A bit of maths and physics

Tag: vectors

  • Why, exactly, do glass and liquids refract light?

    Why, exactly, do glass and liquids refract light?

    Summer has arrived, and you have been served a gorgeous-looking cocktail. Condensation droplets on the glass reveal you are set for a much needed particularly refreshing indulgence. However, just as you were about to soak up the colourful fluid of blissful gratification, your shockingly intelligent child asks why the straw seems to be broken inside your drink. And if not about that, then it’s about why this bear’s head is in the wrong place. Sure, the answer is light refraction, but why, exactly, do glass and liquids refract light?

    People looking at a bear in his habitat in the zoo. The side is transparent, so people can see the bear standing in the water from the side, partially submerged. Due to the light refraction caused by the water and the glass, the bear's head is located at a different place than his submerged body. Dramatically displaced.

    We will provide you with the answer. However, before we begin, we need to ask, TL;DR? Rather not see formulas? Scroll down to the last section, the Quick summary. More curious? Then by all means, read on. I promise, not a single calculation will be done. And if you do read on, you will know actual physics. Shockingly more than most.

    For your convenience, here’s a little table of contents:
    A few incorrect explanations
    Why are they incorrect?
    Step 1. Not particles, not waves: it’s all fields
    Step 2. Maxwell’s field equations
    Step 3. Draw the vectors
    Step 4. The electric field inside of materials
    Quick summary: why, exactly, do glass and liquids refract light?

    A few incorrect explanations

    What would you answer? Here are just a few bad examples which other people (but not you) tend to tell their offspring.

    1. Light takes the fastest route. As its speed differs per material, it needs to change direction. Or: light takes the path of the least amount of action. Same reasoning.
    2. When light enters the glass and the liquid, it bounces back and forth between the molecules and atoms of the material. Due to their crystalline or liquid arrangement, the overall direction of light changes. Hence, light is refracted.
    3. Light consists of particles, so-called photons, which get absorbed by the atoms of the material, causing their electrons to temporarily increase their orbital radius around the nucleus. The instant they fall back to their original orbit, they emit another photon in a direction which depends on and is consistent with the type of atom, i.e. material. The overall result is that the beam of particles has changed direction. Hence, light is refracted.
    4. Huygens’ Principle. Light is a wave. Every point on its wavefront can be a source for a circular wavelet. Draw them, connect the dots and you’ll see: light gets refracted.

    Why are they incorrect?

    1. Okay, this is not incorrect, however, while light does that, it doesn’t explain what really happens. It’s an answer to a different kind of question. So, to be ‘that person’ here, in terms of answer-to-the-question-asked, it’s incorrect after all.
    2. By this logic, light should appear much more spread out due to the probabilistic nature of the supposed bouncing back and forth between chaotically moving or vibrating molecules and atoms. The specific direction of bouncing light is not guaranteed to be as consistent as we nevertheless observe in the real world. The resulting image should be a blur. It is not.
    3. Here too, light should appear much more spread out. The direction of the re-released photon is not guaranteed to be in the direction we observe in the real word. A photon could be re-emitted in any direction, regardless of the type of atom. The frequency of the photon correlates with the atomic configuration, not its direction. Moreover, ‘getting absorbed’ and ‘re-emitted’ are not well defined. What does that even mean?
    4. This is a sophisticated one. At first glance, it does produce an angle for the outbound light beam. However, Huygens’ Principle only corresponds to observations if you cherry pick from multiple possibilities. See Figure 1 for a brief explanation.
    (a) The vertical lines represent the crests of the light wave. The blue area is the glass or liquid. As light only bends in these materials at an angle, the diagram shows a beam of light approaching the surface of the material at an angle. Huygens proposed that at every instance 'wavelets' (drawn here as segments of dotted circles) can be thought emanating at every point in space, growing over time. Connecting the wavefronts of those wavelets predicts the course of the next wave (crest). As the bottom of the incoming crests hit the surface first, those wavelets will have had time to grow larger before the top of the incoming crests hit the surface. (b) Over time, multiple wavelets can be thought to have developed. (c) Where the wavelets intersect each other wave crests can be drawn. The result seems to be the predicted new progression of the light beam inside of the material. (d) However, over time, multiple intersections will have developed. By Huygens' logic, multiple wave crests could be drawn. This, however, would result in a diffuse light wave, spreading out its light instead of a distinct bending of the one beam. Stating that a situation as sketched in (c) will occur is selectively choosing a preferred scenario while (d) shows multiple would occur. Hence, Huygens' principle seems right at first but ultimately breaks down over time.
    (a) The vertical lines represent the crests of the light wave. The blue area is the glass or liquid. As light only bends in these materials at an angle, the diagram shows a beam of light approaching the surface of the material at an angle. Huygens proposed that at every instance ‘wavelets’ (drawn here as segments of dotted circles) can be thought emanating from the wavefront at every point in space, growing over time. Connecting the wavefronts of those wavelets predicts the course of the next wave (crest). As the bottom of the incoming crests hit the surface first, those wavelets will have had time to grow larger before the top of the incoming crests hit the surface. (b) Over time, multiple wavelets can be thought to have developed. (c) Where the wavelets intersect each other wave crests can be drawn. The result seems to be the predicted new progression of the light beam inside of the material. (d) However, over time, multiple intersections will have developed. By Huygens’ logic, multiple wave crests could be drawn. This, however, would result in a diffuse light wave, spreading out its light instead of a distinct bending of the one beam. Stating that a situation as sketched in (c) will occur is selectively choosing a preferred scenario while (d) shows multiple would occur. Hence, Huygens’ principle seems right at first but ultimately breaks down over time.

