Can a Fish Move a Ball in Water?

[jbriggs444]

The batteries and motor are within the fish and equivalent to the fish's muscles.

If the arm extends outside the tank then you have an external force. At this point, your responses are going well past tiresome and silly into the arena of obnoxious.

 

[metastable]

The point was the center of mass within the tank doesn't change but both carriages still derive thrust. Changing the center of mass within the tank isn't necessary for both carriages to move in opposite directions.

 

[.Scott]

The fish swims in a circle near the middle of the bowl. To balance the angular momentum, the remainder of the water and bowl turns in the opposite direction.

 

[A.T.]

Changing the center of mass within the tank isn't necessary...

Nobody says it is. See post #5 and #103. They show how to do this with a closed rigid container without changing the center of mass relative to the container.

But your robotic arm violates the closed rigid container condition, so it's trivial.

 

[metastable]

...the fish crawls with its fins along the edge of the glass, applying tangential force directly to the glass via the friction with its fins. the “normal force” holding the fish against the glass in this case would the fish’s forward velocity combined with the curvature of the glass...

This is a workable mechanism. Simplify and turn the fish into an octopus that attaches its tentacles to the wall and pulls itself along. The result is that we have the fish (and water due to viscous drag) rotating one way and a resulting torque tending to rotate the ball the other.

Nobody says it is. See post #5 and #103. They show how to do this with a closed rigid container without changing the center of mass relative to the container.

But your robotic arm violates the closed rigid container condition, so it's trivial.

It's already been acknowledged that the fin action on the right circled in red (applying tangential force directly to the glass via the friction with its fins) causes the ball to roll.

Can someone explain why the fin action circled on the left (applying direct force with its fin against the left side of the ball) is not exactly equivalent to applying tangential force via friction with its fin at the top of the ball?

 

[jbriggs444]

It's already been acknowledged that the fin action on the right circled in red (applying tangential force directly to the glass via the friction with its fins) causes the ball to roll.

Can someone explain why the fin action circled on the left (applying direct force with its fin against the left side of the ball) is not exactly equivalent to applying tangential force via friction with its fin at the top of the ball?

Conservation of momentum. No external forces means no motion of the center of mass. That's the easy argument. It is bulletproof.

This one of the nifty things about conservation laws. They let you skip past the pesky details and go straight to the conclusion.

The harder part is seeing how this is enforced mechanically. It is especially hard if one hand-waves the water away as being negligible. Because the ball rigidly encloses the fish and water, any forward motion by the fish relative to the ball must also involve an equal and opposite motion of the water relative to the ball. In order to get the water to flow rearward, either the fish or the ball (or both) must push the water rearward.

1. If it's the fish pushing the water rearward, we just have the fish swimming. No push-off from the ball. No motion of the ball.

2. If it's the ball pushing the water rearward, we have equal and opposite forces on the ball. No motion of the ball.

Either way and in any combination, the ball does not move.

 

[metastable]

It is especially hard if one hand-waves the water away as being negligible. Because the ball rigidly encloses the fish and water, any forward motion by the fish relative to the ball must also involve an equal and opposite motion of the water relative to the ball.

Are you sure it's not an assumption that if the fish moves from the left side to the right side, that an equivalent amount of water to the mass of the fish must move the same distance?

Suppose the fish is shaped like a square dowel. 1 meter long, 1cm square with curved ends. If the bowl is 1 meter and 1 centimeter wide, then after moving 1 centimeter, the fish has gotten to the other side. About 1 cubic centimer of water was moved "sideways" or "up and down" to accommodate the advance of the fish, while another cubic centimeter moved sideways and up and down between the fish and tank on the opposite side.

After the fish pushes off the glass, the fish moves forward, "parting" the water molecules as it goes. Some water moves left, some right, some up, some down to accommodate the fish. It does not necessarily follow that an equivalent amount of water to the fish's mass was transported all the way to the other side of the tank. This is the same concept as a wave moving through water doesn't necessarily mean the water molecules are translating at the velocity of the wave.

The ~100 cubic centimeter fish only displaced 2 centimeters of water (internally) during the entire "journey."

In the scenario I described above (only 2 cubic centimeters water displaced by 100 cubic centimeter fish): can you prove the following statement:

any forward motion by the fish relative to the ball must also involve an equal and opposite motion of the water relative to the ball.

The formula for calculating the joules (relative to the glass) per second transferred to the water by the fish (where rho is the fluid density of water is):

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

This energy per second is proportional to the frontal area, the measured drag coefficient of the fish in the tank, and the cube of the fish's velocity through the liquid.

