Posts Tagged ‘University of Namibia’

Some of you may be aware that there is an annular eclipse of the Sun on Sunday 26 February, which is why I am posting this blog a few days before it. Annular eclipses occur when the Moon is a little too far away to block the Sun out entirely, so instead we see a ring of light around the Moon, as this picture below shows. This particular picture was taken during the May 20 2012 annular eclipse


An annular eclipse happens when the Moon is slightly too far away to block out the Sun entirely. This is a picture of the May 20 2012 annular eclipse.

The Moon’s elliptical orbit about the Earth

The diagram below shows an exaggerated cartoon of the Moon’s orbit about the Earth. The Moon’s orbit is an ellipse, it has an eccentricity of 0.0549 (a perfect circle has an eccentricity of 0). The average distance of the Moon from the Earth (actually, the distance between their centres) is 384,400 kilometres. The point at which it is furthest from the Earth is called the apogee, and is at a distance of 405,400 km. The point at which it is closest is called the perigee, and it is at a distance of 362,600 km.


The Moon orbits the Earth in an ellipse, not a circle. The furthest it is from the Earth in its orbit (the apogee) is at a distance of 405,400 km, the nearest (the perigee) is at a distance of 362,600 km.

The angular size of the Moon

It is pure coincidence that the Moon is the correct angular size to block out the Sun. The Moon is slightly oblate, but has a mean radius of 1,737 km. With its average distance of 384,400, this means that from the Earth’s surface (the Earth’s mean radius is 6,371 km) the Moon has an angular size on the sky of

2 \times \tan^{-1} \left( \frac{ (1.737 \times 10^{6}) }{ (3.84 \times 10^{8} - 6.371 \times 10^{6}) } \right) = 2 \times \tan^{-1} (4.59975 \times 10^{-3} )

= 2 \times 0.2635 = \boxed{ 0.527 ^{\circ} \text{ or } 31.62 \text{ arc minutes} }
So, just over half a degree on the sky. But, this of course will vary depending on its distance. When it is at apogee (furthest away), its angular size will be

\boxed{ \text{ at apogee } 29.93 \text{ arc minutes } }

and when it is at perigee (closest) it will be

\boxed{ \text{ at perigee } 33.53 \text{ arc minutes } }

The angular size of the Sun

The Sun has an equatorial radius of 695,700 km, and its average distance from us is 149.6 million km (the Astronomical Unit – AU). So, at this average distance the Sun has an angular size of

2 \times \tan^{-1} \left( \frac{ (6.957 \times 10^{8}) }{ (1.496 \times 10^{11} - 6.371 \times 10^{6} ) } \right) = 2 \times 0.266 = \boxed {0.533^{\circ} }

Converting this to arc minutes, we get that the angular size of the Sun at its average distance is

\boxed{ 31.97 \text{ arc minutes} }

Compare this to the angular size of the Moon at its average distance, which we found to be 31.62 \text{ arc minutes}.

The angular size of the Sun varies much less than the variation in the angular size of the Moon, at aphelion (when we are furthest) from the Sun, we are at a distance of 152.1 million km, so this gives an angular size of

\boxed{ \text{ at aphelion } 31.44 \text{ arc minutes } }

and, at perielion, when the distance to the Sun is 147.095 million km, the angular size of the Sun is

\boxed{ \text{ at perihelion } 32.52 \text{ arc minutes } }

Annular Eclipses

So, from the calculations above one can see that, if the Moon is at or near perigee, its angular size of 33.53 \text{ arc minutes } is more than enough to block out the Sun. When the Moon is at its average distance, its angular size is 31.62 \text { arc minutes }, which is enough to block out the Sun unless we are near perihelion. But, when the Moon is near apogee, its angular size drops to 29.93 \text{ arc minutes }, and this is not enough to block out the Sun, even if we are at aphelion.

The Earth is currently at perihelion in early January (this year it was on January 4), so the Sun is slightly larger in the sky that it will be in August for the next solar eclipse. This, combined with the Moon being near its apogee, which occurred on February 18, (for a table of the dates of the Moon’s apogees and perigees in 2017 follow this link) means that the solar eclipse on Sunday February 26 is annular, and not total.

