Are astronauts, in Earth orbit, without weight?
First, I should mention that there are two types of weight: actual weight and apparent weight. An Earth-bound object's actual weight is the downward force exerted upon it by the Earth's gravity. The object's apparent weight is the upward force, typically transmitted through the ground, that opposes gravity and prevents the object from falling through the floor or ground (towards the center of the Earth). When you stand on your bathroom scale, it's measuring your apparent weight (i.e. how hard it's having to push up on you to prevent you from accelerating downwards through the scale, crushing it). This doesn't necessarily have to be equal to your actual weight. In your bathroom, your apparent weight is equal to your actual weight, but if you carry the scale into an elevator and weigh yourself while accelerating upwards, the scale will register an apparent weight that is greater than your actual weight. Since a cable is pulling the elevator up rapidly, the elevator's floor is pushing up on the scale, and the scale is, in turn, pushing up on your feet. In order to force you upwards, against gravity, the scale is having to push on you harder than if it (and you) were stationary. Your actual weight doesn't change here, because it's dependent on your mass (which isn't changing) and your distance from the center of the Earth (which is changing only a negligible amount). But to the scale, it feels like you're growing heavier, because it's having to not only support you but push you (accelerate you) upwards. So it registers a heavier "weight", which we now know to be your apparent weight. What about when the elevator accelerates downwards? Your apparent weight becomes less than your actual weight. And now, what if the elevator's cable were to break and the elevator, scale, and you were all to freefall towards the ground below? The scale would be falling at the same rate you were falling, and so it wouldn't be supporting any of your weight. It would indicate a weight of zero. That is, your apparent weight would be zero. This is the definition of weightlessness. Weightlessness means without apparent weight; it has nothing to do with your actual weight. So ... an astronaut in orbit, in a constant state of freefall, kept to a near circular path around the Earth by gravity, is weightless, but only in the sense that he or she has no apparent weight. Astronauts most definitely do have actual weight!
Apparent weight can change fairly easily, we see. All you have to do is take a ride on an elevator, or rollercoaster, or some other device that accelerates you in a vertical direction. Does actual weight ever change (ignoring the effects of food)? Yes, it does. The force of gravity on you (which determines your actual weight) is dependent on your mass, the mass of the Earth, the gravitational constant G, and the distance between you and the center of the Earth. Assuming your mass is held constant, you can reduce your actual weight by increasing your distance from the center of the Earth. So astronauts have actual weight, but an actual weight that is slightly less than their actual weight back on Earth. How much less? In orbit around 300 km (185 miles) above the surface of the Earth, astronauts' actual weight will be about 8.8% less than back on Earth.
What makes you feel weightless when you're falling, even though you still have an actual weight? Or one could ask, what makes you feel heavy (or with weight) when you're standing on the ground? It's not gravity. It's the force of the surface you're standing on, pushing against you. If you're standing on a sidewalk, the concrete is pressing against your feet, which are in turn pressing against your ankles, which are pressing against your lower legs, which are pressing against your upper legs, and on and on. The feet are supporting your entire mass. Your chest, for example, only supports the mass of the body above the chest. You don't feel pressure evenly distributed throughout your body. (Well, you're used to the feeling of standing on a surface, and so you may have a hard time sensing this uneven pressure distribution, but it's there.) Your sense of weight also comes from your arms pulling down on your shoulders. When in freefall, this pressure gradient (or change over space) disappears. Each section of your body, each cell, is falling at the same rate. Therefore, your upper body isn't pushing on your lower body. Your ankles aren't pushing on your feet. There is no pushing at all. Neither are your arms pulling down on your shoulders. The absence of these sensations is what one equates to feeling weightless.
How does NASA simulate a weightless environment for astronaut training? They could put their astronauts in an elevator, take it to the top of a tall building, cut the supporting cable, and allow the elevator and its inhabitants to freefall for several seconds. But the impact upon hitting the ground would be extreme and most unpleasant. Instead, NASA sends its astronauts up in an airplane, and the airplane flies in the parabolic trajectories of freely falling objects. Soaring over the Gulf of Mexico, pilots level off at about 26,000 feet. They then shoot the plane upward at about a 45-degree angle. At this point, the apparent weight of the people inside the nearly empty, padded fuselage increases to about 1.8 times their actual weight. Half a minute later, pilots push the aircraft's nose over the top of this "parabola", and the plane falls some 8,000 feet or so until its pointing downward at about 30 degrees. During this freefall, the aircraft's acceleration matches Earth's acceleration of gravity, making everything inside weightless for 17 to 25 seconds. (Parts of the movie Apollo 13 were filmed on this aircraft.) Over a two-hour flight, the aircraft may fly through some 40 of these parabolas. NASA used two KC-135 Stratotanker aircraft for these sessions from 1973 until 2005, when they were retired and replaced with a McDonnell Douglas C-9. The plane, not too surprisingly, earned the nickname "Vomit Comet."
Sunday, November 22, 2009
Monday, November 2, 2009
Why Do Golf Balls Have Dimples?
Why do golf balls have dimples? The dimples enable the ball to fly much farther through the air. A swing, driving a smooth golf ball 70 yards, could drive a dimpled ball perhaps 250 yards. Why?
First, air pressure is the force exerted by air molecules divided by the area on which the force is exerted. That is, force per unit area. The force comes from the countless collisions of the air molecules (i.e. nitrogen and oxygen molecules, as well as a very small number of carbon dioxide molecules and argon atoms) against the surface in question. Keep in mind that a net force on an object causes that object to accelerate (or decelerate). If the net force acts in the direction of the object's motion, it accelerates the object; acting in the direction opposite the object's motion, it decelerates the object.
Daniel Bernoulli was born in the Dutch Republic (now known as The Netherlands) in the year 1700. He's perhaps best known for discovering a relationship between the pressure, velocity (speed in a certain direction), and height (above some arbitrary reference level) of an incompressible fluid in perfect steady-state flow. Water being pumped through a pipe can fit this description. It's virtually incompressible, and the pump can keep it moving at a steady rate through the pipe. Air, while not incompressible, is close enough to an incompressible fluid in steady-state flow under certain conditions (velocity less than 300 km/h and no pressure differences of more than one tenth of an atmosphere) that we can use Bernoulli's equation to understand its behavior. So what's the relationship? For a fluid as described above, the pressure, plus one-half times the density times the velocity squared, plus the density times the acceleration due to gravity times the height above some arbitrary reference level, is constant. In equation form, P + 1/2 dv2 + dgh = constant. So what happens if I increase the velocity (v) of the fluid? Either the pressure (P) must decrease or the height (h) must decrease, so that the left side of the equation remains equal to the constant. You should take from this equation the following: for an incompressible, steady-state flow liquid, of a particular density (d), and at a set height (that doesn't change), pressure and velocity always move in opposite directions. If pressure decreases, velocity increases. If pressure increases, velocity decreases.
When the path of a fluid in steady-state flow bends, the pressure on the outside of the bend is always higher than the pressure on the inside of the bend. It's this pressure imbalance that causes the fluid to bend. This pressure change indicates a change in the fluid's velocity. So does the fluid on the outside of the bend speed up or slow down? It slows down. And the fluid on the inside of the bend? It speeds up, of course.
When a ball is hurtling through the air, the air it encounters is forced to flow around it. Some of the air flows over the top of the ball, some flows beneath the ball, and some air flows around each side. Air pressure above, beneath, and aside the ball is not everywhere the same. As the air encounters the front of the ball, it bends away from the ball, moving out of the way. (The ball is on the outside of the bend.) This creates a high-pressure zone in front of the ball. And the air here slows down. The air then curves back towards the ball, on all sides of the ball, hugging its surface as it moves towards the back of the ball. This puts the ball on the inside of many curved paths (or bends). Therefore, the air around the ball's middle is at low pressure and high speed. As the air reaches the back of the ball, it peels away from the ball and straightens back out. This bending of the air away from the ball creates a high-pressure zone behind the ball. Low speed air. Now you ask, how can the low-pressure air along the sides of the ball move into the high-pressure zone behind the ball? Doesn't air always move from a high-pressure zone into a low-pressure zone? Normally, yes. Here, the low-pressure air is definitely moving against the tide, so to speak. It's fighting its way into the high-pressure zone, slowing down (decelerating) as the high-pressure air pushes on it. But it has enough energy to successfully make the trip. It does reach the back of the ball. Now, these pressure imbalances are symmetric about the ball; they balance one another and produce no net force on the ball. They don't accelerate or decelerate the ball itself. Air resistance does exist, but it's a result of air near the ball's surface rubbing against the surface, producing a type of friction. Viscous drag, it's called. The air resistance is not a result of the pressure variations just described. Okay, now for a qualifier! The behavior of the air about the ball, as described in this paragraph, applies to balls traveling at slow speeds. This is important. The air behaves differently when it encounters a ball moving at high speed.
