Some of my readers, JeffId, Gavin and others are discussing the possibility that GCM’s do not properly account for condensation on local pressure. I am agnostic on this (I’m trying to think about it.) However, plowing through the comments at The Air Vent, I came across an argument between a number of people (Carrick, SteveF and others). I think the question that triggered the argument is the following one:
94. Gavin said
Consider a thought experiment. Take a closed container filled with super-saturated air and place it on a scale. Will you be able to detect the moment of condensation by monitoring it’s weight (i.e. the pressure on the scale)?
I bet you all think you know the answer to this question. Right? I’m going to answer this question in agonizing nit-picky detail.
But before I do, I’m going to show discuss a gedanken problem that contains some of the physics we need to understand before we can fully answer Gavin’s question. This will serve as a preliminary to the full answer to Gavin’s question. The gedanken will also address a debate going on in comments on that thread. By showing this problem we will be able to go forward and figure out if the nit-picky details discussed here could ever have any practical importance in any application of interest (and more particularly GCMs.)
The Thought Experiment:
Preliminaries:
Suppose we had a very tall massless container full of some viscous fluid, with density Ïf and weight Wf which for the purpose of this discussion will be 1N. The container placed on the read out on the scale will balance the weight of the fluid, W=1N.
Next, suspend a small diameter, heavy sphere with weight Wp from the ceiling in the room; we select this sphere so that its density is much larger than the fluid Ïf< < Ïp. After immersing the sphere we wait for everything to come to static equilibrium, then read weight registered on the scale.
Because the weight supported by the scale is the sum of the fluid and the weight of water displaced by the sphere, the scale will now read W = 1N + Fb where Fb is the weight of buoyancy force, and is equal to the weight of displaced fluid. To facilitate discussion, I’ll break this out into a numbered equation:
where g is the acceleration due to gravity and Vp is the volume of the sphere (or particle).
Note in the figure to the right, the scale reads W~1N. This is because given the assumption Ïf< < Ïp, the weight of water displaced by the sphere is very small relative to the weight of the solid sphere and my illustration neglects that small weight.
Begin experiment: Break support
At a time we will call t=0, we will break the support and simultaneously read the weight on the scale.
First question: What should the scale read? The answer is very, very close to 1N+Fb. That’s the same weight we read just prior to cutting the support.
Why?
The sphere is motionless. In this circumstance, the upward force exerted by the fluid on the sphere is approximately equal to that due to buoyancy and equal to the weight of displaced fluid
The particle exerts an equal and opposite force on the fluid. So, for all practical purposes, at this point, the pressure field in the fluid obeys the laws of hydrostatics. The pressure at the bottom of the cylinder must result in a force that balances the the sum of the weight of the fluid only and the weight of fluid displaced by the particle. It need not balance the weight of the particle. (Note: For those who are aware of funky forces like added mass or the Basset force, I’m both neglecting and not discussing them. However I am including ~ in place of = for that reason.)
Next question: What happens to the sphere? The answer is simple: It starts to accelerate downward. This happens because while the force of gravity on the particle is Fg = Ïp g Vp, and acts downward.
So, the net downward force on the particle is:
So, at exactly t=0, when the velocity of the particle is exactly 0, the particle will accelerate at
where Mp is the mass of the sphere and U is the vertical velocity with positive values pointing down.
Owing to this acceleration, the velocity of the particle will increase; so at 0< t, the 0< U, with positive values of U corresponding to the downward direction.
What does the scale register at 0<t?
To figure out what the scale registers after the sphere is released, we must discuss the viscous force acting on the particle. For the purpose of further discussion, we will assume that other than in the region near the surface of the particle the fluid velocity is zero, and neglect any acceleration of the bulk fluid when discussing the weight registering on the scale. The magnitude of the fluid acceleration will be small if Ïf< <Ïp as previously assumed.
Force at time 0< t.
Once the sphere’s velocity U is not zero, the force exerted by the fluid on the particle will include both the buoyancy force acting upward and the viscous force, in this case also acting upward. To avoid specifying the constitutive relation for the viscous force, we will represent this as Fv and observe that it is a function of the velocity of the particle relative to the velocity at some point far from the surface of the particle. Given previous assumptions, the viscous force acting on the particle will be a function of particle velocity only, Fv(U).
Because this viscous force is exerted on the fluid by the particle, an equal and opposite force is exerted by the particle on the fluid. If the bulk of the fluid does not accelerate, the weight exerted by the bottom of the container on the fluid will be equal to the sum of the mass of fluid, the mass of fluid displaced by the sphere and force exerted by the particle on the fluid. So, the scale now registers a weight W of :
This is larger than the previous read out of W=1N+Fb. How much larger will depend on the magnitude of Fv, which will depend on the velocity of the sphere.
Final state:
If the container is sufficiently tall (i.e. infinitely), the particle will eventually accelerate to its terminal velocity. At that point, the viscous force acting on the particle will exactly balance the weight of the particle. So
At this point, the weight registered by the scale will be:
That is: Once the particles have achieved their terminal velocity, the weight on the scale will equal the sum of the weight of the fluid and the weight of the solids. Of course, that will also be the weight when the sphere comes to rest at the bottom of the container.
Observation
For this gedanken experiment, the weight registered by the scale would vary over time, increasing from approximately 1N when the velocity of the sphere is motionless to 3N when the sphere has either reached its terminal velocity or come to rest at the bottom of the container.