    Step 1. Not particles, not waves: it’s all fields

    So, what does make light refract then? We need to take a few mental steps. Here is the first one, which you’ll just have to get used to.

    Space throughout the entire observable Universe is filled with fields. In fact, fields are a property of space. Space without fields does not exist. With space come fields. Points in most fields not only have a value, they also have a direction. They are called vector fields. Some are called scalar fields; their points have no direction, they only have values. There are more types of fields, such as tensor fields and fermionic fields. This is quantum field theory (QFT), the most successful and accurate theory to date. Has been for well over ninety years (including a renaissance in the 1970s).

    Next question is, what concrete fields are we talking about? You probably heard of or read about the Higgs field(beginfootnote)It just so happens this is not a vector field; its points have no direction, just values, and so, it is a scalar field.(endfootnote). In 2012, the Large Hadron Collider at CERN produced an oscillation in the Higgs field or rather an excitation. That excitation is what we call the Higgs particle. The energy produced inside the LHC was more than enough to cause an excitation of the Higgs field, which we perceive as a particle(beginfootnote)The Higgs particle itself was indirectly observed as its lifespan is too short. It decays quickly into other particles, or excitations, in other fields. Those, however, live long enough for the detectors to observe.(endfootnote). The field was proven to be a real thing. Two Nobel Prizes were awarded to François Englert and Peter Higgs for having proposed the existence of the Higgs field forty-eight years earlier. It proved how humans with their shockingly tiny brains were able to probe the depths of the subatomic world, a thousand times smaller than the atomic nucleus, and the entire observable Universe at the same time. By using maths and, forty-eight years later, by building ingenious experiments.

    There are more fields. There is an electron field. Most of the time, the field has value zero. But when the values of a tiny part of that field oscillate at a distinct frequency, we call that an electron.

    There is also an electromagnetic field. A stream of billions of local oscillations of a range of frequencies is what we call a beam of visible light. It’s practical to sometimes talk about it as it being particles (called photons) as well as it being waves (electromagnetic radiation). It depends on what you’re calculating.

    You could say there is also a proton field, although there are more fundamental fields than this, for instance the quark and gluon fields (protons aren’t elementary particles, they consist of quarks and gluons). However, for the purpose of this post, we will work with the simpler notion of a proton field. A fairly local oscillation is a proton, which we usually perceive as a particle.

    There are many more fields but to discuss them all would justify a separate article. Or several books. And a couple of years of study.

    So, what is light, what are electrons, what are protons or quarks? Are they both particle and wave? No, that’s an old and misleading question. Are they sometimes particles, sometimes waves then? No, also not that.

    ‘Particles’ aren’t actual particles like tiny silver ball bearings or something like that. They are best described as a mathematical function, which we call the wave function (denoted by the symbol Ψ). When measured they are fairly local oscillations or excitations at specific frequencies in fields pervading through all of space, almost behaving like particles. Sometimes, it’s practical to mathematically model them as either particles or waves, depending on the situation. However, it’s meaningless to state they are either or both at the same time. It’s more accurate to just treat them as mathematical wave functions instead of anything elseA personal conviction is currently that the wave function is all there is. Elementary ‘particles’ are wave functions. Nothing more, nothing less. Favouring ‘tangible objects’ over ‘mere mathematical descriptions’, which, nevertheless have been proven to be incredibly accurate after billions and billions of experimental runs, is really just exposing our limited understand of quantum physics as confined by everyday, large-scale experiences such as playing with base-, basket- and footballs. There is no reason, however, to assume the latter are a measure to gauge the subatomic foundation of our Universe. In my view, that should be the wave function. — KJ.

    Figure 2. Three fields of space are drawn stacked. In reality, they are three-dimensional and not stacked and separated as depicted here. Instead, they are occupying the same space, completely blended with each other.
    Figure 2. Three fields of space are drawn stacked. They are two-dimensional here, but in reality they are three-dimensional and fill the same three-dimensional space, completely immersed in and blended with each other. The problem is that drawing mixed and blended three-dimensional stuff is hard on a two-dimensional screen. In this diagram, no ‘particles’ are present at the moment. There are no oscillating excitations in the fields. In other words, the field values are zero, there are no particles, but the fields are still there. Filling space. (Just to be entirely precise, in reality, the fields do always oscillate a little bit as predicted by Heisenberg’s uncertainty principle.)

