• Posted by Konstantin 21.03.2017 No Comments

    Ever since Erwin Schrödinger described a thought experiment, in which a cat in a sealed box happened to be "both dead and alive at the same time", popular science writers have been relying on it heavily to convey the mysteries of quantum physics to the layman. Unfortunately, instead of providing any useful intuition, this example has instead laid solid base to a whole bunch of misconceptions. Having read or heard something about the strange cat, people would tend to jump to profound conclusions, such as "according to quantum physics, cats can be both dead and alive at the same time" or "the notion of a conscious observer is important in quantum physics". All of these are wrong, as is the image of a cat, who is "both dead and alive at the same time". The corresponding Wikipedia page does not stress this fact well enough, hence I thought the Internet might benefit from a yet another explanatory post.

    The Story of the Cat

    The basic notion in quantum mechanics is a quantum system. Pretty much anything could be modeled as a quantum system, but the most common examples are elementary particles, such as electrons or photons. A quantum system is described by its state. For example, a photon has polarization, which could be vertical or horizontal. Another prominent example of a particle's state is its wave function, which represents its position in space.

    There is nothing special about saying that things have state. For example, we may say that any cat has a "liveness state", because it can be either "dead" or "alive". In quantum mechanics we would denote these basic states using the bra-ket notation as |\mathrm{dead}\rangle and |\mathrm{alive}\rangle. The strange thing about quantum mechanical systems, though, is the fact that quantum states can be combined together to form superpositions. Not only could a photon have a purely vertical polarization \left|\updownarrow\right\rangle or a purely horizontal polarization \left|\leftrightarrow\right\rangle, but it could also be in a superposition of both vertical and horizontal states:

        \[\left|\updownarrow\right\rangle + \left|\leftrightarrow\right\rangle.\]

    This means that if you asked the question "is this photon polarized vertically?", you would get a positive answer with 50% probability - in another 50% of cases the measurement would report the photon as horizontally-polarized. This is not, however, the same kind of uncertainty that you get from flipping a coin. The photon is not either horizontally or vertically polarized. It is both at the same time.

    Amazed by this property of quantum systems, Schrödinger attempted to construct an example, where a domestic cat could be considered to be in the state

        \[|\mathrm{dead}\rangle + |\mathrm{alive}\rangle,\]

    which means being both dead and alive at the same time. The example he came up with, in his own words (citing from Wikipedia), is the following:

    Schrodingers_cat.svgA cat is penned up in a steel chamber, along with the following device (which must be secured against direct interference by the cat): in a Geiger counter, there is a tiny bit of radioactive substance, so small, that perhaps in the course of the hour one of the atoms decays, but also, with equal probability, perhaps none; if it happens, the counter tube discharges and through a relay releases a hammer that shatters a small flask of hydrocyanic acid. If one has left this entire system to itself for an hour, one would say that the cat still lives if meanwhile no atom has decayed. The first atomic decay would have poisoned it.

    The idea is that after an hour of waiting, the radiactive substance must be in the state

        \[|\mathrm{decayed}\rangle + |\text{not decayed}\rangle,\]

    the poison flask should thus be in the state

        \[|\mathrm{broken}\rangle + |\text{not broken}\rangle,\]

    and the cat, consequently, should be

        \[|\mathrm{dead}\rangle + |\mathrm{alive}\rangle.\]

    Correct, right? No.

    The Cat Ensemble

    Superposition, which is being "in both states at once" is not the only type of uncertainty possible in quantum mechanics. There is also the "usual" kind of uncertainty, where a particle is in either of two states, we just do not exactly know which one. For example, if we measure the polarization of a photon, which was originally in the superposition \left|\updownarrow\right\rangle + \left|\leftrightarrow\right\rangle, there is a 50% chance the photon will end up in the state \left|\updownarrow\right\rangle after the measurement, and a 50% chance the resulting state will be \left|\leftrightarrow\right\rangle. If we do the measurement, but do not look at the outcome, we know that the resulting state of the photon must be either of the two options. It is not a superposition anymore. Instead, the corresponding situation is described by a statistical ensemble:

        \[\{\left|\updownarrow\right\rangle: 50\%, \quad\left|\leftrightarrow\right\rangle: 50\%\}.\]

    Although it may seem that the difference between a superposition and a statistical ensemble is a matter of terminology, it is not. The two situations are truly different and can be distinguished experimentally. Essentially, every time a quantum system is measured (which happens, among other things, every time it interacts with a non-quantum system) all the quantum superpositions are "converted" to ensembles - concepts native to the non-quantum world. This process is sometimes referred to as decoherence.

