You didn’t read my post. Let me clarify a bit. Think of it like a Frisbee: it’s spinning, so one side is moving forward, while the other is moving backward, so of course there is not net forward movement inherent in the rotation (the equivalent of “flow” within the vortex you mention). BUT the system as a whole, also has a forward momentum component. The flow of fluid within the vortex is independent of its forward motion.
It’s an object which has the same density as the air, and that’s what is giving me conceptual problems.
OK, imagine a closed mathematical surface. Fire a vortex through that surface. Does the vortex carry momentum into the closed surface? I would assume so, but… As the “vortex pattern” passes through the surface, for every fluid parcel moving in, there is another nearby fluid parcel simultaneously moving out. In other words, no net volume of air was pumped across the boundary, and I think this means that no net momentum crossed the boundary either. I may be wrong about this, but I can’t see my error yet.
A vortex SHOULD carry a hunk of momentum with it as it goes, but when I try to crudely add up the momentum of a simplified cylinder-shaped vortex, I get zero. There is net movement of the air outside the vortex. The air doesn’t just “open up” to let the vortex pass, it also moves backwards to fill the space behind the vortex, and the momentum of this backwards-moving air is opposite to the forward momentum carried by the vortex. Maybe I’m not supposed to add these together. If so, I don’t understand why not.
PS
How did all this get started? It’s actually some fallout from the great “Bernoulli vs. Newton airfoil controversy.” Airplanes generate lift because they create a downward-moving vortex pair. Here are some excellent photos:
http://www.efluids.com/efluids/gallery/cessnajet_1.jpg
http://www.efluids.com/efluids/gallery/Trailing_vortices_1.html
But then one professional physicist pointed out that these vortices cannot act as “downward moving air.” The net momentum carried by the vortex is zero unless we arbitrarily ignore the backwards-moving air which surrounds each vortex.
I haven’t been able to clear up this problem, so I thought some fresh minds from the ranks of doper physicists might shed some light.
I can almost SEE those bird wings and those flapping fish tails throwing off vortices which carry momentum. But then I get confused when I try to estimate the amount of momentum.
You need to not think about the flow of fluid within the vortex, and only consider its forward motion component. Yes, the flow of fluid contained in the vortex nets to zero forward momentum. It’s the forward motion of the mass of the air contained in the vortex that’s carrying the momentum. You’re overthinking this, as others have suggested. Try graphing out the motion of a single molecule of the fluid as it moves around in the vortex AS the vortex moves forward through the air. You can do this easily if you have one of those Spirograph toys with a straight rack-type gear, and run one of the round gears along it with your pen. That ought to convince you there’s a net forward momentum in a moving vortex.
When a Frisbee in the air penetrates a mathematical surface, it carries mass through that surface, so I have no problem with that. Much more frisbee mass goes forward than displaced air mass flows backwards. I conclude that the plastic frisbee is a misleading analogy.
OK, lets make a Frisbee out of ice and then take it underwater. (Assume for a moment that the density of ice is exactly the same as that of water.) Now when the Frisbee penetrates an invisible surface, it will be surrounded by a layer of water which flows backwards to occupy the space vacated by the frisbee, so the net amount of liquid and frozen water which crosses that surface will be zero. To make this resemble a vortex-ring, just use two rotating frisbees with edges touching (like a pair of adjacent gears.) As they penetrate the surface, an equal amount of water flows the other way and fills the space which would otherwise be left behind the frisbees. When a volume of Frisbee moves forward, an equal volume of water moves backward, and since the two materials have the same density, no net mass crosses the invisible surface.
Maybe we’re supposed to ignore the momentum of the water which flows backwards relative to the forward-moving frisbees? If so, then my question is WHY? Ignoring that water clears up the conceptual tangle, but ignoring it seems totally arbitrary. I need a better reason to ignore it than “it clears up the mess.”