    Step 2. Maxwell’s field equations

    James Clerk Maxwell was, besides Scottish, a scientist in the field of mathematical physics. Having studied the previous work of Faraday, Gauss, and Ampère, he showed that an electric field and a magnetic field were the same thing, just different aspects of it. That thing is what we now call the aforementioned, space-filling electromagnetic field. He also showed that light was an electromagnetic phenomenon. A disturbance in the field.

    Engraving of James Clerk Maxwell by G. J. Stodart from a photograph by Fergus of Greenock. Frontpiece in James Maxwell, The Scientific Papers of James Clerk Maxwell. Ed: W. D. Niven. New York: Dover, 1890. Public domain.

    He formulated a set of four differential equations. This set bears his name. We will only ‘use’ two of the four:

    \begin{align}
    \mathbf{\nabla} \cdot \mathbf{E} &= \frac{\rho}{\varepsilon_0}, \\
    \mathbf{\nabla} \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t}.
    \end{align}

    I say, ‘use’, but don’t worry, we’re not going to do any complicated calculations.

    Equation (1) is called Gauss’s law and shows how the electric field (which is one aspect of the electromagnetic field), denoted by E, is influenced by a charge $\rho$, such as the negative charge of an electron or the positive charge of a proton. The symbol $\varepsilon_0$ denotes a constant, which differs depending on the material. The subscript 0 denotes it’s the constant of the vacuum of space. This physical constant $\varepsilon_0$ has different names such as vacuum permittivity, permittivity of free space or the electric constant. In this article, we will use different values for this constant, however. We will use

    \[ \varepsilon_\text{air} \text{ and } \varepsilon_\text{mat}. \]

    ‘Mat’ is short for ‘material’ which could be glass or liquid, for example. The precise numerical values we won’t use, because that’s not important for understanding why light bends. These two epsilons will turn out to play a pivotal role in the bending of light by materials, however. Do read on, I’d say.

    In Figure 3, the three fields in space are again depicted. This time, you can see the elevated values as blobs in the proton field. They are protons. They are surrounded by electron blobs as is depicted in the electron field. Both ‘particles’ influence the electromagnetic field, or, rather, the electric subfield thereof. Note that the blobs in the latter do not constitute ‘particles’, merely influences in the electric field.

    Figure 3. Protons and electrons, together constituting atoms, influence the electromagnetic field.
    Figure 3. Protons and electrons, together constituting atoms, influence the electromagnetic field.

    Step 3. Draw the vectors

    Have a look at Figure 4. The orange arrow or vector denotes the direction of the light inside whatever material we have, such as glass or a liquid. Maxwell showed that light, being an oscillation in the electromagnetic field, has an oscillatory component in the electric subfield of the electromagnetic field. And that electric field is orientated perpendicular to the direction of light. This is represented by the green vector.

    Figure 4. Light inside of the material falls at an angle onto the surface of the material. It has an electric field oscillation perpendicular to its direction.
    Figure 4. Light inside of the material falls at an angle onto the surface of the material. It has an electric field oscillation perpendicular to its direction.

    It is important to note that the electric field vector has two fundamental components, namely a component vector parallel to the surface, denoted by the symbol $\parallel$, and a component vector perpendicular to the surface, denoted by the symbol $\perp$. This is depicted in Figure 5.

    Figure 5. The electric field inside the material, caused by the light, has two vector components: parallel and perpendicular to the surface.
    Figure 5. The electric field inside the material, caused by the light, has two vector components: parallel and perpendicular to the surface.

    Exactly at the surface, the transition from the material to air, the electric field of the material and the electric field of the air will have to ‘slide’ to an equal value (or else we would have a tear in our universe). This means that

    \begin{align}
    \varepsilon_\text{air}(\mathbf{\nabla} \cdot \mathbf{E}_\text{air}) &= \varepsilon_\text{mat}(\mathbf{\nabla} \cdot \mathbf{E}_\text{mat}), \\
    \mathbf{\nabla} \times \mathbf{E}_\text{air} &= \mathbf{\nabla} \times \mathbf{E}_\text{mat}.
    \end{align}

    If we do a little bit of calculus, we come to the following equations:

    \begin{align}
    \mathbf{E}_{\text{mat}\parallel} &= \mathbf{E}_{\text{air}\parallel} \\
    \varepsilon_\text{mat}\mathbf{E}_{\text{mat}\perp} &= \varepsilon_\text{air}\mathbf{E}_{\text{air}\perp}.
    \end{align}

    So, the vector components of both air and the material parallel to the surface are equal. However, the vector components perpendicular to the surface are not. This is the crux:

    \[ \varepsilon_\text{mat} \neq \varepsilon_\text{air}. \]

    In fact,

    \[ \varepsilon_\text{mat} > \varepsilon_\text{air}. \]

    This means that the perpendicular vector component of air has to be larger than that of the material in order to satisfy equation (6). This is depicted in Figure 6.