    Now recall the Schrödinger's cat. For the cat to die, a Geiger counter must register a decay event, triggering a killing procedure. The registration within the Geiger counter is effectively an act of measurement, which will, of course, "convert" the superposition state into a statistical ensemble, just like in the case of a photon which we just measured without looking at the outcome. Consequently, the poison flask will never be in a superposition of being "both broken and not". It will be either, just like any non-quantum object should. Similarly, the cat will also end up being either dead or alive - you just cannot know exactly which option it is before you peek into the box. Nothing special or quantum'y about this.

    The Quantum Cat

    "But what gives us the right to claim that the Geiger counter, the flask and the cat in the box are "non-quantum" objects?", an attentive reader might ask here. Could we imagine that everything, including the cat, is a quantum system, so that no actual measurement or decoherence would happen inside the box? Could the cat be "both dead and alive" then?

    Indeed, we could try to model the cat as a quantum system with |\mathrm{dead}\rangle and |\mathrm{alive}\rangle being its basis states. In this case the cat indeed could end up in the state of being both dead and alive. However, this would not be its most exciting capability. Way more suprisingly, we could then kill and revive our cat at will, back and forth, by simply measuring its liveness state appropriately. It is easy to see how this model is unrepresentative of real cats in general, and the worry about them being able to be in superposition is just one of the many inconsistencies. The same goes for the flask and the Geiger counter, which, if considered to be quantum systems, get the magical abilities to "break" and "un-break", "measure" and "un-measure" particles at will. Those would certainly not be a real world flask nor a counter anymore.

    The Cat Multiverse

    There is one way to bring quantum superposition back into the picture, although it requires some rather abstract thinking. There is a theorem in quantum mechanics, which states that any statistical ensemble can be regarded as a partial view of a higher-dimensional superposition. Let us see what this means. Consider a (non-quantum) Schrödinger's cat. As it might be hopefully clear from the explanations above, the cat must be either dead or alive (not both), and we may formally represent this as a statistical ensemble:

        \[\{\left|\text{dead}\right\rangle: 50\%, \quad\left|\text{alive}\right\rangle: 50\%\}.\]

    It turns out that this ensemble is mathematically equivalent in all respects to a superposition state of a higher order:

        \[\left|\text{Universe A}, \text{dead}\right\rangle + \left|\text{Universe B}, \text{alive}\right\rangle,\]

    where "Universe A" and "Universe B" are some abstract, unobservable "states of the world". The situation can be interpreted by imagining two parallel universes: one where the cat is dead and one where it is alive. These universes exist simultaneously in a superposition, and we are present in both of them at the same time, until we open the box. When we do, the universe superposition collapses to a single choice of the two options and we are presented with either a dead, or a live cat.

    Yet, although the universes happen to be in a superposition here, existing both at the same time, the cat itself remains completely ordinary, being either totally dead or fully alive, depending on the chosen universe. The Schrödinger's cat is just a cat, after all.

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  • Posted by Konstantin 17.11.2016 No Comments

    Mass on a spring

    Imagine a weight hanging on a spring. Let us pull the weight a bit and release it into motion. What will its motion look like? If you remember some of your high-school physics, you should probably answer that the resulting motion is a simple harmonic oscillation, best described by a sinewave. Although this is a fair answer, it actually misses an interesting property of real-life springs. A property most people don't think much about, because it goes a bit beyond the high school curriculum. This property is best illustrated by

    The Slinky Drop

    The "slinky drop" is a fun little experiment which has got its share of internet fame.

    The Slinky Drop

    The Slinky Drop

    When the top end of a suspended slinky is released, the bottom seems to patiently wait for the top to arrive before starting to fall as well. This looks rather unexpected. After all, we know that things fall down according to a parabola, and we know that springs collapse according to a sinewave, however neither of the two rules seem to apply here. If you browse around, you will see lots of awesome videos demonstrating or explaining this effect. There are news articles, forum discussions, blog posts and even research papers dedicated to the magical slinky. However, most of them are either too sketchy or too complex, and none seem to mention the important general implications, so let me give a shot at another explanation here.