The February 26 2017 Annular Eclipse

Here is a map of the path of the eclipse, it is taken from the wonderful NASA Eclipse website. If you follow this link, you can find interactive maps of all the eclipses from -1999 BC to 3000 AD! If you have about 6 years to waste, this is an ideal place to do it!


The February 26 2017 annular eclipse will start in the southern Pacific ocean, sweep across Chile and Argentina, then across the Atlantic Ocean, before reaching Angola, Zambia and the Democratic Republic of Congo (Congo-Kinshasa)

The eclipse finishes in Africa and, as luck would have it, I am going to be in Namibia on the day of the eclipse. In fact, if you are reading this anytime in the week before the eclipse, I am already there. I am in Namibia for a week as part of Cardiff University’s Phoenix Project, and I will be giving a public lecture at the University of Namibia about the eclipse on Wednesday 22 February. I also hope to give a public lecture to the Namibian Scientific Society on the Friday, and on the Sunday I will be helping University of Namibia astronomers with a public observing session in Windhoek.


The February 20 2017 annular eclipse will finish in Africa, passing through Angola, Zambia and the Democratic Republic of Congo (Congo Kinshasa)

The interactive map to this eclipse, which you can find by following this link, allows you to click on any place and find out the eclipse details for that location. So, for Windhoek, the eclipse begins at 15:09 UT (which will be 17:09 local time), with the maximum of the partial eclipse being at 16:16 UT (18:16 local time), and the eclipse ending at 17:16 UT (19:16 local time). Because Windhoek is to the south of the path which will experience an annular eclipse, it will be a partial eclipse, with a coverage of 69%.


As seen from Windhoek, where I will be for the annular eclipse, the obscuration will be 69%.

So, if you are anywhere Chile, Argentina, in western South Africa, in Namibia, in Angola, or the western parts of Congo-Kinshasa and Congo-Brazzaville, look out for this wonderful astronomical event this coming Sunday. And, remember to follow the safety advise when viewing an eclipse; never look directly at the Sun and only look through a viewing device that has correct filtration. Failure to follow these precautions can result in permanently damaging your eyesight.

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On Tuesday of last week I blogged about the H.E.S.S. telescopes in Namibia which detect Cherenkov radiation produced by high energy rays (cosmic rays or gamma rays) entering the Earth’s atmosphere from outer space. These high energy rays can hit atoms in the Earth’s atmosphere; it is these collisions which can give rise to Cherenkov radiation. Today I am going to give more detail about Cherenkov radiation; this is based on my lecture notes from teaching High Energy Astrophysics at Cardiff University, a Masters level astrophysics course.

Who was Cherenkov and why does he have a type of radiation named after him?

Cherenkov radiation is named in honour of Pavel Alekseyevich Cherenkov, a Soviet scientist who won the 1958 Nobel Prize for his experimental observation of this type of radiation.


Figure 1: Pavel Alekseyevich Cherenkov (1904-1990), who won the 1958 Nobel Prize in Physics for his theoretical prediction of what we now call Cherenkov radiation

Cherenkov worked at the Lebedev Physical Institute, which is based in Moscow. His observation of Cherenkov radiation was made in 1934, and 24 years later, in 1958, he won the Nobel Prize for this work. He shared the prize with fellow Russian-scientists Igor Tamm and Ilya Frank who worked out the theory of the radiation. Cherenkov was also awarded two Stalin prizes, the first in 1946 and the second in 1952.


Figure 2: The characteristic blue light produced by Cherenkov radiation in a nuclear reactor

What Cherenkov observed in 1934 was a blue light being give off by a bottle of water which was being subjected to bombardment by radioactivity (what we call irradiated). By the 1950s, with the advent of nuclear reactors, this blue glow had become a common sight – an example is shown in Figure 2. It shows some nuclear fuel rods submerged in water (the moderator), and  the interaction of the high-energy radiation from the fuel rods with electrons in the water causes the water to glow blue – this is Cherenkov radiation.

