Physics

Supernovae – An Introduction

The following is an excerpt from my book “Introduction to Stars: Spectra, Formation, Evolution, Collapse”, available here for Kindle. Enjoy!

A vast number of written recordings show that people watching the night sky on 4 July 1054 had the opportunity to witness the sudden appearance of a new light source in the constellation Taurus that shone four times brighter than our neighbor Venus. The “guest star”, as Chinese astronomers called the mysterious light source, shone so bright that it even remained visible in daylight for over three weeks. After around twenty-one months the guest star disappeared from the night sky, but the mystery remained. What caused the sudden appearance of the light source? Thanks to the steady growth of knowledge and the existence of powerful telescopes, astronomers are now able to answer this question.

The spectacular event of 4 July 1054 was the result of a supernova (plural: supernovae), an enormous explosion that marks the irreversible death of a massive star. Pointing a modern telescope in the direction of the 1054 supernova, one can see a fascinatingly beautiful and still rapidly expanding gas cloud called the Crab Nebula at a distance of roughly 2200 parsecs from Earth. At the center of the Crab Nebula cloud lies a pulsar (Crab Pulsar or PSR B0531+21) having a radius of 10 km and rotating at a frequency of roughly 30 Hz.

CrabNebulaHubble

Image of the Crab Nebula, taken by the Hubble Space Telescope.

Let’s take a look at the mechanisms that lead to the occurrence of a supernova. At the end of chapter two we noted that a star having an initial mass of eight solar masses or more will form an iron core via a lengthy fusion chain. We might expect that the star could continue its existence by initiating the fusion of iron atoms. But unlike the fusion of lighter elements, the merging of two iron atoms does not produce energy and thus the star cannot fall back on yet another fusion cycle to stabilize itself. Even worse, there are several mechanisms within the core that drain it of much needed energy, the most important being photodisintegration (high-energy photons smash the iron atoms apart) and neutronization (protons and electrons combine to form neutrons). Both of these processes actually speed up the inevitable collapse of the iron core.

Calculations show that the collapse happens at a speed of roughly one-fourth the speed of light, meaning that the core that is initially ten thousand kilometers in diameter will collapse to a neutron star having a radius of only fifteen kilometers in a fraction of a second – literally the blink of an eye. As the core hits the sudden resistance of the degenerate neutrons, the rapid collapse is stopped almost immediately. Because of the high impact speed, the core overshoots its equilibrium position by a bit, springs back, overshoots it again, springs back again, and so on. In short: the core bounces. This process creates massive shock waves that propagate radially outwards, blasting into the outer layers of the star, creating the powerful explosion known as the (collapsing core) supernova.

Within just a few minutes the dying star increases its luminosity to roughly one billion Suns, easily outshining smaller galaxies in the vicinity. It’s no wonder then that this spectacular event can be seen in daylight even without the help of a telescope. The outer layers are explosively ejected with speeds in the order of 50,000 kilometers per second or 30,000 miles per second. As time goes on, the ejected layers slow down and cool off during the expansion and the luminosity of the supernova steadily decreases. Once the envelope has faded into deep space, all that remains of the former star is a compact neutron star or an even more exotic remnant we will discuss in the next section. Supernovas are relatively rare events, it is estimated that they only occur once every fifty years in a galaxy the size of the Milky Way.

At first sight, supernovae may seem like a rather destructive force in the universe. However, this is far from the truth for several reasons, one of which is nucleosynthesis, the creation of elements. Scientists assume that the two lightest elements, hydrogen and helium, were formed during the Big Bang and accordingly, these elements can be found in vast amounts in any part of the universe. Other elements up to iron are formed by cosmic rays (in particular lithium, beryllium and boron) or fusion reactions within a star.

But the creation of elements heavier than iron requires additional mechanisms. Observations indicate that such elements are produced mainly by neutron capture, existing atoms capture a free neutron and transform into a heavier element, either within the envelope of giant stars or supernovae. So supernovae play an important role in providing the universe with many of the heavy elements. Another productive aspect of supernovae is their ability to trigger star formation. When the enormous shock wave emitted from a supernova encounters a giant molecular cloud, it can trigger the collapse of the cloud and thus initiate the formation of a new cluster of stars. Far from destructive indeed.

The rapid collapse of a stellar core is not the only source of supernovae in the universe. A supernova can also occur when a white dwarf locked in a binary system keeps pulling in mass from a partner star. At a critical mass of roughly 1.38 times the mass of the Sun, just slightly below the Chandrasekhar limit, the temperature within the white dwarf would become high enough to re-ignite the carbon. Due to its exotic equilibrium state, the white dwarf cannot make use of the self-regulating mechanism that normally keeps the temperature in check in main sequence stars. The result is thus a runaway fusion reaction that consumes all the carbon and oxygen in the white dwarf within a few seconds and raises the temperature to many billion K, equipping the individual atoms with sufficient kinetic energy to fly off at incredible speeds. This violent explosion lets the remnant glow with around five billion solar luminosities for a very brief time.

Type_Ia_supernova_simulation_-_Argonne_National_Laboratory

Simulation of the runaway nuclear fusion reaction within a white dwarf that became hot enough to re-ignite its carbon content. The result is a violent ejection of its mass.

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Stellar Physics – Gaps in the Spectrum

The following is an excerpt from my e-book “Introduction to Stars: Spectra, Formation, Evolution, Collapse” available here for Kindle.

When you look at the spectrum of the heat radiation coming from glowing metal, you will find a continuous spectrum, that is, one that does not have any gaps. You would expect to see the same when looking at light coming from a star. However, there is one significant difference: all stellar spectra have gaps (dark lines) in them. In other words: photons of certain wavelengths seem to be either completely missing or at least arriving in much smaller numbers. Aside from these gaps though, the spectrum is just what you’d expect to see when looking at a heat radiator. So what’s going on here? What can we learn from these gaps?

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Spectrum of the Sun. Note the pronounced gaps.

To understand this, we need to delve into atomic physics, or to be more specific, look at how atoms interact with photons. Every atom can only absorb or emit photons of specific wavelengths. Hydrogen atoms for example will absorb photons having the wavelength 4102 A, but do not care about photons having a wavelength 4000 A or 4200 A. Those photons just pass through it without any interaction taking place. Sodium atoms prefer photons with a wavelength 5890 A, when a photon of wavelength 5800 A or 6000 A comes by, the sodium atom is not at all impressed. This is a property you need to keep in mind: every atom absorbs or emits only photons of specific wavelengths.

Suppose a 4102 A photon hits a hydrogen atom. The atom will then absorb the photon, which in crude terms means that the photon “vanishes” and its energy is transferred to one of the atom’s electrons. The electron is now at a higher energy level. However, this state is unstable. After a very short time, the electron returns to a lower energy level and during this process a new photon appears, again having the wavelength 4102 A. So it seems like nothing was gained or lost. Photon comes in, vanishes, electron gains energy, electron loses energy again, photon of same wavelength appears. This seems pointless, why bother mentioning it? Here’s why. The photon that is created when the electron returns to the lower energy level is emitted in a random direction and not the direction the initial photon came from. This is an important point! We can understand the gaps in a spectrum by pondering the consequences of this fact.