    The Slinky Drop Explained Once More

    Let us start with the classical, "high school" model of a spring. The spring has some length L in the relaxed state, and if we stretch it, making it longer by \Delta L, the two ends of the spring exert a contracting force of k\Delta L. Assume we hold the top of the spring at the vertical coordinate y_{\mathrm{top}}=0 and have it balance out. The lower end will then position at the coordinate y_{\mathrm{bot}} = -(L+mg/k), where the gravity force mg is balanced out exactly by the spring force.

    How would the two ends of the spring behave if we let go off the top now? Here's how:

    The falling spring, version 1

    The horozontal axis here denotes the time, the vertical axis - is the vertical position. The blue curve is the trajectory of the top end of the spring, the green curve - trajectory of the bottom end. The dotted blue line is offset from the blue line by exactly L - the length of the spring in relaxed state.

    Observe that the lower end (the green curve), similarly to the slinky, "waits" for quite a long time for the top to approach before starting to move with discernible velocity. Why is it the case? The trajectory of the lower point can be decomposed in two separate movements. Firstly, the point is trying to fall down due to gravity, following a parabola. Secondly, the point is being affected by string tension and thus follows a cosine trajectory. Here's how the two trajectories look like separately:

    They are surprisingly similar at the start, aren't they? And indeed, the cosine function does resemble a parabola up to o(x^3). Recall the corresponding Taylor expansion:

        \[\cos(x) = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \dots \approx 1 - \frac{x^2}{2}.\]

    If we align the two curves above, we can see how well they match up at the beginning:

    Consequently, the two forces happen to "cancel" each other long enough to leave an impression that the lower end "waits" for the upper for some time. This effect is, however, much more pronounced in the slinky. Why so?

    Because, of course, a single spring is not a good model for the slinky. It is more correct to regard a slinky as a chain of strings. Observe what happens if we model the slinky as a chain of just three simple springs:

    Each curve here is the trajectory of one of the nodes inbetween the three individual springs. We can see that the top two curves behave just like a single spring did - the green node waits a bit for the blue and then starts moving. The red one, however, has to wait longer, until the green node moves sufficiently far away. The bottom, in turn, will only start moving observably when the red node approaches it close enough, which means it has to wait even longer yet - by that time the top has already arrived. If we consider a more detailed model, the movement  of a slinky composed of, say, 9 basic springs, the effect of intermediate nodes "waiting" becomes even more pronounced:

    To make a "mathematically perfect" model of a slinky we have to go to the limit of having infinitely many infinitely small springs. Let's briefly take a look at how that solution looks like.

    The Continuous Slinky

    Let x denote the coordinate of a point on a "relaxed" slinky. For example, in the two discrete models above the slinky had 4 and 10 points, numbered 1,\dots, 4 and 1,\dots, 10 respectively. The continuous slinky will have infinitely many points numbered [0,1].

    Let h(x,t) denote the vertical coordinate of a point x at time t. The acceleration of point x at time t is then, by definition \frac{\partial^2 h(x,t)}{\partial^2 t}, and there are two components affecting it: the gravitational pull -g and the force of the spring.

    The spring force acting on a point x is proportional to the stretch of the spring at that point \frac{\partial h(x,t)}{\partial x}. As each point is affected by the stretch from above and below, we have to consider a difference of the "top" and "bottom" stretches, which is thus the derivative of the stretch, i.e. \frac{\partial^2 h(x,t)}{\partial^2 x}. Consequently, the dynamics of the slinky can be described by the equation:

        \[\frac{\partial^2 h(x,t)}{\partial^2 t} = a\frac{\partial^2 h(x,t)}{\partial^2 x} - g.\]

    where a is some positive constant. Let us denote the second derivatives by h_{tt} and h_{xx}, replace a with v^2 and rearrange to get:

    (1)   \[h_{tt} - v^2 h_{xx} = -g,\]

    which is known as the wave equation. The name stems from the fact that solutions to this equation always resemble "waves" propagating at a constant speed v through some medium. In our case the medium will be the slinky itself. Now it becomes apparent that, indeed, the lower end of the slinky should not move before the wave of disturbance, unleashed by releasing the top end, reaches it. Most of the explanations of the slinky drop seem to refer to that fact. However when it is stated alone, without the wave-equation-model context, it is at best a rather incomplete explanation.

    Given how famous the equation is, it is not too hard to solve it. We'll need to do it twice - first to find the initial configuration of a suspended slinky, then to compute its dynamics when the top is released.