The details of Cherenkov radiation

Cherenkov radiation is only one type of interaction between either high-energy particles, or high-energy radiation, and matter. Other such interactions include when a high-energy nucleon (proton or neutron) interacts with an electron, or a high-energy electron interacts with an electron, or a high-energy electron interacts with a gamma-ray. But, in this blogpost I am going to concentrate on Cherenkov radiation.

Cherenkov radiation is produced when a charged particle travels faster than the speed of light in that medium. As nothing can travel faster than the speed of light in a vacuum, Cherenkov radiation can only happen in a medium such as air, water or glass. Many of you are probably familiar with the idea of the refractive index n of a medium. It crops up for example in Snell’s law, which allows us to calculate the angle through which radiation deviates when it travels from one medium to another. Snell’s law states that

\frac{ n_{2} }{ n_{1} } = \frac{ sin \theta_{1} }{ sin \theta_{2} }

where n_{1} \text{ and } n_{2} are the refractive indices of the two media and \theta_{1} \text{ and } \theta_{2} are the angles between the direction of the ray and the normal, as shown in Figure 3.


Figure 3: Snell’s law gives the relationship between the angles between the direction of the rays and the refractive indices n of the two media

Snell’s law is usually taught in A-level physics, but what is not taught is the relationship between the refractive index and the speed of light. It is a very simple relationship; if n_{1} is the refractive index in medium 1, we can simply write

n_{1} = \frac{ c }{ v_{1} }

where v_{1} is the speed of light in that medium (and c is, of course, the speed of light in a vacuum). We should note that, in general, the refractive index n is a function of wavelength; this means that red light and blue light travel at different speeds in e.g. glass (or water). This is why these media disperse (split up) light and you see the colours of the rainbow when light passes through a prism, the blue and red lights are refracted different amounts in the glass because they travel at different speeds.

This page gives a list of refractive indices for different media. For pure water at 20 Celsius, n=1.3325, whereas for heavy water (deuterium oxide, each hydrogen atom is replaced with deuterium which has a neutron in the nucleus in addition to the proton present in hydrogen), the refractive index according to this page is n=1.328, so slightly less than for pure water. Heavy water is used as a moderator in nuclear reactors as the additional neutron helps slow down the fast neutrons from the nuclear reactions more effectively than pure water.

Assuming we are dealing with heavy water, this means that the speed of light in heavy water is v = 3 \times 10^{8} / 1.328 = 2.259 \times 10^{8} metres per second. If a charged particle travels faster than this speed in heavy water, it will emit Cherenkov radiation. A cone of constructive interference will form at an angle \theta to the direction of travel of the charged particle.

If v is the speed of the particle in the medium and v_{1} is the speed of light in that medium (where v_{1} = c/n), we can write that

\boxed{ cos \theta = \frac{ v_{1} }{ v } = \frac{ c }{ nv } } \; \; \text{(1)}

Clearly, as cos \theta can never be greater than 1, v (the speed of the particle) cannot be less than c/n. This is the threshold, Cherenkov radiation will only happen if v > c/n.


Figure 4: In Cherenkov radiation, a wavefront forms at an angle \theta to the direction of travel of the charged particle. In this figure, the particle is travelling at a velocity v horizontally; the radiation in the medium travels at a velocity v_{1}


This cone of light is very analogous to the cone of sound which forms when an object travels faster than the speed of sound in that medium. In Figure 5 we show an aeroplane travelling at less than the speed of sound (on the left), at the speed of sound (in the middle) and in excess of the speed of sound (on the right). The cone of sound produced at speeds greater than the speed of sound has the same relationship for the angle as with Cherenkov radiation, it is given by the ratio of the speed of the object to the speed of the waves in that medium.


