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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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More pertinent today than ever…..

thecuriousastronomer

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?

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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…

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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.

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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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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

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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.

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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!

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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.

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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%.

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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 Monday (6th February) the sad news was announced that the great South African scrum-half Joost van der Westhuizen (JVD) had died at the age of 45. In 2011 JVD was diagnosed with motor neurone disease. Over the next few years he did much to raise awareness of and money to conduct research into this cruel disease; showing the same fighting spirit which led to his being one of the true greats of rugby of any era.

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Joost van der Westhuizen was one of the greats of world rugby. In 2011 he was diagnosed with motor neurone disease. He died on Monday (6th February) at the age of 45.

 

I heard it said this week that JVD was the first muscular scrum  half and the first large scrum half (a position traditionally played by smaller men). I would disagree with both of these statements. I grew up watching Gareth Edwards, often considered the greatest Welsh rugby player, who was a strong, muscular and dynamic scrum half. The only thing he lacked was height, but in the early 1980s Terry Holmes played for Wales, and he was 1m87, the same height as JVD. So, I would not agree that JVD was the first muscular scrum half or the first scrum half who was as large as a back-row forward.

It is sometimes easy when someone has died far too early to overstate their greatness. But, JVD was a great scrum half, there is no denying that. He was an inspiration to his team, and someone that other teams feared. In the 1995 World Cup, he was the first player to successfully tackle Jona Lomu, who had run rampant through every team against which the All Blacks had played.

But, JVD showed his true greatness in the way with which he dealt with his motor neurone disease (MND). He took it as an another challenge, and spent the rest of his life raising awareness of MND and raising money for researching in to it. In the video below is an excerpt from an interview which JVD did with the BBC in late 2014 or early 2015. It was replayed on Monday evening, the day of his death. Listen to his final words, when he is asked whether his MND may be considered a “blessing”

In a way I am glad I had MND. I now know what life is about

RIP Joost van der Westhuizen.

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Reblog of my 2013 list of 5 Christmas songs, this is the 5th and last one……

thecuriousastronomer

The fifth and final Christmas song I thought I’d share with you is “Happy Christmas (War is Over)” by John Lennon, released in 1971 in the USA and 1972 in the DUK.

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This song was part of a long running anti-war campaign by John & Yoko which included the “bed-ins” for peace in Amsterdam and Toronto, the 1969 single Give Peace a Chance, and a billboard campaign which is shown in the video. Lennon said in a 1980 interview that he wanted to write a Christmas song so that there was an alternative to White Christmas. John & Yoko are joined in the song by the Harlem Community Choir.

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Reblog of the 4th of my five Christmas songs from 2013. I think the YouTube link has stopped working. If I get time today I will see if I can replace it. 

thecuriousastronomer

The fourth Christmas song I thought I’d share is “Stop the Cavalry” by Jona Lewie. It got to number 3 in December 1980.

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