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## In My Life – The Beatles (song)

On this day in 1980 John Lennon was gunned down in front of his apartment in New York City.

John Lennon in 1965, the year “In My Life” was recorded.

Here is one of his more beautiful songs, “In My Life” from the 1965 album Rubber Soul.

(Words & music by Lennon & McCartney)

There are places I remember
All my life, though some have changed
Some forever not for better
Some have gone and some remain
All these places have their moments
With lovers and friends I still can recall
Some are dead and some are living
In my life I’ve loved them all

But of all these friends and lovers
There is no one compares with you
And these memories lose their meaning
When I think of love as something new
Though I know I’ll never lose affection
For people and things that went before
I know I’ll often stop and think about them
In my life I love you more

Though I know I’ll never lose affection
For people and things that went before
I know I’ll often stop and think about them
In my life I love you more
In my life I love you more

John – you’re missed 😦

## Studying the Universe using gravitational waves

The European Space Agency announced last week (28th of November) that a space-based gravitational wave observatory will form one of its next two large science missions. This is exciting news for Cardiff University, as it has a very active gravitational waves research group headed up by Professor B S Sathyaprakash (known to everyone as “Sathya”). The Cardiff group, along with others in the Disunited Kingdom, played an important role in persuading ESA to make this observatory one of its next large science missions. Just over a year ago I blogged about some theoretical modelling of gravitational waves the Cardiff group had done.

The ESA plan is to launch a space-based gravitational wave observatory in 2034, which will have much more sensitivity than any current or even future ground-based gravitational wave observatory. Although NASA also had plans to launch a space-based gravitational wave observatory, there is currently none in existence, so this is really ground-breaking technology that ESA is announcing. From what I can understand, the current announcement by ESA is their commitment to the proposed LISA (Laser Interferometer Space Antenna) gravitational wave observatory. It would seem NASA has withdrawn their commitment to what was originally going to be a joint NASA/ESA mission.

ESA has chosen a space-based gravitational wave detector to be funded as one of its two key future missions. It should go into operation in 2034.

## What are gravitational waves?

Gravitational waves are ripples in the fabric of space. They were predicted by Einstein as part of his theory of general relativity, the best theory we currently have to describe gravity. In his theory, events which involve extreme gravitational forces (such as two neutron stars orbiting each other (or merging), or the creation of a black hole) will lead to the emission of these gravitational waves. As they spread out from the source at the speed of light, they literally deform space as they pass by, just as ripples deform the surface of a pond as they spread out from a dropped stone.

An artist’s impression of gravitational waves being produced as two black holes orbit each other.

## Current gravitational wave observatories

There are several current gravitational wave observatories, all ground-based. These include VIRGO (in Italy) and LIGO (Laser Interferometer Gravitational Wave Observatory) which is in the United States. LIGO is the most sensitive of the current generation of gravitational wave detectors. LIGO actually comprises three separate detectors; one in Livingston, Louisiana and two in Hanford, Washington State. Each of the three separate detectors consists of two long arms at right angles to each other, forming a letter “L”. The idea behind these detectors is that, if a gravitational wave were to pass the detector, each of the two arms would have its length changed differently by the deformation of space as the gravitational wave passes through. Thus the detectors work on the principle of an interferometer, looking for tiny changes in the relative length of the two arms. And, when I say tiny, I mean tiny. In a 4km arm they are looking for changes of the order of $10^{-18} \text{ m}$, or about one thousandth the size of a proton!

The principle of a gravitational wave detector. They are essentially “interferometers”, with two arms at right angles to each other. As the gravitational waves pass the detector, space will be deformed and alter the relative lengths of the two arms.

## LIGO (Laser Interferometer Gravitational Wave Detector)

Currently the most sensitive gravitational wave detector is LIGO. LIGO consists of three separate detectors, one in Livingston, Louisiana and two in Hanford, Washington State. The detector in Louisiana is shown below.

The LIGO detector in Livingstone, Louisiana. Each arm is 4km in length, and can detect changes in the relative length of the two arms of less than the size of a proton.

