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Q & A: How do magnets work?
How do magnets work?
- Marsha Borski (age 9)
Marsha,
That's an excellent question.
Try the following experiment with a magnet, some needles (the plain kind used to hold stuff together as you sew it), and a screwdriver. It will help you understand the explanation.
Touch a needle with a screwdriver and nothing unusual should happen (unless the screwdriver is already magnetized). Now, take some of the needles and hold them against the end of a magnet for a while (the stronger the better). After doing this, you will have turned each needle into a little magnet, and when you touch them with a screwdriver they will stick to it.
Some kinds of metals (like steel that the needles are made of) are made up of billions and billions of individual atoms that each have the properties of a microscopic magnet. The atoms in steel naturally tend to get together in tiny little groups called domains, and within each domain the atoms tend to point in the same direction, which makes the domains behave like a tiny little bar magnets just like the kind you have probably played with at school. The needle of a compass is also a bar magnet, and we know what this does: it points north because it likes to line itself up with the magnetic field of the earth.
When you make a piece of steel (like a needle), all of the tiny domain-magnets inside tend to get stuck pointing in different directions, which means that they more or less cancel each-other out, so to begin with the needle does not behave much like a magnet at all.
However, if you bring the needle close to another magnet of some kind, (like the ones on your fridge, or better yet some stronger ones you can find in your classroom) something interesting can happen: Because of the other magnet you are holding it near, the little domain magnets in your needle will tend to line up so that they are pointing along the same direction. (You can visualize this by holding two bar magnets near each other and noticing how they like to line up a certain way). The domains in the needle will do the same thing, and after you take the needle away from the other magnet, the domains in the needle will tend to stay pointing this way (they sort of get stuck pointing in the same direction). Now, since you have lots of these little domain magnets in the needle all pointing in the same direction, the needle will itself have become a small magnet. (This is exactly how bar magnets are made).
Magnets attract other magnets (as you can see), and also attract some kinds of metal like many kinds of steel. If you touch your magnetic needle with a steel screwdriver it will stick. However, stainless steel is not a very good magnetic material, so if you touch your magnetic needle with something made of stainless steel it will probably not stick (try it).
Follow-Up #1: why not more magnets?
I need a more in-depth answer to Marsha's question. I know that natural magnets have magnetic fields becuase of the electrons moving aroung inside thier atoms(while electric fields are present so are magnetic fields), but why then doenst matter like water and tree bark have natural magnetism if they also have electrons inside thier atons?
- -Anonymous
Excellent question.
In most materials almost all the electrons form pairs, with the magnetism from the two paired electrons exactly canceling. The result ultimately is due to something called the Pauli exclusion principle, which says that no more than one electron can exist in any particular quantum state. So if there's some nice low-energy state waveform for an electron to sit in in some molecule, it tends to get two electrons for the two possible quantum states with that form: one spin up, the other spin down.
This very fact is the origin of most of the properties of chemical bonding. Remember how a typical covalent bond consists of two shared electrons? The reason is the exclusion principle plus spin. In ionic bonds, typically one electron will mostly move from one atom with an odd number of electrons over to another atom with an odd number, leaving each with an even number.
So the tendency of most materials not to be magnetic is closely tied to the same quantum mechanical facts which account for the prevalence of pair-bonds and ionic bonds in chemistry.
Follow-Up #2: sad truth
why are most materials non-magnetic? (no quantum physics please!)
- ivan brown (age 13)
Newport, Wales, Britain
Ivan- I sympathize with your question, but the sad fact is that there's absolutely nothing honest one can say about the origin of magnetism in materials without using a quantum mechanical basis. Sorry.
Follow-Up #3: pauli exclusion principle
what is pauli exclusion principle?
- starlai (age 18)
zamboanga, philippines
It states that for particles called 'fermions', including electrons, protons, neutrons, etc., no two particles can be in exactly the same quantum state. There must be something different about them, either that they're in different places or have different spins or something. There's another class of particles, called 'bosons', including photons, for which this principle does not apply.
Follow-Up #4: non-magnetic trees
i cant exactly get the answer of why magnets has no effect in trees example where infact it has also an electron inside thier atoms... please expain it in a simle way for me thanks!
- lai (age 18)
The basic reason is that almost every electron is paired up with one other one that is in the same state except that it is spinning the opposite way. (see the Pauli exclusion answer above) That means that the magnetism of the two just cancels.
In leaves, however, as photosynthesis occurs some electrons get shuttled around individually as part of the light-driven chemical process. So their magnetism isn't automatically canceled, and magnetic resonance signals from them can be measured.
Follow-Up #5: domain interactions
Do domains work agnaist each other
I think what you're asking is whether the forces between magnetic domains push them toward pointing opposite ways. That actually depends on whether the domains are side-by-side or end-to-end. If you picture the domains as little arrows pointing up or down, side-by-side ones tend to point opposite directions but end-to-end ones tend to line up the same way.
Think of it this way. If you have two bar magnets, each has a North pole and a South pole.
If the two magnets have their two N poles pointing up then when they are side by side they will repel each other but when they are head-to-tail they will attract. In the other case when one has a N pole pointing up and the other has a S pole pointing up then they will attract each other when they are side by side but repel when they are head-to-tail.
Lee H
Follow-Up #6: permanent magnets
So what makes a magnet magnetic? You told me how iron was magnetic and i understand that but what is the difference between a magnet and something that is magnetic?
- Anonymous (age 13)
Most strongly magnetic materials, like iron, have magnetic domains which act like small magnets themselves. However, the directions of those magnets are all scrambled up, so the total magnetism cancels out. If you put them in a magnetic field, the domains tend to line up with the field, so you get something which really has net magnetism and can feel a force from the field.
To make a permanent magnet, you need some way to keep those domains lined up even after the applied field is removed. In certain special materials, it's hard for the domains to change which way they point. Usually those materials have some mixture of non-magnetic atoms, which tend to catch and trap the domain walls, the regions where the magnetism changes from one direction to another. Trapping the walls keeps them from moving, and keeps the magnetic direction from changing.
If you heat up a permanent magnet, the walls tend to come loose and the net magnetism will go away.
Follow-Up #7: magnetic vs. electric fields
So it sounds like you're saying that what makes something magnetic is the presence of these "single" particles or particles that are not in a pair (if so they would cancel each others magnetism)and there are regions of these single particles that are called domains? I have two question. How can one particle be magnetic? and Is the source of magnetism the attraction/repulsion relation between positive and negative charges? And if so, what makes magnetic fields different from electric fields? (detailed please) Thanks
Actually, for something to be very strongly magnetic it takes more than just some unpaired particles. There has to also be a tendency for the magnetism of different particles to line up together, and that tendency is present in only some materials.
One of the sources of magnetism is electrical current. That's why a current through a copper coil, for example, creates a magnetic field. Now (and this isn't quite being honest) imagine a charged particle as a little ball of charged stuff. If it's spinning, then that's like having currents go around in a loop, and you should get fields like those when current goes around in a coil. Elementary particles like electrons do indeed have "spin", measurable angular momentum. However, if you took this story too literally you'd calculate a magnetic moment of an electron which is only half the actual value. On a small scale, that spin is really something quantum mechanical, not something you can picture as charges running around in circles.
Magnetism is related to the electrical fields which pull on charged particles. However, it's connected not just with the positions of the particles but also their velocities. If you look at a charged particle (without spin) from a frame where you say it's standing still, it will only have electrical fields. If you look from another point of view and say it's moving, it will also have magnetic fields. Of course the net physical effects on its neighbors will be the same in either case. We obviously cannot fully introduce this area here, but if you want to see a beautiful book on the topic, I recommend E. Purcell's "Electricity and Magnetism"..
To add one more twist to the tale, Maxwell's equations are consistent with terms involving magnetic monopoles. None have ever been seen. For more information check out:
LeeH
And for another twist, if those magnetic monopoles exist, then an observer who sees one moving will say it produces electric as well as magnetic fields. mike w
Follow-Up #8: making magnets and demagnetizing them
So magnets are made by putting a magnetic object under a magnetic field, this makes all the domains spin in the same direction, which is what defines a magnet. Also electricity running through a wire creates a magnetic field because it has a flow of electrons in one direction. so does that mean that if you try creating a magnet off of an electrical current, then the domains of the magnet will spin in the direction of the current, and if so, does that mean that making a magnet using AC current would be very ineffective?
- Kassem
The direction the domains will point is set by the current in the electromagnet, although it is not the same as the current direction. If the magnet is in a coil, running the current one way around the coil will make the magnetism point one way; reversing the current makes it point the other.
So you're absolutely right that an ac current is no way to magnetize something. In fact, when you want to demagnetize something (e.g. erase a magnetic recording) the standard way is by applying a big ac field and gradually decreasing the magnitude. That pretty much scrambles up the domains. It's often called "de-Gaussing" since one standard unit of magnetic strength is a Gauss.
Follow-Up #9: Magnets do not bend light, however...
I was wondering if Magnets have any effect on light. I know that light is really electromagnetic waves, but I was wondering if these waves could be attracted/repelled against a magnet. Thanks!
- Joe Shock-Ra
In a vacuum there is no effect of a magnet on light. It doesn't bend it or anything.
There is a subtle effect, however, when light passes through a dielectric medium like glass or sugar water solution. It turns out that in the presence of a magnetic field the light beam is not bent but the plane of polarization can be rotated a bit. This is called Faraday rotation. You can find out more about it at
Follow-Up #10: first magnets
First, thank you for answering for our questions. My question is, you say that to obtain a magnet we need to have a magnet. Then how was the first magnet made, when there were no any other magnets? How to make a magnet if i have only a piece of iron and nothing else??? Thank you again.
- Mehdi
Really nice questions!
Some stones are magnetic as found. They are magnetized because they cooled in the Earth's magnetic field. The Earth is magnetic for complicated reasons which I don't really understand, but have something to do with currents stirred up as heat flows from the hot core to the cooler surface, creating convection patterns.
You don't need a permanent magnet to magnetize a piece of iron. It's usually done using an electromagnet, so what you really need is a current source to drive the electromagnet. Although many current sources are ultimately based on generators which themselves use magnets, there are other current sources which do not need magnets. For example batteries can be made using chemicals and no magnets.
Follow-Up #11: magnetic water?
can a magnet work when there is ice between the two magnets and if there is water between them ( the icce and water will be between them at different times ) if so will the magnetic force be different
- ted
Water (whether liquid or frozen) has so little magnetic susceptibility that you will not notice its effects unless you\'re deliberately making a sensitive experiment.
Follow-Up #12: magnetic forces
What causes magnetics to attract to each other? And why do they stick for so long?
- Tiffany (age 13)
Richmond hill, GA, US
Those are great questions but I can't think of any way to answer the first one without giving a little course. As for the second one, it's normal for the forces which hold things together to last forever.
Think of the gravity that holds you and the Earth stuck together. As long as you and the Earth are around, that gravity will be there. The same rule applies to magnets. As long as they stay magnets, the force between them will be there. There is one difference, however. You can think of a magnet as made up of little magnets, called domains. Each one has a North and a South pole. Over a very long time, those can jiggle around so the little magnets don't line up with each other. Then they won't add up to make a big magnet. Gravity doesn't have poles, the gravity from every part just adds together. So even as parts of the Earth shift around, the gravity won't go away.
Follow-Up #13: compass physics
The South Pole is magnetized, correct? How come a compass never points south. If we switched the charge in our compasses would they point south? Also, how come compasses become more inaccurate as you near the north pole? That seems counter-intuitive. I'd think they'd become more accurate since the force is stronger.
- John (age 33)
It's just a convention to label the north-pointing end of the magnet in a compass by sharpening it or painting it with an arrow head. You could do that to the other end if you wished.
The inaccuracy near the north pole arises because the magnetic north pole isn't actually at the same place as the rotational north pole. The difference becomes important in that vicinity. If you stood in between them, your compass would point south.
Follow-Up #14: Do two bar magnets attract or repel each other?
If two magnets are put end to end with the current patterns lined up in the same direction, the two poles next to each other would do what?
- Drew (age 17)
____________________ __________________
| | | | Currents aligned
| N A S | | N B S | Attraction
|___________________| |__________________|
| | | | Currents anti-aligned
| N C S | | S D N | Repulsion
Just to be clear, in the picture above A and B attract, C and D repel. You might also wonder about what happens between A and C and between B and D, in the positions shown. A and C repel. B and D attract. If you imagine each magnet as a tube of current going around the axis, the ones that attract are the ones where the nearby parts of the current go the same way in each tube. Mike W.
Follow-Up #15: magnets attract or repel
How do magnets stick together only sometimes ?
- Isabel (age 10)
Hudson,OH
Hello Isabel-
Bar magnets only sometimes stick together and sometimes repel because each has a north pole and a south pole. Opposite poles attract, and like poles repel. If you were to put two north poles together, they would repel, but a north and a south would stick together. The picture above shows an example.
Chris F. and Mike W.
Follow-Up #16: are magnets permanent?
In permanent magnets what sort of energy do they use to push or pull other magnets if any at all? I was also wondering does a permanent magnet ever lose its magnetism?
A permanent magnet will lose its magnetism fairly quickly when it's heated enough. A good permanent magnet stays that way for a very long time at room temperature. In the extremely long run, its little domains should lose their alignment, but that's not in our lifetime.
You also ask, if I understand right, what sort of energy is involved in the magnetic forces between magnets. There is energy stored in the magnetic field itself. The density of that energy is proportional to the square of the field strength. When magnets move near each other, that field energy generally changes.
Follow-Up #17: shielding a magnet with steel
I know a magnet can attract steel paper clips through paper, cardboard, plastics, cloth, etc. But can a magnet attract steel paper clips through a sheet of sheet? I have beeen confused as some sources say it will not. I know you can make a iron nail a temporary magnet by attracting it to a magnet at one end and have paper clips attracted to the nail. Shouldn't it be the case for the sheet of steel?
- SARAH (age 26)
I assume that was a typo, and you're asking about a sheet of steel. it's an interesting question.
Here's a nice physical way to picture what happens. Picture the magnetic field in the usual way, as a set of field lines coming out of one pole of the magnet and returning to the other. There's a tendency for those field lines to get sucked into pieces of iron or steel or other easily magnetized material. That's exactly the reason that the paperclips stick to the magnet. Some of the field lines get stuck in them.
Now what happens if you put a big sheet of steel in between the magnet and the paper clips? The field lines mostly get drawn into the sheet, spread out in the sheet, and return without going past the sheet to the paperclip. So the clip doesn't stick much if it's just behind the sheet.
This effect depends a lot on the shape of the sheet. A small piece of steel can pull field lines into it leaving more field just behind it than were there before. That's just what happens when you chain paperclips together hanging from a magnet. Each clip tends to steer field lines on to the next one. So the steel redirects the field, with a sheet redistributing it away from the clip just on the other side. It would tend to grab clips right near the edge of the sheet, where some of the field lines exit.