Suppose both of us observe a certain heat source. The light from this source reaches me directly while you see the light through a cloud of hydrogen. Both of us are equipped with a device that generates the spectrum of the incoming light. Comparing the resulting spectra, we would see that they are for the most part the same. This is because most photons pass through the hydrogen cloud without any interaction. Consider for example photons of wavelength 5000 A. Hydrogen does not absorb or emit photons of this wavelength, so we will both record the same light intensity at 5000 A. But what about the photons with a 4102 A wavelength?

Imagine a directed stream of these particular photons passing through the hydrogen cloud. As they get absorbed and re-emitted, they get thrown into random directions. Only those photons which do not encounter a hydrogen atom and those which randomly get thrown in your direction will reach your position. Unless the hydrogen cloud is very thin and has a low density, that’s only a very small part of the initial stream. Hence, your spectrum will show a pronounced gap, a line of low light intensity, at λ = 4102 A while in my spectrum no such gap exists.

What if it were a sodium instead of hydrogen cloud? Using the same logic, we can see that now your spectrum should show a gap at λ = 5890 A since this is the characteristic wavelength at which sodium atoms absorb and emit photons. And if it were a mix of hydrogen and sodium, you’d see two dark lines, one at λ = 4102 A due to the presence of hydrogen atoms and another one at λ = 5890 A due to the sodium atoms. Of course, and here comes the fun part, we can reverse this logic. If you record a spectrum and you see gaps at λ = 4102 A and λ = 5890 A, you know for sure that the light must have passed through a gas that contains hydrogen and sodium. So the seemingly unremarkable gaps in a spectrum are actually a neat way of determining what elements sit in a star’s atmosphere! This means that by just looking at a star’s spectrum we can not only determine its temperature, but also its chemical composition at the surface. Here are the results for the Sun:

– Hydrogen 73.5 %

– Helium 24.9 %

– Oxygen 0.8 %

– Carbon 0.3 %

– Iron 0.2 %

– Neon 0.1 %

There are also traces (< 0.1 %) of nitrogen, silicon, magnesium and sulfur. This composition is quite typical for other stars and the universe as a whole: lots of hydrogen (the lightest element), a bit of helium (the second lightest element) and very little of everything else. Mathematical models suggest that even though the interior composition changes significantly over the life time of a star (the reason being fusion, in particular the transformation of hydrogen into helium), its surface composition remains relatively constant in this time.

New Release for Kindle: Introduction to Stars – Spectra, Formation, Evolution, Collapse

I’m happy to announce my new e-book release “Introduction to Stars – Spectra, Formation, Evolution, Collapse” (126 pages, $ 2.99). It contains the basics of how stars are born, what mechanisms power them, how they evolve and why they often die a spectacular death, leaving only a remnant of highly exotic matter. The book also delves into the methods used by astronomers to gather information from the light reaching us through the depth of space. No prior knowledge is required to follow the text and no mathematics beyond the very basics of algebra is used.

If you are interested in learning more, click the cover to get to the Amazon product page:

Screenshot_4

Here’s the table of contents:

Gathering Information
Introduction
Spectrum and Temperature
Gaps in the Spectrum
Doppler Shift

The Life of a Star
Introduction
Stellar Factories
From Protostar to Star
Main Sequence Stars
Giant Space Onions

The Death of a Star
Introduction
Slicing the Space Onion
Electron Degeneracy
Extreme Matter
Supernovae
Black Holes

Appendix
Answers
Excerpt
Sources and Further Reading

Enjoy the reading experience!

The Weirdness of Empty Space – Casimir Force

(This is an excerpt from The Book of Forces – enjoy!)

The forces we have discussed so far are well-understood by the scientific community and are commonly featured in introductory as well as advanced physics books. In this section we will turn to a more exotic and mysterious interaction: the Casimir force. After a series of complex quantum mechanical calculations, the Dutch physicist Hendrick Casimir predicted its existence in 1948. However, detecting the interaction proved to be an enormous challenge as this required sensors capable picking up forces in the order of 10^(-15) N and smaller. It wasn’t until 1996 that this technology became available and the existence of the Casimir force was experimentally confirmed.

So what does the Casimir force do? When you place an uncharged, conducting plate at a small distance to an identical plate, the Casimir force will pull them towards each other. The term “conductive” refers to the ability of a material to conduct electricity. For the force it plays no role though whether the plates are actually transporting electricity in a given moment or not, what counts is their ability to do so.

The existence of the force can only be explained via quantum theory. Classical physics considers the vacuum to be empty – no particles, no waves, no forces, just absolute nothingness. However, with the rise of quantum mechanics, scientists realized that this is just a crude approximation of reality. The vacuum is actually filled with an ocean of so-called virtual particles (don’t let the name fool you, they are real). These particles are constantly produced in pairs and annihilate after a very short period of time. Each particle carries a certain amount of energy that depends on its wavelength: the lower the wavelength, the higher the energy of the particle. In theory, there’s no upper limit for the energy such a particle can have when spontaneously coming into existence.

So how does this relate to the Casimir force? The two conducting plates define a boundary in space. They separate the space of finite extension between the plates from the (for all practical purposes) infinite space outside them. Only particles with wavelengths that are a multiple of the distance between the plates fit in the finite space, meaning that the particle density (and thus energy density) in the space between the plates is smaller than the energy density in the pure vacuum surrounding them. This imbalance in energy density gives rise to the Casimir force. In informal terms, the Casimir force is the push of the energy-rich vacuum on the energy-deficient space between the plates.

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(Illustration of Casimir force)

It gets even more puzzling though. The nature of the Casimir force depends on the geometry of the plates. If you replace the flat plates by hemispherical shells, the Casimir force suddenly becomes repulsive, meaning that this specific geometry somehow manages to increase the energy density of the enclosed vacuum. Now the even more energy-rich finite space pushes on the surrounding infinite vacuum. Trippy, huh? So which shapes lead to attraction and which lead to repulsion? Unfortunately, there is no intuitive way to decide. Only abstract mathematical calculations and sophisticated experiments can provide an answer.

We can use the following formula to calculate the magnitude of the attractive Casimir force FCS between two flat plates. Its value depends solely on the distance d (in m) between the plates and the area A (in m²) of one plate. The letters h = 6.63·10^(-34) m² kg/s and c = 3.00·10^8 m/s represent Plank’s constant and the speed of light.

FCS = π·h·c·A / (480·d^4) ≈ 1.3·10^(-27)·A / d^4

(The sign ^ stands for “to the power”) Note that because of the exponent, the strength of the force goes down very rapidly with increasing distance. If you double the size of the gap between the plates, the magnitude of the force reduces to 1/2^4 = 1/16 of its original value. And if you triple the distance, it goes down to 1/3^4 = 1/81 of its original value. This strong dependence on distance and the presence of Plank’s constant as a factor cause the Casimir force to be extremely weak in most real-world situations.

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Example 24:

a) Calculate the magnitude of the Casimir force experienced by two conducting plates having an area A = 1 m² each and distance d = 0.001 m (one millimeter). Compare this to their mutual gravitational attraction given the mass m = 5 kg of one plate.

b) How close do the plates need to be for the Casimir force to be in the order of unity? Set FCS = 1 N.

