating vortices?

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#### metastable

Just to be clear, if the ball is floating in the space station, and the ball is filled with vacuum except for the water density fish and the fish extends its many elbowed arm to push its body from one side of the glass to the other side where its nose impacts the other side, I think we can agree an astronaut in the same compartment would observe the outside of the ball move slightly...

I think you are saying you believe if the same process occurs, but the ball is filled with water, then astronaut won't observe the outside of the ball move.

#### A.T.

... the ball is filled with vacuum except for the water density fish ...
Here the CoM is not fixed relative to the ball. Hamster method would work on Earth.

I think you are saying you believe if the same process occurs, but the ball is filled with water, then astronaut won't observe the outside of the ball move.
Depends on whether the CoM can move relative to the ball.

#### jbriggs444

Homework Helper
I think you are saying you believe if the same process occurs, but the ball is filled with water, then astronaut won't observe the outside of the ball move.
Correct.

Assumptions: Rigid, spherically symmetric ball. Ball starts at rest, filled with incompressible fluid of uniform density. Fish is also incompressible, uniform density, same density as water. No leaks, voids or air bubbles. No external forces penetrate the ball to act directly on the fish or water.

Flashlights out the back are deemed to weak to be effective.

We are neglecting, at least for the moment, any unbalanced external forces on the ball. This includes thermal induced pressure gradients. If you wanted to turn the ball into a Crooke's radiometer then that would be an innovative solution, but let's rule it out for the moment.

Caveat: While the astronaut should not see the ball move linearly, it is possible for the fish to cause a rotation. That possibility has already been explored as early as post #5 in this thread.

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#### metastable

So what happens when I make the fish more streamlined than it was initially (lower the drag coefficient but same mass and frontal area), and using the many elbowed arm accelerate the fish to the same velocity at the same midpoint before freezing the elbows... according to $F=(1/2)*C_d*rho*A_f*V^2$ the water exerts less force on the fish with the same initial velocity in the second case, so it has more velocity relative to the space station than the less streamlined fish at every point along its journey, correct?

#### jbriggs444

Homework Helper
So what happens when I make the fish more streamlined than it was initially (lower the drag coefficient but same mass and frontal area), and using the many elbowed arm accelerate the fish to the same velocity at the same midpoint before freezing the elbows... according to $F=(1/2)*C_d*rho*A_f*V^2$ the water exerts less force on the fish in the second case, so it has more velocity relative to the space station than the less streamlined fish at every point along its journey, correct?

Yes, you can make the fish accelerate more strongly or make its drift last longer. But the water flowing rearward past the fish then accelerates more strongly or lasts longer. It accomplishes nothing. No matter how the fish pushes off, no matter how he coasts, no matter how hard his nose bumps the glass, no net momentum changes result due to any of it.

Internal force pairs do not affect total momentum.

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"Can the fish move the ball?"

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