Follow-Up #18: magnetized nail
We made electromagnets in class with wire, a large nail and a 6 volt battery. The nails were not magnetic, but after we disconnected the batteries from the nails, the nails would pick up staples. How did the battery or the electricity magnetize the nail?
- Karen (age 10)
The nail started off already magnetized in small regions, called domains. At room temperature the little magnets of the electrons in the iron tend to line up with other, making those magnetic domains. However, the magnetism of the different domains points all different directions, so overall it cancels out.
When you put the nail in the electromagnet made with the coiled wire powered by the battery, it lines up a lot of those domains to point the same way. After you take the nail away, some of the domains stay stuck, so you're still left with more of the magnetism pointing one way than any other. That means the nail is a bit magnetized.
This is temporarily posted without the usual check, until Lee gets back from Paris.
Follow-Up #19: magnets and their fields
Two questions: 1!: Can you make a 'doughnut' magnet so as to have, say, all N charge on the outside and S on the inside, in the hole of the doughnut?? 2!: what exactly is the field itself? the magnetic field, i mean. The magnetic fields' COMPOSITION. The answer I usually get is something like: " well, the magnet makes the field" "...I am aware of that. But what exactly is the field MADE-OUT-OF?" "It is the magnetic force in the magnet interacting with things around it" "...*facepalm*" But you seem thorough. so please answer in that manner,I'll understand everything right up to the point where there needs to be calculus to understand. I'm still in Algebra 2. Thank You.
- Atticus (age 14)
Salt Lake City, Utah,
Your first question is easy.
Sure, you can make a donut-shaped magnet with the N poles on the outer part and the S poles near the donut hole. It's not a standard form, but it's certainly possible. (Those poles aren't really "charges" though.)
Your second question is extremely hard to answer. I think for now perhaps the best non-answer is this. Whatever the universe consists of, at some point our description has to get down to the most basic ingredients, or, if every level of ingredient should be thought of as made of something deeper, at some point we'll just get to the deepest level we know about. At that point, all you can do is describe the properties, not say what things are made of. Electromagnetic fields are pretty close to that point. We could give maybe a step or so deeper description, but it would be in terms of quantum fields. Those are even more abstract mathematical entities than magnetic fields.
Follow-Up #20: magnetism from moving charges
Well, i know that moving or spinning charged particles can cause a magnetic field. But how does it do that? And why don't stationary charged particles create magnetic fields, even though they create electrostatic forces? And what causes those forces in the first place? For example, gravity is caused by mass curving or warping space time or something of the sort, what causes magnetic and electrostatic forces? I know very basic quantum mechanics. Thank you!
- Rahul (age 15)
Rahul- You're asking just the right questions to start in on Purcell's Electricity and Magnetism book, from the Berkeley series. The outline of Purcell's presentation looks something like this.
1. Start with the existence of simple electrostatic forces following Coulomb's law.
2. Assume that special relativity gives the right rules for how things look in different reference frames.
3. Look at a moving charge near a neutral, current-carrying wire. There's no electrical force.
4. But if you now look in the rest frame of the charge, the wire is no longer neutral, thanks to the different Lorentz contractions of the differently-moving plus and minus charges.
5. So there's an electrical force in this frame.
6. So there must have been a velocity-dependent force back in the frame where the wire was neutral.
7. We call that velocity-dependent force magnetism.
Ok, that handles the part about why magnetism, given electricity. The part about why electricity in the first place is unfortunately over my head. Here's a few words to get you started. Purcell derives the magnetism from electricity from assuming a symmetry, special relativity, a rule about how the world obeys the same laws of physics even as you represent its contents in different ways. There are some other subtle symmetries (gauge symmetries) in relativistic quantum field theory, and I've heard that they require the electrostatic force.
Follow-Up #21: how do magnets act at a distance
What a neat site. These kids are so smart! As a follow up to Atticus' question, please explain how magnets achieve action at a distance. Also, to Atticus and the other brilliant kids who ask the right questions, give them more questions that they can access and imagine! One that I have often pondered (please answer): What do you see when looking through an infinitely powerful microscope? Or Telescope? Something? Nothing? What is a particle made of (the smallest one)? It's all just organised energy right? Then the question is (of course) what is energy? Is there an answer to this question? Or is it unanswerable?
- Steve (age 28)
Victor, MT
We were just discussing doing more to encourage our readers to do experiments. Your suggestion about raising follow-up questions for them is along the same lines. We'll do our best.
On "action-at-a-distance", we don't really think of it that way. We think of the magnetic field as a real thing, all spread out in space, acting where it is. But how did that field get all spread-out? Not by sudden action-at-a-distance. It had to work its way over from the source, by an electromagnetic wave traveling at the speed of light. That's not much of a coincidence, since light itself is an electromagnetic wave.
We don't really know what anything is made of at the deepest level. I guess the two basic approaches are that everything is some sort of differential equation (a little bit like Maxwell's equations for electromagnetism) or that everything is some sort of cellular automaton, kind of like a bunch of little digital bits. Right now, the deepest things we know are all differential equations, but some people suspect that may change.
Saying "It's all just organized energy" doesn't really convey a definite enough idea for us to say it's true or false.
Follow-Up #22: Adding bar magnets together
If you place two equally strong magnets together so they attract do the act as a single magnet with one magnetic field? Will the resulting magnet be twice as strong as one of the original magnets?
- Mark Basham (age 50)
If you have two thin bar magnets and put them together in parallel the resulting strength is about double. If you keep adding more and more of them eventually the resulting sum is not the sum of the number of magnets due to the over all geometry of the combination. It involves some integrals.
If you add them lengthwise you don't get a factor of two in the end-on direction but you would get a factor of two if you placed them face down on an iron surface.
Far away, you do get a factor of two in the fields, even though close by it's more complicated. /mw
Follow-Up #23: Does positive attract positive?
I read all your post to look for my answer, but still couldn't find an answer. People say Positive energy attracts Positive. But how is this possible if they repel each other? Or is that a bunch of bull? My mom bought 2 expensive bar magnets to stand on while exercising. She said it will help her heal w/positive energy?? Is she nuts or is this true? Thanks, Moe
- Moe (age 16)
Positive energy does attract positive energy by ordinary gravity. However positive electric charge repels positive electric charge. This old physics has nothing to do with various new-age ideas.
Magnets are still another matter. No magnetic charges ("monopoles") have ever been found. However, like poles of magnets (e.g. North and North) repel.
Your mom may not be nuts because many medical conditions are known to be helped by something called the placebo effect. People who believe that something (e.g. an alleged medicine) will help them tend to do better, sometimes in real objective measured traits, not always just in how they feel. () Magnets are truly harmless. Maybe having to pay a lot for them made her feel like they're doing more good, and maybe that feeling makes them do more good. I have a lot of trouble motivating myself to exercise enough. Maybe it's time to invest in some magnets.
Follow-Up #24: Storing energy in magnets
Is there a way to create energy from magnets? For example, when putting the same end of a magnet together, the magnets push away from each other. Where does this energy come from? Would it be possible to create giant magnets or contained magnetic fields in order to create energy and power mechanisms? Understanding that all energy is simply transferred but not newly created, where is this energy coming from?
As you point out, you can't create energy with magnets, but you can store energy. When you push two same-sign poles together it requires energy. This energy can then be released and do useful work when you let go of the magnets.
Follow-Up #25: natural magnetism
Can non-magnetized iron or steel have a weak inherent magnetic field? When reading William Gilbert "on the loadstone" I performed an experiment described at the end of the first book using a non-magnetizad iron or steel to attract each other, by putting a nail not previously magnetized floating on water or suspended by a thread , then another non magnetized iron would act as a weak magnet clearly atractiing each other. Does this mean these metals were somehow weakly magnetized in he mining and manufacturing process?or can there be a weak magnetic field in non purposefully magnetized metals?
- Juan Preciado-Riestra (age 42)
Waupaca WI
Yes, ferromagnetic materials typically do have a weak net magnetization. The Earth is a weak magnet (that's why compasses work) so the ferromagnetic material will magnetize a little as it cools into the magnetic state.
Follow-Up #26: what is a magnetic field?
The words magnetism, field and energy seem even more abstract and elusive to me now that I have read this series of Q&A. So, I pose a few questions.One of the answers mentioned that a magnetic field represents stored energy. (1)If this is true, then is there not a way to remove that energy for real use? (2)For example, if the magnet is heated to the loss of the magnetic field, where does that energy "go"?(3)Concerning a bar magnet, exactly what IS the magnetic field "made of", or how can it be physically described? (4)Is the field in motion about the magnet? (5)Or is it static with most domains spinning in alignment without any other movement? Since I see the field described with terms similar to DC electrical circuits, I get this idea that the flux is "flowing" like an electrical current. But I see no reason for the field to flow. (7)So, how did it get there if it didn't "flow" there.And (8)how is it possible that the field extends so far from a magnet, like an inch from a half-inch neodymium? (9)Is there a theoretical maximum for a given lump of magnet?Thanks for any thoughts you can share!
- Dennis (age 45)
1) You could always remove the energy for use once, just by letting two oppositely polarized magnets attract and do work as they approach each other. That's just like how you can get work out of letting some object fall in a gravitational field or from the attraction of opposite electric charges. Once it's gone,though, it's gone. There are no perpetual motion machines.
2) Just that fact that you have to heat the magnet tells you you're mostly adding energy. That's because the magnetized region has negative interaction energy between its spins. It also has a little positive field energy. As the material demagnetizes by individual spins flipping over, each emits an electromagnetic wave. A little energy can leave the material that way, if it's not absorbed and turned into general thermal shaking of parts.
3) See this: .
4) No.
5) The spins within each domain are mostly aligned with each other. The field isn't going anywhere.
6) ??
7) When the field was first set up, energy did have to flow there. The flow density is described by the Poynting vector, proportional to E X B. So during that process there was also an electrical field. You wouldn't see the reason for that if you came along after the process was over and everything was static.
8) How does the Sun's gravitational field extend out to us? These fields propagate very fast, at the speed of light. Your question seems to assume that things are made of little parts that interact only when they touch each other. Michael Faraday struggled with pictures like that before working his way over to something like our modern idea of fields.
9) I suppose there's a theoretical maximum at which the field energy is enough to collapse the object into a black hole. Maybe there's some other theoretical maximum that's not so huge, but I'm not aware of it. For the types of magnets we're familiar with on Earth, the practical maximum would have a few spins, and thus a few Bohr magnetons, for each atom. That magnetization corresponds to about 10,000 Gauss, which is around what the most intense permanent magnets reach. Big superconducting electromagnets can reach over 100,000 Gauss. Higher values can be reached briefly. For very different types of matter (neutron stars) fields can reach over 1014 Gauss. () At that point the field energy density is pretty significant, about a million times higher than the rest mass energy density of water. That's still a lot less than the energy density of the magnetic neutron star itself.
Follow-Up #27: magnetic domains
In my chemistry book it is stated: "Iron, nickel, and cobalt are ferromagnetic; the unpaired d electrons in large numbers of atoms line up with parallel spins in regions called domains." What causes the formation of these domains? Is this simply a property of some materials in that they tend to form domains? Also, I don't quite understand how electrons can "line up" with parallel spins without violating the Heisenberg Uncertainty Principle. If "line up" is not meant literally, can you please explain how the electrons actually form domains?Thanks.
- Kevin Jin (age 15)
First, on the "line up" issue, you raise a good point. Quantum mechanics does not allow the spin to be fully lined up in any direction. If the average spin is in the z direction, there will be an uncertain spread of possible spins in the xy plane. Still, that allows the average values of different spins to line up with each other, despite the leftover spread.
The domains form because the electron waves can lower their energy when nearby spins line up. That's not a general property of all materials but rather depends on the quantum states of the electrons in the atoms, on the types of neighbors, and on the distances between the neighbors. In iron, for example, different densities of different amorphous alloy forms can either favor alignment or anti-alignment.
So if the material happens to be one in which alignment is favored, why doesn't it form one big domain? As the domains get bigger, the plain old magnetic interaction, like you see when you play with magnets, makes adjacent regions want to anti-align if they are sitting lateral to the field direction. Ones displaced along the field direction, in contrast, lower their magnetic field energy by lining up the same way. Thus ordinary magnetic material is a patchwork of domains pointing different directions. A thin needle, however, can form a single domain with the field along the needle direction.
Follow-Up #28: magnetic direction
What do you mean by "line up" exactly? From my understanding an atom is made up of a positive nucleus orbited by negative electrons, forming a spherical shape when viewing all possible electron locations, how can this sphere point in any direction when the net charge is zero?
- Conner (age 16)
Quite aside from the shape of the electron cloud, the electron itself has an intrinsic property called "spin". That gives it some angular momentum and magnetic moment, which point in some direction. So even a nice simple hydrogen atom with a spherical electron cloud already has magnetism with a direction.
Follow-Up #29: electron spin in magnets
Mr Feynman states that the 'attraction' and 'repulsion' of magnets is caused because of electrons spinning in the same direcrion which coincidentally amplifies the force, this is why you can feel it at a distance... Now, the spin is in one direction in one pole and opposite direction on the other pole; however if the electrons of one north pole face those of another north pole they are both spinning in the same direction opposite to the other so there will be 'repulsion', for this example lets say that they both spin counterclockwise... If you flip one magnet 180 degrees on its side and you leave the other magnet alone, the spin on one north pole will be clockwise and the other counterclockwise.... Why do they not attract?
- Freddy (age 24)
The electrons in the north and south poles of a bar magnet are spinning the same way, not opposite ways.
Follow-Up #30: magnet repel or attract?
Hi mike, thanks for the reply but unfortunatelt because of my bad explanation i dont think you grasped my question. Let me rephrase, and for this example lets forget about the south pole since the question is in regard to the north poles of 2 separate magnets facing eachother; below is a crude diagram[S|N] [N|S] (birds eye view)The above diagram shows the magnets' north pole facing from a birds eye view... Here we can assume that the spin of their respective north poles is causing the repulsion, however if one of the two magnets is flipped 180 degrees on its side (see crude diagram #2 below) then by virtue of the geometry of spin (180 degrees opposite to what it was), they should attract. Why do they not? Crude diagram #2:[S|N] [N|S] (initial position)[S|N] [N|S] (position after one of the two magnets has been flipped 180 degrees) - notice that both north poles are still facing however the spin of the electrons of one of the two magnets is 180 degrees opposite to the "initial position"If we are to believe that it is the spin of the electrons that is causing "repulsion" then why, after the spin of the electrons of one magnet were "flipped opposite" to that of the initial position do they still not attract?
Hi Fred- Unfortunately pictures don't come through in our question sysem, which has some filters designed to stop various hacks. The internet has turned into a dangerous place, and that interferes with ordinarye life.
I'll try to reconstruct what I think you're asking. Spin directions will represented by little arrowheads.
N<<<<<<<<< S S>>>>>>>>>>N. These two repel.
Flip the second around the middle of its long direction and you get:
N<<<<<<<<< S N<<<<<<<<< S These two attract.
But maybe you were asking about
N<< S S>>N
These two repel.
Flip the second around the middle of its long direction and nothing changes, they still repel. Perhaps that's the picture you were describing.
Flip it around the middle of its skinny direction and you get:
N<< S N<< S
N<< S N<< S
These now attract.
I suspect that you took the spins from one picture and the N,S labels from the other.