Solution:

a)

Inserting the given values into the formula for the Casimir force leads to (units not included):

FCS = 1.3·10^(-27)·A/d^4
FCS = 1.3·10^(-27) · 1 / 0.0014
FCS ≈ 1.3·10^(-15) N

Their gravitational attraction is:

FG = G·m·M / r²
FG = 6.67·10^(-11)·5·5 / 0.001²
FG ≈ 0.0017 N

This is more than a trillion times the magnitude of the Casimir force – no wonder this exotic force went undetected for so long.  I should mention though that the gravitational force calculated above should only be regarded as a rough approximation as Newton’s law of gravitation is tailored to two attracting spheres, not two attracting plates.

b)

Setting up an equation we get:

FCS = 1.3·10^(-27)·A/d^4
1 = 1.3·10^(-27) · 1 / d^4

Multiply by d4:

d4 = 1.3·10^(-27)

And apply the fourth root:

d ≈ 2·10^(-7) m = 200 nanometers

This is roughly the size of a common virus and just a bit longer than the wavelength of violet light.

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The existence of the Casimir force provides an impressive proof that the abstract mathematics of quantum mechanics is able to accurately describe the workings of the small-scale universe. However, many open questions remain. Quantum theory predicts the energy density of the vacuum to be infinitely large. According to Einstein’s theory of gravitation, such a concentration of energy would produce an infinite space-time curvature and if this were the case, we wouldn’t exist. So either we don’t exist (which I’m pretty sure is not the case) or the most powerful theories in physics are at odds when it comes to the vacuum.

All about the Gravitational Force (For Beginners)

(This is an excerpt from The Book of Forces)

All objects exert a gravitational pull on all other objects. The Earth pulls you towards its center and you pull the Earth towards your center. Your car pulls you towards its center and you pull your car towards your center (of course in this case the forces involved are much smaller, but they are there). It is this force that invisibly tethers the Moon to Earth, the Earth to the Sun, the Sun to the Milky Way Galaxy and the Milky Way Galaxy to its local galaxy cluster.

Experiments have shown that the magnitude of the gravitational attraction between two bodies depends on their masses. If you double the mass of one of the bodies, the gravitational force doubles as well. The force also depends on the distance between the bodies. More distance means less gravitational pull. To be specific, the gravitational force obeys an inverse-square law. If you double the distance, the pull reduces to 1/2² = 1/4 of its original value. If you triple the distance, it goes down to 1/3² = 1/9 of its original value. And so on. These dependencies can be summarized in this neat formula:

F = G·m·M / r²

With F being the gravitational force in Newtons, m and M the masses of the two bodies in kilograms, r the center-to-center distance between the bodies in meters and G = 6.67·10^(-11) N m² kg^(-2) the (somewhat cumbersome) gravitational constant. With this great formula, that has first been derived at the end of the seventeenth century and has sparked an ugly plagiarism dispute between Newton and Hooke, you can calculate the gravitational pull between two objects for any situation.

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(Gravitational attraction between two spherical masses)

If you have trouble applying the formula on your own or just want to play around with it a bit, check out the free web applet Newton’s Law of Gravity Calculator that can be found on the website of the UNL astronomy education group. It allows you to set the required inputs (the masses and the center-to-center distance) using sliders that are marked special values such as Earth’s mass or the distance Earth-Moon and calculates the gravitational force for you.

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Example 3:

Calculate the gravitational force a person of mass m = 72 kg experiences at the surface of Earth. The mass of Earth is M = 5.97·10^24 kg (the sign ^ stands for “to the power”) and the distance from the center to the surface r = 6,370,000 m. Using this, show that the acceleration the person experiences in free fall is roughly 10 m/s².

Solution:

To arrive at the answer, we simply insert all the given inputs into the formula for calculating gravitational force.

F = G·m·M / r²
F = 6.67·10^(-11)·72·5.97·10^24 / 6,370,000² N ≈ 707 N

So the magnitude of the gravitational force experienced by the m = 72 kg person is 707 N. In free fall, he or she is driven by this net force (assuming that we can neglect air resistance). Using Newton’s second law we get the following value for the free fall acceleration:

F = m·a
707 N = 72 kg · a

Divide both sides by 72 kg:

a = 707 / 72 m/s² ≈ 9.82 m/s²

Which is roughly (and more exact than) the 10 m/s² we’ve been using in the introduction. Except for the overly small and large numbers involved, calculating gravitational pull is actually quite straight-forward.

As mentioned before, gravitation is not a one-way street. As the Earth pulls on the person, the person pulls on the Earth with the same force (707 N). However, Earth’s mass is considerably larger and hence the acceleration it experiences much smaller. Using Newton’s second law again and the value M = 5.97·1024 kg for the mass of Earth we get:

F = m·a
707 N = 5.97·10^24 kg · a

Divide both sides by 5.97·10^24 kg:

a = 707 / (5.97·10^24) m/s² ≈ 1.18·10^(-22) m/s²

So indeed the acceleration the Earth experiences as a result of the gravitational attraction to the person is tiny.

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Example 4:

By how much does the gravitational pull change when the  person of mass m = 72 kg is in a plane (altitude 10 km = 10,000 m) instead of the surface of Earth? For the mass and radius of Earth, use the values from the previous example.

Solution:

In this case the center-to-center distance r between the bodies is a bit larger. To be specific, it is the sum of the radius of Earth 6,370,000 m and the height above the surface 10,000 m:

r = 6,370,000 m + 10,000 m = 6,380,000 m

Again we insert everything:

F = G·m·M / r²
F = 6.67·10^(-11)·72·5.97·10^24 / 6,380,000² N ≈ 705 N

So the gravitational force does not change by much (only by 0.3 %) when in a plane. 10 km altitude are not much by gravity’s standards, the height above the surface needs to be much larger for a noticeable difference to occur.

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With the gravitational law we can easily show that the gravitational acceleration experienced by an object in free fall does not depend on its mass. All objects are subject to the same 10 m/s² acceleration near the surface of Earth. Suppose we denote the mass of an object by m and the mass of Earth by M. The center-to-center distance between the two is r, the radius of Earth. We can then insert all these values into our formula to find the value of the gravitational force:

F = G·m·M / r²

Once calculated, we can turn to Newton’s second law to find the acceleration a the object experiences in free fall. Using F = m·a and dividing both sides by m we find that:

a = F / m = G·M / r²

So the gravitational acceleration indeed depends only on the mass and radius of Earth, but not the object’s mass. In free fall, a feather is subject to the same 10 m/s² acceleration as a stone. But wait, doesn’t that contradict our experience? Doesn’t a stone fall much faster than a feather? It sure does, but this is only due to the presence of air resistance. Initially, both are accelerated at the same rate. But while the stone hardly feels the effects of air resistance, the feather is almost immediately slowed down by the collisions with air molecules. If you dropped both in a vacuum tube, where no air resistance can build up, the stone and the feather would reach the ground at the same time! Check out an online video that shows this interesting vacuum tube experiment, it is quite enlightening to see a feather literally drop like a stone.

