Feeds:
Posts
Comments

## 2017 6 Nations – 3rd weekend review

The 3rd weekend of the 2017 6 Nations saw home victories in all three matches. Scotland beat Wales at Murrayfield, Ireland beat France in Dublin, and England beat Italy at Twickenham.

## Scotland v Wales

Scotland beat Wales for the first time since 2007, with a well-deserved win that saw them score 20 unanswered points in the 2nd half. Wales went into half-time with a slim 13-9 lead. It could have been more, with Wales missing a chance to go 16-6 up with only a few minutes left of the first half. Instead, Scotland scored a penalty just before half-time and instead of 16-6 it was 13-9.

If ever there was a game of two halves this was it. The second half saw Scotland score 20 unanswered points, including two well deserved tries. It is true that Wales did have their chances to score points in the 2nd half, but Scotland were by far the better team after the break and fully deserved their victory over a largely ineffective Welsh team.

Scotland beat Wales 29-13 in a well-deserved victory, their first win over Wales since 2007.

Whereas the Scottish team took their chances, Wales were unable to come away with points when they were in the Scottish 22. It was a poor performance by Wales, but also a very good one by Scotland. After being uncompetitive for most of the last 10 years, it is good to see Scotland back mixing it with the other countries in the 6 Nations.

## Ireland v France

This was a cracker of a match. For a neutral, it had everything. Ireland went into half-time with a narrow 7-6 lead, and it was not clear well into the second half who was going to win this keenly contested match. In the end the better play of Irish half-backs Connor Murray and Johnny Sexton proved decisive, Ireland emerging victorious 19-9. With first Ireland, and then France, being the last two sides that Wales will face, the performance of both teams in this match does not bode well for Wales’ chances of winning either match, in my opinion.

## England v Italy

I have not seen this match, but am looking forward to seeing the highlights, mainly for the tactics which Italy adopted. By not committing any men to the breakdown, they avoided any rucks forming and thus the off-side line which normally exists did not apply. This allowed them to have men standing between the English backs, completely disrupting the game that England hoped to play. In fact, so effective was this tactic that Italy were 10-5 ahead at half-time. It is, apparently, only in the second half that England got to grips with this novel tactic, and eventually ran away to win 36-15.

## The 4th weekend

The 6 Nations now has another brief hiatus, before the next round of matches in just under two weeks. Ireland come to Cardiff to play Wales on Friday evening (10th March), then on the Saturday Italy take on France before the big game of the weekend, England v Scotland. Not only is this for the Calcutta Cup, but both countries are going for the Triple Crown. It will  be the first time in probably some 20 years that Scotland will go to Twickenham with even the slightest hope of winning.

Based on our performance against Scotland, I cannot see Wales beating Ireland, which leaves us with the real possibility of finishing the 2017 6 Nations with only one win. Quite where this will leave the Welsh team and management is a good question,  but it is clear from our performances this season that Wales have not developed the creativity in attack which is needed to win matches. Scotland showed us how it should be done; all we could do was look on and marvel at how much they have developed and how much we have regressed.

Advertisements

Read Full Post »

## 2017 6 Nations – 3rd weekend preview

This weekend sees the 2017 6 Nations come out of its hiatus for the 3rd weekend of matches. The weekend kicks off with Scotland v Wales at Murrayfield, followed by Ireland v France in Dublin, and Sunday sees England v Italy at Twickenham. It is an important weekend in the Championships, particularly for Wales, Scotland, Ireland and France. Win this weekend and their records will be 2 wins from 3, but lose and it will be 1 win from 3, a big difference.

This weekend is an important one in the 2017 6 Nations. For Wales, Scotland, Ireland and France it will make the difference between being 2 wins from 3 or 1 win from 3.