text:
\section{A lower bound for the first eigenvalue of the drift Laplacian}

Recall that  $(M, g, f)$, a triple consisting of a manifold $M$, a Riemannian metric $g$ and a smooth function $f$,   is called a  gradient Ricci soliton if the Ricci curvature and the Hessian of $f$ satisfy:
\begin{equation}\label{sgs}
{\rm Rc}_{ij}+f_{ij}=a g_{ij}.
\end{equation}
It is called shrinking, steady, or expanding soliton if $a>0$, $a=0$ or $a<0$ respectively.
In this paper we apply the modulus of continuity estimates developed in \cites{AC1,AC2, AC3} to give an eigenvalue estimate on gradient  solitons for the operator $\Delta_f\doteqdot \Delta -\langle \nabla(\cdot), \nabla f \rangle$ on strictly convex $\Omega\subset M$ with diameter $D$ and smooth boundary. In fact the result works for manifolds with lower bound on the  so-called Bakry-Emery Ricci tensor, namely ${\rm Rc}_{ij}+f_{ij}\ge a g_{ij}$ for some $a\in \mathbb{R}$. An earlier result of this kind was obtained in \cite{FS} for shrinking solitons.

In this  section, we extend a comparison theorem of \cite{AC3} on the modulus of continuity to manifolds with lower bound on the co-called Bakry-Emery Ricci tensor. The eigenvalue comparison result for manifolds with lower bound on the Bakry-Emery tensor generalizes the earlier lower estimates of Payne-Weinberger\cite{PW}, Li-Yau\cite{LY} and Zhong-Yang\cite{ZY}. A consequence of this is a lower diameter estimate for nontrivial gradient shrinking solitons, which improves \cite{FS} with a different approach and a rather short argument.  We remark here that the eigenvalue estimate we obtain is sharp for $(M,g,f)$ satisfying the Bakry-Emery-Ricci lower bound ${\rm Rc}_{ij}+f_{ij}\geq ag_{ij}$, but presumably is not so for Ricci solitons where the Bakry-Emery-Ricci tensor is constant, and so we expect that our diameter bound is also not sharp.  We discuss the sharpness of the eigenvalue inequality in section \ref{sec:sharp}.

