Astronomy

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.

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.

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 …

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.

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.

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)

Released Today for Kindle: Physics! In Quantities and Examples

I finally finished and released my new ebook … took me longer than usual because I always kept finding new interesting topics while researching. Here’s the blurb, link and TOC:

This book is a concept-focused and informal introduction to the field of physics that can be enjoyed without any prior knowledge. Step by step and using many examples and illustrations, the most important quantities in physics are gently explained. From length and mass, over energy and power, all the way to voltage and magnetic flux. The mathematics in the book is strictly limited to basic high school algebra to allow anyone to get in and to assure that the focus always remains on the core physical concepts.

(Click cover to get to the Amazon Product Page)

cover

Table of Contents:

Length
(Introduction, From the Smallest to the Largest, Wavelength)

Mass
(Introduction, Mass versus Weight, From the Smallest to the Largest, Mass Defect and Einstein, Jeans Mass)

Speed / Velocity
(Introduction, From the Smallest to the Largest, Faster than Light, Speed of Sound for all Purposes)

Acceleration
(Introduction, From the Smallest to the Largest, Car Performance, Accident Investigation)

Force
(Introduction, Thrust and the Space Shuttle, Force of Light and Solar Sails, MoND and Dark Matter, Artificial Gravity and Centrifugal Force, Why do Airplanes Fly?)

Area
(Introduction, Surface Area and Heat, Projected Area and Planetary Temperature)

Pressure
(Introduction, From the Smallest to the Largest, Hydraulic Press, Air Pressure, Magdeburg Hemispheres)

Volume
(Introduction, Poisson’s Ratio)

Density
(Introduction, From the Smallest to the Largest, Bulk Density, Water Anomaly, More Densities)

Temperature
(Introduction, From the Smallest to the Largest, Thermal Expansion, Boiling, Evaporation is Cool, Why Blankets Work, Cricket Temperature)

Energy
(Introduction, Impact Speed, Ice Skating, Dear Radioactive Ladies and Gentlemen!, Space Shuttle Reentry, Radiation Exposure)

Power
(Introduction, From the Smallest to the Largest, Space Shuttle Launch and Sound Suppression)

Intensity
(Introduction, Inverse Square Law, Absorption)

Momentum
(Introduction, Perfectly Inelastic Collisions, Recoil, Hollywood and Physics, Force Revisited)

Frequency / Period
(Introduction, Heart Beat, Neutron Stars, Gravitational Redshift)

Rotational Motion
(Extended Introduction, Moment of Inertia – The Concept, Moment of Inertia – The Computation, Conservation of Angular Momentum)

Electricity
(Extended Introduction, Stewart-Tolman Effect, Piezoelectricity, Lightning)

Magnetism
(Extended Introduction, Lorentz Force, Mass Spectrometers, MHD Generators, Earth’s Magnetic Field)

Appendix:
Scalar and Vector Quantities
Measuring Quantities
Unit Conversion
Unit Prefixes
References
Copyright and Disclaimer

As always, I discounted the book in countries with a low GDP because I think that education should be accessible for all people. Enjoy!

NASA’s O-Ring Problem and the Challenger Disaster

In January 1986 the world watched in shock as the Challenger Space Shuttle, on its way to carry the first civilian to space, exploded just two minutes after lift-off. A presidential commission later determined that an O-ring failure in the Solid Rocket Booster (SRB) caused the disaster. This was not a new problem, there’s a long history of issues with the O-rings leading up to Challenger’s loss.

Before the Space Shuttle was declared operational, it performed four test flights to space and back. The first O-ring anomaly occurred on the second test flight, named STS-2 (November 1981). After each flight Thiokol, the company in charge of manufacturing the SRBs, sent a team of engineers to inspect the retrieved boosters. The engineers found that the primary O-ring had eroded by 0.053”. The secondary O-ring, which serves as a back-up for the primary O-ring, showed no signs of erosion. On further inspection the engineers also discovered that the putty protecting the O-rings from the hot gas inside the SRB had blow-holes.

Luckily, the O-rings sealed the SRB despite the erosion. Simulations done by engineers after the STS-2 O-ring anomaly showed that even with 0.095” erosion the primary O-ring would perform its duty up to a pressure of 3000 psi (the pressure inside the SRB only goes up to about 1000 psi). And if the erosion was even stronger, the second O-ring could still finish the job. So neither Thiokol nor NASA, neither engineers nor managers considered the problem to be critical. After the putty composition was slightly altered to prevent blow-holes from forming, the problem was considered solved. The fact that no erosion occurred on the following flights seemed to confirm this.