This may seem perplexing to those whose eyes are focused on the outside of the container which is motionless giving the false impression the container and its contents are in static equilibrium. If that were so, the weight registered on the scale jump to 3N the exact instant the cable supporting the 2N sphere was broken.
However, contrary to one’s initial impression, the contents of the container are not in static equilibrium. The heavy sphere is accelerating downward. Because the sphere is denser than fluid, this means the center of mass of the contents of the container are accelerating downward. This can only occur if the weight exerted on the container by the scale is less than the weight of the container’s contents.
This analysis will represent the second part of an analysis in my next more complicated, gedanken experiment which may require magic to occur at time t=0. The result will have some consequences if one wishes to provide a full, thorough, complete, true to the tiniest nit answer to Gavin’s question. We’ll see that the answer to his question is: Under circumstances (possibly those that involve magic at time t=0), it might be hypothetically possible to detect condensation by monitoring weight with a scale.
Will the result of this or the second analysis have any practical consequences for climate models, or the issue of properly modeling condensation? I have no idea. It may very well have no real consequences. But, recognizing why the answer to Gavin’s question is not simply, “Flat out no, not under any circumstances, no”, will help people do scaling analyses to either show the effect doesn’t matter or show that it does.
Fascinating, fun, and educational! Thanks Lucia, very well presented.
Hmmm… How about detecting the relativistic loss of mass when the energy of condensation radiates out of the system ? 🙂
toto–
I’ve assumed radiative effects are small, as I always do in engineering analyses. I’m also assuming the speeds are small relative to the speed of light. But you are welcome to extend to the more general case.
Lucia, for many engineering problems, acoustic radiation is one of the dominant sources of energy loss. While it doesn’t affect the mass, it still has an effect on the system.
One of the conundrums I pointed out to SteveF is, if you take the weight of the air in the atmosphere and divide it by the surface area of the Earth, the answer is larger than the average static pressure on the surface of the Earth (the effect does not depend on rotation, it is purely a geometrical effect).
The naive idea that the pressure on the surface of the Earth equals the restoring force per unit area on the column of the air above it only applies to horizontal, stratified, stationary media… which is to say, not the Earth.
You lost me at equation 5.
As the problem is stated, it could be satisfied by having a small sphere with a volume of 1 cm3, made of uranium (density = 19g/cm3). Or, it might be made of wonderflonioum (density = 9g/cm3).
Surely, the final state for the uranium gedanken will be
W = 1N + ((19 grams) * g)
while for wonderflonium, it will be
W = 1N + ((9 grams) * g)
Sorry if I am missing something.
Where g = 9.8 m/sec2
(to get an answer in Newtons rather than grams)
What a fascinating thought experiment! As a slight variation (a bit closer to the original question), you could have the fluid cylinder completely closed and sealed, with the weight suspended from the roof on a string. At time t=0 the string breaks. The weight recorded on the scale goes from 3N for t<0 to 1N at t=0 and then gradually back up to 3N again.
Carrick–
I saw the geometry issue, but this doesn’t address that. The next step only addresses the simpler question: Even if we had the simple planar geometry, will the pressure at the bottom of the column balance the weight of the contents at all times.
The answer is not necessarily. Sometimes it does not.
The analysis above doesn’t deal with energy. Only momentum. 🙂
Of course, if we want to figure out whether energy dissipates to heat or understand any irreverisible effects, we will want to look at factors affecting energy losses. We may need to look at the constitutive equation for force more carefully etc.
But that’s not going to change the answer to the simpler problem which is that, no matter what happens with energy and dissipation, the pressure at the bottom of the container is not required to offset the weight of the material in the container at all times.
Carrick– I should add: I am looking at artificial problems to isolate issues. In this one, I am trying to isolate the issue Gavin brought up with his hypothetical. Can we detect condensation from the weight. The answer in a very simple example will be that hypothetically in a very clean system we might be able to do so.
Paul M–
Does the container hold fluid with a density equal to the sourrounding fluid? And does it hold the same volume as the particle?
If so, it’s neutrally bouyant. The fluid cylnder won’t weigh 2N. The fluid cylinder will weigh Ïf g Vp, which given the assumptions above is much less than both 1N (because the suspended cylinder is small compared to the container). It will weight Ïf/Ïb ) which is <<2N because Ïf/Ïb <<1. In fact, the fluid is air and the particle is a water filled balloon, Ïf/Ïb ~10^3.
But anyway, if the fluid cylinder is neutrally bouyant, it will never accelerate and nothing happens when you cut the chord.
Amac–
I specified the weight, not its density or volume. I also didn’t specify the planet.
Just to have numbers, I said the of the sphere is 2N. So, the mass is not 9 grams * g– unless g= 222 m/2^2.
It’s a gedanken could be a huge planet like jupiter, but in some other solar system, so warmer or colder than jupiter (if those choices make us be able to better locate fluids with viscosities you want to fiddle with.)
So, g could be 200 m/s^2. The fluid could be a air, a heavy gas whatever you like. The container can be really, really tall.
For this exposition, I figure it’s easier to just say it’s 2N rather than specifying the densities, size of the particle etc. That way, at the end of the problem we have the weight is 1N +2N with no additional calculations. I figure that’s easier for the more math averse than saying Wp and Wf.
Amac-
I should add:
Yes. For your choices, that’s the weight in the final state. It’s
W=Wf + Wp.