To describe the path of air flowing around a fast-moving ball, I must introduce the term boundary layer. A thin layer of air moving very close to the surface of the ball is called the boundary layer, and it behaves differently from air farther from the surface. It moves more slowly and has less total energy than the freely flowing air farther out. Why? Because friction with the ball's surface (i.e. viscous drag) slows it down and robs it of energy.
Hmmm. So you're thinking, it's hard for the air along the sides of the ball to push into the high-pressure zone behind the ball. Okay. But it sounds like it can do it anyways. Guess it has enough energy to do so. And that boundary layer. It has less energy than the air just a bit farther out. But, well, it seems that it, too, is able to push into the high-pressure zone. At least when the ball is moving slowly. (Good. You're right so far.) And so does this change when the ball is moving rapidly? Yes. When the ball is moving rapidly, this lower-energy boundary layer of air is no longer able to push into the high-pressure zone behind the ball. In fact, it is pushed back towards the sides of the ball by the adverse pressure gradient, cutting like a wedge between the ball and the freely flowing air outside this boundary layer. No longer does the air curve around behind the ball. This leaves us with an air pocket behind the ball; a turbulent wake, in other words. In this wake, the air pressure is roughly atmospheric. There goes the symmetry of pressure forces on the ball. Now there is no high-pressure zone behind the ball to cancel the high-pressure zone in front of the ball. There is a large pressure drag, a force on the ball in the direction of downwind, slowing the ball down. Decelerating it. This pressure drag is what limits the range of a smooth golf ball. Yes, there is also viscous drag, but it's not nearly as significant as the large pressure drag caused by the turbulent wake.
(turbulent wake behind ball, which is moving to the left)
So dimpled golf balls travel farther than smooth golf balls. Do the dimples somehow reduce the size (and severity) of this turbulent wake, reducing the pressure drag on the ball, preventing the ball from slowing so much as it arcs through the air? Yes, indeed. The dimples, or surface irregularities, cause the air in the boundary layer to tumble about. This tumbling about gives the boundary-layer air more energy, and more forward momentum. It now has a much better chance of pushing around to the back side of the ball, into the high-pressure zone. Alas, it still doesn't make it, but it comes much closer. It travels partially around the back of the ball before its progress is stopped and it separates from the surface. The air outside the boundary layer, following along, hugs the ball for a longer time, as well. It separates from the ball at the same spot where the boundary layer separates, this being a fair ways down the backside of the ball. The result is a smaller air pocket. A small turbulent wake. A less dramatic variation in air pressure between the front of the ball and the back of the ball. A more modest force of pressure drag. And this reduction in pressure drag is what enables the dimpled ball to soar some 200 yards farther than a smooth ball.
First, air pressure is the force exerted by air molecules divided by the area on which the force is exerted. That is, force per unit area. The force comes from the countless collisions of the air molecules (i.e. nitrogen and oxygen molecules, as well as a very small number of carbon dioxide molecules and argon atoms) against the surface in question. Keep in mind that a net force on an object causes that object to accelerate (or decelerate). If the net force acts in the direction of the object's motion, it accelerates the object; acting in the direction opposite the object's motion, it decelerates the object.
Daniel Bernoulli was born in the Dutch Republic (now known as The Netherlands) in the year 1700. He's perhaps best known for discovering a relationship between the pressure, velocity (speed in a certain direction), and height (above some arbitrary reference level) of an incompressible fluid in perfect steady-state flow. Water being pumped through a pipe can fit this description. It's virtually incompressible, and the pump can keep it moving at a steady rate through the pipe. Air, while not incompressible, is close enough to an incompressible fluid in steady-state flow under certain conditions (velocity less than 300 km/h and no pressure differences of more than one tenth of an atmosphere) that we can use Bernoulli's equation to understand its behavior. So what's the relationship? For a fluid as described above, the pressure, plus one-half times the density times the velocity squared, plus the density times the acceleration due to gravity times the height above some arbitrary reference level, is constant. In equation form, P + 1/2 dv2 + dgh = constant. So what happens if I increase the velocity (v) of the fluid? Either the pressure (P) must decrease or the height (h) must decrease, so that the left side of the equation remains equal to the constant. You should take from this equation the following: for an incompressible, steady-state flow liquid, of a particular density (d), and at a set height (that doesn't change), pressure and velocity always move in opposite directions. If pressure decreases, velocity increases. If pressure increases, velocity decreases.
When the path of a fluid in steady-state flow bends, the pressure on the outside of the bend is always higher than the pressure on the inside of the bend. It's this pressure imbalance that causes the fluid to bend. This pressure change indicates a change in the fluid's velocity. So does the fluid on the outside of the bend speed up or slow down? It slows down. And the fluid on the inside of the bend? It speeds up, of course.
When a ball is hurtling through the air, the air it encounters is forced to flow around it. Some of the air flows over the top of the ball, some flows beneath the ball, and some air flows around each side. Air pressure above, beneath, and aside the ball is not everywhere the same. As the air encounters the front of the ball, it bends away from the ball, moving out of the way. (The ball is on the outside of the bend.) This creates a high-pressure zone in front of the ball. And the air here slows down. The air then curves back towards the ball, on all sides of the ball, hugging its surface as it moves towards the back of the ball. This puts the ball on the inside of many curved paths (or bends). Therefore, the air around the ball's middle is at low pressure and high speed. As the air reaches the back of the ball, it peels away from the ball and straightens back out. This bending of the air away from the ball creates a high-pressure zone behind the ball. Low speed air. Now you ask, how can the low-pressure air along the sides of the ball move into the high-pressure zone behind the ball? Doesn't air always move from a high-pressure zone into a low-pressure zone? Normally, yes. Here, the low-pressure air is definitely moving against the tide, so to speak. It's fighting its way into the high-pressure zone, slowing down (decelerating) as the high-pressure air pushes on it. But it has enough energy to successfully make the trip. It does reach the back of the ball. Now, these pressure imbalances are symmetric about the ball; they balance one another and produce no net force on the ball. They don't accelerate or decelerate the ball itself. Air resistance does exist, but it's a result of air near the ball's surface rubbing against the surface, producing a type of friction. Viscous drag, it's called. The air resistance is not a result of the pressure variations just described. Okay, now for a qualifier! The behavior of the air about the ball, as described in this paragraph, applies to balls traveling at slow speeds. This is important. The air behaves differently when it encounters a ball moving at high speed.
To describe the path of air flowing around a fast-moving ball, I must introduce the term boundary layer. A thin layer of air moving very close to the surface of the ball is called the boundary layer, and it behaves differently from air farther from the surface. It moves more slowly and has less total energy than the freely flowing air farther out. Why? Because friction with the ball's surface (i.e. viscous drag) slows it down and robs it of energy.
Hmmm. So you're thinking, it's hard for the air along the sides of the ball to push into the high-pressure zone behind the ball. Okay. But it sounds like it can do it anyways. Guess it has enough energy to do so. And that boundary layer. It has less energy than the air just a bit farther out. But, well, it seems that it, too, is able to push into the high-pressure zone. At least when the ball is moving slowly. (Good. You're right so far.) And so does this change when the ball is moving rapidly? Yes. When the ball is moving rapidly, this lower-energy boundary layer of air is no longer able to push into the high-pressure zone behind the ball. In fact, it is pushed back towards the sides of the ball by the adverse pressure gradient, cutting like a wedge between the ball and the freely flowing air outside this boundary layer. No longer does the air curve around behind the ball. This leaves us with an air pocket behind the ball; a turbulent wake, in other words. In this wake, the air pressure is roughly atmospheric. There goes the symmetry of pressure forces on the ball. Now there is no high-pressure zone behind the ball to cancel the high-pressure zone in front of the ball. There is a large pressure drag, a force on the ball in the direction of downwind, slowing the ball down. Decelerating it. This pressure drag is what limits the range of a smooth golf ball. Yes, there is also viscous drag, but it's not nearly as significant as the large pressure drag caused by the turbulent wake.
(turbulent wake behind ball, which is moving to the left)So dimpled golf balls travel farther than smooth golf balls. Do the dimples somehow reduce the size (and severity) of this turbulent wake, reducing the pressure drag on the ball, preventing the ball from slowing so much as it arcs through the air? Yes, indeed. The dimples, or surface irregularities, cause the air in the boundary layer to tumble about. This tumbling about gives the boundary-layer air more energy, and more forward momentum. It now has a much better chance of pushing around to the back side of the ball, into the high-pressure zone. Alas, it still doesn't make it, but it comes much closer. It travels partially around the back of the ball before its progress is stopped and it separates from the surface. The air outside the boundary layer, following along, hugs the ball for a longer time, as well. It separates from the ball at the same spot where the boundary layer separates, this being a fair ways down the backside of the ball. The result is a smaller air pocket. A small turbulent wake. A less dramatic variation in air pressure between the front of the ball and the back of the ball. A more modest force of pressure drag. And this reduction in pressure drag is what enables the dimpled ball to soar some 200 yards farther than a smooth ball.