Before we state the result, we define a corresponding $1$-dimensional eigenvalue problem. On $[-\frac{D}{2}, \frac{D}{2}]$ we consider the functionals
$$\mathcal{F}(\psi)=\int_{\frac D 2}^{\frac D 2} \mathrm{e}^{-\frac{a}{2} s^2} (\psi')^2\, ds, \quad \text{ and } \quad    \mathcal{R}(\psi)=\frac{\mathcal{F}(\psi)}{\int \mathrm{e}^{-\frac{a}{2}s^2}\psi^2ds}, $$  namely the Dirichlet energy with weight $\mathrm{e}^{-\frac{a}{2} s^2}$ and its Rayleigh quotient. The  associated elliptic operator is  $\mathcal{L}_{a}=\frac{d^2}{ds^2}-a\, s \frac{d}{ds}$. Let $\bar{\lambda}_{a,D}$ be the first non-zero Neumann eigenvalue of $\mathcal{L}_{a}$, which is the minimum of $\mathcal{R}$ among $W^{1, 2}$-functions with zero average.

\begin{theorem}\label{sgs-thm1} Let $\Omega$ be a compact manifold $M$,  or a bounded strictly convex domain inside a complete manifold $M$, satisfying that ${\rm Rc}_{ij}+f_{ij}\ge a\, g_{ij}$. Assume that $D$ is the diameter of $\Omega$. Then the first non-zero Neumann eigenvalue $\widetilde{\lambda}_1$ of the operator $\Delta_f$ is at least $\bar{\lambda}_{a,D}$.
\end{theorem}
\begin{proof} First we extend Theorem 2.1 of \cite{AC3} to this setting.  Recall that $\omega$ is a modulus of continuity for a function $f$ on $M$ if for all $x$ and $y$ in $M$, $|f(y)-f(x)|\leq 2\omega\left(\frac{d_M(x,y)}{2}\right)$.

\begin{proposition}\label{ex-AC-com}Let $v(x,t)$ be  a solution to
 \begin{equation}\label{heat-drift}
\frac{\partial v}{\partial t}=\Delta v-2\langle X, \nabla v\rangle
\end{equation}
  with $2X=\nabla f$. Assume also $v(x,t)$ satisfies the Neumann boundary condition. Suppose that $v(x, 0)$ has a modulus of continuity $\varphi_0(s): [0, \frac{D}{2}]\to \mathbb{R}$ with $\varphi_0(0)=0$ and $\varphi_0'>0$ on $[0, \frac{D}{2}]$. Assume further that there exists a
function $\varphi(s, t):[0, \frac{D}{2}]\times \mathbb{R}_+\to \mathbb{R}$ such that
 \begin{itemize}
\item[(i)] $\varphi(s,0) = \varphi_0(s)$ on $ [0,D/2]$;
\item[(ii)] $\frac{\partial \varphi}{\partial t}\ge \varphi'' -a\, s\, \varphi'$ on
$[0, D/2]\times \mathbb{R}_+$;
\item[(iii)] $\varphi'(s,t)>0$ on $[0, \frac{D}{2}]$;
\item
[(iv)] $\varphi(0, t) \ge 0$ for each $t\ge 0$.
\end{itemize}
Then $\varphi(s, t)$ is a modulus of the continuity of $v(x, t)$ for $t>0$.
 \end{proposition}

 The proof to the proposition is a modification of the argument to Theorem 2.1 of \cite{AC3}. Precisely, consider $$
 \mathcal{O}_\epsilon(x, y, t)\doteqdot v(y, t)-v(x, t)-2\varphi\left(\frac{r(x, y)}{2}, t\right)-\epsilon \mathrm{e}^t
  $$
  and it suffices to prove that the maximum of $\mathcal{O}_\epsilon(\cdot, \cdot, t)$ is non-increasing in $t$. The strictly convexity, the Neumann boundary condition satisfied by $v(x, t)$, and the positivity of $\varphi'$ rule out the possibility that the maximum can be attained at $(x_0, y_0)\in \partial (\Omega\times \Omega)$. For the interior pair $(x_0, y_0)$ where the maximum of  $\mathcal{O}_\epsilon(\cdot, \cdot, t)>0$ is attained, pick a frame $\{e_i\}$ as before at $x_0$ and parallel translate it along a minimizing geodesic $\gamma(s):[0, d]\to M$ joining $x_0$ with $y_0$. Still denote it by $\{e_i\}$. Let $\{E_i\}$ be the frame at $(x_0, y_0)$ (in $T_{(x_0, y_0)}\Omega\times \Omega$) as in Section 2. Direct calculations show that at $(x_0, y_0)$,
  \begin{eqnarray*}
  \left(\frac{\partial}{\partial t}-\sum_{j=1}^n \nabla^2_{E_j E_j} \right)\mathcal{O}_\epsilon (x, y, t)
  &=& -\left(\langle \nabla f(y), \gamma'\rangle-\langle \nabla f(x), \gamma'\rangle \right) \varphi' \\
  &\,&+\varphi'\sum_{i=1}^{n-1}\nabla^2_{E_i E_i} r(x, y) -2\varphi_t+2\varphi''-\epsilon \mathrm{e}^t.
  \end{eqnarray*}
Here we have used the first variation $\nabla \mathcal{O}(\cdot, \cdot, t)=0$ at $(x_0, y_0)$ which implies the identities
$$
(\nabla v)(y, t)=\varphi' \, \gamma'(d),\quad \quad  (\nabla v)(x, t)=\varphi'\,  \gamma'(0).
$$
Now choose the variational vector field $V_i(s)=e_i(s)$, the parallel transport of $e_i$ along $\gamma(s)$, along $\gamma(s)$, the second variation computation gives  that
$$
\sum_{i=1}^{n-1}\nabla^2_{E_i E_i} r(x, y) \le -\int_0^d \operatorname{Rc}(\gamma', \gamma')\, ds.
$$
Hence at $(x_0, y_0)$ we have that
\begin{eqnarray*}
 \left(\frac{\partial}{\partial t}-\sum_{j=1}^n \nabla^2_{E_j E_j} \right)\mathcal{O}(x, y, t)
&\le& -\varphi'\int_0^d (\nabla^2 f+\operatorname{Rc})(\gamma', \gamma')\, ds -2\varphi_t+2\varphi''\\
&\le& -\varphi' a r(x, y)-2\varphi_t+2\varphi''\\
&\le & 0.
\end{eqnarray*}
Here we have used $d=r(x, y)$ and $s=\frac{r(x, y)}{2}$.
This is enough to prove the proposition.

To prove the theorem, let $\bar{\omega}(s)$ be the first non-constant eigenfunction for $\mathcal{L}_{a}$, which can be chosen to be positive on $(0, \frac{D}{2})$. To apply the proposition we consider $\bar{\omega}^{D'}(s)$ be the eigenfunction on $[-\frac{D'}{2},\frac{D'}{2}]$ with the corresponding eigenvalue $\bar{\lambda}_a^{D'}$. Let $\varphi(x, t)=C \mathrm{e}^{-\bar{\lambda}^{D'}_a t} \bar{\omega}^{D'}(s)$. Let $w(x)$ be the first non-constant eigenfunction of $\Delta_f$ and let $v(x, t)=e^{-\widetilde{\lambda}_1 t}w(x)$. Since $\bar{\omega}^{D'}(s)$ is an odd function (by adding an eigenfunction $\psi(s)$ with $\psi(-s)$ one can always obtain one), we do have $\varphi(0, t)=0$. The possibility of choosing $(\bar{\omega}^{D'})'(s)> 0$ on $[0,\frac{D}{2}]$ can be proved as follows. By the uniqueness, we have that  $(\bar{\omega}^{D'})'(0)>0$. It suffices to show that  $(\bar{\omega}^{D'})'(s)>0$ for $s\in (0, \frac{D'}{2})$.
 Also observe that $\bar{\omega}^{D'}(\frac{D'}{2})>0$. By the ODE $\mathcal{L}_a\bar{\omega}^{D'}=-\bar{\lambda} \bar{\omega}^{D'}$ we can conclude that $(\bar{\omega}^{D'})'>0$ on $[\frac{D'}{2}-\epsilon, \frac{D'}{2})$. Let $y(s)=(\bar{\omega}^{D'})'(s)$.
The ODE also  forces $y>0$ for $s\in (0, \frac{D'}{2})$ since otherwise we assume $s_1<\frac{D'}{2}$ is the biggest zero. Note that $\bar{\omega}^{D'}(s_1)<0$ and it is strictly convex near $s_1$. Now clearly near $s_1$ one can raise the value of $\bar{\omega}^{D'}$ by replacing part of the graph with a line interval parallel to the $x$-axis, hence lower the energy $\mathcal{F}$. This contradicts  the fact that  $\bar{\lambda}$ is the minimum of the quotient $\mathcal{R}(\psi)$   among all nonzero $W^{1,2}(\mathrm{e}^{-\frac{a}{2}s^2}ds)$ function with zero average.

 Finally as before the proposition implies that for sufficient large $C$, $\varphi(s, t)$ is a modulus of continuity of $v(x, t)$. Hence $\widetilde{\lambda}_1\ge \bar{\lambda}_{a,D'}$. The claimed result follows by letting $D'\to D$.
 \end{proof}

\section{Sharpness of the lower bound}\label{sec:sharp}

In this section we show that (for $n\geq 3$ for any $a$ or for $n\geq 2$ for $a\leq 0$) the lower bound $\tilde\lambda_1\geq \bar\lambda_a$ given in Theorem \ref{sgs-thm1} is sharp:  Precisely, for each $\varepsilon>0$ we construct a Bakry-Emery manifold $(M,g,f)$ with diameter $D$ and $\tilde\lambda_1<\bar\lambda_{a,D}+\varepsilon$.

We will construct a smooth manifold $M$ which is approximately a thin cylinder with hemispherical caps at each end.  Let $\gamma$ be the curve in $\mathbb{R}^2$ with curvature $k$ given as function of arc length as follows for suitably small positive $r$ and $\delta>0$ small compared to $r$:
\begin{equation}\label{eq:k}
k(s) = \begin{cases}
\frac{1}{r},&s\in\left[0,\frac{\pi r}{2}-\delta\right];\\
\varphi\left(\frac{s-\frac{\pi r}{2}}{\delta}\right)\frac1r,&s\in\left[\frac{\pi r}{2}-\delta, \frac{\pi r}{2}+\delta\right];\\
0,&s\in\left[\frac{\pi r}{2}+\delta,\frac{D}{2}\right],
\end{cases}
\end{equation}
extended to be even under reflection in both $s=0$ and $s=D/2$.  This corresponds to a pair of line segments parallel to the $x$ axis, capped by semicircles of radius $r$ and smoothed at the joins.  We write the corresponding embedding $(x(s),y(s))$.
Here $\varphi$ is a smooth nonincreasing function with $\varphi(s)=1$ for $s\leq -1$, $\varphi(s)=0$ for $s\geq 1$, and satisfying $\varphi(s)+\varphi(-s)=1$.
We choose the point corresponding to $s=0$ to have $y(0)=0$ and $y'(0)=1$.  The manifold $M$ will then be the hypersurface of rotation in $\mathbb{R}^{n+1}$ given by
$\{(x(s),y(s)z):\ s\in\mathbb{R},\ z\in \mathbb{S}^{n-1}\}$.  On $M$ we choose the function $f$ to be a function of $s$ only, such that
\begin{equation}\label{eq:f}
f''=\begin{cases}
a\left(1-\frac{D}{\pi r}\right),& s\in\left[0,\frac{\pi r}{2}-\delta\right];\\
\varphi\left(\frac{s-\frac{\pi r}{2}}{\delta}\right)a\left(1-\frac{D}{\pi r}\right)
+\left(1-\varphi\left(\frac{s-\frac{\pi r}{2}}{\delta}\right)\right)a,&s\in\left[\frac{\pi r}{2}-\delta, \frac{\pi r}{2}+\delta\right];\\
a,&s\in\left[\frac{\pi r}{2}+\delta,\frac{D}{2}\right],
\end{cases}
\end{equation}
with $f'(0)=0$ (the value of $f(0)$ is immaterial).  Note that this choice gives $f'(D/2)=0$.  We extend $f$ to be even under reflection in $s=0$ and $s=D/2$.