That is: the sum of the weight of the fluid and particles. In my case, Wp=2N and Wf=1 because I said they were.
But the more general is W=Wf + Wp with the weights being whatever they are in the contraption you set up on your gedanken planet.
Well I was thinking of a sealed container of water surrounded by air, with the sphere inside the cylinder suspended from the lid by a dodgy string. In this case the measured weight goes from 3 to 1 to 3. But if the fluids are the same inside and out it would go from 2 to 1 to 2.
Sorry that should say 2 to 0 to 2.
Paul–
Are there two strings? One suspending the sphere inside a capsule and another suspending the capsule from the ceiling? Then both break? Or only one breaks? I think I need a picture of what you are describing.
The main reason I wanted to bring up the geometry is to dissuade people from the notion that air is a column of bricks. It’s simply not the case that the pressure has to be equal to the weight of the column of air above it divide by its mass in order to have hydrostatic equilibrium. In fact for the Earth it’s in general not true.
Carrick– I agree.
Right now, I’m more interested in the condensation issue, so I’m looking at the simplified geometry.
Out of curiosity, is treating the atmosphere as a plane for the sake of computing hydrostatic pressure a mistake the models make? Seems like they get that geometric feature right. It’s can’t be computationally intensive to recognize the earth’s is not flat. Or am I mistaken about that?
Lucia, thanks for the clarification. (Wipes away egg,) and sorry to have muddied the waters.
Amac–
Not a problem. I’m deferring writing the 2nd gedanken problem until I see what people ask on this one. That way, I’m more likely to write the second one in a way my audience grasps.
The 2nd example will include an order or magnitude estimate. (I don’t know what it will come out to.) With luck, someone who knows more about weather than I do will also criticize the order of magnitude estimate.
Lucia,
I took Gavin’s notion as Paul interpreted – the string is attached not to the sky but to the lid of the container, so the tension of the string initially adds to the measured weight. Then the final weight is the same as the pre-cut wt.
The scales deviation of the free-fall period is brief – with water droplets very much so. Far too small to affect models, and probably anything one could practically measure.
In fact, since the droplets grow from near-zero mass, it’s arguable that they are at terminal velocity at all stages.
Are you _sure_ about this analysis ?
If we imagine the fluid is very viscous, such that the initial acceleration is very slow, it’s pretty obvious the weight reading at the start would be 3W.
The question is then why should less viscous fluids be any different ?
Obviously gases behave differently due to compressibility issues, ie the gas will locally be compressed under the ball, rather than passing the forces directly to the scale.
It’s amazing how seemingly simple physics rapidly gets complicated (at least in my head).
Nick Stokes:
Actually there is radiative sound associated with it (at least with the change of state of matter).
Radiative sound can be important as a source of energy loss.
Also, Nick, if the media is not stratified along horizontal planes (as it won’t be), we already know all bets are off.
Just as with the Jet plane, the surface area of “support” for the falling water drops increases as you go farther below the falling (terminal velocity) particle.
This is neglecting effects of entrainment of the falling water drops on the nearby air molecules of course.
Lucia,
Am I thinking about this correctly?
The force that keeps the water droplets suspended in a free-fall (vapor) has to be external to the system, i.e. a string attached to the ceiling not to the inside of the container. Condensation is the removal of the force (string being cut).
Nick/Tamara–
This analysis doesn’t map into the full condensation problem. It’s a preliminary to a 2nd gedankan, which would be closer. So, we’ll get there.
I just want people to see this part first and see if for this problem the agree this would happen to the weight.
I’ll get to the difference between the string attached to the lid on the container and attached to the ceiling later.
None
In a very viscous fluid, the terminal velocity will be lower, but the initial acceleration at exactly time t=0 will not be affected by viscosity.
For a very viscous fluid and a small sphere with radius “a”, the viscous force on the particle obeys
F=6 π μ a U where μ is viscosity and U is the velocity of the particle relative to the fluid. If the velocity is zero, the force due to viscosity is zero.
The net force is as in (2) above and the acceleration is as in 3.
Notice when U=0, the constitutive law for the force doesn’t affect the acceleration. (An added mass term might…. but I’m neglecting that, and it won’t matter given my assumption about the density ratio.)
Nick–
Nearly. Were going to get to scaling analysis later. We may not even need to get this far. The hypothetical ‘issue’ may be shown small because of a different issue that doesn’t require us to even know the size of the drops when the magically appear.
The thing is we can’t do the scaling analysis if we assume the problem away from the outset. So, to show it’s small, we first have to show what could, hypothetically, happen in some weird situation. Then, we can specify all sorts of dimensions to figure out if there is any real situation where this “matters”.
So we are going to get to the size issue latter– it may even take 2-3 gedankan’s each progressing.
Oh– for the next one. I’m getting a scale for something that’s about 0.1 for an effect that I thought would be smaller in “real” situations. I totally wagged numbers for a first “look-see”
For this: Can you tell me typical values of pressure gradients in various weather situations in locations where water could condense? I swagged:
I figure hurricanes might have some location where water condenses. I swagged: Pressure drop from center to edge of large hurricane: 0.2 atm.
Extent from center to edge: 90 miles.
PaulM–
Somehow my brain wasn’t engaged. I was thinking you meant something more complicated– but yes. This will be more like real condensation.
