In the last post, I finally finished the first "thread" about galaxy rotation curves. My dad (who apparently also reads this blog, although not as consistently as my mom) wasn't quite sure how everything tied together (I believe he missed some entries in the middle). So to briefly recap: I started by explaining the Doppler effect, which was then followed by a series of posts on the wave/particle nature of light. I then discussed the Bohr model of the atom, because it provides a nice framework for understanding the emission of light by atoms. Combining all those posts, we can now understand how to measure the speed of rotation of a galaxy - hydrogen in stars emits light at known wavelengths/frequencies which are then shifted by the Doppler effect. Knowing the math behind the Doppler effect, we can determine how fast the galaxy is rotating. Next, we talked about Newtonian gravity, which led to a prediction for what we expect the rotation of galaxies to look like. In the last post, I described what we actually see, providing evidence for dark matter.
This is the goal of this blog - to try and describe all the pieces that go into a physics argument in a way that's understandable. My hope is that interested readers will see that while specific parts of physics may be esoteric or complicated (i.e. high level math), in a general sense we're making deductions in a way that is very similar to those made in almost any other field of study.
Saturday, May 16, 2009
Monday, May 11, 2009
Galaxy rotation curves
Ok, so finally I think we can look at rotation curves. We'll make the simplifying assumption that the objects we are interested in are in a perfectly circular orbit about the center of the galaxy, an assumption which doesn't really change anything so it's ok (another larger point about physics: quite often [in fact, almost always], we take a complicated problem and approximate it into something smaller that we can solve [often called the "spherical cow" approach - we would approximate a cow to be a sphere and go from there]. The question then often becomes "how good was the approximation?" In this case, there is no real difference between circular and elliptical orbits, so the approximation is fine and the conclusions are valid).
We know the equation of circular motion, F=mv2/r. And by hypothesis, the only force acting on the object in orbit is the force of gravity, F=G*m1*m2/r2. In this case, m1 is the mass of the galaxy, and m2 is the mass of the object. We equate the forces, so mv2/r = G*m1*m2/r2. Now, the mass from the circular motion equation is just the mass of the object in orbit, so m2 will cancel. All that remains is to solve for the velocity, since that's what we measure using the Doppler effect and red shift.
m2*v2/r = G*m1*m2/r2
First, divide both sides by m2
v2/r = G*m1/r2
Next, multiply through by r
v2 = G*m1/r
Now, take the square root of both sides
v = Sqrt(G*m1/r)
And that's it. We have derived that the velocity of an object in orbit about a galaxy should be proportional to the square root of the mass of the galaxy divided by the orbital radius. There is one more thing we should be aware of, which is that I haven't made any assumptions yet about the size of the mass of the galaxy. A galaxy is a very large thing, and what happens if you're inside part of it? For example, the Sun and the Earth are somewhere inside the Milky Way galaxy. We do orbit the Milky Way center, but part of the Milky Way is outside our orbit. The answer is that in this case, m1 refers the total mass inside the orbit. It doesn't matter how spatially extended it is in space, as long as the object in which we're interested is outside of the galaxy, the equations are fine. And since I'm particularly interested in the mass of the bright part of the galaxy, it's easy to know when we're outside that part, so everything holds.
Now, let's look at a plot. If all the mass were in the bright part of the galaxy, then outside the bright part (say a radius of 100, just to make the plot look right), from the last equation, we would expect the velocity to fall like 1/Sqrt(r) (G and m1 would be constant). That would look like this:
Instead, we measure a flat line, like this:
Therefore, by deduction, we know that either Newtonian gravity is wrong (a possibility, I'll admit), or that there is more mass than we thought, mass that is not contained in the bright part of the galaxy. In fact, we know the distribution of that mass, as it has to increase like 1/Sqrt(r) or else the the velocity would not be flat.
