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## Expanding spheres of light

In this blogpost here, I derived the Lorentz transformations from first principles. The derivation uses a simple thought experiment; two reference frames $S \text{ and } S^{\prime}$ are moving relative to each other with a speed $v$. The origins of $S \text{ and } S^{\prime}$ coincide spatially at time $t=t^{\prime}=0$. At this moment, a flash of light is created at their origins, and expands as a sphere.

The radius of this sphere of light will be $r \text{ and } r^{\prime}$ in $S \text{ and } S^{\prime}$ respectively. So, we can write

$x^{2} + y^{2} + z^{2} = r^{2} \text{ in } S \text{ (1) }$
and
$x^{\prime}{^2} + y^{\prime}{^2} +z^{\prime}{^2} = r^{\prime}{^2} \text{ in } S^{\prime} \text{ (2) }$

But, in reference frame $S$, the distance $r$ that light travels in time $t$ is related to the speed of light $c$, as $c = r/t$, so for $S$ we can write that

$r = ct$

We can do the same thing for reference frame $S^{\prime}$. The distance $r^{\prime}$ that light travels in time $t^{\prime}$ in $S^{\prime}$ is related to the speed of light $c^{\prime}$ in $S^{\prime}$, $c^{\prime} = r^{\prime} / t^{\prime}$, so for $S^{\prime}$ we can write

$r^{\prime} = c^{\prime} t^{\prime}$.

However, crucially, Einstein said that the speed of light $c$ is the same for all inertial observers. This is the most important principle which underpins special relativity. It means that $c^{\prime} = c$, and so we can write

$r^{\prime} = c t^{\prime}$

which allows us to write Equations (1) and (2) as

$x^{2} + y^{2} + z^{2} -c^{2}t^{2} = 0 \text{ in } S \text{ (3) }$

and

$x^{\prime}{^2} + y^{\prime}{^2} + z^{\prime}{^2} - c^{2}t^{\prime}{^2} = 0 \text{ in } S^{\prime} \text{ (4) }$

Using the Galilean Transformations, which can be written as

we can substitute $for$latex x^{\prime} (=x – vt)\$, $y^{\prime}=y$, $z^{\prime}=z$ and $t^{\prime} =t$ in Equ. (4) to give us

$(x-vt)^{2} + y^{2} + z^{2} - c^{2}t^{2} = x^{2} - 2vxt +v^{2}t^{2} + y^{2} + z^{2} - c^{2}t^{2}$

which is the same  as Equ. (3), except for the two extra terms $-2vxt \text{ and } v^{2}t^{2}$. This shows that Equations (3) and (4) are not equal under a Galilean transformation, even though the two equations should be equal (as both are equal to zero).

Let us now show that Equations (3) and (4) are equal using the Lorentz transformations. These can be written as

where $\gamma$ is the Lorentz factor, and is defined as

$\gamma = \frac{ 1 }{ \sqrt { \left( 1 - v^{2}/c^{2} \right) } }$

We will substitute these expressions for $x^{\prime},y^{\prime},z^{\prime}$ and $t^{\prime}$ into Equ. (4). When we do this, we have

$\left( \gamma (x-vt) \right)^{2} + y^{2} + z^{2} - c^{2} \left( \gamma (t - vx/c^{2}) \right)^{2}$

Multiplying this out, and dropping the $y^{2}$ and $z^{2}$ terms, we get

$\gamma^{2}(x^{2} -2vxt +v^{2}t^{2}) -c^{2}\gamma^{2}\left(t^{2} - \frac{2vxt}{c^{2}} + \frac{v^{2}x^{2}}{c^{4}}\right)$

$\gamma^{2}x^{2} -2\gamma^{2}vxt + \gamma^{2}v^{2}t^{2} - c^{2}\gamma^{2}t^{2} +2\gamma^{2}vxt - (\gamma^{2}v^{2}x^{2})/c^{2}$

The two terms $2\gamma^{2} vxt$ disappear and, gathering terms in $x^{2}$ and $t^{2}$ together, we can write

$\gamma^{2}x^{2} - \left( \frac{ (\gamma^{2}v^{2}) }{ c^{2} } \right) x^{2} + \gamma^{2}v^{2}t^{2} - \gamma^{2}c^{2}t^{2}$

$\gamma^{2}x^{2} (1 - v^{2}/c^{2}) + \gamma^{2}t^{2}(v^{2}-c^{2})$

Remembering that $\gamma^{2} = \frac{ 1 }{ ( 1 - v^{2}/c^{2} ) }$, and changing the sign of the $t^{2}$ term, we can write

$\frac{ 1 }{ (1 - v^{2}/c^{2} ) } \cdot (1 - v^{2}/c^{2}) \cdot x^{2} - \frac{ 1 }{ (1 - v^{2}/c^{2} ) } \cdot (c^{2} - v^{2}) \cdot t^{2}$

which is

$x^{2} - \frac{ 1 }{ (1 - v^{2}/c^{2} ) } \cdot c^{2}(1 - v^{2}/c^{2}) \cdot t^{2}$

which is

$x^{2} - c^{2}t^{2}$, exactly as in Equ. (3).