Figure 5: The sound waves created as an aeroplane travels at different speeds. When it exceeds the speed of sound (figure on the right), a cone of sound is formed around the aeroplane, the angle depends on the ratio of the aeroplane to the speed of sound, just like the angle of the cone of light in Cherenkov radiation

Another way to illustrate the radiation given off in Cherenkov radiation is shown in figure 6. It is just a different way of showing the same thing as shown in Figure 4, but may be clearer to some of you. In Figure 6 the wavefronts are shown after a time t, where the particle will have travelled a distance vt.

c/n in Figure 6 is the same as v_{1} in Figure 4, the speed of the radiation in the medium. \beta is defined as v/c which means that \beta c = v, the speed of the particle in the medium. Figure 6 shows nicely why a cone of light develops which shines forwards at an angle of \theta to the direction of travel of the charged particle.


Figure 6: This is a different way of illustrating the emission of Cherenkov radiation.


The energy emitted per unit length travelled by an electron per angular frequency interval d\omega is given by the Frank-Tamm formula (see here)

\boxed{ \frac{ dE }{ dx d\omega } = \frac{ e^{2}}{ 4 \pi } \mu(\omega) \omega \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\omega) } \right) }

where v is the speed of the particle, e is the charge on an electron, \mu(\omega) is the magnetic permeability of the medium (which may be frequency dependent), n(\omega) is the refractive index of the medium (which may be frequency dependent) and c/n(\omega) is the speed of light in the medium. Frank and Tamm are the two other physicists who shared the 1958 Nobel with Cherenkov.

As we saw for Equation (1), which gave the angle of the wavefronts to the direction of the charged particle, the specific intensity will go to zero if the speed of the particle v is equal to c/n_{\nu}, and becomes negative if v becomes less than c/n_{\nu}.

Rather than show the energy, it is more usual to calculate the number of photons emitted per particle length per frequency interval or wavelength interval. We are going to show it per wavelength interval d\lambda. We start with Equation (2), and first we change from angular frequency \omega to temporal frequency \nu. To do this we remember that \omega = 2 \pi \nu, and so d \omega = 2 \pi d\nu. This gives us

\frac{ dE }{ dx d\nu } = \frac{ e^{2}}{ 4 \pi } \mu(\nu) 2 \pi \nu \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\nu) } \right) \cdot 2 \pi

which simplifies to

\frac{ dE }{ dx d\nu } = \frac{ \pi e^{2}}{ 1 } \mu(\nu) \nu \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\nu) } \right)

We are now going to assume that the permeability \mu(\nu) is unity. To go from the energy to the number of photons we remember that the energy of each photon is h \nu, where h is Planck’s constant, so dE = N h \nu, where N is the number of photons. So, we can write

\frac{ d^{2}N }{ dx d\nu } = \frac{ \pi e^{2} }{ h \nu } \nu \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\nu) } \right)

We are now going to express this per wavelength interval. Remember c = \lambda \nu so \nu = c/\lambda and d\nu = - (c/\lambda^{2})d\lambda. We can ignore the minus sign, this is just telling us that wavelength decreases as frequency increases, so then we can write

\frac{ d^{2}N }{ dx d\lambda } = \frac{ \pi e^{2} }{ h } \frac{ \lambda }{ c } \frac{ c }{ \lambda }\left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\lambda) } \right) \frac{ c }{ \lambda^{2} }

which simplifies to

\frac{ d^{2}N }{ dx d\lambda } = \frac{ \pi e^{2} }{ hc \lambda^{2} } \left( 1 - \frac{ c^{2} }{ v^{2} n^{2}(\lambda) } \right)

Finally, we multiply by 4 \pi, the solid angle in a sphere, to give us our final expression

\boxed{ \frac{ d^{2}N }{dxd\lambda} = \frac{ 4 \pi^{2} e^{2} }{ h c \lambda^{2} } \left( 1 - \frac{ c^{2} }{ v^{2} n_{\lambda}^{2} } \right) } \; \; (3)

In Figure 7 I use Equation (3) to show the numbers for two different particle speeds, (a) v=2.8 \times 10^{8} m/s and (b) 2.3 \times 10^{8} m/s. I have illustrated the visual part of the spectrum by the shaded box.