Wednesday, October 14, 2009
Superconductivity
When Thomas Edison went about providing electric power to New York City in the late 1800s, he knew that energy was dissipated as heat in the wires that delivered electric current to his customers. This reduced the amount of power that made it to the homes of his customers, and presented Edison with the problem of trying to minimize this power loss. (I talked about this in a previous blog entry.)
The issue faced by Edison was one of electrical resistance, which is a measure of the degree to which an object opposes an electric current through it. When current flows through an object with resistance, electrical energy is converted to heat at a rate equal to the square of the current times the resistance. This rate is a measure of power loss.
While Edison had means to lessen this loss of power, he couldn't escape it completely. That's because conductors (i.e. materials that conduct electricity) naturally heat up as an electric current moves through them. The electrons that comprise this current, as they snake forward through the material, are constantly bumping into the atoms (ions) of the conductor. At each collision, an electron loses a bit of kinetic energy to an ion, increasing the kinetic energy of the ion, generating heat and increasing the temperature of the conductor. While conductors exhibit less resistance at lower temperatures, ordinary conductors can never be cooled enough to achieve zero resistance.
It was in 1911 that a scientist, Heike Kamerlingh Onnes, discovered that certain unordinary conductors, under certain conditions, do possess zero electrical resistance. That is, passing an electric current through these materials does not result in the heating of the materials and, therefore, no power is lost in them. The reason why no one had seen such behavior before: it only takes place in certain materials, and these materials have to be unimaginably cold. It was only just prior to 1911 that such cold temperatures were achieved in the laboratory (by Onnes). Onnes had taken helium gas and got it so cold (down to 4.2 degrees above absolute zero) that it condensed into a liquid. Using this liquid helium as a refrigerant, he tested the electrical resistance of mercury and was amazed to find that it actually dropped to ZERO! Such behavior was a completely new phenomenon, never before witnessed. Onnes labeled it "superconductivity."
Onnes didn't understand what was going on inside the superconducting material. How could the electrons avoid bumping into the material's ions, passing kinetic energy to them? Why did such behavior occur only below a certain temperature, labeled the critical temperature? Twenty-two years later, in 1933, the answer was still unknown. But in this year, Walter Meissner and Robert Ochsenfeld made an important new discovery about superconducting materials (which, as a class, had expanded to include materials other than mercury). They found that superconductors expelled applied magnetic fields. Magnetic field lines that passed through a sample of material were, in a sense, pushed out of the material (or more accurately, cancelled within the material) when the material was cooled below its critical temperature. This finding, now known as the Meissner effect, provided evidence that superconductivity was, most fundamentally, a magnetic phenomenon. Such a finding also changed the mindset that the fundamental property of a superconductor was zero resistance.
A theory explaining the phenomenon of superconductivity was proposed in 1957 by John Bardeen, Leon Cooper, and Robert Schrieffer. It became known as the BCS Theory, after their initials. It had to do with phonons (not photons) and Cooper pairs. Phonons are quantized crystal lattice vibrations. What does this mean? Certain materials exist as crystals, which means "the constituent atoms, molecules or ions [which are atoms or molecules with a net electric charge] are packed in a regularly ordered, repeating pattern in all three spatial dimensions." (Wikipedia) The graphic below is an example of a unit cell, which is periodically repeated in three dimensions to form a crystal. Each sphere represents an atom and the tubes represent bonds between atoms.
A lattice is a sort of framework upon which, at each point, there exists a unit cell like you see pictured above. So the crystal looks the same when viewed from any lattice point. As an electron moves through a crystal, it exerts a force (i.e. it pulls) on the positively charged lattice ions, distorting them towards its (the electron's) path. As the electron then moves away from that point on the lattice, the lattice ions return to their original position. Because all atoms in a crystal are connected, "the displacement of one or more atoms from their equilibrium positions will give rise to a set of vibration waves propagating through the lattice." (Wikipedia) Finally, these vibration waves are quantized, which means they can't possess just any amount of energy but only certain discrete numerical values.
What happens as an electron moves through a crystal, generating a phonon? Let's picture an electric current flowing through the material. One electron after another. An electron zips past a point in the crystal lattice, distorting the lattice through the creation of a phonon. The lattice is pulled inward towards the negatively-charged electron, but the electron quickly moves away, faster than the lattice can relax back to its original position. This creates a region of positive charge, as the lattice ions that are pulled inward are positively charged. Here's the cool part. A second electron can be attracted to the region of positive charge along the path of the first electron. And these two electrons, which would normally repel one another (because they are both negatively charged), can become bound to one another. "If this binding energy is higher than the energy provided by kicks from oscillating atoms in the conductor (which is true at low temperatures), then the electron pair will stick together and resist all kicks, thus not experiencing resistance." (Wikipedia) These electron pairs are called Cooper pairs, and they lie at the heart of the BCS Theory. They are what allow for superconductivity; they carry the superconducting current. But, as noted just above, the temperature has to be low. Above a critical temperature, the atoms in the crystal are jostling around too much, bumping into the electron pairs with enough force to knock them apart. This breaking apart of the Cooper pairs destroys superconductivity in the material, and the material becomes "normal." What's the highest temperature at which a known material will superconduct? A special ceramic material comprised of many different atoms has been observed to superconduct at -135 degrees C. Notice the negative sign. The holy grail of those working in the field is to find a material that superconducts at room temperature. (Obviously, no material yet identified would have helped Edison ... although there are techniques, which I addressed in a previous blog entry, that lessen the problem.)
The material that superconducts at -135 degrees C (or 138 K), like all materials that superconduct above around -243 degrees C (or 30 K), is called a "high-temperature" superconductor. This is obviously a relative term. Such materials are not consistent with the BCS Theory and there is no good theory to describe how these high-temperature superconductors work.
The issue faced by Edison was one of electrical resistance, which is a measure of the degree to which an object opposes an electric current through it. When current flows through an object with resistance, electrical energy is converted to heat at a rate equal to the square of the current times the resistance. This rate is a measure of power loss.
While Edison had means to lessen this loss of power, he couldn't escape it completely. That's because conductors (i.e. materials that conduct electricity) naturally heat up as an electric current moves through them. The electrons that comprise this current, as they snake forward through the material, are constantly bumping into the atoms (ions) of the conductor. At each collision, an electron loses a bit of kinetic energy to an ion, increasing the kinetic energy of the ion, generating heat and increasing the temperature of the conductor. While conductors exhibit less resistance at lower temperatures, ordinary conductors can never be cooled enough to achieve zero resistance.
It was in 1911 that a scientist, Heike Kamerlingh Onnes, discovered that certain unordinary conductors, under certain conditions, do possess zero electrical resistance. That is, passing an electric current through these materials does not result in the heating of the materials and, therefore, no power is lost in them. The reason why no one had seen such behavior before: it only takes place in certain materials, and these materials have to be unimaginably cold. It was only just prior to 1911 that such cold temperatures were achieved in the laboratory (by Onnes). Onnes had taken helium gas and got it so cold (down to 4.2 degrees above absolute zero) that it condensed into a liquid. Using this liquid helium as a refrigerant, he tested the electrical resistance of mercury and was amazed to find that it actually dropped to ZERO! Such behavior was a completely new phenomenon, never before witnessed. Onnes labeled it "superconductivity."
Onnes didn't understand what was going on inside the superconducting material. How could the electrons avoid bumping into the material's ions, passing kinetic energy to them? Why did such behavior occur only below a certain temperature, labeled the critical temperature? Twenty-two years later, in 1933, the answer was still unknown. But in this year, Walter Meissner and Robert Ochsenfeld made an important new discovery about superconducting materials (which, as a class, had expanded to include materials other than mercury). They found that superconductors expelled applied magnetic fields. Magnetic field lines that passed through a sample of material were, in a sense, pushed out of the material (or more accurately, cancelled within the material) when the material was cooled below its critical temperature. This finding, now known as the Meissner effect, provided evidence that superconductivity was, most fundamentally, a magnetic phenomenon. Such a finding also changed the mindset that the fundamental property of a superconductor was zero resistance.
A theory explaining the phenomenon of superconductivity was proposed in 1957 by John Bardeen, Leon Cooper, and Robert Schrieffer. It became known as the BCS Theory, after their initials. It had to do with phonons (not photons) and Cooper pairs. Phonons are quantized crystal lattice vibrations. What does this mean? Certain materials exist as crystals, which means "the constituent atoms, molecules or ions [which are atoms or molecules with a net electric charge] are packed in a regularly ordered, repeating pattern in all three spatial dimensions." (Wikipedia) The graphic below is an example of a unit cell, which is periodically repeated in three dimensions to form a crystal. Each sphere represents an atom and the tubes represent bonds between atoms.
A lattice is a sort of framework upon which, at each point, there exists a unit cell like you see pictured above. So the crystal looks the same when viewed from any lattice point. As an electron moves through a crystal, it exerts a force (i.e. it pulls) on the positively charged lattice ions, distorting them towards its (the electron's) path. As the electron then moves away from that point on the lattice, the lattice ions return to their original position. Because all atoms in a crystal are connected, "the displacement of one or more atoms from their equilibrium positions will give rise to a set of vibration waves propagating through the lattice." (Wikipedia) Finally, these vibration waves are quantized, which means they can't possess just any amount of energy but only certain discrete numerical values.