Lucia, Carrick,
The issue raised by Gavin is (I think) not terribly important (Lucia, I wrote you a brief email message about this). The point of the paper is really that the local pressure under a region of rising moist air must be lowered because the mas of the column will always be lower than a comparable column of dry air due to the substantial heating effect of condensation. The exaggerated expansion pushes air aloft away, and the lower pressure below forces a flow into base of the rising column. The “engine” of circulation (like the Hadley circulation) is driven by thermal expansion due primarily to condensation, not by sensible heat… which is a minor player.
By the way Carrick, few people think the atmosphere is like a column of bricks; I certainly do not. But there are significant differences in surface pressure (eg. the pressure is usually lower when it rains). As the paper points out, the shape of the atmosphere on the scale of the atmospheric circulation is very shallow and very wide, which allows significant pressure differentials to exist, even though the average atmospheric pressure does not change if averaged over enough time.
Condensation driven circulation seems perfectly consistent with the observations. I am impressed with the logic of the paper.
SteveF
But I think Gavin things they capture the heat addition due to phase change during condensation. So, I’m going through baby steps, looking at extreme cases to try to clarify for myself what Anastasia is saying is correct vs. what models supposedly do.
The assumption of local pressure being equal to the pressure of everything above it is turns out to to be not strictly correct. I suspect though possibly incorrectly, that Gavin intended us to all answer “we expect the weight on the scale to always be equal to the weight of the contents of the container”. But that answer would be wrong.
Does it matter that answer is wrong? Maybe not. But if I know what the effect is, I can later do a scaling analysis. If I just assume the error is small relative to relevant physics and don’t look at it for that reason, I can’t later test to see whether the assumption that it is small is true.
BTW: In a tube, this effect would also cause the air in a tube to rise after the expansion. So, if the extra rise Anastasia is describing isn’t due to this then we will have two or more mechanisms for air rising. But…. I’m going to wait until I get to something 3 d before worrying about that.
Lucia,
The falling particle analysis is completely correct for the case of the string holding up the particle. In the case of the closed container with a droplet that instantly condenses, I believe there will have to be a very small ,very brief effect, since the conversion of vapor into an initially stationary droplet means that the mass of the remaining vapor must be reduced (the scale briefly shows a blip downward) until the speed of the droplet reaches Stokes’ terminal velocity. The negative blip on the scale when the droplet forms is exactly counter balanced (in a force times time sense) by a brief positive blip when the droplet hits the bottom of the container.
Lucia,
“But I think Gavin things they capture the heat addition due to phase change during condensation.”
I also think he does, but I suspect he may be mistaken about that. The question is really if the models handle the physics robustly in a rising/moist column of air. Gavin’s initial comments suggested to me that they may not. Gavin’s claim that the models duplicate the Hadley circulation velocity is certainly correct; the real issue is if they get the right answer for the wrong reasons.
lucia – the bold tag was not closed for this heading:
What does the scale register at 0
Which on my work browser (ie – bleh) makes the rest bold too.
I think the paper has caused a few headaches and revising at nasa central – i.e. a lot of motion under the surface while all looks serne up top.
I’m not saying they are wrong – just double checking if they have this affect covered.
SteveF>
“In the case of the closed container with a droplet that instantly condenses, I believe there will have to be a very small ,very brief effect, since the conversion of vapor into an initially stationary droplet means that the mass of the remaining vapor must be reduced (the scale briefly shows a blip downward) until the speed of the droplet reaches Stokes’ terminal velocity.”
You need to think much more carefully about actual droplet formation mechanics, if you want to go down that route. When a droplet forms in a container of super-saturated air, there must be a localised pressure change, rapidly counter-acted by movement of gas within the container (because the water vapour will have been removed). The only simple situation is one in which the droplet condenses on the container. Otherwise, you’d have to factor in the location of the formed droplet and the shape of the container, and work out the movement of the gas as well as the drop.
The rising column of moist air exists in a 3-D world, so pressure differences are affected by what is also going on around a given volume. This causes mixing and circulation. The thing to look at is that a given volume of moist air (say 1 percent by mass) suddenly condensing, loses only 1 percent initial volume, but gives off about 2240 J/g (which goes to heat the surrounding air). The air has a Cp about 1 J/g, so the 100X as much air mass is driven to a temperature rise of 22.4 C. This results in about a 7% or so net volume increase at constant pressure (7% or so density drop), even including the loss of the water vapor partial volume. The large temperature and volume increase would drive large mixing and circulation to surrounding air. This is a major component of transporting energy from the surface to higher altitudes, and along with other drivers of convective circulation (day/night differences, latitude differences, and rotation of the Earth) cause the mixing that drives continual maintaining of the dry or wet adiabatic lapse rate.
pete m– Of course they are going to read and try to figure out if a) the paper is correct, incorrect, or ambiguous and b) if their model accounts for things correctly.
Sorry Lucia, I confused you, delete the word ceiling, replace it with lid of container. Only one string, inside the cylinder.
You have to weigh the mass of a closed system to ask the question. Fail.
Eli–
Did you take lessons to learn how to sound stupid? Or does it just come naturally?
Lucia,
He has a Phd in sounding stupid. remember those who can do, those who cant teach, those who cant teach (read what students have to say about Eli) blog with stupid monikers
Re: lucia (Oct 25 14:44),
That’s too large. The record low pressure in the eye of a hurricane in the Atlantic Basin is 882 mbar for Wilma in 2005. Typhoon Tip in the Pacific reached 870 mbar. Typical storms have a much lower pressure drop.