With these choices we compute the Bakry-Emery-Ricci tensor and verify that the eigenvalues are no less than $a$ for suitable choice of $r$.  The eigenvalues of the second fundamental form are
$k(s)$ (in the $s$ direction) and $\frac{\sqrt{1-(y')^2}}{y}$ in the orthogonal directions.  Therefore the Ricci tensor has eigenvalues $(n-1)k\frac{\sqrt{1-(y')^2}}{y}$ in the $s$ direction, and $k\frac{\sqrt{1-(y')^2}}{y}+(n-2)\frac{1-(y')^2}{y^2}$ in the orthogonal directions.
We can also compute the eigenvalues of the Hessian of $f$:  The curves of fixed $z$ in $M$ are geodesics parametrized by $s$, the the Hessian in this direction is just $f''$ as given above.  Since $f$ depends only on $s$ we also have that $\nabla^{2}f(\partial_{s},e_{i})=0$ for $e_{i}$ tangent to $\mathbb{S}^{n-1}$, and $\nabla^{2}f(e_{i},e_{j})=\frac{y'}{y}f'\delta_{ij}$.

The identities $y(s)=\int_0^s\cos\left(\theta(\tau)\right)\,d\tau$ and $y'(s) = \cos\left(\theta(s)\right)$ where $\theta(s)=\int_0^sk(\tau)\,d\tau$ applied to \eqref{eq:k} imply that
$$
y=\begin{cases}r\sin(s/r),&s\in\left[0,\frac{\pi r}{2}-\delta\right];\\
r(1+o(\delta)),&s\in\left[\frac{\pi r}{2}-\delta, \frac{D}{2}\right],
\end{cases}\qquad
y'=\begin{cases}
\cos(s/r),&s\in\left[0,\frac{\pi r}{2}-\delta\right];\\
o(\delta),&s\in\left[\frac{\pi r}{2}-\delta, \frac{\pi r}{2}+\delta\right];\\
0,&s\in\left[\frac{\pi r}{2}+\delta,\frac{D}{2}\right],
\end{cases}
$$
as $\delta$ approaches zero.
This gives the following expressions for the Bakry-Emery Ricci tensor $\text{\rm Rc}_f=\text{\rm Rc}+\nabla^2f$:
$$
\text{\rm Rc}_f(\partial_s,\partial_s)
=\begin{cases}
a+\frac{n-1}{r^2}-\frac{aD}{\pi r},&s\in\left[0,\frac{\pi r}{2}-\delta\right];\\
a+\varphi\left(\frac{2-\frac{\pi r}{2}}{\delta}\right)\left(\frac{n-1}{r^2}(1+o(\delta))-\frac{aD}{\pi r}\right)&s\in\left[\frac{\pi r}{2}-\delta, \frac{\pi r}{2}+\delta\right];\\
a,&s\in\left[\frac{\pi r}{2}+\delta,\frac{D}{2}\right],
\end{cases}
$$
and
$$
\text{\rm Rc}_f(e,e)=
\begin{cases}
\frac{n-1}{r^2}+\frac{as\left(1-\frac{D}{\pi r}\right)}{r\tan\left(s/r\right)},&s\in\left[0,\frac{\pi r}{2}-\delta\right];\\
\frac{n-2}{r^2}+o(\delta),&s\in\left[\frac{\pi r}{2}-\delta, \frac{\pi r}{2}+\delta\right];\\
\frac{n-2}{r^2}(1+o(\delta))&s\in\left[\frac{\pi r}{2}+\delta,\frac{D}{2}\right],
\end{cases}
$$
while $\text{\rm Rc}_f(\partial_s,e)=0$, for any unit vector $e$ tangent to $\mathbb{S}^{n-1}$.  In particular we have $\text{\rm Rc}_f\geq ag$ for sufficiently small $r$ and $\delta$ for any $a\in\mathbb{R}$ if $n\geq 3$, and for $a<0$ if $n=2$.  Note also that the diameter of the manifold $M$ is $D(1+o(\delta))$.

Having constructed the manifold $M$, we now prove that for this example the first non-trivial eigenvalue $\tilde\lambda_1$ of $\Delta_f$ can be made as close as desired to $\bar\lambda_{a,D}$ by choosing $r$ and $\delta$ small.   Theorem \ref{sgs-thm1} gives the upper bound $\tilde\lambda_1\geq \bar\lambda_{a,D(1+o(\delta))}=\bar\lambda_{a,D}+o(\delta)$.  To prove an upper bound we can simply find a suitable test function to substitute into the Rayleigh quotient which defines $\tilde\lambda_1$:  We set
$$
\psi(s,z) = \begin{cases}w(s-D/2),& \frac{\pi r}{2}+\delta\leq s\leq D-(\frac{\pi r}{2}+\delta);\\
w\left(\frac{D}{2}-\frac{\pi r}{2}-\delta\right),& 0\leq s\leq \frac{\pi r}{2}+\delta, \mbox{ and } D-\frac{\pi r}{2}-\delta \le s\le D.
\end{cases}
$$
where $w$ is the solution of $w''-asw'+\bar\lambda_{a,D-\pi r-2\delta}w=0$ with $w(0)=0$ and $w'\left(\frac{D}{2}-\frac{\pi r}{2}-\delta\right)=0$ and $w'(0)=1$.  This choice gives
$$
{\mathcal R}(\psi) = \frac{
\bar\lambda_{a,D-\pi r-2\delta}\int_{\{|s-D/2|\leq \frac{D}{2}-\frac{\pi}{2r}-\frac{\delta}{2}\}}\psi^2\mathrm{e}^{-f}dVol(g)
}{
\int_{\{|s-D/2|\leq D/2\}}\psi^2\mathrm{e}^{-f}dVol(g)
}\leq \bar\lambda_{a,D-\pi r-2\delta}.
$$
It follows that $\tilde\lambda_1\to\bar\lambda_{a,D}$ as $r$ and $\delta$ approach zero, proving the sharpness of the lower bound in Theorem \ref{sgs-thm1}.

\begin{remark}
If we allow manifolds with boundary the construction is rather simpler:  Simply take a cylinder $r\mathbb{S}^{n-2}\times[-D/2,D/2]$ for small $r$, with quadratic potential $f=\frac{a}{2}s^2$, and substitute the test function $\psi(z,s)=w(s)$ defined above.
\end{remark}

\section{A linear lower bound}\label{sec:linear}

Concerning the lower estimate of $\bar{\lambda}_a$, at least for $a\ge 0$, note that  $y=(\bar{\omega}^{D'})'$ satisfies that
$$
\left(\mathrm{e}^{-\frac{a}{2} s^2}y'\right)'=(a-\bar{\lambda}_a)\mathrm{e}^{-\frac{a}{2} s^2}y.
$$
This together with the maximum principle applying to $y^2$ implies that $\bar{\lambda}_a\ge a$. Applying Proposition \ref{ex-AC-com} to the trivial case $M=\mathbb{R}$ with $\varphi(x, t)=\mathrm{e}^{-(\frac{\pi}{D'})^2t} \sin (\frac{\pi}{D'} s)$, and letting $D'\to D$ we also get $\bar{\lambda}_a\ge \frac{\pi^2}{D^2}$.

On the other hand, a normalization procedure reduces the problem of finding/estimation of the first nontrivial Neumann eigenvalue for the involved ODE to finding the first nontrivial Neumann eigenvalue $\bar{\lambda}_{2, \sqrt{\frac{a}{2}} D}$ for the Hermite equation: $\frac{d^2}{ds^2}-2s\frac{d}{ds}$ on the interval $[-\sqrt{\frac{a}{2}}\frac{D}{2}, \sqrt{\frac{a}{2}}\frac{D}{2}]$ since $\bar{\lambda}_a=\frac{a}{2}\bar{\lambda}_{2, \sqrt{\frac{a}{2}} D}$. The Neumann eigenvalue for the Hermite equation is then related to the  eigenvalue of the harmonic oscillator: $\frac{d^2}{ds^2}-s^2 $ with a certain  Robin boundary condition. The following result and its consequence improve the main results of \cite{FS}.

\begin{proposition}\label{ODE1} When $a>0$, the first nonzero Neumann eigenvalue $\bar{\lambda}_a$ is bounded from below by $\frac{a}{2}+\frac{\pi^2}{D^2}$. In particular $\widetilde{\lambda}_1(\Omega)$, with $\Omega$ being a convex domain in any Riemannian manifold with ${\rm Rc}_{ij}+f_{ij}\ge a g_{ij}$, is bounded from below by $\frac{a}{2}+\frac{\pi^2}{D^2}$.
\end{proposition}
\begin{proof} By the above renormalization procedure, it is enough to prove the result for the operator
$\frac{d^2}{ds^2}-2s\frac{d}{ds}$ on interval $[-\frac{D}{2}, \frac{D}{2}]$. Let $\phi$ be the first eigenfunction which is  odd. Let $y=\phi'$ and denote $\lambda$ the first (nonzero) Neumann value. Then direct calculation shows that
$$
\left(\mathrm{e}^{-s^2}y'\right)'=-(\lambda-2)\mathrm{e}^{-s^2 }y.
$$
Multiply $y$ on both sides  of the above equation and integrate the resulting equation on $[-\frac{D}{2}, \frac{D}{2}]$.
The fact $y=0$ on the boundary implies that
$$
\int_{-D/2}^{D/2} \mathrm{e}^{-s^2}(y')^2=(\lambda-2)\int_{-D/2}^{D/2}  \mathrm{e}^{-s^2}y^2.
$$
In the view that $y$ vanishes on the boundary, it implies that
$\lambda-2\ge \lambda_0$, the first Dirichlet eigenvalue of the operator $\frac{d^2}{ds^2}-2s\frac{d}{ds}$. Now we may introduce the the tranformation $w=\mathrm{e}^{-\frac{s^2}{2}}\varphi$. Direct calculation shows that  $\varphi$ is the first Dirichlet eigenfunction if and only if
$$
\frac{d^2}{ds^2} w- s^2 w=-(\lambda_0+1)w
$$
with $w$ vanishes on the boundary. By  Corollary 6.4 of \cite{N-expandMo} we have that
$$
\lambda_0+1\ge \frac{\pi^2}{D^2}.
$$
Combining them together we have that $\lambda\ge 1+\frac{\pi^2}{D^2}$. Scaling will give the claimed result.
\end{proof}

\begin{corollary}\label{con-FS-improve} If $(M, g, f)$ is a nontrivial gradient shrinking soliton satisfying (\ref{sgs}) with $a>0$. Then
$$
\operatorname{Diameter}(M, g)\ge \sqrt{\frac{2}{3a}}\pi.
$$
\end{corollary}
\begin{proof}
The result follows from the above lower estimate on the first Neumann eigenvalue, applying to the case that $\Omega=M$,  and the observation, Lemma 2.1 of \cite{FS}, that $2a$ is an eigenvalue of the operator $\Delta -\langle \nabla f, \nabla (\cdot)\rangle$.
\end{proof}

This result clearly is not sharp.   A better eigenvalue lower bound (and hence a better diameter lower bound) will follow from a better understanding of the first Dirichlet eigenvalue of the harmonic oscillator.  We investigate this in the next section.

We  remark that Proposition \ref{ODE1} also implies that for a compact Riemannian manifold $(M, g)$ satisfying ${\rm Rc}\ge (n-1)K$ for some $K>0$, the estimate $\lambda_1(M)\ge \frac{n-1}{2}K+\frac{\pi^2}{D^2}$ holds with $D$ being its diameter. This improves the earlier corresponding works in \cite{Ln}, \cite{Ya}, etc.