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(All bodies are subject to the same gravitational acceleration)

Since all objects experience the same acceleration near the surface of Earth and since this is where the everyday action takes place, it pays to have a simplified equation at hand for this special case. Denoting the gravitational acceleration by g (with g ≈ 10 m/s²) as is commonly done, we can calculate the gravitational force, also called weight, an object of mass m is subject to at the surface of Earth by:

F = m·g

So it’s as simple as multiplying the mass by ten. Depending on the application, you can also use the more accurate factor g ≈ 9.82 m/s² (which I will not do in this book). Up to now we’ve only been dealing with gravitation near the surface of Earth, but of course the formula allows us to compute the gravitational force and acceleration near any other celestial body. I will spare you trouble of looking up the relevant data and do the tedious calculations for you. In the table below you can see what gravitational force and acceleration a person of mass m = 72 kg would experience at the surface of various celestial objects. The acceleration is listed in g’s, with 1 g being equal to the free-fall acceleration experienced near the surface of Earth.

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So while jumping on the Moon would feel like slow motion (the free-fall acceleration experienced is comparable to what you feel when stepping on the gas pedal in a common car), you could hardly stand upright on Jupiter as your muscles would have to support more than twice your weight. Imagine that! On the Sun it would be even worse. Assuming you find a way not get instantly killed by the hellish thermonuclear inferno, the enormous gravitational force would feel like having a car on top of you. And unlike temperature or pressure, shielding yourself against gravity is not possible.

What about the final entry? What is a neutron star and why does it have such a mind-blowing gravitational pull? A neutron star is the remnant of a massive star that has burned its fuel and exploded in a supernova, no doubt the most spectacular light-show in the universe. Such remnants are extremely dense – the mass of several suns compressed into an almost perfect sphere of just 20 km radius. With the mass being so large and the distance from the surface to the center so small, the gravitational force on the surface is gigantic and not survivable under any circumstances.

If you approached a neutron star, the gravitational pull would actually kill you long before reaching the surface in a process called spaghettification. This unusual term, made popular by the brilliant physicist Stephen Hawking, refers to the fact that in intense gravitational fields objects are vertically stretched and horizontally compressed. The explanation is rather straight-forward: since the strength of the gravitational force depends on the distance to the source of said force, one side of the approaching object, the side closer to the source, will experience a stronger pull than the opposite side. This leads to a net force stretching the object. If the gravitational force is large enough, this would make any object look like a thin spaghetti. For humans spaghettification would be lethal as the stretching would cause the body to break apart at the weakest spot (which presumably is just above the hips). So my pro-tip is to keep a polite distance from neutron stars.

Antimatter Production – Present and Future

When it comes to using antimatter for propulsion, getting sufficient amounts of the exotic fuel is the biggest challenge. For flights within the solar system, hybrid concepts would require several micrograms of antimatter, while pure antimatter rockets would consume dozens of kilograms per trip. And going beyond the solar system would demand the production of several metric tons and more.

We are very, very far from this. Currently around 10 nanograms of anti-protons are produced in the large particles accelerators each year. At this rate it would take 100 years to produce one measly microgram and 100 billion years to accumulate one kilogram. However, the antimatter production rate has seen exponential growth, going up by sixteen orders of magnitude over the past decades, and this general trend will probably continue for some time.

Even with a noticeable slowdown in this exponential growth, gram amounts of anti-protons could be manufactured each year towards the end of the 21st century, making hybrid antimatter propulsion feasible. With no slowdown, the rate could even reach kilograms per year by then. While most physicists view this as an overly optimistic estimate, it is not impossible considering the current trend in antimatter production and the historic growth of liquid hydrogen and uranium production rates (both considered difficult to manufacture in the past).

There is still much to be optimized in the production of antimatter. The energy efficiency at present is only 10-9, meaning that you have to put in one gigajoule of pricey electric energy to produce a single joule of antimatter energy. The resulting costs are a staggering 62.5 trillion USD per gram of anti-protons, making antimatter the most expensive material known to man. So if you want to tell your wife how precious she is to you (and want to get rid of her at the same time), how about buying her a nice anti-matter scarf?

Establishing facilities solely dedicated to antimatter production, as opposed to the by-product manufacturing in modern particle accelerators, would significantly improve the situation. NASA experts estimate that an investment of around 5 billion USD is sufficient to build such a first generation antimatter factory. This step could bring the costs of anti-protons down to 25 billion USD per gram and increase the production rate to micrograms per year.

While we might not see kilogram amounts of antimatter or antimatter propulsion systems in our lifetime, the production trend over the next few decades will reveal much about the feasibility of antimatter rockets and interstellar travel. If the optimists are correct, and that’s a big if, the grandchildren of our grandchildren’s grandchildren might watch the launch of the first spacecraft capable of reaching neighboring stars. Sci-fi? I’m sure that’s what people said about the Moon landing and close-up pictures from Mars and Jupiter just a lifetime ago.

For more information, check out my ebook Antimatter Propulsion.

New Release for Kindle: Antimatter Propulsion

I’m very excited to announce the release of my latest ebook called “Antimatter Propulsion”. I’ve been working working on it like a madman for the past few months, going through scientific papers and wrestling with the jargon and equations. But I’m quite satisfied with the result. Here’s the blurb, the table of contents and the link to the product page. No prior knowledge is required to enjoy the book.

Many popular science fiction movies and novels feature antimatter propulsion systems, from the classic Star Trek series all the way to Cameron’s hit movie Avatar. But what exactly is antimatter? And how can it be used accelerate rockets? This book is a gentle introduction to the science behind antimatter propulsion. The first section deals with antimatter in general, detailing its discovery, behavior, production and storage. This is followed by an introduction to propulsion, including a look at the most important quantities involved and the propulsion systems in use or in development today. Finally, the most promising antimatter propulsion and rocket concepts are presented and their feasibility discussed, from the solid core concept to antimatter initiated microfusion engines, from the Valkyrie project to Penn State’s AIMStar spacecraft.

Section 1: Antimatter

The Atom
Dirac’s Idea
Anti-Everything
An Explosive Mix
Proton and Anti-Proton Annihilation
Sources of Antimatter
Storing Antimatter
Getting the Fuel

Section 2: Propulsion Basics

Conservation of Momentum
♪ Throw, Throw, Throw Your Boat ♫
So What’s The Deal?
May The Thrust Be With You
Acceleration
Specific Impulse and Fuel Requirements
Chemical Propulsion
Electric Propulsion
Fusion Propulsion

Section 3: Antimatter Propulsion Concepts

Solid Core Concept
Plasma Core Concept
Beamed Core Concept
Antimatter Catalyzed Micro-Fission / Fusion
Antimatter Initiated Micro-Fusion

Section 4: Antimatter Rocket Concepts

Project Valkyrie
ICAN-II
AIMStar
Dust Shields

You can purchase “Antimatter Propulsion” here for $ 2.99.

The Problem With Antimatter Rockets

The distance to our neighboring star Alpha Centauri is roughly 4.3 lightyears or 25.6 trillion km. This is an enormous distance. It would take the Space Shuttle 165,000 years to cover this distance. That’s 6,600 generations of humans who’d know nothing but the darkness of space. Obviously, this is not an option. Do we have the technologies to get there within the lifespan of a person? Surprisingly, yes. The concept of antimatter propulsion might sound futuristic, but all the technologies necessary to build such a rocket exist. Today.

What exactly do you need to build a antimatter rocket? You need to produce antimatter, store antimatter (remember, if it comes in contact with regular matter it explodes, so putting it in a box is certainly not a possibility) and find a way to direct the annihilation products. Large particle accelerators such as CERN routinely produce antimatter (mostly anti-electrons and anti-protons). Penning-Traps, a sophisticated arrangement of electric and magnetic fields, can store charged antimatter. And magnetic nozzles, suitable for directing the products of proton / anti-proton annihilations, have already been used in several experiments. It’s all there.