    Figure 6. Because the epsilon (electric constant) of the material is larger than that of air, the perpendicular component vector of air has to be larger due to satisfy the equation. Here, a new resultant vector of the electric field in air has been drawn superimposed on the old electric field vector to emphasise the difference.
    Figure 6. Because the epsilon (electric constant) of the material is larger than that of air, the perpendicular component vector of air has to be larger due to satisfy the equation. Here, a new resultant vector of the electric field in air has been drawn superimposed on the old electric field vector to emphasise the difference.

    The only thing left to do, is to draw the new direction of the light in the air. As we know that the direction of the electric field is perpendicular to the direction of light, courtesy to Maxwell and colleagues, we can easily construct the new course of the light outside in the open air as is depicted in Figure 7.

    Figure 7. The new direction of light in air is refracted relative to the original angle inside of the material, just as vector calculus predicted.
    Figure 7. The new direction of light in air is refracted relative to the original angle inside of the material as predicted by vector calculus, just as observed in reality.

    Step 4. The electric field inside of materials

    The question is now: what causes the electric constant of materials to be so different?

    Figure 8 shows a schematic depiction of what light, being the cause for disturbances in the electric field itself, does to electrically charged ‘particles’ inside a material in terms of quantum field perturbations. Note how the alignment of charges and thus the oscillations in the electromagnetic field have changed in such a way that the electric subfield as a whole, inside of the material, has to have changed values as well. These changes are encapsulated in the electric constant, $\varepsilon_\text{mat}$.

    Light changes a material’s electromagnetic configuration, which then influences the trajectory of that same light.

    Figure 8. Light changes the electromagnetic configuration inside a material, which then influences the trajectory of that same light.
    Figure 8. Light changes the electromagnetic configuration inside a material, which then influences the trajectory of that same light.

    Quick summary: why, exactly, do glass and liquids refract light?

    The Universe, i.e. space itself(beginfootnote)With ‘space’, we don’t mean ‘outer space’ but rather the thing we move in, the volume, the expanse, the invisible yet essential thing allowing us to move back-and-forth, up-and-down, left-and-right.(endfootnote), contains an omnipresent electromagnetic field. Mathematically, we can divide this field up into two components: its electric (sub)field and its magnetic (sub)field.

    Electrons in glass and liquids as well as light are influenced by the electric field. At the same time, they influence that same electric field. When light hits the material, it changes the electric field inside the material. This makes electrons bring about opposite electric field changes in turn. The net electric field inside the glass changes the light’s direction of propagation in a perfectly predictable way. Courtesy of vector calculus.

    Or, if you prefer:

    Light pushes on electrons via the electric field. Electrons push back a bit via the same field. Light says, ‘Okay, okay, relax!’, and takes a slightly different route.

  • Finding the normal force in planar non-uniform circular motion using polar coordinates

    Finding the normal force in planar non-uniform circular motion using polar coordinates

    In this post, we will derive an expression for the normal force on a uniform mass which is in planar non-uniform circular motion using polar coordinates. Finding this expression is enormously useful to calculate under which circumstances a mass would be slung off its orbital path. Of course, there are numerous situations for which we should be able find the normal force. Here, we will look at a system as shown in Figure 1. Sometimes, obtaining an expression in terms of the variables given is not straightforward. You will find a useful trick in step 7 to arrive at an expression in terms of a simple $\theta$ instead of its secondary-order derivative $\ddot\theta$ which we initially obtain.

    This could be seen as an undergraduate-physics-level post. Download this article

    Notation

    We will apply Newton’s notation (the dot notation) whenever possible as this is the most compact form. For instance, if $\mathbf{x}$ is a vector, then its first-order and its second-order derivative with respect to time $t$ are denoted by

    \[ \dot{\mathbf{x}}\text{ and }\ddot{\mathbf{x}}, \]

    respectively. Where needed, in order to state explicitly that we are dealing with a time-derivative and to help in solving a time-integral for example, we will use Leibniz’s notation, i.e.

    \[ \frac{\text{d}\mathbf{x}}{\text{d}t}\text{ and }\frac{\text{d}^2\mathbf{x}}{\text{d}t^2}. \]

    Assignment

    Look at the system as sketched in Figure 1. Imagine we stand in front of this system. Mass $m$ is attached to a model string. At $t=0$, it rests at level with the centre of the cylinder with radius $R$ with the string draped over the top. A constant force $\mathbf{P}$ pulls the string downwards. At a later time $t$, mass $m$ has slid over the top with a coefficient of friction $\mu$. Let $\theta$ denote the angle between its initial and its current position, subtended at the centre of the cylinder. Calculate the normal force on $m$, and, hence, proof that the radius of the cylinder is irrelevant.

    Figure 1. The system

    Step 1. Force diagrams and unit vectors

    It is essential to draw force diagrams and unit vectors to define the acting forces and parameters. We choose the unit vectors to be the radial and the tangential vectors. This makes calculating most forces a lot easier. This is done in Figure 2.