This is what some of the actual data looks like:
These are measurements of galaxy rotation curves (Begeman, Broeils and Sanders, Mon. Not. R. astr. Soc., 1991, 249, 523). I apologize for the image quality, but velocity is on the y-axis and radius is on the x-axis, and all the black points are actual measurements. You can see at small radius the velocity increases. This is where the bright part of the galaxy is, and as the radius increases, we are containing more mass in the orbit. At larger radius, we would expect to see the velocity drop. But instead, it stays constant. The dashed and dotted lines are the the components of the mass, the bright part and the dark part. This data provides evidence that there is matter that we are not seeing, that is not interacting with light, but is dark matter.
We know the equation of circular motion, F=mv2/r. And by hypothesis, the only force acting on the object in orbit is the force of gravity, F=G*m1*m2/r2. In this case, m1 is the mass of the galaxy, and m2 is the mass of the object. We equate the forces, so mv2/r = G*m1*m2/r2. Now, the mass from the circular motion equation is just the mass of the object in orbit, so m2 will cancel. All that remains is to solve for the velocity, since that's what we measure using the Doppler effect and red shift.
m2*v2/r = G*m1*m2/r2
First, divide both sides by m2
v2/r = G*m1/r2
Next, multiply through by r
v2 = G*m1/r
Now, take the square root of both sides
v = Sqrt(G*m1/r)
And that's it. We have derived that the velocity of an object in orbit about a galaxy should be proportional to the square root of the mass of the galaxy divided by the orbital radius. There is one more thing we should be aware of, which is that I haven't made any assumptions yet about the size of the mass of the galaxy. A galaxy is a very large thing, and what happens if you're inside part of it? For example, the Sun and the Earth are somewhere inside the Milky Way galaxy. We do orbit the Milky Way center, but part of the Milky Way is outside our orbit. The answer is that in this case, m1 refers the total mass inside the orbit. It doesn't matter how spatially extended it is in space, as long as the object in which we're interested is outside of the galaxy, the equations are fine. And since I'm particularly interested in the mass of the bright part of the galaxy, it's easy to know when we're outside that part, so everything holds.
Now, let's look at a plot. If all the mass were in the bright part of the galaxy, then outside the bright part (say a radius of 100, just to make the plot look right), from the last equation, we would expect the velocity to fall like 1/Sqrt(r) (G and m1 would be constant). That would look like this:
Instead, we measure a flat line, like this:
Therefore, by deduction, we know that either Newtonian gravity is wrong (a possibility, I'll admit), or that there is more mass than we thought, mass that is not contained in the bright part of the galaxy. In fact, we know the distribution of that mass, as it has to increase like 1/Sqrt(r) or else the the velocity would not be flat.
This is what some of the actual data looks like:
These are measurements of galaxy rotation curves (Begeman, Broeils and Sanders, Mon. Not. R. astr. Soc., 1991, 249, 523). I apologize for the image quality, but velocity is on the y-axis and radius is on the x-axis, and all the black points are actual measurements. You can see at small radius the velocity increases. This is where the bright part of the galaxy is, and as the radius increases, we are containing more mass in the orbit. At larger radius, we would expect to see the velocity drop. But instead, it stays constant. The dashed and dotted lines are the the components of the mass, the bright part and the dark part. This data provides evidence that there is matter that we are not seeing, that is not interacting with light, but is dark matter.
Saturday, May 9, 2009
Newton's theory of gravity (part 2)
I have not posted (I can't quite bring myself to use "blogged" as a verb in the past tense, but I should probably get over such squeamishness) in almost a month now, for which I apologize. My excuses are standard - my work intruded. In the last month I have started looking at data from a new run and been to a conference (I've also been to my 10th high school reunion), and I am now visiting the mine in Canada I mentioned in one of the original posts to help run an experiment underground. Enough about my current activities, however, as it's time that I finally finished off the series on galactic rotation curves and Newtonian gravity. Unfortunately, I've realized that I still can't finish this in one post, but we'll get there eventually.