    In the beginning the slinky must satisfy h_t(x, t) = 0 (because it is not moving at all), h(0, t) = 0 (because the top end is located at coordinate 0), and h_x(1, t) = 0 (because there is no stretch at the bottom). Combining this with (1) and searching for a polynomial solution, we get:

        \[h(x, t) = h_0(x) = \frac{g}{2v^2}x(x-2).\]

    Next, we release the slinky, hence the conditions h_t(x,t)=0 and h(0,t)=0 disappear and we may use the d'Alembert's formula with reflected boundaries to get the solution:

        \[h(x,t) = \frac{1}{2}(\phi(x-vt) + \phi(x+vt)) - \frac{gt^2}{2},\]

        \[\text{ where }\phi(x) = h_0(\mathrm{mod}(x, 2)).\]

    Here's how the solution looks like visually:

    Notice how the part of the slinky to which the wave has not arrived yet, stays completely fixed in place. Here are the trajectories of 4 equally-spaced points on the slinky:

    Note how, quite surprisingly, all points of the slinky are actually moving with a constant speed, changing it abruptly at certain moments. Somewhat magically, the mean of all these piecewise-linear trajectories (i.e. the trajectory of the center of mass of the slinky) is still a smooth parabola, just as it should be:

    The Secret of Spring Motion

    Now let us come back to where we started. Imagine a weight on a spring. What will its motion be like? Obviously, any real-life spring is, just like the slinky, best modeled not as a Hooke's simple spring, but rather via the wave equation. Which means that when you let go off the weight, the weight will send a deformation wave, which will move along the spring back and forth, affecting the pure sinewave movement you might be expecting from the simple Hooke's law. Watch closely:

    Here is how the movement of the individual nodes looks like:

    The fat red line is the trajectory of the weight, and it is certainly not a sinewave. It is a curve inbetween the piecewise-linear "sawtooth" (which is the limit case when the weight is zero) and the true sinusoid (which is the limit case when the mass of the spring is zero). Here's how the zero-weight case looks like:

    And this is the other extreme - the massless spring:

    These observations can be summarized into the following obviously-sounding conclusion: the basic Hooke's law applies exactly only to the the massless spring. Any real spring has a mass and thus forms an oscillation wave traveling back and forth along its length, which will interfere with the weight's simple harmonic oscillation, making it "less simple and harmonic". Luckily, if the mass of the weight is large enough, this interference is negligible.

    And that is, in my opinion, one of the interesting, yet often overlooked aspects of spring motion.

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  • Posted by Konstantin 24.07.2015 No Comments

    Holography is a fascinating way of storing and displaying visual information - one which will, hopefully, one day become as commonplace as photography and video is today. Unfortunately, at the current moment it is still a largely exotic trade, limited to artistic exhibitions or science museum displays. While the explanation of how photography works is usually part of the standard school curriculum, holography, at the same time, is often regarded as some mysterious sci-fi technology. The problem is made worse by the fact that there do not seem to be too many explanations around in the Internet, which would aim to properly dispel the aura of mystery. Here is my take at fixing that problem.

    Let us start by considering the following illustration, where two hypothetical eyes are observing two hypothetical light sources.

    Figure 1

    The light from the two lamps reaches the eyes at different angles and intensities, which depend on the positions of the eyes relative to the lights. Because of that difference in angles and intensities, the brain, attached to the eyes, is capable of "triangulating" the location of the lamps, thus forming an enjoyable "three-dimensional" perception of the whole scene for itself.

    Figure 2

     

    Let us position a piece of glass in front of the light sources. We can regard this glass as a set of "pixels", through which light rays fly from the lamps towards the eyes. The figure below illustrates three such randomly selected "pixels" on the glass along with the directions of rays that leave them.

    Figure 3

     

    What if we could design a piece of glass which would fully consist of such hypothetical "pixels", with each pixel emitting two light rays in the appropriate directions? The eyes, observing such a piece of glass, would not know that the rays are really produced "in the pixels". Indeed, they perceive exactly the same light rays as if the light was coming from two lamps, positioned somewhere behind the glass. Hence, the brain, attached to the eyes, has to perform the same triangulation as the one it did when the lamps were real, and is not be able to distinguish the "fake glass" from a real scene with the lamps.

    Although this idea of faking reality might look nice, the task of manufacturing small enough "pixels", which would be able to emit light in a number of given directions with given intensities and colors, is, unfortunately, technologically too complex to be practical.