    Figure 2. Force diagram and unit vectors at time $t>0$

    We identify the following forces on $m$:

    • $\mathbf{P}$ is the vector denoting the constant force pulling the model string,
    • $\mathbf{N}$ is the vector denoting the normal force acted on $m$ by the cylinder,
    • $\mathbf{F}$ is the vector denoting the frictional force,
    • $\mathbf{W}$ is the vector denoting the weight of $m$ as a result of the gravitational field of whatever planet the system is located,
    • $\mathbf{e}_r$ is the radial unit vector,
    • $\mathbf{e}_\theta$ is the tangential unit vector.

    Step 2. Apply Newton’s second law

    As this is a dynamical system, where $m$ is in non-uniform circular motion, we apply Newton’s second law, more specifically in the following form:

    \begin{equation}
    \sum\mathbf{F} = m\ddot{\mathbf{r}},
    \end{equation}

    where $\ddot{\mathbf{r}}$ is the rate of change of the rate of change over time, that is, the second time-derivative of the displacement vector $\mathbf{r}$ of mass $m$. We can now easily identify the constituents of the vector sum as we did that already in Step 1. And so, equation (1) becomes

    \begin{equation}
    m\ddot{\mathbf{r}} = \mathbf{P} + \mathbf{N} + \mathbf{F} + \mathbf{W}.
    \end{equation}

    Step 3. Rewrite the forces in terms of their magnitudes and unit vectors

    As pulling force $\mathbf{P}$ with magnitude $|\mathbf{P}|$ acts in the direction of tangential unit vector $\mathbf{e}_\theta$, we can write for $\mathbf{P}$:

    \begin{equation}
    \mathbf{P} = |\mathbf{P}|\mathbf{e}_\theta.
    \end{equation}

    Since we don’t have any other information regarding this force, we leave it at that.

    Normal force $\mathbf{N}$ points in the direction of radial unit vector $\mathbf{e}_r$, so, we write:

    \begin{equation}
    \mathbf{N} = |\mathbf{N}|\mathbf{e}_r.
    \end{equation}

    Friction $\mathbf{F}$ is in the opposite direction of the tangential unit vector $\mathbf{e}_\theta$, so, we need to place a minus-sign in its expression. Furthermore, as (dry) friction is usually modelled by the product of the coefficient of friction and the magnitude of the normal force, we can write:

    \begin{equation}
    \mathbf{F} = \mu|\mathbf{N}|(-\mathbf{e}_\theta).
    \end{equation}

    Lastly, weight is the force due to gravity, $|\mathbf{W}|=mg$, where $g$ is the gravitational constant. However, we need to express this force in terms of its components. In this case, those components are directed parallel to the radial and tangential unit vectors. As the latter are pointed (partly) upwards, as opposed to the downwards-pointing weight, we already know that both its components carry a minus-sign, i.e. $(-\mathbf{e}_r)$ and $(-\mathbf{e}_\theta)$. What remains, is the correct expression for the magnitude of the weight in terms of its respective unit vectors.

    To clearly show how we get an expression for $\mathbf{W}$ in terms of its components along the directions of $\mathbf{e}_r$ and $\mathbf{e}_\theta$, have a look at Figure 3.

    Figure 3. Finding the components of $\mathbf{W}$

    What you see is just the weight vector $\mathbf{W}$ from our force diagram in Figure 2, including the radial and tangential unit vectors $\mathbf{e}_r$ and $\mathbf{e}_\theta$. For visual clarity, we subtended them on mass $m$. Also added are the two component vectors in the opposite direction of the unit vectors for which we need to find expressions.

    Let component vector $\mathbf{v}_r = a(-\mathbf{e}_r)$ and component vector $\mathbf{v}_\theta = b(-\mathbf{e}_\theta)$, where $a$ and $b$ are some magnitude value such that the vector sum of $\mathbf{v}_r$ and $\mathbf{v}_\theta$ equals $\mathbf{W}$. In other words,

    \begin{equation}
    \mathbf{W} = \mathbf{v}_r + \mathbf{v}_\theta = a(-\mathbf{e}_r) + b(-\mathbf{e}_\theta).
    \end{equation}

    To find the values of the magnitude of $a$ and $b$, we use the fact that the magnitude $|\mathbf{W}| = mg$. So, using high school trigonometry, we deduce that

    \begin{align}
    a &= mg\sin\theta, \\
    b &= mg\cos\theta.
    \end{align}

    Now, we can write $\mathbf{W}$ in terms of its components by substituting equations (7) and (8) into (6):

    \begin{equation}
    \mathbf{W} = mg\sin\theta(-\mathbf{e}_r) + mg\cos\theta(-\mathbf{e}_\theta).
    \end{equation}

    And so, if we substitute equations (3), (4), (5), and (9) into equation (2), we get:

    \begin{align}
    m\ddot{\mathbf{r}} &= |\mathbf{P}|\mathbf{e}_\theta + |\mathbf{N}|\mathbf{e}_r + \mu|\mathbf{N}|(-\mathbf{e}_\theta)\nonumber \\
    &\hspace{2em}+ mg\sin\theta(-\mathbf{e}_r) + mg\cos\theta(-\mathbf{e}_\theta).
    \end{align}

    Step 4. Express the Cartesian $\ddot{\mathbf{r}}$ in polar coordinates

    As we know that the expression for the second time derivative of non-uniform circular motion is

    \begin{equation}
    \ddot{\mathbf{r}} = -R\dot{\theta}^2\mathbf{e}_r + R\ddot{\theta}\mathbf{e}_\theta,
    \end{equation}

    where $R$ is the radius of the circular motion, i.e. the cylinder. We proceed to substitute this into equation (10).