The detector in Louisiana, and one of the two detectors in Washington State, consist of two 4km long arms at right angles to each other. An event like the collapse of a 10 solar-mass star into a black hole is expected to produce a change in length in a 4km arm of about $10^{-18} \text{ m}$, which is about one thousandth the size of a proton. This is just at the limit of the detection capabilities of LIGO, which is why astrophysicists are wanting more sensitive detectors to be placed into space. The other detector in Washington State has arms which are 2km in length, but just as sensitive as the detector with 4km arms at frequencies above 200 Hz, due to a different design.

## LISA – Laser Interferometer Space Antenna

The ESA plans just announced will be based on the NASA/ESA plans for LISA, which have been on the drawing board for most of the last 10 years. ESA’s plan is to build two space-based interferometers, which will be in the form of equilateral triangles as this artist’s description shows.

An artist’s impression of LISA, the “Laser Interferometer Space Antenna”. Each interferometer will consist of an equilateral triangle, with each side 5 million km in length.

The plan for LISA is to have arms which are 5 million km long! Compare this to the 4km long arms of LIGO. The changes in the length of a 5 million km long arm would be roughly one million times more than for LIGO, so rather than $10^{-18} \text{ m it would be } 10^{-12} \text{ m}$, which should be well within the capabilities of the detectors. This means that less energetic events than the collapse of a 10-solar mass star into a black hole would be detectable by LISA. All kinds of astrophysical events which involve large changes in gravitational fields should be detectable by LISA, including the afore-mentioned creation of black holes, but also the merging of neutron stars, and even the merging of less massive stars.

But, possibly most exciting is the opportunity that gravitational waves provide to probe the very earliest moments after the Big Bang. With normal electromagnetic radiation (light, x-rays, infrared light etc.), we can only see as far back as about 300,000 years after the Big Bang. This is when the Cosmic Microwave Background Radiation was produced. Prior to this time, the Universe was opaque to EM radiation of any wavelength, because it was full of unbound electrons, and the photons would just scatter off of them and not get anywhere (see my blog here about the CMB). But, it was not opaque to gravitational waves, so they provide a way for us to see back beyond the CMB, and a unique way to learn about the conditions of the Universe in its earliest moments.

## The 500 greatest albums – no. 6 – “What’s Going On” (Marvin Gaye)

At number 6 in Rolling Stone Magazine’s 500 greatest albums is “What’s Going On” by Marvin Gaye.

At no. 6 in Rolling Stone Magazine’s 500 greatest albums is “What’s Going On” by Marvin Gaye.

This is the only other album in the top 10 (along with “London Calling” by The Clash) which I did not own prior to seeing this list. So, I bought it about 18 months ago, and since then I have listened to it about two dozen times. The only songs on the album I knew before buying the album were “Mercy Mercy Me” and “What’s Going On”, both songs I like a lot.

I must say I am not sure what to make of this album, I find it very dated. I can appreciate that in the early 1970s it was very groundbreaking, and is possibly one of the first environmentally aware albums. Certainly Marvin Gaye has a message to get across. But I cannot imagine an album sounding like this being made now, it sounds as if it comes from the early 1970s and does not sound timeless. To be controversial, is this album the token album by a black artist in the top ten? Is it in the top ten just so there is a “black/soul album” in the top ten? Or does it deserve to be in the top ten on its own merits? I don’t know, I am just raising these questions as a talking point.

I am a big fan of Motown music in general, and have quite a number of Motown compilation albums. And, to my years, much of Motown’s music sounds timeless, but of all the stuff Marvin Gaye has done with which I am familiar, this album has the most songs which I cannot imagine being recorded at any other time but the early 1970s. What do you think?

Here is a live performance from 1980 of “Mercy Mercy Me” and “What’s Going On”. Enjoy!

Is this album Marvin Gaye’s greatest work? Does it deserve to be in the top ten of the 500 greatest albums of all time?

## “Perfect Day” (song)

Normally I don’t post two songs by the same artist in the same week, but as Lou Reed died last Sunday (27th of October 2013) I thought I would make an exception this week. On Tuesday I blogged about his song “Walk on the Wild Side”, which was the first Lou Reed song I remember hearing when I was a teenager.