Dewitt– Standard atmspheric pressure is 1013.25 millibars. So.. yeah… I guess I used I rounded up?
I’ll be putting up some sort of estimate, then maybe others can supply more reasonable ones based on real weather. The difficulty is that I don’t know reasonable levels of condensation and I don’t know what contexts to place them in.
Mosher–
I can’t even figure out if he thinks he is being ironic or what?
Normally, idiots who don’t understand the 2nd law post things like that on analyses involving the 2nd law. So, if this post was about the 2nd law, I might think he was pretending to be an idiot who didn’t understand the 2nd law, and leaving a stupid comment.
But this post doesn’t use the 2nd law, it uses conservation of momentum. Idiots don’t usually drop that.
But of course, we can apply cons. mass, cons. momen, cons. energy and the 2nd law in open systems and routinely do.
So, is Eli pretending to be an idiot who thinks I need to add something like “assume the top of the fluid is covered by thin impermialble membrane that can transmit no force, but prevents fluid from suddenly being propelled out the top like fluid from a rocket propulsion system.”
Since there is no jet engine at the top, a piece of saran wrap would do.
So, is Eli trying to make an obscure joke? Just doesn’t know what he is talking about? What?!!!!
I don’t think Eli understands that readers don’t view him as a commenter who is giving his/her honest opinion, like (the majority of) the rest of us. He can be as clever and as esoteric as makes him feel good, but at the end of the day, who cares?
I invite Eli to come down from the clouds and have an honest conversation with someone.
Andrew
Lucia,
I checked the Atlantic Basin pressures for all tropical cyclone observations 1979-2008 where the wind speed was 35 knots or higher.
For 6789 observations the median was 993 millibars.
882 – 969 = 16.2%
970 – 989 = 26.7%
990 – 999 = 24.2%
higher = 32.6%
Der Gedanke (sing.), die Gedanken (pl.)
die gendanken sind frei…
😉
Regarding the original question, I have been wondering how much of the energy loss during condensation goes into measurable infrared photons. (all energy exchanges are through photons, on shell or virtual of course).
I googled an article with an interesting abstract behind a pay wall:
http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6V62-4YR33HC-1&_user=10&_coverDate=07%2F31%2F2010&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_searchStrId=1514868635&_rerunOrigin=google&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=0233e44ca09ea8a57b7296d51866542b&searchtype=a
Based on experimental data, the author proposes a model explaining the appearance of a window of transparency for the characteristic radiation in the substances when first order phase transitions take place.
It would be interesting to see the processes and see how much smaller the mass of the condensate is after the photons radiate away. Maybe a very accurate scale could see the difference.
( you know the joke, “if you are a hammer everything looks like a nail”, so if you are a particle physicist ….)
Alex–Thanks. Oddly, I got it right in the body of the discussion, but wrong in the title!
Lucia,
far be it from Eli to engage sensibly. He could learn something from Gavin. maybe he’s used to bullying students. breeds contempt and a sense of self importance that is inconsistent with the pay scale.
read what students have to say about Eli
.
This bunny would love to read that. Do you have a link, oh two-tongued one?
“read what students have to say about Eli”
Steven
There’s a lot of provocation around, and I understand your frustration with the wabbit, but I don’t think it’s worth impugning personal reputations. That just sinks to the level of the Wegman attackers.
No, he really should post a link, because I can’t seem to find anything on Google, and you would think I would quickly have been led to a skeptic blog with someone saying something along Mosher’s line with a link for everyone to enjoy.
.
The reputation has already been impugned. Now all I need is a link, so I can see for myself if what Eli’s students write about him is more awful than what I wrote about my teachers.
.
That just sinks to the level of the Wegman attackers.
.
You still do not get it, do you? The attackers attack not Wegman, but the Denial Machine of which Wegman has been but a small part. You know, the thing that can do whatever it wants to your thinking and lets you cry for the moon. Shine on…
Neven–
I found “Rate your professor” here:
recent: http://www.ratemyprofessors.com/ShowRatings.jsp?tid=543236&page=1
past: http://www.ratemyprofessors.com/ShowRatings.jsp?tid=543236&page=5
It looks like back in 2005, his students thought he s*cked. More recently, the reviewers are better.
I don’t know how students find that site, so I wouldn’t assume the reviews are all unbiased. (A group of friends who all hate or love a prof could act together etc. Who knows?)
Neven (Comment#56840) October 26th, 2010 at 6:45 pm
No, he really should post a link, because I can’t seem to find anything on Google, and you would think I would quickly have been led to a skeptic blog with someone saying something along Mosher’s line with a link for everyone to enjoy.
.
The reputation has already been impugned. Now all I need is a link, so I can see for myself if what Eli’s students write about him is more awful than what I wrote about my teachers.
###############
“He lecutres off the notes that he types and gets off of the internet. There isn’t much problem working/solving during class. Recitation doesn’t help that much either, you might be better getting help from your peers.”
“HORRIBLE! HIS LECTURES ARE COPIED AND PASTED FROM THE INTERNET. FALL 2004 WAS THE FIRST SEMESTER OF CHEMISTRY HE TAUGHT IN 15 YEARS. AVERAGES OF ALL HIS EXAMS WERE BETWEEN 46-53. HE IS HEARTLESS AND WOULD BE HAPPY TO FAIL YOU. NOT HELPFULL AT ALL. IF YOU WANT TO ACTUALLY “LEARN” CHEMISTRY, TAKE ZUK.”