    And so, we get

    \begin{align*}
    m(-R\dot{\theta}^2\mathbf{e}_r + R\ddot{\theta}\mathbf{e}_\theta) &= |\mathbf{P}|\mathbf{e}_\theta + |\mathbf{N}|\mathbf{e}_r + \mu|\mathbf{N}|(-\mathbf{e}_\theta) \\
    &\hspace{2em}+ mg\sin\theta(-\mathbf{e}_r) + mg\cos\theta(-\mathbf{e}_\theta),
    \end{align*}

    which, of course, after expansion, becomes

    \begin{align}
    -mR\dot{\theta}^2\mathbf{e}_r + mR\ddot{\theta}\mathbf{e}_\theta &= |\mathbf{P}|\mathbf{e}_\theta + |\mathbf{N}|\mathbf{e}_r + \mu|\mathbf{N}|(-\mathbf{e}_\theta) \nonumber \\
    &\hspace{2em}+ mg\sin\theta(-\mathbf{e}_r) + mg\cos\theta(-\mathbf{e}_\theta).
    \end{align}

    Step 5. Resolve radially and tangentially

    We can now resolve equation (12) into its radial and tangential components.

    \begin{align}
    \mathbf{e}_r &: -mR\dot{\theta}^2 = N – mg\sin\theta, \\
    \mathbf{e}_\theta &: mR\ddot{\theta} = P – \mu N – mg\cos\theta.
    \end{align}

    Step 6. Write down the equation of motion (in polar coordinates)

    Rearranging equation (14), we can write down the second-order differential equation of motion:

    \begin{equation}
    \ddot{\theta} = \frac{P – \mu N – mg\cos\theta}{mR}.
    \end{equation}

    While we could have solved equation (14) for $N$, this would still leave us with the second time-derivative of $\theta$. Instead, we want an expression of $N$ in terms of a simple $\theta$. This means that we need to get rid of $\ddot{\theta}$ in some way. It is not immediately clear how equation (14) or (15) should be operated on to achieve this. However, here is a neat trick.

    Step 7. The trick

    Have a look at the following equation where we apply the chain rule:

    \begin{equation}
    \frac{\text{d}\dot{\theta}^2}{\text{d}t} = \frac{\text{d}\dot{\theta}^2}{\text{d}\dot{\theta}}\frac{\text{d}\dot{\theta}}{\text{d}t} = 2\dot{\theta}\frac{\text{d}\dot{\theta}}{\text{d}t} = 2\dot{\theta}\ddot{\theta}.
    \end{equation}

    So, if we substitute equation (15) into (16), we get

    \begin{equation}
    \frac{\text{d}\dot{\theta}^2}{\text{d}t} = 2\dot{\theta}\left(\frac{P – \mu N – mg\cos\theta}{mR}\right).
    \end{equation}

    If we now integrate both sides with respect to time, we get

    \begin{align}
    \int \frac{\text{d}\dot{\theta}^2}{\text{d}t}\text{d}t &= \int 2\dot{\theta}\left(\frac{P – \mu N – mg\cos\theta}{mR}\right)\text{d}t, \nonumber \\
    \dot{\theta}^2 + A &= 2 \int \frac{\text{d}\theta}{\text{d}t}\left(\frac{P – \mu N – mg\cos\theta}{mR}\right)\text{d}t, \nonumber \\
    &\text{where $A$ is an arbitrary constant}, \nonumber \\
    \dot{\theta}^2 + A &= 2 \int \left(\frac{P – \mu N – mg\cos\theta}{mR}\right)\text{d}\theta, \nonumber \\
    \dot{\theta}^2 + A &= \frac{2}{mR} \int (P – \mu N – mg\cos\theta)\,\text{d}\theta, \nonumber \\
    \dot{\theta}^2 + A &= \frac{2}{mR} \left( P\int 1\,\text{d}\theta – \mu N\int 1\,\text{d}\theta – mg\int \cos\theta\,\text{d}\theta\right), \nonumber \\
    \dot{\theta}^2 + A &= \frac{2P\theta}{mR} – \frac{2\mu N\theta}{mR} – \frac{2mg\sin\theta}{mR} + B, \nonumber \\
    &\text{where $B$ is an arbitrary constant}, \nonumber \\
    \dot{\theta}^2 &= \frac{2P\theta}{mR} – \frac{2\mu N\theta}{mR} – \frac{2g\sin\theta}{R} + B – A, \nonumber \\
    \dot{\theta}^2 &= \frac{2P\theta}{mR} – \frac{2\mu N\theta}{mR} – \frac{2g\sin\theta}{R} + C, \\
    &\text{where $C=B-A$} \nonumber.
    \end{align}