So why are we not on the way to Alpha Centauri? We should be making sweet love with green female aliens, but instead we’re still banging our regular, non-green, non-alien women. What’s the hold-up? It would be expensive. Let me rephrase that. The costs would be blasphemous, downright insane, Charlie Manson style. Making one gram of antimatter costs around 62.5 trillion $, it’s by far the most expensive material on Earth. And you’d need tons of the stuff to get to Alpha Centauri. Bummer! And even if we’d all get a second job to pay for it, we still couldn’t manufacture sufficient amounts in the near future. Currently 1.5 nanograms of antimatter are being produced every year. Even if scientists managed to increase this rate by a factor of one million, it would take 1000 years to produce one measly gram. And we need tons of it! Argh. Reality be a harsh mistress …

New Release for Kindle: Math Shorts – Integrals

Yesterday I released the second part of my “Math Shorts” series. This time it’s all about integrals. Integrals are among the most useful and fascinating mathematical concepts ever conceived. The ebook is a practical introduction for all those who don’t want to miss out. In it you’ll find down-to-earth explanations, detailed examples and interesting applications. Check out the sample (see link to product page) a taste of the action.

Important note: to enjoy the book, you need solid prior knowledge in algebra and calculus. This means in particular being able to solve all kinds of equations, finding and interpreting derivatives as well as understanding the notation associated with these topics.

Click the cover to open the product page:

cover

Here’s the TOC:

Section 1: The Big Picture
-Anti-Derivatives
-Integrals
-Applications

Section 2: Basic Anti-Derivatives and Integrals
-Power Functions
-Sums of Functions
-Examples of Definite Integrals
-Exponential Functions
-Trigonometric Functions
-Putting it all Together

Section 3: Applications
-Area – Basics
-Area – Numerical Example
-Area – Parabolic Gate
-Area – To Infinity and Beyond
-Volume – Basics
-Volume – Numerical Example
-Volume – Gabriel’s Horn
-Average Value
-Kinematics

Section 4: Advanced Integration Techniques
-Substitution – Basics
-Substitution – Indefinite Integrals
-Substitution – Definite Integrals
-Integration by Parts – Basics
-Integration by Parts – Indefinite Integrals
-Integration by Parts – Definite Integrals

Section 5: Appendix
-Formulas To Know By Heart
-Greek Alphabet
-Copyright and Disclaimer
-Request to the Reader

Enjoy!

New Release for Kindle: Introduction to Differential Equations

Differential equations are an important and fascinating part of mathematics with numerous applications in almost all fields of science. This book is a gentle introduction to the rich world of differential equations filled with no-nonsense explanations, step-by-step calculations and application-focused examples.

Important note: to enjoy the book, you need solid prior knowledge in algebra and calculus. This means in particular being able to solve all kinds of equations, finding and interpreting derivatives, evaluating integrals as well as understanding the notation that goes along with those.

Click the cover to open the product page:

cover

Here’s the TOC:

Section 1: The Big Picture

-Population Growth
-Definition and Equation of Motion
-Equilibrium Points
-Some More Terminology

Section 2: Separable Differential Equations

-Approach
-Exponential Growth Revisited
-Fluid Friction
-Logistic Growth Revisited

Section 3: First Order Linear Differential Equations

-Approach
-More Fluid Friction
-Heating and Cooling
-Pure, Uncut Mathematics
-Bernoulli Differential Equations

Section 4: Second Order Homogeneous Linear Differential Equations (with Constant Coefficients)

-Wait, what?
-Oscillating Spring
-Numerical Example
-The Next Step – Non-Homogeneous Equations

Section 5: Appendix

-Formulas To Know By Heart
-Greek Alphabet
-Copyright and Disclaimer
-Request to the Reader

Note: With this book release I’m starting my “Math Shorts” Series. The next installment “Math Shorts – Integrals” will be available in just a few days! (Yes, I’m working like a mad man on it)

Motion With Constant Acceleration (Examples, Exercises, Solutions)

An abstraction often used in physics is motion with constant acceleration. This is a good approximation for many different situations: free fall over small distances or in low-density atmospheres, full braking in car traffic, an object sliding down an inclined plane, etc … The mathematics behind this special case is relatively simple. Assume the object that is subject to the constant acceleration a (in m/s²) initially has a velocity v(0) (in m/s). Since the velocity is the integral of the acceleration function, the object’s velocity after time t (in s) is simply:

1) v(t) = v(0) + a · t

For example, if a car initially goes v(0) = 20 m/s and brakes with a constant a = -10 m/s², which is a realistic value for asphalt, its velocity after a time t is:

v(t) = 20 – 10 · t

After t = 1 second, the car’s speed has decreased to v(1) = 20 – 10 · 1 = 10 m/s and after t = 2 seconds the car has come to a halt: v(2) = 20 – 10 · 2 = 0 m/s. As you can see, it’s all pretty straight-forward. Note that the negative acceleration (also called deceleration) has led the velocity to decrease over time. In a similar manner, a positive acceleration will cause the speed to go up. You can read more on acceleration in this blog post.

What about the distance x (in m) the object covers? We have to integrate the velocity function to find the appropriate formula. The covered distance after time t is:

2) x(t) = v(0) · t + 0.5 · a · t²

While that looks a lot more complicated, it is really just as straight-forward. Let’s go back to the car that initially has a speed of v(0) = 20 m/s and brakes with a constant a = -10 m/s². In this case the above formula becomes:

x(t) = 20 · t – 0.5 · 10 · t²

After t = 1 second, the car has traveled x(1) = 20 · 1 – 0.5 · 10 · 1² = 15 meters. By the time it comes to a halt at t = 2 seconds, it moved x(2) = 20 · 2 – 0.5 · 10 · 2² = 20 meters. Note that we don’t have to use the time as a variable. There’s a way to eliminate it. We could solve equation 1) for t and insert the resulting expression into equation 2). This leads to a formula connecting the velocity v and distance x.

3) Constant acceleration_html_b85f3ec

Solved for x it looks like this:

3)’ Constant acceleration_html_m23bb2bb3

It’s a very useful formula that you should keep in mind. Suppose a tram accelerates at a constant a = 1.3 m/s², which is also a realistic value, from rest (v(0) = 0 m/s). What distance does it need to go to full speed v = 10 m/s? Using equation 3)’ we can easily calculate this:

Constant acceleration_html_m11de6604

————————————————————————————-

Here are a few exercises and solutions using the equations 1), 2) and 3).

1. During free fall (air resistance neglected) an object accelerates with about a = 10 m/s. Suppose the object is dropped, that is, it is initially at rest (v(0) = 0 m/s).

a) What is its speed after t = 3 seconds?
b) What distance has it traveled after t = 3 seconds?
c) Suppose we drop the object from a tower that is x = 20 meters tall. At what speed will it impact the ground?
d) How long does the drop take?

Hint: in exercise d) solve equation 1) for t and insert the result from c)

2. During the reentry of space crafts accelerations can be as high as a = -70 m/s². Suppose the space craft initially moves with v(0) = 6000 m/s.

a) What’s the speed and covered distance after t = 10 seconds?
b) How long will it take the space craft to half its initial velocity?
c) What distance will it travel during this time?