No, I’m trying to construct a counterargument after someone (long ago) argued that a travelling ring-vortex carries zero net momentum since it is surrounded by an equal rearward flow.
Your answer, simplifed, seems to be that we should ignore the backwards-flowing fluid outside the vortex “separatrix” surface. The “separatrix” is the imaginary surface which contains the fluid which travels along with the vortex. I think the separatrix of a ring-vortex resembles a sphere. This “sphere” moves forwards but it is surrounded by fluid which moves backwards. Why should we ignore the backwards-moving fluid? I need a good, solid reason.
Again, that would be arbitrarily ignoring the backwards-moving fluid outside the separatrix surface. Personally, I really want to ignore that fluid, but I’ve never hit on a detailed and sensible explanation of why I’m allowed to do so.
Here’s a little article I wrote years ago which offers an alternative way to explain aircraft flight:
http://www.amasci.com/wing/rotbal.html
According to one physicist that article is wrong, since according to him vortices don’t carry momentum, since the net mass transported by a vortex is zero. To make my article work, I must ignore the counterflow of air outside those “massive disk-balloons.” But at present I’m ignoring it arbitrarily because it makes the physics work right. This has always bugged me, so I’m trying to do something about it.
Is there not mass in the vortex as it travels forward? Instead of the plastic in the frisbee, there is a net movement of air particles forward. It doesn’t matter that the density of the air in the vortex is the same as that of the air, there is still a net movement of particles going forward. As Q.E.D. pointed out, if you put smoke in the Wham-O gun, it acts as a tracer for the air moving forward. Air still moves forward, despite it’s density being the same as the surrounding air. When you are considering the amount of momentum imparted, my guess is it’s equal to the volume of the vortex multiplied by the air density (hence the mass of the moving air), times the velocity of the vortex.
I’m not sure I follow you with regard to “the backwards-moving fluid outside the separatrix surface.” This is just the still fluid ouside the vortex. It isn’t moving at all, except to "get out of the way"of the moving vortex, then flow back in after the vortex has passed. Clearly, empirical evidence shows that a vortex does carry momentum, and no number of physicists saying they don’t is gonna make it so.
Okay. You have some mass coming out of your jet engine or whatever, with a momentum of m[sub]1[/sub]v[sub]1[/sub].
Some other air replaces it. A reasonable assumption is:
m[sub]1[/sub] = m[sub]2[/sub]
Your implied assuption that
v[sub]1[/sub] = v[sub]2[/sub]
is unwarranted.
If that air simply parted to let the vortex pass, that would answer my question. Yet as far as I know, it does not simply part. Some air “opens up” ahead of the vortex. This leaves extra air, but that’s OK, since other air is “closing up” behind the vortex, and the extra air from the front can just rush backwards around the outside of the vortex to fill the space behind. That’s what this animation tries to show:
The red beads go forward, but they’re accompanied by a pattern of backwards-flowing dark beads.
Imagine using your hand to push a baseball through a water-filled pipe. Water rushes around the rim of the baseball to get from front to back. Make the pipe bigger, and that region of backwards flow does not go away. It’s the rushing backwards that screws everything up!
One other viewpoint: if the vortex did not carry net momentum, then it could not travel forwards at all, instead it would just sit in place and twirl internally. (But this still does not tell me WHY we’re allowed to ignore the momentum in that pattern of backwards-moving fluid which moves along with the vortex.)
Yes, that’s essentially what I said before. But it’s not physicists asking the question, it’s ME. It’s ok to say “sorry, I don’t know the answer”, but it’s not ok to say “the question itself is not important.”
My question can be reduced to this: The vortex pattern is composed of forward-flowing air and backwards-flowing air which have opposite values of momentum, so WHY are we allowed to ignore the backwards flowing air when calculating the total momentum?
I don’t know.