Today I am blogging about one of his other most famous solo songs, “Perfect Day”. This song was released in 1972, actually as the B-side to “Walk on the Wild Side”, both songs taken from his Transformer album. The song conveys what appears to be a romantic message, with lines such as

It’s such a perfect day
So glad I spent it with you.

However, many commentators have argued that the “you” refers not to a person but to heroin, and is a song about Reed’s addiction to the drug. This interpretation has been further strengthened by its inclusion in the 1996 movie Trainspotting, a movie about a group of Scottish youngsters addicted to heroin.

“Perfect Day” was the B-side of “Walk on the Wild Side”

Here is the original version.

In 1997, the BBC released a charity version, using an all-star line-up including Reed himself (for the opening and closing lines) and Bono, David Bowie, Elton John, Emilou Harris, Tom Jones and many others. Here is that version.

How many of the singers do you recognise in the BBC version? Which version do you prefer?

## “Walk on the Wild Side” – Lou Reed (song)

On Sunday (27th of October 2013) I heard the sad news that Lou Reed had died. He was 71, and died of a “liver related ailment” according to a spokesperson. Reed is probably best known for fronting the highly influential “underground” band The Velvet Underground, I blogged about their album “The Velvet Underground” here in my series of Rolling Stone Magazine’s 500 greatest albums.

Personally I prefer Reed’s solo work. One of my favourites from his solo period is “Perfect Day”, but I will blog about that song at a future date. Here I will include the first song of his that I remember hearing as a teenager – “Walk on the Wild Side”.

Enjoy!

Which is your favourite Lou Reed song?

## The redshift men

And Hale created Yerkes Observatory in the 1890s, with the then World’s largest telescope, the 40-inch refractor.

The redshift men

The Mount Wilson Observatory in California had been built around a telescope with a 60-inch reflecting mirror, which came into operation in 1908. Just ten years later, this was joined on the mountain by the 100-inch Hooker Telescope (named after the benefactor who paid for it), which was to be the most powerful astronomical telescope on Earth for nearly 30 years, until the completion of the famous 200-inch Hale Telescope (named after George Ellery Hale, the astronomer who created both the Mount Wilson and the Mount Palomar observatories), at Mount Palomar, near Los Angeles (not far from Pasadena), in 1947. There were two people who would push the 100-inch to its limits in the 1920s.
The first of those pioneers, Milton Humason, was born in Dodge Center, Minnesota, on 19 August 1891; but his parents moved the family to the West Coast when he was a child…

View original post 5,136 more words

## A simple pendulum and circular motion

Legend has it that it was Galileo who first noticed that the period of a pendulum’s swing does not depend on how large the swing is, but only on the length of the pendulum. The swinging back and forth of a pendulum is an example of a very important type of motion which crops up in many places in Nature, so called Simple Harmonic Motion (SHM). In this blog I will derive the basic equations of SHM, and then go on and talk about the deep connection between SHM and circular motion.

## A swinging pendulum

If we start off by looking at a simple pendulum which has been displaced so that the bob is to the right of the vertical position, the angle the line of the pendulum makes with the vertical is given by $\theta$, and for this derivation to work $\theta$ needs to be small.

A simple pendulum will swing back and forth, exhibiting Simple Harmonic Motion.

The force restoring the pendulum bob back to the middle, which I have called $F$ in the diagram above, is given by $F = -T\sin(\theta)$ (the minus sign comes about because the the force is back towards the centre, even though the angle $\theta$ increases as we move the bob to the right).

The restoring force, $F$, can be written using Newton’s 2nd law as $F=ma=-T\sin(\theta)$. The angle $\theta$ is measured in radians (see my blog here for a tutorial on radians). When $\theta$ is small, $\sin(\theta) \approx \theta$ and so we can write that

$\sin(\theta) \approx \theta = \frac{x}{l}$

where $l$ is the length of the pendulum and $x$ is the horizontal displacement of the pendulum bob.

Finally, the tension $T$ in the pendulum chord can be written as $T \approx mg$ where $m$ is the mass of the bob and $g$ is the acceleration due to gravity (9.8 m/s/s for the Earth).