    This is where Physics comes to help us out. "Hey, guys," - she says, - "you are forgetting that light is not really a set of rays flying from from point A to point B along straight lines. Light is a wave. Hence, assuming that your light sources are emitting coherent light of a given frequency, a correct illustration of what happens in your scene with two lamps would be the following:"

    Animation 1

     

    If we now consider an arbitrary point-"pixel" on our glass, the electromagnetic waves which pass through it are perceived there in the form of some oscillation. The whole situation above could be modeled by considering waves on a water surface. To simulate the "glass" (the blue line) let us have a horizontal row of buoys float on the surface. Each of the buoys will be moving up and down in its place due to the waves, and we can record this movement.

    Here comes a question: what if we remove the original two oscillation sources and instead will start "replaying" the recorded movements of the buoys buy forcing them to move up and down and thus form their own waves? The example below simulates this situation using 19 "buoys":

    Animation 2

     

    Does not look nice at all, does it? Now Mathematics comes along to help: "Try harder!" - she says. "If your buoys are dense enough, everything will work out!". Hence, we increase the number of buoys and try again:

    Animation 3

     

    Indeed, now we observe how a row of oscillating buoys (an analogue of a "glass" in our model) is generating a wave in front of it, which is completely identical to the one that was produced when two sources were oscillating behind it. No matter where we position ourselves in front of the buoy row, our perception will happily fool us into thinking that the wave we see is really a result of two point sources oscillating somewhere in a distance. Our brain will even help us to estimate the exact position of those sources somewhere behind the row. He does not know he is being fooled. Try covering the upper half of the animation above to get a better understanding of this illusion.

    Hence, if we can make a glass, which consists of "pixels" each of which could "record" a profile of a light wave and then be able to reproduce it, the "image" observed by looking at such glass would be perfectly identical to an image that would be observed if the glass would simply pass through it the light from some sources located behind it. But is it possible? Isn't asking each "pixel" to record an arbitrary oscillation too much?

    Here is when Mathematics comes along to help again. "Do you want to see a magic trick?" - she asks. "Yes!" - we answer.

    Remember that you have two light sources emitting waves behind your piece of glass. As we agreed, they are emitting coherent light, i.e. light of the same frequency (for example, you could shine a laser light on the two points and have them reflect this light). Now let us see what happens at one of the pixels-"buoys" of your glass.

    This point receives light waves from both sources. As the distances to the two sources are different, the two light waves reach the pixel with different intensities and phases. Formally, suppose the wave from the first source reaches the pixel with amplitude a_1 and phase b_1 and the wave from the second source reaches the pixel with amplitude a_2 and phase b_2. The two waves add up, so the pixel "feels" the overall oscillation of the form

        \[a_1 \sin(ft + b_1) + a_2 \sin(ft + b_2).\]

    Now if you have not two, but three, four, ten, or a million of point sources behind the glass, the arriving signal will be, correspondingly, a sum of three, four, ten, or even a million sinewaves with different amplitudes and phases.

    Now here comes the trick, hold your breath: Sum of any number of such sinewaves is itself just a single sinewave with some amplitude and phase.

    In other words, even if there is a million point sources emitting coherent light from behind the glass, each "pixel-buoy" of our glass will still perform a simple oscillation of the form A \sin(ft + B). Hence, to "record" the oscillation of each "pixel-buoy" we must store only two parameters: its intensity and phase.

    Now everything starts to look much simpler. Storing light intensity of a pixel is simple - any photograph does that. Dark spots on a photo or a screen emit (or reflect) low intensity light, white spots emit high intensity. It only remains to store the phase. Several methods have been proposed for that. Conceptually the simplest one (but technologically the most advanced) makes use of particular crystals which can shift the phase of light that passes through them. Each "pixel" could then be made out of such a crystal with an appropriate phase shift coded in it. When a "glass" made of such crystals is lit from the back by a coherent light source, each pixel would introduce the necessary phase shift into the light which passes through it, and thus reproduce the necessary wave front. I presume that this HoloVisio project is relying on this idea to design a proper holographic monitor.

    The conventional holograms, the ones you typically see in science museums, are made using a simpler principle, though. Omitting some technical details, the idea behind it is the following. Let us add to the scene a reference ray of coherent light.