The forwards-flowing molecules stay with the vortex, while the backwards-flowing molecules are part of a “travelling pattern” which follows the vortex, yet this pattern is made of constantly different molecules. The backwards-flow pattern follows the vortex, but it doesn’t carry any molecules along with the pattern. Maybe this fact contains a hint, but I’ve not been able to follow it up and show myself why the “real” momentum of the vortex is not simply zero.
Take a breif photo of all the molecules in the vortex pattern and sum up their momentum. It’s zero. (The photo doesn’t let us tell the difference between molecules which follow the vortex and molecules which are temporarily part of the flow patterns.) Let the vortex travel along, and take another flash picture. Net momentum still zero.
Maybe the solution is to figure out what happens when a Wham-O air blaster gun creates a vortex from nothing… or what happens when a vortex is destroyed by striking a flat surface. The gun obviously deposits momentum into the atmosphere, and the flat plate obviously extracts momentum from the atmosphere. It’s what happens while the vortex is “on the fly” which has me all confused.
You’re thinking about it wrong, There is no backward flow around the vortex, unless you consider the frame of reference to be the vortex itself, in which case the vortex is stationary and everything else is moving backwards, which is a bit silly. I think you’re trying to incorporate two different frames of reference in one visualization, which is throwing you off. In one component of your thought experiment you are thinking about the vortex travelling forward relative to a fixed point in space, while in another you are simultaneously thinking about the motion of the air relative to the vortex itself, and that’s just plain wrong. Imagine what happens to the air around the vortex relative to a fixed point in space and I think you’ll see the point more clearly.
The water is NOT rushing backwards. There is nothing in the system to give the water a velocity in the direction opposite to the motion of the baseball. The water is trying very hard to stay still (object at rest stays at rest…) but is being pushed forward by the baseball. The water you see as “rushing backward” is actually either stationary or moving forward at a velocity less than the baseball. It is only moving backward in the reference frame of the baseball itself.
Try this yourself. Get a pan of water and pull a large spoon in a straight line. You’ll get some localized “backward” flow (in the direction opposite the spoon) caused by the vortices in the wake, but the vortices themselves and the bulk motion of the fluid is all in the same direction the spoon is moving. From the reference of the spoon, the water is “rushing backwards”, but don’t let that change of reference frames trick you into thinking that the fluid is actually moving backward in a stationary reference frame.
You’re not ignoring anything. Feel free to sum the momentums in the vortex and come up with zero. However, if the vortex itself is moving, then there is a bulk motion of the fluid in addition to the rotational motion. Q.E.D’s analogy to a frisbee was quite apt; the fluid in the vortex has a rotational component around its axis, but it also has another component which accounts for the overall motion of the vortex. The momentum of the former (rotational components) cancel but the momentum of the latter (bulk motion) is what balances the drag on the oar.
You can enhance this nicely by sprinkling black pepper or talcum powder on the surface of the water first. Pepper is especially good, since you can see the individual granules and note that they don’t move much with respect to the pan.
So if I repeatedly fire a Wammo gun at the other end of the room, I’ll end up with a vacuum on my end? Air must most certainly move backwards, as well. If you say that the air the vortex displaces just “gets out of the way”, then where does it go? Does it just compress transversely? Somehow, I don’t buy that.
I would think that it’s temporarily displaced to allow the vortex flow to pass. Using mere speculation, the air particles are forced to the side when the vortex passes and then forced back to replace the void left by the vortex after it has passed. The air shot by the Whammo gun stays together and acts as a mass of particles, however, since the air was originally in the gun, there is no loss of particles in the surrounding air. Even if the gun were to operate using the surrounding air, certainly air flow would replace the displaced air.
That’s exactly what I’m saying. I don’t see what’s so inconceivable about that.
It was mentioned earlier, but since this thread now seems to be about the physics of vortices, it needs to be repeated:
Vortices, in both rowing and wings, are lost energy and undesirable. They represent a fraction of the total energy involved, and have nothing to do with why your boat moves forward or plane stays in the sky. They result from air/water flows around the wing/oar because there is an end to that wing/oar. They are an evil side-effect.