There used to be more at other sites, one particularly unpleasant incident. The comments in general comport with the “personality” we all witness here on the internet. FWIW. Further, the test is not whether the comments are worse that what you wrote because you too are a hopeless judge of things and unless you can produce your comments we really are left just believing you. not happening today. As a person who had to read student reviews for an entire university ( to select a “best faculty award) I’ll take my judgement over yours.
In any case you can go back to CA circa 2007 ( dont try to use google it wont help you) and search for the links for other material. But what we have here from these two anonymous students ( kinda like DC) is enough to get one interested. I dunno maybe his university does lecture notes. That would be hella funny.
Thanks for the links!
Well, of course I am a hopeless judge with a confirmation bias to boot, but it wasn’t as black and white as I expected based on Mr Mosher’s objective judgement (I expected far worse).
.
It looks like the stupid and lazy students don’t like the no-nonsense Eli Rabett. Stupid because they probably flunked some test and have to blame somebody. Lazy because they can’t be bothered to explain beyond ‘he’s the worst teacher evah’. What a surprise. The karma point people such as Mr Mosher don’t like him either. You fill in the rest, little bunnies.
AnyColourYouLike (Comment#56837) October 26th, 2010 at 6:19 pm
The ‘attackers’ are pointing out that his attack on climate scientists is incompetent. We have the ridiculous spectacle of the star statistician demanding that climate scientists should call upon the expertise of people like him, because they are incompetent and can’t do the stats right, and then proceeds to make a mess of his own paper making the accusations. For bonus hypocrite points, he has not released the code he promised to release. McIntyre has demanded condemnation and public humiliation of scientists for such actions. His super powers of auditing also seem to fail him whenever he comes near a claim or paper that agrees with his point of view.
If you are so concerned with keeping the debate civil and showing respect for other’s points of view, I suggest you wind back time and undo the creation of sites such as climate audit and wuwt.
Lucia gave us:
“… this means the center of mass of the contents of the container are accelerating downward.”
That probably gives you what you need to show that the weight must decrease during the condensation experiment.
For the centre of gravity is definitely lower finally than it was initially. To preserve continuity it must accelerate downwards and then upwards to get between the two equilibrium states.
Alex
Hmm,
Interestingly that implies that it would also loose weight if the condensation only occurs at the bottom surface.
Bizarre, but true, (for one thing the molecules that condense fail to bounce, hence their change in momentum is halved).
Alex
The strategy of personalizing the scientific and policy issues is unhelpful when the A-Listers do it.
Equally distasteful and counterproductive here.
It takes almost superhuman effort for the target of a mean-spirited attack to say, “I’m still smarting from the poison-pen insults that [star blogger X] and his acolytes directed at me concerning Issue A. But they’ve marshalled good evidence and insightful arguments on Issue B.”
The conspiracy-themed worldviews of some people are mostly silliness. Both sides make some good points and some lame points.
My two cents.
Higher density material accelerates down; lower density fluid accelerates up. The net is for the center of mas to accelerate down. In this problem, I neglect the density of the fluid to simplify presentation; it could be easily incorporated.
Yes if you mean “near”. But it’s a transient until the heavy stuff rests on the bottom.
Lucia,
No I mean at.
To begin with the water is on average in the middle, at the end it is at the bottom. It has descended as has the centre of mass, at some point the container must have lost weight, irrespective of the mechanism.
Alex
Lucia,
It acts like a propulsion system for the can and dry air part. If it was not for the gravity field, condensation on just one surface would move the can to maintain the centre of mass, it has to.
Alex
If there are no forces acting on the total and the density distribution changes within, a propulsion of the whole thing would violate momentum conservation. If there is a momentum vector towards the side an equal and opposite motion, a recoil of the air, will maintain momentum conservation.Re: Alexander Harvey (Oct 27 06:59),
Bugs:
“McIntyre has demanded condemnation and public humiliation of scientists for such actions.”
really. mcIntyre demanded this.
“If you are so concerned with keeping the debate civil and showing respect for other’s points of view, I suggest you wind back time and undo the creation of sites such as climate audit and wuwt.”
As I mentioned in my essay on “blackmail” it is entirely predictable for people who engage in the kind of truth bending you do to take the next step and engage in fantasy. Those fantasy’s are part of your human make up. You cannot help but experience and express these fantasies once you start down the path you do. “cant I just make my opponents disappear? cant we go back to a time when they didnt exist? I wish they never were born. They started it, so I am justified.
When you spend years and years reading how people progress in this mind set you will see that you really have no control over your thoughts. You dont even know that the fanatasy you express is not your own. The meme has a mind of its own. My opponent wont listen to my arguments. my opponent is stupid, my opponent is irrational. There is no point in talking to them. They arent human. I wish they never existed. lets joke about blowing them up. no, lets really do something. It’s in your genes. you are the progeny of generations who have survived by this pattern of thinking and behaving.