    Animation 4

     

    Now at some pixels on our "glass" the phase of the reference ray will coincide with the phase of the light coming from the scene. The overall wave oscillation at those points will be increased. At other points the phases of the reference and the scene light will mismatch and the signal will be weakened. We can record the resulting interference pattern by storing light spots on the glass at the pixels where the phases matched and dark spots where the phases did not match.

    Now, whenever we shine the reference ray again through the resulting plate, only the parts of the ray with the "correct" phase will be let through, as those are exactly the places where the glass is transparent to the reference ray. As a result, you'll obtain a decent holographic reproduction. Here's a figure from Wikipedia, illustrating this principle of recording and reproducing holograms:

    Hologram

     

    Considering the speed at which our photocameras have been transitioning over the recent decades from megapixel resolutions into gigapixels and observing the fast spread of stereoscopic "3D" from movie theaters into TV-sets and portable devices, it is both surprising and sad to see no notable signs of proper holographic display technology being developed in parallel. I do hope to see mass-produced holographic displays along with appropriate software within my own lifetime, though.

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  • Posted by Konstantin 19.02.2009 6 Comments

    Everyone, who flies reasonably often has a chance to observe a remarkable effect: no matter how steep the roll angle of an airliner is during a turn, the tea in your cup will always stay parallel to the floor, and not to the ground. In the course of one recent unexpectedly long discussion on this topic, it turned out there is no much easily googleable material out there to refer to. I thought I'd create some, hence a somewhat longer post on an otherwise simple matter of minor relevance to my main affairs.

    So, firstly, why doesn't tea orient itself parallel to the ground? To understand that, consider a pair of insightful examples. They should rid you of the incorrect intuitive assumption that it is gravity, which plays a defining role in the orientation of your tea.

    The bucket experiment

    Example 1: Water in the bucket

    Consider the experiment described in the previous post about a water bucket on a string. No matter what you do with the bucket, as long as the string is at strain, the water will stay parallel to bottom of the bucket, and not to the ground. The explanation for that is the following. Water is affected by gravity G. The bucket is affected by gravity G and the strain of the string S. Therefore, from the frame of reference of the bucket, water experiences force G-(G+S) = -S, which pulls it towards the bottom of the bucket.

    Example 2: Paragliders

    A paraglider

    You have probably seen paragliders in the air. If not, just search for "extreme paragliding" on youtube. Note that whatever the angle of the wing, the person is always hanging perpendicularly to it. Now if you'd hand him a bottle with water, the water would, naturally, also orient itself in parallel to the wing. The reason of that behaviour is the same as in the water bucket experiment, with the lift of the wing playing the same role as the strain of the string.

    The main idea of both examples is that as long as the container and the object inside it are equally affected by gravity, it is not gravity that orients the object with respect to the container. More precisely, for the water to incline to the side, there must be some force acting on the container from that side.

    The Airplane: A Simple Model
    Let us now consider an idealized airplane. According to a popular model, an airplane in flight is affected by four forces: the thrust of the engines T, the drag due to air resistance D, the lift of the wings L and the gravitational weight of the airplane G. The glass of tea inside the airplane is, of course, only affected by gravity G. By applying the same logic as before, we can easily compute, that in the frame of reference of the airplane, the tea is experiencing acceleration F = G-(T+D+L+G) = -(T+D+L).

    However, once the airplane has attained constant speed (which is true for most of the duration of the flight), its thrust is completely cancelled by the air resistance, i.e. T+D = 0, in which case F = -L. It now remains to note that the lift force of the wings is always approximately perpendicular to the wings and thus to the floor of the airliner. The tea in your cup must therefore indeed be parallel to the floor.

    The Airplane: A More Realistic Model
    The model considered in the previous section is somewhat too simplistic. According to it there are no forces acting on the sides of the airplane whatsoever, and it should therefore be absolutely impossible to incline the tea in the cup to the side, which is, of course, not true for a real airplane. So, assuming a pilot would want to incline the tea and spill it, what would be his options?