 

[jbriggs444]

Are you sure it's not an assumption that if the fish moves from the left side to the right side, that an equivalent amount of water to the mass of the fish must move the same distance?

Quite sure, yes.

Suppose the fish is shaped like a square dowel. 1 meter long, 1cm square with curved ends. If the bowl is 1 meter and 1 centimeter wide, then after moving 1 centimeter, the fish has gotten to the other side. About 1 cubic centimer of water was moved "sideways" or "up and down" to accommodate the advance of the fish, while another cubic centimeter moved sideways and up and down between the fish and tank on the opposite side.

Rather than hand-wave, let's do the calculation. Let us skip past the algebra (trivial) and use numbers instead.

The tank is 100 cm top to bottom, 100 cm front to back and 101 cm left to right.
The rod is 1 cm top to bottom, 1 cm front to back and 100 cm left to right.
The rod begins touching the left side of the tank. One second later it contacts the right side of the tank.

We lay out coordinates with the origin at the front left bottom corner and the x-axis oriented positive rightward.

The rod masses a total of 100 grams. Its center of mass starts at x=50 cm and ends at x=51 cm. It moves at one centimeter per second. Momentum = 100 gm cm/sec during the traversal.

There are many possible ways the water could flow in response to this motion. We need not identify any particular way. It is enough to consider the motion of the center of mass of the water.

On the left face of the volume there are 9999 square centimeters that extend as columns all the way to the right face. The center of mass of this portion is at x=50.5 cm. There is an additional cubic centimeter of water that starts at the right face with its center of mass at x=100.5 cm and ends at the left face with its center of mass at x=0.5 cm.

The combined center of mass of the two at the beginning is at $$\frac{999900 \times 50.5\ +\ 1 \times 100.5}{999901} = 50,495,050.5/m ~= 50.50005000...$$

The combined center of mass of the two at the end is at $$\frac{999900 \times 50.5\ +\ 1 \times 0.5}{999901} = 50,494,950.5/m ~= 50.49995000...$$

At this point, an astute observer might note that the huge volume of water could be ignored. Effectively it is only the one cubic centimeter that is moving 100 cm in one second. Such an observer would quickly calculate the momentum of this piece as -100 gram cm/sec. However, let us proceed with brute force calculation according to the original plan.

The difference between the two centers of mass is approximately one micron. 0.0001 cm. But multiplying by the mass, that displacement in one second amounts to 0.0001... cm * 999901 grams / 1 second = 100 gram cm/sec. [That's not an approximation. It's exact]

100 gram cm/sec for the "fish". -100 gram cm/sec for the water. As expected.

Can you lay off the energy calculations already? They are irrelevant. Just because you know a formula does not mean that the results are useful.

 

[metastable]

Effectively it is only the one cubic centimeter that is moving 100 cm in one second.

If you are relying on this one point for your proof, I have to say I don’t think it’s been effectively proven.

I consider the minimum displacement necessary to allow the rod to move a centimeter.

The cubic centimeter of water at the front has to move a centimeter to the side. A cubic centimeter of water at the rear has to move a centimeter. So far that’s 2 cubic centimeters of water moving by 1 centimeter in one second not 1 cubic centimeter moving 100 cubic centimeters in one second. Perhaps a 3rd cubic centimeter of water along the skin of the fish shifts by 1 centimeter during the passage. Still we’re only up to 3 cubic centimeters of water moving 1 centimeter in one second not 1 cubic centimeter moving 100 centimeters in one second.

Another way to think of it is the work the rod does lifting the water column at the front is almost completely offset by lowering the water column at the the rear... almost completely because there are friction forces between the water molecules that over time convert into heat.

In any case the energy required to move the water for one second will be proportional to the drag equation:

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

 

[jbriggs444]

If you are relying on this one point for your proof, I have to say I don’t think it’s been effectively proven.

Not at all. In the post you are responding to I gave no proof. I merely responded to your request for a worked example.

A general proof follows immediately from conservation of momentum: The momentum of the whole is conserved and is the sum of the momenta of the parts.

Here endeth the lesson.

 

[metastable]

momentum of the whole is conserved and is the sum of the momenta of the parts.

Right, I'm thinking the ground has some type of oil on it so the friction at the point of contact between the ball and the ground is effectively zero.

The 5kg fish pushes off the glass acquiring 10m/s relative to the ground measured at one nanometer change in distance from the glass after glass separation. Glass + Water weighs 5kg. Fish weighs 5kg.