Figure 7: the number of photons emitted in Cherenkov radiation per unit length per wavelength interval for two different particle velocities – v=2.8\times 10^{8} m/s (green) and (b) v=2.3 \times 10^{8} m/s (blue), both for heavy water with n_{\lambda}=1.328. The shaded area represents the visible part of the spectrum.


As can be seen from Figure 7, (a) the number of photons is more when the charged particles are moving quicker and (b) the number of photons increases with shorter wavelength, which is why Cherenkov radiation has its characteristic blue appearance. As the number goes as 1/\lambda^{2}, the number of blue photons is roughly four times as many as the number of red photons. In fact, the intensity is even higher in the ultraviolet, but of course we cannot see this light with our eyes.

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I am currently visiting Namibia, giving talks and meeting the astronomers here at the University of Namibia. My university, Cardiff University, has a formal university-wide collaboration with the University of Namibia through something called the Phoenix project. The level of collaboration between different departments goes from very strong to zero; for the Physics and Astronomy department it is non-zero but could be stronger. Hopefully, my visit here can help make it stronger.

Although the physics department at the University of Namibia is involved in many exciting areas of research, the one that interests me most is its involvement in a high-energy radiation system of telescopes called H.E.S.S. (High Energy Stereoscopic System). The name H.E.S.S., however, is no coincidence; it is also the name of Victor Hess, one of the founders of studies of cosmic rays. I blogged about him here.

H.E.S.S. (which I am now going to type as just HESS) is a system of telescopes which detect Cherenkov radiation produced when high energy particles (cosmic rays) or gamma rays from space strike atoms or molecules in the Earth’s atmosphere. This provides an indirect way of detecting this high energy radiation, as the radiation itself does not reach the ground; but the Cherenkov radiation it produces does reach the ground.


HESS (High Energy Stereoscopic System) is an system of telescopes which detect Cherenkov radiation, produced when cosmic rays or gamma rays from space strike atoms and molecules in the Earth’s atmosphere

But, what kind of astronomical sources does HESS detect? What is Cherenkov radiation? And, why has HESS been sited in Namibia?

What kind of astronomical sources does HESS detect?

Radiation from astronomical sources falls broadly into four categories,

  • thermal continuum radiation (also known as blackbody radiation)
  • non-thermal continuum radiation
  • non-thermal emission which is not a continuum
  • line emission

I have blogged several times about blackbody radiation. Here I derived Planck’s radiation law using his original arguments of 1900, here I blogged about the fact that the Cosmic Microwave Background is a perfect blackbody (which is a very important fact in its interpretation as being due to radiation from the hot, early Universe), here I blogged about blackbody radiation and the ultraviolet catastrophe, and here I showed how we can use the fact that stars radiate as blackbodies to determine their sizes. And blackbody radiation has cropped up in several others of my blogposts.

Non-thermal continuum radiation (also known as synchrotron radiation) is something I have been planning to blog about for a while. I mentioned synchrotron radiation here when I derived the reason that accelerated electrons emit electromagnetic radiation. I will blog about the details of synchrotron radiation, as planned, in the near future; but for now I will just say that is produced when electrons spiral along magnetic field lines. As they spiral they accelerate (they are moving in a circle around the magnetic field line and moving along the field line at the same time) and, as I showed in that blogpost, accelerated electrons emit EM radiation.

Line emission is also something I have blogged about. For example, here in a blog entitled Emission Line Spectra, and here in my basic explanation of the three kinds of spectra we see in nature. It is the kind of emission given off by e.g. the Orion nebula, Messier 42.

The HESS telescope is looking for radiation which falls into the fourth category, non-thermal radiation which is not a continuum. In particular, it is looking for the high-energy end of this radiation, which is going to come from cosmic rays or gamma rays. As explained in my blogpost here, the source of cosmic rays is still hotly debated. They are not rays as such, put rather high energy charged particles. As I discussed in my blogpost here, some 89% of cosmic rays are high-energy protons (hydrogen nuclei), some 10% are high-energy helium nuclei (alpha particles), and the remaining 1% are high-energy electrons (beta particles).