We have therefore shown that Equations (3) and (4) are equal if we use the Lorentz transformations.

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## The Beatles rooftop concert (January 30th, 1969)

Today (January 30th) marks the 50th anniversary of the last time The Beatles played live together, in the infamous “rooftop” concert in 1969. Although they would go on to make one more studio album, Abbey Road in the summer of 1969; due to contractual and legal wranglings the rooftop concert, which was meant to be the conclusion to the movie they were shooting, would not come out until 1970 in the movie Let it Be.

It is also true to say that some of the songs on Abbey Road were performed “live” in the studio with very little overdubbing (as opposed to separate instrument parts being recorded separately as was done on e.g. Sgt. Pepper). But, the rooftop concert was the last time the greatest band in history were seen playing together, and has gone down in infamy. It has been copied by many, including the Irish band U2 who did a similar thing to record the video for their single “Where the Streets Have no Name” in 1987 in Los Angeles.

The Beatles were trying to think of a way to finish the movie that they had been shooting throughout January of 1969. They had discussed doing a live performance in all kinds of places; including on a boat, in the Roundhouse in London, and even in an amphitheatre in Greece. Finally, a few days before January 30th 1969, the idea of playing on the roof of their central-London offices was discussed. Whilst Paul and Ringo were in favour of this idea, and John was neutral, George was against it.

The decision to go ahead with playing on the roof was not made until the actual day. They took their equipment up onto the roof of their London offices at 3, Saville Row, and just start playing. No announcement was made, only The Beatles and their inner circle knew about the impromptu concert.

The concert consisted of the following songs :

1. “Get Back” (take one)
2. “Get Back” (take two)
3. “Don’t Let Me Down” (take one)
4. “I’ve Got a Feeling” (take one)
5. “One After 909”
6. “Dig a Pony”
7. “I’ve Got a Feeling” (take two)
8. “Don’t Let Me Down” (take two)
9. “Get Back” (take three)

People in the streets below initially had no idea what the music (“noise”) coming from the top of the building was, but of course younger people knew the building was the Beatles’ offices. However, they would not have recognised any of the songs, as these were not to come out for many more months. After the third song “Don’t Let Me Down”, the Police were called and came to shut the concert down. The band managed nine songs (five different songs, with three takes of “Get Back”, two takes of “Don’t Let Me Down”, and two takes of “I’ve Got a Feeling”) before the Police stopped them. Ringo Starr later said that he wanted to be dragged away from his drums by the Police, but no such dramatic ending happened.

At the end of the set John said

I’d like to thank you on behalf of the group and ourselves, and I hope we’ve passed the audition.

You can read more about the rooftop concert here.

Here is a YouTube video of “Get Back” (which may get taken down at any moment)

and here is a video on the Daily Motion website of the whole rooftop concert (again, it may get taken down at any moment).

Enjoy watching the greatest band ever perform live for the very last time!

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## First ever asteroid from another solar system detected

In late October 2017, astronomers announced the first ever discovery of an asteroid (or comet?) coming into our Solar System from another stellar system. The object was first spotted on 19 October by the University of Hawaii’s Pan-STARRS telescope, during its nightly search for near-earth objects. Based on its extreme orbit and its rapid speed, it was soon determined that the object has come into our Solar System from somewhere else, and this makes it the first ever asteroid/comet with an extra-solar origin to have been discovered. Originally given the designation A/2017 U1, the International Astronomical Union (IAU) have now renamed it 1I/2017 U1, with the I standing for “interstellar”.

The object, given the designation A/2017 U1, was deemed to be extra-solar in origin from an analysis of its motion.

In addition to its strange trajectory, observations suggest that the object also has quite an unusual shape. It is very elongated, being ten times longer than it is wide. It is thought to be at least 400 metres long but only about 40 metres wide. This was determined by the rapid and dramatic changes in its brightness, which can only be explained by an elongated object tumbling rapidly.

The object has also been given the name Oumuamua (pronounced oh MOO-uh MOO-uh), although this is not its official name (yet).  This means “a messenger from afar arriving first” in Hawaiian. In other respects, it seems to be very much like asteroids found in our own Solar System, and is the confirmation of what astronomers have long suspected, that small objects which formed around other stars can end up wandering through space, not attached to any particular stellar system.