The fish acquires 10m/s relative to the ground so the only thing that can translate in the opposite direction is the glass + water as a whole, also 10m/s initially relative to the ground because the same mass as the fish.

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, therefore the water and glass must still be moving in the opposite direction until either a) the fish hits the opposite wall or the b) work done via the drag equation transfers all the fish's kinetic energy relative to the glass into the water.

In any case 1 nanosecond isn't enough time to transfer the fish's kinetic energy relative to the glass into the water, and because the initial push caused the 5kg fish mass to translate relative to the ground in one direction, and the ball has no friction with the ground, conservation of momentum requires the water and glass to move in the opposite direction until the fish transfers its kinetic energy relative to the glass into either the water and glass, but the process takes time, so in that time, the ball and water as a whole shifts to one side (with some of this side motion converted over time to internal motion in the water, which later turns to heat) while the fish shifts or translates to the other side.

 

[Vanadium 50]

Metastable, the reason you are still confused is that you are not paying attention to what other people are saying. This is a problem present on multiple threads. Think momentum, not energy.

 

[metastable]

Think momentum, not energy.

Right I'm using the same momentum equation $M_1V_1=M_2V_2$, that someone would use to calculate the speed of an iceskater in a push off with another ice skater. In this case I am comparing the acquired momentum of the fish to $m_1v_1$ and the acquired momentum of the glass+water as a whole on the frictionless ground as $m_2v_2$ with a correction applied due to the drag of the fish through the water which makes the situation not frictionless.

 

[jbriggs444]

Right I'm using the same momentum equation $M_1V_1=M_2V_2$, that someone would use to calculate the speed of an iceskater in a push off with another ice skater. In this case I am comparing the acquired momentum of the fish to $m_1v_1$ and the acquired momentum of the glass+water as a whole on the frictionless ground as $m_2v_2$ with a correction applied due to the drag of the fish through the water which makes the situation not frictionless.

The water plus fish cannot move relative to the globe because it does not leak [and is rigid and spherically symmetric, fluid and fish density is uniform, both are incompressable and there are no voids]. Any attempt to "push off" is doomed to failure. The water is constrained by the globe to circulate back behind the fish. The force required to accelerate this circulation into existence is equal and opposite to the force of the push off. It has to be in order to conserve momentum.

I already formalized this fact into a proof that you did not read. Feel free to find it.

 

[A.T.]

1 Nanosecond later, based on the work done in one whole second by the drag equation $W=(1/2)*C_d*rho*A_f*V^3$

This equation is for a steady movement through the fluid, not for accelerated movement like during push-off.

See Eq. 3.2 here:
https://www.cambridge.org/core/serv..._force_on_an_accelerating_submerged_plate.pdf

 

[metastable]

The water plus fish cannot move relative to the globe because it does not leak [and is rigid and spherically symmetric, fluid and fish density is uniform, both are incompressable and there are no voids]. Any attempt to "push off" is doomed to failure. The water is constrained by the globe to circulate back behind the fish. The force required to accelerate this circulation into existence is equal and opposite to the force of the push off. It has to be in order to conserve momentum.

So then a swimmer (who is neutrally buoyant) can't push off the side of a pool with their feet and travel any distance underwater, if the pool has a flat metal cover on top of it?

 

[metastable]

https://youtu.be/u9rx2lzF_5Q?t=18

 

[jbriggs444]

So then a swimmer (who is neutrally buoyant) can't push off the side of a pool with their feet and travel any distance underwater, if the pool has a flat metal cover on top of it?

The swimmer can push off. But the water will fill in behind. The momentum of the water plus swimmer totals zero at all times. It follows that there can be no non-zero total horizontal net force on the pair from the pool top, bottom and sides. Any unbalanced force you specify on one side will be accompanied by a balancing force on the other.

To re-emphasize, nobody has said that the swimmer cannot push off. Post #117 is your argument against a straw man of your own construction.

To be extra clear, this applies for pools with rigid metal tops. Pools that can slosh back and forth will not have contents with zero total momentum at all times. A free-standing pool with open top in the middle of a skating rink could move in reaction to a swimmer push-off. Or in reaction to sloshing water. Or in reaction to a complicated combination of both.

 

[metastable]

The swimmer can push off. But the water will fill in behind. The momentum of the water plus swimmer totals zero at all times. It follows that there can be no non-zero total horizontal net force on the pair from the pool top, bottom and sides. Any unbalanced force you specify on one side will be accompanied by a balancing force on the other.