Mother, I'm sorry to say that we're going to need equations, but hopefully it won't be so bad. We'll start with Newton's Second Law of Motion, which says that F=ma. That's it. Force equals mass times acceleration. This law governs almost all macroscopic kinematics (a fancy word for the physics of motion). When you apply a force to something (say by pushing a book across a table), the book will accelerate with a particular acceleration depending on its mass. As with a lot of physics, this is intuitive, especially if we rewrite the equation as a=F/m. If I push a very light object, I can accelerate it rapidly (m is small, so the acceleration is big). If I try to push a heavy object like a car, I can barely move it at all (m is very large, so the acceleration is small). Likewise, the harder I push (more force), the faster it will accelerate.
Ok, next we'll look at circular motion where I'm presented with a pedagogical dilemma. Ultimately, I only really want the equation of circular motion (F=mv2/r), but I can't very well just state the equation without trying to explain it. I condescendingly fear, however, that any attempts at explaining it with a single paragraph in this blog will only result in confusion (which raises a question concerning the wisdom of the entire endeavor, but I'll ignore that). But here goes. In a circle, there are two important directions: the radial dire
ction, which points from the center of the circle to a point on the circle itself, and the tangential direction, which is perpendicular to the radial direction. At any point on the circle, these directions have the same relationship to each other.
The key to circular motion (i.e. when an object just travels in a perfect circle forever and ever) is that the instantaneous velocity is always tangential to the circle while the acceleration is always radially inward. Imagine that my mother is on a ferris wheel. At point "A" on the ferris wheel, she is travelling straight to the left. At point "B" on the wheel, she is travelling straight down (for example, if she were to jump off the ferris wheel at that point, she would drop straight down). Likewise at points "C" and "D," she is travelling to the right and straight up, respectively. The point being that she is always moving tangentially and never radially.
The strange thing is that the force she experiences (the "centri
petal force") is always radially inward. At this point, I think I may just have to revert to a "trust me." It's a matter of mathematical fact that to keep her motion circular, she must experience only radial forces. And if you work out the math, the necessary force is F=mv2/r; in other words, the force required to keep her in circular motion is proportional to the square of her velocity divided by the radius. This does make sense intuitively. The faster she is moving, the more force is required to get her to go around the circle. And, the larger the circle, the less change is required to keep her in orbit.
I'm very unsatisfied with that explanation, but I'll leave it for now. The final equation we need is Newton's equation of gravity. This is an empirical law (keeping in mind the discussion in the prior post). Newton discovered that two masses (call them m1 and m2) separated by a distance (call it r) will exert an attractive force on each other, F=G*m1*m2/r2, where G is a constant that is experimentally determined and set by nature. Every mass in the universe exerts such a force on every other mass and vice versa. However, the value of G is very small, so it takes either an extraordinarily large mass (like the Earth or the Sun) or an extremely small separation to observe this force. But that's all there is to Newtonian gravity.
Mother, I'm sorry to say that we're going to need equations, but hopefully it won't be so bad. We'll start with Newton's Second Law of Motion, which says that F=ma. That's it. Force equals mass times acceleration. This law governs almost all macroscopic kinematics (a fancy word for the physics of motion). When you apply a force to something (say by pushing a book across a table), the book will accelerate with a particular acceleration depending on its mass. As with a lot of physics, this is intuitive, especially if we rewrite the equation as a=F/m. If I push a very light object, I can accelerate it rapidly (m is small, so the acceleration is big). If I try to push a heavy object like a car, I can barely move it at all (m is very large, so the acceleration is small). Likewise, the harder I push (more force), the faster it will accelerate.
Ok, next we'll look at circular motion where I'm presented with a pedagogical dilemma. Ultimately, I only really want the equation of circular motion (F=mv2/r), but I can't very well just state the equation without trying to explain it. I condescendingly fear, however, that any attempts at explaining it with a single paragraph in this blog will only result in confusion (which raises a question concerning the wisdom of the entire endeavor, but I'll ignore that). But here goes. In a circle, there are two important directions: the radial dire
ction, which points from the center of the circle to a point on the circle itself, and the tangential direction, which is perpendicular to the radial direction. At any point on the circle, these directions have the same relationship to each other.The key to circular motion (i.e. when an object just travels in a perfect circle forever and ever) is that the instantaneous velocity is always tangential to the circle while the acceleration is always radially inward. Imagine that my mother is on a ferris wheel. At point "A" on the ferris wheel, she is travelling straight to the left. At point "B" on the wheel, she is travelling straight down (for example, if she were to jump off the ferris wheel at that point, she would drop straight down). Likewise at points "C" and "D," she is travelling to the right and straight up, respectively. The point being that she is always moving tangentially and never radially.