What happens as an electron moves through a crystal, generating a phonon? Let's picture an electric current flowing through the material. One electron after another. An electron zips past a point in the crystal lattice, distorting the lattice through the creation of a phonon. The lattice is pulled inward towards the negatively-charged electron, but the electron quickly moves away, faster than the lattice can relax back to its original position. This creates a region of positive charge, as the lattice ions that are pulled inward are positively charged. Here's the cool part. A second electron can be attracted to the region of positive charge along the path of the first electron. And these two electrons, which would normally repel one another (because they are both negatively charged), can become bound to one another. "If this binding energy is higher than the energy provided by kicks from oscillating atoms in the conductor (which is true at low temperatures), then the electron pair will stick together and resist all kicks, thus not experiencing resistance." (Wikipedia) These electron pairs are called Cooper pairs, and they lie at the heart of the BCS Theory. They are what allow for superconductivity; they carry the superconducting current. But, as noted just above, the temperature has to be low. Above a critical temperature, the atoms in the crystal are jostling around too much, bumping into the electron pairs with enough force to knock them apart. This breaking apart of the Cooper pairs destroys superconductivity in the material, and the material becomes "normal." What's the highest temperature at which a known material will superconduct? A special ceramic material comprised of many different atoms has been observed to superconduct at -135 degrees C. Notice the negative sign. The holy grail of those working in the field is to find a material that superconducts at room temperature. (Obviously, no material yet identified would have helped Edison ... although there are techniques, which I addressed in a previous blog entry, that lessen the problem.)
The material that superconducts at -135 degrees C (or 138 K), like all materials that superconduct above around -243 degrees C (or 30 K), is called a "high-temperature" superconductor. This is obviously a relative term. Such materials are not consistent with the BCS Theory and there is no good theory to describe how these high-temperature superconductors work.
Sunday, October 4, 2009
Blackbodies
What is a blackbody?
A blackbody is an idealized type of object that absorbs ALL electromagnetic radiation that falls on it. It therefore reflects no light. Now let's hold that the blackbody is in thermal equilibrium with its surroundings. (This means that it's at the same temperature as its surroundings.) With a bit of physics background, it becomes apparent that the object must not only absorb all radiation incident upon it (which is what makes it a blackbody) but it must also emit radiation at an equal rate, otherwise the net inflow or outflow of radiation would cause its temperature to change. (It should seem reasonable that radiation incident on an object can alter its temperature ... think of a microwave oven.) This emission of radiation may be in the form of visible light, so we acknowledge that even though no light is reflected from the blackbody, the body may still give off light. In other words, it may not actually be black in color.
(If you're familiar with the term "electromagnetic radiation" and you know what wavelength is, you can skip this paragraph.) Electromagnetic radiation is the collective term for radiation in its many guises. Microwaves, radio waves, visible light, X-rays. These are all examples of radiation, which can be viewed as a wave, with electric and magnetic components, propagating through space, carrying along energy. Waves can vary from one another in various ways, with a notable example being wavelength, or the distance between two adjacent crests or troughs. (More precisely, wavelength is the distance between adjacent maximums in the oscillating electric field.) Frequency, which is inversely proportional to wavelength, is another distinguishing characteristic of waves. It's a measure of the rate of oscillation of the wave. As a wave passes through a point in space, the shorter the wavelength, the more rapidly crests (or troughs) pass through that point. And vice-versa. This inverse proportionality holds, by the way, because radiation travels at a constant speed -- namely, the speed of light. In what wavelengths can radiation come? Any and all. A millionth of a centimeter, or two centimeters, or two meters, or bigger or smaller. People have arbitrarily divided this wavelength continuum into sections and given names to the different regions. Radiation that has a wavelength anywhere between 1 mm and 10 cm is called microwave radiation. Radiation with a wavelength between 400 and 700 nanometers (or one billionth of a meter) is called visible light. And, within the range of visible light, wavelength furthermore determines the color. As an example, light at 500 nm is green. You get the idea.
Funny thing about a blackbody in thermal equilibrium: it will emit a specific radiation spectrum (or a specific distribution of energy spread over all the possible wavelengths of radiation) that is characteristic NOT of the shape or size of the object, or even of what it is made of, but ONLY of the temperature of the object. Therefore, any two objects at 300 K (I'm using the Kelvin temperature scale, where the temperature in Kelvin is always 273.15 degrees higher than what it is in Celsius) will have the same radiation spectrum, and two objects at 3000 K will also share the same radiation spectrum, though one that is different than that shared by the objects at 300 K. (The radiation spectrum is generally referred to as a thermal spectrum, but I'll stick with the first term.) Here's a graph of spectrums at 4 different temperatures. Each has a peak, with a rather sharp tapering of the shorter-wavelength side and a more gradual tapering of the longer-wavelength side.
Quick question: How hot should the filament in a light bulb be, so that is will produce the same "white" light spectrum produced by the Sun? Answer: the same temperature as the surface of the Sun, which is about 5800 K. From the graph above, we can see that something with this temperature (or 6000 K) produces most of its radiation in and around the visible portion of the spectrum but also produces X-rays and microwaves and other types of waves in lesser quantities. And because the spectrum peak at about 5800 K is in the middle of the visible region, we get from the Sun fairly equal amounts of all different colors of visible light. The colors of the rainbow (in roughly equal amounts) blend to form a nice "white" light. (But, you object, the Sun is yellow! Actually, it only looks yellow from the surface of the Earth because of the distorting effects of the atmosphere.) What would we see, however, if the Sun's surface was only 3000 K? Well, the Sun would then emit a lot more red light (which corresponds to the right side of the visible band) than blue light (which is nearer the left side of the band), and sunlight would have a reddish hue (though I don't think it would be very obvious to the naked eye). No doubt you've seen something glowing "red hot", like, say, the heating element on the stove. The color is an indication that the stove is hot enough to produce a radiation spectrum that has a sufficient bit of energy allocated to the visible red region but little or no energy allocated to the other, shorter-wavelength colors of light, which would make the stove more orange or even white in appearance, depending on its exact temperature. For the same stove, guess which section of the radiation spectrum is best represented, so to speak. That would be infrared radiation, which our bodies perceive as heat. (A hot stove burner that is dull red in color is about 800 K; if you can raise the temperature high enough, it will turn orange at about 1150 K.)
(Another graph to look at.)
Normal incandescent bulbs don't get close to the temperature of the Sun, so they fall short of producing the same pleasant white light that emanates from our star. These bulbs contain tungsten filaments that reach temperatures of about 2500 K. So the light coming from them is redder than that of the Sun. Tungsten has the highest melting point of all metals. So to better mimic the color spectrum of the Sun in a light bulb, we can't try to heat a metal filament to 5800 K. It would simply melt well before reaching that high temperature. We must turn to a bulb that produces light not by getting hot, but by a different mechanism. Fluorescent bulbs are a case in point.
Is a blackbody black in color? It can be, but it doesn't have to be, as we saw in the first paragraph. The Sun is very nearly a blackbody, meaning it absorbs very nearly all radiation incident upon it. It also remains fairly constant in temperature because, even though its emitting a lot of radiation, its creating more of it deep within its core. And it's most definitely not black in color. Actually, to a reasonable approximation, all matter in thermal equilibrium behaves like a blackbody. A book does. A car does. Even a person does. As an example, the actress Halle Berry is a blackbody. Now, if she were a true ideal blackbody, she wouldn't reflect light (and she wouldn't have made the movie Catwoman), so we wouldn't be able to see her in color. She would appear pitch black. She would absorb all light and emit a radiation spectrum characteristic of something at 98.6 F (or 310 K). But she, and all people, approximate blackbodies. We reflect some of the light falling on us, which makes us visible, and absorb the rest. And we all emit a radiation spectrum that peaks in the middle of the infrared region of the spectrum (like the stove), producing such little visible light (as well as certain other wavelengths) that we don't shine in a dark room (unlike the stove, which is hotter). Want to be able to "see" someone that isn't reflecting visible light? Try infrared goggles. These pick up the infrared radiation produced by the person (and other warm things around the person) and convert it to visible light that your eyes can detect.
A blackbody is an idealized type of object that absorbs ALL electromagnetic radiation that falls on it. It therefore reflects no light. Now let's hold that the blackbody is in thermal equilibrium with its surroundings. (This means that it's at the same temperature as its surroundings.) With a bit of physics background, it becomes apparent that the object must not only absorb all radiation incident upon it (which is what makes it a blackbody) but it must also emit radiation at an equal rate, otherwise the net inflow or outflow of radiation would cause its temperature to change. (It should seem reasonable that radiation incident on an object can alter its temperature ... think of a microwave oven.) This emission of radiation may be in the form of visible light, so we acknowledge that even though no light is reflected from the blackbody, the body may still give off light. In other words, it may not actually be black in color.