We know what the cosmic rays are, but where they come from in terms of what kind of astronomical sources emit them is still a mystery. Also, we know that cosmic rays do not come from thermal sources; the energies are just too high. There is some kind of acceleration mechanism (often called cosmic accelerators) which are accelerating these charged particles to nearly the speed of light. Importantly for HESS’s work, gamma rays almost always accompany the cosmic accelerators, so we can use gamma rays to learn more about these mysterious phenomena.

Detecting gamma rays rather than cosmic rays has a distinct advantage; gamma rays travel in straight lines whereas cosmic rays, being charged particles, are bent by any magnetic fields. This has always been the main problem in determining the source of cosmic rays, as they do not travel in straight lines identifying their origin has been nigh-on impossible. I discussed that problem in more detail here.

A list of the sources detected by HESS since it saw first light in 2002 is given here. As you can see from this list, some are associated with supernovae remnants (such as the Crab nebula and the Vela nebula), some are from so-called active galactic nuclei (such as NGC 253), some are associated with Quasi-Stellar Objects (QSOs), and some have yet to be identified. The image below, taken from the HESS website, shows very high energy (VHE) gamma-ray emission from RCW 86, another supernova remnant.


A very high-energy gamma-ray image of RCW 86, a supernova remnant. To understand what 3, 5 or 7\sigma significance means, read my blogpost here.

What is Cherenkov radiation?

I will blog in more detail about Cherenkov radiation, showing the derivation of the formulae involved. But, for now, let me give a brief non-technical explanation.

Cherenkov radiation is produced when a high-energy charged particle (usually an electron) travels faster than the speed of light in that medium. You may think that you have heard that nothing can travel faster than the speed of light. This is true, in the sense that nothing can travel faster than the speed of light in a vacuum. But, it is possible for something to travel faster than the speed of light in a non-vacuum, where the speed of light is reduced by the medium through which the light is travelling.

So, for example, Cherenkov radiation is the preferred way of detecting neutrinos; the neutrinos strike a sub-atomic particle in a liquid, often heavy water (a rare event, but it does happen), and the accelerated sub-atomic particle may produce Cherenkov radiation if it is charged  (so, if it is either an electron or a proton) and if that charged particle travels faster than the speed of light in that liquid.

When high-energy cosmic or gamma rays enter the Earth’s atmosphere, Cherenkov radiation may be produced when this radiation (a gamma ray or a cosmic ray) strikes a sub-atomic particle in the atmosphere. This collision may produce an electron-positron pair, and the Cherenkov radiation occurs if this pair travel faster than the speed of light in the atmosphere.

The HESS telescopes detect this Cherenkov radiation, and so are able to pin-point the place in the atmosphere where the collision took place. Through this, we can trace back to from where in space the cosmic or gamma rays entered the Earth’s atmosphere; and hence indirectly ‘see’ the cosmic or gamma rays. In the case of gamma rays, which are not bent by magnetic fields, it allows us to construct a gamma-ray image of the astronomical source.

Why is HESS in Namibia?

Parts of Namibia are ideal to site a telescope which is looking for Cherenkov radiation from the atmosphere. The country has some of the clearest and driest skies in the World, and the driest climate of any country in sub-Saharan Africa. The HESS telescopes are located about 100km to the south-west  of the capital Windhoek, near the Gamsberg mountain in the Khomas Highland, which is a plateau at an elevation of nearly 2km above sea level.


HESS is located some 100 km south-west of Windhoek, near the Gamsberg mountain in the Khomas Highland. This part of Namibia enjoys a warm desert climate, which with its elevation of nearly 2km above sea level makes it ideal for observing Cherenkov radiation from cosmic rays and gamma rays

The HESS telescopes saw first light in September 2002, with more telescopes being added to the system in 2004. In 2012 a much larger telescope, HESS II, went into operation which allows detection of lower energy cosmic and gamma rays.

HESS and HESS II are a collaboration between scientists from 32 scientific institutions in 12 countries, including the UK. It is an extremely exciting project, and one in which I hope my department in Cardiff can be more involved.

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