The strange thing is that the force she experiences (the "centri
petal force") is always radially inward. At this point, I think I may just have to revert to a "trust me." It's a matter of mathematical fact that to keep her motion circular, she must experience only radial forces. And if you work out the math, the necessary force is F=mv2/r; in other words, the force required to keep her in circular motion is proportional to the square of her velocity divided by the radius. This does make sense intuitively. The faster she is moving, the more force is required to get her to go around the circle. And, the larger the circle, the less change is required to keep her in orbit.I'm very unsatisfied with that explanation, but I'll leave it for now. The final equation we need is Newton's equation of gravity. This is an empirical law (keeping in mind the discussion in the prior post). Newton discovered that two masses (call them m1 and m2) separated by a distance (call it r) will exert an attractive force on each other, F=G*m1*m2/r2, where G is a constant that is experimentally determined and set by nature. Every mass in the universe exerts such a force on every other mass and vice versa. However, the value of G is very small, so it takes either an extraordinarily large mass (like the Earth or the Sun) or an extremely small separation to observe this force. But that's all there is to Newtonian gravity.
Tuesday, April 14, 2009
Newton's theory of gravity (part 1)
I fear that I got a bit ahead of myself at the end of the last post on the spectral lines of hydrogen. To fully close the circle between dark matter and everything I've been talking about in the last few entries, we do need to cover Newton's theory of gravity. Therefore, I will try to do so now, so that we can put this particularly sequence to rest.
First, on talking with a friend earlier today, I was asked, "I always hear the term "Newtonian gravity: is there any other kind?" This is exactly the kind of question I need to be asked, because otherwise I forget that I've been studying this stuff for 10 years. The answer is yes, there is another kind of gravity; more precisely, the answer is that there is a more complete form of gravity, namely Einstein's Theory of General Relativity (and possible string theory is a yet more complete form, although I'll leave that argument to Brian Greene's The Elegant Universe). Newton observed the effects of gravity and described those effects using math. However, he didn't describe how gravity works. To borrow an analogy directly from Greene's book, when my mom (and no, Greene didn't refer explicitly to my mom in his book [he did address himself directly to "you," but I don't think my mom has read it]) uses her computer, she doesn't know how it actually works (i.e. little electrical signals flashing through a chip). She only knows how to use it to write an article or to read this blog (sometimes she is unsure even how to do those tasks, at which times, coincidentally enough, she often checks in with her favorite son). Newton gave us a personal computer (gravity) and told us how to use it to write articles and check email (the equations), but he didn't tell us how it actually works (how is gravity transmitted, or how does the apple "know" it has to fall to the ground?).
In addition,
Newton's formulation was only an approximation that was incorrect on certain scales (which are largely inaccessible to us and unobservable in our day to day lives). For example, I'll poach another commonly used illustration; from far away, a Seurat painting looks like a smooth picture. Newton's theory of gravity is accurate when viewed "from far away," which more or less is the point from which we all experience gravity. Up close, however, and all the dots become clear. While Newton got the big picture right, he did not describe the dots. General relativity can handle both the smooth "far away" view, but also the rough "up close" view.
That's why we can talk about "Newtonian gravity." For all intents and purposes, we could just talk about gravity and we'd all be referring to Newton, but since this is a physics blog, I figure I should try to be more precise in my language. In the next post, I'll actually use Newton's equations to look at the rotation of galaxies.