    Solving the initial condition problem to find $C$, we use the fact that at $t=0$, angle $\theta = 0$, thus $\dot{\theta} = \ddot{\theta} = 0$. This renders $C = 0$ in equation (18), and so, we have

    \begin{equation}
    \dot{\theta}^2 = \frac{2P\theta}{mR} – \frac{2\mu N\theta}{mR} – \frac{2g\sin\theta}{R}.
    \end{equation}

    Note, we now have obtained an expression for $\dot{\theta}^2$ which already appeared in equation (13). We can, therefore, substitute equation (19) in (13), and we obtain:

    \begin{equation}
    -mR\left(\frac{2P\theta}{mR} – \frac{2\mu N\theta}{mR} – \frac{2g\sin\theta}{R}\right) = N – mg\sin\theta.
    \end{equation}

    Expanding and rearranging this, we get

    \begin{align}
    N – mg\sin\theta &= -2P\theta + 2\mu N\theta + 2mg\sin\theta, \nonumber \\
    N – 2\mu N\theta &= -2P\theta + 2mg\sin\theta + mg\sin\theta, \nonumber \\ N(1 – 2\mu \theta) &= -2P\theta + 3mg\sin\theta, \nonumber \\
    N &= \frac{3mg\sin\theta – 2P\theta}{1-2\mu\theta}.
    \end{align}

    So, now we have an expression of $N$ in terms of the gravitational constant $g$, the variables $m$, $\mu$, and $P$, and the more reasonable $\theta$ instead of $\dot\theta^2$.

    And so, if we want to calculate when a mass would be slung out of its orbital path, we write $N = 0$ as this means, in physical terms, that the mass isn’t resting on the cylinder anymore (since it doesn’t exert a normal force on the mass). In other words, find the roots of equation (21) to find the one unknown variable. Note, $R$ does not play a role. Of course, bear in mind that $m$ is a point mass.

  • Why your coffee does not have tides

    Why your coffee does not have tides

    The Moon orbits the earth and its gravity is causing the tides. But why don’t swimming pools have tides? Or a cup of coffee? Human bodies consist of water, mostly. Aren’t they tidally influenced by the Moon? If you’re asking all these beautiful questions, then what you thought is causing the tides is probably wrong, and here’s why.

    Remember, back in high school, when the science or physics teacher had all the air sucked out of a large, transparent tube which contained a feather and a little steel ball or something like that? And that she asked you to predict which would drop to the bottom first if she would turn the tube upside down?

    Of course, both objects turned out to fall to the bottom at the exact same speed. We learnt it did not matter if the steel ball had more mass than the feather. Earth’s gravity works the same on both. In fact, anything which is being ‘pulled down’ by our planet’s gravity gets to be pulled down at the same rate, no matter how much mass these things have (provided we ignore any form of friction).

    Lunar gravity

    Even though the Moon’s gravity is smaller than Earth’s, the principle is the same. Irrespective of an object’s mass, it falls straight to the lunar surface at precisely the same rate as any other thing. On 2 August 1971, NASA Commander David Scott demonstrated that a feather and a hammer hit the Moon’s soil simultaneously.

    Photo: NASA

    The Moon’s gravity is strong enough to have a noticeable effect on Earth, as we all know. Indeed, it is the reason why our oceans have tides. However, if gravity, whether on our planet or on the Moon, acts the same way on every object irrespective of their mass, how come our bathtub does not experience tides, for instance? Yes, it has less mass, but by Cmdr David Scott’s experiment, that shouldn’t matter. And if the Moon’s gravity is capable of pulling on vast bodies of water such as oceans causing them to rise literally meters high, why does our rubber duck not start levitating up in the air as soon as the Moon rushes past our homes?

    The answer sounds both obvious and contradictory: because the force of the Moon’s gravity is negligibly small, except when it is not.

    The wrong picture

    Let’s have a look at the simplified drawing of Figure 1. Just to make things a little less complicated, we imagine our planet to be covered by water entirely. There are no continents for now.

    We see a schematic drawing of earth and the moon. Earth is covered with water with bulges left and right, representing the two high tides. Point A is the point closest to the moon on the right, located on the surface of the earth in the middle of the bulge on the right. Point B is at exactly the opposite location on the far side of the earth, the most distant point from the moon.
    Figure 1. Earth’s tides and the Moon. (Not to scale!)

    First misconception. Even though, intuitively, it may seem to be the case, the bulge at point A is not because the Moon’s gravity is tugging at it, contrary to popular belief.

    And in many texts, you might encounter the following incorrect explanation for the bulge at point B. ‘The Moon’s pull is smaller at point B than at point A, so, point B stays more or less where it is, while point A gets pulled more towards the Moon. Everything in between A and B gets stretched like chewing gum. So, from the perspective of someone standing (on land) at point B, the water rises there as well.’

    This is also mostly incorrect. It is true, the Moon’s gravitational pull is smaller at B than it is at A. But that is not what is causing the bulge at point B. Not in the direct way as stated here, that is.