(If you're familiar with the term "electromagnetic radiation" and you know what wavelength is, you can skip this paragraph.) Electromagnetic radiation is the collective term for radiation in its many guises. Microwaves, radio waves, visible light, X-rays. These are all examples of radiation, which can be viewed as a wave, with electric and magnetic components, propagating through space, carrying along energy. Waves can vary from one another in various ways, with a notable example being wavelength, or the distance between two adjacent crests or troughs. (More precisely, wavelength is the distance between adjacent maximums in the oscillating electric field.) Frequency, which is inversely proportional to wavelength, is another distinguishing characteristic of waves. It's a measure of the rate of oscillation of the wave. As a wave passes through a point in space, the shorter the wavelength, the more rapidly crests (or troughs) pass through that point. And vice-versa. This inverse proportionality holds, by the way, because radiation travels at a constant speed -- namely, the speed of light. In what wavelengths can radiation come? Any and all. A millionth of a centimeter, or two centimeters, or two meters, or bigger or smaller. People have arbitrarily divided this wavelength continuum into sections and given names to the different regions. Radiation that has a wavelength anywhere between 1 mm and 10 cm is called microwave radiation. Radiation with a wavelength between 400 and 700 nanometers (or one billionth of a meter) is called visible light. And, within the range of visible light, wavelength furthermore determines the color. As an example, light at 500 nm is green. You get the idea.
Funny thing about a blackbody in thermal equilibrium: it will emit a specific radiation spectrum (or a specific distribution of energy spread over all the possible wavelengths of radiation) that is characteristic NOT of the shape or size of the object, or even of what it is made of, but ONLY of the temperature of the object. Therefore, any two objects at 300 K (I'm using the Kelvin temperature scale, where the temperature in Kelvin is always 273.15 degrees higher than what it is in Celsius) will have the same radiation spectrum, and two objects at 3000 K will also share the same radiation spectrum, though one that is different than that shared by the objects at 300 K. (The radiation spectrum is generally referred to as a thermal spectrum, but I'll stick with the first term.) Here's a graph of spectrums at 4 different temperatures. Each has a peak, with a rather sharp tapering of the shorter-wavelength side and a more gradual tapering of the longer-wavelength side.
Quick question: How hot should the filament in a light bulb be, so that is will produce the same "white" light spectrum produced by the Sun? Answer: the same temperature as the surface of the Sun, which is about 5800 K. From the graph above, we can see that something with this temperature (or 6000 K) produces most of its radiation in and around the visible portion of the spectrum but also produces X-rays and microwaves and other types of waves in lesser quantities. And because the spectrum peak at about 5800 K is in the middle of the visible region, we get from the Sun fairly equal amounts of all different colors of visible light. The colors of the rainbow (in roughly equal amounts) blend to form a nice "white" light. (But, you object, the Sun is yellow! Actually, it only looks yellow from the surface of the Earth because of the distorting effects of the atmosphere.) What would we see, however, if the Sun's surface was only 3000 K? Well, the Sun would then emit a lot more red light (which corresponds to the right side of the visible band) than blue light (which is nearer the left side of the band), and sunlight would have a reddish hue (though I don't think it would be very obvious to the naked eye). No doubt you've seen something glowing "red hot", like, say, the heating element on the stove. The color is an indication that the stove is hot enough to produce a radiation spectrum that has a sufficient bit of energy allocated to the visible red region but little or no energy allocated to the other, shorter-wavelength colors of light, which would make the stove more orange or even white in appearance, depending on its exact temperature. For the same stove, guess which section of the radiation spectrum is best represented, so to speak. That would be infrared radiation, which our bodies perceive as heat. (A hot stove burner that is dull red in color is about 800 K; if you can raise the temperature high enough, it will turn orange at about 1150 K.)
(Another graph to look at.)
Normal incandescent bulbs don't get close to the temperature of the Sun, so they fall short of producing the same pleasant white light that emanates from our star. These bulbs contain tungsten filaments that reach temperatures of about 2500 K. So the light coming from them is redder than that of the Sun. Tungsten has the highest melting point of all metals. So to better mimic the color spectrum of the Sun in a light bulb, we can't try to heat a metal filament to 5800 K. It would simply melt well before reaching that high temperature. We must turn to a bulb that produces light not by getting hot, but by a different mechanism. Fluorescent bulbs are a case in point.
Is a blackbody black in color? It can be, but it doesn't have to be, as we saw in the first paragraph. The Sun is very nearly a blackbody, meaning it absorbs very nearly all radiation incident upon it. It also remains fairly constant in temperature because, even though its emitting a lot of radiation, its creating more of it deep within its core. And it's most definitely not black in color. Actually, to a reasonable approximation, all matter in thermal equilibrium behaves like a blackbody. A book does. A car does. Even a person does. As an example, the actress Halle Berry is a blackbody. Now, if she were a true ideal blackbody, she wouldn't reflect light (and she wouldn't have made the movie Catwoman), so we wouldn't be able to see her in color. She would appear pitch black. She would absorb all light and emit a radiation spectrum characteristic of something at 98.6 F (or 310 K). But she, and all people, approximate blackbodies. We reflect some of the light falling on us, which makes us visible, and absorb the rest. And we all emit a radiation spectrum that peaks in the middle of the infrared region of the spectrum (like the stove), producing such little visible light (as well as certain other wavelengths) that we don't shine in a dark room (unlike the stove, which is hotter). Want to be able to "see" someone that isn't reflecting visible light? Try infrared goggles. These pick up the infrared radiation produced by the person (and other warm things around the person) and convert it to visible light that your eyes can detect.
Monday, September 21, 2009
A Bit on Work: Part III of III
Hold a permanent magnet above a paper clip and the paper clip "jumps" up to the magnet. How does this happen given that the magnetic field is not doing any work on the paper clip?
First off, the paper clip is made of steel, which contains a large amount of iron. Iron is a ferromagnetic material, meaning it can become magnetized when placed in a magnetic field and remain magnetized when removed from that field. Non ferromagnetic materials, on the other hand, would lose their magnetization upon removal of the external magnetic field. (In this case, we're not removing the magnetic field, but it's still nice that we're working with a ferromagnetic material. You'll see why later.) When we place a permanent magnet above a paper clip, the magnetic field produced by the magnet induces magnetism in the paper clip by applying a torque to the magnetic dipoles in the iron, lining them up.
What's a magnetic dipole? A small current loop (say, electrons flowing around a tiny loop of wire) is, more or less, a magnetic dipole. We call the small current loop a magnetic dipole because it produces a magnetic field, at some distance, that is strictly "dipolar" in nature. Not all systems produce a magnetic field that is dipolar in nature. Some systems produce a field that is not dipolar at all, but perhaps "quadrupolar" or "octopolar." Other systems might produce fields that are largely dipolar but a little quadrupolar, too. What does it mean for the field to be strictly "dipolar" in nature? It means that, as you move away from the system, the strength of the magnetic field drops off as one over the distance cubed. There's no component of the field that drops off as one over the distance or one over the distance to the fourth power, etc. Not too many systems can actually produce a field that is strictly dipolar. It's easy to produce one that is largely dipolar, but quite difficult to produce one that is strictly dipolar. A tiny current loop does the trick, however. But it has to be really tiny, as in infinitesimally small. How do you make such a thing in the lab? You don't. Luckily, I suppose, they already exist in nature as electrons whizzing about nuclei inside of atoms. (No wires necessary.) Previously, I said that the magnetic field of a permanent magnet applies a torque to the magnetic dipoles in a sample of iron, lining them up. Now it should be fairly clear that it is atomic electrons (acting in their capacity as magnetic dipoles) that are doing the lining up. Note: not every electron in a sample of iron experiences this torque. Only the unpaired electrons do. (Each iron atom has 4 unpaired electrons.)
OK, so how can an electron point in a particular direction? Really, it can't. What I mean is that this whirrling electron, this magnetic dipole, points its magnetic dipole moment in a particular direction. This so-called dipole moment is a property of the dipole and can act to represent the physical dipole. It's a vector (with, of course, a magnitude and a direction) that quantifies the contribution of a system's internal magnetism to the external dipolar magnetic field produced by the system. That is, a measure of how much what's going on inside the system is effecting the magnetic field observed outside the system. The moment may be non-physical, but it often proves useful to picture an electron as a little vector when doing calculations or thinking through problems like the one we're addressing here. Therefore, to say that dipoles are lined up is to say that their dipole moments are lined up (or parallel to one another).