You say the swimmer can push off. (which gives the swimmer momentum relative to the ground)

(But if the pool and walls of the pool itself has no friction with the ground, the pool has significantly less mass than the Earth and can't be treated as having effectively infinite mass as in the commonly used surface vehicle approximation between the vehicle and the mass of the earth...)

(and especially if the pool and walls of the pool have the same combined mass as the swimmer)...

We are in agreement any unbalanced force I specify (such as the swimmer moving relative to the ground) will be accompanied by a balancing force on the other (such as the walls of the pool and water overall experiencing a force minus the "drag force transferred to the water" by the swimmer via the drag equation)

 

[jbriggs444]

We are in agreement any unbalanced force I specify (such as the swimmer moving relative to the ground) will be accompanied by a balancing force on the other (such as the walls of the pool and water overall experiencing a force minus the drag force transferred to the water by the swimmer via the drag equation)

Sorry, no. What you have written here is gibberish. Equating force with momentum, seriously?

The force balance that I refer to is based on Newton's second law, not Newton's third.

 

[metastable]

@jbriggs444 your proof relies on the exact same same water atoms in front of the 1cm*1cm*100cm rod being transported all the way behind the same rod after it moves one cm. In practice this doesn't happen according to the drag equation. It isn't necessary to move that cubic centimeter of water that distance in that time to accommodate the motion of the rod, according to the drag equation. This can be shown simply by varying the length of the rod and looking at the difference in energy transfer according to rod length in transit from one wall to the opposite wall.

 

[Wes Tausend]

I see the problem as similar to a person standing on a simple furniture cart (the type with a wood frame and four caster wheels) and propelling themselves forward in lurches by jerking motions. This is possible and it is the result of the difference between higher static friction and lesser kinetic friction.

After the first lurch ahead, the idea is to keep backing up easy, in order to keep from going the wrong way after each cart stop, i.e. during static contact, then start the next easy acceleration forward only stop with a sudden hard deceleration to shoot the cart ahead again during lesser kinetic friction. It could even be done with a dry surfboard if one's feet were strapped to it.

In a similar method, it seems to me that if the fish starts and accelerates easy-peasy enough, it won't overcome the static friction of the ball rolling in the reverse direction, so the ball will remain still. But when it suddenly impacts the front of the sphere wall and stops, the mass impact may transmit enough kinetic energy to allow breaking the static limit and allowing the energy to lurch the entire ball/water/fish ahead until the rolling friction stops it. A really savvy fish (like ours) might even blow a mouthful of water out at the wall immediately following contact, which will add a bit more to forward impact and subsequent increased roll distance. Voila! Our fish also slowly drifts rearward from veracious spitting, set up for the next run. (My Cichlids spit sand.)

Parts of this are similar, but not exactly like metastable's #7 post. It's where I thought he was going.

Without the difference between starting and rolling frictions in an accelerated (or magnetic?) environment none of this would work. For example, periodic motion to and fro of any demeanor standing on a free-floating surfboard (feet strapped on) in gravity-free outer space would be a wash without friction. So imagine a hapless crewmember of ISS hovering in the center of a module trying to get going. He could try to swim in the air. Still, it would be like a fish in a ball filled with water in outer space... without the help of external friction.

Wes

 

[Vanadium 50]

Metastable, the situation was clearly and succinctly explained earlier.

Conclusion: The only way the ball moves is if you can contrive to get an external force to move it. This is do-able as has been described earlier.

Your posts are all over the map, mixing up forces and energy and momenta, so I can't quite figure out what you are trying to say. Do you disagree with the above? If so, exactly where?

These increasingly complicated (and, frankly confused) scenarios remind me of the inventor of the overbalanced wheel who knows that he's just one additional complication from making it work.

 

[metastable]

@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?

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 amount 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.

 

[A.T.]

In practice this doesn't happen according to the drag equation.

The drag equation you are using is invalid during the push-off. See post #115.

pressure and sound waves have a finite speed in water

Not under the assumption that water is incompressible. If you drop that assumption, then it again becomes trivial, because then the CoM can obviously move relative to 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]

@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 amount 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]

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?

 

[jbriggs444]

@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]

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]

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?

Did you not read #130?

 

[Vanadium 50]

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]

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]

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]

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:

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

 

[jbriggs444]

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?

 

[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]

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.

 

[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]

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?

Adding complexities to your do-nothing machine again? How tedious.

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.