The image of Seurat's Sunday Afternoon on the Island of La Grande Jatte was scanned by Mark Harden
First, on talking with a friend earlier today, I was asked, "I always hear the term "Newtonian gravity: is there any other kind?" This is exactly the kind of question I need to be asked, because otherwise I forget that I've been studying this stuff for 10 years. The answer is yes, there is another kind of gravity; more precisely, the answer is that there is a more complete form of gravity, namely Einstein's Theory of General Relativity (and possible string theory is a yet more complete form, although I'll leave that argument to Brian Greene's The Elegant Universe). Newton observed the effects of gravity and described those effects using math. However, he didn't describe how gravity works. To borrow an analogy directly from Greene's book, when my mom (and no, Greene didn't refer explicitly to my mom in his book [he did address himself directly to "you," but I don't think my mom has read it]) uses her computer, she doesn't know how it actually works (i.e. little electrical signals flashing through a chip). She only knows how to use it to write an article or to read this blog (sometimes she is unsure even how to do those tasks, at which times, coincidentally enough, she often checks in with her favorite son). Newton gave us a personal computer (gravity) and told us how to use it to write articles and check email (the equations), but he didn't tell us how it actually works (how is gravity transmitted, or how does the apple "know" it has to fall to the ground?).
In addition,
Newton's formulation was only an approximation that was incorrect on certain scales (which are largely inaccessible to us and unobservable in our day to day lives). For example, I'll poach another commonly used illustration; from far away, a Seurat painting looks like a smooth picture. Newton's theory of gravity is accurate when viewed "from far away," which more or less is the point from which we all experience gravity. Up close, however, and all the dots become clear. While Newton got the big picture right, he did not describe the dots. General relativity can handle both the smooth "far away" view, but also the rough "up close" view. That's why we can talk about "Newtonian gravity." For all intents and purposes, we could just talk about gravity and we'd all be referring to Newton, but since this is a physics blog, I figure I should try to be more precise in my language. In the next post, I'll actually use Newton's equations to look at the rotation of galaxies.The image of Seurat's Sunday Afternoon on the Island of La Grande Jatte was scanned by Mark Harden
Wednesday, April 8, 2009
Spectral analysis
As mentioned in the last post, the Bohr atom is not correct, quantum mechanically speaking. It does, however, do an excellent job in modeling the simplest atom, that of hydrogen. Hydrogen is the lightest element, consisting of one proton with one electron in orbit. With the Bohr atom model, we know that the orbiting electron can exist in various discrete orbits, corresponding to different energies. In addition, we know that when the electron jumps between these levels, it emits or absorbs a photon. Finally, we know that the energy of a photon is proportional to its frequency, which is related to its wavelength or color. Putting all this together, we can predict that a hydrogen atom will emit or absorb very specific colors of light.
We are now talking about "spectral analysis." The OED (do you like the use of the OED, mom?) defines spectrum in a couple of ways, but I'll print two of them here. First, a general definition for physicists: "An actual or notional arrangement of the component parts of any phenomenon according to frequency, energy, mass, or the like." Physicists often talk about an energy spectrum, a frequency spectrum, etc., and what we mean is exactly the definition given by the OED - breaking up some group or data set into its components.
A second definition is this: "The coloured band into which a beam of light is decomposed by means of a prism or diffraction grating. Also, a dark band containing bright lines produced similarly; such a (coloured or dark) band, or the pattern of lines in it, as characteristic of the light source; hence, the pattern of absorption or emission of light or other electromagnetic radiation over any range of wavelengths exhibited by a body or substance." Now this is exactly what I'm talking about. The general idea is familiar - anyone who has seen a rainbow has seen light broken up into its various colors. When applied to the hydrogen atom, the spectrum is the characteristic colors of light that can be emitted or absorbed. Using the Bohr model, we can predict which wavelengths can interact with hydrogen (this may be the subject of another quantitative post).
At right is an image of a tube filled with hydrogen that is being excited by high voltage (taken from here). The light passes through grating to separate the spectral lines, with the result being the smaller lines shown to the right. All hydrogen, anywhere in the universe, will emit these colors when excited.