## Scotland v Wales

Wales go to Murrayfield having not lost to Scotland since 2007, either home or away. George North comes back into the team that lost to England, after injury kept him out of that game. Otherwise the Welsh team is unchanged, with perhaps the biggest surprise being that Ross Moriarty keeps his starting position, forcing Taulupe Faletau to start on the bench. Given how well Moriarty played against England I think it is the correct call by Rob Howley.

This will be the sternest test Wales have had against Scotland in the last 10 years, and they will need to be at their best to beat a resurgent Scotland. It will be a fascinating encounter, and I am going to be on the edge of my seat watching it.

## Ireland v France

Ireland will be wanting to continue getting their 2017 campaign back on track after their opening loss to Scotland. They beat Italy easily in the 2nd weekend of matches, so if they can beat France their campaign to win the 2017 Championship will still be possible. Lose and they can forget about being crowned champions. France too will be wanting to make it 2 wins from 3, and an away win against Ireland would be a scalp to bolster their confidence of being in contention for the Championship title come the final weekend.

## England v Italy

Sunday’s game seems England play Italy at home. This should be a rout, it is really just a question of how many points England can put on Italy. There has been a lot of debate this Championships as to whether Italy deserve to be part of the 6 Nations, and this match will probably do nothing to quell that debate. I expect a cricket score of 60+ points for England.

Read Full Post »

## Imaging the Galaxy’s supermassive blackhole – part 1

Last week, I blogged about the theoretical arguments for the Galaxy harbouring a supermassive black hole at its centre, and here I blogged about the observational evidence. The work done by the UCLA and MPE teams, discussed here, has led to a determination that the central black hole has a mass of between 4.4 and 4.5 million solar  masses. I am going to take the upper end  of this range, just for convenience.

An artist’s impression of Sgr A*, showing the central supermassive black hole and the accretion disk which surrounds it.

## The size of the event horizon

In this blog here I showed that the radius of a blackhole’s event horizon can be calculated by using the equation for the escape velocity $v_{esc}$ when that velocity is equal to the speed of light $c$. That is

$v_{esc} = c = \sqrt{ \frac{2GM }{ R } }$

where $M$ is the mass of the blackhole, $G$ is the universal gravitational constant, and $R$ is the size of the object, which in this case is the radius of the event horizon (also known as the Swarzchild radius $R_{s}$). So, we can write

$R_{s} = \frac{ 2GM }{ c^{2} }$

Putting in a mass of 4.5 million solar masses, we find

$R_{s} = 1.33 \times 10^{10} \text{ metres}$

Converting this to AUs, we find the radius of the event horizon is 0.09 AUs, much smaller than the radius of Mercury’s orbit, which is about 0.3 AUs.

At the distance of the Galactic centre, 8 kpc, this would subtend an angle of
$\theta = 6.17 \times 10^{-9} \text{ degrees}$ (remember to double $R_{s}$ to get the diameter of the event horizon). This is the same as

$\boxed{ \theta = 22.22 \text{ micro arc seconds} }$

Converting this to radians, we get

$\theta ( \text{in radians}) = 1.08 \times 10^{-10}$

In fact, we do not need to resolve the event horizon itself, but rather the “shadow” of the event horizon, which is about four times the size, so we need to resolve an angle of

$\theta ( \text{in radians}) \approx 4 \times 10^{-10}$

## The resolution of a telescope

There is a very simple formula for the resolving power of a telescope, it is given by

$\theta( \text{in radians}) = \frac{ 1.22 \lambda }{ D }$

where $D$ is the diameter of the telescope and $\lambda$ is the wavelength of the observation. Let us work out the diameter of a telescope necessary to resolve an object with an angular size of $50 \times 10^{-4} \text{ radians }$ at various wavelengths.

For visible light, assuming $\lambda = 550 \text{ nanometres}$

$D = \frac{ 1.22 \times 550 \times 10^{-9} }{ 4 \times 10^{-10 } }, \boxed{ D = 1.68 \text { km} }$

There is no visible light telescope this large, nor will there ever be. At the moment, visible-light interferometry is still not technically feasible over this kind of a baseline, so imaging the event horizon of the Galaxy’s supermassive blackhole is not currently possible at visible wavelengths.