Magnets come in different strengths, which we quantify through the concept of magnetization. Something with a large magnetization is both strongly affected by external magnetic fields and the source of its own strong magnetic field. We define magnetization as the amount of magnetic dipole moment per unit volume. Therefore, given a unit volume, we perform a vector sum of all the little moments (vectors) in that volume, and we see how strong our magnet is. Two vectors of equal magnitude pointing in opposite directions sum to zero. Likewise, a large number of arbitrarily directed vectors also sums to zero. This explains why the paper clip, before being magnetized by the permanent magnet, isn't magnetic. It has plenty of little moments (or vectors) inside, but they are arbitrarily directed (well, sorta) and so the net sum of these moments, per unit volume, is pretty close to zero. Once the permanent magnet acts on the dipole moments in the paper clip and lines them up, the vector sum no longer equals zero. Rather, it is now rather large, and the paper clip now has a large magnetization and acts outwardly like a magnet.
Electrons, bound to atoms, move in two ways. This leads to two magnetic dipoles or, better put, two contributions to a single magnetic dipole. (It's a simple vector sum.) Firstly, an electron "orbits" the nucleus. Even though it's not accurate, people often picture this motion as being like a planet orbiting the sun. That's a good way to think of it at the present. Secondly, an electron "spins." You might think of this as an electron spinning about its own axis, just as the Earth spins about its axis once every 24 hours. But this is a rather horrible and misleading analogy, because this "spin" is not really a physical rotation about an axis. The electron is a point particle with no physical size, so there's really no way it could have a component off-axis that could move around some central point. For this reason, physicists say the electron has an intrinsic magnetic dipole moment that originates with its spin. It just exists.
Let's step back now and look at what we have. A permanent magnet (which itself is comprised of magnetic dipoles, with the moments all pointing in the same direction, hence its large magnetization) produces a magnetic field that exists in the space around the magnet. This space includes the paper clip, sitting on a table. The magnetic field interacts with the electrons in the iron/steel paper clip, changing the magnetic dipole moments of these electrons, and inducing magnetism in the paper clip. How exactly?
The magnetic field produced by the permanent magnet has the property, determined through experimentation, that it can exert a force on a moving charged particle, like our atomic electrons. (This force, called the Lorentz force, was defined in the previous blog entry.) Acting on each magnetic dipole, this force (actually, torque) acts to twist the dipole moments such that they line up parallel to the field. The result: countless magnetic dipole moments in our paper clip are now pointing in the direction of the field. (It is at this point we appreciate the paper clip being made of iron, a ferromagnetic material. If it was not, the magnetic force would have a more difficult time turning all of the dipole moments and would ultimately manage to turn only some of them, diminishing the strength of the magnetization induced in the paper clip.) The paper clip is now a magnet.
How else does the magnetic field interact with the atomic electrons? Surprisingly, it can change the speed with which the electrons orbit their nuclei! Before the magnetic field enters the picture, the electrons are held in their orbits by electrical forces alone. (Unlike charges, i.e. protons and electrons, attract.) When the magnetic field shows up, it produces a force that acts in the opposite direction as the electrical force, at the location of each orbiting electron, and serves to weaken the pull on the electron towards the center of the orbit. The electron no longer needs to travel so quickly to maintain its orbital radius and it slows down. Now, think back to the previous blog entry and the example of you holding on to a string, at the other end of which is attached a ball. You're whirling this ball about your head. In this case, there is a centripetal force pulling the ball towards the center of its circular path, thereby acting, at all times, perpendicular to the direction of the ball's motion. (This is like the electric force holding the electron in its orbit about the atom's nucleus.) Likewise, the magnetic force, once introduced, acts perpendicular to the circling electron. The magnetic force, however, is acting not towards the center of the circle but radially outward, away from the center. As stated before, this diminished centripetal force causes the electron to slow down. We'll soon see that this slowing of the electrons in the atoms of the paper clip is key to the lifting of the paper clip by the permanent magnet.
Now is the time to point out that the magnetic field produced by the permanent magnet is non-uniform. It generally is pointing downwards, assuming the north pole of the magnet is nearer the paper clip than the south pole, but it also flares out. It's the vertical component of the field which acts to slow down the electrons but the horizontal component, existing in the plane of the orbiting electrons, that provides an upward force. Adding together these two components we end up with a force (represented by a vector) that is pointing up and out (away from the center of the electron's circular orbit). We know this force must be perpendicular to the motion of the electron, and indeed it is, as the electron begins to move upwards along a helical path.
The motion of an electron in its orbit constitutes stored kinetic energy. It's this energy that is tapped to lift the paper clip off the table. As the paper clip (acting as a magnet) rises, the unpaired electrons inside slow down and the stored kinetic energy decreases. The magnetic force redirects this energy into lifting the paper clip/magnet against the force of gravity. The net magnetic force, like the normal force mentioned in the previous blog entry, is responsible for the vertical motion of the object (previously a box and now an electron) despite the fact that it doesn't do work on that object. Both of these forces (the magnetic and the normal) are redirecting work done by another agent. In the case of the box being pushed up the incline, the other agent is a person. And in this case, it's ... Well, who or what is this agent?
Uhhhhhh. Well, the agent is whatever got all those electrons circling around all those nuclei to begin with. Whatever it was, it did work and imparted kinetic energy to each little electron. Trying to trace the formation of these iron atoms back to the original source of energy would lead us back to the Big Bang. So it was God, I guess. "God" lifted the paper clip.
First off, the paper clip is made of steel, which contains a large amount of iron. Iron is a ferromagnetic material, meaning it can become magnetized when placed in a magnetic field and remain magnetized when removed from that field. Non ferromagnetic materials, on the other hand, would lose their magnetization upon removal of the external magnetic field. (In this case, we're not removing the magnetic field, but it's still nice that we're working with a ferromagnetic material. You'll see why later.) When we place a permanent magnet above a paper clip, the magnetic field produced by the magnet induces magnetism in the paper clip by applying a torque to the magnetic dipoles in the iron, lining them up.
What's a magnetic dipole? A small current loop (say, electrons flowing around a tiny loop of wire) is, more or less, a magnetic dipole. We call the small current loop a magnetic dipole because it produces a magnetic field, at some distance, that is strictly "dipolar" in nature. Not all systems produce a magnetic field that is dipolar in nature. Some systems produce a field that is not dipolar at all, but perhaps "quadrupolar" or "octopolar." Other systems might produce fields that are largely dipolar but a little quadrupolar, too. What does it mean for the field to be strictly "dipolar" in nature? It means that, as you move away from the system, the strength of the magnetic field drops off as one over the distance cubed. There's no component of the field that drops off as one over the distance or one over the distance to the fourth power, etc. Not too many systems can actually produce a field that is strictly dipolar. It's easy to produce one that is largely dipolar, but quite difficult to produce one that is strictly dipolar. A tiny current loop does the trick, however. But it has to be really tiny, as in infinitesimally small. How do you make such a thing in the lab? You don't. Luckily, I suppose, they already exist in nature as electrons whizzing about nuclei inside of atoms. (No wires necessary.) Previously, I said that the magnetic field of a permanent magnet applies a torque to the magnetic dipoles in a sample of iron, lining them up. Now it should be fairly clear that it is atomic electrons (acting in their capacity as magnetic dipoles) that are doing the lining up. Note: not every electron in a sample of iron experiences this torque. Only the unpaired electrons do. (Each iron atom has 4 unpaired electrons.)
OK, so how can an electron point in a particular direction? Really, it can't. What I mean is that this whirrling electron, this magnetic dipole, points its magnetic dipole moment in a particular direction. This so-called dipole moment is a property of the dipole and can act to represent the physical dipole. It's a vector (with, of course, a magnitude and a direction) that quantifies the contribution of a system's internal magnetism to the external dipolar magnetic field produced by the system. That is, a measure of how much what's going on inside the system is effecting the magnetic field observed outside the system. The moment may be non-physical, but it often proves useful to picture an electron as a little vector when doing calculations or thinking through problems like the one we're addressing here. Therefore, to say that dipoles are lined up is to say that their dipole moments are lined up (or parallel to one another).
Magnets come in different strengths, which we quantify through the concept of magnetization. Something with a large magnetization is both strongly affected by external magnetic fields and the source of its own strong magnetic field. We define magnetization as the amount of magnetic dipole moment per unit volume. Therefore, given a unit volume, we perform a vector sum of all the little moments (vectors) in that volume, and we see how strong our magnet is. Two vectors of equal magnitude pointing in opposite directions sum to zero. Likewise, a large number of arbitrarily directed vectors also sums to zero. This explains why the paper clip, before being magnetized by the permanent magnet, isn't magnetic. It has plenty of little moments (or vectors) inside, but they are arbitrarily directed (well, sorta) and so the net sum of these moments, per unit volume, is pretty close to zero. Once the permanent magnet acts on the dipole moments in the paper clip and lines them up, the vector sum no longer equals zero. Rather, it is now rather large, and the paper clip now has a large magnetization and acts outwardly like a magnet.