\section{The harmonic oscillator on bounded symmetric intervals}
\label{sec:EHO}
In this section we will investigate the sharp lower bound given by the eigenvalue of the one-dimensional harmonic oscillator on a bounded symmetric interval:   Recall from section \ref{sec:linear} that the first Neumann eigenvalue $\bar\lambda_{2,D}$ is equal to $\hat\lambda_{1,D}+1$, where $\hat\lambda_{b,D}$ is defined by the existence of a solution of the eigenvalue problem
\begin{align*}
w''+\left(\hat\lambda_{b,D}-bs^2\right)w&=0,\ s\in[-D/2,D/2];\\
w(D/2)=w(-D/2) &=0;\\
w(x)&>0,\ s\in(-D/2,D/2).
\end{align*}
The solution of the ordinary differential equation $w''-s^2w+\lambda w=0$ (with $w'(0)=0$), which is also called Weber's equation,  can be written in terms of confluent hypergeometric functions:  We have
$$
w(s) = \mathrm{e}^{-s^2}U\left(\frac14-\frac{\lambda}{8},\frac12,2s^2\right)
$$
where $U$ is the confluent hypergeometric function of the first kind.  Thus $\hat\lambda_{1,D}$ is the first root of the equation $U\left(\frac14-\frac{\lambda}{8},\frac12,\frac{D^2}{2}\right)=0$.  Since $U$ is strictly monotone in the first argument, the solution is an analytic function of $D$.

Noting that $\hat\lambda_{1,D}=\frac{\pi^2}{D^2}\hat\lambda_{\frac{D^4}{\pi^4},\pi}$,
we use a perturbation argument to compute the Taylor expansion for $\hat\lambda_{b,\pi}$ as a function of $b=\frac{a^2}{4}$ about $b=0$ (this provides an expansion for $\hat\lambda_{1,D}$ about $D=0$).
That is, we consider the solution $u$ of the eigenvalue problem
\begin{align*}
u''+\left(\lambda-bs^2\right)u&=0,\ s\in[-\pi/2,\pi/2];\\
u(\pi/2)=u(-\pi/2)&=0;\\
u(s)&>0,\ s\in(-\pi/2,\pi/2).
\end{align*}
The solution for $b=0$ is of course given by $u(s)=\cos(s)$.  The perturbation expansion produces a solution of the form
$$
u(s,b) = \sum_{k=0}^\infty b^k\sum_{j=0}^{3k}\left(\alpha_{k,j}s^j\cos s+\beta_{k,j}s^k\sin s\right),
$$
with $\lambda=\sum_{k=0}^\infty\lambda_kb^k$.  This expansion is unique provided we specify that $u$ is even, $\alpha_{0,0}=1$, $\beta_{0,1}=0$, and $\alpha_{k,0}=\beta_{k,0}=0$ for $k>0$.  The first few terms in the expansion for $\lambda$ are given by
\begin{align*}
\hat\lambda_{b,\pi} &= 1+\left(\frac{\pi^2}{12}-\frac12\right)b
+\left(\frac{\pi^4}{720}-\frac{5\pi^2}{48}+\frac78\right)b^2
+\left(\frac{\pi^6}{30240}-\frac{\pi^4}{48}+\frac{31\pi^2}{32}-\frac{121}{16}
\right)b^3\\
&\quad\null+\left(\frac{\pi^8}{362880}-\frac{\pi^6}{270}+\frac{683 \pi^4}{1280}-\frac{14573 \pi^2}{768}+\frac{17771}{128}\right)b^4+O(b^5).
\end{align*}

\begin{figure}\label{fig:Weber}
\includegraphics[scale=0.5]{WeberEV.eps}
\caption{The eigenvalue $\hat\lambda_{b,\pi}$ for Weber's equation $y''+(\lambda-bs^2)y=0$, $y(\pm\pi/2)=0$ [solid curve]; shown also are the lower bounds $\hat\lambda\geq 1$ and $\hat\lambda\geq\sqrt{b}$ [dashed curves].}
\end{figure}

We note that there is also a useful lower bound for $\hat\lambda_{b,\pi}$, which we can arrive at as follows:  The inclusion of $[-D/2,D/2]$ in $\mathbb{R}$ implies $\hat\lambda_{1,D}\geq \lim_{d\to\infty}\hat\lambda_{1,d}=1$, with eigenfunction $u(s) = \mathrm{e}^{-s^2/2}$.  Therefore we also have
$$
\hat\lambda_{b,\pi}=\sqrt{b}\hat\lambda_{1,\pi b^{1/4}}\geq\sqrt{b}.
$$

This translates to an estimate for the drift eigenvalue $\bar\lambda_{a,\pi}$ appearing in Theorem \ref{sgs-thm1}:  We have $\bar\lambda_{a,\pi} = \frac{a}{2}+\hat\lambda_{a^2/4,\pi}$, giving the following Taylor expansion:
\begin{align*}
\bar\lambda_{a,\pi} &= 1+\frac{a}{2}+\left(\frac{\pi^2}{12}-\frac12\right)\frac{a^2}{4}
+\left(\frac{\pi^4}{720}-\frac{5\pi^2}{48}+\frac78\right)\frac{a^4}{16}
+\left(\frac{\pi^6}{30240}-\frac{\pi^4}{48}+\frac{31\pi^2}{32}-\frac{121}{16}
\right)\frac{a^6}{64}\\
&\quad\null+\left(\frac{\pi^8}{362880}-\frac{\pi^6}{270}+\frac{683 \pi^4}{1280}-\frac{14573 \pi^2}{768}+\frac{17771}{128}\right)\frac{a^8}{256}+O(a^{10}).
\end{align*}
In particular the lower bound $\hat\lambda\geq 1$ translates to $\bar\lambda\geq 1+a/2$, and the lower bound $\hat\lambda\geq\sqrt{b}$ translates to $\bar\lambda\geq a/2+\sqrt{a^2/4}=a$.  Finally, by scaling we obtain the following:
\begin{align*}
\bar\lambda_{a,D} &= \frac{\pi^2}{D^2}+\frac{a}{2}+\left(\frac{\pi^2}{12}-\frac12\right)\frac{D^2a^2}{4\pi^2}
+\left(\frac{\pi^4}{720}-\frac{5\pi^2}{48}+\frac78\right)\frac{D^6a^4}{16\pi^6}\\
&\quad\null
+\left(\frac{\pi^6}{30240}-\frac{\pi^4}{48}+\frac{31\pi^2}{32}-\frac{121}{16}
\right)\frac{D^{10}a^6}{64\pi^10}\\
&\quad\null+\left(\frac{\pi^8}{362880}-\frac{\pi^6}{270}+\frac{683 \pi^4}{1280}-\frac{14573 \pi^2}{768}+\frac{17771}{128}\right)\frac{D^{14}a^8}{256\pi^{14}}\\
&\quad\null+O(D^{18}a^{10})\quad\text{\rm as}\ D^2a\to 0.
\end{align*}

An interesting consequence of the Taylor  expansion (combined with the fact that the estimate $\tilde\lambda_1\geq \bar\lambda_{a,D}$ is sharp as proved in section \ref{sec:sharp}) is the following:

\begin{proposition}
The constant $a/2$ in the lower bound $\tilde\lambda_1\geq \frac{\pi^2}{D^2}+\frac{a}{2}$ is the largest possible.
\end{proposition}

This follows from the Taylor expansion for small values of $aD^2$.

We note that the sharp diameter bound (given by the value of $a$ where the dotted line $\lambda=2a$ intersects with the solid curve in Figure \ref{fig:drift}) is not dramatically different
from the one given in Corollary \ref{con-FS-improve} (where the dotted line intersects the dashed line $\lambda=1+a/2$).  Since the eigenvalue estimate $\tilde\lambda_1\geq \bar\lambda_{a,D}$ appears from the examples in section \ref{sec:sharp} to be sharp only in situations which are far from gradient solitons, we expect that neither of these diameter bounds is
close to the sharp lower diameter bound for a nontrivial gradient Ricci soliton.

\begin{figure}
\includegraphics[scale=0.5]{DriftEV.eps}
\caption{The eigenvalue $\bar\lambda_{a,\pi}$ for the drift Laplacian equation $y''-asy'+\lambda y=0$, $y'(\pm\pi/2)=0$ [solid curve]; shown also are the lower bounds $\bar\lambda\geq 1+\frac{a}{2}$ and $\bar\lambda\geq a$ [dashed lines], and the line $\bar\lambda=2a$ corresponding to non-Einstein gradient Ricci solitons [dotted line].}\label{fig:drift}
\end{figure}
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redpajama_set_name: RedPajamaArXiv


text:
Kilkivan  is a town and locality in the Gympie Region of Queensland, Australia. At the , Kilkivan had a population of 713.

Geography
The town is situated on the Wide Bay Highway,  north of the state capital, Brisbane and  west of Gympie. One Mile Creek () meanders through the town, east of the Wide Bay Highway. The town lies in the Mary River drainage basin.

Rossmore is a neighbourhood within the locality to the south-west of the town of Kilkivan ().

History

Kilkivan was first inhabited by the Wakka Wakka tribe of the Australian Aboriginal peoples.  The town was first settled by Europeans in the 1840s. Queensland's first gold discovery was at Kilkivan in 1852 and subsequent findings escalated into a gold rush in the 1860s. The town was named for a pastoral run owned by pastoralist John Daniel MacTaggart (1823–1871) after his father's farm name near Drumlemble, Kintyre, Scotland.  The nearby Australian 'Glenbarr' property owned by MacTaggart was also named for a nearby Scottish farm.

Kilkivan Post Office opened on 8 June 1868. It was known as Neureum between 1876 and 1881.

Kilkivan State School opened in March 1876.

In 1883, a court house, police quarters and lock-up was built at a total cost of £748 10s. It was a wooden building with an iron roof and consisted of the court house, two bedrooms, sitting room, kitchen, verandahs, and two cells.

The Nanango railway line opened in December 1886 as far as Kilkivan. The locality was also served by Mouingba and Bular railway stations on the same line. Anticipation of the line opening brought an increase in the desire to build near the station, which was located on a goldfield.  The  line was officially closed in early 2010.

In 1887, two parcels of land were resumed from the Kilkivan pastoral run: one of  (described as "Murray's debatable land") and the other . The land was offered for selection for the establishment of small farms on 17 April 1887.

St Matthew's Anglican Church was dedicated in 1888 by Reverend E. C. Osbourne of Gympie. Its closure circa 2013 was approved by Bishop Rob Nolan.

Fairfield State School opened circa 1915. In 1916  it was renamed Rossmore State School. It was closed circa 1943.

The Kilkivan Library opened in 1967.

The Kilkivan War Memorial commemorates those of the district who served in the wars. It was dedicated on 28 March 1996.

At the , Kilkivan had a population of 713.

Heritage listings
Kilkivan has a number of heritage-listed sites, including:
 Rossmore Road: Mount Clara smelter

Economy

Major industries today include tourism, beef, dairying and forestry.

Education

Kilkivan State School is a primary and secondary (P-10) school for boys and girls operated by the Queensland Government at 6 Council Street. In 2016, the school had an enrolment of 97 students with 14 teachers (12 full-time equivalent) and 10 non-teaching staff (7 full-time equivalent).