And now we finally get back to dark matter. In the first post, I talked about how we can tell how fast the galaxies are spinning using redshift or the Doppler effect. That's because any hydrogen in those distant galaxies will have the same spectrum as hydrogen on Earth, which means the hydrogen in the distant galaxy is emitting the same wavelengths of light that we can measure here on Earth. As discussed in the Doppler effect posts, since light is a wave, its wavelength will be shifted via the Doppler effect depending on the speed of the source. Because we can measure the relationship between the different lines, we can use the observed shifts to deduce the rotational speed of the source galaxy (one side spins away from us, one side spins towards us). And voila, now we know that the galaxies are spinning too fast and that there must be matter we aren't seeing (well, we would see that if we understood Newtonian gravity, which will be the topic a future post, I'm sure).
The above plot taken from the website of an MIT experiment, FIRE, shows the redshift for three real objects. Why is it called "redshift?" Most objects in the universe are moving away from us, and when the source of light is moving away, the light shifts towards the red end of the spectrum (or towards lower energy).
We are now talking about "spectral analysis." The OED (do you like the use of the OED, mom?) defines spectrum in a couple of ways, but I'll print two of them here. First, a general definition for physicists: "An actual or notional arrangement of the component parts of any phenomenon according to frequency, energy, mass, or the like." Physicists often talk about an energy spectrum, a frequency spectrum, etc., and what we mean is exactly the definition given by the OED - breaking up some group or data set into its components.
A second definition is this: "The coloured band into which a beam of light is decomposed by means of a prism or diffraction grating. Also, a dark band containing bright lines produced similarly; such a (coloured or dark) band, or the pattern of lines in it, as characteristic of the light source; hence, the pattern of absorption or emission of light or other electromagnetic radiation over any range of wavelengths exhibited by a body or substance." Now this is exactly what I'm talking about. The general idea is familiar - anyone who has seen a rainbow has seen light broken up into its various colors. When applied to the hydrogen atom, the spectrum is the characteristic colors of light that can be emitted or absorbed. Using the Bohr model, we can predict which wavelengths can interact with hydrogen (this may be the subject of another quantitative post).
At right is an image of a tube filled with hydrogen that is being excited by high voltage (taken from here). The light passes through grating to separate the spectral lines, with the result being the smaller lines shown to the right. All hydrogen, anywhere in the universe, will emit these colors when excited.
And now we finally get back to dark matter. In the first post, I talked about how we can tell how fast the galaxies are spinning using redshift or the Doppler effect. That's because any hydrogen in those distant galaxies will have the same spectrum as hydrogen on Earth, which means the hydrogen in the distant galaxy is emitting the same wavelengths of light that we can measure here on Earth. As discussed in the Doppler effect posts, since light is a wave, its wavelength will be shifted via the Doppler effect depending on the speed of the source. Because we can measure the relationship between the different lines, we can use the observed shifts to deduce the rotational speed of the source galaxy (one side spins away from us, one side spins towards us). And voila, now we know that the galaxies are spinning too fast and that there must be matter we aren't seeing (well, we would see that if we understood Newtonian gravity, which will be the topic a future post, I'm sure).
The above plot taken from the website of an MIT experiment, FIRE, shows the redshift for three real objects. Why is it called "redshift?" Most objects in the universe are moving away from us, and when the source of light is moving away, the light shifts towards the red end of the spectrum (or towards lower energy).
Tuesday, March 31, 2009
The Bohr atom
In this post, I hope to describe the Bohr model of the atom. In the next post, I will use the Bohr model (together with the nature of light discussed in the last few posts) to predict the existence of "spectral lines," which will finally bring me back to dark matter by explaining exactly how we measure the speed of those rotating galaxies (see the Dark Matter Intro link at the right if this is not familiar). Historically speaking, I'm presenting this material backwards, as the observation of spectral lines came first and the explanation came later, but I will proceed anyway.
Niels Bohr is in many ways the father of quantum physics, if not its prime mover. He came of age before the revolutionary wave of the 1920s, but almost all of the physicists involved in developing quantum mechanics (Heisenberg, Dirac, Pauli to name a few) spent some time at the Institute of Theoretical Physics he founded in 1921. His model of the atom was a precursor to all of the discoveries of quantum physics to follow.