Electrons, bound to atoms, move in two ways. This leads to two magnetic dipoles or, better put, two contributions to a single magnetic dipole. (It's a simple vector sum.) Firstly, an electron "orbits" the nucleus. Even though it's not accurate, people often picture this motion as being like a planet orbiting the sun. That's a good way to think of it at the present. Secondly, an electron "spins." You might think of this as an electron spinning about its own axis, just as the Earth spins about its axis once every 24 hours. But this is a rather horrible and misleading analogy, because this "spin" is not really a physical rotation about an axis. The electron is a point particle with no physical size, so there's really no way it could have a component off-axis that could move around some central point. For this reason, physicists say the electron has an intrinsic magnetic dipole moment that originates with its spin. It just exists.
Let's step back now and look at what we have. A permanent magnet (which itself is comprised of magnetic dipoles, with the moments all pointing in the same direction, hence its large magnetization) produces a magnetic field that exists in the space around the magnet. This space includes the paper clip, sitting on a table. The magnetic field interacts with the electrons in the iron/steel paper clip, changing the magnetic dipole moments of these electrons, and inducing magnetism in the paper clip. How exactly?
The magnetic field produced by the permanent magnet has the property, determined through experimentation, that it can exert a force on a moving charged particle, like our atomic electrons. (This force, called the Lorentz force, was defined in the previous blog entry.) Acting on each magnetic dipole, this force (actually, torque) acts to twist the dipole moments such that they line up parallel to the field. The result: countless magnetic dipole moments in our paper clip are now pointing in the direction of the field. (It is at this point we appreciate the paper clip being made of iron, a ferromagnetic material. If it was not, the magnetic force would have a more difficult time turning all of the dipole moments and would ultimately manage to turn only some of them, diminishing the strength of the magnetization induced in the paper clip.) The paper clip is now a magnet.
How else does the magnetic field interact with the atomic electrons? Surprisingly, it can change the speed with which the electrons orbit their nuclei! Before the magnetic field enters the picture, the electrons are held in their orbits by electrical forces alone. (Unlike charges, i.e. protons and electrons, attract.) When the magnetic field shows up, it produces a force that acts in the opposite direction as the electrical force, at the location of each orbiting electron, and serves to weaken the pull on the electron towards the center of the orbit. The electron no longer needs to travel so quickly to maintain its orbital radius and it slows down. Now, think back to the previous blog entry and the example of you holding on to a string, at the other end of which is attached a ball. You're whirling this ball about your head. In this case, there is a centripetal force pulling the ball towards the center of its circular path, thereby acting, at all times, perpendicular to the direction of the ball's motion. (This is like the electric force holding the electron in its orbit about the atom's nucleus.) Likewise, the magnetic force, once introduced, acts perpendicular to the circling electron. The magnetic force, however, is acting not towards the center of the circle but radially outward, away from the center. As stated before, this diminished centripetal force causes the electron to slow down. We'll soon see that this slowing of the electrons in the atoms of the paper clip is key to the lifting of the paper clip by the permanent magnet.
Now is the time to point out that the magnetic field produced by the permanent magnet is non-uniform. It generally is pointing downwards, assuming the north pole of the magnet is nearer the paper clip than the south pole, but it also flares out. It's the vertical component of the field which acts to slow down the electrons but the horizontal component, existing in the plane of the orbiting electrons, that provides an upward force. Adding together these two components we end up with a force (represented by a vector) that is pointing up and out (away from the center of the electron's circular orbit). We know this force must be perpendicular to the motion of the electron, and indeed it is, as the electron begins to move upwards along a helical path.
The motion of an electron in its orbit constitutes stored kinetic energy. It's this energy that is tapped to lift the paper clip off the table. As the paper clip (acting as a magnet) rises, the unpaired electrons inside slow down and the stored kinetic energy decreases. The magnetic force redirects this energy into lifting the paper clip/magnet against the force of gravity. The net magnetic force, like the normal force mentioned in the previous blog entry, is responsible for the vertical motion of the object (previously a box and now an electron) despite the fact that it doesn't do work on that object. Both of these forces (the magnetic and the normal) are redirecting work done by another agent. In the case of the box being pushed up the incline, the other agent is a person. And in this case, it's ... Well, who or what is this agent?
Uhhhhhh. Well, the agent is whatever got all those electrons circling around all those nuclei to begin with. Whatever it was, it did work and imparted kinetic energy to each little electron. Trying to trace the formation of these iron atoms back to the original source of energy would lead us back to the Big Bang. So it was God, I guess. "God" lifted the paper clip.
Wednesday, September 9, 2009
A Bit on Work: Part II of III
This entry logically follows the preceding entry, so you should read that one first unless you already have a good understanding of the physicist's concept of work.
In that preceding entry we spoke of a person pushing a box up an incline, such that the person's arms were always horizontal or parallel to the ground (and not the incline). That is, the force imparted by the person on the box was entirely horizontal. To find the work done by the person on the box, we had to find the component of the force in the direction of the box's motion. That is, we had to break up (mathematically) the force into a component parallel to the incline and a component perpendicular to the incline, and then we took the component parallel to the incline and multiplied that by the distance up the incline that the box moved. We also reasoned that we could just as easily multiply the total horizontal force applied by the person by the horizontal displacement of the box and arrive at the same answer.
But if we take the second approach for calculating the work done on the box, how do we explain the increase in the vertical position of the box? Was work done on the box in moving it higher above the ground? And by what force? (Really, we face the same question in the first approach to calculating work, but it's easier to conceptualize using the second approach.) Indeed, work was done, and it was done by the person. But it was not the force imparted by the person that moved the box higher. Here, the normal force (if you don't know what the normal force is, you should look it up before continuing), while doing no work itself, redirects the efforts of the person from horizontal to vertical. We say the normal force does no work, because the normal force is always perpendicular to the incline and the box never moves off the inclined plane. But the normal force does have a vertical component that lifts the box by redirecting some of the work done by the person. This is a bit tricky, but understanding it will come in handy later.
The question at the end of the last entry was: What force never does any work? And here is the answer: the magnetic force. There is a law, called the Lorentz force law, which tells us how to compute the magnetic force on a charged particle. It says to take the value of the charge (say, the charge associated with a single electron) and to multiply that value by a vector that is perpendicular to both the magnetic field at the location of the particle and to the velocity vector describing the instantaneous motion of the particle. The magnitude of this mutually-perpendicular vector is equal to the magnitude of the velocity times the magnitude of the magnetic field times sine of the angle between the two vectors. This long description can be written simply in mathematical notation: F = q (v x B), where q represents charge and the "x" signifies the cross product between the velocity v and the magnetic field B. (F, v, and B are all vectors, here.) If you're not familiar with a "cross product," it's more or less explained in the wordy description above, so just digest that and ignore the formula.
In summary, the magnetic force on a charged particle is perpendicular to the particle's direction of motion (as well as the direction of the magnetic field). In other words, no component of the force is ever in the same direction as that in which the particle is moving, and therefore no work can ever be done by this force. (This makes perfect sense if you remember the definition of work, which was stated in the previous blog entry.) As an example, if you had an electron (somehow made visible) that was traveling from the front towards the back of your desk, and then you turned on a magnetic field that was uniform and pointing in the direction of "down", i.e. it is perpendicular to the desk and points down right through the table top, then the electron would immediately get tugged towards the left (while continuing to move forward) and fall into a circular path. Note that as soon as the electron moves a bit forward and to the left along this circular path, the force is not longer exclusively left but is now left and a bit down. I don't know if I've explained this clearly or not, so I'll continue a bit and then move on. As an analogy, picture yourself whirling a ball tied to a string around your head. You hold on to one end of the string and raise your arm and get the ball swinging about in a circular path. The force here, exerted by the string on the ball, like in the case of the magnetic field acting on the electron, is always perpendicular to the direction of motion of the ball. It's a centripetal force, in other words. It does alter the direction of the electron, but it certainly doesn't do work on it. The math here is straightforward, but the idea can seem hard to swallow in certain physical examples.
Let's say I take a magnet from the refrigerator and place it over a paper clip lying on my desk. The paper clip "jumps" to the magnet and sticks to it. Is it really true that the magnet's magnetic force did no work in moving the paper clip the few centimeters from the desk top to the magnet in my hand? Yes, because magnetic forces never do work. So then what did the work? What we're going to find is that, like the normal force mentioned earlier, the magnetic force redirects work done by another agent. But what is this other agent and how does the magnetic force redirect its work?
I'm now going to attempt an explanation of what it is that actually does the work in this example. Now, no good teacher would ever introduce a concept and then quickly jump to a difficult and confusing example involving that concept. They would cover some easy examples first and work their way up to a non-intuitive, challenging problem. But I want to jump right into the difficult explanation as to how the paper clip is pulled upwards towards the magnet. I'll begin this explanation in the next blog entry because this one is long enough.
In that preceding entry we spoke of a person pushing a box up an incline, such that the person's arms were always horizontal or parallel to the ground (and not the incline). That is, the force imparted by the person on the box was entirely horizontal. To find the work done by the person on the box, we had to find the component of the force in the direction of the box's motion. That is, we had to break up (mathematically) the force into a component parallel to the incline and a component perpendicular to the incline, and then we took the component parallel to the incline and multiplied that by the distance up the incline that the box moved. We also reasoned that we could just as easily multiply the total horizontal force applied by the person by the horizontal displacement of the box and arrive at the same answer.