    1. Thrust. As noted above, tea must only be parallel to the floor when thrust is perfectly cancelled by drag and the plane is moving with constant speed. A rapid increase or decrease in thrust could therefore incline the tea towards or against the direction of flight. But of course, there is usually no need for such a maneuver during a passenger flight except for takeoff and landing.
    2. Rotation. The model above does not consider the fact, that the pilot may use the ailerons and rudder to turn the airplane, rotating it around its axes. And of course, when the resulting rotation is abrupt enough, the tea could incline somewhat. However, during passenger flights the turns are always performed very smoothly, with rotation speeds around 1 degree per second. In fact it is risky, if not impossible to perform any hasty rotationary maneuvers on an airliner travelling at about 800km/h.
    3. Turn with a skid

      Turn with a skid

      Slip. The simple model above considered the case when the air is flowing directly along the main axis of the airplane, which need not necessarily always be the case. The condition may be violated either due to a strong sidewind, or during a peculiar kind of a turn, where the airplane "slips" or "skids" on the side. In both cases, the airflow is exerting pressure on the side of the airplane's hull, which generates the so-called body lift. It is usually incomparably smaller than the lift of the wings, but nonetheless, it can incline the water to the side.
      It is interesting to understand why you should almost never experience slip in an airliner. There are two reasons for that. Firstly, most airplanes have a degree of weathercock stability. Like a weathercock, an airplane with a vertical tail stabilizer tends to automatically orient itself into the direction of airflow and thus avoid slip. This effect is especially strong at the speed of a commercial airliner.
      Secondly, if the weathercock effect is not enough to prevent slip, the pilot himself will always ensure that the slip is never too large by watching the slip-indicator (aka inclinometer) and coordinating the turn.
      Why that? Because the airplane is not constructed, aerodynamically, to fly sideways. When the plane is moving sideways, the body of the plane blocks airflow over the trailing wing. So the wing loses lift and begins to drop. This naturally will put the airplane into a bank in the direction of the turn, but it does so at great cost in drag. When the slip is too large the lifting properties of the wings change so drastically that this might put the airplane at the risk of a crash.

    Summary
    Q: Why is the tea in an airliner parallel to the floor even when the airplane is turning?
    A: It follows from the following three conditions, that must all be satisfied for a safe flight:

    1. The forces of thrust and drag cancel each other and the airplane moves at a constant speed.
    2. The turns are performed smoothly.
    3. There is no slip: the air flows directly along the main axis of the airplane.

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  • Posted by Konstantin 12.02.2009 3 Comments
    Bucket swing experiment

    Most of us probably remember this experiment from high school physics lessons: you take a bucket on a string filled with water, spin it around your head and the water does not spill. "But how?" - you would ask in amazement. And the teacher would explain then:
    "You see, the bucket is spinning, and this creates the so-called centrifugal force acting on the water, which cancels out gravity and thus keeps the water in the bucket". And you will have a hard time finding any other explanation. At least I failed no matter how hard I googled it.

    Unfortunately, this explanation looses an essential point of the experiment and I have seen people irreparaply braindamaged by the blind belief that it is only due to rotation and the resulting virtual centrifugal force that the water does not spill.

    However, it is not quite the case. Let us imagine that the bucket has accidentally stopped right over your head and as a result, all centrifugal force has been immediately lost. Would the water spill? It will certainly fall down on your head, but it will do so together with the bucket. Thus, technically, the water will stay inside the bucket.

    In fact, the proper way to enjoy the true magic of the experiment is not to swing the bucket in full circles, but rather let it swing back and forth as a pendulum (if you have a string and a beverage bottle nearby, you can do an experiment right now). One will then observe that even at the highest points of the swing, where the bottom of the bucket is at its steepest angle and the centrifugal force is nonexistant, the water stays strictly parallel to the bottom of the bucket, as if no gravity would act upon it. Why doesn't it spill? Clearly, the argument of centrifugal force cancelling gravity is inappropriate.

    Forces

    The proper explanation is actually quite simple and much more generic. We have two objects here: the bucket and the water in it. There is one (real) force acting on the water: gravity G. There are two (real) forces acting on the bucket: gravity G and the strain S of the string pulling the bucket perpendicularly to its bottom. (Note that the centrifugal force is not "real" and I do not consider it here, but if you wish, you may. Just remember then that it acts both on the water and the bucket.)
    Now the question of interest is, how does water behave with respect to the bucket? That is, what force "pulls" the water towards the bucket and vice-versa. This can be easily computed by subtracting all forces acting on the bucket from all forces acting on the water. And the result is, of course, G - (S+G) = -S, i.e. a force, pulling the water directly towards the bottom of the bucket.

    A magical consequence of this argument is that gravity does not matter inside the bucket, as long as it can act on the bucket freely in the same way as on anything inside it. Nothing special about rotation here, really. It takes a while to realize.

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