Putting all of this together, we can write that the restoring force $F$ can be written as

$F = - mg\theta = -mg\frac{x}{l}$

This means that the acceleration $a$ can be written as

$\vec{a} = -\frac{g}{l} \vec{x}$

It is more common to write this as

$\boxed{ \vec{a} = -\omega^{2} \vec{x} }$

where $\omega^{2} = \frac{g}{l}$ for the pendulum. $\omega$ is called the angular frequency and it is related to how long the pendulum takes to complete one full swing, the period, by the equation $T = \frac{2\pi}{\omega}$. The frequency is just the reciprocal of the period, so we can write $\nu = \frac{\omega}{2\pi}$ and so the angular frequency is related to the time frequency as $\omega = 2\pi \nu$.

Whenever the acceleration can be written as being proportional to the displacement, and in the opposite direction to the displacement, we have Simple Harmonic Motion. Other examples of SHM are an object bouncing vertically on a spring, or moving horizontally back and forth due to a spring attached at one end, even the vibrations of atoms in molecules.

## Solutions to the SHM equation

What are the solutions to the second order differential equation $\frac{d^{2}\vec{x}}{dt^{2}} = \vec{a} = - \omega^{2} \vec{x}$? We have a displacement, $\vec{x}$, and it is proportional to the acceleration $\vec{a}$, but the acceleration acts in the opposite direction to the displacement.

We differentiate the displacement twice with respect to time to produce the acceleration (remember $\vec{a} = \frac{d^2}{dx^2} \vec{x}$), and for SHM, when we do this, the acceleration is proportional to the displacement and in the opposite direction.

Let us suppose we try the displacement

$x = A \sin(\omega t) \text{ (remember that } \theta = \omega t)$

If we differentiate this once with respect to time we get the velocity

$v = \frac{dx}{dt} = \frac{d}{dt} A \sin(\omega t) = A\omega \cos(\omega t)$

To get the acceleration we need to differentiate the velocity with respect to time so

$a = \frac{dv}{dt} = \frac{d}{dt} A\omega \cos(\omega t) = - A \omega^{2} \sin(\omega t) = -\omega^{2}x$

Loh and behold, we now have that $a \propto - x$, so we have shown that, if $x = A \sin(\omega t)$ that an object which has this displacement as a function of time will display SHM.

As the acceleration is proportional to the displacement, it will be at its maximum when the displacement is maximum, so for a pendulum when the bob is at its extreme positions. The acceleration at the centre is zero, as the displacement at the centre is zero.

Conversely the velocity behaves in the opposite sense. Remember, as $\vec{v} = \frac{d}{dt}\vec{x} = A\omega\cos(\omega t)$ it means that the velocity and displacement are $90^{\circ}$ out of phase with each other, when the displacement is a maximum the velocity is zero, and when the displacement is zero the velocity is a maximum. So, the bob will be travelling at its quickest when it passes through the centre, and at its extremes the velocity is (temporarily) zero.

## SHM and circular motion

If an object is moving at a constant speed in a circle in the x-y plane we can write that it’s position at any time is given by

$x = A \cos(\theta) \text{ and } y = A \sin(\theta)$

For an object moving with a constant speed in a circle, its x-position and y-position can be written in terms of the radius A and the angular velocity $\omega$.

But notice that the expression for $x$ is exactly the same as the expression which we had above, so an object moving in a circle performs SHM. But how? Surely, if it is moving with a constant speed (and hence constant angular velocity $\omega$), how is it also displaying SHM?

The SHM comes in when we look at the object’s x or y-position as a function of time. So, for example, if we look at the circle from below, as if we were looking along the plane of the screen from below, we would only see the x-displacement of the object, as the y-displacement would be invisible to us. The x-displacement is given by $x=A \cos(\theta) = A \cos(\omega t)$, so the x-position will move back and forth about the central point, displaying SHM. This regular “wobble” is one of the things we look for in trying to find exoplanets – planets around other stars. If we see a regular, rhythmic wobble in the position of the parent star, it’s a pretty good bet that it has a planet going around it with the force of gravity between the host star and its planet producing the apparent SHM.