On STS-41-B (February 1984), the tenth Space Shuttle mission including the four test flights, the anomaly surfaced again. This time two primary O-rings were affected and there were again blow-holes in the putty. However, the erosion was within the experience base (the 0.053” that occurred on STS-2) and within the safety margin (the 0.095” resulting from simulations). So neither Thiokol nor NASA was alarmed over this.

Engineers realized that it was the leak check that caused the blow-holes in the putty. The leak check was an important tool to confirm that the O-rings are properly positioned. This was done by injecting pressurized air in the space between the primary and secondary O-ring. Initially a pressure of 50 psi was used, but this was increased to 200 psi prior to STS-41-B to make the test more reliable. After this change, O-ring erosion occurred more frequently and became a normal aspect of Space Shuttle flights.

On STS-41-C (April 1984), the eleventh overall mission, there was again primary O-ring erosion within the experience base and safety margin. The same was true for STS-41-D (August 1984), the mission following STS-41-C. This time however a new problem accompanied the known erosion anomaly. Engineers found a small amount of soot behind the primary O-ring, meaning that hot gas was able to get through before the O-ring sealed. There was no impact on the secondary O-ring. This blow-by was determined to be an acceptable risk and the flights continued.

The second case of blow-by occurred on STS-51-C (January 1985), the fifteenth mission. There was erosion and blow-by on two primary O-rings and the blow-by was worse than before. It was the first time that hot gas had reached the secondary O-ring, luckily without causing any erosion. It was also the first time that temperature was discussed as a factor. STS-51-C was launched at 66 °F and the night before the temperature dropped to an unusually low 20 °F. So the Space Shuttle and its components was even colder than the 66 °F air temperature. Estimates by Thiokol engineers put the O-ring temperature at launch at around 53 °F. Since rubber gets harder at low temperatures, low temperatures might reduce the O-rings sealing capabilities. But there was no hard data to back this conclusion up.

Despite the escalation of O-ring anomalies, the risk was again determined to be acceptable, by Thiokol as well as by NASA. The rationale behind this decision was:

  • Experience Base: All primary O-ring erosions that occurred after STS-2 were within the 0.053” experience base.

  • Safety Margin: Even with 0.095” erosion the primary O-ring would seal.

  • Redundancy: If the primary O-ring failed, the secondary O-ring would seal.

The following missions saw more escalation of the problem. On STS-51-D (early April 1985), carrying the first politician to space, primary O-ring erosion reached an unprecedented 0.068”. This was outside the experience base, but still within the safety margin. And on STS-51-B (late April 1985) a primary O-ring eroded by 0.171”, significantly outside experience base and safety margin. It practically burned through. On top of that, the Space Shuttle saw its first case of secondary O-ring erosion (0.032”).

Post-flight analysis showed that the burnt-through primary O-ring on STS-51-B was not properly positioned, which led to changes in the leak check procedure. Simulations showed that O-ring erosion could go up to 0.125” before the ability to seal would be lost and that under worst case conditions the secondary O-ring would erode by no more than 0.075”. So it seemed impossible that the secondary O-ring could fail and the risk again was declared acceptable. Also, the fact that the O-ring temperature at STS-51-B’s launch was 75 °F seemed to contradict the temperature effect.

Despite these reassurances, concerns escalated and O-ring task forces were established at Thiokol and Marshall (responsible for the Solid Rocket Motor). Space Shuttle missions continued while engineers were looking for short- and long-term solutions.

On the day of STS-51-L’s launch (January 1986), the twenty-fifth Space Shuttle mission, the temperature was expected to drop to the low 20s. Prior to launch a telephone conference was organized to discuss the effects of low temperatures on O-ring sealing. Present at the conference were engineers and managers from Thiokol, Marshall and NASA. Thiokol engineers raised concerns that the seal might fail, but were not able to present any conclusive data. Despite that, Thiokol management went along with the engineer’s position and decided not to recommend launch for temperatures below 53 °F.

The fact that there was no conclusive data supporting this new launch criterion, that Thiokol did not raise these concerns before and just three weeks ago recommended launch for STS-61-C at 40 °F caused outrage at Marshall and NASA. Thiokol then went off-line to discuss the matter and management changed their position despite the warnings of their engineers. After 30 minutes the telcon resumed and Thiokol gave their go to launch the Challenger Space Shuttle. Shortly after lift-off the O-rings failed, hot gas leaked out of the SRB and the shuttle broke apart.