To read more about this fascinating object, follow this link.

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## Derivation of E=mc2

There are quite a few ways to derive Einstein’s famous equation $E=mc^{2}$. I am going to show you what I consider to be the simplest way.  Feel free to comment if you think you know of an easier way.

We will start off with the relationship between energy, force and distance. We can write

$dE = F dx \text{ (1) }$

Where $dE$ is the change in energy, $F$ is the force and $dx$ is the distance through which the object moves under that force.  But, force can also be written as the rate of change of momentum,

$F = \frac{dp}{dt}$

Allowing us to re-write Equation (1) as

$dE = \frac{dp}{dt}dx \rightarrow dE = dp \frac{dx}{dt} = vdp \text{ (2) }$

Remember that momentum $p$ is defined as

$p =mv$

In classical physics, mass is constant. But this is not the case in Special Relativity, where mass is a function of velocity (so-called relativistic mass).

$m = \frac{ m_{0} }{ \sqrt{ ( 1 - v^{2}/c^{2} ) } } \text{ (3) }$

where $m_{0}$ is defined as the rest mass (the mass of an object as measured in a reference frame where it is stationary).

Assuming that both $m \text{ and } v$ can change, we can therefore write

$dp =mdv + vdm$

This allows us to write Equ. (2) as

$dE = vdp = v(mdv + vdm) = mvdv + v^{2}dm \text{ (4) }$

Differentiating Equ. (3) with respect to velocity we get

$\frac{dm}{dv} = \frac{d}{dv} \left( \frac{ m_{0} }{ \sqrt{ (1 - v^{2}/c^{2}) } } \right) = m_{0} \frac{d}{dv} (1 - v^{2}/c^{2})^{-1/2}$

Using the chain rule to differentiate this, we have

$\frac{dm}{dv} = m_{0} \cdot - \frac{1}{2} (1 - v^{2}/c^{2})^{-3/2} \cdot (-2v/c^{2}) = m_{0} (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-3/2} \text{ (5) }$

But, we can write

$(1 - v^{2}/c^{2})^{-3/2}$ as $(1-v^{2}/c^{2})^{-1/2} \cdot (1-v^{2}/c^{2})^{-1}$

This allows us to write Equ. (5) as

$\frac{dm}{dv} = m_{0} (v/c^{2}) \cdot (1 - v^{2}/c^{2})^{-1} \cdot (1 - v^{2}/c^{2})^{-1/2}$

From the definition of the relativistic mass in Equ. (3), we can rewrite this as

$\frac{dm}{dv} = \frac{ m v }{ c^{2} }(1-v^{2}/c^{2})^{-1}$

Which is

$\frac{dm}{dv} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}} - \frac{ v^{2}}{c^{2} } \right)^{-1} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}-v^{2}}{c^{2}} \right)^{-1} = \frac{ m v }{ c^{2} } \left( \frac{c^{2}}{c^{2}-v^{2}} \right)$

$\frac{dm}{dv} = \frac{ m v }{ (c^{2}-v^{2}) } \text{ (6) }$

So we can write

$c^{2}dm - v^{2}dm = mvdv$

Substituting this expression for $mvdv$ into Equ. (4) we have

$dE = vdp = vd(mv) = mvdv + v^{2}dm = c^{2}dm - v^{2}dm + v^{2}dm$

So

$dE = c^{2} dm$

Integrating this we get

$\int_{E_{0}}^{E} dE = c^{2} \int_{m_{0}}^{m} dm$

So

$E - E_{0} = c^{2} ( m - m_{0} ) = mc^{2} - m_{0}c^{2}$

$E - E_{0} = mc^{2} - m_{0}c^{2}$

This tells us that an object has rest mass energy $E_{0} = m_{0}c^{2}$ and that its total energy is given by

$\boxed{ E = mc^{2} }$

where $m$ is the relativistic mass.

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## Should Justin Gatlin be allowed to compete?

More pertinent today than ever…..

It was announced a few days ago that the American sprinter Justin Gatlin is on the shortlist for the IAAF (International Association of Athletics Federations) “Athlete of the Year” award for 2014. This is largely due to his having set the fastest times over both 100m and 200m this year; faster than Usain Bolt, faster than Yohan Blake, faster than anyone. In fact, he has set 6 of the 7 fastest times over 100m in 2014! Also, he has run faster over both 100m and 200m than anyone one else in their 30s (he is 32). Ever. But, should Gatlin be considered by the IAAF for such a prestigious award? Should he be even allowed to compete at all?

For those of you not familiar with Gatlin’s athletics career, he has twice been banned for failing drugs tests. In 2001 he failed a doping test, testing positive for amphetamines. He…

View original post 600 more words

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## 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.

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## 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.

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