For 21cm radio radiation (the neutral hydrogen line)

$D = \frac{ 1.22 \times 21 \times 10^{-2} }{ 4 \times 10^{-10 } }, \boxed{ D = 640,000 \text { km} }$

This is more than the distance to the Moon (which is about 400,000 km away). So, until we have a radio dish in space, we cannot resolve the supermassive blackhole at 21cm either.

For millimetre waves, we have

$D = \frac{ 1.22 \times 1 \times 10^{-3} }{ 4 \times 10^{-10 } }, \boxed{ D = 3,100 \text { km} }$

which is feasible with very long baseline interferometry (VLBI). So, with current technology, imaging the event horizon of the Milky Way’s supermassive blackhole is only feasible at millimetre wavelengths. Millimetre waves lie in a niche between visible light and radio waves. They are long enough that we can do VLBI, but they are short enough that the baseline to image the supermassive black hole’s event horizon is small enough to be possible with telescope on the Earth.

Next week I will talk about a project to do just that!

Read Full Post »

## Free As A Bird – The Beatles (song)

Today I thought I would share this lesser known song by The Beatles – “Free As A Bird”. It was released in 1996, when The Beatles put together their Anthology series of CDs, book and TV series. By this time, of course, John Lennon had been dead for 16 years. The other Beatles got in touch with Yoko who gave them a recording that Lennon had made of this song. They added their own voices and musical accompaniment to his original track, to create the closest thing possible to a new Beatles song. There are also some lyrics added by Paul McCartney, the parts that he sings on his own, rather than in harmony to Lennon’s voice.

The Beatles single “Free As A Bird” was released in 1996, to coincide with the release of the Anthology CDs, book and TV series

Free as a bird
It’s the next best thing to be
Free as a bird

Home, home and dry
Like a homing bird I’ll fly
As a bird on wings

Whatever happened to
The life that we once knew?
Can we really live without each other?

Where did we lose the touch
That seemed to mean so much?
It always made me feel so…

Free as a bird
Like the next best thing to be
Free as a bird

Home, home and dry
Like a homing bird I’ll fly
As a bird on wings

Whatever happened to
The life that we once knew?
Always made me feel so free

Ah…
Ah…
Ah…

Free as a bird
It’s the next best thing to be
Free as a bird
Free as a bird
Free as a bird
Oooooo

Free…

[Turn out nice again, mother.]

The video to accompany “Free As A Bird” is fascinating. For aficionados of Beatles trivia, there are all kinds of obvious and less obvious references to Beatles songs in the video. See how may you can spot?

Enjoy!

Read Full Post »

## The February 2017 annular solar eclipse

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.

Read Full Post »

## Beyond Here Lies Nothin’ – Bob Dylan (song)

Today I will continue with my series of blogposts of some Bob Dylan songs, in celebration of his winning the 2016 Nobel prize for literature. I am concentrating on songs which are on  his official Vevo channel, as other songs of his which are uploaded to YouTube are almost always swiftly removed.

The song I am sharing today is “Beyond Here Lies Nothin'”, which appears on his 2009 album Together Through Life. According to his website, he first performed this live in July 2009 and the most recent live performance was earlier this year, in July 2016. As of my writing this, he has performed it 398 times!

“Beyond Here Lies Nothin'” appears on Dylan’s 2009 album Together Through Life.

Here are the lyrics to this song, which you can find here on Dylan’s official website.