If you’d like to know more, check out this great book (which served as the source for this post):

The Challenger Launch Decision: Risky Technology, Culture, and Deviance at NASA

Here you can find a thorough accident investigation report by NASA:

For the broader picture you can check out this great documentary:

More on Space Shuttles in general can be found here: Space Shuttle Launch and Sound Suppression.

Space Shuttle Launch and Sound Suppression

The Space Shuttle’s first flight (STS-1) in 1981 was considered a great success as almost all the technical and scientific goals were achieved. However, post flight analysis showed one potentially fatal problem: 16 heat shield tiles had been destroyed and another 148 damaged. How did that happen? The culprit was quickly determined to be sound. During launch the shuttle’s main engine and the SRBs (Solid Rocket Boosters) produce intense sound waves which cause strong vibrations. A sound suppression system was needed to protect the shuttle from acoustically induced damage such as cracks and mechanical fatigue. But how do you suppress the sound coming from a jet engine?

Let’s take a step back. What is the source of this sound? When the hot exhaust gas meets the ambient air, mixing occurs. This leads to the formation of a large number of eddies. The small-scale eddies close to the engine are responsible for high frequency noise, while the large-scale eddies that appear downstream cause intense low-frequency noise. Lighthill showed that the power P (in W) of the sound increases with the jet velocity v (in m/s) and the size s (in m) of the eddies:

P = K * D * c-5 * s2 * v8

with K being a constant, D the exhaust gas density and c the speed of sound. Note the extremely strong dependence of acoustic power on jet velocity: if you double the velocity, the power increases by a factor of 256. Such a strong relationship is very unusual in physics. The dependence on eddy size is also significant, doubling the size leads to a quadrupling in power. The formula tells us what we must do to effectively suppress sound: reduce jet velocity and the size of the eddies. Water injection into the exhaust gas achieves both. The water droplets absorb kinetic energy from the gas molecules, thus slowing them down. At the same time, the water breaks down the eddies.

During the second Space Shuttle launch (STS-2) a water injection system was used to suppress potentially catastrophic acoustic vibrations. This proved to be successful, it reduced the sound level by 10 – 20 dB (depending on location), and accordingly was used during every launch since then. But large amounts of water are needed to accomplish this reduction. The tank at the launch pad holds about 300,000 gallons. The flow starts at T minus 6.6 seconds and last for about 20 seconds. The peak flow rate is roughly 15,000 gallons per seconds. That’s a lot of water!

The video below shows a test run of the sound suppression system:

Sources and further reading:

http://www.scielo.cl/pdf/ingeniare/v14n3/art09.pdf

http://www-pao.ksc.nasa.gov/nasafact/count4ssws.htm

http://dodo.inm.ras.ru/russia-china/wp-content/uploads/2013/09/CAE_XUYue_Investigation-of-Flow-Control-with-Fluidic-injection-for-Jet-Noise-Reduction.pdf

Comets: Visitors From Beyond

The one thing we love the most in the world of astronomy is a good mystery. And if there was ever a mysterious and yet very powerful force of nature that we witness in the night skies, it is the coming of the mighty comet.

The arrival of a comet within view of Earth is an event of international importance. Witness the huge media attention that the Haley or Hale-Bopp have had when they have come within view The sight of these amazing space objects is simultaneously frightening and awe inspiring.

Image

Above all, it is during these comet viewings that the astronomer comes out in all of us. But what is a comet? Where did it come from? And how does it get that magnificent tail?

We should never confuse comets with asteroids. Asteroids are small space rocks that come from an asteroid belt between Mars and Jupiter. While still quite stunning to see, they pale in comparison to the arrival of a comet. Asteroids also have received considerable study by the scientific community.

Not as much is known about comets. As a rule, comets are considerably larger than asteroids. The composition of a comet is a mixture of nebulous, gasses, ice, dust and space debris. One scientist called the composition of a comet as similar to a “dirty snowball” because the composition is so diverse and changeable. The center or nucleus of a comet is usually quiet solid but the “snowball” materials often create a “cloud” around that nucleus that can become quite large and that extends at great lengths behind the comet as it moves through space. That trailing plume is what makes up the comet’s magnificent tail that makes it so exciting to watch when a comet comes within view of Earth.