Amenities

The Gympie Regional Council operates a public library at 31 Bligh Street, Kilkivan.

The Kilkivan branch of the Queensland Country Women's Association meets at the CWA hall at 6 Park Street. The building was built as the council chambers of the Division of Kilkivan (later the Shire of Kilkivan) at 26 Bligh Street in 1890, but was purchased and relocated to its present site in 1958. The QCWA branch in Kilkivan was established in 1926.

Kilkivan Bowls Club is in Crescent Street.

Attractions
The Kilkivan and District Museum and information centre in the main street features extensive displays which pay tribute to the region's early pioneers.

The country town features parks and restored historical buildings.  Kilkivan has antique stores, a historical walk, and a cafe and B&B housed in the town's original bank.

Kilkivan is one of the few towns on the Bicentennial National Trail. A stone plaque commemorates the opening of the trail in 1988.

The Kilkivan Great Horse Ride is held annually in April.  Over 1,000 horses and riders start from five points around Kilkivan and travel  along parts of the National Trail to Kilkivan for the Grand Parade down the main street.

Mudlo Gap is a popular forest park  north of Kilkivan.

The Mount Clara chimney of Mount Clara Copper Smelter, thought to be the oldest surviving mining industry chimney in Queensland, and also among the first to have been built, is located  by road outside town, and is now conserved as a tourist attraction.

The Kilkivan to Kingaroy Rail Trail follows the old railway line between the two towns. It is  long and passes through Goomeri, Murgon, Wondai, Tingoora, Wooroolin, Memerambi, and Crawford. The trail from Kilkivan to Murgon is unsealed. The first creek crossing is of Coppermine Creek about  south west of Kilkivan.  A low level bridge was opened across Wide Bay Creek in October 2021.  Short distances between towns means coffee is never far away.

References

External links

Towns in Queensland
Wide Bay–Burnett
Gympie Region
Localities in Queensland

redpajama_set_name: RedPajamaWikipedia


Sherry Burri has been a longtime resident of Blenheim. Shortly after her birth in Simcoe, her parents made the decision to move back to the beautiful town of Blenheim where they had both grown up. Sherry's grandfather, L. Stewart Paling married a Blenheim girl, Maude Holmes, and then he started a career as one of the first Real Estate Agents in Blenheim, he was on the Town Council for many years, and even served a term as Mayor. Her paternal grandparents Frank & Bessie Pokornik emigrated from Czechoslovakia and made their home in Blenheim as well. Sherry's parents, Jerry and Shirley Pokornik owned and operated a tool & die shop "The Toolroom" in Blenheim, so Sherry's roots in the Blenheim community are very deep. In her early years, Sherry went to the Talbot Street School, then attended W.J. Baird Public School, and went on to graduate from the Blenheim District High School.
Sherry married her husband Peter in 1979 and during the first few years of marriage they worked in the family restaurant, "Mamma's". Then their lives took a different path when they started a signs and graphics business, "Auto Trim Design", still currently working for many of the local automotive dealerships in Kent and Essex counties.
When her children Alaina and Mitchell were raised, she worked part time for several years in office administration for Jon Ford who owned the J.L. Ford Funeral Home in Blenheim. Then in 2007 when Rob McKinlay purchased the Funeral Home, Sherry was "adopted" into the McKinlay Funeral Home family and has worked full-time ever since. "Working at the Funeral Home has never felt like just a job to me, it's really a way that I can help the community that I love, and I have the privilege of working in one of the most beautiful buildings in Blenheim" she says.
Her family all live in Blenheim and she has enjoyed seeing her children and grandchildren attend the same schools that she did. She loves spending time with her grandchildren Avia, Levi, and little Everly, her volunteer work as one of Jehovah's Witnesses, and summer vacations on Manitoulin Island where she can catch her breath and enjoy the sunsets.


Heart diseases are broadly known as Cardiovascular Disease which is further classified into Coronary Artery Disease (CAD), Peripheral Artery Disease (PAD), Valvular Heart Disease, Congenital Heart Disease and Rheumatic Heart Disease.
Angioplasty is a surgical procedure which is performed to widen the narrowed or obstructed arteries or veins to clean the accumulation of white blood corpuscles in the arteries.
While performing the surgery the doctor passes an empty balloon, commonly called balloon catheter, over a wire into the narrowed location and then swells it to a fixed size. The force exerted by the inflated balloon on the stenosis inside the vessel and the muscular wall, broadens the blood vessel again for improved flow. The balloon is then deflated and taken out. Depending on the condition of the patient, a stent may or may not be introduced at the time of ballooning.
When is an angioplasty performed?
An angioplasty is performed when the patient is having a coronary artery disease, more specifically, narrowing of the arteries due to the accumulation of fat, cholesterol, calcium or atherosclerosis.
The cost of angioplasty starts from USD 5,000 in India, if it includes a single stent. For more than one stent an additional USD 2,000 is charged.
The arteries that bring blood to the heart can get clogged at times with plaque (an accumulation of cholesterol, fat and other substances) that prevents the blood to flow smoothly. Coronary artery bypass surgery or simply CABG is a surgical procedure that is used to bypass blood through those clogged arteries to improve the flow of blood and oxygen in the heart.
CABG is performed in two ways: the "off-pump" and the "on-pump".
In "off-pump" the surgery is performed on a beating heart whereas in "on-pump" the surgery is performed with the heart stopped. Arteries and veins from the other parts of the body are sewn to one of the coronary arteries and the other end is attached to the aorta (the largest artery in the body) to bypass the clogged arteries.
A CABG is performed on patients suffering from coronary artery diseases.
The cost of CABG or bypass surgery starts from USD 4400 in India.
Heart valve replacement surgery is the replacement of one or more heart valves. It can be replaced with either an artificial heart valve or a bioprosthesis, an alternative for valve repair.
When is a valve replacement performed?
A valve replacement is performed when any of the four major valves of the heart (aortic, mitral, tricuspid and pulmonary) are damaged, that is, if there is a narrowing of the valve (valve stenosis) or if the valve is unable to close properly (valve incompetence or regurgitation or a leaky valve).
The cost of a valve replacement starts from USD 6,500 in India.
CAG or Coronary Angiography is an X-Ray test used to see the flow of blood through your heart. CAG is performed to locate if there are any blockage in the coronary arteries, how many arteries have been blocked, where are the blockages and the severity of it.
A CAG lasts for 30-60 minutes in which the patient is treated under local anesthesia. A catheter is inserted through the groin or the arm of the patient and is placed in the heart through the artery. After placing it, the cardiologist injects a special dye into the bloodstream of the patient that makes the coronary arteries visible in the X-Ray. The x-ray images are then observed to see how the dye moves through the arteries so as to locate the blockages. The catheter is removed after completion of the test.
When is a CAG performed?
The cost of a CAG surgery in India starts from USD 350 and may go up till USD 500.
In hospital: 1 day in the hospital.
In India: It depends on the doctor's diagnosis.
AICD or Automatic Implantable Cardioverter Defibrillator is a device that is used to monitor and regulate the irregular heartbeat of the patient. It can speed up or slow down the heart depending upon the heart rate.
Before performing the AICD implantation, the doctor performs an Electrophysiological study to see whether an AICD is necessary or the treatment can be done with proper medication. The implantation surgery is performed both under local and general anesthesia. The patient will be under constant sedation. To prevent movement during the AICD test, his arms are loosely strapped.
After making an incision in the chest, below the collarbone, a wire will be inserted in the heart through a vein. After creating a pocket in his chest, the doctor will insert the AICD (connected to the wires) inside his heart.
He will create an arrhythmia to test the functioning of the device that is necessary for the treatment. Later on it will be again tested with the help of a computer known as "programmer". This test procedure is known as "non-invasive programmed stimulation" or "pre-disharge testing".
When is AICD implantation surgery performed?
When the patient has irregular heart beat or arrhythmia, then AICD implantation surgery is performed.
The cost of an AICD implantation surgery starts from USD 6500 in India.
A heart transplant is performed to remove a diseased heart and replace it with a healthy heart of another patient who has been declared to be brain dead by at least two or three hospitals.
The surgeon will make an incision down to the chest from just below the Adam's apple. The breastbone is cut in half and spread apart. To pump the blood through the body the surgeon put tubes into the patient's chest by a heart-lung machine, cardiopulmonary bypass.
After the blood has been completely pumped to the bypass machine, the diseased heart of the patient is removed. The donor heart is then sewn into place and blood vessels are connected carefully to prevent leaks.
Blood circulating through the bypass machine is allowed to flow back into the heart while the tubes connected to the machine are removed. The heartbeat is revived by mild shocks. The healthcare team, thereafter, keeps it under surveillance to see its functionality and to make sure there are no further leaks.
When is a heart transplant performed?
A heart transplant is performed when the patient experiences an end stage heart failure.
A heart transplant surgery cost starts from USD 50,000 in India.
Coronary Artery Disease- All you need to know !!


Cozart, who started KU's first four games, appeared to suggest Tuesday that he would be playing at some point Saturday.
Cozart was asked afterwards if he expected to play against TCU.
Willis replaced Cozart against Texas Tech, completing 14 of 26 passes for 142 yards with a touchdown. Cozart, meanwhile, was 9 of 20 for 97 yards with a touchdown and interception.
Willis received congratulatory texts from friends and high school coaches following Beaty's announcement Monday.
Beaty's conviction in naming a starter was a change in tone from Thursday. Following his team's 55-19 loss to Texas Tech, the coach indicated that he and the staff had evaluated both Cozart and Willis every day, and that neither had emerged to take over the role.
Five days later, Beaty was eager to see how Willis handled his new status as KU starting quarterback.


text:
Home » Cal Poly AppointsRaymond Fernando to New Coatings Chair
Industry NewsBreaking News
Cal Poly AppointsRaymond Fernando to New Coatings Chair
SAN LUIS OBISPO, CA - California Polytechnic University-San Luis Obispo recently appointed Raymond Fernando, a former research chemist with Air Products and Chemicals Inc., as the first holder of the Arthur C. Edwards Endowed Chair for Coatings Technology and Ecology.
The professorship, created with an endowment from Dunn-Edwards Corp. and contributions from 21 donors from the polymers and coatings industries, was created as part of the polymers and coatings program in the university's Chemistry and Biochemistry Department. The endowment fund was created with $1 million in gifts.
The professorship is named for the co-founder of Dunn-Edwards Corp. Arthur Edwards' son, Ken, and Dunn-Edwards Co. gave the primary gift of $500,000 to establish the chair.
Fernando was formerly a lead research chemist for Air Products' Corporate Science and Technology Center, and has 15 years of experience in the coatings field. He is considered an authority on waterborne coatings technology, and his doctoral work at North Dakota State University focused on the study of coatings rheology.
Fernando is currently chairman of the Technical Program Committee of the Federation of Societies for Coatings Technology.
Cal Poly said Fernando will help lead the development of the university's new master's degree program in polymers and coatings science, which began this fall, and will be involved in expanding the existing undergraduate polymers and coatings program. He is also working to establish student internships in the coatings industry, in a variety of product and technology areas.
Cal Poly Launches Campaign to Build Coatings Tech. Center
Cal Poly to Offer Winter Short Course on Coatings Science and Technology
Behr Process/Masco Corp. Give $300,000 to Cal Poly's Western Coatings Technology Center
Dunn-Edwards Makes Additional Gift to Cal Poly Coatings Technology Center

redpajama_set_name: RedPajamaCommonCrawl


9. Scutellaria obtusifolia Hemsley, J. Linn. Soc., Bot. 26: 296. 1890.
Herbs perennial, rhizomatous. Stems 20-60 cm tall, ascending, ± woody and leafless at base, unbranched, densely pubescent. Leaves elliptic to obovate or subcircular, 2-8 × 1.3-3.5 cm, ± leathery, adaxially glabrous or very sparsely pubescent, abaxially densely gray pubescent, 3-veined at base, whitish along veins abaxially, base cuneate-attenuate to rounded, margin entire to inconspicuously few crenate or 4-8-undulate-dentate, apex obtuse to acuminate. Racemes terminal, 4-18 cm; bracts rhombic-obovate, to slightly longer than pedicel, puberulent. Pedicel 4 mm, glandular puberulent. Calyx ca. 2.5 mm, ca. 4 mm in fruit, glandular puberulent; scutellum open, semicircular, ca. 1.8 mm, to 5 mm in fruit, conspicuously veined. Corolla purple or purple-blue, 2-3 cm, sparsely puberulent; tube 2-2.3 cm, bent at base, throat to 7 mm wide; upper lip ovate, ca. 5 mm wide; middle lobe of lower lip subcircular, ca. 4.5 mm in diam., apex emarginate; lateral lobes ovate, ca. 3 mm wide. Nutlets tuberculate. Fl. May-Jun, fr. Jun-Jul.