I love you pretty baby
You’re the only love I’ve ever known
Just as long as you stay with me
The whole world is my throne
Beyond here lies nothin’
Nothin’ we can call our own

I’m movin’ after midnight
Down boulevards of broken cars
Don’t know what to do without it
Without this love that we call ours
Beyond here lies nothin’
Nothin’ but the moon and stars

Down every street there’s a window
And every window made of glass
We’ll keep on lovin’ pretty baby
For as long as love will last
Beyond here lies nothin’
But the mountains of the past

My ship is in the harbor
And the sails are spread
Listen to me pretty baby
Lay your hand upon my head
Beyond here lies nothin’
Nothin’ done and nothin’ said

Here is the official video of “Beyond Here Lies Nothin'” from Dylan’s Vevo channel. Enjoy!

Read Full Post »

## Evidence for the Galaxy’s supermassive black hole

On Tuesday, I blogged about the theoretical work done in the early 1970s by Martin Rees, and others, which proposed that there may be a supermassive black hole at the centre of our Galaxy and most spiral galaxies.

## What about the observational evidence?

In the early 1980s two teams set about observing the orbits of stars near Sgr A*. The two teams, working separately, were at UCLA and The Max Planck Institute For Extra-terrestrial Physics (MPE). The team at UCLA is known as the UCLA Galactic Center Group, the team at the MPE doesn’t have a snazzy name, but their website can be found here. Gradually, over many years, each of the two teams has determined the orbits of several dozens of stars, and hence have been able to use the laws of gravity to determine the mass of the enclosed mass which the stars are orbiting.

Below is an image of Sgr A* taken by the MPE team using the NACO near-infrared camera on the VLT with adaptive optics. The entire image is only 30 arc seconds across.

A combined H, K and L-band near infrared image of the Galactic Centre obtained by the NACO camera on the VLT using adaptive optics. This image is from the MPE website.

Here is a paper, published in 2009, entitled “Monitoring Stellar Orbits Around the Massive Black Hole in the Galactic Center”, published by the MPE group in The Astrophysical Journal. Here is a link to the paper.

This paper, published in The Astrophysical Journal in 2009, is one of several showing overwhelming evidence for a supermassive blackhole at the centre of the Milky Way galaxy.

In this paper, entitled “The Galactic Center massive black hole and nuclear star cluster”, Reinhard Genzel (the director of the MPE) and colleagues summarise the evidence from their studies of their being a supermassive blackhole at the centre of the Milky Way, with a calculated mass of about 4.4 million solar masses. Here is a link to the paper.

In a paper entitled “The Galactic Center massive black hole and nuclear star cluster”, Genzel etal. summarise their finding that the Galaxy harbours a massive black hole with a mass of about 4.4 million solar masses.

The UCLA group published this paper “Measuring Distance and Properties of the Milky Way’s Central Supermassive Black Hole With Stellar Orbits”, in 2008 in The Astrophysical Journal (here is a link to the paper). In it,they calculate the mass of the supermassive black hole to be 4.5 million solar masses, with an error of plus or minus 0.4 million solar masses.

Ghez etal. (2008) find the  mass of the supermassive black hole to be 4.5 million solar masses, slightly higher than the MPE group, but well within the errors of the two groups’ measurements.

The mass of this black hole is about 4.45 million times the mass of the Sun (the two groups calculate different masses, with the UCLA group calculating about 4.5 million solar masses, the MPE group about 4.4 million solar masses). Let us assume it is 4.5 million solar masses, just to round up to the nearest half a million solar masses.

As some of you may know, blackholes are observable in certain ways. They clearly affect the orbit of nearby objects (this is how the UCLA and MPE teams have garnered the evidence for the supermassive blackhole), but also the accretion disk which usually surrounds a blackhole has very hot gas spiralling into the blackhole. This very hot gas emits radiation as a blackbody, so most of it comes out in the X-ray part of the spectrum due to the very high temperature of several millions of Kelvin.

But, a blackbody will also radiate at other wavelengths (see my blog here to remind yourself of the shape of a blackbody curve), so such accretion disks will also radiate visible light, infrared light, and even radio emission. The question then arises, is it possible to observe the accretion disk way in towards the event horizon of the Galaxy’s supermassive blackhole?

I will answer that question next week.

Read Full Post »

Older Posts »