3. An investigator arrives at the scene of a car crash. From the skid marks he deduces that it took the car a distance x = 55 meters to come to a halt. Assume full braking (a = -10 m/s²). Was the car initially above the speed limit of 30 m/s?

————————————————————————————-

Solutions to the exercises:

Exercise 1

a) 30 m/s
b) 45 m
c) 20 m/s
d) 2 s

Exercise 2

a) 5,300 m/s and 56,500 m
b) 42.9 s (rounded)
c) 192,860 m (rounded)

Exercise 3

Yes (he was initially going 33.2 m/s)

————————————————————————————-

To learn the basic math you need to succeed in physics, check out the e-book “Algebra – The Very Basics”. For an informal introduction to physics, check out the e-book “Physics! In Quantities and Examples”. Both are available at low prices and exclusively for Kindle.

Decibel – A Short And Simple Explanation

A way of expressing a quantity in relative terms is to do the ratio with respect to a reference value. This helps to put a quantity into perspective. For example, in mechanics the acceleration is often expressed in relation to the gravitational acceleration. Instead of saying the acceleration is 22 m/s² (which is hard to relate to unless you know mechanics), we can also say the acceleration is 22 / 9.81 ≈ 2.2 times the gravitational acceleration or simply 2.2 g’s (which is much easier to comprehend).

The decibel (dB) is also a general way of expressing a quantity in relative terms, sort of a “logarithmic ratio”. And just like the ratio, it is not a physical unit or limited to any field such as mechanics, audio, etc … You can express any quantity in decibels. For example, if we take the reference value to be the gravitational acceleration, the acceleration 22 m/s² corresponds to 3.5 dB.

To calculate the decibel value L of a quantity x relative to the reference value x(0), we can use this formula:

MIXING_html_m6e565f15

In acoustics the decibel is used to express the sound pressure level (SPL), measured in Pascal = Pa, using the threshold of hearing (0.00002 Pa) as reference. However, in this case a factor of twenty instead of ten is used. The change in factor is a result of inputting the squares of the pressure values rather than the linear values.

MIXING_html_297973dd

The sound coming from a stun grenade peaks at a sound pressure level of around 15,000 Pa. In decibel terms this is:

MIXING_html_m75e9ffd9

which is way past the threshold of pain that is around 63.2 Pa (130 dB). Here are some typical values to keep in mind:

0 dB → Threshold of Hearing
20 dB → Whispering
60 dB → Normal Conversation
80 dB → Vacuum Cleaner
110 dB → Front Row at Rock Concert
130 dB → Threshold of Pain
160 dB → Bursting Eardrums

Why use the decibel at all? Isn’t the ratio good enough for putting a quantity into perspective? The ratio works fine as long as the quantity doesn’t go over many order of magnitudes. This is the case for the speeds or accelerations that we encounter in our daily lives. But when a quantity varies significantly and spans many orders of magnitude (which is what the SPL does), the decibel is much more handy and relatable.

Another reason for using the decibel for audio signals is provided by the Weber-Fechner law. It states that a stimulus is perceived in a logarithmic rather than linear fashion. So expressing the SPL in decibels can be regarded as a first approximation to how loud a sound is perceived by a person as opposed to how loud it is from a purely physical point of view.

Note that when combining two or more sound sources, the decibel values are not simply added. Rather, if we combine two sources that are equally loud and in phase, the volume increases by 6 dB (if they are out of phase, it will be less than that). For example, when adding two sources that are at 50 dB, the resulting sound will have a volume of 56 dB (or less).

(This was an excerpt from Audio Effects, Mixing and Mastering. Available for Kindle)

Temperature – From The Smallest To The Largest

For temperature there is a definite and incontrovertible lower limit: 0 K. Among the closest things to absolute zero in the universe is the temperature of supermassive black holes (10-18 K). At this temperature it will take them 10100 years and more to evaporate their mass. Yes, that’s a one with one-hundred zeros. If the universe really does keep on expanding as believed by most scientist today, supermassive black holes will be the last remaining objects in the fading universe. Compared to their temperature, the lowest temperature ever achieved in a laboratory (10-12 K) is a true hellfire, despite it being many orders of magnitudes lower than the background temperature of the universe (2.73 K and slowly decreasing).

In terms of temperature, helium is an exceptional element. The fact that we almost always find it in the gaseous state is a result of its low boiling point (4.22 K). Even on Uranus (53 K), since the downgrading of Pluto the coldest planet in the solar system and by far the planet with the most inappropriate name, it would appear as a gas. Another temperature you definitely should remember is 92 K. Why? Because at this temperature the material Y-Ba-Cu-oxide becomes superconductive and there is no material known to man that is superconductive at higher temperatures. Note that you want a superconductor to do what it does best at temperatures as close to room temperature as possible because otherwise making use of this effect will require enormous amounts of energy for cooling.

The lowest officially recorded air temperature on Earth is 184 K ≈ -89 °C, so measured in 1983 in Stántsiya Vostók, Antarctica. Just recently scientists reported seeing an even lower temperature, but at the time of writing this is still unconfirmed. The next two values are very familiar to you: the melting point (273 K ≈ 0 °C) and the boiling point (373 K ≈ 100 °C) of water. But I would not advise you to become too familiar with burning wood (1170 K ≈ 900 °C) or the surface of our Sun (5780 K ≈ 5500 °C).

Temperatures in a lightning channel can go far beyond that, up to about 28,000 K. This was topped on August 6, 1945, when the atomic bomb “Little Boy” was dropped on Hiroshima. It is estimated that at a distance of 17 meters from the center of the blast the temperature rose to 300,000 K. Later and more powerful models of the atomic bomb even went past the temperature of the solar wind (800,000 K).

If you are disappointed about the relatively low surface temperature of the sun, keep in mind that this is the coldest part of the sun. In the corona surrounding it, temperatures can reach 10 million K, the center of the Sun is estimated to be at 16 million K and solar flares can be as hot as 100 million K. Surprisingly, mankind managed to top that. The plasma in the experimental Tokamak Fusion Test Reactor was recorded at mind-blowing 530 million K. Except for supernova explosions (10 billion K) and infant neutron stars (1 trillion K), there’s not much beyond that.

Overtones – What They Are And How To Compute Them

In theory, hitting the middle C on a piano should produce a sound wave with a frequency of 523.25 Hz and nothing else. However, running the resulting audio through a spectrum analyzer, it becomes obvious that there’s much more going on. This is true for all other instruments, from tubas to trumpets, basoons to flutes, contrabasses to violins. Play any note and you’ll get a package of sound waves at different frequencies rather than just one.

First of all: why is that? Let’s focus on stringed instruments. When you plug the string, it goes into its most basic vibration mode: it moves up and down as a whole at a certain frequency f. This is the so called first harmonic (or fundamental). But shortly after that, the nature of the vibration changes and the string enters a second mode: while one half of the string moves up, the other half moves down. This happens naturally and is just part of the string’s dynamics. In this mode, called the second harmonic, the vibration accelerates to a frequency of 2 * f. The story continues in this fashion as other modes of vibration appear: the third harmonic at a frequency 3 * f, the fourth harmonic at 4 * f, and so on.