So what is that model? Although it turns out not to be accurate, it's a pretty good start, and I would guess that th
e Bohr model is generally the picture most of us have in our minds for the atom today. Analogous to our picture of the solar system, the Bohr model imagines a dense, very small nucleus, surrounded by orbiting electrons (I took the picture from a website at Jefferson Lab).
By itself, that isn't all that interesting - the real theoretical interest of the Bohr atom was that the electrons are constrained to lie at specific orbits. In the solar system, the planets could theoretically lie at any radius - we could take the Earth, move it a little farther away from the sun, and it would still orbit contentedly (we might all be a bit colder, but the orbit would be fine). In Bohr's atom, an electron can't move to a slightly larger orbit; instead, it would have to jump to the next available orbit. The analogy is that the Earth can only be at our orbit or Mars' orbit, but nowhere in between.
To take this a step further, we can add energy considerations. The farther away from the nucleus, the more energy an electron has. Therefore, whenever an electron switches orbits, it gives up or takes in energy (depending on whether it's heading out or heading in). If it is only allowed to be in certain orbits, then the possible energy steps are discrete - it can only take in or give up a very specific amount of energy. For one more analogy, suppose my mom is on an elevator on the ground floor. If she goes up in the elevator, she can only get off on floors, she can't get off in between floors. And, as she goes up, she picks up a specific amount of energy (which she could give back if she were to jump out a window - she would be rather more regretful if she jumped out a 4th floor window as opposed to a 1st floor window because of all the energy she picked up by going up the elevator). The electrons in the orbit of the Bohr atom are like my mom on the elevator - they can't get off between floors and the amount of energy they can pick up or give out is discrete and fixed.
The very astute reader might see a similarity between this model and the photoelectric effect of Einstein from a few posts ago, when light could carry a specific and discrete amount of energy depending on its frequency. In fact, this is exactly the same thing - how does the electron gain or give up energy? By absorbing or emitting photons (in the photoelectric effect, an electron absorbs enough energy to jump off the surface of the metal, or be "freed" from the orbit). This has significant results for observations of atoms as we'll see in the next post. For now, though, I'll put up one more diagram, similar to the one above, containing the positively charged nucleus at the center, and an electron that can be in one of three "energy levels" (or floors) by emitting or absorbing a photon (the wavy line). Also, for a little interactive demonstration of the same thing, try this.
Niels Bohr is in many ways the father of quantum physics, if not its prime mover. He came of age before the revolutionary wave of the 1920s, but almost all of the physicists involved in developing quantum mechanics (Heisenberg, Dirac, Pauli to name a few) spent some time at the Institute of Theoretical Physics he founded in 1921. His model of the atom was a precursor to all of the discoveries of quantum physics to follow.
So what is that model? Although it turns out not to be accurate, it's a pretty good start, and I would guess that th
e Bohr model is generally the picture most of us have in our minds for the atom today. Analogous to our picture of the solar system, the Bohr model imagines a dense, very small nucleus, surrounded by orbiting electrons (I took the picture from a website at Jefferson Lab).By itself, that isn't all that interesting - the real theoretical interest of the Bohr atom was that the electrons are constrained to lie at specific orbits. In the solar system, the planets could theoretically lie at any radius - we could take the Earth, move it a little farther away from the sun, and it would still orbit contentedly (we might all be a bit colder, but the orbit would be fine). In Bohr's atom, an electron can't move to a slightly larger orbit; instead, it would have to jump to the next available orbit. The analogy is that the Earth can only be at our orbit or Mars' orbit, but nowhere in between.
To take this a step further, we can add energy considerations. The farther away from the nucleus, the more energy an electron has. Therefore, whenever an electron switches orbits, it gives up or takes in energy (depending on whether it's heading out or heading in). If it is only allowed to be in certain orbits, then the possible energy steps are discrete - it can only take in or give up a very specific amount of energy. For one more analogy, suppose my mom is on an elevator on the ground floor. If she goes up in the elevator, she can only get off on floors, she can't get off in between floors. And, as she goes up, she picks up a specific amount of energy (which she could give back if she were to jump out a window - she would be rather more regretful if she jumped out a 4th floor window as opposed to a 1st floor window because of all the energy she picked up by going up the elevator). The electrons in the orbit of the Bohr atom are like my mom on the elevator - they can't get off between floors and the amount of energy they can pick up or give out is discrete and fixed.