Not owning a whammo air gun (need to get one of those), I can only theorize about their operation. I assume they set in motion a mass of air (reaction felt in the kick)…the air “bullet” is in friction with the still air it moves through. This would cause essentially toroidal dissipation of the energy around the path of the bullet. (or does the gun actually induce spin in the bullet to create a vortex – doesn’t really matter because the toroidal dissipation would still happen) Note that oar and wing vortices are NOT toroids, but there is a common theme…this air movement represents losses in the system; it is a constraint on the operation of the system, not part of the system itself.
There’s certainly a counterflow set up, but it’s large scale and not terribly useful in the overall momentum balance. Do the spoon-in-water experiment I suggested. As the spoon creates a bulk fluid motion in one direction, that flow will hit the side of the pan, deflect down the sides and form a large circular counterflow which fills in the wake of the spoon. Even in a very large pan (e.g. the ocean) fluid is just pulled in from the side to even out the pressure variations caused by the bulk flow. No vacuum. No magic. Nothing counterintuitive as far as I can tell. If you push something through a fluid, it induces a motion in the fluid along with it. This in turn induces a motion in the fluid to even out pressure gradients.
This list (fish, birds, helicopters, propellers) may be confusing the issue. Some of these things use lift-based mechanisms, others use drag-based mechanisms - the difference being whether they are producing forward propulsion by using the pressure induced by differential velocities of fluid over an airfoil (lift), or by pushing against a relatively incompressible fluid (drag).
What you’re seeing in a vortex, (and thank you QuantGuy for the clear explanation), is the undesireable side effects of a reaction. The real fluid movement in a vortex, the one that produces the forward movement, is the linear flow of fluid through the center of the doughnut. The fluid which has been linearly accelerated drags nearby fluid into its flow, producing the vortex. This vortex then takes a while to dissipate. But it is the linear flow of fluid - through the center of the vortex - which will knock over objects at a distance.
For how some of this applies to fish swimming you might start with this paper: http://darwin.bio.uci.edu/~edrucker/home/Other%20PDFs/NIPS.pdf
As for propellors of all sorts, they don’t push - they pull. Since they are made up of twisted airfoils, they generate a lift vector which pulls the propellor forward. They don’t screw through the fluid pushing it backward behind them. The same is true of bird wings.
In Chronos’s experiment, it would seem that the only reason the air moves ‘backwards’ is because it’s constrained by the walls. Consider this set-up :
You have a sufficiently wide and infinitely long room. At the start, there is a removable wall creating a small enclosed section, which contains an air blaster and some air. The rest of the room is vacuum. You remove the wall (instantaneously) and fire the gun at the same time. If the gun’s powerful enough, why shouldn’t the mass of air fired by the gun move ahead of the rest of the air advancing into the room and never stop?
I think it’s true that, for fairly reasonable boundary conditions, an incompressible fluid in a rigid, stationary container can’t have nonzero momentum. (This is just the gradient of the conservation of the position of the center of mass of the liquid.) So how can you gain momentum rowing across a lake? Well, think of your rowboat as displacing an equal mass of water. Now it’s certainly possible to create a pair of counterrotating vortices, one on either side of this missing volume of water. (These are pure vortices, transferring no total momentum to the system.) The water on either side of the boat moves in one direction; in the center, the water, and your rowboat displacing some of the water, moves forwards. All your rowboat is doing, and all your efforts in rowing, go toward maintaining this pair of vortices next to your rowboat, where they’ll propel the boat, rather than at a fixed position in the lake, where all they’ll do is spin the water around.
Just think of replacing your boat at any moment by an equal mass of water, and imagine the net forces on that fluid element due to the two vortices. Of course your boat is, hopefully, rigid, so when that fluid element starts to separate the boat’s internal forces will cause more eddies as the water compensates, but this is a secondary consideration.