String Vibrating Modes Overtones

A note is determined by the frequency. As already stated, the middle C on the piano should produce a sound wave with a frequency of 523.25 Hz. And indeed it does produce said sound wave, but it is only the first harmonic. As the string continues to vibrate, all the other harmonics follow, producing overtones. In the picture below you can see which notes you’ll get when playing a C (overtone series):

(The marked notes are only approximates. Taken from http://legacy.earlham.edu)

Quite the package! And note that the major chord is fully included within the first four overtones. So it’s buy a note, get a chord free. And unless you digitally produce a note, there’s no avoiding it. You might wonder why it is that we don’t seem to perceive the additional notes. Well, we do and we don’t. We don’t perceive the overtones consciously because the amplitude, and thus volume, of each harmonic is smaller then the amplitude of the previous one (however, this is a rule of thumb and exceptions are possible, any instrument will emphasize some overtones in particular). But I can assure you that when listening to a digitally produced note, you’ll feel that something’s missing. It will sound bland and cold. So unconsciously, we do perceive and desire the overtones.

If you’re not interested in mathematics, feel free to stop reading now (I hope you enjoyed the post so far). For all others: let’s get down to some mathematical business. The frequency of a note, or rather of its first harmonic, can be computed via:

(1) f(n) = 440 * 2n/12

With n = 0 being the chamber pitch and each step of n one half-tone. For example, from the chamber pitch (note A) to the middle C there are n = 3 half-tone steps (A#, B, C). So the frequency of the middle C is:

f(3) = 440 * 23/12 = 523.25 Hz

As expected. Given a fundamental frequency f = F, corresponding to a half-step-value of n = N, the freqency of the k-th harmonic is just:

(2) f(k) = k * F = k * 440 * 2N/12

Equating (1) and (2), we get a relationship that enables us to identify the musical pitch of any overtone:

440 * 2n/12 = k * 440 * 2N/12

2n/12 = k * 2N/12

n/12 * ln(2) = ln(k) + N/12 * ln(2)

n/12 = ln(k)/ln(2) + N/12

(3) n – N = 12 * ln(k) / ln(2) ≈ 17.31 * ln(k)

The equation results in this table:

k

n – N (rounded)

1 0
2 12
3 19
4 24
5 28

And so on. How does this tell us where the overtones are? Read it like this:

  • The first harmonic (k = 1) is zero half-steps from the fundamental (n-N = 0). So far, so duh.
  • The second harmonic (k = 2) is twelve half-steps, or one octave, from the fundamental (n-N = 12).
  • The third harmonic (k = 3) is nineteen half-steps, or one octave and a quint, from the fundamental (n-N = 19).
  • The fourth harmonic (k = 4) is twenty-four half-steps, or two octaves, from the fundamental (n-N = 24).
  • The fifth harmonic (k = 5) is twenty-wight half-steps, or two octaves and a third, from the fundamental (n-N = 28).

So indeed the formula produces the correct overtone series for any note. And for any note the same is true: The second overtone is exactly one octave higher, the third harmonic one octave and a quint higher, and so on. The corresponding major chord is always contained within the first five harmonics.

The Doppler Effect in Pictures

The siren of an approaching police car will sound at a higher pitch, the light of an approaching star will be shifted towards blue and a passing supersonic jet will create a violent thunder. What do these phenomenon have in common? All of them are a result of the Doppler effect. To understand how it arises, just take a look at the animations below.

Stationary Source: The waves coming from the source propagate symmetrically.

Subsonic Source (moving below sound speed): Compression of waves in direction of motion.

Sonic Source (moving at sound speed): Maximum compression.

Supersonic Source (moving beyond sound speed): Source overtakes its waves, formation of Mach cone and sonic boom.

WOWC_IFUSA_16rotated

(Pictures taken from http://www.acs.psu.edu)

Einstein’s Special Relativity – The Core Idea

It might surprise you that a huge part of Einstein’s Special Theory of Relativity can be summed up in just one simple sentence. Here it is:

“The speed of light is the same in all frames of references”

In other words: no matter what your location or speed is, you will always measure the speed of light to be c = 300,000,000 m/s (approximate value). Not really that fascinating you say? Think of the implications. This sentence not only includes the doom of classical physics, it also forces us to give up our notions of time. How so?

Suppose you watch a train driving off into the distance with v = 30 m/s relative to you. Now someone on the train throws a tennis ball forward with u = 10 m/s relative to the train. How fast do you perceive the ball to be? Intuitively, we simply add the velocities. If the train drives off with 30 m/s and the ball adds another 10 m/s to that, it should have the speed w = 40 m/s relative to you. Any measurement would confirm this and all is well.

Now imagine (and I mean really imagine) said train is driving off into the distance with half the light speed, or v = 0.5 * c. Someone on the train shines a flashlight forwards. Obviously, this light is going at light speed relative to the train, or u = c. How fast do you perceive the light to be? We have the train at 0.5 * c and the light photons at the speed c on top of that, so according to our intuition we should measure the light at a velocity of v = 1.5 * c. But now think back to the above sentence:

“The speed of light is the same in all frames of references”

No matter how fast the train goes, we will always measure light coming from it at the same speed, period. Here, our intiution differs from physical reality. This becomes even clearer when we take it a step further. Let’s have the train drive off with almost light speed and have someone on the train shine a flashlight forwards. We know the light photons to go at light speed, so from our perspective the train is almost able to keep up with the light. An observer on the train would strongly disagree. For him the light beam is moving away as it always does and the train is not keeping up with the light in any way.

How is this possible? Both you and the observer on the train describe the same physical reality, but the perception of it is fundamentally different. There is only one way to make the disagreement go away and that is by giving up the idea that one second for you is the same as one second on the train. If you make the intervals of time dependent on speed in just the right fashion, all is well.

Suppose that one second for you is only one microsecond on the train. In your one second the distance between the train and the light beam grows by 300 meter. So you say: the light is going 300 m / 1 s = 300 m/s faster than the train.

However, for the people in the train, this same 300 meter distance arises in just one microsecond, so they say: the light is going 300 m / 1 µs = 300 m / 0.000,001 s  = 300,000,000 m/s faster than the train – as fast as it always does.

Note that this is a case of either / or. If the speed of light is the same in all frames of references, then we must give up our notions of time. If the light speed depends on your location and speed, then we get to keep our intiutive image of time. So what do the experiments say? All experiments regarding this agree that the speed of light is indeed the same in all frames of references and thus our everyday perception of time is just a first approximation to reality.

Earth’s Magnetic Field

You have been in a magnetic field all your life. The Earth, just like any other planet in the solar system, spawns its own magnetic field. The strength of the field is around B =0.000031 T, but research has shown that this value is far from constant. Earth’s magnetic field is constantly changing. How do we know this? When rocks solidify, they store the strength and direction of the magnetic field. Hence, as long it is possible to figure out the orientation of a rock at the time of solidification, it will tell us what the field was like back then

For rocks that are billions of years old, deducing the original orientation is impossible. Continental drifting has displaced and turned them too often. But thanks to the incredibly low speed of drifting continents, scientists were able to recreate the magnetic field of Earth for the past several million years. This revealed quite a bit.