The very astute reader might see a similarity between this model and the photoelectric effect of Einstein from a few posts ago, when light could carry a specific and discrete amount of energy depending on its frequency. In fact, this is exactly the same thing - how does the electron gain or give up energy? By absorbing or emitting photons (in the photoelectric effect, an electron absorbs enough energy to jump off the surface of the metal, or be "freed" from the orbit). This has significant results for observations of atoms as we'll see in the next post. For now, though, I'll put up one more diagram, similar to the one above, containing the positively charged nucleus at the center, and an electron that can be in one of three "energy levels" (or floors) by emitting or absorbing a photon (the wavy line). Also, for a little interactive demonstration of the same thing, try this.
(Courtesy of the Department of Energy)
Friday, March 27, 2009
The Double-Slit Experiment (quantitative) - Part 2
(click on the picture for a larger view)
Now that we know what a sine wave is, we can understand the double-slit experiment. I need to start with a few definitions that I probably should have put in the last post: the wavelength is the length between two successive peaks in the wave (often represented by λ) and the amplitude is the height of the wave (we'll call it A). There is a symmetry property of the wave; if you shifted the wave to the right or left by its wavelength, it would look exactly the same. In fact, you could shift the wave by any integer times the wavelength, and you wouldn't be able to tell the difference. This will be important later on.
To understand the double slit experiment, we need to ask what happens when two waves overlap. The answer depends on their "phase," or where each wave is in its oscillation relative to the other. For example, suppose two waves are perfectly "in phase," so that when one wave is peaking, so is the other. When you add these two waves together, you'll get a wave that is twice as big in amplitude.
What about when the waves are "out of phase" so that one is all the way up when the other is all the way down? In that case, the waves destructively interfere so that the addition contains no wave at all.
This interference is the key to the double-slit experiment and allows us to predict the shape of the light pattern on the screen. When the light impinges on the slit, the waves that come come out the other side are initially in phase. If you look at the the screen directly across from the slit, you would see a dark spot. However, that bright spot will be banded by bright spots, which will in turn be banded by two more dark spots in a fringe pattern. To understand why, let's zoom in on the slit right where the light passes through (on this lovely diagram I stole from here).
Here, the distance between the slit and the wall is L and the slit separation is d. Where on the wall do you expect to see bright or dark spots? If we want there to be a bright spot on the wall, then we know the two waves must interfere constructively (the first case discussed above) or be in phase. A dark spot will appear when the waves are out of phase and interfere destructively. Let θ (again, angles are always θs) be the angle between the horizontal and the position of a given fringe on the wall). If we look high on the wall, the light that came out of the top slit doesn't have to go as far as the light that came out of the bottom slit (in other words, r1 is bigger than r2). This means that the bottom ray of light will have more time to trace out its wavelength and will drop out of phase with the top ray of light. The extra distance traveled by the bottom ray is equal to d*sin θ (remember that the sine function also related the sides of a triangle and notice that the light paths r1 and r2 form a triangle with the slit). Now, remember the symmetry of the wave - a wave that is shifted by its wavelength looks the same. So if the extra distance traveled by the bottom wave equaled exactly its wavelength, it would look identical to the top wave, and the waves would interefere constructively - a bright spot would appear on the wall. If, on the other hand, the extra distance traveled was exactly half a wavelength, so that the bottom wave had just enough time to get out of phase, the two waves would interfere destructively and a dark spot would appear on the wall.
This is exactly what happens - bright spots appear if d*sin θ = λ or some multiple of λ, while dark spots appear if d*sin θ = λ/2. Wave properties predict exactly the patterns that appear in the double-slit experiment, confirming that light is like a wave.
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