I will say the same thing. If somebody doesnt release their code and data I am under no obligation to believe their work. wegman, scafetta, skeptic or believer. So, if you want to start a blog dedicated to getting wegman to release his code, I’ll support you. If you want to organize a campaign to pressure him to release it, I’ll help. If you dont, then I have no time for hypocrits. If you want me or somebody else to head up that effort, then I got no time for lazy punks. get a wordpress blog, start blogging or shut…
BOSTFU
Ok this is hella cool
http://www.hpcinthecloud.com/features/SC10-Disruptive-Technology-Preview–The-First-Cloud-Portal-to-R-and-Beyond-105776458.html
I see cutting the string as effectively adding a mass to the fluid. This mass is then accelerated through the fluid by the force gravity exerts on it, minus the force its bouyancy exerts upwards. Its acceleration will reduce as the drag force it experiences increases, and I agree with Lucia that the weight of the total cylinder and contents will equalise when the particle reaches terminal velocity. I also think that the pressure at a nominal horizontal plane through the cylinder will reduce as the particle passes from above it to below it. IMO the actual real weight is the final position of particle on the bottom of the cycliner and the starting condition has a reduced downward force due to the tension in the string. Once this is cut the “restraining” force on the fall of the centre of mass is reduced allowing it to accelerate as described above. The reduction in tension in the string is accompanied by a reduction in compressive load for whatever is holding the other end of the string up. On getting the bottom of the cylinder it will give deliver an impulse causing the scale to register a blip increase before returning to register the total weight of the cyclinder and contents. There are no violations of any simple laws that I can see and I’d expect viscosity and dissipative effects to all follow standard derivations. Could be wrong though – awaiting part 2! 🙂
It is not my fantasy, I was just stating the absurdity of ever trying to get a rational, unemotional debate happening. You can’t unwind time, and McIntyre will always have existed, along with Watts and Marohasy and Morano.
Science has spent many years trying to have a rational, unemotional debate, since that is one of the requirements for the advancement of science. McIntyre and his upsidedownmann level of debate is the antisethis of that. How do you fight that? If you ignore him, he gets the status of ‘most important’, although he does not understand the ambiguity of that status, if you take him on, he sets the level of the fight, which includes the juvenile, irrational and emotional. It’s a lose/lose situation.
Wow, bugs. Just wow…
That screed causes me to be truly concerned about you.
I don’t know what you have been smoking, but I never said what you have put in quotes.
I said
Which is not a fantasy, but clearly impossible. ACYL said he doesn’t like the mudpit we find ourselves in. The mudpit was created by McIntyre and his friends. You just have to accept that is the situation we are in now. The standards of science were created to avoid exactly the predicament in which we now find ourselves. Those standards were trashed by juvenile behavior such as ‘upsidedownmann’. If you want to get out of that situation, some apologies would be a good start.
anna v:
I did not say the whole thing.
“It acts like a propulsion system for the can and dry air part.”
The can and air would move to counter balance the movement of the water part.
The air being much less dense than liquid water will not move sufficiently to act as a counter.
Alex
Anna V —
The scenario which Alexander Harvey was considering, involves condensation *at just one end*, and in this case, the cylinder will change position such that the condensing end moves toward the (constant) center of mass.
I prefer to think of the sequence in zero-g as follows:
1) Water vapor near one end condenses. At this point, the mass distribution remains constant along the cylinder’s axis, but
2) Pressure near the condensing end drops with the phase change.
3) The remaining atmosphere redistributes itself to equalize pressure.
As step #3 completes, it’s clear that the center of mass of the remaining air has moved towards the condensation surface, to compensate for the “hole” in pressure. There being no external forces, the center of mass of the entire system must be stationary, hence the cylinder (as a whole, including the liquid water) must have moved in the opposite direction.
Or one can think that, just after the moment of condensation, the force of internal air pressure on the condensing end of the cylinder — which will tend to move that end further from the center of mass — is less than the corresponding force on the opposite end; hence there is a net force on the cylinder moving the condensing end towards the center of mass.
With the cylinder on a scale in a gravitational field, there’s no motion of the cylinder, but the effect will show up as a dip in the scale’s reading, assuming the condensation is at the bottom end.
If condensation occurs symmetrically in the cylinder, then one can’t make the above argument, and the cylinder remains stationary in zero-g.
Hope this doesn’t confuse the issue further.
Re: HaroldW (Oct 28 08:00),
I am sorry, but if the cylinder as a whole moves in a force free environment, there is violation of momentum conservation. I cannot put my finger on where the arguments fail, but for sure, momentum is a conserved quantity.
Anna –
I fully agree that momentum is conserved, but that’s what tells me that the cylinder moves (in zero-g). The center of mass (c.m.) of the cylinder does *not* move — that’s the conservation of momentum part. But the internal constituents are relocated — after air pressure equalizes, there’s an excess of mass at the condensing end, where the condensate is.
The c.m. of the initial system is (from symmetry) at the center of the cylinder. After condensation & pressure equalization, the c.m. of the air remains at the center of the cylinder, but there’s also condensate at one end. The only way to keep the entire system’s c.m. in the same spot, is for the c.m. of the air to be on the opposite side of the [constant] system’s c.m. from the condensate.
Re: HaroldW (Oct 28 16:46),
The center of mass of the whole is at (m1″x1 +m2*x2)/(m1+m2).
The solid cylinder’s center of mass cannot change, thus because of momentum conservation the gas must move so as to keep its center of mass constant. I see it as a pressure wave hitting the opposite side of the container.
A spin could start to pick up the kinetic energies released by condensation, the container in the opposite direction to the gas so as to conserve angular momentum.
The question in my mind is which way the cylinder begins to spin in the Northern Hemisphere…
Okay not really, but I AM waiting for Lucia’s next post on this topic!
Anna v:
“The solid cylinder’s center of mass cannot change, …”
Why do you assume that?