    Many a little makes a mickle

    Why don’t we have a look at points C and D in two different, little patches of water in Figure 2? The Moon’s force of gravity acts on these points at a certain angle as is represented by the blue arrows, or vectors. At the same time, the entire earth experiences a slight force towards the Moon as is modelled by the red vector.

    Same schematic as the previous one, but more points are added. Point C is located more or less on top of the earth, a little to the right of the North Pole. Point D is located between the North Pole and point B. Little blue arrows, called vectors, are drawn from points C and D, pointing towards the centre of the moon. A little red vector is drawn at the centre of the earth, pointing to the centre of the moon. These vectors represent the forces acted on these points caused by the moon's gravity.
    Figure 2. The force of the Moon’s gravity acting on points C and D and the entire earth

    So, point C and D undergo two simultaneous forces as is explicitly shown in Figure 3. Note that the blue and red vectors have different directions. Our high school physics or maths teacher then taught us that two or more forces acting on the same point can be modelled as one resultant force.

    Now we need to take two important steps: 1. Newton taught us that a force is an acceleration, so, from now on we will regard the arrows in Figure 3 as being accelerations. 2. To determine the acceleration of the patches of water at points C and D relative to Earth’s surface, we subtract the red vector from the blue vector. What’s left is the green vector, the resultant.

    We see a close-up of points C and D. The red vector, representing the force of the moon exerted on the earth, originates here from point C. The blue vector, representing the force of the moon exerted on point C, still originates from point C. So, we have two vectors coming from point C. A little green vector is drawn between the heads of the other vectors, pointing down, towards the location of where the bulge closest to the moon will emerge. And so, the combination of the two real forces exerted by the moon results in a net force. The exact same procedure has been applied to point D. Only here, the green net force vector is pointed the other way, towards where the other bulge, on the other side of the planet, will emerge.
    Figure 3. The resultant forces are represented by the green vector

    In Figure 3, it is shown how the combination of the two gross forces blue and red yield a net force as represented by the green vectors. Do note, the net forces are what is called apparent forces. Think of a car suddenly accelerating. Relative to the ground, your head is standing still for an instant of time. However, from within the car, your head seems like it is being pushed back by some invisible force. Tides are thus being caused by so-called tidal forces, which are apparent forces.

    So, if we do the same for many other points, you get many green vectors as they are shown in Figure 4. And guess what, all the green (now black) arrows point in a way that look a lot like bulges in the water.

    We see the earth where all the net force vectors are lined up in a way that, together, result in a picture exactly the same as our tides: two bulges on either side.
    Figure 4. An array of net forces (the green arrows are here the black arrows)

    This shows that what actually happens is that every minuscule patch of water gets influenced by a tiny bit of net force in the direction of the places where the bulges will emerge, pushing every other patch in front of it towards the bulges, thereby creating the bulges in the first place.

    Now, in the drawing, all arrows are relatively massive, so we can actually see them. In reality, however, the net forces are tiny. Microscopically tiny.

    And this is the key to solving the paradox. Even though a net force, resulting from the Moon’s gravitational influences, is utterly insignificant on a single patch of the water, the amount of ocean on Earth is quite the opposite of negligible, rendering the sum of all net forces on every cubic patch within the oceanic liquid highly significant, and in some cases, depending on the shape of the land, dangerously significant.

    Conclusion

    The Moon’s influence on tiny things is tiny. It does not noticeably influence your cup of coffee, your body, your bathtub, ponds, and lakes. Any tidal height difference in a cup of coffee could be thinner than a bacterium, the significance of which is immediately squashed by the mere presence of, well, a bacterium in your coffee, practising its back crawl. If your coffee starts to display any tidal effects, prepare for the Apocalypse and/or escaped dinosaurs. Either case, something is really wrong then.

    Even a lake the size of Lake Michigan will only rise a couple of centimetres—easily negated by its murmuring surface on a sunny day in May. So, you can imagine, your body does not feel a thing. The pressure needed for delivering oxygen to your brains alone squashes out every single tidal influence by the Moon, which would have been smaller than a hair’s thickness anyway. If you feel less capable of rational thought, you now know it’s not the Moon. But do check your blood pressure.

    However, in the case of an ocean, a body with many, many, many tiny, watery parts which can roll, slip and slide freely on top of one another, there will be bulges about where the Moon whizzes. However, the swelling occurs by virtue of pushing not pulling, directly.

    In short, a quindecillion minuscule little net forces on every cubic piece of the ocean cause an upward push so the two bulges emerge. Lakes, ponds, bathtubs, human bodies, and coffee mugs do not come close to even a little bit of that amount.

    EDIT: the original article omitted to mention how tidal force is an apparent force, resulting in a fundamental misinterpretation of Figures 3 and 4. This has been corrected.

    Figure 4 is an adapted version (cropped) of the original made by Krishnavedala under CC BY-SA 3.0.

    We did not consider the rotation of the earth, the Coriolis effect, the presence of the sun, the presence of land, etc., just to keep it simple. This changes the situation somewhat, but does not change the gist of it all.