But if we take the second approach for calculating the work done on the box, how do we explain the increase in the vertical position of the box? Was work done on the box in moving it higher above the ground? And by what force? (Really, we face the same question in the first approach to calculating work, but it's easier to conceptualize using the second approach.) Indeed, work was done, and it was done by the person. But it was not the force imparted by the person that moved the box higher. Here, the normal force (if you don't know what the normal force is, you should look it up before continuing), while doing no work itself, redirects the efforts of the person from horizontal to vertical. We say the normal force does no work, because the normal force is always perpendicular to the incline and the box never moves off the inclined plane. But the normal force does have a vertical component that lifts the box by redirecting some of the work done by the person. This is a bit tricky, but understanding it will come in handy later.
The question at the end of the last entry was: What force never does any work? And here is the answer: the magnetic force. There is a law, called the Lorentz force law, which tells us how to compute the magnetic force on a charged particle. It says to take the value of the charge (say, the charge associated with a single electron) and to multiply that value by a vector that is perpendicular to both the magnetic field at the location of the particle and to the velocity vector describing the instantaneous motion of the particle. The magnitude of this mutually-perpendicular vector is equal to the magnitude of the velocity times the magnitude of the magnetic field times sine of the angle between the two vectors. This long description can be written simply in mathematical notation: F = q (v x B), where q represents charge and the "x" signifies the cross product between the velocity v and the magnetic field B. (F, v, and B are all vectors, here.) If you're not familiar with a "cross product," it's more or less explained in the wordy description above, so just digest that and ignore the formula.
In summary, the magnetic force on a charged particle is perpendicular to the particle's direction of motion (as well as the direction of the magnetic field). In other words, no component of the force is ever in the same direction as that in which the particle is moving, and therefore no work can ever be done by this force. (This makes perfect sense if you remember the definition of work, which was stated in the previous blog entry.) As an example, if you had an electron (somehow made visible) that was traveling from the front towards the back of your desk, and then you turned on a magnetic field that was uniform and pointing in the direction of "down", i.e. it is perpendicular to the desk and points down right through the table top, then the electron would immediately get tugged towards the left (while continuing to move forward) and fall into a circular path. Note that as soon as the electron moves a bit forward and to the left along this circular path, the force is not longer exclusively left but is now left and a bit down. I don't know if I've explained this clearly or not, so I'll continue a bit and then move on. As an analogy, picture yourself whirling a ball tied to a string around your head. You hold on to one end of the string and raise your arm and get the ball swinging about in a circular path. The force here, exerted by the string on the ball, like in the case of the magnetic field acting on the electron, is always perpendicular to the direction of motion of the ball. It's a centripetal force, in other words. It does alter the direction of the electron, but it certainly doesn't do work on it. The math here is straightforward, but the idea can seem hard to swallow in certain physical examples.
Let's say I take a magnet from the refrigerator and place it over a paper clip lying on my desk. The paper clip "jumps" to the magnet and sticks to it. Is it really true that the magnet's magnetic force did no work in moving the paper clip the few centimeters from the desk top to the magnet in my hand? Yes, because magnetic forces never do work. So then what did the work? What we're going to find is that, like the normal force mentioned earlier, the magnetic force redirects work done by another agent. But what is this other agent and how does the magnetic force redirect its work?
I'm now going to attempt an explanation of what it is that actually does the work in this example. Now, no good teacher would ever introduce a concept and then quickly jump to a difficult and confusing example involving that concept. They would cover some easy examples first and work their way up to a non-intuitive, challenging problem. But I want to jump right into the difficult explanation as to how the paper clip is pulled upwards towards the magnet. I'll begin this explanation in the next blog entry because this one is long enough.
Tuesday, September 1, 2009
A Bit on Work: Part I of III
The concept of work has a specific meaning in the sciences. It is best described by stating how it is calculated, which is the force applied to an object in the direction of its motion times the distance it moves. A force is, more or less, a push or pull. Newton stated that force, mass, and acceleration are linked through the equation F = m a. (He stated it in different terms, but this is how it is normally written today.) Some prefer to write it as a = F/m, to emphasize that a force causes acceleration and not the other way around.
If I push on something (say, a piece of paper taped to a wall) and it does not move, am I applying a force to the paper? Yes, but in this case I am not the only thing applying a force. The wall is pushing on the paper just as hard as I am but in the opposite direction. The net force F on the paper is zero and both the left and right sides of the equation a = F/m are zero and all is well.
Am I doing work on the paper? I'm applying a force but there is no displacement of the paper, so the answer is no. If I manage to push the paper through the wall then indeed I have done work (and I will feel very stupid). New example. Let's say a heavy box is at rest on an incline. I push on it with my arms parallel to the ground (not the incline), and it moves a bit up the incline. Let's say I push with a force of 20 newtons. (Newtons are the SI units of force. One newton is equal to the amount of force required to give a mass of one kilogram an acceleration of one meter per second squared.) Let's also say that the box moves up the incline a meter. Is the work done equal to 20 newtons x 1 meter? No, it's not, because we are concerned not with the overall force but with the force in the direction of the object's motion. We therefore need to mathematically subdivide this overall force (exerted by myself) into a component in the direction of the object's motion (which is of interest to us) and one in the direction perpendicular to this motion (which is not of interest to us). We then multiply the force component in the direction of the object's motion (which is going to be the overall force times cosine of the angle of the incline, with respect to the ground) with 1 meter to get the amount of work done. Yes, you can also calculate work by taking the total force I exert and multiplying this by the horizontal displacement of the object, which will be less than 1 meter. By the way, the SI unit of work is the joule, which is equal to, of course, a newton times a meter.
Let's see if I can confuse you. You push a bag of flour across the kitchen counter. According to Newton's third law, the bag of flour pushes back on you with an equal (in magnitude) but oppositely directed force. Why does the bag of flour move if the forces cancel out?
If you need a hint, take another look at the equation a = F/m. The forces exerted by you and by the bag may be equal in magnitude but your masses surely are not. Therefore you should have different accelerations, which indeed you do. This problem is complicated by the force of friction, which is robbing you of your acceleration and reducing the acceleration of the bag of flour. If you could get rid of the friction, you too would accelerate backwards (but not very quickly because of your relatively large mass). Do you see the difference between this example and the paper against the wall?
I want to talk more about work and a special kind of force that, oddly enough, never does any work. There aren't many forces in nature, so perhaps you can figure out which one I'm talking about. I'll discuss it in my next blog entry. Thanks for reading.
If I push on something (say, a piece of paper taped to a wall) and it does not move, am I applying a force to the paper? Yes, but in this case I am not the only thing applying a force. The wall is pushing on the paper just as hard as I am but in the opposite direction. The net force F on the paper is zero and both the left and right sides of the equation a = F/m are zero and all is well.
Am I doing work on the paper? I'm applying a force but there is no displacement of the paper, so the answer is no. If I manage to push the paper through the wall then indeed I have done work (and I will feel very stupid). New example. Let's say a heavy box is at rest on an incline. I push on it with my arms parallel to the ground (not the incline), and it moves a bit up the incline. Let's say I push with a force of 20 newtons. (Newtons are the SI units of force. One newton is equal to the amount of force required to give a mass of one kilogram an acceleration of one meter per second squared.) Let's also say that the box moves up the incline a meter. Is the work done equal to 20 newtons x 1 meter? No, it's not, because we are concerned not with the overall force but with the force in the direction of the object's motion. We therefore need to mathematically subdivide this overall force (exerted by myself) into a component in the direction of the object's motion (which is of interest to us) and one in the direction perpendicular to this motion (which is not of interest to us). We then multiply the force component in the direction of the object's motion (which is going to be the overall force times cosine of the angle of the incline, with respect to the ground) with 1 meter to get the amount of work done. Yes, you can also calculate work by taking the total force I exert and multiplying this by the horizontal displacement of the object, which will be less than 1 meter. By the way, the SI unit of work is the joule, which is equal to, of course, a newton times a meter.
Let's see if I can confuse you. You push a bag of flour across the kitchen counter. According to Newton's third law, the bag of flour pushes back on you with an equal (in magnitude) but oppositely directed force. Why does the bag of flour move if the forces cancel out?
If you need a hint, take another look at the equation a = F/m. The forces exerted by you and by the bag may be equal in magnitude but your masses surely are not. Therefore you should have different accelerations, which indeed you do. This problem is complicated by the force of friction, which is robbing you of your acceleration and reducing the acceleration of the bag of flour. If you could get rid of the friction, you too would accelerate backwards (but not very quickly because of your relatively large mass). Do you see the difference between this example and the paper against the wall?
I want to talk more about work and a special kind of force that, oddly enough, never does any work. There aren't many forces in nature, so perhaps you can figure out which one I'm talking about. I'll discuss it in my next blog entry. Thanks for reading.
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