The origins of comets is similarly mysterious. There are a number of theories about where they come from but it is clear that they originate from outside our solar system, somewhere in deep space. Some have speculated they are fragments left over from the organization of planets that get loose from whatever gravitational pull and are sent flying across space to eventually get caught up in the gravity of our sun bringing them into our solar system.

Another theory is that they come from a gaseous cloud called the Oort cloud which is cooling out there after the organization of the sun. As this space debris cools, it gets organized into one body which then gathers sufficient mass to be attracted into the gravity of our solar system turning into a fast moving comet plummeting toward our sun. However, because of the strong gravitational orbits of the many planets in our solar system, the comet does not always immediately collide with the sun and often takes on an orbit of its own.

The life expectancy of comets varies widely. Scientists refer to a comet that is expected to burn out or impact the sun within two hundred years as a short period comet whereas a long period comet has a life expectancy of over two hundred years. That may seem long to us as earth dwellers but in terms of stars and planets, this is a very short life as a space object indeed.

Scientists across the globe have put together some pretty impressive probes to learn more about comets to aid our understanding of these visitors from beyond. In 1985, for example, the United States put a probe into the path of the comet Giacobini-Zinner which passed through the comets tail gathering tremendous scientific knowledge about comets. Then in 1986, an international collation of scientists were able to launch a probe that was able to fly close to Haley’s comet as it passed near Earth and continue the research.

While science fiction writers and tabloid newspapers like to alarm us with the possibility of a comet impacting the earth, scientists who understand the orbits of comets and what changes their paths tell us this is unlikely. That is good because some comets reach sizes that are as big as a planet so that impact would be devastating. For now, we can enjoy the fun of seeing comets make their rare visits to our night sky and marvel at the spectacular shows that these visitors from beyond put on when they are visible in the cosmos.

The Fourth State of Matter – Plasmas

From our everyday lifes we are used to three states of matter: solid, liquid and gas. When we heat a solid it melts and becomes liquid. Heating this liquid further will cause it to evaporate to a gas. Usually this is what we consider to be the end of the line. But heating a gas leads to many surprises, it eventually turns into a state, which behaves completely different than ordinary gases. We call matter in that state a plasma.

 To understand why at some point a gas will exhibit an unusual behaviour, we need to look at the basic structure of matter. All matter consists of atoms. The Greeks believed this to be the undivisible building blocks of all objects. Scientists however have discovered, that atoms do indeed have an inner structure and are divisible. It takes an enormous amount to split atoms, but it can be done.

 Further research showed that atoms consist of three particles: neutrons, protons and electrons. The neutrons and protons are crammed into the atomic core, while the electrons surround this core. Usually atoms are not charged, because they contain as much protons (positively charged) as electrons (negatively charged). The charges balance each other. Only when electrons are missing does the atom become electric. Such charged atoms are called ions.

 In a gas the atoms are neutral. Each atom has as many protons as electrons, they are electrically balanced. When you apply a magnetic field to a gas, it does not respond. If you try to use the gas to conduct electricity, it does not work.

 Remember that gas molecules move at high speeds and collide frequently with each other. As you increase the temperature, the collisions become more violent. At very high temperatures the collisions become so violent, that the impact can knock some electrons off an atom (ionization). This is where the plasma begins and the gas ends.

 In a plasma the collisions are so intense that the atoms are not able to hold onto their outer electrons. Instead of a large amount of neutral atoms like in the gas, we are left with a mixture of free electrons and ions. This electric soup behaves very differently: it responds to magnetic fields and can conduct electricity very efficiently.

plasma1

 (The phases of matter. Source: NASA)

Most matter in the universe is in plasma form. Scientist believe that only 1 % of all visible matter is either solid, liquid or gaseous. On earth it is different, we rarely see plasmas because the temperatures are too small. But there are some exceptions.

 High-temperature flames can cause a small volume of air to turn into a plasma. This can be seen for example in the so called ionic wind experiment, which shows that a flame is able to transmit electric currents. Gases can’t do that. DARPA, the Pentagon’s research arm, is currently using this phenomenon to develop new methods of fire suppression. Other examples for plasmas on earth are lightnings and the Aurora Borealis.

plasma2

 (Examples of plasmas. Source: Contemporary Physics Education Project)

The barrier between gases and plasmas is somewhat foggy. An important quantity to characterize the transition from gas to plasma is the ionization degree. It tells us how many percent of the atoms have lost one or more electrons. So an ionization degree of 10 % means that only one out of ten atoms is ionized. In this case the gas properties are still dominant.

plasma3

 (Ionization degree of Helium over Temperature. Source: SciVerse)