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In the age of smartphones, air travel and ride sharing, Americans are more mobile than ever — except, perhaps, in the one way that matters most.
According to Robert Archibald, William & Mary chancellor professor of economics, and David Feldman, professor of economics at W&M, when it comes to measures of social mobility — a concept referring to the generational movement in social status and income over time — the U.S. seems to be falling short.
That issue was top of mind in the W&M School of Education on April 21-22, when educators, economists and policymakers from around the country gathered to share research, ideas and discussion at the Conference on Higher Education and Social Mobility. The brainchild of W&M Provost Michael R. Halleran, the event was co-organized by Archibald and Feldman, co-authors of the books Why Does College Cost so Much? and the forthcoming The Road Ahead for America's Colleges and Universities.
The conference kicked off with a welcome address by W&M President Taylor Reveley, who stressed the importance of a society in which social mobility can be achieved by everyone and offered ways to think about tackling the issue from a university standpoint.
In a series of six sessions, speakers presented a wealth of solutions to these and more challenges, ranging in topic from federal and state policy options to the role community colleges and online education can play in advancing social mobility. Sarah Turner, university professor of economics and education at the University of Virginia, shared research on access, or how much high ability, low-income students really know about college, while Sandy Baum, senior fellow with the Urban Institute, focused on how students can navigate their options in federal financial aid.
In the first session, Archibald discussed two solutions for making public investment in education more progressive and more conducive to social mobility.
"You can move the low-income students to where the resources are, or you can move the resources to where the low-income students are," Archibald said.
Citing research done in conjunction with Feldman and Peter McHenry, associate professor of economics at W&M, Archibald showed the effects of low social mobility in a sample of high school graduates from 2004. Their study analyzed, among other things, student socioeconomic status and academic potential (measured by GPA and test performance) and compared that to the selectiveness of the college the students attended (if any) and their graduation rates.
The results put a spotlight on educational inequality, with students from the upper end of the socioeconomic pool more likely to attend competitive institutions and more likely to graduate. Lower income students, on the other hand, though often testing equally as high in measures of academic potential, are less likely to attend the same schools.
Whether that's accomplished through outreach or through swapping merit-based for need-based financial aid (or a combination of several things), conference participants left brimming with ideas to mull over back home in their respective roles.


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NASA Ames Research Center
01/2015 - 12/2019 (CAN 7)
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The Evolution of Prebiotic Chemical Complexity and the Organic Inventory of Protoplanetary Disks and Primordial Planets
Life on Earth is tied to the abiotic formation and chemical history of compounds based mainly on the elements C, H, N, O, S, and P. Apart from H, the evolution of these biogenic elements starts with their nucleosynthesis in stars and ejection into the interstellar medium (ISM). Once in the ISM, the ejecta are modified by physical and chemical processes such as UV and cosmic ray irradiation, gas-phase reactions, accretion and reaction on grain surfaces, and shock waves. Within molecular clouds, the birthplace of stars and planets, most of these elements are frozen onto small, cold dust grains where they are processed into more complex molecules by UV photons, cosmic rays, and catalytic surface reactions. For the past thirty years, laboratories around the world have simulated UV photochemistry and cosmic-ray bombardment processes on interstellar and Solar System ice analogs that reveal the central role these ices play in the astrobiology and astrochemistry in star and planet-forming regions. From this work it is clear that ices play important roles not only as repositories of reactants, but also as structures that mediate the reactions that occur on and in them. Studies of the role of surface chemistry on astrophysically relevant solids have just begun, with laboratory work investigating surface reactions of atoms and simple molecules. However, processes involving a wider range of astrophysically-relevant molecules on relevant minerals and ices have been largely unexplored.
The NAI Ames team seeks a greater understanding of chemical processes at every stage in the evolution of organic chemical complexity, from quiescent regions of dense molecular clouds, through all stages of cloud collapse, protostellar disk, and planet formation, and ultimately to the materials that rain down on planets, and understanding how these depend on environmental parameters like the ambient radiation field and the abundance of H2O. This team is structured as an integrated, coherent program of astrochemical experiments, quantum chemical computations, disk modeling, and observations of astronomical sources. The ultimate goal is to move beyond identifying particular molecules, ions, or radicals present in astronomical objects to understanding their place in the dynamics of the chemical evolution of different astrophysical environments, with particular attention to the formation sites of new planetary systems and our own Solar System.
The research of the NASA Ames team consists of three Modules, each of which has well defined tasks but that integrates into the other two. The Modeling and Observations of Disks and Exoplanets Module leverages the abundance of available observational data, including line and continuum spectra from protoplanetary disks and data from exoplanet surveys, to understand the formation and evolution of organics during planet formation. Members of their team are conducting a thorough investigation of the evolution of organic molecules using theoretical disk models that include a new, specific, thermo-chemical network constructed with inputs from laboratory experiments and quantum chemical theory, and chemo-dynamical models that follow the transport and irradiation of these species as disks evolve, form planets, and eventually disperse.
Exoplanet surveys suggest that planet formation may occur in situ and thus may not depend on large-scale Type I migration. The mode of planet formation has strong consequences for the compositional mixes and catalytic processes that are most astrobiologically relevant. The Ames team, therefore, combines statistical modeling of the observed planetary census with the results of structural models to determine the distribution of planetary core-envelope fractions, which encode key information regarding the disk environment in which the planets formed. They also carry out dynamical simulations that test migratory and in situ planet formation theories, and determine disk parameters that are consistent with exoplanetary architectures. Chemical evolution of these prototypical extrasolar nebulae are followed to track the incorporation of materials from different parts of the full condensation sequence, and radiative transfer modeling are conducted to predict line emission signatures that stem from molecular processes in planet-forming disks.
The Laboratory Module utilizes conditions gleaned from the literature and obtained from the theoretical modeling described above, to conduct laboratory studies with mixtures of relevant minerals, organic molecules, and ices to study radiation- and surface-mediated (i.e., catalysis) chemistry. Experimental series are designed to investigate specific regions of protoplanetary disks. The analyses used to study the organic products are carried out with several goals in mind: (i) characterize the products and compare them to organics found in relevant regions of protoplanetary disks, exoplanets, as well as meteorites and samples from sample return missions; (ii) search for products of specific astrobiological interest, including amino acids, amphiphiles, quinones, sugar-like compounds, and nucleobases; (iii) search for specific compounds that are under consideration as potential biomarkers to test if they are robust bio-indicators or can be formed abiotically; and (iv) determine whether the organic residues produced have interesting astrobiological behaviors (for example, determine whether they form membranes).
The third Module of the Ames team makes use of computational quantum chemistry techniques, which have become vital tools to elucidate the detailed mechanisms of chemical reactions both in the gas and condensed phases, as well as diverse catalytic processes. Using ab initio and density functional theory approaches, including quantum mechanical/molecular mechanics methods, it is possible to determine which reaction products are more feasible or even possible under various physical conditions.
The Ames team also investigates the reactions of organic molecules that are present in the different environments of protoplanetary disks, including gas-phase, gas surface and condensed-phase chemical reactions both in H2O-rich ('hydrous') and H2O-free ('anhydrous') environments. The formation mechanisms of small biogenic molecules, such as amino acids and nucleic acids, are being examined first, starting from simple C,N,O-bearing precursors. Mechanisms that lead to the formation of these molecules reveal key information about the environments in which these reactions can or cannot occur. Similar to their prior studies, the catalytic role played by ices (mainly H2O) and by anhydrous grains is also being investigated.
The three modules of the Ames team are not carried out in isolation, but are part of an interacting whole. The results of disk and exoplanet modeling are used to establish the environmental conditions (temperature, composition, radiation environment, etc.) that are examined in the laboratory studies and quantum computational chemistry modules. Results from the laboratory studies (reaction rates, chemical products, etc.) are fed back into the disk models to improve their prediction of organic synthesis and are used to test the results of the quantum chemistry approach. Quantum chemistry results (reaction mechanisms, reaction rates, product ratios, etc.) are used to help interpret the laboratory results, and feed into improved parameters for the disk modeling studies. In addition, validation of the quantum chemistry results will allow the team to address environments and conditions that may be predicted by disk models, but that are outside the range of experimental conditions that the laboratory effort can address.
Figure 1. Overview for the interactions among the three modules. Module1-Modeling & Observational Studies, Module 2-Laboratory Studies, and Module 3-Quantum Chemistry Studies.
TEAM LOGO:
Scott Sandford
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Team Website:
http://amesteam.arc.nasa.gov/
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A Syrian national is warning fellow Canadians to stand up for secular values and not make unreasonable accommodations for Islamists.
Although her grandmother was a practicing Muslim, her parents were devout Atheists and her father was a militant activist with the Parti de l'Avant-Garde Socialiste (PAGS), a left-wing political movement that operated mostly underground in Algeria and fought Islamist interference in politics. There was never a Koran in her home, she recalled, and was taught at a very young age that women deserved the same rights as men.
Benhabib questions the Islamist obsession with headscarves for women; she said she grew up around many observant Muslims who did not wear them. It is sometimes a symbol of modesty, sometimes one of repression, she explained.
Describing Islamists as "enemies of equality," she was inspired to write the book after watching the Bouchard-Taylor commission on "reasonable accommodation" and being disappointed with the final report. She saw many Quebecers sharing concerns over the integration of immigrants and wanted to share her perspectives as an Algerian native who seeks no accommodation.


GLENDALE, ARIZONA – Phoenix Coyotes General Manager Don Maloney announced today that the Coyotes have assigned two players to the Portland Pirates, the club's American Hockey League affiliate.
The players assigned to Portland are forwards Andy Miele and Jordan Szwarz. The Coyotes also announced that they have released forward Gilbert Brule from his professional tryout agreement.
There are now 25 players remaining in the Coyotes veterans' training camp.