For one, the poles don’t stand still, but rather wander across the surface with around 50 km per year. The strength of the field varies from practically zero to 0.0001 T (about three times the current strength). And even more astonishingly: the polarity of the field flips every 300,000 years or so. The north pole then turns into the south pole and vice versa. The process of pole reversal takes on average between 1000 and 5000 years, but can also happen within just 100 years. There is no indication that any of these changes had a noticeable impact on plants or animals.

Where does the magnetic field come from? At present there’s no absolute certainty, but the Parker Dynamo Model developed in the sixties seems to be provide the correct answer. The inner core of Earth is a sphere of solid iron that is roughly equal to the Moon in size and about as hot as the surface of the Sun. Surrounding it is the fluid outer core. The strong temperature gradient within the outer core leads to convective currents that contain large amounts of charged particles. According to the theory, the motion of these charges is what spawns the field. Recent numerical simulations on supercomputers have shown that this model is indeed able to reproduce the field in most aspects. It explains the intensity, the dipole structure, the wandering of the poles (including the observed drifting speed) and the pole reversals (including the observed time spans).

It is worth noting that the pole which lies in the geographic north, called the North Magnetic Pole, is actually a magnetic south pole. You might recall that we defined the north pole of a magnet as the pole which will point northwards when the magnet is allowed to turn freely. Since unlike poles attract, this means that there must be a magnetic south pole in the north (talk about confusing). By the same logic we can conclude that the Earth’s South Magnetic Pole is a magnetic north pole.

Hollywood and Physics

We’ve all seen these kinds of movies. After a fast and dramatic chase, the bad guy jumps out of the car, determined to end the good guy once and for all. His evil plans have been thwarted for the last time! In self-defense, the good guy is forced to take a shot and when the bullet hits, the evildoer is thrown violently backwards as a result of the impact and through the nearest shop window. Once the hero is reunited with the love of his life, the credits roll and we are left to wonder if that’s really how physics work.

In a previous example we calculated the momentum of a common 9 mm bullet (p = 5.4 kg m/s). Suppose the m = 75 kg evildoer gets hit by just this bullet. Since the bullet practically comes to a halt on impact, this momentum has to be transferred to the unlucky antihero for the conservation of momentum to hold true. Accordingly, this is the speed at which the bad guy is thrown back:

5.4 kg m/s = 75 kg · v’

v’ ≈ 0.07 m/s ≈ 0.26 km/h ≈ 0.16 mph

This is not even enough to topple a person, let alone make him fly dramatically through the air. From a kinematic point of view, the impact is not noticeable. The same is true for more massive and faster bullets as well as for a series of impacts. The only thing that can make a person fall instantly after getting shot is a sudden drop in blood pressure and the resulting loss of consciousness. But in this case, the evildoer would simply drop where he stands instead of being thrown backwards.

This is not the only example of Hollywood bending the laws of physics. You’ve probably heard the weak “fut” sound a Hollywood gun makes when equipped with a silencer. This way the hero can take out an entire army of bad guys without anyone noticing. But that’s not how pistol silencers work. At best, they can reduce the the sound level to about 120 dB, which is equivalent to what you hear standing near a pneumatic hammer or right in front of the speakers at a rock concert. So unless the hero is up against an army of hearing impaired seniors (which wouldn’t make him that much of a hero), his coming will be noticed.

This was an excerpt from my Kindle book: Physics! In Quantities and Examples

For more interesting physics, check out my Best of Physics selection.

Wavelength (And: Why Is The Sky Blue?)

A very important type of length is wavelength, usually symbolized by the Greek letter λ (in m). It is defined as the distance from crest to crest (one complete cycle) and can easily be calculated for any wave by dividing the speed of the wave c (in m/s) by its frequency f (in Hz):

λ = c / f

What are typical wavelengths for sound? At room temperature, sound travels with a speed of c = 343 m/s. The chamber pitch has a frequency of f = 440 Hz. According to the equation, the corresponding wavelength is:

λ = 343 / 440 ≈ 0.8 m ≈ 2.6 ft

Are you surprised? I bet most people would greatly underestimate this value. Bass sounds are even longer than that. The lowest tone on a four-string bass guitar has a frequency of f = 41.2 Hz, which leads to the wavelength:

λ = 343 / 41.2 ≈ 8.3 m ≈ 27 ft

So the wave coming from the open E string of a bass guitar doesn’t even fit in a common room. In the case of light, the situation is very different. As noted in the introduction, the wavelength of light ranges between 4000 A (violet light) and 7000 A (red light), which is just below the size of a bacterium.

Wavelength plays an important role in explaining why the sky is blue. When light collides with a particle, parts of it are deflected while the rest continues along the initial path. This phenomenon is known as scattering. The smaller the wavelength of the light, the stronger the effect. This means that scattering is particularly pronounced for violet and blue light.

Unless you are looking directly at the Sun, all the light you see when looking at the sky is scattered light coming from the particles in the atmosphere. Since blue light tends to scatter so easily, the sky ends up in just this color. But why not violet? This is a legitimate question. After all, due to its smaller wavelength, violet light is even more willing to scatter. While this is true, it is also important to note that the sun’s rays don’t contain all the colors in the same ratio. In particular, they carry much less violet than blue light. On top of that, our eyes are less sensitive to violet light.

(This is an excerpt from my Kindle book: Physics! In Quantities and Examples)

The Jeans Mass, or: How are stars born?

No, this has nothing to do with pants. The Jeans mass is a concept used in astrophysics and its unlikely name comes from the British physicist Sir James Jeans, who researched the conditions of star formation. The question at the core is: under what circumstances will a dark and lonely gas cloud floating somewhere in the depth of space turn into a shining star? To answer this, we have to understand what forces are at work.

One obvious factor is gravitation. It will always work towards contracting the gas cloud. If no other forces were present, it would lead the cloud to collapse into a single point. The temperature of the cloud however provides an opposite push. It “equips” the molecules of the cloud with kinetic energy (energy of motion) and given a high enough temperature, the kinetic energy would be sufficient for the molecules to simply fly off into space, never to be seen again.

It is clear that no star will form if the cloud expands and falls apart. Only when gravity wins this battle of inward and outward push can a stable star result. Sir James Jeans did the math and found that it all boils down to one parameter, the Jeans mass. If the actual mass of the interstellar cloud is larger than this critical mass, it will contract and stellar formation occurs. If on the other hand the actual mass is smaller, the cloud will simply dissipate.

The Jeans mass depends mainly on the temperature T (in K) and density D (in kg/m³) of the cloud. The higher the temperature, the larger the Jeans mass will be. This is in line with our previous discussion. When the temperature is high, a larger amount of mass is necessary to overcome the thermal outward push. The value of the Jeans mass M (in kg) can be estimated from this equation:

M ≈ 1020 · sqrt(T³ / D)

Typical values for the temperature and density of interstellar clouds are T = 10 K and D = 10-22 kg/m³. This leads to a Jeans mass of M = 1.4 · 1032 kg. Note that the critical mass turns out to be much greater than the mass of a typical star, indicating that stars generally form in clusters. Rather than the cloud contracting into a single star, which is the picture you probably had in your mind during this discussion, it will fragment at some point during the contraction and form multiple stars. So stars always have brothers and sisters.

(This was an excerpt from the Kindle book Physics! In Quantities and Examples)