The only thing that must remain stationary is the centre of mass. If the centre of mass of its contents change the cylinder’s must change; and the centre of mass of the contents do change in this case.
You only need to think about the intitial and the final states. The particular dynamics, the detailed description of the path from one to the other, are not relevent.
If the centre of mass of the water part moves the rest must move to counter act this.
Alex
Re: Alexander Harvey (Oct 29 03:45),
I will assume there is a semantics misunderstanding.
The center of mass of a solid cannot change unless there are deformations, something that is not under discussion here.
Anna
What if it moves? Translation or rotation not around the center of mass will change the center of mass of a solid. Otherwise, you’re left with deformations (or maybe weird internal nuclear reactions? 🙂 )
Re: lucia (Oct 30 06:54),
We were discussing a force free environment, free of external forces. The center of mass of a solid is a fixed point in the geometry of the solid.
example
It will move if there are external forces.
If there are no external forces acting, momentum conservation says that the (cylinder + gas) center of mass cannot move. Since the cylinder center of mass will not move, cylinder is a solid, the gas center of mass cannot move.If
If m1 is the mass of the cylinder and x1 its center of mass, and m2 is the mass of the gas and x2 its center of mass, the combined center of mass is
x= (m1*x1 +m2*x2)/(m1+m2).
If part of the gas condenses, the center of mass of the condensate and the rest of the gas is constant, which means that the rest of the gas has to recoil to keep the constancy.
I hear the question: x is the constant not x1 or x2 and in a coordinate system outside the cylinder x1 could compensate for x2.
It comes down to this:
Take a closed space capsule in a force free environment. Can you propel it by firing shots at one end (internally)?
p.s to my previous.
I had been thinking in a coordinate system attached to the cylinder. In this case the gas will be vibrating around the center of mass (of condensate+left over gas), until the energies turn to heat.
Came to me that if the coordinate system is outside the cylinder there will be vibration, back and forth from x, the total center of mass, which will be diminishing as the internal energies turn to heat.
Anna V (#57572)
“Propel” may be too strong a word — the cylinder doesn’t end up moving in one direction indefinitely. But consider the following case — a man is at the center of a hollow cylinder in zero-g. As the man walks (or pulls himself) towards one end, the cylinder moves in the other direction in reaction. If the man’s mass >> the cylinder’s, then the man stays in the same inertial position (more or less), while the cylinder moves from under him. The max motion of the cylinder is half its length, at which point the man bumps into one of the ends and stops moving (and the cylinder stops too). As one reduces the ratio of man’s mass to cylinder’s, the cylinder’s motion also decreases; and if that ratio << 1, then one can hardly notice the cylinder's motion.
Similarly with the condensation example. The contents of the cylinder are redistributed so there's an imbalance at one end; this is analogous to the man moving to that end. The cylinder moves in reaction — the amount depends on the ratio of the mass being redistributed to the total mass of the system.
As I said in my p.s. the true picture comes from the extra degrees of freedom about the center of mass, rotations and vibrations. I had mentioned rotation in a post above, but had forgotten about vibrations.
Any internal motion in the type of problem we are discussing, a solid shell in a force free environment with variable position contents, will generate about the overall center of mass rotations to conserve angular momentum and/or vibrations . Vibrations are 0 phase rotations and conserve momentum.
Your example with the man sees only half the cycle. When he hits the end the momentum moves the end + man back, the cyinder does not stop for an outside coordinate system : vibration in slow motion.
i don’t mean vibrations are zero phase rotations , do I? That seems wrong since they are an extra degree of freedom.
What an interesting thought experiment! I have an off-the-cuff response that undoubtedly considers only some part of the relevant variables, and so I’ll phrase it as a question rather than an assertion. Is it possible that the moment of condensation will be marked by a very slight reduction in the weight recorded on the scale?
For simplicity, my arbitrary assumption is to ignore condensation occurring anywhere except within the air phase (i.e., not on the walls). I would suggest that when all the water is in a vapor phase, the water molecules can be considered to be in a state of “terminal velocity” downward in that there is no net acceleration over time (the downward trend driven by gravity is balanced by collisions that keep the molecules suspended as a gas). In that sense, the scale is registering the full effect of gravity acting on those water molecules, so that pressure on the bottom exceeds that on the top of the container.
At condensation, some of the molecules combine to form droplets. This reduces the number of molecules in a “terminal velocity” capacity, and substitutes droplets that will fall. However, because of scaling, their surface to mass ratio declines vis-a-vis individual molecules. As a result, the air molecules below them are no longer capable of keeping them in suspension, and they start to accelerate. At this point, therefore, gravity is divided between the downward pressure exerted on the air and the acceleration it is causing. The downward pressure will therefore be less than the pressure the gas exerted when it had its full complement of water vapor.
I’m sure there are other factors, but what I suggest may be one part of the process.
I notice that Alexander Harvey (comment 56884) made the same point earlier as in my comment 57818, and managed to do it more succinctly. My only contribution is to relate the reduced ability of droplets to stay suspended to their increased mass/surface ratio compared with water molecules in the vapor phase.
Alexander (56885) also points out that condensation occurring only on the bottom would have the same effect. This raises an interesting point. If condensation is initiated by an externally imposed cooling, how does the location of the cooling source affect the change in weight? If the container is cooled from above, so that more condensation occurs on the upper lid, won’t the container temporarily gain rather than lose weight, at least initially?